MULTIPLE CHOICE QUESTION Electronics & Communication Engineering Fifth Edition

GATE EC BY RK Kanodia

MULTIPLE CHOICE QUESTION

Electronics & Communication Engineering Fifth Edition R. K. Kanodia B.Tech.

NODIA & COMAPNY JAIPUR

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CHAPTER

1.1 BASIC CONCEPTS

1. A solid copper sphere, 10 cm in diameter is deprived of 10

20

electrons by a charging scheme. The charge on

the sphere is (A) 160.2 C

(B) -160.2 C

(C) 16.02 C

(D) -16.02 C

(A) 1 A

(B) 2 A

(C) 3 A

(D) 4 A

6. In the circuit of fig P1.1.6 a charge of 600 C is delivered to the 100 V source in a 1 minute. The value of v1 must be v1

2. A lightning bolt carrying 15,000 A lasts for 100 ms. If the lightning strikes an airplane flying at 2 km, the charge deposited on the plane is (A) 13.33 mC

(B) 75 C

(C) 1500 mC

(D) 1.5 C

60 V

20 W

100 V

3. If 120 C of charge passes through an electric

Fig. P.1.1.6

conductor in 60 sec, the current in the conductor is (A) 0.5 A

(B) 2 A

(C) 3.33 mA

(D) 0.3 mA

(A) 240 V

(B) 120 V

(C) 60 V

(D) 30 V

4. The energy required to move 120 coulomb through 7. In the circuit of the fig P1.1.7, the value of the

3 V is (B) 360 J

(C) 40 J

(D) 2.78 mJ

voltage source E is 0V +

+

(A) 25 mJ

–

–

2V

5. i = ?

+ – +

5V

–

–

4V

i

E

+

1A

1V

10 V

2A

5A

Fig. P.1.1.7 3A

4A

Fig. P.1.1.5

(A) -16 V

(b) 4 V

(C) -6 V

(D) 16 V

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Page 3


UNIT 1

8. Consider the circuit graph shown in fig. P1.1.8. Each

Networks

12. v1 = ?

branch of circuit graph represent a circuit element. The

v1 –

+

value of voltage v1 is

1 kW

2

7V

+ 105 V –

– 15 V +

+

+

55 V + v1 –

–

8V

35 V

–

–

5V

3

6V

kW

+

–

100 V

– 10 V + +

65 V

kW

V

30

V

30

4 kW

+

–

Fig. P1.1.12 Fig. P.1.1.8

(A) -30 V

(B) 25 V

(C) -20 V

(D) 15 V

9. For the circuit shown in fig P.1.1.9 the value of

(A) -11 V

(B) 5 V

(C) 8 V

(D) 18 V

13. The voltage vo in fig. P1.1.11 is always equal to 1A

voltage vo is 5W + vo

4W

+ vo –

1A

15 V

5V

Fig. P1.1.11

–

Fig. P.1.1.9

(A) 10 V

(B) 15 V

(C) 20 V

(D) None of the above

(A) 1 V

(B) 5 V

(C) 9 V


14. Req = ? 5W

10 W

10 W

10 W

10. R1 = ? 60 W

Req

10 W

+

10 W

10 W

up to ¥

R1 100 V R2

+ 20 V –

70 V

Fig. P1.1.14

–

Fig. P.1.1.10

(A) 11.86 W

(B) 10 W

(C) 25 W

(D) 11.18 W

15. vs = ? (A) 25 W

(B) 50 W

(C) 100 W

(D) 2000 W

180 W + 60 W

11. Twelve 6 W resistor are used as edge to form a cube.

vs

The resistance between two diagonally opposite corner of the cube is 5 (A) W 6 (C) 5 W

Page 4

(B)

90 W

6 W 5

(D) 6 W

40 W

20 V –

Fig. P.1.1.15

(A) 320 V

(B) 280 V

(C) 240 V

(D) 200 V

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180 W


UNIT 1

24. Let i( t) = 3te-100 t A and v( t) = 0.6(0.01 - t) e -100 t V for

Networks

28. vab = ?

a

the network of fig. P.1.1.24. The power being absorbed

2 8

–

b 0.3i1

A 2

W

W

N

0.2i1

i1

A

6

+

W

R

i v

2

by the network element at t = 5 ms is

Fig. P.1.1.24

(A) 18.4 mW

(B) 9.2 mW

(C) 16.6 mW

(D) 8.3 mW

Fig. P.1.1.28

25. In the circuit of fig. P.1.1.25 bulb A uses 36 W when lit, bulb B uses 24 W when lit, and bulb C uses 14.4 W

(A) 15.4 V

(B) 2.6 V

(C) -2.6 V

(D) 15.4 V

29. In the circuit of fig. P.1.1.29 power is delivered by

when lit. The additional A bulbs in parallel to this

500 W

circuit, that would be required to blow the fuse is

400 W

ix

20 A

200 W

2ix

40 V 12 V C

B

A

Fig. P.1.1.29

(A) dependent source of 192 W

Fig. P.1.1.25

(A) 4

(B) 5

(B) dependent source of 368 W

(C) 6

(D) 7

(C) independent source of 16 W

26. In the circuit of fig. P.1.1.26, the power absorbed by

(D) independent source of 40 W

the load RL is

30. The dependent source in fig. P.1.1.30

i1

5W RL = 2 W

2i1

1W

1V

v1

20 V

v1 5

5W

Fig. P.1.1.26

(A) 2 W

(B) 4 W

(C) 6 W

(D) 8 W

Fig. P.1.1.30

27. vo = ?

(A) delivers 80 W

(B) delivers 40 W

(C) absorbs 40 W

(D) absorbs 80 W

31. In the circuit of fig. P.1.1.31 dependent source 0.2 A

5W

+ v1 0.3v1 –

8W

+ v2 –

5v2

18 W

+ vo –

+ 8V – ix

Page 6

(A) 6 V

(B) -6 V

(C) -12 V

(D) 12 V

2ix

4A

Fig. P.1.1.27

Fig. P.1.1.31

(A) supplies 16 W

(B) absorbs 16 W

(C) supplies 32 W

(D) absorbs 32 W

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Basic Concepts

Chap 1.1

32. A capacitor is charged by a constant current of 2 mA

36. The waveform for the current in a 200 mF capacitor

and results in a voltage increase of 12 V in a 10 sec

is shown in fig. P.1.1.36 The waveform for the capacitor

interval. The value of capacitance is

voltage is

(A) 0.75 mF

(B) 1.33 mF

(C) 0.6 mF

(D) 1.67 mF

i(mA) 5

33. The energy required to charge a 10 mF capacitor to 100 V is

Fig. P. 1.1.36

(A) 0.10 J (C) 5 ´ 10

t(ms)

4

(B) 0.05 J

-9

v

(D) 10 ´ 10

J

-9

J

v 50m

5

34. The current in a 100 mF capacitor is shown in fig.

t(ms)

4

P.1.1.34. If capacitor is initially uncharged, then the

4

(A)

(B)

v

waveform for the voltage across it is

t(ms)

v 50m

250m

i(mA) t(ms)

4

2

4

(C)

t(ms)

t(ms)

(D)

Fig. P. 1.1.34 v

37. Ceq = ?

v 10

10

2

4

t(ms)

2

(A)

v

2.5 mF

4

t(ms) 1.5 mF

(B)

v

2

4

1 mF

Ceq

0.2

0.2 t(ms)

2

(C)

4

2 mF

t(ms)

(D) Fig. P.1.1.37

35. The voltage across a 100 mF capacitor is shown in fig. P.1.1.35. The waveform for the current in the

(A) 3.5 mF

(B) 1.2 mF

capacitor is

(C) 2.4 mF

(D) 2.6 mF

v 6

38. In the circuit shown in fig. P.1.1.38 1

2

3

t(ms)

iin ( t) = 300 sin 20 t mA, for t ³ 0.

Fig. P.1.1.35 i(mA) 6

C2

C2

C2

C2

+

i(mA) 600

vin C1

iin 1

2

3

t(ms)

1

2

3

C1

C1

C1

60 mF

t(ms)

–

(A)

(B)

i(mA) 6

Fig. P. 1.1.38

i(mA) 600 2 1

(C)

3

t(ms)

2 1

3

t(ms)

Let C1 = 40 mF and C2 = 30 mF. All capacitors are initially uncharged. The vin ( t) would be

(D)

(A) -0.25cos 20t V

(B) 0.25cos 20t V

(C) -36cos 20t mV

(D) 36cos 20t mV

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Basic Concepts

Chap 1.1

v3 - 30 = v2

SOLUTIONS

Þ

v3 = 65 V

105 + v4 - v3 - 65 = 0 Þ

v4 = 25 V

v4 + 15 - 55 + v1 = 0 Þ v1 = 15 V

1. (C) n = 10 20 , Q = ne = e10 20 = 16.02 C Charge on sphere will be positive.

9. (B) Voltage is constant because of 15 V source.

2. (D) DQ = i ´ D t = 15000 ´ 100m = 15 . C

10. (C) Voltage across 60 W resistor = 30 V 30 Current = = 0.5 A 60

3. (B) i =

dQ 120 = =2 A dt 60

4. (B) W = Qv = 360 J

Voltage across R1 is = 70 - 20 = 50 V 50 = 100 W R1 = 0.5

6. (A)

11. (C) The current i will be distributed in the cube branches symmetrically 1A

i=1A

a

5A

2A 3A

4A

i 3

i i 3

6A

1A

i 3

2A

i 6

Fig. S 1.1.5

6. (A) In order for 600 C charge to be delivered to the 100 V source, the current must be anticlockwise. dQ 600 i= = = 10 A dt 60 Applying KVL we get

0V

Fig. S. 1.1.11

+

+

1V

–

–

+ –

12. (C) If we go from +side of 1 kW through 7 V, 6 V and 5 V, we get v1 = 7 + 6 - 5 = 8 V

E

13. (D) It is not possible to determine the voltage across

–

–

+

+

5V

1 A source.

10 V

Fig. S 1.1.7

14. (D) Req = 5 +

10 + 5 + E + 1 = 0 or E = -16 V 8. (D) 100 = 65 + v2 Þ

v2 = 35 V

+ 105 V –

+

30 +

Req

Fig. S 1.1.14

–

–

10 W

–

V

v2

Req

55 V + v1 –

5W

– 10 V +

30

V

–

+ v3 –

10 + 5 + Req 5W

+

v4

10 ( Req + 5)

+

– +

– 15 V +

+

65 V 100 V

b

6i 6i 6i + + = 5 i, 3 6 3 v Req = ab = 5 W i

7. (A) Going from 10 V to 0 V

4V

i

vab =

v1 + 60 - 100 = 10 ´ 20 or v1 = 240 V

2V

i 3

Þ Req2 + 15 Req = 5 Req + 75 + 10 Req + 50 Fig. S 1.1.8

Þ

Req = 125 = 1118 . W

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Page 9


Basic Concepts

vo - 20 vo 20 + = 5 5 5 Power is P = vo ´

Þ

We can say Cd = 20 mF, Ceq = 20 + 40 = 60 mF 1 1 æ 300 ö cos 20 t ÷ ´ 10 -3 = - 0.25 cos 20 t V vC = ò idt= ç60m è 20 C ø

vo = 20 V

20 v1 = 20 ´ = 80 W 5 5

39. (C) iC1 =

31. (D) Power P = vi = 2 ix ´ ix = 2 ix2 ix = 4 A, P = 32 W (absorb) 32. (D) vt 2 - vt1 = Þ

1 C

t2

ò idt Þ

iin C1 = 0.8 sin 600 t mA C1 + C2

At t = 2 ms, iC1 = 0.75 mA

Þ

12 =

t1

12 C = 2m ´ 10

Chap 1.1

1 2m ( t2 - t1 ) C

40. (B) vC1 =

4 vin vin C2 = C1 + C2 6 + 4

Þ

vc1 = 0.4 vin

C = 1.67 mF 41. (D) V = 2 + 3 + 5 = 10, Q = 1 C, C =

1 33. (B) E = Cv 2 = 5 ´ 10 -6 ´ 100 2 = 0.05 J 2

42. (A) vL = L

di Þ dt

This 0.2 V increases linearly from 0 to 0.2 V. Then

43. (B) vL = L

di = 0.01 ´ 2( 377 cos 377 t) V dt

current is zero. So capacitor hold this voltage.

= 7.54 cos 377 t V

1 34. (D) vc = c

35. (D) i = C

-3

10 ´ 10 -3 ò0 idt = 100 ´ 10 -6 (2 ´ 10 ) = 0.2 V

2m

dv dt

For 0 < t < 1 , C

dv 6 -0 = 100 ´ 10 -6 ´ = 600 mA dt 10 -3 - 0

1 1 12000 120 cos 3t dt = vdt = sin 377 t 0.01 ò Lò 377 12000 ´ 120 P = vi = (sin 377 t)(cos 377 t) 377

45. (D) vL = L

1 1 idt = ò 200 ´ 10 -6 C

5m

ò 4m tdt = 3125 t

vC = 3vL Þ

iC = 3 LC

46. (B) vL = L

It will be parabolic path. at t = 0 t-axis will be tangent.

combination is in series with 1.5 mF. 15 . (2 + 1) = 1mF, C1 is in parallel with 2.5 mF C1 = 15 . +2+1 . mF Ceq = 1 + 2.5 = 35

diL dt æ -100 - 0 ö vL = (0.05)ç ÷ = - 2.5 V 2 è ø

For 4 < t £ 8,

æ 100 + 100 ö vL = (0.05)ç ÷ = 2.5 V 4 è ø

For 8 < t £ 10,

æ 0 - 100 ö vL = (0.05)ç ÷ = - 2.5 V 2 è ø

Thus (B) is correct option.

30 ´ 60 30(20 + 40) 38. (A) Ca = = 20 mF, Cb = = 20 mF 30 + 60 30 + 20 + 0 C2

Cc

C2

Cb

C2

Ca

47. (C) Algebraic sum of the current entering or leaving a cutset is equal to 0. 6 16 i2 + i4 + i3 = 0 Þ + + i3 = 0 2 4

C2

+

i3 = - 7 A, C1

d 2 iL = - 9.6 sin 4 t A dt

For 2 < t £ 4,

37. (A) 2 mF is in parallel with 1 mF and this

vin

diL dv , iC = C C dt dt

2

At t = 4 ms, vc = 0.05 V

iin

L = 2 mH

= 1910 sin 754 t W

36. (B) For 0 £ t £ 4,

Cd

Þ

44. (A) i =

For 1 ms < t < 2 ms, dv 0-6 C = 100 ´ 10 -6 ´ = - 600 mA ( 3 - 2)m dt

vC =

100m = L

200m 4m

Q = 0.1 F V

C1

C1

C1

v3 = - 7 ´ 3 = - 21 V

60 mF

–

Fig. S 1.1.38

*********

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Page 11


CHAPTER

1.2 GRAPH THEORY

1. Consider the following circuits :

Non-planner graphs are (A) 1 and 3

(B) 4 only

(C) 3 only

(D) 3 and 4

3. A graph of an electrical network has 4 nodes and 7 branches. The number of links l, with respect to the (1)

(2)

chosen tree, would be (A) 2

(B) 3

(C) 4

(D) 5

4. For the graph shown in fig. P.1.1.4 correct set is

(3)

(4) Fig. P.1.1.4

The planner circuits are (A) 1 and 2

(B) 2 and 3

(C) 3 and 4

(D) 4 and 1

2. Consider the following graphs

Node

Branch

Twigs

Link

(A)

4

6

4

2

(B)

4

6

3

3

(C)

5

6

4

2

(D)

5

5

4

1

5. A tree of the graph shown in fig. P.1.2.5 is c

(1)

(2)

b

3 g

e

d

4

f

2

a

1

5

h

Fig. P.1.2.5

(3) Page 12

(4)

(A) a d e h

(B) a c f h

(C) a f h g

(D) a e f g

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UNIT 1

é 1 -1 0 ù (A) ê-1 0 1ú ê ú 1 -1úû êë 0 é-1 -1 0 ù (C) ê 0 1 1ú ê ú êë 1 0 -1ûú

é 1 0 -1ù (B) ê-1 -1 0 ú ê ú 1 1úû êë 0 1ù é-1 0 ê (D) 0 1 -1ú ê ú êë 1 -1 0 úû

2

2

4

1

3

4

1

5

13. The incidence matrix of a graph is as given below

3

5

(C)

é-1 1 1 0 0 0 ù ê 0 0 -1 1 1 0 ú ú A =ê ê 0 -1 0 -1 0 -1ú ê 1 0 0 0 -1 -1ú ë û

(D)

15. The incidence matrix of a graph is as given below é-1 1 1 0 0 0 ù ê 0 0 -1 1 1 0 ú ú A =ê ê 0 -1 0 -1 0 0 ú ê 1 0 0 0 -1 -1ú ë û

The graph is 2

Networks

2

The graph is 4

4 3

1

3

1

(A)

1

2

3

1

2

3

(B)

2

2

4

4

(A) 4

(B)

4 3

1

3

1

(C)

1

2

3

1

2

3

(D)

14. The incidence matrix of a graph is as given below é- 1 0 0 - 1 1 0 0 ù ê 0 0 0 1 0 1 1ú ê ú A =ê 0 0 1 0 0 0 -1ú ê 0 1 0 0 -1 -1 0 ú ê ú ëê 1 -1 -1 0 0 0 0 úû

4

4

(C)

(D)

16. The graph of a network is shown in fig. P.1.1.16. The number of possible tree are

The graph is 2

2

4

1

3

4

1

3

Fig. P.1.1.16

5

(A) Page 14

5

(A) 8

(B) 12

(C) 16

(D) 20

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UNIT 1

Networks

22. The fundamental cut-set matrix of a graph is é1 -1 0 0 0 ê0 -1 1 0 1 QF = ê 0 0 0 0 1 ê ê0 0 0 1 -1 ë

0ù 0ú ú 1ú 0 úû

The oriented graph of the network is

2

1

(C)

24. A graph is shown in fig. P.1.2.24 in which twigs are

2

1

3

3

4

solid line and links are dotted line. For this tree

4

5

5

6

fundamental loop matrix is given as below é1 1 1 0 ù BF = ê ú ë1 0 1 1û

6

(A)

(D)

(B)

1

2

1

2

1

3

3

2

4

3

4

4

5

5 6

6

Fig. P.1.2.24

(C)

(D)

The oriented graph will be

23. A graph is shown in fig. P.1.2.23 in which twigs are solid line and links are dotted line. For this chosen tree fundamental set matrix is given below. é 1 1 BF = ê 0 -1 ê êë 0 0

0 1 0

0 1 1

1 0 1

0ù 0ú ú 1úû

(A)

(B)

3 2

1

4

5

6

(C)

Fig. P. 1.2.23


(D)

25. Consider the graph shown in fig. P.1.2.25 in which twigs are solid line and links are dotted line.

1

5

6

(A)

2

4

3

(B)

Fig. P. 1.2.25

Page 16

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Graph Theory

A fundamental loop matrix for this tree is given as below é 1 BF = ê 0 ê êë 0

0 1 0

1ù 0 -1 -1 0 ú ú 1 0 1 -1úû 0

1

0

é 1 ê 0 (D) ê ê 0 ê-1 ë

Chap 1.2

1

0

0

0

0

1 -1

1

0

0 0

1 0

1 1

0 1

0

1

0

1

0

1 0

0 0

0 0

0ù 0ú ú 0ú 1úû

27. Branch current and loop current relation are


expressed in matrix form as

(A)

é i1 ù é 0 êi ú ê 0 ê 2ú ê ê i3 ú ê 1 êi ú ê-1 ê 4ú = ê ê i5 ú ê 1 êi ú ê 0 6 ê ú ê i ê 7ú ê 0 êë i8 úû êë 0

(B)

1 -1 0 -1 0 0 1

0

0 1

0 0

0 0

1 0

0ù 1ú ú -1ú é I1 ù 0 ú êI 2 ú úê ú 0 ú êI 3 ú 0 ú êëI 4 úû ú 0ú 1úû

where i j represent branch current and I k loop current. The number of independent node equation are

(C)

(D)

(A) 4

(B) 5

(C) 6

(D) 7

28. If the number of branch in a network is b, the

26. In the graph shown in fig. P.1.2.26 solid lines are twigs and dotted line are link. The fundamental loop

number of nodes is n and the number of dependent loop is l, then the number of independent node equations will be

matrix is i c

a

e

(A) n + l - 1

(B) b - 1

(C) b - n + 1

(D) n - 1

Statement for Q.29–30: h

b

Branch current and loop current relation are

f

d

expressed in matrix form as 1 0ù é i1 ù é 0 0 êi ú ê-1 -1 -1 0 ú ê 2ú ê ú 1 0 0 ú é I1 ù ê i3 ú ê 0 êi ú ê 1 0 0 0 ú ê I ú ê 4ú = ê ú ê 2ú i 0 0 1 1 ê 5ú ê ú êI 3 ú ê i ú ê 1 1 0 -1ú êI ú ë 4û 6 ê ú ê ú ê i7 ú ê 1 0 0 0 ú êë i8 úû êë 0 0 0 1úû

g

Fig. P.1.2.26

é 1 ê 0 (A) ê ê 0 ê-1 ë

é-1 1 0 0 ê 0 -1 -1 -1 (B) ê ê 0 0 0 -1 ê 1 0 1 0 ë é ê (C) ê ê ê ë

1 0 0 1

0 0 1 0

0 1 0 0

1 0 0 0


0 0 0 0 1 -1

0 1 0

1 0 0

1 0 0 0 1 -1 -1 0 0 0 1 -1 0 -1 0 -1

1 1 0 0

0 1 0 1

1

0

0

0


0 0 1 0 1 -1 0 -1

0 0 1 0

0 -1 1 0 0 0 0 0


where i j represent branch current and I k loop current. 29. The rank of incidence matrix is (A) 4

(B) 5

(C) 6

(D) 8

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Page 17


UNIT 1

30. The directed graph will be

Networks

33. The oriented graph for this network is

8

8

5

5

6

2

3

1

2

7

1

6

3

1

4

3

2

7

4

1

5

8 5

6

3

2

4

1

2

6 5

5

(C) 2

3

1

4

3

4

5

8

5

3

4 1

2

7

3

1

4

(D)

7

4

(C)

************

(D)

31. A network has 8 nodes and 5 independent loops. The number of branches in the network is (A) 11

(B) 12

(C) 8

(D) 6

32. A branch has 6 node and 9 branch. The independent loops are (A) 3

(B) 4

(C) 5

(D) 6

Statement for Q.33–34: For a network branch voltage and node voltage relation are expressed in matrix form as follows: 1ù é v1 ù é 1 0 0 êv ú ê 0 1 0 0ú ê 2ú ê ú 1 0 ú é V1 ù ê v3 ú ê 0 0 êv ú ê 0 0 0 1ú êV2 ú ê 4ú = ê úê ú ê v5 ú ê 1 -1 0 0 ú ê V3 ú êv ú ê 0 1 -1 0 ú êëV4 úû 6 ê ú ê ú 1 -1ú ê v7 ú ê 0 0 êë v8 úû êë 1 0 -1 0 úû where vi is the branch voltage and Vk is the node voltage with respect to datum node. 33. The independent mesh equation for this network are (A) 4

(B) 5

(C) 6

(D 7

Page 18

2

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Graph Theory

Chap 1.2

7. (D) D is not a tree

SOLUTIONS 1. (A) The circuit 1 and 2 are redrawn as below. 3 and 4 can not be redrawn on a plane without crossing other branch.

(A) (1)

(B)

(2) Fig. S1.2.1

2. (B) Other three circuits can be drawn on plane without crossing

(C) (1)

(D)

(2)

Fig. S .1.2.7

8. (D) it is obvious from the following figure that 1, 3, and 4 are tree 2

2

(3) a

Fig. S1.2.1

c

1

3. (C) l = b - ( n - 1) = 4.

a

b

e

(1) 2

a c

1 g

e

a

b

f

a

(2)

2

c

d

4

4

5. (C) From fig. it can be seen that a f h g is a tree of

b

f

f

l=b-n+1=3

given graph

3 e

d

4. (B) There are 4 node and 6 branches. t = n - 1 = 3,

c

1

3

d

b

c

1

3 e

d

b 3 e

d

f

f 4

4

h

(3)

Fig. S 1.2.5

(4) 2

6. (B) From fig. it can be seen that a d f is a tree. c b

a

a e

d

b c

1

e

d

f

3

f 4

(5) Fig. S. 1.2.6

Fig. S. 1.2.8

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Page 19


UNIT 1

So independent mesh equation =Number of link.

i l1

a

34. (D) We know that [ vb ] = ArT [ Vn ]

e

c

So reduced é1 0 ê0 1 Ar = ê ê0 0 ê0 0 ë

l4

l2 h

b

l3

Networks

d

f

g

Fig. S 1.2.26

incidence matrix is 0 0 1 0 0 1ù 0 0 -1 1 0 0 ú ú 1 0 0 -1 1 -1ú 0 1 0 0 -1 0 úû

At node-1, three branch leaves so the only option is (D).

This in similar to matrix in (A). Only place of rows has been changed. 27. (A) Number of branch =8

***********

Number of link =4 Number of twigs =8 - 4 = 4 Number of twigs =number of independent node equation. 28. (D) The number of independent node equation are n - 1. 29. (A) Number of branch b = 8 Number of link l = 4 Number of twigs t = b - l = 4 rank of matrix = n - 1 = t = 4 30. (B) We know the branch current and loop current are related as [ ib ] = [ B T ] [ I L ] So fundamental loop matrix is 1 1 0ù é0 -1 0 1 0 ê0 -1 1 0 0 1 0 0ú ú Bf = ê ê1 -1 0 0 -1 0 0 0 ú ê0 0 0 0 -1 -1 0 1ú û ë f-loop 1 include branch (2, 4, 6, 7) and direction of branch–2 is opposite to other (B only). 31. (B) Independent loops =link l = b - ( n - 1) Þ

5 = b - 7, b = 12

32. (B) Independent loop =link l = b - ( n - 1) = 4 33. (A) There are 8 branches and 4 + 1 = 5 node Number of link = 8 - 5 + 1 = 4 Page 22

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CHAPTER

1.3 METHODS OF ANALYSIS

1. v1 = ?

6R

6R

4vs

+ v1 –

(B) -120 V (D) -90 V

(A) 120 V (C) 90 V

3R

vs

4. va = ?

10 W 4W

12 V

10 V

Fig. P1.3.1

(A) 0.4vs

(B) 1.5vs

(C) 0.67vs

(D) 2.5vs

va 4A

1W

2. va = ?

2W

Fig. P1.3.4

3A

(A) 4.33 V (C) 8.67 V 2W

5. v2 = ?

va

3W

(B) 4.09 V (D) 8.18 V

20 W

1A

+ 30 W

60 W

v2 – 10 V

0.5 A

30 W

Fig. P1.3.2

(A) -11 V

(B) 11 V

(C) 3 V

(D) -3 V

Fig. P1.3.5

(A) 0.5 V (C) 1.5 V

3. v1 = ? 10 W

6. ib = ?

30 V

3A

(B) 1.0 V (D) 2.0 V

64 W

30 W

37 W ib

20 W

0.5 A

10 V

60 W

69 W

36 W

– v1 +

9A

6A

60 W

Fig. P1.3.6

Fig. P1.3.3

(A) 0.6 A (C) 0.4 A www.gatehelp.com

(B) 0.5 A (D) 0.3 A Page 23


UNIT 1

7. i1 = ? 6A

8W

(A) 20 mA

(B) 15 mA

(C) 10 mA

(D) 5 mA

11. i1 = ?

2W

i1

Networks

50 W

10 W 5W

3W

4W

75 V

100 W i1

6.6 V

Fig. P1.3.7

(A) 3.3 A

(B) 2.1 A

(C) 1.7 A

(D) 1.1 A

0.1 A

40 W

0.06 A

60W

8. i1 = ? 0.1A i2

90 kW 7.5mA

i1

10 kW

Fig. P1.3.11

10 kW

75 V

(A) 0.01 A

(B) -0.01 A

(C) 0.03 A

(D) 0.02 A

90 kW

12. The value of the current measured by the ammeter in Fig. P1.3.12 is

Fig. P1.3.8

(A) 1 mA

(B) 1.5 mA

(C) 2 mA

(D) 2.5 mA

2A

Ammeter

4W

7W

9. i1 = ? 2A

6W

3A 5W

3V

4W

2W

Fig. P1.3.12 3W

4A

2W i1

(B) 3 A

(C) 6 A

(D) 5 A

2 A 3

(C) -

Fig. P1.3.9

(A) 4 A

(A)

5 A 6

(B)

5 A 3

(D)

2 A 9

13. i1 = ? 200 W

10. i1 = ? 2 kW

45 V

i1

40 mA

500 W

Fig. P1.3.10

Page 24

15 mA

100 W

50 W

i1

Fig. 1.3.13

(A) 10 mA

(B) -10 mA

(C) 0.4 mA

(D) -0.4 mA

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10 mA


Methods of Analysis

Chap 1.3 5W

8W

4W

14. The values of node voltage are va = 12 V, vb = 9.88 V and vc = 5.29 V. The power supplied by the voltage

i1

12 V

source is

2W

i3

2W

i2

20 V

8V 6W

va

4W

3W

vb

12 V

1A

vc

Fig. 1.3.17

0 ù éi1 ù é12 ù é 4 -2 (A) ê-2 8 -2 ú êi2 ú = ê-8 ú ê úê ú ê ú 5 úû êëi3 úû êë20 úû êë 0 -2

2W

-2 0 ù éi1 ù é12 ù é6 (B) ê2 -12 2 ú êi2 ú = ê 8 ú ê úê ú ê ú 2 -7 úû êëi3 úû êë20 úû êë0

Fig. 1.3.14

(A) 19.8 W

(B) 27.3 W

(C) 46.9 W

(D) 54.6 W

0 ù éi1 ù é12 ù é 6 -2 (C) ê-2 12 -2 ú êi2 ú = ê 8 ú ê úê ú ê ú 7 úû êëi3 úû êë20 úû êë 0 -2

15. i1 , i2 , i3 = ? 2W

3W

15 V

9W

i1

6W

i2

0 ù éi1 ù é12 ù é 4 -2 ê (D) 2 -8 2 ú êi2 ú = ê 8 ú ê úê ú ê ú 2 -5 úû êëi3 úû êë20 úû êë 0 18. For the circuit shown in Fig. P1.3.18 the mesh

21 V

i3

equation are 6 kW

Fig. P1.3.15 6 kW

(A) 3 A, 2 A, and 4 A

(B) 3 A, 3 A, and 8 A

(C) 1 A, 3 A, and 4 A

(D) 1 A, 2 A, and 8 A

i3

6 kW

i1 6V

i2 5 mA

6 kW

16. vo = ? Fig. 1.3.18 4 mA

2 kW

2 mA

é 6 k -12 k -12 k ù éi1 ù é-6 ù ê ú ê ú (A) -6 k 6 k -18 k i2 = ê 0 ú ê úê ú ê ú -1k 0 k úû ëêi3 úû êë -1k êë 5 úû

1 kW

1 kW

+ 1 kW

2 kW

1 mA

vo

é 6 k 12 k -12 k ù éi1 ù é-6 ù (B) ê-6 k -6 k 18 k ú êi2 ú = ê 0 ú ê úê ú ê ú 1k 0 k úû êëi3 úû êë -1k êë 5 úû

–

Fig. P1.3.16

(A)

6 V 5

(B)

8 V 5

(C)

6 V 7

(D)

5 V 7

é-6 k -12 k 12 k ù éi1 ù é-6 ù (C) ê 6 k -6 k 18 k ú êi2 ú = ê 0 ú ê úê ú ê ú 1k 0 k úû êëi3 úû êë 1k êë 5 úû

17. The mesh current equation for the circuit in Fig. P1.3.17 are

é-6 k 12 k -12 k ù éi1 ù é-6 ù (D) ê-6 k 6 k -18 k ú êi2 ú = ê 0 ú ê úê ú ê ú 1k 0 k úû êëi3 úû êë -1k êë 5 úû

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Page 25


Methods of Analysis

R1

i3

R2

Chap 1.3

(A) 66.67 mA

(B) 46.24 mA

(C) 23.12 mA

(D) 33.33 mA

R3

29. va = ?

i1

R4

i2

10 W

v2

v1 25i2

4A

50 W

va

40 W

Fig. P1.3.25 200 W

10 A

The value of R4 is (A) 40

(B) 15

(C) 5

(D) 20

2.5 kW 10 kW

20 V

10 kW

(A) 342 V

(B) 171 V

(C) 198 V

(D) 396 V

30. ia = ?

va

5 kW

20 A

Fig. P1.3.29

26. va = ?

10 kW

20 W

100 W

5A

50 W

150 W

4 mA

225 W

100 W

200 W

ia

Fig. P1.3.26

2V

(A) 26 V

(B) 19 V

(C) 13 V

(D) 18 V

75 W

2A

50 W

(A) 14 mA

(B) -6.5 mA

(C) 7 mA

(D) -21 mA

31. v2 = ?

20 W

v

8V

Fig. P1.3.30

27. v = ?

10 W

4V

50 W – v2 +

4A

100 W

10 V

15 W

0.04v2

5W

Fig. P.3.1.27

Fig. P1.3.31

(A) 60 V

(B) -60 V

(A) 5 V

(B) 75 V

(C) 30 V

(D) -30 V

(C) 3 V

(D) 10 V

32. i1 = ?

28. i1 = ?

2W

300 W

40 V

i1

500 W

8V

0.4i1

0.5i1

4A

4W

6V

i1

Fig. P1.3.32

Fig. P1.3.28

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Page 27


UNIT 1

(A) -1.636 A

(B) -3.273 A

(C) -2.314 A

(D) -4.628 A

Networks

37. va = ? 0.8va

16 A

33. vx = ?

+ 100 W

1.6 A

2.5 W

5W

10 A

vx

50 W

0.02vx

2W

va

–

Fig. P1.3.37 Fig. P1.3.33

(A) 32 V

(B) -32 V

(C) 12 V

(D) -12 V

(A) 25.91 V

(B) -25.91 V

(C) 51.82 V

(D) -51.82 V

38. For the circuit of Fig. P1.3.38 the value of vs , that

34. ib = ?

will result in v1 = 0, is 1 kW

va

3A

3 kW

2A

ib

0.1v1 6V

4va

2 kW

10 W

20 W +

vs

40 W

Fig. P1.3.34

(A) 4 mA

(B) -4 mA

(C) 12 mA

(D) -12 mA

4 kW

vb

(A) 28 V

(B) -28 V

(C) 14 V

(D) -14 V

39. i1 , i2 = ?

2ix

4W 2V

48 V

Fig. P1.3.38

35. vb = ? ia

v1 –

2W

5ia

2 kW

ix 15 V

6W

i1

i2

18 V

Fig. P1.3.35

(A) 1 V

(B) 1.5 V

(C) 4 V

(D) 6 V

Fig. P1.3.39

36. vx = ?

(A) 2.6 A, 1.4 A

(B) 2.6 A, -1.4 A

(C) 1.6 A, 1.35 A

(D) 1.2 A, -1.35 A

40. v1 = ?

50 W

3W

iy + 2A

100 W

25iy

50 W

vx

+ vy –

0.2vx

–

3W

2A 6W

14 V + v1 –

Fig. P1.3.36

Page 28

(A) -3 V

(B) 3 V

(C) 10 V

(D) -10 V

7A

Fig. P1.3.40

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2W

2vy


Methods of Analysis

(A) 10 V

(B) -10 V

(C) 7 V

(D) -7 V

SOLUTIONS 1. (B) Applying the nodal analysis v 4 vs + s R R 6 3 v1 = = 15 . vs 1 1 1 + + 6 R 3R 6 R

41. vx = ? – vx +

500 W

0.5vx

500 W 900 W

600 W

0.6 A

Chap 1.3

2. (C) va = 2( 3 + 1) + 3 (1) = 11 V

0.3 A

3. (D) -

v1 -v + 1 + 6 =9 60 60

Þ

v1 = - 90 V

Fig. P1.3.41

(A) 9 V

(B) -9 V

(C) 10 V

(D) -10 V

42. The power being dissipated in the 2 W resistor in the

4. (C)

va - 10 va + =4 4 2

5. (D)

v2 v2 + 10 + = 0.5 20 30

Þ

va = 8.67 V

Þ

v2 = 2 V

circuit of Fig. P1.3.42 is 5W

6. (B) Using Thevenin equivalent and source transform

ia

3W

8W 3

2W

i1

2W

10 W

va

60 V

2A 2.5 A

4W

6ia

3W

25 V

30 V

Fig. P1.3.42

Fig. S.1.3.6

(A) 76.4 W

(B) 305.6 W

(C) 52.5 W

(D) 210.0 W

43. i1 = ? 500 W + vx – 100 W

180 V 400 W

5W

25 60 + + 2 15 va = = 15.23 V 3 1 1 + + 14 3 15 25 - 15.23 = 2.09 A i1 = 14 3 8 3

0.6 A

+ vy –

7. (A) ib = 0.001vy

100 W

10 + 0.5 = 0.6 A 64 + 36

8. (B) 75 = 90 ki1 + 10 k( i1 - 7.5m) 150 = 100 ki1

Þ

i1 = 15 . mA

9. (B) 3 = 2 i1 + 3( i1 - 4)

0.005vy

Þ

i1 = 3 A

Fig. P1.3.43

10. (B) 45 = 2 ki1 + 500 ( i1 + 15m) (A) 0.12 A

(B) 0.24 A

(C) 0.36 A

(D) 0.48 A

Þ

i1 = 15 mA

11. (D) 6.6 = 50 i1 + 100( i1 + 0.1) + 40( i1 - 0.06) + 60( i1 - 0.1)

*****************

i1 = 0.02 A www.gatehelp.com

Page 29


UNIT 1

38. (D) If v1 = 0, the dependent source is a short circuit v1 v - vs v1 - 48 + 1 + =2 - 3 Þ 40 10 20 48 v - s = - 1 Þ vs = - 14 V 10 20 3A

v1 = 0

ia =

Networks

30 + 7.5 + 2 = 329 . A 12

Þ

6 ia = 19.75 V

voltage across 2 W resistor 30 - 19.75 = 10.25 V, P=

2A

(10.25) 2 = 52.53 W 2

43. (A) vx = 500 i1 10 W 40 W

vs

v y = 400( i1 - 0.001vx ) = 400 ( i1 - 0.5 i1 ) = 200 i1

20 W + v1 –

180 = 500 i1 + 100( i1 - 0.6) + 200 i1 + 100( i1 + 0.005 v y)

48 V

180 = 900 i1 - 60 + 100 ´ 0.005 ´ 200 i1 i1 = 0.12 A

Fig. S1.3.38 ************

39. (D) ix = i1 - i2 15 = 4 i1 - 2( i1 - i2 ) + 6( i1 - i2 ) 8 i1 - 4 i2 = 15

Þ

K(i)

-18 = 2 i2 + 6( i2 - i1 ) 3i1 - 4 i2 = 9

Þ

K(ii)

. A, i2 = -1.35 A i1 = 12 40. (B) 14 = 3i1 + v y + 6( i1 - 2 - 7) + 2 v y + 2( i1 - 7) v y = 3( i1 - 2) 14 = 3i1 + 9( i1 - 2) + 6( i1 - 9) + 2( i1 - 7) 14 = 20 i1 - 18 - 54 - 14

Þ

i1 = 5 A

v1 = 6(5 - 2 - 7) + 2 ´ 3(5 - 2) + 2(5 - 7) = -10 V 3W + vy

3W

–

2A 6W

14 V

i1

+ v1 –

2vy

7A

2W

Fig. S1.3.40

41. (D) Let i1 and i2 be two loop current 0.5 vx = 500 i1 + 500( i1 - i2 ), vx = -500 i1 Þ

5 i1 - 2 i2 = 0

K(i)

500( i2 - i1 ) + 900( i2 + 0.3) + 600( i2 - 0.6) = 0 -5 i1 + 20 i2 = 0.9

K(ii)

i1 = 20 mA, vx = -500 ´ 20m = -10 V 42. (C) 30 = 5 ia + 3( ia - 2.5) + 4( ia - 2.5 + 2)

Page 32

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CHAPTER

1.4 NETWORKS THEOREM

1. vTH , RTH = ?

3W

4. A simple equivalent circuit of the 2 terminal network

2W

6W

6V

shown in fig. P1.4.4 is

vTH, RTH

R

i

Fig. P.1.4.1

(A) 2 V, 4 W

(B) 4 V, 4 W

(C) 4 V, 5 W

(D) 2 V, 5 W

2. i N , R N = ?

2W

v

Fig. P.1.4.4

R

2W

R v

4W

15 V

iN, RN

(A)

(B)

Fig. P.1.4.2

10 W 3

(A) 3 A,

(B) 10 A, 4 W

(C) 1,5 A, 6 W

2W

3W

i i

(D) 1.5 A, 4 W

3. vTH , RTH = ?

2A

R R

(C)

(D)

5. i N , R N = ? 1W

2W

vTH, RTH 6A

4W

3W

iN RN

Fig. P.1.4.3

(A) -2 V,

6 W 5

5 (C) 1 V, W 6

(B) 2 V,

5 W 6

6 (D) -1 V, W 5

Fig. P.1.4.5

(A) 4 A, 3 W

(B) 2 A, 6 W

(C) 2 A, 9 W

(D) 4 A, 2 W

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Page 33


UNIT 1

6. vTH , RTH = ?

Networks

The value of the parameter are 25 W

30 W 20 W

vTH, RTH

5V 5A

vTH

RTH

iN

RN

(A)

4 V

2 W

2 A

2 W

(B)

4 V

2 W

2 A

3 W

(C)

8 V

1.2 W

30 3

A

1.2 W

8 V

5 W

8 5

A

5 W

(D) Fig. P.1.4.6

10. v1 = ? (A) -100 V, 75 W

(B) 155 V, 55 W

(C) 155 V, 37 W

(D) 145 V, 75 W

2W

3W

1W

6W

2W

8V

7. RTH = ?

1W

+ v1

6W

18 V

–

6W

Fig. P.1.4.10

6W

2A

RTH 5V

Fig. P.1.4.7

(A) 6 V

(B) 7 V

(C) 8 V

(D) 10 V

11. i1 = ?

(A) 3 W

(B) 12 W

(C) 6 W

(D) ¥

4 kW

8. The Thevenin impedance across the terminals ab of

12 V

i1

20 V

4 kW

6 kW

24 V

3 kW

4 kW

the network shown in fig. P.1.4.8 is a 3W

6W

2A

Fig. P.1.4.11

2V

8W

(A) 3 A

(B) 0.75 mA

(C) 2 mA

(D) 1.75 mA

8W

Statement for Q.12–13:

b

Fig. P.1.4.8

(A) 2 W

(B) 6 W

(C) 6.16 W

4 (D) W 3

A circuit is given in fig. P.1.4.12–13. Find the Thevenin equivalent as given in question.. 10 W

16 W

x

y

40 W

5V

8W

9. For In the the circuit shown in fig. P.1.4.9 a network and its Thevenin and Norton equivalent are given 2W

x’

3W

y’

Fig. P.1.4.12–13 RTH

4V

iN

2A vTH

RN

12. As viewed from terminal x and x¢ is (A) 8 V, 6 W

(B) 5 V, 6 W

(C) 5 V, 32 W

(D) 8 V, 32 W

Fig. P.1.4.9

Page 34

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1A


Network Theorems

13. As viewed from terminal y and y¢ is (A) 8 V, 32 W

(B) 4 V, 32 W

(C) 5 V, 6 W

(D) 7 V, 6 W

Chap 1.4

19. vTH , RTH = ? 6W

3i1

14. A practical DC current source provide 20 kW to a

i1

iN,

4W

RN

50 W load and 20 kW to a 200 W load. The maximum power, that can drawn from it, is

Fig. P1.4.19

(A) 22.5 kW

(B) 45 kW

(C) 30.3 kW

(D) 40 kW

Statement for Q.15–16: In the circuit of fig. P.1.4.15–16 when R = 0 W , the

(A) 0 W

(B) 1.2 W

(C) 2.4 W

(D) 3.6 W

20. vTH , RTH = ?

current iR equals 10 A. 2W

4W

4V

2W

+

E

4W

2W

R

5W

0.1v1

4A

vTH RTH

v1

iR

–

Fig. P.1.4.20

Fig. P.1.4.15–16.

15. The value of R, for which it absorbs maximum

(A) 8 V, 5 W

(B) 8 V, 10 W

power, is

(C) 4 V, 5 W

(D) 4 V, 10 W

(A) 4 W

(B) 3 W

(C) 2 W


21. RTH = ? 3W

2W

+

16. The maximum power will be (A) 50 W

(B) 100 W

(C) 200 W

(D) value of E is required

vx 4

4V

vx

RTH

–

17. Consider a 24 V battery of internal resistance

Fig. P.1.4.21

r = 4 W connected to a variable resistance RL . The rate of heat dissipated in the resistor is maximum when the current drawn from the battery is i . The current drawn form the battery will be i 2 when RL is equal to

(A) 3 W

(B) 1.2 W

(C) 5 W

(D) 10 W

(A) 2 W

(B) 4 W

22. In the circuit shown in fig. P.1.4.22 the effective

(C) 8 W

(D) 12 W

resistance faced by the voltage source is

18. i N , R N = ?

4W 5W

10 W i1

20i1

iN,

30 W

vs

RN

i 4

i

Fig. P.1.4.22 Fig. P.1.4.18

(A) 2 A, 20 W

(B) 2 A, -20 W

(A) 4 W

(B) 3 W

(C) 0 A, 20 W

(D) 0 A, -20 W

(C) 2 W

(D) 1 W

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Page 35


UNIT 1

23. In the circuit of fig. P1.4.23 the value of RTH at

Networks

26. The value of RL will be

terminal ab is

ix

0.75va 16 V

–

a

Fig. P.1.4.26–27

va +

4W

b

Fig. P.1.4.23

(A) -3 W

9 W 8

(B)

8 W 3

(C) -

RL

2W

8W

9V

3W

0.9 A

(A) 2 W

(B) 3 W

(C) 1 W


27. The maximum power is


(A) 0.75 W

(B) 1.5 W

(C) 2.25 W

(D) 1.125 W

28. RTH = ?

24. RTH = ?

-2ix

200 W – va 100

va +

100 W

50 W

RTH

100 W

300 W

Fig. P.1.4.24

(A) ¥ (C)

(D)

125 W 3

maximum power if RL is equal to

(B) 136.4 W

(C) 200 W

(D) 272.8 W

29. Consider the circuits shown in fig. P.1.4.29 ia

100 W

200 W

2W 6W

6W 2W

2W

3i

–

(A) 100 W

i 6V

RTH

Fig. P.1.4.28

25. In the circuit of fig. P.1.4.25, the RL will absorb 40 W

+ vx

800 W

ix

(B) 0

3 W 125

100 W

0.01vx

RL

12 V 12 V

8V

Fig. P.1.4.25

6W

(A)

400 W 3

(B)

2 kW 9

(C)

800 W 3

(D)

4 kW 9

ib

2W 6W

6W

2W


18 V

2W

6W

3A

In the circuit shown in fig. P1.4.26–27 the maximum power transfer condition is met for the load RL . Page 36

www.gatehelp.com

Fig. P.1.4.29a & b

12 V


Network Theorems

Chap 1.4

33. If vs1 = 6 V and vs 2 = - 6 V then the value of va is

The relation between ia and ib is (A) ib = ia + 6

(B) ib = ia + 2

(A) 4 V

(B) -4 V

(C) ib = 15 . ia

(D) ib = ia

(C) 6 V

(D) -6 V

34. A network N feeds a resistance R as shown in fig.

30. Req = ? 12 W

P1.4.34. Let the power consumed by R be P. If an

4W

identical network is added as shown in figure, the Req

6W

power consumed by R will be

2W

18 W

6W 9W

R

N

N

N

R

Fig. P.1.4.30 Fig. P.1.4.34

72 (B) W 13

(A) 18 W 36 (C) W 13

(D) 9 W

31. In the lattice network the value of RL for the maximum power transfer to it is

(A) equal to P

(B) less than P

(C) between P and 4P

(D) more than 4P

35. A certain network consists of a large number of ideal linear resistors, one of which is R and two constant ideal source. The power consumed by R is P1

7W

when only the first source is active, and P2 when only

6 W

the second source is active. If both sources are active RL

5

W

simultaneously, then the power consumed by R is

9W

(A) P1 ± P2

(B)

(C) ( P1 ± P2 ) 2

(D) ( P1 ± P2 ) 2

P1 ± P2

Fig. P.1.4.31

(A) 6.67 W

(B) 9 W

36. A battery has a short-circuit current of 30 A and an

(C) 6.52 W

(D) 8 W

open circuit voltage of 24 V. If the battery is connected to an electric bulb of resistance 2 W, the power dissipated by the bulb is

Statement for Q.32–33: A circuit is shown in fig. P.1.4.32–33. 12 W 1W

3W

3W

1W

(A) 80 W

(B) 1800 W

(C) 112.5 W

(D) 228 W

37.

+

The

following

results

were

obtained

from

measurements taken between the two terminal of a vs1

1W

va

vs2

resistive network

–

Terminal voltage

12 V

0V

Terminal current

0A

1.5 A

Fig. P.1.4.32–33

32. If vs1 = vs 2 = 6 V then the value of va is (A) 3 V

(B) 4 V

(C) 6 V

(D) 5 V

The Thevenin resistance of the network is (A) 16 W

(B) 8 W

(C) 0

(D) ¥

www.gatehelp.com

Page 37


UNIT 1

SOLUTIONS

38. A DC voltmeter with a sensitivity of 20 kW/V is used to find the Thevenin equivalent of a linear network. Reading on two scales are as follows

1. (B) vTH =

(a) 0 - 10 V scale : 4 V

The Thevenin voltage and the Thevenin resistance

(C) 18 V,

32 V, 3

1 MW 15

(D) 36 V,

200 kW 3

(B)

2 MW 15

( 6)( 6) = 4 V, 3+ 6

RTH = ( 3||6) + 2 = 4 W

(b) 0 -15 V scale : 5 V

of the network is 16 200 (A) V, kW 3 3

Networks

2. (A)

2W

isc

Fig. S.1.4.2

R N = 2 ||4 + 2 = + RL

4W

15 V

39. Consider the network shown in fig. P.1.4.39.

Linear Network

2W

v1

10 W, 3

15 2 v1 = =6 V 1 1 1 + + 2 2 4 v isc = i N = 1 = 3 A 2

vab –

Fig. P.1.4.39

The power absorbed by load resistance RL is shown in table :

(2)( 3)(1) = 1 V, 3+ 3 5 = 1||5 = W 6

3. (C) vTH =

RL

10 kW

30 kW

P

3.6 MW

4.8 MW

RTH

4. (B) After killing all source equivalent resistance is R The value of RL , that would absorb maximum power, is (A) 60 kW

(B) 100 W

(C) 300 W

(D) 30 kW

Open circuit voltage = v1 5. (D) The short circuit current across the terminal is 2W isc

40. Measurement made on terminal ab of a circuit of

6A

4W

3W

fig.P.1.4.40 yield the current-voltage characteristics shown in fig. P.1.4.40. The Thevenin resistance is Fig. S1.4.5

i(mA) +

30 Resistive Network

20 10 -4

-3 -2

-1

0

vab –

1

2

v

Fig. P.1.4.40

a

b

isc =

6´ 4 = 4 A = iN , 4+2

R N = 6 ||3 = 2 W 6. (B) For the calculation of RTH if we kill the sources then 20 W resistance is inactive because 5 A source will

(A) 300 W

(B) -300 W

be open circuit

(C) 100 W

(D) -100 W

RTH = 30 + 25 = 55 W, vTH = 5 + 5 ´ 30 = 155 V

***********

Page 38

www.gatehelp.com


Network Theorems

Chap 1.4

12. (B) We Thevenized the left side of xx¢ and source

7. (C) After killing the source, RTH = 6 W

transformed right side of yy¢

6W

8W

16 W

x

y

8W

6W RTH 4V

8V

x’

Fig. S.1.4.7

y’

Fig. S1.4.12

8. (B) After killing all source, RTH = 3||6 + 8 ||8 = 6 W

4 8 + = 8 24 = 5 V, 1 1 + 8 24

vxx ¢ = vTH

9. (D) voc = 2 ´ 2 + 4 = 8 V = vTH RTH = 2 + 3 = 5 W = R N ,

iN =

RTH = 8 ||(16 + 8) = 6 W

vTH 8 = A RTH 5

13. (D) Thevenin equivalent seen from terminal yy¢ is 4 8 + 24 8 = 7 V, = 1 1 + 24 8

10. (A) If we solve this circuit direct, we have to deal with three variable. But by simple manipulation

v yy¢ = vTH

variable can be reduced to one. By changing the LHS and RHS in Thevenin equivalent 1W

1W

1W

RTH = ( 8 + 16)||8 = 6 W

2W

14. (A)

+ 6W

4V

12 V

v1 –

i

RL

r

Fig. S1.4.10 Fig. S1.4.14

4 12 + v1 = 1 + 1 1 + 2 = 6 V 1 1 1 + + 1+1 6 1+2

2

æ ir ö ç ÷ 50 = 20 k, è r + 50 ø

( r + 200) 2 = 4( r + 50) 2

11. (B) If we solve this circuit direct, we have to deal with three variable. But by simple manipulation

Þ

r = 100 W

i = 30 A,

Pmax =

variable can be reduced to one. By changing the LHS and RHS in Thevenin equivalent 2 kW

i1

4 kW

2

æ ir ö ç ÷ 200 = 20 k è r + 200 ø

20 V

6V

( 30) 2 ´ 100 = 22.5 kW 4

15. (C) Thevenized the circuit across R, RTH = 2 W 4W

2 kW

2W

2W

4W

8V

2W

Fig. S1.4.15 Fig. S1.4.11

i1 =

20 - 6 - 8 = 0.75 mA 2k + 4k + 2k

16. (A) isc = 10 A, RTH = 2 W, 2

æ 10 ö Pmax = ç ÷ ´ 2 = 50 W è 2 ø www.gatehelp.com

Page 39


UNIT 1

Now in this circuit all straight-through connection

Networks

i 2 R = ( P1 ± P2 ) 2

have been cut as shown in fig. S1.4.32b 6W

36. (C) r =

1W 3W 2W

+

P=

voc = 1. 2 W isc

24 2 ´ 2 = 112.5 W (1. 2 + 2) 2

6V

va

37. (B) RTH =

–

voc 12 = =8W isc 15 .

Fig. S.1.4.32b

va =

6 ´ (2 + 3) =5 V 2 + 3+1

38. (A) Let

33. (B) Since both source have opposite polarity, hence short circuit the all straight-through connection as shown in fig. S.1.4.33 6W

2W

For 0 -10 V scale Rm = 10 ´ 20 k = 200 kW For 0 -50 V scale Rm = 50 ´ 20 k = 1 MW 4 For 4 V reading i = ´ 50 = 20 mA 10 vTH = 20mRTH + 20m ´ 200 k = 4 + 20mRTH 5 For 5 V reading i = ´ 50m = 5 mA 50

1W 3W

1 1 = = 50 mA sensitivity 20 k

+

vTH = 5m ´ RTH + 5m ´ 1M = 5 + 5mRTH 6V

va

Solving (i) and (ii) 16 200 V, RTH = vTH = kW 3 3

–

Fig. S1.4.33

39. (D) v10 k = 10 k ´ 3.6m = 6

6 ´ ( 6 ||3) va = = -4 V 2 +1

v30 k = 30 k ´ 4.8m = 12 V

34. (C) Let Thevenin equivalent of both network RTH

RTH

vTH

R

vTH

RTH

R

vTH

6 =

10 vTH 10 + RTH

12 =

30 vTH 30 + RTH

Þ

Þ

10 vTH = 6 RTH + 60

5 vTH = 2 RTH + 60

RTH = 30 kW 40. (D) At v = 0 , isc = 30 mA

Fig. S1.4.34

At i = 0, voc = - 3 V v -3 RTH = oc = = - 100 W isc 30m

2

æ VTH ö ÷÷ R P = çç è RTH + R ø æ ç V TH P¢ = ç çç R + RTH 2 è

2

ö ÷ æ VTH ÷ R = 4çç ÷÷ è 2 R + RTH ø

2

ö ÷÷ R ø

************

Thus P < P ¢ < 4 P 35. (C) i1 =

P1 P2 and i2 = R R

using superposition i = i1 + i2 =

Page 42

P1 ± R

P2 R www.gatehelp.com

...(i)

...(ii)


CHAPTER

1.6 THE RLC CIRCUITS

1. The natural response of an RLC circuit is described

(A) iL¢¢ ( t) + 1100 i¢¢L ( t) + 11 ´ 108 iL ( t) = 108 is ( t)

by the differential equation

(B) i¢¢L ( t) + 1100 i¢¢L ( t) + 11 ´ 108 iL ( t) = 108 is ( t)

d 2v dv dv(0) +2 + v = 0, v(0) = 10, = 0. dt 2 dt dt The v( t) is (A) 10(1 + t) e - t V

(B) 10(1 - t) e - t V

(C) 10e - t V

(D) 10te - t V

vs

2W

1 mH

. i¢¢L ( t) iL¢¢ ( t) 11 . iL ( t) = is ( t) + + 11 108 10 4

(D)

i¢¢L ( t) 11i¢L ( t) + + 11iL ( t) = is ( t) 108 10 4

4. In the circuit of fig. P.1.6.4 vs = 0 for t > 0. The initial

2. The differential equation for the circuit shown in fig. P1.6.2. is

(C)

condition are v(0) = 6 V and dv(0) dt = -3000 V s. The v( t) for t > 0 is

v

1H

10 mF

100 W

vs

25 mF

80 W

+ vC –

Fig. P1.6.2

(A) v¢¢( t) + 3000 v¢( t) + 102 . ´ 108 v( t) = 108 vs ( t)

Fig. P1.6.4

(B) v¢¢( t) + 1000 v¢ ( t) + 102 . ´ 10 v( t) = 10 vs ( t) 8

8

(A) -2 e -100 t + 8 e -400 t V

(B) 6 e -100 t + 8 e -400 t V (D) None of the above

(C)

2 v¢ ( t) v¢¢( t) . v( t) = vs ( t) + + 102 108 10 5

(C) 6 e -100 t - 8 e -400 t V

(D)

2 v¢ ( t) v¢¢( t) . v( t) = vs ( t) + + 198 8 10 10 5

5. The circuit shown in fig. P1.6.5 has been open for a long time before closing at t = 0. The initial condition is

3. The differential equation for the circuit shown in fig.

v(0) = 2 V. The v( t) for t > is

P1.6.3 is

t=0 1H

10 W is

3 4W

1 3F

+ vC –

10 mF

100 W

Fig. P.1.6.5 iL

Fig. P.1.6.3

Page 54

-t

(A) 5 e - 7 e

-3t

V

(C) - e - t + 3e -3t V www.gatehelp.com

(B) 7 e - t - 5 e -3t V (D) 3e - t - e -3t V


The RLC Circuits


Chap 1.6

10. The switch of the circuit shown in fig. P1.6.10 is

Circuit is shown in fig. P.1.6. Initial conditions are

opened at t = 0 after long time. The v( t) , for t > 0 is

i1 (0) = i2 (0) = 11 A

t=0

3W

i1

i2

2H

1W

1 2H

1W

6V

+ vC –

1 4F

2W

3H

Fig. P1.6.10 Fig. P1.6.6–7

6. i1 (1 s) = ?

(A) 4 e -2 t sin 2 t V

(B) -4 e -2 t sin 2 t V

(C) 4 e -2 t cos 2 t V

(D) -4 e -2 t cos 2 t V

(A) 0.78 A

(B) 1.46 A

11. In the circuit of fig. P1.6.23 the switch is opened at

(C) 2.56 A

(D) 3.62 A

t = 0 after long time. The current iL ( t) for t > 0 is 4H

iL

7. i2 (1 s) = ? (A) 0.78 A

(B) 1.46 A

(C) 2.56 A

(D) 3.62 A

t=0 2W

1 4F

8W

8. vC ( t) = ? for t > 0

4W

7A

25 mH

Fig. P1.6.11

10 mF

100 W

30u(-t) mA

+ vC –

(A) e -2 t (2 cos t + 4 sin t) A

(B) e -2 t ( 3 sin t - 4 cos t) A

(C) e -2 t ( -4 sin t + 2 cos t) A

(D)e -2 t (2 sin t - 4 cos t) A

Statement for Q.12–14: Fig. P1.6.8

In the circuit shown in fig. P1.6.12–14 all initial

(A) 4 e -1000 t - e -2000 t V

(B) ( 3 + 6000 t) e -2000 t V

(C) 2 e -1000 t + e -2000 t V

(D) ( 3 - 6000 t) e -2000 t V

condition are zero. iL

9. The circuit shown in fig. P1.6.9 is in steady state with switch open. At t = 0

10 mH

1 mF

+ vL –

the switch is closed. The

output voltage vC ( t) for t > 0 is

Fig. P1.5.12-14

12. If is ( t) = 1 A, then the inductor current iL ( t) is

0.8 H

250 W

100 W 65

isu(t) A

500 W t=0 9V

5 mF

+ vC –

(A) 1 A

(B) t A

(C) t + 1 A

(D) 0 A

13. If is ( t) = 0.5 t A,

Fig. P1.6.9

A

(B) 2 t - 3250 A

-3

A

(D) 2 t + 3250 A

(A) 0.5 t + 3. 25 ´ 10 (C) 0.5 t - 0. 25 ´ 10

then iL ( t) is

-3

(B) e

-400 t

[ 3 cos 300 t + 4 sin 300 t ]

14. If is ( t) = 2 e -250 t A then iL ( t) is 4000 -250 t 4000 -250 t (A) A (B) A te e 3 3

(C) e

-300 t

[ 3 cos 400 t + 4 sin 300 t ]

(C)

(A) -9 e

-400 t

+ 12 e

-300 t

(D) e -300 t [ 3 cos 400 t + 2. 25 sin 300 t ]

200 -250 t A e 7

www.gatehelp.com

(D)

200 -250 t A te 7

Page 55


UNIT 1

15. The forced response for the capacitor voltage v f ( t) is

Networks

19. In the circuit shown in fig. P 1.5.19 v( t) for t > 0 is

100 W

2u(-t) A iL

– vx +

avx

50 W

1H

20 mH

0.04 F + vC – 2W

4W

Fig. P1.6.15

(A) 0. 2 t + 117 . ´ 10 -3 V

(B) 0. 2 t - 117 . ´ 10 -3 V

(C) 117 . ´ 10 -3 t - 0. 2 V

(D) 117 . ´ 10 -3 t + 0. 2 V

50u(t) V

Fig. P1.6.19

16. For a RLC series circuit R = 20 W , L = 0.6 H, the value of C will be

(B) 50 + ( 46.5 sin 3t + 62 cos 3t) e -4 t V

[CD =critically damped, OD =over damped,

(C) 50 + ( 62 cos 4 t + 46.5 sin 4 t) e -3t V

UD =under damped]. CD

OD

UD

(A) C = 6 mF

C >6 mF

C <6 mF

(B) C = 6 mF

C < 6 mF

C > 6 mF

(C) C >6 mF

C = 6 mF

C < 6 mF

(D) C < 6 mF

C =6 mF

C > 6 mF

(A) 50 - ( 46.5 sin 3t + 62 cos 3t) e -4 t V

(D) 50 - ( 62 cos 4 t + 46.5 sin 4 t) e -3t V 20. In the circuit of fig. P1.6.20 the switch is closed at t = 0 after long time. The current i( t) for t > 0 is 1 16 F + vC –

17. The circuit shown in fig. P1.6.17 is critically damped. The value of R is

20 V

t=0

1 4H

Fig. P1.6.20

120 W

R

5W

iL

10 mF 4H

(A) -10 sin 8 t A

(B) 10 sin 8 t A

(C) -10 cos 8 t A

(D) 10 cos 8 t A

21. In the circuit of fig. P1.6.21 switch is moved from 8 Fig. P1.6.17

(A) 40 W

(B) 60 W

(C) 120 W

(D) 180 W

V to 12 V at t = 0. The voltage v( t) for t > 0 is 2W t=0

18. The step response of an RLC series circuit is given

8V

12 V

by

1H

d 2 i( t) 2 di( t) di(0 + ) = 4. + + 5 i( t) = 10, i(0 + ) = 2, dt dt dt

Fig. P1.6.21

(A) 12 - ( 4 cos 2 t + 2 sin 2 t) e - t V

The i( t) is (A) 1 + e - t cos 4 t A

(B) 4 - 2 e - t cos 4 t A

(B) 12 - ( 4 cos 2 t + 8 sin 2 t) e - t V

(C) 2 + e - t sin 4 t A

(D) 10 + e - t sin 4 t A

(C) 12 + ( 4 cos 2 t + 8 sin 2 t) e - t V (D) 12 + ( 4 cos 2 t + 2 sin 2 t) e - t V

Page 56

www.gatehelp.com

1 F 6

+ vC –


The RLC Circuits

22. In the circuit of fig. P1.5.22 the voltage v( t) is

5W

3u(t) A

25. In the circuit shown in fig. P1.6.25 a steady state has been established before switch closed. The i( t) for

1W

5H

t > 0 is 5W

+ vC –

0.2 F

20 V

Chap 1.6

20 W

1H

2W

i

t=0

1 F 25

5W

100 V

Fig. P1.6.22

(A) 40 - (20 cos 0.6 t + 15 sin 0.6 t) e -0 .8 t V (B) 35 + (15 cos 0.6 t + 20 sin 0.6 t) e

-0 .8 t

(C) 35 - (15 cos 0.6 t + 20 sin 0.6 t) e (D) 35 - 15 cos 0.6 t e

-0 .8 t

-0 .8 t

Fig. P1.6.25

V

(A) 0.73e -2 t sin 4.58 t A

V

(B) 0.89 e -2 t sin 6.38 t A

V

(C) 0.73e -4 t sin 4.58 t A

23. In the circuit of fig. P1.6.23 the switch is opened at t = 0 after long time. The current i( t) for t > 0 is

(D) 0.89 e -4 t sin 6.38 t A 26. The switch is closed after long time in the circuit of fig. P1.6.26. The v( t) for t > 0 is

2A

2A 3 4H 10 W

1 3F

t=0

1W

1H 6W

5W

10 W

4V

t=0

1 25 F

+ vC –

Fig. P1.6.23

(A) e-2 .306 t + e-0 . 869t A

Fig. P1.6.26

(B) -e -2 .306 t + 2 e -0 . 869t A

(A) -8 + 6 e -3t sin 4 t V

(C) e -4 .431 t + e -0 .903t A

(B) -12 + 4 e -3t cos 4 t V

(D) 2 e -4 .431 t - e -0 .903t A

(C) -12 + ( 4 cos 4 t + 3 sin 4 t) e -3t V

24. In the circuit of fig. P1.6.24 switch is moved from position a to b at t = 0. The iL ( t) for t > 0 is 0.02 F

(D) -12 + ( 4 cos 4 t + 6 sin 4 t) e -3t V 27. i( t) = ? 2 kW

14 W

b

i

2 H t=0 2W

12 V

5 mF

12u(t) V

8 mH

a iL

6W

Fig. P1.6.27

(A) 6 - ( 6 cos 500 t + 6 sin 5000 t) e -50 t mA (B) 8 - ( 8 cos 500 t + 0.06 sin 5000 t) e -50 t mA

4A

Fig. P1.6.24

(A) ( 4 - 6 t) e 4 t A

(B) ( 3 - 6 t) e -4 t A

(C) ( 3 - 9 t) e -5t A

(D) ( 3 - 8 t) e -5t A

(C) 6 - ( 6 cos 5000 t + 0.06 sin 5000 t) e -50 t mA (D) 6 e -50 t sin 5000 t mA www.gatehelp.com

Page 57


UNIT 1

Networks

SOLUTIONS

28. In the circuit of fig. P1.6.28 i(0) = 1 A and v(0) = 0. The current i( t) for t > 0 is i 4u(t) A

1. (A) s 2 + 2 s + 1 = 0 2W

1H

s = -1, - 1,

v( t) = ( A1 + A2 t) e dv(0) v(0) = 10 V, = 0 = - 1 A1 + A2 dt

0.5 F

A1 = A2 = 10

Fig. P1.6.28

(A) 4 + 6.38 e -0 .5t A

Þ

-t

(B) 4 - 6.38 e -0 .5t A

(C) 4 + ( 3 cos 1.32 t + 113 . sin 1.32 t) e -0 .5t A

2. (A) iL =

(D) 4 - ( 3 cos 1.32 t + 113 . sin 1.32 t) e -0 .5t A

v dv + 10 ´ 10 -6 100 dt iL

29. In the circuit of fig. P1.6.29 a steady state has been

2W

1 mH

v

established before switch closed. The vo( t) for t > 0 is 10 W

vs

t=0 5W

3A

+

10 mF

1H

Fig. S1.6.2

vo –

vs = 2 iL + 10 -3

Fig. P1.6.29

(A) 100 te -10 t V

(B) 200 te -10 t V

(C) 400 te -50 t V

(D) 800 te -50 t V

108 vs ( t) = v¢¢( t) + 3000 v¢( t) + 102 . v( t) t > 0 is

established before switch closed. The i( t) for i 1H 1 4F

2W t=0

(B) -e

sin 2 t A

(C) -2(1 - t) e

-2 t

-2 t

sin 2 t A

(D) 2(1 - t) e -2 t A

A

4. (A)

31. In the circuit of fig. P1.6.31 a steady state has been established. The i( t) for t > 0 is i 3A

10 W

6u(t) A

diL di d 2 iL + iL + 10 -4 L + 10 -8 dt dt dt 2 . i¢¢L ( t) 11 . iL ( t) = is ( t) + iL¢ ( t) + 11 108 10 4

Þ

Fig. P1.6.30

(A) 2 e

vC dvC + iL + 10m 100 dt di vC = 10 iL + 10 -3 L dt di d di is = 0.1iL + 10 -5 L + iL + 10 -5 (10 iL + 10 -3 L ) dt dt dt 3. (C) is =

= 0.1iL + 10 -5

6V

-2 t

diL +v dt

æ 1 dv dv ö d2 v ö æ v ÷+ v ÷ + 10 -3 çç = 2ç + 10 -6 ´ 10 - t + 10 ´ 10 -6 dt ø dt 2 ÷ø è 100 è 100 dt

30. In the circuit of fig. P1.6.30 a steady state has been

1W

10 mF

100 W

40 W

10 mF

4H

v dv + 25m + ( v - vs ) dt = 0 80 dt ò

d 2v dv + 500 + 40000 = 0 2 dt dt

Þ

s 2 + 500 s + 40000 = 0 Þ

s = -100, - 400,

v( t) = Ae -100 t + Be -400 t A + B = 6, -100 A - 400 B = -3000

Fig. P1.6.31

(A) 9 + 2 e

-10 t

- 8e

-2 .5t

(B) 9 - 8 e

10 t

A

(C) 9 + (2 cos 10 t + sin 10 t) e

-2 .5t

A

(D) 9 + (cos 10 t + 2 sin 10 t) e -2 .5t A ***************

Page 58

+ 2e

-2 .5t

Þ

B = 8, A = -2

A 5. (C) The characteristic equation is s 2 + After putting the values, v( t) = Ae - t + Be -3t , www.gatehelp.com

s2 + 4 s + 3 = 0

1 s + =0 RC LC


The RLC Circuits

v(0 + ) = 2 V

Þ

iL (0 + ) = 0Þ

iR (0) =

-C

9. (B) vC (0 + ) = 3 V , iL (0 + ) = -12 mA

A + B =2

dv(0 + ) 8 = dt 3

vC dvC + iL + 5 ´ 10 -6 =0 250 dt

2 8 = , 34 3

dvC (0 + ) 3 dvC (0 + ) - 12m + 5 ´ 10 -6 =0 Þ =0 dt 250 dt 1 s + =0 s2 + -6 250 ´ 5 ´ 10 0.8 ´ 5 ´ 10 -6

dv(0 + ) = - 8, dt

Þ

- A - 3B = -8, B = 3, A = -1

Þ

di1 di - 3 2 = 0, dt dt 3di2 di 2 i2 + - 3 1 =0 dt dt 6. (D) i1 + 5

Þ

i1 = A e

+ Be

-2 t

10. (B) v(0 + ) = 0, s2 + 4 s + 8 = 0 A1 = 0,

, i(0) = A + B = 11

-

1 6

i2 = - e

2W

1 4F

-

t 6

+ De

+ 12 e

di2 (0) -143 C = = - - 2D dt 6 6

A, i2 (1 s) = e

+ vC –

8W

diL (0 + ) diL (0 + ) = 8 - ( -4) ´ 8 Þ = 10 dt dt 1 s vC + vC + iL = 0, vC = 4 siL + 8 iL 4 2

4

-2 t

s 2 iL + 4 siL + 5 = 0, s = -2 ± j

and D = 12 -2 t

4H

Fig. S1.6.11

-

1 6

+ 12 e

-2

= 0.78 A

iL ( t) = e-2 t ( A1 cos t + A2 sin t) A1 = -4,

8. (B) vC (0 + ) = 30m ´ 100 = 3 V dvC (0 - ) dvC (0 + ) = iL (0 - ) = 0 = iL (0 + ) = C dt dt 100 1 s2 + s+ 25 ´ 10 -3 25 ´ 10 -3 ´ 10 ´ 10 -6 C

Þ

vC (0 + ) = 8 V iL

+ 8 e -2 = 3.62 A

i2 (0) = 11 = C + D,

t 6

11. (D) iL (0 + ) = -4,

+ 8e ,

7. (A) i2 = Ce

-

dvC (0 + ) = -8 = -2 (0 + 0) + (0 + 2 A2 ), A2 = -4 dt

-2 t

i1 (1 s) = 3e

C = -1

s = - 2 ± j2

vC ( t) = e ( A1 cos 2 t + A2 sin 2 t)

di1 (0 + ) 33 di2 (0 + ) 143 ==, dt 2 dt 6 A 33 - 2 B = - , Þ A = 3, B = 8, 6 2 i1 = 3e

Þ

1 dvC (0 + ) = -2 4 xdt

iL (0 + ) = 2 A,

-2 t

In differential equation putting t = 0 and solving

t 6

s = -400 ± j 300

vC ( t) = e-400 t ( A1 cos 300 t + A2 sin 300 t) dvC (0) A1 = 3, = -400 A1 + 300 A2 , A2 = 4 dt

6 s 2 + 13s + 2 = 0 1 s = - , -2 6 1 - t 6

s 2 + 800 s + 25 ´ 10 4 = 0 Þ

(1 + 5 s) i1 - 3si2 = 0, -3si1 + (2 + 3s) i2 = 0 ( 3s)( 3s) i1 (1 + 5 s) i1 =0 2 + 3s Þ

Chap 1.6

s = -2000, -2000

diL (0 + ) = 10 = -2 ( A1 + 0) + A2 , A2 = 2 dt

12. (A) is = is =

v dv di + 10 -3 + iL , v = 10 ´ 10 -3 L 100 65 dt dt

di d 2 iL 65 (10 ´ 10 -3) L + 10 -3(10 ´ 10 -3) + iL = 0 dt dt 100

d 2 iL di + 650 L + 10 5 iL = 10 5 is dt dt

vC ( t) = ( A1 + A2 t) e -2000 t dvC ( t) = A2 e -2000 t + ( A1 + A2 t) e -2000 t ( -2000) dt dvC (0) = A2 - 2000 ´ 3 = 0 vC (0 + ) = A1 = 3, dt

0 + 0 + 10 5 B = 10 5,

Þ

0 + 650 A + ( At + B)10 5 = 10 5(0.5 t), A = 0.5

A2 = 6000

Trying iL ( t) = B B = 1,

iL = 1 A

13. (A) Trying iL ( t) = At + B, www.gatehelp.com

Page 59


The RLC Circuits

Chap 1.6

di(0 + ) -16 = = -4.431 A - 0.903B dt 3

28.(D) i(0 + ) = 1 A, v(0 + ) =

A = 1, B = 1

a=

Ldi(0 + ) dt

1 = 0.5, 2 ´ 2 ´ 0.5

4´ 6 =3 6 +2 dvC (0) dv (0) = 150 0.02 C = iL (0) = 3 Þ dt dt 6 + 14 1 a= = 5, wo = =5 2´2 2 ´ 0.02

1 = 4 + A, Þ A = -3 di(0) = 0 = 0.5 A + 1.32 B, dt

a = wo critically damped

29. (B) Vo(0 + ) = 0 ,iL (0 + ) = 1 A

24. (C) vC (0) = 0,

v( t) = 12 + ( A + Bt) e 0 = 12 + A,

v( t) = 12 + (90 t - 12) e

A = -12,

iL ( t) = 0.02( -5) e (90 t - 12) + 0.02(90) e

-5t

= ( 3 - 9 t) e

-5t

diL (0 + ) = v1 (0) = 0 dt 1 a= = 10, 2 ´ 5 ´ 0.01

B = -113 .

Wo =

1 1 ´ 0.01

= 10

a = Wo, so critically damped response

100 ´ 5 50 , 25. (A) v(0 ) = = 5 + 5 + 20 3 +

+

iL (0 ) = 0

s = -10, - 10 i( t) = 3( A + Bt) e -10 t ,

if = 0 A

i(0) = 1 = 3 + A

+

di(0 ) = -10 A + B dt

diL (0 + ) 50 10 = 20 = dt 3 3 4 1 a= = 2, wo = =5 2´1 1 1´ 25

iL ( t) = 3 - (2 + 20 t) e -10 t ,

i( t) = ( A cos 4.58 t + B sin 4.58 t) e -2 t 26.(A) iL (0 + ) = 0, vL (0 + ) = 4 - 12 = -8 +

LdiL ( t) = 200 te -10 t dt

di(0 + ) -6 vc (0 + ) = 2 ´ 1 = 2 = = -2 A, dt 1+2 1 1 1 a= = = 2, Wo = =2 2 RC 2 ´ 1 ´ 0.25 LC a = Wo, critically damped response s = -2 , -2

1 dvL (0 ) = iL (0 + ) = 0 25 dt 6 1 a = = 3, Wo = =5 2 1 ´ 1 / 25

A = -2 i( t) = ( A + Bt) e -2 t , di( t) = ( -2 + Bt) e 2 t ( -2) + (0 + B) e -2 t dt

b = -3 ± 9 - 25 = -3 ± j 4 v1 ( t) = -12 + ( A cos 4 t + B sin 4 t) e

vo =

30. (C) i(0 + ) =

s = -2 ± 4 - 25 = -2 ± j 4.58

At t = 0, Þ B = -2 -3t

vL (0) = -8 = 12 + A, Þ A =4 dvL (0) = 0 = -3 A + 4 B, Þ B=3 dt 1 1 = = 50 2 RC 2 ´ 2 k ´ 54 1 1 = = 5000 LC 8 m ´ 5m

27. (C) a = Wo =

B = 90

-5t

-5t

= 2

i( t) = 4 + ( A cos 1.32 t + B sin 1.32 t) e -0 .5t

-5t

Þ

1 - 0.5

s = -0.5 ± 0.5 2 - 2 = 0.5 ± j1.323

iL (0) =

150 = -5 A + B

1

Wo =

31. (A) i(0 + ) = 3 A, vC (0 + ) = 0 V = is = 9 A, R = 10||40 = 8 W 1 1 a= = = 6.25 2 RC 2 ´ 8 ´ 0.01 1 1 Wo = = =5 LC 4 ´ 0.01 a > Wo, so overdamped response

a < Wo, underdamped response.

s = -6.25 ± 6.25 2 - 25 = -10, -2.5

s = -50 ± 50 2 - 5000 2 = -50 ± j5000

i( t) = 9 + Ae -10 t + Be -2 .5t

i( t) = 6 + ( A cos 5000 t + B sin 5000 t) e -50 t mA

3 = 9 + A + B,

i(0) = 6 = 6 + A, Þ A = -6 di(0) = -50 A + 5000 B = 0, B = -0.06 dt

4 di(0 + ) dt

0 = -10 A - 2.5 B

On solving, A = 2, B = -8 ************ www.gatehelp.com

Page 61


CHAPTER

1.7 SINUSOIDAL STEADY STATE ANALYSIS

1. i( t) = ?

20cos 300t V

(A)

3W

i

(C)

~

25 mH

1 2 1 2

cos (2 t - 45 ° ) V

(B)

sin (2 t - 45 ° ) V

(D)

1 2 1 2

cos (2 t + 45 ° ) V sin (2 t + 45 ° ) V

4. vC ( t) = ? Fig. P1.7.1

3H

(A) 20 cos ( 300 t + 68. 2 ° ) A (B) 20 cos( 300 t - 68. 2 ° ) A

8cos 5t V

(C) 2.48 cos( 300 t + 68. 2 ° ) A

~

+ vC –

50 mF 9W

(D) 2.48 cos( 300 t - 68. 2 ° ) A

Fig. P1.7.4

2. vC ( t) = ? (A) 2. 25 cos (5 t + 150 ° ) V 3

cos 10 t A

~

2W

+ vC –

1 mF

(B) 2. 25 cos (5 t - 150 ° ) V (C) 2. 25 cos (5 t + 140.71° ) V (D) 2. 25 cos (5 t - 140.71° ) V

Fig. P1.7.2

(A) 0.89 cos (10 3 t - 63.43° ) V

5. i( t) = ?

(B) 0.89 cos (10 3 t + 63.43° ) V

1W

4W

(C) 0.45 cos (10 t + 26.57 ° ) V 3

i

(D) 0.45 cos (10 t - 26.57 ° ) V 3

10cos 2t V

~

0.25 F

4H

3. vC ( t) = ? 5W

cos 2t V

~

0.1 F

Fig. P1.7.5 + vC –

(A) 2 sin (2 t + 5.77 ° ) A

(B) cos (2 t - 84. 23° ) A

(C) 2 sin (2 t - 5.77 ° ) A

(D) cos (2 t + 84. 23° ) A

Fig. P1.7.3

Page 62

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UNIT 1

13. In the bridge shown in fig. P1.7.13, Z1 = 300 W, Z 2 = ( 300 - j 600) W, Z 3 = (200 + j 100)W. The Z 4

Networks

Statement for Q.17-18:

at

The circuit is as shown in fig. P1.7.17-18

balance is

1W

1W

1H

Z

Z

3

1

1H

1W

~

i1

o

10cos (4t-30 ) V

1F

Z

Z

~

i2

4

2

5cos 4t V

Fig. P1.7.17–18

~

17. i1 ( t) = ? (A) 2.36 cos ( 4 t - 4107 . °) A

Fig. P1.7.13

(A) 400 + j 300 W

(B) 400 - j 300 W

(C) j100 W

(D) - j900 W

(B) 2.36 cos ( 4 t + 4107 . °) A (C) 1.37 cos ( 4 t - 4107 . °) A (D) 2.36 cos ( 4 t + 4107 . °) A

14. In a two element series circuit, the applied voltage

and

the

resulting

current

are

v( t) = 60 + 66 sin (10 3 t) V, i( t) = 2.3 sin (10 3 t + 68.3° ) A. The nature of the elements would be (A) R - C

(B) L - C

(C) R - L

(D) R - R

(A) 2.04 sin ( 4 t + 92.13° ) A (B) -2.04 sin ( 4 t + 2.13° ) A (C) 2.04 cos ( 4 t + 2.13° ) A (D) -2.04 cos ( 4 t + 92.13° ) A 19. I x = ?

15. Vo = ? 40 W

j20

120Ð-15o V

18. i2 ( t) = ?

~

~

50 W

-j30

Ix 6Ð30 A

10Ð30 V

o

o

~

Fig. P1.7.15

-j2 W

(B) 223Ð56 ° V

(C) 124 Ð - 154 ° V

(D) 124 Ð154 ° V

16. vo( t) = ?

(A) 394 . Ð46. 28 ° A

(B) 4.62 Ð97.38 ° A

(C) 7.42 Ð92.49 ° A

(D) 6.78 Ð49. 27 ° A

20. Vx = ?

3W

j10 W

20 W 10sin (t+30o) V

~

1F

+ vo –

j3 W

Fig. P1.7.19

(A) 223Ð - 56 ° V

1H

0.5Ix

4W

Vo

~

o

20cos (t-45 ) V

4Vx

3Ð0 A o

~

20 W

+ Vx –

Fig. P1.7.16 Fig. P1.7.20

(A) 315 . cos ( t + 112 ° ) V (B) 43. 2 cos ( t + 23° ) V

(A) 29.11Ð166 ° V

(B) 29.11Ð - 166 ° V

(C) 315 . cos ( t - 112 ° ) V

(C) 43. 24 Ð124 ° V

(D) 43. 24 Ð - 124 ° V

(D) 43. 2 cos ( t - 23° ) V Page 64

www.gatehelp.com


UNIT 1


Networks

35. In the circuit shown in fig. P1.7.35 power factor is

Determine the complex power for hte given values

-j2

4W

in question. 27. P = 269 W, Q = 150 VAR (capacitive) (A) 150 - j269 VA

(B) 150 + j269 VA

(C) 269 - j150 VA

(D) 269 + j150 VA

-j2

j5

Fig. P1.7.35

28. Q = 2000 VAR, pf = 0.9 (leading) (A) 4129.8 + j2000 VA

(B) 2000 + j 4129.8 VA

(C) 2000 - j 4129 . .8 VA

(D) 4129.8 - j2000 VA

(A) 56.31 (leading)

(B) 56.31 (lagging)

(C) 0.555 (lagging)

(D) 0.555 (leading)

36. The power factor seen by the voltage source is 29. S = 60 VA, Q = 45 VAR (inductive) (A) 39.69 + j 45 VA

(B) 39.69 - j 45 VA

(C) 45 + j 39.69 VA

(D) 45 - j 39.69 VA

4W

1W

+ v1 – 10cos 2t V

30. Vrms = 220 V, P = 1 kW, |Z |= 40 W (inductive) (A) 1000 - j 68125 . VA

(B) 1000 + j 68125 . VA

(C) 68125 . + j1000 VA

(D) 68125 . - j1000 VA

31. Vrms = 21Ð20 ° V, Vrms = 21Ð20 ° V, I rms = 8.5 Ð - 50 ° A (A) 154.6 + j 89.3 VA

(B) 154.6 - j 89.3 VA

(C) 61 + j167.7 VA

(D) 61 - j167.7 VA

~

1 3F

3v 4 1

Fig. P1.7.36

(A) 0.8 (leading)

(B) 0.8 (lagging)

(C) 36.9 (leading)

(D) 39.6 (lagging)

37. The average power supplied by the dependent source is

32. Vrms = 120 Ð30 ° V, Z = 40 + j 80 W

Ix

(A) 72 + j144 VA

(B) 72 - j144 VA

(C) 144 + j72 VA

(D) 144 - j72 VA

2Ð90 A o

~

j1.92

4.8 W

8W

1.6Ix

33. Vo = ? Fig. P1.7.37

+

6Ð0 A o

~

16 kW 0.9 pf lagging

VO

20 kW 0.8 pf lagging

(A) 96 W

(B) -96 W

(C) 92 W

(D) -192 W

–

38. In the circuit of fig. P1.7.38 the maximum power

Fig. P1.7.33

absorbed by Z L is (A) 7.1Ð32. 29 ° kV

(B) 42.59 Ð32.29 ° kV

(C) 38.49 Ð24.39 ° kV

(D) 38.49 Ð32. 29 ° kV

10 W

120Ð0o V

34. A relay coil is connected to a 210 V, 50 Hz supply. If

~

j15

-j10

it has resistance of 30 W and an inductance of 0.5 H, the Fig. P1.7.38

apparent power is

Page 66

(A) 30 VA

(B) 275.6 VA

(C) 157 VA

(D) 187 VA

(A) 180 W

(B) 90 W

(C) 140 W

(D) 700 W

www.gatehelp.com

ZL


Sinusoidal Steady State Analysis

Chap 1.7

39. The value of the load impedance, that would

The line impedance connecting the source to load is

absorbs the maximum average power is

0.3 + j0.2 W. If the current in a phase of load 1 is

j100

3Ð20o A

~

I = 10 Ð20 ° A rms , the current in source in ab branch is

-j40

80 W

ZL

(A) 15 Ð - 122 ° A rms

(B) 8.67 Ð - 122 ° A rms

(C) 15 Ð27.9 ° A rms

(D) 8.67 Ð - 57.9 ° A rms

45. Fig. P1.7.39

(A) 12.8 - j 49.6 W

(B) 12.8 + j 49.6 W

(C) 339 . - j 86.3 W

(D) 339 . + j 86.3 W

An

abc

phase

sequence

3-phase

balanced

Y-connected source supplies power to a balanced D –connected load. The impedance per phase in the load is 10 + j 8 W. If the line current in a phase is I aA = 28.10 Ð - 28.66 ° A rms and the line impedance is zero, the load voltage V AB is

Statement for Q.40–41: In a balanced Y–connected three phase generator Vab = 400 Vrms 40. If phase sequence is abc then phase voltage Va , Vb , and Vc are respectively

(A) 207.8 Ð - 140 ° Vrms

(B) 148.3Ð40 ° Vrms

(C) 148.3Ð - 40 ° Vrms

(D) 207.8 Ð40 ° Vrms

46. The magnitude of the complex power supplied by a 3-phase balanced Y-Y system is 3600 VA. The line voltage is 208 Vrms . If the line impedance is negligible

(A) 231Ð0 ° , 231Ð120 ° , 231Ð240 °

and the power factor angle of the load is 25°, the load

(B) 231Ð - 30 ° , 231Ð - 150 ° , 231Ð90 °

impedance is

(C) 231Ð30 ° , 231Ð150 ° , 231Ð - 90 °

(A) 5.07 + j10.88 W

(B) 10.88 + j5.07 W

(D) 231Ð60 ° , 231Ð180 ° , 231Ð - 60 °

(C) 432 . + j14.6 W

(D) 14.6 + j 432 . W

41. If phase sequence is acb then phase voltage are (A) 231Ð0 ° , 231Ð120 ° , 231Ð240 ° (B) 231Ð - 30 ° , 231Ð - 150 ° , 231Ð90 °

***********

(C) 231Ð30 ° , 231Ð150 ° , 231Ð - 90 ° (D) 231Ð60 ° , 231Ð180 ° , 231Ð - 60 ° 42. A balanced three-phase Y-connected load has one phase voltage Vc = 277 Ð45 ° V. The phase sequence is abc. The line to line voltage V AB is (A) 480 Ð45 ° V

(B) 480 Ð - 45 ° V

(C) 339 Ð45 ° V

(D) 339 Ð - 45 ° V

43. A three-phase circuit has two parallel balanced D loads, one of the 6 W resistor and one of 12 W resistors. The magnitude of the total line current, when the line-to-line voltage is 480 Vrms , is (A) 120 A rms

(B) 360 A rms

(C) 208 A rms

(D) 470 A rms

44. In a balanced three-phase system, the source has an abc phase sequence and is connected in delta. There are two parallel Y-connected load. The phase impedance of load 1 and load 2 is 4 + j 4 W and 10 + j 4 W respectively. www.gatehelp.com

Page 67


UNIT 1

SOLUTIONS

æ -j ö 7. (C) Z = çç ÷÷||( 6 + j(27m) w) è w(22m ) ø

1. (D) Z = 3 + j(25m)( 300) = 3 + j7.5 W = 8.08 Ð68. 2 ° I=

20 Ð0 = 2.48 Ð - 68. 2 ° A 8.08 Ð68. 2 °

i( t) = 2.48 cos ( 300 t - 68. 2 ° ) A 1 + j(1m)(10 3) = 0.5 + j = 112 . Ð63.43° 2 (1Ð0) VC = = 0.89 Ð - 63.43° V 112 . Ð63.43° 2. (A) Y =

vC ( t) = 0.89 cos (10 3 t - 63.43° ) V

(1Ð0) (5 Ð - 90 ° )

vC ( t) =

5 2 Ð - 45 ° 1 2

1

=

2

27 ´ 10 3 j 6 ´ 106 - j106 ( 6 + j27 ´ 10 -3 w) 22 22 w = 22 w = 6 10 æ 106 ö 6 + j(27mw ) ÷ 6 + jwçç 27m 22 w 22 w2 ÷ø è j27 ´ 10 3 - j 36 ´ 106 22 w22

æ 106 wçç 27m 22 w2 è

Þ w = 1278 1278 w Hz = = 203 Hz f = 2p 2p

V1 = Vs - V2 = 7.68 Ð47° - 7.51Ð35° = 159 . Ð125° 9. (B) vin = 32 + (14 - 10) 2 = 5

Ð - 45 ° V

10. (C) I1 = 744 Ð - 118 ° mA,

cos (2 t - 45 ° ) V

I 2 = 540 Ð100 ° mA

4. (D) Z = 9 + j( 3)(5) +

I = I1 + I 2 = 744 Ð - 118 ° + 540.5 Ð100 °

-j = 9 + j11 (50m) (5)

= 460 Ð - 164 ° i( t) = 460 cos ( 3t - 164 ° ) mA

Þ

Z = 14.21Ð50.71° W ( 8 Ð0)( 4 Ð - 90 ° ) VC = = 2. 25 Ð140.71° V 14. 21Ð50.71°

11. (A)

2 Ð45° =

VC V - 20 Ð0 + C - j4 j5 + 10

vC ( t) = 2. 25 cos (5 t - 140.71° ) V

j50

10 Ð0 10 Ð0 1 5. (B) Va = V = 1 1 1 105 . + j0.4 + + 1 - j2 4 + j 8 I=

ö ÷÷ = 0 ø

8. (C) Vs = 7.68 Ð47° V, V2 = 7.51Ð35°

-j 3. (A) Z = 5 + = 5 - j5 = 5 5 Ð - 45 ° (0.1)(2) VC =

Networks

Ö2Ð45 A o

~

+ -j4

Va 10 Ð0 = = 1Ð - 84. 23 A 4 + j 8 1 + j10 1W

Va

10Ð0 V

~

–

~

20Ð0 V o

Fig. S1.7.11 4W I

o

VC

10 W

-j2

j8

(1 + j)( - j 4)(10 + j5) = VC (10 + j5 - j 4) + j 8 Þ

60 - j100 = VC (10 + j) Þ

VC = 11.6 Ð - 64.7°

12. (D) X = X L + X C = 0 Fig. S1.7.5

i( t) = cos (2 t - 84.23° ) A 6. (D) w = 2 p ´ 10 ´ 10 3 = 2 p ´ 10 4 -j 1 Y = j(1m )(2 p ´ 10 4 ) + + 4 (160m )(2 p ´ 10 ) 36 = 0.0278 - j0.0366 S 1 Z= = 1316 . + j17.33 W Y

Page 68

So reactive power drawn from the source is zero. 13. (B) Z1 Z 4 = Z 3Z 2 300 Z 4 = ( 300 - j 600)(200 + j100) Þ Z 4 = 400 - j 300 14. (A) R - C causes a positive phase shift in voltage Z =|Z |Ðq , -90 ° < q < 0 , I=

V V = Ð-q Z |Z |

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120 Ð15 ° - 6 Ð30 ° 40 + j20 15. (C) Vo = = 124 Ð - 154 ° 1 1 1 + + 40 + j20 - j 30 50

20. (B) Let Vo be the voltage across current source Vo - 4 Vx Vo - Vx + =3 20 j10 Vo(20 + j10) - (20 + j 40) Vx = j 600 Vo(20) V Vx = Þ Vo = x (2 + j) 20 + j10 2

16. (C) 10 sin ( t + 30 ° ) = 10 cos ( t - 60 ° ) 10 Ð - 60 ° 20 Ð - 45 ° + j 3 Vo = 1 1 1 + + j -j 3

æ (2 + j)(20 + j10) ö Vx = ç - 20(1 + j2) ÷ = j 600 2 è ø j 600 Vx = = 29. 22 Ð - 166 ° -5 - j20

= 30 Ð - 150 °+20 Ð - 45 ° . Ð - 112 ° V Vo = 315

10Ð-60 V

æ j ö V - V2 21. (A) I1 = V3ç ÷ + 3 = j0.1V2 + j0.4 V3 j10 è2 ø

3W

j1

o

+

~

-j1 W

~

Vo –

= (0.1Ð90 ° )(0.757 Ð66.7 ° ) + (0.4 Ð90 ° )(0.606 Ð - 69.8 ° ) Þ

20Ð-45 V o

I1 = 0.196 Ð35.6 °

22. (A) Fig. S.1.7.16

1W

5Ð0 V o

~

1W

10Ð-30 V o

12Ð0 V o

-j0.25 W

j4

+

~

2W

I1

Vo

I2

–

2W

Þ

( 8 + j15) I1 - ( 4 - j) I 2 = 20 Ð0 j j -10 Ð - 30 ° = I 2 (1 + j 4 + 1 - ) - I1 (1 - ) 4 4 ( 4 - j) I1 - ( 8 + j15) I 2 = 40 Ð - 30 °

~

...(i)

2Ð0o A

Fig. P1.7.23

...(ii) 12 Ð0 ° = I1 ( - j 3 + 2 + 2) + 8 Ð90 °-4 Ð0 °

= (20 Ð0)( 8 + j15) - ( 40 Ð - 30 ° )( 4 - j)

Þ

I1 ( -176 + j248) = 41.43 + j 414.64

. + j0.64 + j 4) = 11.65 Ð52.82 ° V Vo = 2( 352

Þ

2

. - j0.9 = 1.37 Ð - 4107 . I1 = 103

. + j0.64 I1 = 352

24. (D) I 2 = 3Ð0 ° A , I 4 - I 3 = 6 Ð0 ° A

( 8 + j15)(103 . - j0.9) - 20 Ð0 ° 18. (B) I 2 = 4-j = -0.076 + j2.04

Þ

o

2W

I1 [( 8 + j15) - ( 4 - j) ] 2

4Ð90 V

~

Fig. S.1.7.17

Þ

Vo = 5.65 Ð - 75 °

-j3

~

I2

Þ

23. (D) I 2 = 4 Ð90 ° , I 3 = 2 Ð0 °

j4

1W I1

Vo Vo - 3Vo + = 4 Ð - 30 ° 2 j4

Vo(0.5 + j0.5) = 3.46 - j2

jö jö æ æ 17. (C) 5 Ð0 ° = I1 ç j 4 + 1 + 1 - ÷ - I 2 ç 1 - ÷ 4ø 4ø è è j4

Chap 1.7

Io

I 2 = 2.04 Ð92.13°

15Ð90o V

~

2W

I2

-j4

j2

~

3Ð0o A

19. (B) 10 Ð30 ° = 4 I1 - 0.5 I x + ( - j2) I x ( - j2) I x = ( I1 - I x ) j 3, I1 = æ4 ö 10 Ð30 ° = ç - 0.5 - j2 ÷I x è3 ø

Ix 3

1W

Þ

Ix =

10 Ð30 ° 2.17 Ð - 67.38 °

I3

~

6Ð0 A I4 o

1W

Fig. S.1.7.24

I 3(1) + ( I 3 - I o)( - j 4) + ( I 4 + I 2 )( j2) + I 4 = 0 www.gatehelp.com

Page 69


UNIT 1

I 3 + ( I 3 - I o)( - j 4) + ( I 3 + 6 Ð0 °+3Ð0 ° )( j2) + I 3 + 60 ° = 0 I 3(2 - j2) + I o( j 4) = -18 j - 6 -I ( j2) - 3 - 9 j I3 = o (1 - j)

S2 = 20 + j

15 Ð90 ° = ( I o + 3Ð0 ° )(2) + ( I o - I 3)( - j 4) Þ

VTH =

|V |2 (120) 2 = = 72 + j144 VA Z* 40 - j 80

20 sin (cos -1 (0.8)) = 20 + j15 0.8

S = VoI * = 6 Vo

( j10)( 8 - j5) = = 9 + j 4.4 8 + j10 - j5

Þ

S=

|V |2 (210) 2 = * Z 30 - j157

...(i)

Apparent power =|S |=

...(ii)

35. (D) Z = 4 +

300 I 2 = 3V1 , V1 = ( - j 300)( I1 - I 2 ) I 2 = - j 3( I1 - I 2 )

Þ

3I1 = ( 3 + j) I 2

Solving (i) and (ii) I 2 = 12.36 Ð - 16 ° mA

36. (A)

q = 25.84 °

Þ

3V 4 1

-j1.5

Fig. S.1.7.36

I1 = 1Ð36.9 ° (1Ð36.9 ° )(10 Ð0 ° ) S= = 5 Ð - 36.9 ° 2

S = 4129.8 - j2000 29. (A) Q = S sin q

1W

~

Q 2000 = = 4588.6 VA sin q sin 25.84 °

P = S cos q = 4129.8,

V1 = 4 Ð36.9 °,

+ V1 –

I1 o

S=

Þ

4W

10Ð0 V

Þ

= 275.6 VA

( - j2)( j5 - j2) - j2 + j5 - j2

10 - V1 V1 3 + V1 = 4 4 1 - j15 .

27. (C) S = P - jQ = 269 - j150 VA

Q = S sin q

30 2 + 152 2

pf = cos 56.31° = 0.555 leading

-2 V1 - V1 = 0 Þ V1 = 0 9 Ð0 ° Þ I sc = = 15 Ð0 ° mA 600 311 . Ð - 16 ° V Z TH = oc = = 247 Ð - 16 ° W I sc 15 Ð0 °´10 -3

Þ

(210) 2

= 4 - j 6 = 7.21Ð - 56.31°,

. Ð - 16 ° Voc = 300 I 2 = 371

28. (D) pf = cos q = 0.9

Vo = 7.1Ð32.29 °

34. (B) Z = 30 + j(0.5)(2 p)(50) = 30 + j157,

( 32 Ð0 ° )( j10) = 339 . Ð58 ° V 8 + j10 - j5

26 (D) ( 600 - j 300) I1 + j 300 I 2 = 9 Þ

16 sin (cos -1 (0.9)) = 16 + j7.75 0.9

S = S1 + S2 = 36 + j2.75 = 42.59 Ð32.29 °

j15 = 2 I o + 6 + ( j 4)( 3 - j 6)

25. (A) Z TH

32. (A) S =

33. (A) S1 = 16 + j

I3 = Io + 3 - j6

Þ

Networks

Þ

sin q =

Q 45 or = S 60

q = 48.59 °,

pf = cos 36.9 ° = 0.8 leading 37. (A) (2 Ð - 90 ° ) 4.8 = - I x ( 4.8 + j192 . ) + 0.6 I x ( 8)

P = S cos q = 39.69,

Ix

j1.92

Va

S = 39.69 + j 45 VA 4.8 W

|V |2 (220 2 ) 30. (B) S = rms = = 1210 |Z | 40

1.6Ix (2Ð90o)4.8 V

P 1000 cos q = = = 0.8264 or q = 34.26 °, S 1210

Fig. S.1.7.37

Q = S sin q = 68125, . S = 1000 + j 68125 . VA * 31. (C) S = Vrms I rms = (21Ð20 ° )( 8.5 Ð50 ° )

I x = 5 Ð0 °, Va = 0.6 ´ 5 ´ 8 = 24 Ð0 °, 1 Pave = ´ 24 ´ 1.6 ´ 5 = 96 2

= 61 + j167.7 VA Page 70

~

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8W



38. (A) Z TH = VTH =

( - j10)(10 + j15) = 8 - j14 W 10 + j15 - j10

120( - j10) = 107.3Ð - 116.6 ° V 10 + j5

40. (B) Va =

3

I aA 3

Ð( q + 30 ° ) = 16.22 Ð1.34 ° A rms

= 207.8 Ð40 ° Vrms 46. (B) |S |= 3VL I L ZY =

( - j 40)( 80 + j100) = 12.8 - j 49.6 W 80 + j 60

400

45. (D) I AB =

V AB = I AB × Z D = (16. 22 Ð1.340 ° )(10 + j 8)

107.3Ð - 116.6 ° IL = = 6.7 Ð - 116.6 ° 16 1 PLmax = ( 6.7) 2 ´ 8 = 180 W 2 39. (B) Z TH =

Chap 1.7

208 10 3

Þ

IL =

3600 208 3

= 10 A rms

Ð25 ° = 12 Ð25 ° = 10.88 + j5.07 W

********

Ð - 30 ° = 231Ð - 30 ° V

Vb = 231Ð - 150 ° V, Vc = 231Ð - 270 ° V 41. (C) For the acb sequence Vab = Va - Vb = Vp Ð0 ° - Vp Ð120 ° æ 1 3 400 = Vp çç 1 + - j 2 2 è 400 Þ Vp = Ð30 ° 3

ö ÷ = Vp 3Ð - 30 ° ÷ ø

Va = Vp Ð0 ° = 231Ð30 ° V, Vb = Vp Ð120 ° = 231Ð150 ° V Vc = Vp Ð240 ° = 231Ð - 90 ° V 42. (B) V A = 277 Ð( 45 °-120 ° ) = 277 Ð - 75 ° V VB = 277 Ð( 45 ° + 120 ° ) = 277 Ð165 ° V V AB = V A - VB = 480 Ð - 45 ° V 43. (C) Z A = 6 ||12 = 4, IP =

480 = 120 A rms 4

I L = 3I P = 208 A rms 44 (B) I =

I aA (10 + j 4) = 10 Ð20 ° (10 + j 4) + ( 4 + j 4) IaA

a

Iac

c

Iab Ibc

IbB b

icC

Fig. S.1.7.44

I aA = 15 Ð - 27.9 ° A rms |I | I ab = - aA Ð( q + 30 ° ) = 8.67 Ð - 122.1° A rms 3 www.gatehelp.com

Page 71


CHAPTER

1.8 CIRCUIT ANALYSIS IN THE S-DOMAIN

s2 + 1 s2 + 2 s + 1 2 s2 + 1 (C) 2 s + 2s + 2

1. Z ( s) = ?

2( s 2 + 1) ( s + 1) 2 s2 + 1 (D) 3s + 2

(A) 1F

2H

Z(s)

1W

1W

(B)

4. Z ( s) = ?

1W

Z(s)

1H

1W

2

s 2 + 3s + 1 s( s + 1) 2 s 2 + 3s + 1 (D) 2 s( s + 1) (B)

W

Fig. P1.8.1

s 2 + 15 . s+1 (A) s( s + 1) 2 s 2 + 3s + 2 (C) s( s + 1)

0.5 F

Fig. P1.8.4

2. Z ( s) = ?

3s 2 + 8 s + 7 s(5 s + 6) 3s 2 + 7 s + 6 (C) s(5 s + 6)

s(5 s + 6) 3s 2 + 8 s + 7 s(5 s + 6) (D) 3s 2 + 7 s + 6 (B)

(A) 1F Z(s)

1W

1H

1W

5. The s-domain equivalent of the circuit of Fig.P1.8.5. is t=0

Fig. P1.8.2

s + s+1 s( s + 1) s( s + 1) (C) 2 s2 + s + 1

2s + s + 1 s( s + 1) s( s + 1) (D) 2 s + s+1

2

3W

2

(A)

(B)

6V

3F

+ vC –

Fig. P1.8.5

3. Z ( s) = ?

3W

3W + 1 3s

1H Z(s)

2W 1F

Fig. P1.8.3

6 V s

(A) (C) Both A and B

Page 72

1 3s

VC(s)

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+ VC(s) -

-

(B) (D) None of these

2A


Circuit Analysis in the s-Domain

Chap 1.8

6. The s–domain equivalent of the circuit shown in Fig.

9. For the network shown in fig. P1.8.9 voltage ratio

P1.8.6 is

transfer function G12 is t=0

+ 12 W

2A

2H

1H

1H

1F

1F

vL v1

–

1F

1F

1F

+ v2 -

Fig. P1.8.6 Fig. P1.8.9

(A)

(B) +

+

(A)

( s 2 + 2) 5 s + 5 s2 + 1

(B)

s2 + 1 5 s + 5 s2 + 1

(C)

( s 2 + 2) 2 5 s4 + 5 s2 + 1

(D)

( s 2 + 1) 2 5 s4 + 5 s2 + 1

2s 12 W

12 W

VL

2A s

VL

2s

4

4

4V -

-

(C) Both A and B

10. For the network shown in fig. P1.8.10, the admittance transfer function is

(D) None of these

Y12 = Statement for Q.7-8:

K ( s + 1) ( s + 2)( s + 4)

3 W 2

The circuit is as shown in fig. P1.8.7–8. Solve the

i1

problem and choose correct option.

i2

1W

+ +

is

1W

1H

io

v1

+ vs

1F

1F

1W

2F

2F

3

-

Fig. P1.8.10

The value of K is (A) -3

V ( s) 7. H1 ( s) = o =? Vs ( s)

(C)

(A) s( s3 + 2 s2 + 3s + 1) -1

(B) 3

1 3

(D) -

1 3

11. In the circuit of fig. P1.8.11 the switch is in position

(B) ( s 3 + 3s 2 + 2 s + 1) -1

1 for a long time and thrown to position 2 at t = 0. The

(C) ( s 3 + 2 s 2 + 3s + 2) -1

equation for the loop currents I1 ( s) and I 2 ( s) are

(D) s( s 3 + 3s 2 + 2 s2 + 2) -1

1F

1

I o( s) =? Vs ( s)

2

t=0

12 V

(A)

-s ( s + 3s + 2 s + 1)

(B) -( s 3 + 3s 2 + 2 s + 1) -1

(C)

-s 3 2 ( s + 2 s + 3s + 1)

(D) ( s 3 + 2 s 2 + 3s + 2) -1

2

v2

-

Fig. P1.8.7–8

3

6

vo –

8. H 2 ( s) =

1W

i1

3H

2W i2 1F

Fig. P1.8.11

1 é ê2 + 3s + s (A) ê ê -3s ë

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ù - 3s ú 1 ú 2+ ú s û

é12 ù é I1 ( s) ù ê s ú êI ( s) ú = ê ú ë 2 û ê0 ú ë û

Page 73


UNIT 1

1 é ê2 + 3s + s (B) ê ê -3s ë

ù - 3s ú 1 ú 2+ ú s û

é 12 ù é I1 ( s) ù ê- s ú ú êI ( s) ú = ê ë 2 û ê 0 ú ë û

1 é -3s ê2 + 3s + s (C) ê 2 + 3s + ê -3s ë

é 12 ù ù ú é I1 ( s) ù ê- s ú = ú 1 ú êëI 2 ( s) úû ê ê 0 ú ú ë û sû

1 é -3s ê2 + 3s + s (D) ê 2 + 3s + ê -3s ë

ù é12 ù ú é I1 ( s) ù ê s ú = 1 ú êëI 2 ( s) úû ê ú ú ê0 ú sû ë û

Networks

3F

S1 -

4F

Va +

5V

+

10 V

2F

1V

S2

-

+ 6V

5V

-

Fig. P1.8.14

(A)

9 t

(B) 9e - t V

V

(C) 9 V

(D) 0 V

15. A unit step current of 1 A is applied to a network whose driving point impedance is

12. In the circuit of fig. P1.8.12 at terminal ab

1

The steady state and initial values of the voltage

s

a

4 A

+ Vo(s) -

2W

(s+1)

developed across the source would be respectively

2Vo(s)

b

(A)

3 4

(C)

3 4

V, I V

1 4

V,

3 4

V

(D) 1 V,

3 4

V

(B)

V, 0 V

16. In the circuit of Fig. P1.8.16 i(0) = 1 A, vC (0) = 8

Fig. P1.8.12

4

V and v1 = 2 e -2 ´10 t u( t). The i( t) is

(A) VTH ( s) =

-8( s + 2) -(2 s + 1) , Z TH ( s) = 3s( s + 1) 3s

(B) VTH ( s) =

8( s + 2) (2 s + 1) , Z TH ( s) = 3s( s + 1) 3s

(C) VTH ( s) =

4( s + 3) (2 s + 1) , Z TH ( s) = 3s( s + 1) 6s

50 W

i

1m H

2.5 mF

v1

+ vC –

Fig. P1.8.16 4

-4( s + 3) -(2 s + 1) (D) VTH ( s) = , Z TH ( s) = 3s( s + 1) 6s

(A)

1 15

[10 e-10

(B)

1 15

[ -10 e -10

13. In the circuit of fig. P1.8.13 just before the closing of

(C) 13 [10 e-10

switch at t = 0, the initial conditions are known to be

(D) 13 [ -10 e -10

-

V ( s) ( s + 3) = I ( s) ( s + 2) 2

Z ( s) =

Thevenin equivalent is

-

vC1 (0 ) = 1 V, vC2 (0 ) = 0. The voltage vC1 ( t) is

4

- 3e-2 ´10

t 4

t

4

t

4

- 22 e-4 ´10 t ]u( t) A

t

+ 3e -2 ´10

+ 3e-2 ´10

t

4

4

4

4

+ 22 e -4 ´10 t ]u( t) A 4

+ 22 e-4 ´10 t ]u( t) A

t

+ 3e -2 ´10

t

4

t

4

- 22 e -4 ´10 t ]u( t) A

17. In the circuit shown in Fig. P1.8.18 v(0 - ) = 8 V and + vC1 -

iin ( t) = 4d( t). The vC ( t) for t ³ 0 is

t=0 1F

1F

+ vC2 -

iin

50 W

20 mF

Fig. P1.8.13

(A) u( t) V (C) 0.5 e

-t

(B) 0.5 u( t) V V

Fig. P1.8.17

(D) e - t V -

14. The initial condition at t = 0 of a switched capacitor

(A) 164e- t V

(B) 208e- t V

(C) 208(1 - e -3t ) V

(D) 164 e -3t V

circuit are shown in Fig. P1.8.14. Switch S1 and S2 are closed at t = 0. The voltage va ( t) for t > 0 is Page 74

+ vC –

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18. The driving point impedance Z ( s) of a network has

Chap 1.8

The steady state voltage across capacitor is

the pole zero location as shown in Fig. P1.8.18. If

(A) 6 V

(B) 0 V

Z(0) = 3, the Z ( s) is

(C) ¥

(D) 2 V

jw

23.

1

-3

transformed

VC ( s) = -1

4( s + 3) s2 + s + 1

(B)

2( s + 3) (C) 2 s + 2s + 2

2( s + 3) 2 s + 2s + 2

4( s + 3) (D) 2 s + s+2

20 s + 6 (10 s + 3)( s + 4) (B) - 0.12 mA

(C) 0.48 mA

(D) - 0.48 mA

24. The current through an 4 H inductor is given by I L ( s) =

The circuit is as shown in the fig. P1.8.19–21. All initial conditions are zero. io

1H

60 mF

the

(A) 0.12 mA


iin

across

The initial current through capacitor is

Fig. P1.8.18

(A)

voltage

capacitor is given by

s

-1

The

1W

1F

10 s( s + 2)

The initial voltage across inductor is (A) 40 V

(B) 20 V

(C) 10 V

(D) 5 V

25. The amplifier network shown in fig. P1.8.25

1W

is stable if 4W

1F

1H

Fig. P1.8.19–21

19.

I o( s) =? I in ( s)

(A)

( s + 1) 2s

+ + v1 Amplifier v 2 gain=K -

2W

(B) 2 s( s + 1) -1

(C) ( s + 1) s -1

Fig.P1.8.25

(D) s( s + 1) -1 (A) K £ 3

20. If iin ( t) = 4d( t) then io( t) will be

(C) K £

(A) 4d( t) - e- t u( t) A (B) 4 d( t) - 4 e- t u( t) A

(B) K ³ 3

1 3

(D) K ³

1 3

26. The network shown in fig. P1.8.26 is stable if

(C) 4 e - t u( t) - 4 d( t) A

2F

1W

(D) e - t u( t) - d( t) A

+

21. If iin ( t) = tu( t) then io( t) will be (A) e - t u( t) A

(B) (1 - e - t ) u( t) A

(C) u( t) A

(D) (2 - e - t ) u( t) A

Kv2

1W

1F

v2 -

Fig.P1.8.26

22. The voltage across 200 mF capacitor is given by VC ( s) =

2s + 6 s( s + 3)

5 2 2 (C) K ³ 5 (A) K ³

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5 2 2 (D) K £ 5 (B) K £

Page 75


UNIT 1

27. A circuit has a transfer function with a pole s = - 4 and a zero which may be adjusted in position as s = - a The response of this system to a step input has a term of form Ke -4 t . The K will be aö æ (A) H ç 1 - ÷ 4ø è

(H= scale factor) aö æ (B) H ç 1 + ÷ 4ø è

aö æ (C) H ç 4 - ÷ 4ø è

aö æ (D) H ç 4 + ÷ 4ø è

io( t) = 2 sin 2 t u( t) A. The circuit had no internal stored energy at t = 0. The admittance transfer function is 2 s (A) (B) s 2 (D)

31. The current ratio transfer function

Io is Is

(A)

s( s + 4) s + 3s + 4

(B)

s( s + 4) ( s + 1)( s + 3)

(C)

s 2 + 3s + 4 s( s + 4)

(D)

( s + 1)( s + 3) s( s + 4)

2

32. The response is

28. A circuit has input vin ( t) = cos 2 t u( t) V and output

(C) s

Networks

1 s

(A) Over damped

(B) Under damped

(C) Critically damped

(D) can’t be determined

33. If input is is 2u( t) A, the output current io is (A) (2 e - t - 3te -3t ) u( t) A

(B) ( 3te - t - e -3t ) u( t) A

(C) ( 3e - t - e -3t ) u( t) A

(D) ( e -3t - 3e - t ) u( t) A

34. In the network of Fig. P1.8.34, all initial condition are zero. The damping exhibited by the network is

29. A two terminal network consists of a coil having an inductance L and resistance R shunted by a capacitor

1F 4

C. The poles of the driving point impedance function Z of this network are at - 12 ± j

3 2

and zero at -1. If

+

Z(0) = 1 the value of R, L, C are 1 (A) 3 W, 3 H, F 3

2H

vs

2W

vo -

1 (B) 2 W, 2 H, F 2 Fig. P1.8.34

1 (C) 1 W, 2 H, F 2

(D) 1 W, 1 H, 1 F

(A) Over damped (B) Under damped

30. The current response of a network to a unit step input is

(C) Critically damped (D) value of voltage is requires

10( s + 2) Io = 2 s ( s + 11s + 30)

35. The voltage response of a network to a unit step input is

The response is (A) Under damped

(B) Over damped

(C) Critically damped


Vo( s) =

10 s( s 2 + 8 s + 16)

The response is Statement for Q.31-33:

(A) under damped

(B) over damped

(C) critically damped

(D) can’t be determined

The circuit is shown in fig. P1.8.31-33. 36. The response of an initially relaxed circuit to a 1H

io

(

signal vs is e -2 t u( t). If the signal is changed to vs + , the response would be

1F

is

3

4W

(A) 5 e -2 t u( t)

(B) -3e -2 t u( t)

(C) 4 e -2 t u( t)

(D) -4 e -2 t u( t)

Fig. P1.8.31-33 Page 76

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2 dvs dt

)



37. Consider the following statements in the circuit shown in fig. P1.8.37 i

4W

2H

1W

10 V

1F

2

+ vC –

42. The network function

(B) RL admittance

(C) LC impedance function

(D) LC admittance

43. A valid immittance function is ( s + 4)( s + 8) s( s + 1) (A) (B) ( s + 2)( s - 5) ( s + 2)( s + 5) s( s + 2)( s + 3) ( s + 1)( s + 4)

(C) 1. It is a first order circuit with steady state value of 10 5 , i= A vC = 3 3 2. It is a second order circuit with steady state of vC = 2 V , i = 2 A V ( s) has one pole. I ( s)


V ( s) has two poles. I ( s)

(D)


(B) 1 and 4

(C) 2 and 3

(D) 2 and 4


(B) RC admittance

(C) LC admittance

(D) Above all

Z ( s) =

3W

s 2 + 10 s + 24 represent a s 2 + 8 s + 15

1F


(C) LC impedance

(D) None of these

3W

8F

1F

(A)

(B)

1F

1W

3

1H

1F

8

s( 3s + 8) 40. The network function represents an ( s + 1)( s + 3)

1W

3W

3

(C)

(A) RL admittance

(B) RC impedance

(C) RC admittance



1W

3

8

s( s + 4) represents ( s + 1)( s + 2)( s + 3) (B) RL impedance

1W

3

1F

(C) LC impedance

(A) RC impedance

3( s + 2)( s + 4) s( s + 3)

The network for this function is

(B) RL impedance

an

s 2 + 8 s + 15 is a s2 + 6 s + 8

45. A impedance function is given as

(A) RC admittance


s( s + 2)( s + 6) ( s + 1)( s + 4)

(A) RLadmittance

The true statements are (A) 1 and 3

s2 + 7 s + 6 is a s+2

(A) RL impedance function

Fig. P1.8.37


Chap 1.8

( s + 1)( s + 4) is a s( s + 2)( s + 5)

1H

8

3W

(D)

************

(A) RL impedance function (B) RC impedance function (C) LC impedance function (D) Above all

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Page 77


UNIT 1

18. (B) Z ( s) =

K ( s + 3) K ( s + 3) = 2 ( s - ( -1 + j))( s - (1 - j)) s + 2 s + 2

3K Z (0) = =3 2

Þ

I ( s) 19. (D) o = I in ( s)

26. (B) Let v1 be the node voltage of middle node V1 ( s) =

K =2 s s+1

s 1 + s+1 s+1

=

4s 4 =4s+1 s+1

Þ

s s+1

Þ (2 s + 1) V2 ( s) = 2 sV1 ( s) Þ ( 3s + 1)(2 s + 1) = 2 s(2 s + K ) 2 s 2 + (5 - 2 K ) s + 1 = 0, 5 5 - 2 K > 0, K < 2

io( t) = 4 d( t) - 4 e - t u( t)

H ( s + a) s+4

27. (A) H ( s) =

1 , s2 1 1 1 I o( s) = = s( s + 1) s s + 1 21. (B) I in ( s) =

aö æ Hç 1 - ÷ H ( s + a) Ha 4ø R( s) = = + è s( s + 4) 4s s+4

io( t) = u( t) - e - t u( t) = (1 - e - t ) u( t) 22. (D) vC ( ¥) = lim sVC ( s) = lim s ®0

KV2 ( s) + 2 sV2 ( s) 1 + 2s + s

Þ ( 3s + 1) V1 ( s) = (2 s + K ) V2 ( s) 2 sV1 ( s) Þ V2 ( s) = 2s + 1

20. (B) I in ( s) = 4 I o( s) =

Networks

s ®0

r ( t) =

2s + 6 =2 V s+3

Ha aö æ u( t) + H ç 1 - ÷e - 4 t 4 4ø è

28. (A) Vin ( s) = 23. (D) vC (0 + ) = lim sVC ( s)= s ®¥

CdvC iC = dt

Þ

s(20 s + 6) =2 V (10 s + 3)( s + 4)

I C ( s) = C[ sVC ( s) - vC (0 )]

æ s(20 s + 6) ö -480 ´ 10 -6 (10 s + 3) = 60 ´ 10 -6 çç - 2 ÷÷ = 10 s 2 + 43s + 12 è (10 s + 3)( s + 4) ø s ®¥

Rö 1æ 1 çs + ÷ C Lø è sC = 29. (D) Z ( s) = R 1 1 s2 + sL + R + + L LC sC K ( s + 1) K ( s + 1) Z ( s) = = 2 ( s + s + 1) æ 1 3 öæ 1 3ö çs + + j ÷ç s + - j ÷ ç ÷ ç ÷ 2 2 øè 2 2 ø è ( sL + R)

+

iC (0 + ) = lim sI C ( s) = - 480 ´ 10 -6 = - 0.48 mA

I o( s) 2 s 2 , I o( s) = 2 , = s2 + 1 s + 1 Vin ( s) s

d iL Þ VL ( s) = L [ sI L ( s) - iL (0 + )] dt 10 iL (0 + ) = lim sI L ( s) = =0 s ®¥ s+2

24. (A) vL = L

VL ( s) =

s ®¥

s 40 = 40 s+2

25. (A) V2 ( s) = KV1 ( s) V1 ( s) V1 ( s) - KV1 ( s) + =0 1 2 4+s+ s 1 4 + s + + 2 - 2K = 0 s Þ

s2 + ( 6 - 2 K ) s + 1 = 0

(6 - 2K ) > 0

Page 80

Þ

1 Cs R

40 s 40 = s( s + 2) s + 2

vL (0 + ) = lim sVL ( s) =

Þ

sL Z(s)

Fig. S1.8.29

Since Z (0) = 1, thus K = 1 1 R 1 = 1, = 1, =1 C L LC Þ

C = 1, L = 1, R = 1

30. (B) The characteristic equation is s 2 ( s 2 + 11s + 30) = 0

Þ

s 2 ( s + 6) ( s + 5) =0

s = -6, - 5, Being real and unequal, it is overdamped.

K <3

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31. (B)

Io s+4 s( s + 4) = = 3 Is s + 4 + ( s + 1)( s + 3) s

39. (D) Poles and zero does not interlace on negative real axis so it is not a immittance function.

32. (A) The characteristic equation is ( s + 1) ( s + 3) = 0. Being real and unequal root, it is overdamped response. 33. (C) is = 2 u( t) I o ( s) =

40. (C) The singularity nearest to origin is a zero. So it may be RL impedance or RC

admittance function.

Because of (D) option it is required to check that it is a valid RC

2 I s ( s) = s

Þ

Chap 1.8

admittance function. The poles and zeros

interlace along the negative real axis. The residues of YRC ( s) are real and positive. s

2( s + 4) 3 1 = ( s + 1)( s + 3) s+1 s+ 3

io = ( 3e- t - e-3t ) u( t) A

41. (B) The singularity nearest to origin is a pole. So it may be RC impedance or RL admittance function.

V ( s) 2 1 34. (B) o = = 2 4 Vs ( s) + 2s + 2 s + s + 2 s

42. (A)

The roots are imaginary so network is underdamped.

s 2 + 7 s + 6 ( s + 1)( s + 6) = s+2 ( s + 2)

The singularity nearest to origin is at zero. So it may be

35. (C) The characteristic equation is

RC admittance or RL impedance function.

s( s 2 + 8 s + 16) = 0, ( s + 4) 2 = 0, s = -4, - 4

43. (D)

Being real and repeated root, it is critically damped.

(A) pole lie on positive real axis

36. (B) vo = e -2 t u( t) v¢s = vs +

2 dvs dt

Þ

Þ

Vo( s) = H ( s) Vs ( s) =

(B) poles and zero does not interlace on axis.

1 s+2

(C) poles and zero does not interlace on axis. (D) is a valid immittance function.

Vs¢( s) = (1 + 2 s) Vs ( s)

44. (A)

Vo¢( s) = H ( s) Vs¢( s) = (1 + 2 s) Vs ( s) H ( s) Vo¢( s) =

1 + 2s 3 =2 s+2 s+2

Þ

v¢o = 2 d( s) - 3e -2 t u( t)

The singularity nearest to origin is a pole. So it may be a RL admittance or RC impedance function.

37. (C) It is a second order circuit. In steady state i=

10 =2 A , v =2 ´ 1 =2 V 4+1

I ( s) =

10 2s + 4 +

V ( s) =

1 1 1+ s 2

=

10 1 1+ s 2 1 (2 s + 4) + 1 1+ s 2

s 2 + 8 s + 15 ( s + 3) ( s + 5) = s 2 + 6 s + 8 ( s + 2) ( s + 4)

45. (A) The singularity nearest to origin is a pole. So this is RC impedance function. 8 1 8 13 Z ( s) = 3 + + =3+ + s s+3 s 1+ s 3

5( s + 2) ( s + 2) 2 + 1

**************

=

10 ( s + 2) 2 + 1

V ( s) 2 , It has one pole at s = -2 = I ( s) s + 2 38. (D)

s 2 + 10 s + 24 ( s + 4)( s + 6) = s 2 + 8 s + 15 ( s + 3)( s + 5)

The singularity near to origin is pole. So it may be RC impedance or RL admittance function.

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Page 81


CHAPTER

1.9 MAGNETICALLY COUPLED CIRCUITS

4. If i1 = e -2 t V and i2 = 0, the voltage v2 is

Statement for Q.1-2: In the circuit of fig. P1.9.1-2 i1 = 4 sin 2 t A, and i2 = 0. i2

i1 1H

+ 2H

v1

+

(A) -6 e -2 t V

(B) 6 e -2 t V

(C) 15 . e -2 t V

(D) -15 . e -2 t V

Statement for Q.5-6: Consider the circuit shown in fig. P19.5-6

v2

1H

i2

i1

-

-

2H

+

+

Fig. P1.9.1-2 v1

2H

3H

v2

1. v1 = ? (A) -16 cos 2 t V

(B) 16 cos 2 t V

(C) 4 cos 2 t V

(D) -4 cos 2 t V

Fig. P1.9.5-6

5. If current i1 = 3 cos 4 t A and i2 = 0, then voltage v1 and

2. v2 = ? (A) 2 cos 2 t V

(B) -2 cos 2 t V

v2 are

(C) 8 cos 2 t V

(D) -8 cos 2 t V

(A) v1 = -24 sin 4 t V,

v2 = -24 sin 4 t V

(B) v1 = 24 sin 4 t V,

v2 = -36 sin 4 t V

(C) v1 = 15 . sin 4 t V,

v2 = sin 4 t V

(D) v1 = -15 . sin 4 t V,

v2 = -sin 4 t V

Statement for Q.3-4: Consider the circuit shown in Fig. P1.9.3-4 i2

i1 3H

+ v1

3H

+ 4H

v2 -

-

Fig. P1.9.5-6

3. If i1 = 0 and i2 = 2 sin 4 t A, the voltage v1 is

Page 82

-

-

(A) 24 cos 4 t V

(B) -24 cos 4 t V

(C) 15 . cos 4 t V

(D) -15 . cos 4 t V

6. If current i1 = 0 and i2 = 4 sin 3t A, then voltage v1 and v2 are (A) v1 = 24 cos 3t V,

v2 = 36 cos 3t V

(B) v1 = 24 cos 3t V,

v2 = -36 cos 3t V

(C) v1 = -24 cos 3t V,

v2 = 36 cos 3t V

(D) v1 = -24 cos 3t V,

v2 = -36 cos 3t V

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Magnetically Coupled Circuits


Chap 1.9

12. Leq = ?

In the circuit shown in fig. P1.9.7-8, i1 = 3 cos 3t A

3.6 H

and i2 = 4 sin 3t A.

Leq 1H

+ v1

1H

i2

i1

2H

1.4 H

+ v2

2H

Fig. P1.9.12

-

-

(A) 4 H

(B) 6 H

(C) 7 H

(D) 0 H

Fig. P1.9.7-8

7. v1 = ? (A) 6( -2 cos t + 3 sin t) V

(B) 6( 2 cos t + 3 sin t) V

(C) -6(2 cos t + 3 sin t) V

(D) 6(2 cos t - 3 sin t) V

13. Leq = ? 4H Leq

2H

8. v2 = ?

2H

(A) 3( 8 cos 3t - 3 sin t) V

(B) 6(2 cos t + 3 sin t) V

(C) 3( 8 cos 3t + 3 sin 3t) V

(D) 6(2 cos t - 3 sin t) V

Statement for Q.9-10: In the circuit shown in fig. P1.9.9-10, i1 = 5 sin 3t A and i2 = 3 cos 3t A 3H

+

(A) 2 H

(B) 4 H

(C) 6 H

(D) 8 H

14. Leq = ?

i2

i1

Fig. P1.9.13

4H

+ Leq

v1

3H

v2

4H

6H

4H

-

-

Fig. P1.9.14

Fig. P1.9.9-10

9. v1 =? (A) 9(5 cos 3t + 3 sin 3t) V

(B) 9(5 cos 3t - 3 sin 3t) V

(C) 9( 4 cos 3t + 5 sin 3t) V

(D) 9(5 cos 3t - 3 sin 3t) V

(A) 8 H

(B) 6 H

(C) 4 H

(D) 2 H

15. Leq = ? 2H

10. v2 = ? (A) 9( -4 sin 3t + 5 cos 3t) V

(B) 9( 4 sin 3t - 5 cos 3t) V

(C) 9( -4 sin 3t - 5 cos 3t) V

(D) 9( 4 sin 3t + 5 cos 3t) V

Leq

11. In the circuit shown in fig. P1.9.11 if current

Fig. P1.9.15

i1 = 5 cos (500 t - 20 ° ) mA and i2 = 20 cos (500 t - 20 ° ) mA, the total energy stored in system at t = 0 is i2

i1 k=0.6

+

2H

4H

(A) 0.4 H

(B) 2 H

(C) 1.2 H

(D) 6 H

+

16. The equivalent inductance of a pair of a coupled v1

2.5 H

0.4 H

-

v2

inductor in various configuration are

-

(a) 7 H after series adding connection

Fig. P1.9.11

(A) 151.14 mJ

(B) 45.24 mJ

(C) 249.44 mJ

(D) 143.46 mJ

(b) 1.8 H after series opposing connection (c) 0.5 H after parallel connection with dotted terminal connected together. www.gatehelp.com

Page 83


UNIT 1

Networks

20. Leq = ?

The value of L1 , L2 and M are (A) 3 H, 1.6 H, 1.2 H

(B) 1.6 H, 3 H, 1.4 H

(C) 3.7 H, 0.7 H, 1.3 H

(D) 2 H, 3 H, 3 H

1H 1H Leq

2H

17. Leq = ?

2H 2H

3H Leq

2H

4H

Fig. P1.9.20

Fig. P1.9.17

(A) 0.2 H

(B) 1 H

(C) 0.4 H

(D) 2 H

(A) 1 H

(B) 2 H

(C) 3 H

(D) 4 H

21. Leq = ? 4H

18. Leq = ?

Leq

3H Leq

3H

3H 2H

2H

3H

5H

Fig. P1.9.21

Fig. P1.9.18

41 (A) H 5

(B)

49 H 5

51 H 5

(D)

39 H 5

(C)

(A) 1 H

(B) 2 H

(C) 3 H

(D) 4 H Statement for Q.22-24:

19. In the network of fig. P1.9.19 following terminal are

Consider the circuit shown in fig. P1.9.22–24.

connected together

A

(i) none

(ii) A to B

(iii) B to C

(iv) A to C

B

2H a

A 2H

3H

6t A 20 H

C 5H

4H

3H

B

D

5H

Fig. P1.9.22–24 1H

b

22. The voltage V AG of terminal AD is

C

Fig. P1.9.19

(A) 60 V

(B) -60 V

(C) 180 V

(D) 240 V

The correct match for equivalent induction seen at terminal a - b is

Page 84

4H 6H

15t A

23. The voltage vBG of terminal BD is

(i)

(ii)

(iii)

(iv)

(A) 45 V

(B) 33 V

(A)

1 H

0.875 H

0.6 H

0.75 H

(C) 69 V

(D) 105 V

(B)

13 H

0.875 H

0.6 H

0.75 H

(C)

13 H

7.375 H

6.6 H

2.4375 H

(A) 30 V

(B) 0 V

(D)

1 H

7.375 H

6.6 H

2.4375 H

(C) -36 V

(D) 36 V

24. The voltage vCG of terminal CD is

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UNIT 1

Networks

33. In the circuit of fig. P1.9.33 the w = 2 rad s. The

37. In the circuit of fig. P1.9.37 the maximum power

resonance occurs when C is

delivered to RL is

C

10 W

2H Zin

1 : 4

4W

2H

2H

100 Vrms

~

RL

Fig. P1.9.33

(A) 1 F (C)

1 3

F

Fig. P1.9.37

(B)

1 2

F

(D)

1 6

F

34. In the circuit of fig. P1.9.34, the voltage gain is zero at w = 333.33 rad s. The value of C is 20 W

vin

0.12 H

~

(B) 200 W

(C) 150 W

(D) 100 W

38. The average power delivered to the 8 W load in the circuit of fig. P1.9.38 is

40 W

0.09 H

(A) 250 W

300 W

0.27 H 2F

I1

I2 +

+ vout -

50 Vrms

C

~

+ 8W

V1 -

-0.04V2

Fig. P1.9.34

5 : 1

Fig. P1.9.38

(A) 100 mF

(B) 75 mF

(A) 8 W

(B) 1.25 kW

(C) 50 mF

(D) 25 mF

(C) 625 kW

(D) 2.50 kW

35. In the circuit of fig. P1.9.35 at w = 333.33 rad s, the

39. In the circuit of fig. P1.9.39 the ideal source supplies

voltage gain vout vin is zero. The value of C is

1000 W, half of which is delivered to the 100 W load. The

C

value of a and b are

20 W

4W

40 W k=0.5

vin

~

0.12 H

25 W 1 : a

+ 0.27 H

20 W

100 Vrms

vout

1 : b

~

100 W

-

Fig. P1.9.39

Fig. P1.9.35

(A) 3.33 mF

(B) 33.33 mF

(C) 3.33 mF

(D) 33.33 mF

36. The Thevenin equivalent at terminal ab for the

(A) 6, 0.47

(B) 5, 0.89

(C) 0.89, 5

(D) 0.47, 6

40. I 2 = ? 25 W

network shown in fig. P1.9.36 is 60 W

a

I2

2W

3 : 1

1 : 4

50 Vrms

4 : 3

~

3W

20Ix

20 W

Fig. P1.9.40

Ix b

Fig. P1.9.36

Page 86

V2 -

(A) 6 V, 10 W

(B) 6 V, 4 W

(C) 0 V, 4 W

(D) 0 V, 10 W

(A) 1.65 A rms

(B) 0.18 A rms

(C) 0.66 A rms

(D) 5.90 A rms

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41. V2 = ?

Chap 1.9

SOLUTIONS

50 W 40 W

1. (B) v1 = 2 5 : 2

+

~

80 Vrms

10 W

V2

2. (C) v2 = (1)

-

Fig. P1.9.41

(A) -12.31 V

(B) 12.31 V

(C) -9.231 V

(D) 9.231 V

42. The power being dissipated in 400 W resistor is 1W

~

di2 di di + (1) 1 = 1 = 8 cos 2 t V dt dt dt

3. (B) v1 = 3

di1 di di - 3 2 = - 3 2 = - 24 cos 4 t V dt dt dt

4. (C) v2 = 4

di di2 di - 3 1 = - 3 1 = 6 e -2 t V dt dt dt

di1 di di - 2 2 = 2 1 = -24 sin 4 t V dt dt dt di2 di1 di1 v2 = -3 +2 =2 = -24 sin 4 t V dt dt dt 5. (A) v1 = 2

4W 1 : 2

10 Vrms

di1 di di + 1 2 = 2 1 = 16 cos 2 t V dt dt dt

1 : 5

48 W

400 W

di1 di di - 2 2 = - 2 2 = - 24 cos 3t V dt dt dt di di di v2 = 3 2 + 2 1 = -3 2 = -36 cos 3t V dt dt dt 6. (D) v1 = 2

Fig. P1.9.42

(A) 3 W

(B) 6 W

(C) 9 W

(D) 12 W 7. (A) v1 = 2

43. I x = ? 8W

di1 di +1 2 dt dt

= -18 sin t + 12 cos t = 6 (2 cos t - 3 sin t) V

10 W 2 : 1

100Ð0 V o

8. (A) v2 = 2

~

j6

= 24 cos 3t - 9 sin 3t = 3 ( 8 cos 3t - 3 sin 3t) V

-j4 Ix

9. (A) v1 = 3

Fig. P1.9.43

(A) 1921 . Ð57.4 ° A

(B) 2.931Ð59.4 ° A

(C) 1.68 Ð43.6 ° A

(D) 179 . Ð43.6 ° A

di1 di -3 2 dt dt

= 45 cos 3t + 27 sin 3t = 9 (5 cos 3t + 3 sin 3t) V 10. (D) v2 = -4

44. Z in = ? j16

di2 di +1 1 dt dt

6W 1 : 5

= 36 sin 3t + 45 cos 3t = 9 ( 4 sin 3t + 5 cos 3t) V

6W

24 W 4 : 1

Zin

di2 di +3 1 dt dt

11. (A) W = -j10

1 1 L1 i12 + L2 i22 + Mi1 i2 2 2

At t = 0, i1 = 4 cos ( -20 ° ) = 4.7 mA i2 = 20 cos ( -20 ° ) = 18.8 mA ,

Fig. P1.9.44

(A) 46.3 + j 6.8 W

(B) 432.1 + j0.96 W

(C) 10.8 + j9.6 W

(D) 615.4 + j0.38 W ********************

M = 0.6 2.5 ´ 0.4 = 0.6 W =

1 1 (2.5)( 4.7) 2 + (0.4)(18.8) 2 + 0.6( 4.7)(18.8) 2 2

= 151.3 mJ 12. (C) Leq = L1 + L2 + 2 M = 7 H www.gatehelp.com

Page 87


UNIT 1

13. (A) Leq = L1 + L2 - 2 M = 4 + 2 - 2 ´ 2 = 2 H

Networks

22. (C) v AG = 20

14. (C) Leq =

L1 L2 - M 2 24 - 16 = =4 H L1 + L2 - 2 M 6 + 4 - 8

23. (B) vBG = 3

15. (A) Leq =

L1 L2 - M 2 8 -4 = = 0.4 H L1 + L2 + 2 M 6 + 4

24. (C) vCG = -6

16. (C) L1 + L2 + 2 M = 7, L1 + L2 - 2 M = 1.8 Þ

d( 6 t) d(15 t) +4 = 180 V dt dt

d(15 t) d 6( t) d ( 6 t) +4 -6 = 33 V dt dt dt d( 6 t) = -36 V dt

25. (B) Z = Z11 +

L1 + L2 = 4.4, M = 1.3

w2 M 2 Z 22 2

L1 L2 - M 2 = 0.5, L1 L2 - 1.32 = 0.5 ´ 1.8 L1 + L2 - 2 M

æ1ö (50) 2 ç ÷ æ 1 ö è5 ø = 4 + j (50) ç ÷+ è 10 ø 5 + j (50) 1 2

L1 L2 = 2.59, ( L1 - L2 ) 2 = 4.4 2 - 4 ´ 2.59 = 9 . , L2 = 0.7 L1 - L2 = 3, L1 = 37

= 4.77 + j 115 . W

2

17. (D) Leq = L1 -

M 4 = 4 - =2 H L2 2

26. (B) Vs = j (0.8)10(12 . Ð0) - j (0.2)(10)(2 Ð0) +[ 3 + j(0.5)(10)] (12 . Ð0 + 2 Ð0)

18. (B) Leq = L1 -

= 9.6 + j21.6 = 26.64 Ð66.04 ° V

2

M 9 =5 - =2 H L2 3

27. (A) [ j(100 p) (2) + 10 ]I 2 + j(100 p)(0.4) (2 Ð0) = 0 19. (A)

Þ -1 H

2H

= 4 Ð - 179.1°

3H

Þ

2H

5H

I 2 = -0.4 - j0.0064,

Vo = 10 I 2 = -4 - j0.064 vo = 4 cos (100 pt - 179.1° ) V

28. (B) 30 Ð30 ° = I ( - j 6 + j 8 - j 4 + j12 - j 4 + 10) Fig. S.1.9.19

Þ

20. (D) VL1 = 1sI + 1sI = 2 sI

30 Ð30 ° = 2.57 - j0.043 (10 + j 6)

Vo = I ( j12 - j 4 + 10)

VL 2 = 2 sI + 1sI - 2 sI = sI ,

= (2.57 - j0.043)(10 + 8 j)

VL 3 = 3sI - 2 sI = sI VL = VL1 + VL 2 + VL 3 = 4 sI

I=

Þ

= 26.067 + j20.14 = 32.9 Ð37.7 ° V

Leq = 4 H

21. (B) Let I1 be the current through 4 H inductor and

29. (A) -j

2W

I 2 and I 3 be the current through 3 H, and 2 H inductor

- Vx +

respectively I1 = I 2 + I 3 , V2 = V3

3Ð-90o A

3sI 2 + 3sI1 = 2 sI 3 + 2 sI1 Þ Þ

Þ 3I 2 + I1 = 2 I 3 4 I1 I 2 = , I 3 = I1 5 5

~

j4

j4

I1

j

4I2 = I3

V = 4 sI1 + 3sI 2 + 2 sI 3 + 3sI 2 + 3s I1 6s 2 ´ 4s = 7 sI1 + I1 + I3 5 5 49 49 H V = sI1 , Leq = 5 5

Page 88

2W

Fig. S1.9.29

( - j + 2 + j 4) I1 - jI 2 = - j 3 ( j 4 + 2) I 2 - jI1 = -12 Ð30 ° V I1 = -1.45 - j0.56, . Vx = -2 I1 = 2.9 + j112 = 311 . Ð2112 . °V

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I2

~

12Ð30 V o



Chap 1.9 20 W

30. (D) j8

j10

0.03 H

j10

j20

j10

0.09 H

Fig. P1.9.34

35. (D) The p equivalent circuit of coupled coil is shown in fig. S1.9.35

= 112 . + j112 . W

M

31. (C) Z in = ( - j 6)||( Z o) (12) 2 Z o = j20 + = 0.52 + j15.7 ( j 30 + j5 - j2 + 4)

L1

L 1L2 - M L1L2 - M

L2

L1 L2 - M 2 = M

w2 M 2 Z L + jwL2

= j250 ´ 10 3 ´ 2 ´ 10 -6 +

L1L2 - M 2 L1 - M

Fig. S1.9.35

M 2 = 160 ´ 10 -12

32. (D) M = k L1 L2 ,

M

2

2

L2 - M

( - j 6)(0.52 + j15.7) = 0.20 - j9.7 W ( - j 6 + 0.52 + j15.7)

Z in = jwL1 +

j30

-j3 1000C

( j14) ( j10 + 2 - j 6) = 10 + j 8 + j14 + j10 + 2 - j 6

Z in =

Vout -

2F

Fig. S.1.9.30

Z eq

0.18 H +

~

Vin j18

40 W

L1 L2 (1 - k2 ) 0.12 ´ 0.27 (10 . .5 2 ) = = 0.27 0.5 k

Output is zero if (250 ´ 10 3)2 ´ 160 ´ 10 -12 2 + j10 + j ´ 250 ´ 10 3 ´ 80 ´ 10 -6

Z in = 0.02 + j0.17 W

C=

-j + jCw = 0 0.27 w

1 = 33.33 mF 0.27 w2

36. (C) Applying 1 V test source at ab terminal,

33. (D)

-j 2C

60 W j2 2

V1

~

I1

j4

1 : 4

j4

I2

4W

20 W

1V

20Ix Ix

Fig. S.1.9.33

V1 = -

jI1 + j 4 I1 + j2 2 I 2 2C

0 = ( 4 + 4 j) I 2 + j2 2 I1 Þ

I2 =

- j 2 I1 2 (1 + j)

2 V1 - j - j + j 8 C + 2 C - j2 C = + j4 + = 1+ j I1 2 C 2C Z in =

- j + j8C + 2C - j 2C 2C

Im ( Z in ) = 0 Þ

C=

Þ

- j + j 8 C - j2 C = 0

1 6

34. (A) j 30 -

3j = 0, C = 100 mF 1000 C

Fig. S1.9.36

Vab = 1 V, I x =

1 = 0.05 A, V2 = 4 V , 20

4 = 60 I 2 + 20 ´ 0.05

Þ

I 2 = 0.05 A

I in = I x + I1 = I x + 4 I 2 = 0.25 A 1 RTH = =4W , VTH = 0 I in 37. (A) Impedance seen by RL = 10 ´ 4 2 = 160 W For maximum power RL = 160 W, Z o = 10 W 2

æ 100 ö PLmax = ç ÷ ´ 10 = 250 W è 10 + 10 ø 38. (B) I 2 = www.gatehelp.com

V2 I V , I1 = 2 = 2 , V1 = 5 V2 8 5 40

Page 89


CHAPTER

1.10 TWO PORT NETWORK

Statement for Q.1-4: The circuit is given in fig. P.1.10.1–4 2W

I1

2W

I2

+

+ 3W

1W

V1

V2

-

-

Fig. P.1.10.1–4

1. [ z ] = ? é 1 ê(A) ê 2 ê- 17 ë 6 é 17 ê(C) ê 6 ê-1 ë 6

3ù 2ú ú 1ú 2û

é1 ê (B) ê 2 ê17 ë6

3ù 2ú ú 1ú 2û

1ù 2ú ú 3ú 2û

é17 ê (D) ê 6 ê1 ë6

1ù 2ú ú 3ú 2û

-

-

3. [ h] = ? 3ù é6 ê17 17 ú (A) ê ú 24 ú ê3 ë17 17 û

é 8 ê (B) ê 3 ê- 1 ë 3

1ù 3ú ú 2ú 3û

é 6 ê (C) ê 17 ê- 3 ë 17

é8 ê (D) ê 3 ê1 ë3

1ù - ú 3 ú 2ú 3û

3ù 17 ú ú 24 ú 17 û

4. [ T ] = ? é17 ù 8ú (A) ê 3 ê ú 3 û ë2

é 17 (B) ê 3 ê ë-2

é 17 ù - 8ú (C) ê 3 ê ú ë 2 - 3û

é17 (D) ê 3 ê ë 2

5. [ z ] = ?

2W

I1

ù - 8ú ú 3û

ù - 8ú ú - 3û

2W

I2

+ 1W

V1

2. [ y ] = ? 1ù é3 ê8 8ú (A) ê ú 1 17 ê ú ë8 24 û

é 3 ê (B) ê 8 ê- 1 ë 8

é17 ê (C) ê 6 ê1 ë2

é 17 ê (D) ê 6 ê- 1 ë 8

1ù 2ú ú 3ú 2û

+

1ù 8ú ú 17 ú 24 û

-

1ù - ú 2 ú 3ú 2û

3W

2W

-

V2 -

Fig. P.1.10.5

é ê (A) ê ê ë

21 16 1 8

é 21 ê (C) ê 16 ê- 1 ë 8 www.gatehelp.com

1 ù 8 ú ú 7 ú 12 û

é ê (B) ê ê ë

7 9 1 6

1ù 6ú ú 7ú 4û

1ù 8ú ú 7 ú 12 û

é ê (D) ê ê ë

7 9 1 3

1ù 3ú ú 7ú 4û

-

Page 91


UNIT 1

6. [ y ] = ?

Networks

11. [ y ] = ? 2W

1W

I1

3W

+

+

V2

V1

-

-

+ 2W

V1

1W

1W

I1

I2

-

+ 2W

1W

2 ù 41 ú ú 19 ú 41 û

é 11 ê (B) ê 41 ê- 2 ë 41

2 ù 41 ú ú 19 ú 41 û

é19 ê (C) ê 41 ê2 ë 41

2 ù 41 ú ú 11 ú 41 û

é 19 ê (D) ê 41 ê- 2 ë 41

2 ù 41 ú ú 11 ú 41 û

é ê (A) ê ê ë

ù 1ú ú -1ú û

1 2 3 2

é 1 ê (C) ê 2 ê- 1 ë 4

é ê (B) ê ê ë

3 2 1 2

ù -1ú ú 1ú û

é 1 ê(D) ê 4 ê 1 ë 2

1ù 2ú ú 3ú 4û

3ù 4ú ú 1ú 2û

12. [ z ] = ?


2W

I1

is

V2

Fig. P.1.10.11

é 11 ê (A) ê 41 ê2 ë 41

port

I1

-

Fig. P.1.10.6

A two

I2

described

by

V1 = I1 + 2 V2 ,

2V1

I2

+

+

I 2 = - 2 I1 + 0.4 V2

1W

V1

7. [ z ] = ?

2W

V2

-

é 11 (A) ê ë-5

-5 ù 2.5 úû

é 11 5 ù (B) ê ú ë 5 2.5 û

é1 (C) ê ë5

-2ù 0.4 úû

é 1 (D) ê ë-2

-

Fig. P.1.10.12

2ù 0.4 úû

é 4 ê (A) ê 3 ê- 2 ë 3

2ù 3ú ú 2ú 3û

é ê (B) ê ê ë

1 2 1 2

1ù - ú 2 ú 1ú û

é 2 ê(C) ê 3 ê 4 ë 3

2ù 3ú ú 2ú 3û

é ê (D) ê ê ë

1 2 1 2

ù 1ú

8. [ y ] = ? é11 5 ù (A) ê ú ë 5 2.5 û

é 1 (B) ê ë-2

-2 ù 4.4 úû

é-2 4.4 ù (C) ê ú ë 4 -2 û

é 11 (D) ê ë- 5

-5 ù 2.5 úû

ú 1 - ú 2û

13. [ y ] = ? 2W

2W

I1

I2

+

+

9. [ h] = ? é3 (A) ê ë4

- 6ù - 4 úû

é 1 (C) ê ë-2

é 4 (B) ê ë-2

-2 ù 4.4 úû

2ù 0.4 úû

é 11 (D) ê ë 5

5ù 2.5 úû

0.5 ù 0.5 úû

é 2. 2 - 0.5 ù (B) ê ú ë 0. 2 - 0.5 û

10. [ T ] = ? é 2. 2 (A) ê ë 0. 2 é 1 (C) ê ë-2 Page 92

2ù 0.4 úû

-2 ù é 1 (D) ê ú ë-2 - 0.4 û

2W

V1

2V1

1W

V2

-

-

Fig. P.1.10.13

é ê (A) ê ê ë

7 4 1 2

é ê (C) ê ê ë

10 19 6 19

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1ù - ú 4 ú 5ú 4û 2 ù 19 ú ú 14 ú 19 û

é ê (B) ê ê ë

7 4 3 4

é ê (D) ê ê ë

6 19 10 19

-

1ù 4ú ú 5ú 4û 14 ù 19 ú ú 2 ú 19 û


Two Port Networks

14. [ z ] = ?

Chap 1.10

17. [ z ] = ? 4V3

I1

2W

I1

2W

I2 +

+ I1

I2

+

2W

2I2

+

1W

V1

V3

2W

V1

+

1W

-

Fig. P.1.10.17

-

é3 (A) ê ê6 ë

Fig. P.1.10.14

é2 (A) ê ë3

3ù 3úû

é-3 (B) ê ë 3

- 2ù 3úû

é3 (C) ê ë3

3ù 2 úû

é 3 (D) ê ë-3

3ù - 2 úû

é ê (C) ê ê ë

15. [ z ] = ? 2V1

2W

I1

2ù 1ú ú 7û ù 1ú ú 3ú û

7 4 1 2

+ 2W

V1

2W 3

3V 2 2

I1

Fig. P.1.1.15

2ù ú 2ú û

é 2 (C) ê ê ë2

3ù 2ú ú 2û

é ê (D) ê ê ë

ù 3ú ú 1ú û

1 2 7 4

1V 5 1

4W

I2 +

1V 10 2

V1

-

é2 (A) ê 3 ê ë2

1ù 7ú ú 2û

+

V2

-

é 6 (B) ê ê ë3

18. [ T ] = ?

I2

+

V2

-

V2

-

2W

4W

V2

-

é -2 (B) ê ê ë 2

3ù 2ú ú -2 û

é 2 -2 ù ú (D) ê 3 ê2ú ë 2 û

-

Fig. P.1.10.18

é 0.35 - 1 ù (A) ê ú ë 2 - 3.33û

- 3.33ù é 2 (B) ê 0 . 35 - 1 úû ë

é 2 (C) ê ë0.35

é 0.35 (D) ê ë 2

3.33ù 1 úû

1 ù 3.33úû

19. [ h] = ? 16. [ y ] = ? I1

1W

2W

I1

I2

+

V2

2W

I2 +

+ +

3W V1

1 2 I2

V1 V2

4W

V2

-

-

V2 -

Fig. P.1.10.19

-

1ù é-1 (A) ê ú ë-1 - 2 û

é 1 - 1ù (B) ê ú ë 1 - 2û

é ê 4 (A) ê ê-2 ë

1ù é 2 ê- 3 3ú (C) ê ú ê- 1 - 1 ú ë 3 3û

1ù é 2 ê- 3 - 3 ú (D) ê ú ê 1 - 1ú ë 3 3û

3ù é ê 4 - 2ú (C) ê ú 1ú ê2 ë 2û

Fig. P.1.10.16

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3ù 2ú ú 1ú 2û

é ê-2 (B) ê ê 4 ë

1ù 2ú ú 3ú 2û

é ê2 (D) ê ê4 ë

1ù 2ú ú 3 - ú 2û

Page 93


Two Port Networks

Z ab ù éZ + Z ab (A) ê a Z b + Z ab úû ë Z ab

éZ - Z ab (B) ê a ë Z ab

éZ + Z ab (C) ê a ë - Z ab

éZ - Z a (D) ê ab ë Z ab

- Z ab

ù Z b + Z ab úû

ù Z b - Z ab úû Z ab

ù - Z b úû

Z ab Z ab

Chap 1.10

The value of

Vo is Vs

(A)

3 32

(B)

1 16

(C)

2 33

(D)

1 17

27. [ y ] = ?

30. The T-parameters of a 2-port network are

Yab

Ya

é2 [T ] = ê ë1

Yb

1ù . 1 úû

If such two 2-port network are cascaded, the Fig. P.1.10.27

éY + Yab (A) ê a ë - Yab

- Yab ù Yb + Yab úû

éY - Yab (B) ê a ë Yab

Yab ù Yb - Yab úû

éY - Ya (C) ê ab ë Yab

Yab ù Yab - Ya úû

éY - Yab (D) ê a ë - Yab

- Yab ù Yb - Yab úû

28. The y-parameters of a 2-port network are é5 [ y] = ê ë1

3ù S 2 úû

z –parameter for the cascaded network is 1ù é 5 - ú é 2 - 2ù ê 3 3 ú (A) ê 1 (B) ê ú 1 2 ê1ú ê ú ë 2 û ë 3 3û é ê (C) ê ê ë

5 3 1 3

1ù 3ú ú 2ú 3û

é2 (D) ê 1 ê ë2

2ù ú 1ú û

31. [ y ] = ? 2W

A resistor of 1 ohm is connected across as shown in

1W

fig. P.1.10.2 8. The new y –parameter would be

1W

1W 2W

1W 2

1W

[ y] = 5 3 1 2

Fig. P.1.10.31

é 19 ê (A) ê 10 ê- 9 ë 10

Fig. P.1.10.28

é6 (A) ê ë2

4ù S 3 úû

é6 (B) ê ë0

2ù S 3 úû

é5 (C) ê ë2

4ù S 2 úû

é4 (D) ê ë2

4ù S 1 úû

é ê (C) ê ê ë

é2 0 ù 29. For the 2-port of fig. P.1.10.29, [ ya ] = ê ú mS ë0 10 û

19 10 9 10

-9 ù 10 ú ú 31 ú 10 û 9 ù 10 ú ú 31 ú 10 û

é 19 ê (B) ê 10 ê- 7 ë 10

-7 ù 10 ú ú 31 ú 10 û

é ê (D) ê ê ë

7 ù 10 ú ú 31 ú 10 û

19 10 7 10

32. [ y ] = ? 2F

I1(s)

60 W

I2(s) +

+

[ ya] Vs 100 W

+ Vo -

300 W

V1(s)

1W 3

2F

2V1(s)

-

1W 4

2F

V2(s) -

Fig. P.1.10.32 Fig. P.1.10.29

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Page 95


UNIT 1

és+3 (A) ê ë2 s + 2

é s+3 (B) ê ë-2 s - 2

2 sù 4 úû

é s + 3 - 2 sù (C) ê ú ë-2 s - 2 4 û

- 2s ù 4 s + 4 úû

Networks

37. The circuit shown in fig. P.1.10.37 is reciprocal if a is

é 3s + 3 - 2 s ù (D) ê ú ë- 2 s - 2 4 s + 4 û

I1

0.5V1

1W

I2

+

+ 2W

33. h21 = ?

V1

I1

R

R

+ V1

V2

I2

aI1

+ R

-

-

Fig. P.1.10.37

V2

-

-

Fig. P.1.10.33

(A) -

3 2

(B)

1 2

(C) -

1 2

(D)

3 2

(A) 2

(B) -2

(C) 1

(D) -1

38. Z in = ? Zin

I2

I1

1 kW

+

+

34. In the circuit shown in fig. P.1.10.34, when the

Vs

[ y] = 4 -0.1 mS 50 1

V1

voltage V1 is 10 V, the current I is 1 A. If the applied

V2

1 kW

-

-

voltage at port-2 is 100 V, the short circuit current Fig. P.1.10.38

flowing through at port 1 will be

V1

Linear Resistive Network

I

(A) 86.4 W

(B) 64.3 W

(C) 153.8 W

(D) 94.3 W

39. V1 , V2 = ? 25 W

Fig. P.1.10.34

(A) 0.1 A

(B) 1 A

(C) 10 A

(D) 100 A

+

+ [ y] = 10 -5 mS 50 20

V1

100 V

V2

100 W

-

-

Fig. P.1.10.39

35. For a 2-port symmetrical bilateral network, if transmission parameters A = 3 and B = 1 W, the value of parameter C is

(A) -68.6 V,

114.3 V

(B) 68.6 V, - 114.3 V

(C) 114.3 V,

- 68.6 V

(D) -114.3 V,

68.6 V

(A) 3

(B) 8 S

(C) 8 W

40. A 2-port network is driven by a source Vs = 100 V in

(D) 9

series with 5 W, and terminated in a 25 W resistor. The impedance parameters are

36. A 2-port resistive network satisfy the condition 3 4 A = D = B = C. The z11 of the network is 2 3 4 3 (A) (B) 3 4 2 (C) 3

Page 96

3 (D) 2

é 20 [z ] = ê ë 40

2ù W 10 úû

The Thevenin equivalent circuit presented to the 25 W resistor is (A) 80 V, 2.8 W

(B) 160 V, 6.8 W

(C) 100 V, 2.4 W

(D) 120 V, 6.4 W

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Two Port Networks

é ê [z ] = ê ê ë

7 4 1 2

Chap 1.10

Z ¢R = n2 4 = 36

ù 1ú ú 3ú û

Z’R

I1=0 +

1 : 3

+ V1

18. (D) Let I 3 be the clockwise loop current in center

I2

4W

+

V’2

9W

V’1 -

-

V2

-

loop I1 =

V2 + I 3, 10

Þ

I1 = 0.35 V2 - I 2

V2 = 4( I 2 + I 3)

Þ

I 3 = 0.25 V2 - I 2 ...(i)

V1 = 4 I1 - 0.2 V1 + V2 Þ V1 = 2 V2 - 3.33I 2

...(ii)

I ö 1 æ 19. (A) V2 = 4ç I 2 + I1 - 2 ÷Þ I 2 = -2 I1 + V2 2 ø 2 è ( V - V2 ) - V2 I V V = -I1 + 2 + 1 - V2 I1 = 2 + 1 2 2 4 2 3 Þ V1 = 4 I1 + V2 2

...(ii)

+

+

+

V’1 -

-

V’2

9 W V2 -

Fig. S1.10.23a

æ4 ö V2 = V2¢ = nV1¢ = 3ç I1 ÷ è5 ø

z11 = Þ

V1 = 0.8 I1 z 21 =

V2 + 0( -I 2 ), 5

...(ii)

I1 = (0) V2 + 5( -I 2 ) ...(i)

27. (A) I1 = ( V1 - V2 ) Yab + V1 Ya

...(ii)

I2=0 1 : 3

Þ

V2 = 2 sI1 + 3sI 2

...(ii)

...(i)

V2 = sI1 + (2 + 2 s) I 2

ZR

4 I1 5

Þ

V2 = ( Z a + Z ab ) I 2 + Z ab I1 = Z ab I1 + ( Z a + Z ab ) I 2

...(ii)

9 9 = =1 n2 9

V1 = ( 4||1) I1 =

V2 = 3sI 2 + 2 sI1

...(i)

...(i)

...(i)

I1 æ1 ö + sI1 + sI 2 = ç + s ÷I1 + sI 2 s ès ø

4W

V1 = 6 sI1 + 2 sI 2

I1 = V1 ( Ya + Yab ) - V2 Yab

V2 = 2.4, I1

....(ii)

...(i)

I 2 = ( V2 - V1 ) Yab + V2 Yb = -V1 Yab + V2 ( Yb + Yab )

I1 = (2 + j2) V1 - jV2 V2 + V1 + j( V2 - V1 ) = (1 - j) V1 + (1 + j) V2 I2 = 1

V1

Þ

Þ

Þ

I1

V2 = 7.2, I2

26. (A) V1 = ( Z a + Z ab ) I1 + Z ab I 2

21. (D) I1 = 2 V1 + jV1 + j( V1 - V2 )

23. (D) Z R =

z 22 =

...(i)

V1 V1 - V2 3 3 + = V1 - V2 1 2 2 2 V V - V1 3 3 I 2 = 2 V1 + 2 + 2 = V1 + V2 1 2 2 2

Þ

Þ

24. (C) V1 = 3sI1 + 3sI1 - 3sI1 + 3sI1 + 2 sI 2

25. (C) V1 =

20. (B) I1 = -V2 +

V2 = 2 I 2 + 2 s I 2 + sI1

V2 = ( 36 || 9) I 2 = 7.2 I 2 z12 = z 21 = 2.4

. V1 = 4(0.35 V2 - I 2 ) + V2 = 2.4 V2 - 4 I 2 12

22. (B) V1 =

Fig. S1.10.23b

...(ii)

28. (B) y-parameter of 1 W resistor network are é 1 ê-1 ë

- 1ù 1úû

é5 New y-parameter = ê ë1

3ù é 1 - 1 ù é6 + = ê 2 úû êë-1 1úû ë0

2ù . 3 úû

-1

0 ù é5000 0 ù é2 mS 29. (A) [ z a ] = ê ú =ê 0 100 úû mS 0 10 ë ë û é5000 [z ] = ê ë 0

0 ù é100 + 100 úû êë100

100 ù é600 = 100 úû êë100

100 ù 200 úû

V1 = 600 I1 + 100 I 2 , V2 = 100 I1 + 200 I 2 Vs = 60 I1 + V1 = 660 I1 + 100 I 2 , V2 = Vo = -300 I 2 2 V Vo = 100 I1 - Vo Þ I1 = o 60 3 Vo 3 V = Vs = 11Vo - o Þ V5 32 3 é2 30. (C) [ TN ] = ê ë1

1ù 1 úû

é2 ê1 ë

1ù é5 = 1 úû êë3

3ù 2 úû

V1 = 5 V2 - 3I 2 , I1 = 3V2 - 2 I 2 www.gatehelp.com

Page 99


UNIT 1

3V1 - 5 I1 = I 2 V2 =

Þ

V1 =

5 1 I1 + I 2 3 3

...(i)

1 2 I1 + I 2 3 3

31. (B)

...(ii)

Networks

I ¢2 = 0.1, I 2¢ = 100 ´ 0.1 = 10 V1¢

Interchanging the port

35. (B) For symmetrical network A = D = 3 For bilateral AD - BC = 1, 9 - C = 1,

1W

2W

C=8 S

2W

36. (A) z11 = 1W

1W 2

1W

A 4 = C 3

37. (A) V1 = 0.5 V1 + I1 + 2( I1 + I 2 ) + aI1 Fig. S.1.10.31a & b

é3 [ za ] = ê ë1

1ù é 2 é 3 ê 5 - 5ú ê 2 1ù , [ ya ] = ê ú , [ yb ] = ê ú 2û 3ú ê- 1 ê- 1 ë 5 ë 2 5û

é 19 ê [ y ] = [ ya ] + [ yb ] = ê 10 ê- 7 ë 10

-

7 ù 10 ú ú 31 ú 10 û

Þ

V1 = ( 6 + 2 a) I1 + 4 I 2

V2 = 2( I1 + I 2 ) + aI1

...(i)

Þ

V2 = (2 + a) I1 + 2 I 2

...(ii)

For reciprocal network z12 = z 21 , 4 = 2 + a

Þ

a =2

38. (C) I1 = 4 ´ 10 3 V1 - 0.1 ´ 10 -3 V2 I 2 = 50 ´ 10 -3 V1 + 10 -3 V2 , V2 = -10 3 I 2 -10 -3 V2 = 50 ´ 10 -3 V1 + 10 -3 V2 , V2 = -25 V1

- 2s ù é 3 , [ yb ] = ê ú 4 sû ë-2

é 3s 32. (D) [ ya ] = ê ë-2 s

1ù - ú 2 ú 5ú 2û

0ù 4 úû

2F

10 3 I1 = 4 V1 + 2.5 V1 ,

V1 10 3 = = 153.8 I1 6.5

39. (B) I1 = 10 ´ 10 -3 V1 - 5 ´ 10 -3 V2 , 100 = 25 I1 + V1

2F

2F

100 - V1 = 0. 25 V1 - 0.125 V2 -3

Þ

800 = 10 V1 - V2

...(i)

-3

I 2 = 50 ´ 10 V1 + 20 ´ 10 V2 , V2 = -100 I 2 V2 = -5 V1 - 2 V2

Fig. S.1.10.32a

1W 3

40. (B) 100 = 5 I1 + V1 , Þ

I2 I1 I1

, -I 2 = V2 = 0

- 2s ù 4 s + 4 úû

R

Þ

I2

R

-

Fig. S.1.10.33

34. (C)

I2 V1

= y21 = V2 = 0

1 = 0.1 10

V2 z 21 = V1 z11

42. (B) I 2 = y21 V1 + y22 V2 , I 2 = -V2 YL y21 V1 + ( y22 + YL ) V2 = 0 ,

+ V1

V2 = 160 + 6.8 I 2

41. (B) V1 = z11 I1 , V2 = z 21 I1 ,

I 1 I1 R , 2 =2 R + R I1

R

V1 = 20 I1 + 2 I 2

VTH = 160 V, RTH = 6.8 W

Fig. S.1.10.32b

33. (C) h21 =

V2 = -114.3 V.

100 = 25 I1 + 2 I 2 , V2 = 40 I1 + 10 I 2

800 - 5 V2 = -34 I 2 é 3s + 3 [ y ] = [ ya ] = [ yb ] = ê ë-2 s - 2

3V2 + 5 V1 = 0

From (i) and (ii) V1 = 68.6 V,

1W 4

2V1

Þ

V2 - y21 = V1 ( y22 + yL )

43. (A) V2 = z 21 I1 + z 22 I 2 ,

V2 = -Z L I 2

æ V ö V2 = z 21 I1 + z 22 çç - 2 ÷÷ è ZL ø V2 ( Z L + z 22 ) = z 21 Z L I1 ,

V2 z Z = 21 L I1 z 22 + Z L **********

Page 100

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...(ii)


CHAPTER

1.11 FREQUENCY RESPONSE


7. A parallel circuit has R = 1 kW , C = 50 mF and L = 10

A parallel resonant circuit has a resistance of 2 kW and half power frequencies of 86 kHz and 90 kHz. 1. The value of capacitor is (A) 6 mF (C) 2 nF

BW = 10 3 rad/s. The value of R and C will be (A) 100 mF, 10 W (C) 100 pF, 10 MW

(B) 43 mH (D) 1.6 mH

The required R for the BW 15.9 Hz is

(A) 22 (C) 48

(A) 0.1 W (C) 15.9 mW

(B) 100 (D) 200


(B) 0.2 W (D) 500 W

10. For the RLC parallel resonant circuit when

resonant

admittance of 25 ´ 10

(B) 100 pF, 10 W (D) 100 mF, 10 MW

9. A series resonant circuit has L = 1 mH and C = 10 mF.

3. The quality factor is

-3

(B) 90.86 (D) None of the above

The resonant frequency wo = 106 rad s and bandwidth is

2. The value of inductor is

A parallel

(A) 100 (C) 70.7

8. A series resonant circuit has an inductor L = 10 mH.

(B) 20 nF (D) 60 mF

(A) 4.3 mH (C) 0.16 mH

mH. The quality factor at resonance is

circuit

has

a

midband

S, quality factor of 80 and a

resonant frequency of 200 krad s.

R = 8 kW, L = 40 mH and C = 0. 25 mF, the quality factor Q is (A) 40

(B) 20

(C) 30

(D) 10

4. The value of R is (A) 40 W (C) 80 W

11. The maximum voltage across capacitor would be

(B) 56.57 W (D) 28. 28 W

0.105v1

5. The value of C is (A) 2 mF (C) 10 mF

+

(B) 28.1 mF (D) 14.14 mF

25 W 3V

6. A parallel RLC circuit has R = 1 kW and C = 1 mF. The quality factor at resonance is 200. The value of inductor is (A) 35.4 mH

(B) 25 mH

(C) 17.7 mH

(D) 50 mH

v1

– 4H

10 W

~

1 mF 4

+ vC –

Fig. P1.11.11

(A) 3200 V (C) -3 V www.gatehelp.com

(B) 3 V (D) 1600 V Page 101


UNIT 1

12. For the circuit shown in fig. P1.1.11 resonant frequency fo is

10 W

Networks

16. H ( w) =

Vo =? Vi

+

v1 -

+

40 W

1 mH

Zin

vs

v1 2

50 nF

~

10 W

0.5 F

vO –

Fig. P1.11.16

Fig. P1.11.12

(A) 346 kHz (C) 196 kHz

(B) 55 kHz (D) 286 kHz

(A) (5 + j20 w) -1

(B) (5 + j 4 w) -1

(C) (5 + j 30 w) -1

(D) 5(1 + j 6 w) -1

13. For the circuit shown in fig. P1.11.13 the resonant

17. The value of input frequency is required to cause a

frequency fo is

gain equal to 1.5. The value is 2 kW

10 mH 600 pF

vs

22 kW

~

60 mF

+ vO –

1.8 W

Fig. P1.11.17

Fig. P1.11.13

(A) 12.9 kHz

(B) 12.9 MHz

(C) 2.05 MHz

(D) 2.05 kHz

14.

The

network

function

of

circuit

shown

fig.P1.11.14 is

in

(A) 20 rad s

(B) 20 Hz

(C) 10 rad s

(D) No such value exists.

18. In the circuit of fig. P1.11.18 phase shift equal to -45 ° is required at frequency w = 20 rad s . The value of R is

4 V H ( w) = o = V1 1 + j0.01w 2 kW

10 W

15 kW Vs

+ vi

~

+ vC –

C

~

1 mF

+ vO –

vo

AvC

–

Fig. P1.11.18 Fig. P1.11.14

The value of the C and A is (A) 10 mF, 6

(B) 5 mF, 10

(C) 5 mF, 6

(D) 10 mF, 10

15. H ( w) = ia

Vo =? Vi

(A) 200 kW

(B) 150 kW

(C) 100 kW

(D) 50 kW

19. For the circuit of fig. P1.11.19 the input frequency is adjusted until the gain is equal to 0.6. The value of the frequency is

20 W

2H

+ vi

~

3ia

4H

0.25 F

vO

vs

~

30 W

+ vO –

–

Fig. P1.11.15

0.6 (A) jw(1 + j0. 2 w) 3 (C) jw(1 + jw) Page 102

0.6 jw(5 + jw) 3 (D) jw(20 + j 4 w)

Fig. P1.11.19

(B)

(A) 20 rad s

(B) 20 Hz

(C) 40 rad s

(D) 40 Hz

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UNIT 1

Networks

31. Bode diagram of the network function Vo Vs for the

SOLUTIONS

circuit of fig. P1.11.30 is 4W

1. (B) BW = w2 - w1 = 2 p(90 - 86)k = 8 p krad s

+ vs

2W

~

BW=

vo

30 mF

Fig.P1.11.30

2. (C) wo =

dB c.

5.56 16.7

dB /d e

0

20

log w

5.56 16.7

=

5.56 16.7 .

0 dB

1 1 = RBW 8 p ´ 10 3 ´ 2 ´ 10 3

( w1 + w2 ) 2 p(90 + 86)k = = 176p krad s 2 2 1 Þ L= 2 C wo

1 = 0.16 mH (176 p ´ 10 ) (20 ´ 10 -9)

log w

wo 176 pk = = 22 8 pk B

-40

-2

dB

B 0d

/de

/d

ec

log w

C =

3 2

3. (A) Q =

c.

5.56 16.7

log w

LC

(B)

(A) 0 dB

1

wo =

c.

dB /de 40

0

Þ

= 19.89 nF

–

dB

1 RC

4. (A) At mid-band frequency Z = R , Y = R=

(C)

(D)

1 = 40 W 25 ´ 10 -3

5. (C) Q = wo RC Q 80 Þ C= = = 10 mF wo R 200 ´ 10 3 ´ 40 ***************

C L

6. (B) Qo = R Þ

8. (B) wo = C=

BW =

1 LC 1 -3

10 ´ 10 ´ (106 ) 2

R L

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= 100 pF

R = 10 ´ 10 -3 ´ 10 3 = 10

Þ R L

R = 15.9 ´ 2 p = 0.1 W 1 ´ 10 -3

10. (B) Q = R

Page 104

10 -6 L

C 50 ´ 10 -6 = 10 3 = 70.7 L 10 ´ 10 -3

9. (A) BW = Þ

200 = 10 3

L = 25 mH

7. (C) Qo = R

Þ

Þ

C L

1 R


Frequency Response

= 8 ´ 10 3

0. 25 ´ 10 -6 = 20 40 ´ 10 -3

15. (A) I a =

11. (A) Thevenin equivalent seen by L-C combination æ v - 0.105 v1 ö 3 = v1 + 10ç 1 ÷ 125 è ø I sc =

Þ

v1 = 100

1100 = 0.8 V 125

Open Circuit : v1 = 0, voc = 3 V RTH =

Chap 1.11

3 = 375 . W, wo = 0.8

1 LC

= 1000

w L 1000 ´ 4 = 1066.67 Qo = o = 375 . R |vC|max = Qo vTH = 1066.67 ´ 3 = 3200 V 12. (B) Applying 1 A at input port V1 = 10 V

Vtest = 10 + jw10

j (5 + 1) w50 ´ 10 -9

Z in = Vtest wo 10 -3 =

6 wo 50 ´ 10 -9

Þ

wo = 346 kHz

fo = 55 kHz

10 10 jw(0.51) 16. (A) Z1 = = 1 + 10 1 + j 3w jw0.5 Vo Z1 = = Vi 40 + Z1

=

= jw6 ´ 10 -10 + 45.45 +

1 1 + 2 ´ 10 3 1.8 + jw10 -5

1.8 - jw10 -5 3. 24 + w2 10 -50

At resonance Im { Y } = 0 wo 6 ´ 10 -10 ( 324 . + w2o 10 -10 ) - wo 10 -5 = 0 3. 24 + w 10 2 o

-10

17. (D) H ( w) =

Vo = AVc

Þ

Þ

18. (D) H ( w) =

= - tan -1 wCR = - 450 °

=

Vi 1 + j2 ´ 10 3 Cw

(15 k) 2 AVc 2 AVi = = 16 k + 30 k 3 3(1 + j2 ´ 10 3 Cw)

Þ

C = 5 mF

20 ´ 1 ´ 10 -6 R = 1 19. (A) H ( w) = gain =

A = 6,

Þ

R = 50 kW.

Vo R = Vs 1 + jwL

R R +w L 2

2

2

=

30 900 + 4 w2 + 0.6 2

50 2 - 30 2 = 20 rad s 2

20. (A) H ( w) =

1 2 ´ 10 3 + jCw

Vo 1 = Vs 1 + jwCR

wCR = 1,

w=

2A Vo 3 = Vi 1 + j2 p ´ 10 3 Cw

2A =4 3

(1 + w2 RC)

For any value of w, R, C gain £ 1.

= 16.67 ´ 10 wo = 12.9 Mrad s

w fo = o = 2.05 MHz 2p

14. (C) VC =

Vo 1 = Vi 1 + jwRC

1

gain =

3

Vi jCw

10 1 + j5 w 10 + 40 1 + j5 w

10 = (5 + j20 w) -1 50 + j200 w

phase shift

13. (C) Y = jw600 ´ 10 -12 +

3I a 0. 25 jw

Thus (D) is correct option.

At resonance Im { Z in } = 0 Þ

Vo =

Vo 3 0.6 = = V1 jw(5 + jw) jw(1 + j0.2 w)

voltage across 1 A source -3

Vi , 20 + j 4w

Vo 1 1 = = Vs 1 + jwCR 1 + j

Phase shift = - tan -1 wCR = - 45 ° gain =

1 1 = = 0.707 | j + 1| 2

21. (B) BW= w2 - w1 = 2 p( 456 - 434) = 44 p wo = 2 pfo = QBW = 20 ´ 44 p fo = 440 Hz

2 ´ 10 3 C = 0.01 22. (C) fo = www.gatehelp.com

1 2 p LC Page 105


UNIT 1

=

1 2 p 360 ´ 10

fo =

-12

´ 240 ´ 10

-6

1 2 p 50 ´ 10

-12

´ 240 ´ 10 -6

Þ

= 541 kHz = 1.45 MHz

R=

Networks

1 . kW = 106 2 p ´ 15 ´ 10 ´ 10 -6

30. (B) 20 log H = 20 log

1 R2 - 2 LC L

1 = -40 log w w2

R 400 10 7 = = -6 L 240 ´ 10 6

1 jw 1+ Vo jw30 ´ 10 -3 .67 16 31. (D) = = jw 1 Vs 6 + 1+ . jw30 ´ 10 -3 356

1 1 1016 = = LC 240 ´ 10 -6 ´ 120 ´ 10 -12 288

20 dB/decade line starting from w = 16.67 rad s

23. (B) fo =

1 2p

2+

-20 dB/decade line starting from w = 5.56 rad s Hence -20 dB/decade line for 5.56 < w < 16.67

R 1 1 < , fo = = 938 kHz L LC 2 p LC

parallel to w axis to w > 16.67

1 , R and C should be as small as possible. RC (1.8) R = ( 3.3) = 1165 . kW 3.3.+1.8 ( 30) C = (10) = 7.5 pF (10 + 30)

24. (B) wo =

w=

1 = 114.5 ´ 106 rad s . 1165 ´ 7.5 ´ 10 -9

25. (D) R¢ = K m R = 800 ´ 12 ´ 10 3 = 9.6 MW L¢ =

800 Km 40 ´ 10 -6 = 32 mF = K f L 1000

C¢ =

C 10 -9 K F = 30 ´ ´ 1000 = 0.375 pF Km 80

26. (A) L¢C¢ =

LC K 2f

Þ

K 2f =

4 ´ 20 ´ 10 -3 ´ 10 -6 1´ 6

Þ K f = 2 ´ 10 -4 (1)(20 ´ 10 -6 ) L¢ L 2 Þ K m2 = = Km (2) ( 4 ´ 10 -3) C¢ C Þ

K m = 0.05 1 RC 1 = 15.9 W R= 2 p ´ 20 ´ 10 3 ´ 0.5 ´ 10 -6

27. (D) wc = 2 pfc = Þ

28. (A) RTH across the capacitor is RTH = (1k + 4 k)||5 k = 2.5 kW 1 . kHz fc = = 106 2 p ´ 2.5 ´ 10 3 ´ 40 ´ 10 -9 29. (B) wc = 2 pfc = Page 106

1 RC

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***********


CHAPTER

2.1 SEMICONDUCTOR PHYSICS

In the problems assume the parameter given in

3. Two semiconductor material have exactly the same

following table. Use the temperature T = 300 K unless

properties except that material A has a bandgap of 1.0

otherwise stated.

eV and material B has a bandgap energy of 1.2 eV. The ratio of intrinsic concentration of material A to that of

Property

Si

Ga As

Ge

material B is

Bandgap Energy

1.12

1.42

0.66

(A) 2016

(B) 47.5

Dielectric Constant

11.7

13.1

16.0

(C) 58.23

(D) 1048

Effective density of 2.8 ´ 1019 states in conduction band N c (cm -3)

4.7 ´ 1017

104 . ´ 1019

Effective density of 104 . ´ 1019 states in valence band N v(cm -3)

7.0 ´ 1018

Intrinsic carrier concertration ni (cm -3) Mobility Electron Hole

15 . ´ 1010

4. In silicon at T = 300 K the thermal-equilibrium concentration of electron is n0 = 5 ´ 10 4 cm -3. The hole concentration is

1.8 ´ 106

6.0 ´ 1018

2.4 ´ 1013

(A) 4.5 ´ 1015 cm -3

(B) 4.5 ´ 1015 m -3

(C) 0.3 ´ 10 -6 cm -3

(D) 0.3 ´ 10 -6 m -3

5. In silicon at T = 300 K if the Fermi energy is 0.22 eV above the valence band energy, the value of p0 is

1350 480

8500 400

3900 1900

(A) 2 ´ 1015 cm -3

(B) 1015 cm -3

(C) 3 ´ 1015 cm -3

(D) 4 ´ 1015 cm -3

6. The thermal-equilibrium concentration of hole p0 in 1. In germanium semiconductor material at T = 400 K

silicon at T = 300 K is 1015 cm -3. The value of n0 is

the intrinsic concentration is

(A) 3.8 ´ 108 cm -3

(B) 4.4 ´ 10 4 cm -3

(C) 2.6 ´ 10 4 cm -3

(D) 4.3 ´ 108 cm -3

(A) 26.8 ´ 1014 cm -3

(B) 18.4 ´ 1014 cm -3

(C) 8.5 ´ 1014 cm -3

(D) 3.6 ´ 1014 cm -3

7. In germanium semiconductor at T = 300 K, the

2. The intrinsic carrier concentration in silicon is to be

acceptor concentrations is N a = 1013 cm -3 and donor

no greater than ni = 1 ´ 1012 cm -3. The maximum

concentration is

temperature allowed for the silicon is ( E g = 112 . eV)

concentration p0 is

(A) 300 K

(B) 360 K

(A) 2.97 ´ 10 9 cm -3

(B) 2.68 ´ 1012 cm -3

(C) 382 K

(D) 364 K

(C) 2.95 ´ 1013 cm -3

(D) 2.4 cm -3

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N d = 0. The thermal equilibrium

Page 109


UNIT 2


Electronics Devices

15. For a particular semiconductor at T = 300 KE g = 15 .

In germanium semiconductor at T = 300 K, the

eV, mp* = 10 mn* and ni = 1 ´ 1015 cm -3. The position of Fermi level with respect to the center of the bandgap is

impurity concentration are N d = 5 ´ 1015 cm -3 and N a = 0

(A) +0.045 eV

(B) - 0.046 eV

(C) +0.039 eV

(D) - 0.039 eV

8. The thermal equilibrium electron concentration n0 is (A) 5 ´ 1015 cm -3 (C) 115 . ´ 10 cm 9

(B) 115 . ´ 1011 cm -3 -3

(D) 5 ´ 10 cm 6

16. A silicon sample contains acceptor atoms at a

-3

concentration of N a = 5 ´ 1015 cm -3. Donor atoms are

9. The thermal equilibrium hole concentration p0 is (A) 396 . ´ 1013

(B) 195 . ´ 1013 cm -3

(C) 4.36 ´ 1012 cm -3

(D) 396 . ´ 1013 cm -3

conduction band edge. The concentration of donors atoms added are

10. A sample of silicon at T = 300 K is doped with boron at a concentration of 2.5 ´ 1013 cm -3 and with arsenic at a concentration of 1 ´ 1013 cm -3. The material is

(B) 4.6 ´ 1016 cm -3

(C) 39 . ´ 1012 cm -3

(D) 2.4 ´ 1012 cm -3

doped with phosphorus atoms at a concentrations of 1015

(B) p - type with p0 = 15 . ´ 10 7 cm -3 (C) n - type with n0 = 15 . ´ 10 cm

(A) 12 . ´ 1016 cm -3

17. A silicon semiconductor sample at T = 300 K is

(A) p - type with p0 = 15 . ´ 1013 cm -3 13

added forming and n - type compensated semiconductor such that the Fermi level is 0.215 eV below the

cm -3. The position of the Fermi level with respect to the

-3

intrinsic Fermi level is

(D) n - type with n0 = 15 . ´ 10 7 cm -3 11. In a sample of gallium arsenide at T = 200 K, n0 = 5 p0 and N a = 0. The value of n0 is

(A) 0.3 eV

(B) 0.2 eV

(C) 0.1 eV

(D) 0.4 eV

(A) 9.86 ´ 10 9 cm -3

(B) 7 cm -3

18. A silicon crystal having a cross-sectional area of

(C) 4.86 ´ 10 3 cm -3

(D) 3 cm -3

0.001 cm 2 and a length of 20 mm is connected to its ends to a 20 V battery. At T = 300 K, we want a current of

12. Germanium at T = 300 K is uniformly doped with an acceptor concentration of N a = 10 cm 15

-3

and a donor

100 mA in crystal. The concentration of donor atoms to be added is

concentration of N d = 0. The position of fermi energy

(A) 2.4 ´ 1013 cm -3

(B) 4.6 ´ 1013 cm -3

with respect to intrinsic Fermi level is

(C) 7.8 ´ 1014 cm -3

(D) 8.4 ´ 1014 cm -3

(A) 0.02 eV

(B) 0.04 eV

(C) 0.06 eV

(D)0.08 eV

19. The cross sectional area of silicon bar is 100 mm 2 . The length of bar is 1 mm. The bar is doped with

13. In germanium at T = 300 K, the donor concentration are N d = 1014 cm -3 and N a = 0. The Fermi energy level with respect to intrinsic Fermi level is

Page 110

(A) 0.04 eV

(B) 0.08 eV

(C) 0.42 eV

(D) 0.86 eV

arsenic atoms. The resistance of bar is (A) 2.58 mW

(B) 11.36 kW

(C) 1.36 mW

(D) 24.8 kW

20. A thin film resistor is to be made from a GaAs film doped n - type. The resistor is to have a value of 2 kW.

14. A GaAs device is doped with a donor concentration of 3 ´ 1015 cm -3. For the device to operate properly, the intrinsic carrier concentration must remain less than 5% of the total concentration. The maximum temperature on that the device may operate is

10 -6 cm 2 . The doping efficiency is known to be 90%. The

(A) 763 K

(B) 942 K

(A) 8.7 ´ 1015 cm -3

(B) 8.7 ´ 10 21 cm -3

(C) 486 K

(D) 243 K

(C) 4.6 ´ 1015 cm -3

(D) 4.6 ´ 10 21 cm -3

The resistor length is to be 200 mm and area is to be mobility of electrons is 8000

cm 2 V - s. The doping

needed is

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Semiconductor Physics

21. A silicon sample doped n - type at 1018 cm -3 have a -6

Chap 2.1

27. For a sample of GaAs scattering time is t sc = 10 -13 s

2

and electron’s effective mass is me* = 0.067 mo . If an

and a length of 10 mm . The doping efficiency of the

electric field of 1 kV cm is applied, the drift velocity

sample is (m n = 800 cm V - s)

produced is

resistance of 10 W . The sample has an area of 10

cm

2

(A) 43.2%

(B) 78.1%

(A) 2.6 ´ 106 cm s

(B) 263 cm s

(C) 96.3%

(D) 54.3%

(C) 14.8 ´ 106 cm s

(D) 482

22.

Six

volts

is

applied

across

a

2

cm

long

semiconductor bar. The average drift velocity is 10 4 cm s. The electron mobility is (A) 4396 cm 2 V - s

(B) 3 ´ 10 4 cm 2 V - s

(C) 6 ´ 10 4 cm 2 V - s

(D) 3333 cm 2 V - s

28. A gallium arsenide semiconductor at T = 300 K is doped with impurity concentration N d = 1016 cm -3. The mobility m n is 7500 cm 2 V - s. For an applied field of 10 V cm the drift current density is

23. For a particular semiconductor material following

(A) 120 A cm 2

(B) 120 A cm 2

(C) 12 ´ 10 4 A cm 2

(D) 12 ´ 10 4 A cm 2

parameters are observed: 29. In a particular semiconductor the donor impurity

m n = 1000 cm 2 V - s ,

concentration is N d = 1014 cm -3. Assume the following

m p = 600 cm V - s , 2

N c = N v = 10 cm 19

parameters,

-3

m n = 1000 cm 2 V - s,

These parameters are independent of temperature.

æ T ö N c = 2 ´ 1019ç ÷ è 300 ø

The measured conductivity of the intrinsic material is s = 10 -6 (W - cm) -1 at T = 300 K. The conductivity at

æ T ö N v = 1 ´ 1019ç ÷ è 300 ø

T = 500 K is (A) 2 ´ 10

-4

(C) 2 ´ 10

-5

(W - cm) (W - cm)

-1

-1

(B) 4 ´ 10

-5

-1

(W - cm)

(D) 6 ´ 10

-3

(W - cm) -1

24. An n - type silicon sample has a resistivity of 5 W - cm at T = 300 K. The mobility is m n = 1350

32

cm -3,

32

cm -3,

. eV. E g = 11 An electric field of E = 10 V cm is applied. The electric current density at 300 K is

cm 2 V - s. The donor impurity concentration is

(A) 2.3 A cm 2

(B) 1.6 A cm 2

(A) 2.86 ´ 10 -14 cm -3

(B) 9.25 ´ 1014 cm -3

(C) 9.6 A cm 2

(D) 3.4 A cm 2

(C) 11.46 ´ 1015 cm -3

(D) 11 . ´ 10 -15 cm -3

25.


In a silicon sample the electron concentration 18

drops linearly from 10

cm

-3

16

to 10

cm

-3

A semiconductor has following parameter

over a length

m n = 7500 cm 2 V - s,

of 2.0 mm. The current density due to the electron diffusion current is ( Dn = 35 cm 2 s).

m p = 300 cm 2 V - s,

(A) 9.3 ´ 10 4 A cm 2

(B) 2.8 ´ 10 4 A cm 2

ni = 3.6 ´ 1012 cm -3

(C) 9.3 ´ 10 9A cm 2

(D) 2.8 ´ 10 9 A cm 2

30.

26. In a GaAs sample the electrons are moving under an

electric

field

of

5

kV cm

and

the

carrier

concentration is uniform at 1016 cm -3. The electron velocity is the saturated velocity of 10

7

(A) 1.6 ´ 10 A cm (C) 1.6 ´ 108 A cm 2

conductivity

is

minimum,

the

hole

(A) 7.2 ´ 1011 cm -3

(B) 1.8 ´ 1013 cm -3

(C) 1.44 ´ 1011 cm -3

(D) 9 ´ 1013 cm -3

cm s. The drift 31. The minimum conductivity is

current density is 4

When

concentration is

2

2

(A) 0.6 ´ 10 -3 (W - cm) -1

(B) 17 . ´ 10 -3 (W - cm) -1

(D) 2.4 ´ 108 A cm 2

(C) 2.4 ´ 10 -3 (W - cm) -1

(D) 6.8 ´ 10 -3 (W - cm) -1

(B) 2.4 ´ 10 A cm 4

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Page 111


UNIT 2

32.

A

particular

intrinsic

semiconductor

has

Electronics Devices

æ -x ö ÷ ç ÷ ç 15 è L p ø

a

Hole concentration p0 = 10 e

resistivity of 50 (W - cm) at T = 300 K and 5 (W - cm) at T = 330 K. If change in mobility with temperature is

cm -3,x ³ 0 ö æ ç -x ÷ ÷ ç

neglected, the bandgap energy of the semiconductor is

Electron concentration n0 = 5 ´ 1014 e è L n ø cm -3,x £ 0

(A) 1.9 eV

(B) 1.3 eV

Hole diffusion coefficient Dp = 10 cm 2 s

(C) 2.6 eV

(D) 0.64 eV

Electron diffusion coefficients Dn = 25 cm 2 s a

Hole diffusion length Lp = 5 ´ 10 -4 cm,

semiconductor. If only the first mechanism were

Electron diffusion length Ln = 10 -3 cm

33.

Three

scattering

mechanism

exist

in

present, the mobility would be 500 cm V - s. If only the 2

second mechanism were present, the mobility would be

The total current density at x = 0 is

750 cm 2 V - s. If only third mechanism were present,

(A) 1.2 A cm 2

(B) 5.2 A cm 2

the mobility would be 1500 cm 2 V - s. The net mobility

(C) 3.8 A cm 2

(D) 2 A cm 2

is (A) 2750 cm 2 V - s

(B) 1114 cm 2 V - s

37. A germanium Hall device is doped with 5 ´ 1015

(C) 818 cm 2 V - s

(D) 250 cm 2 V - s

donor atoms per cm 3 at T = 300 K. The device has the geometry d = 5 ´ 10 -3 cm, W = 2 ´ 10 -2 cm and L = 0.1 cm.

34. In a sample of silicon at T = 300 K, the electron

The current is I x = 250 mA, the applied voltage is

concentration varies linearly with distance, as shown in

Vx = 100

fig. P2.1.34. The diffusion current density is found to be

tesla. The Hall voltage is

J n = 0.19 A cm 2 . If the electron diffusion coefficient is

(A) -0.31mV

(B) 0.31 mV

Dn = 25 cm 2 s, The electron concentration at is

(C) 3.26 mV

(D) -3.26 mV

5´1014 n(cm-3)

Statement for Q.38-39: A silicon Hall device at T = 300 K has the geometry d = 10 -3 cm , W = 10 -2 cm, L = 10 -1 cm. The following parameters are measured: I x = 0.75 mA,

n(0) 0

mV, and the magnetic flux is Bz = 5 ´ 10 -2

0.010

Vx = 15 V, V H = +5.8 mV, tesla

x(cm)

Fig. P2.1.34

38. The majority carrier concentration is (A) 4.86 ´ 10 cm 8

-3

(C) 9.8 ´ 10 26 cm -3

-3

(A) 8 ´ 1015 cm -3, n - type

(D) 5.4 ´ 1015 cm -3

(B) 8 ´ 1015 cm -3, p - type

(B) 2.5 ´ 10 cm 13

35. The hole concentration in p - type GaAs is given by xö æ p = 10 ç 1 - ÷ cm -3 for 0 £ x £ L L è ø

(C) 4 ´ 1015 cm -3, n - type (D) 4 ´ 1015 cm -3, p - type

16

39. The majority carrier mobility is

where L = 10 mm. The hole diffusion coefficient is 10 cm s. The hole diffusion current density at x = 5 mm 2

(A) 430 cm 2 V - s

(B) 215 cm 2 V - s

(C) 390 cm 2 V - s

(D) 195 cm 2 V - s

is (A) 20 A cm 2

(B) 16 A cm 2

40. In a semiconductor n0 = 1015 cm -3 and ni = 1010 cm -3.

(C) 24 A cm 2

(D) 30 A cm 2

The excess-carrier life time is 10 -6 s. The excess hole

36. For a particular semiconductor sample consider following parameters:

Page 112

concentration is dp = 4 ´ 1013 cm -3. The electron-hole recombination rate is (A) 4 ´ 1019 cm -3s -1

(B) 4 ´ 1014 cm -3s -1

(C) 4 ´ 10 24 cm -3s -1

(D) 4 ´ 1011 cm -3s -1

www.gatehelp.com



41. A semiconductor in thermal equilibrium, has a hole concentration

cm -3

p0 = 1016

of

and

an

SOLUTIONS

intrinsic

-3

concentration of ni = 10 cm . The minority carrier life 10

time

4 ´ 10 -7s.

is

The

thermal

equilibrium

recombination rate of electrons is (A) 2.5 ´ 10 22 cm -3 s -1

(B) 5 ´ 1010 cm -3 s -1

(C) 2.5 ´ 1010 cm -3 s -1

(D) 5 ´ 10 22 cm -3 s -1

n-type

1. (C) ni2 = N c N v e

For Ge at 300 K, . ´ 1019, N v = 6.0 ´ 1018 , E g = 0.66 eV N c = 104

silicon

sample

contains

a

donor

concentration of N d = 106 cm -3. The minority carrier hole lifetime is t p 0 = 10 m s.

-3

Þ

(A) 5 ´ 10 cm s

-1

-3

-1

3

-3

-1

(B) 10 cm s

-3

(C) 2. 25 ´ 10 cm s 9

4

-1

(D) 10 cm s

ni = 8.5 ´ 1014 cm -3

2. (C) n = N c N v e 2 i

electron is

T e

(A) 1125 . ´ 10 cm s (C) 8.9 ´ 10 44.

-10

-3

cm s

A n -type

-1

-3

(B) 2.25 ´ 10 cm s 9

-1

sample

concentration of N d = 10

16

contains

a

-13´10 3 T

= 9.28 ´ 10 -8 , By trial T = 382 K

-1

(D) 4 ´ 10 9 cm -3 s -1

silicon

æ 1 .12 e ö ÷÷ kT ø

æ T ö - ççè (1012 ) 2 = 2.8 ´ 1019 ´ 104 . ´ 1019ç ÷ e è 300 ø 3

-3

æ -Eg ö ÷ - çç ÷ è kT ø 3

43. The thermal equilibrium generation rate for 9

æ 0 .66 ö

3

÷÷ - çç æ 400 ö è 0 .0345 ø ni2 = 104 . ´ 1019 ´ 6.0 ´ 1018 ´ ç ÷ é è 300 ø

42. The thermal equilibrium generation rate of hole is 8

æ -Eg ö ÷ - çç ÷ è kT ø

æ 400 ö Vt = 0.0259ç ÷ = 0.0345 è 300 ø

Statement for Q.42-43: A

Chap 2.1

-

donor

E gA

æ E - E gB ö ç gA ÷ ÷ kT ø

-ç n2 e kT 3. (B) iA = E gB = e è 2 niB e kT

= 2257.5

Þ

niA = 47.5 niB

-3

cm . The minority carrier

hole lifetime is t p 0 = 20 m s. The lifetime of the majority carrier is ( ni = 15 . ´ 1010 cm -3) (A) 8.9 ´ 106 s

(B) 8.9 ´ 10 -6 s

(C) 4.5 ´ 10 -17 s

(D) 113 . ´ 10 -7 s

concentration are N a = 10

cm

-3

and N a = 0. The

equilibrium recombination rate is Rp 0 = 1011 cm -3s -1 . A uniform generation rate produces an excess- carrier -3

concentration of dn = dp = 10 cm . The factor, by which 14

the total recombination rate increase is

ni2 (15 . ´ 1010 ) 2 = = 4.5 ´ 1015 cm -3 n0 5 ´ 10 4

5. (A) p0 = N v e

45. In a silicon semiconductor material the doping 16

4. (A) p0 =

6. (B) p0 = N v e

-

-

( EF - Ev ) kT

( EF - Ev ) kT

-0 .22

= 104 . ´ 1019 e 0 .0259 = 2 ´ 1015 cm -3 Þ

At 300 K, N v = 10 . ´ 1019 cm -3 æ 104 . ´ 1019 ö ÷÷ = 0. 239 eV EF - Ev = 0.0259 lnçç 1015 ø è ( Ec - EF )

(A) 2.3 ´ 1013

(B) 4.4 ´ 1013

n0 = N c e

(C) 2.3 ´ 10 9

(D) 4.4 ´ 10 9

At 300 K, N c = 2.8 ´ 1019 cm -3

***********

æN ö EF - Ev = kT ln çç v ÷÷ è p0 ø

kT

Ec - EF = 112 . - 0. 239 = 0.881 eV n0 = 4.4 ´ 10 4 cm -3 2

7. (C) p0 =

Na - Nd N ö æ + ç N a - d ÷ + ni2 2 2 ø è

For Ge ni = 2.4 ´ 10 3 2

æ 1013 ö 1013 ÷÷ + (2.4 ´ 1013) 2 = 2.95 ´ 1013 cm -3 p0 = + çç 2 è 2 ø www.gatehelp.com

Page 113



=

(1.6 ´ 10

Þ n0 =

26. (A) J = evn = (1.6 ´ 10 -19)(10 7)(1016 ) = 1.6 ´ 10 4 A cm 2

0.1 = 11.36 kW )(1100)(5 ´ 1016 )(100 ´ 10 -8 )

-19

L , sA

20. (A) R =

s » e m n n0 ,

R=

27. (A) vd =

L em n n0 A

et sc E (1.6 ´ 10 -19)(10 -13)(10 5) = (0.067)(9.1 ´ 10 -31 ) me*

= 26.2 ´ 10 3 m s = 2.6 ´ 106 cm s

L em n AR

28. (A) N d >> ni

n0 = 0.9 N d =

Chap 2.1

-4

20 ´ 10 = 8.7 ´ 1015 cm -3 (0.9)(1.6 ´ 10 -19)( 8000)(10 -6 )(2 ´ 10 3)

21. (B) s » e m n n0 , R =

L L , n0 = sA em n AR

10 ´ 10 -4 = 7.81 ´ 1017 cm -3 = (1.6 ´ 10 -19)( 800)(10 -6 )(10)

Þ

J = em n n0 E = (1.6 ´ 10 29. (D) ni2 = N c N v e

)(7500)(1016 )(10) = 120 A cm 2

æ -Eg ö ÷ - çç ÷ è kT ø

= (2 ´ 1019)(1 ´ 1019) e Þ

n0 = N d -19

æ 1 .1 ö ÷÷ - çç è 0 .0259 ø

= 7.18 ´ 1019

ni = 8.47 ´ 10 9 cm -3

N d >> ni

Þ

N d = n0

Efficiency =

n0 7.8 ´ 1017 ´ 100 = ´ 100 = 78.1 % Nd 1018

J = sE = em n n0 E

22. (D) E =

V 6 = = 3 V/cm, vd = m n E, L 2

30. (A) s = em n n0 + em p p0 and n0 =

vd 10 4 = 3333 cm 2 V - s = E 3

mn =

= (1.6 ´ 10 -19)(1000)(1014 )(100) = 1.6 A cm 2

Þ

10 -6 = (1.6 ´ 10 -19)(1000 + 600)ni At T = 300 K, ni = 391 . ´ 10 cm 9

ni2 = N c N v e

ni2 + em p p0 , p0

( -1) em n ni2 ds =0 = + em p dp0 p02

23. (D) s1 = eni (m n + m p )

æ Eg ö ÷ - çç ÷ è kT ø

s = em n

ni2 p0

1 -3

æN N ö Þ E g = kT ln çç c 2 v ÷÷ è ni ø

Þ

æm p0 = ni ç n çm è p

1

ö2 2 ÷ = 3.6 ´ 1012 æç 7500 ö÷ ÷ 300 è ø ø

= 7. 2 ´ 1011 cm -3

ö æ 1019 eV ÷ = 1122 Þ E g = 2(0.0259) ln çç . 9 ÷ 391 . ´ 10 ø è

31. (B) smin =

æ 500 ö At T = 500 K , kT = 0.0259ç ÷ = 0.0432 eV, è 300 ø

= 2 ´ 1.6 ´ 10 -19( 3.6 ´ 1012 ) (7500)( 300)

ni2 = (1019) 2 e Þ

æ 1 .122 ö ÷÷ - çç è 0 .0432 ø

32. (B) s =

= (1.6 ´ 10 -19)(2.29 ´ 1013)(1000 + 600) = 5.86 ´ 10 -3(W - cm) -1 24. (B) r = Nd =

1 1 = s em n N d

1 1 = 9.25 ´ 1014 cm -3 = 5(1.6 ´ 10 -19)(1350) rem n

25. (B) J n = eDn

dn dx

æ 1018 - 1016 = (1.6 ´ 10 -19)( 35)çç -4 è 2 ´ 10

mp + mn

= 2 en i m pm n

= 17 . ´ 10 -3(W - cm) -1

cm -3,

ni = 2.29 ´ 1013 cm -3

2 s i m pm n

1 = emni , r

Eg 1 ni1 e 2 kT1 r1 = = Eg 1 ni 2 2 kT2 e r2

0.1 = e

-

Eg æ 1 1 ö÷ ç T2 ÷ø 2 k çè T1

E g æ 330 - 300 ö ç ÷ = ln 10 2 k çè 330 ´ 300 ÷ø E g = 22( k300) ln 10 = 1.31 eV

ö ÷÷ = 2.8 ´ 10 4 A cm 2 ø

33. (D) www.gatehelp.com

1 1 1 1 = + + m m1 m 2 m 3 Page 115


CHAPTER

2.2 THE PN JUNCTION

In this chapter, N d and N a denotes the net donor

5. The Fermi level on p - side is

p-region.

(A) 0.2 eV (C) 0.4 eV

1. An abrupt silicon in thermal equilibrium at T = 300


and acceptor concentration in the individual n and

K is doped such that Ec - EF = 0.21 eV in the n - region

(B) 0.1 eV (D) 0.3 eV

A silicon pn junction at T = 300 K with zero

and EF - Ev = 0.18 eV in the p - region. The built-in

applied bias has doping concentrations of N d = 5 ´ 1016

potential barrier Vbi is

cm -3 and N a = 5 ´ 1015 cm -3.

(A) 0.69 V (C) 0.61 V

(B) 0.83 V (D) 0.88 V

6. The width of depletion region extending into the

2. A silicon pn junction at T = 300 K has N d = 1014

n-region is

cm and N a = 10 cm . The built-in voltage is

(A) 4 ´ 10 -6 cm (C) 4 ´ 10 -5 cm

(A) 0.63 V (C) 0.026 V

7. The space charge width is

-3

17

-3

(B) 0.93 V (D) 0.038 V

3. In a uniformly doped GaAs junction at T = 300 K, at zero bias, only 20% of the total space charge region is to be in the p-region. The built in potential barrier is Vbi = 1.20 V. The majority carrier concentration in n-region is (A) 1 ´ 1016 cm -3 (C) 1 ´ 10

22

cm

-3

(B) 4 ´ 10 16 cm -3 (D) 4 ´ 10

22

cm

(B) 3 ´ 10 -6 cm (D)3 ´ 10 -5 cm

(A) 32 . ´ 10 -5 cm (C) 4.5 ´ 10 -4 cm

(B) 4.5 ´ 10 -5 cm (D) 32 . ´ 10 -4 cm

8. In depletion region maximum electric field | Emax | is (A) 1 ´ 10 4 V cm (C) 3 ´ 10 4 V cm

(B) 2 ´ 10 4 V cm (D) 4 ´ 10 4 V cm

9. An n – n isotype doping profile is shown in fig. P2.2.9.

-3

The built-in potential barrier is

(ni = 15 . ´ 1010 cm -3 )

Nd(cm-3)


10

An abrupt silicon pn junction at zero bias and

10

T = 300 K has dopant concentration of N a = 1017 cm -3

16

15

and N d = 5 ´ 1015 cm -3. 0

4. The Fermi level on n - side is (A) 0.1 eV

(B) 0.2 eV

(C) 0.3 eV

(D) 0.4 eV

Fig. P2.2.9

(A) 0.66 V (C) 0.03 V www.gatehelp.com

(B) 0.06 V (D) 0.33 V Page 117


UNIT 2

Electronics Devices

concentration are N d = 4 ´ 1016 cm -3 and N a = 4 ´ 1017

Statement for Q.10–11: A silicon abrupt junction has dopant concentration

cm -3. The magnitude of the reverse bias voltage is

N a = 2 ´ 1016 cm -3 and N d = 2 ´ 10 15 cm -3. The applied

(A) 3.6 V

(B) 9.8 V

reverse bias voltage is VR = 8 V.

(C) 7.2 V

(D) 12.3 V

10. The maximum electric field | Emax | in depletion region is

17. An abrupt silicon pn junction has an applied

(A) 15 ´ 10 V cm

(B) 7 ´ 10 V cm

reverse bias voltage of VR = 10 V. it has dopant

(C) 35 . ´ 10 4 V cm

(D) 5 ´ 10 4 V cm

concentration of N a = 1018 cm -3 and N d = 1015 cm -3. The

4

4

pn junction area is 6 ´ 10 -4 cm 2 . An inductance of 2.2

11. The space charge region is

mH is placed in parallel with the pn junction. The

(A) 2.5 mm

(B) 25 mm

resonant frequency is

(C) 50 mm

(D) 100 mm

(A) 1.7 MHz

(B) 2.6 MHz

(C) 3.6 MHz

(D) 4.3 MHz

12.

A uniformly

doped

silicon

pn junction

has

N a = 5 ´ 1017 cm -3 and N d = 10 17 cm -3. The junction has

18. A uniformly doped silicon p+ n junction is to be

a cross-sectional area of 10 -4 cm -3 and has an applied

designed such that at a reverse bais voltage of V R = 10

reverse-bias voltage of VR = 5 V. The total junction

V the maximum electric field is limited to Emax = 106

capacitance is

V cm. The maximum doping concentration in the

(A) 10 pF

(B) 5 pF

n-region is

(C) 7 pF

(D) 3.5 pF

(A) 32 . ´ 1019 cm -3

(B) 32 . ´ 1017 cm -3

(C) 6.4 ´ 1017 cm -3

(D) 6.4 ´ 1019 cm -3

Statement for Q.13–14: An ideal one-sided silicon n+ p junction has

19. A diode has reverse saturation current I s = 10 -10 A

uniform doping on both sides of the abrupt junction.

and non ideality factor h = 2. If diode voltage is 0.9 V,

The doping relation is N d = 50 N a . The built-in potential

then diode current is

barrier is Vbi = 0.75 V. The applied reverse bias voltage

(A) 11 mA

(B) 35 mA

is V R = 10.

(C) 83 mA

(D) 143 mA

13. The space charge width is

20. A diode has reverse saturation current I s = 10 -18 A

(A) 1.8 mm

(B) 1.8 mm

and nonideality factor h = 105 . . If diode has current of 70

(C) 1.8 cm

(D) 1.8 m

mA, then diode voltage is

14. The junction capacitance is (A) 3.8 ´ 10 -9 F cm 2

(B) 9.8 ´ 10 -9 F cm 2

(C) 2.4 ´ 10 -9 F cm 2

(D) 5.7 ´ 10 -9 F cm 2

(A) 0.63 V

(B) 0.87 V

(C) 0.54 V

(D) 0.93 V

21. An ideal pn junction diode is operating in the forward bais region. The change in diode voltage, that

+

15. Two p n silicon junction is reverse biased at VR = 5

will cause a factor of 9 increase in current, is

V. The impurity doping concentration in junction A are

(A) 83 mV

(B) 59 mV

(C) 43 mV

(D) 31 mV

N a = 10 cm 18

-3

and N d = 10

-15

-3

cm , and those in junction

B are N a = 1018 cm -3 and N d = 1016 cm -3. The ratio of the 22. An pn junction diode is operating in reverse bias

space charge width is (A) 4.36

(B) 9.8

region. The applied reverse voltage, at which the ideal

(C) 19

(D) 3.13

reverse current reaches 90% of its reverse saturation current, is

16. The maximum electric field in reverse-biased silicon pn junction Page 118

is

| Emax | = 3 ´ 10

5

V cm.

The

doping

(A) 59.6 mV

(B) 2.7 mV

(C) 4.8 mV

(D) 42.3 mV

www.gatehelp.com


The pn Junction

p+ n junction diode the doping

23. For a silicon

cross-sectional area is 10 -3 cm 2 . The minority carrier

-3

cm .

lifetimes are t n 0 = 1 ms and t p 0 = 0.1 m s. The minority

The minority carrier hole diffusion coefficient is Dp = 12

carrier diffusion coefficients are Dn = 35 cm 2 s and

cm 2 s and the minority carrier hole life time is t p 0 = 10 -7

Dp = 10 cm 2 s. The total number of excess electron in

s. The cross sectional area is A = 10 -4 cm 2 . The reverse

the p - region, if applied forward bias is Va = 0.5 V, is

saturation current is

(A) 4 ´ 10 7 cm -3

(B) 6 ´ 1010 cm -3

(C) 4 ´ 1010 cm -3

(D) 6 ´ 10 7 cm -3

concentrations are N a = 10

18

(A) 4 ´ 10

-12

cm

-3

Chap 2.2

and N d = 10

16

(B) 4 ´ 10

A

(C) 4 ´ 10 -11 A

-15

A

(D) 4 ´ 10 -7 A

28. Two ideal pn junction have exactly the same

24. For an ideal silicon pn junction diode

electrical and physical parameters except for the band gap of the semiconductor materials. The first has a

t no = t po = 10 -7 s ,

bandgap energy of 0.525 eV and a forward-bias current

Dn = 25 cm 2 s ,

of 10 mA with Va = 0.255 V. The second pn junction

Dp = 10 cm 2 s

diode is to be designed such that the diode current

The ratio of N a N d , so that 95% of the current in

I = 10 mA at a forward-bias voltage of Va = 0.32 V. The

the depletion region is carried by electrons, is

bandgap energy of second diode would be

(A) 0.34

(B) 0.034

(A) 0.77 eV

(B) 0.67 eV

(C) 0.83

(D) 0.083

(C) 0.57 eV

(D) 0.47 eV


29. A pn junction biased at Va = 0.72 V has DC bias

A ideal long silicon pn junction diode is shown in

current I DQ = 2 mA. The minority carrier lifetime is 1 ms

fig. P2.2.25–26. The n - region is doped with 10

is both the n and p regions. The diffusion capacitance is

organic atoms per cm 3 and the p - region is doped with

in

5 ´ 10

cm . The minority carrier

(A) 49.3 nF

(B) 38.7 nF

lifetimes are D n = 23 cm s and Dp = 8 cm s. The

(C) 77.4 nF

(D) 98.6 nF

16

16

3

boron atoms per 2

2

forward-bias voltage is Va = 0.61 V.

30. A p+ n silicon diode is forward biased at a current of 1 mA. The hole life time in the n - region is 0.1 ms.

W

Neglecting p

Va

n

x=0

x

the

diode

(A) 38.7 + j12.1 W

(B) 235 . + j7.5 W

(C) 38.7 - j12.1 mW

(D) 235 . - j7.5 W

forward-bias current of a p+ n diode is 2.5 ´ 10 -6 F A.

(A) 6.8 ´ 1012 e -246 x cm -3, x ³ 0

The hole lifetime is

(B) 6.8 ´ 1012 e -246 x cm -3, x ³ 0 (C) 3.8 ´ 1014 e -3534 x cm -3, x ³ 0 (D) 3.8 ´ 10 e

capacitance

31. The slope of the diffusion capacitance verses

25. The excess hole concentration is

+3534 x

depletion

impedance at 1 MHz is

Fig. P.2.2.25-26

14

the

(A) 1.3 ´ 10 -7 s

(B) 1.3 ´ 10 -4 s

(C) 6.5 ´ 10 -8 s

(D) 6.5 ´ 10 -4 s

-3

cm , x ³ 0

32. A silicon pn junction with doping profile of N a = 1016

26. The hole diffusion current density at x = 3 mm is (A) 0.6 A cm 2

(B) 0.6 ´ 10 -3 A cm 2

(C) 0.4 A cm 2

(D) 0.4 ´ 10 -3 A cm 2

cm -3 and N d = 10 15 cm -3 has a cross sectional area of 10 -2 cm 2 . The length of the p - region is 2 mm and length of the n - region is 1 mm. The approximately series resistance of the diode is

27. The doping concentrations of a silicon pn junction are

N d = 10

16

cm

-3

and

N a = 8 ´ 10

15

-3

cm .

The

(A) 62 W

(B) 43 W

(C) 72 W

(D) 81 W

www.gatehelp.com

Page 119


UNIT 2

Electronics Devices

33. A gallium arsenide pn junction is operating in

39. A GaAs laser has a threshold density of 500 A cm 2 .

reverse-bias voltage VR = 5 V. The doping profile are

The laser has dimensions of 10 mm ´ 200 mm. The active

N a = N d = 10

16

-3

cm . The minority carrier lifetime are

t p 0 = t n 0 = t 0 = 10

-8

s. The reverse-biased generation (e r = 131 . , ni = 1.8 ´ 10 ) 6

current density is (A) 19 . ´ 10

-8

(C) 1.4 ´ 10

-8

A cm

(B) 19 . ´ 10

2

A cm

-9

(D) 1.4 ´ 10

2

-9

A cm

region is dLas = 100 A °. The electron-hole recombination time at threshold is 1.5 ns. The current density of 5J th is injected into the laser. The optical power emitted, if

2

emitted photons have an energy of 1.43 eV, is

2

(A) 143 mW

(B) 71.5 mW

(C) 62.3 mW

(D) 124.6 mW

A cm

34. For silicon the critical electric field at breakdown is approximately Ecrit = 4 ´ 10 5 V cm. For the breakdown voltage

of 25

V,

maximum n - type

the

doping

+

concentration in an abrupt p n-junction is (A) 2 ´ 1016 cm -3

(B) 4 ´ 10 16 cm -3

(C) 2 ´ 1018 cm -3

(D) 4 ´ 10 18 cm -3

35. A uniformly doped silicon pn junction has dopant profile of N a = N d = 5 ´ 1016 cm -3. If the peak electric field in the junction at breakdown is E = 4 ´ 10 5 V cm, the breakdown voltage of this junction is (A) 35 V

(B) 30 V

(C) 25 V

(D) 20 V

36. An abrupt silicon p+ n junction has an n - region doping

concentration n - region

minimum

of

N d = 5 ´ 10 15

width,

such

that

cm -3.

The

avalanche

breakdown occurs before the depletion region reaches an ohmic contact, is (VB » 100 V) (A) 5.1 mm

(B) 3.6 mm

(C) 7.3 mm

(D) 6.4 mm

37. A silicon pn junction diode has doping profile N a = N d = 5 ´ 1019 cm -3. The space charge width at a forward bias voltage of Va = 0.4 V is (A) 102 A°

(B) 44 A°

(C) 153 A°

(D) 62 A°

38. A

GaAs

pn+

junction LED has following

parameters Dn = 25 cm 2 s, Dp = 12 cm 2 s N d = 5 ´ 1017 cm -3, N a = 1016 cm -3 t n 0 = 10 ns , t p 0 = 10 ns The injection efficiency of the LED is

Page 120

(A) 0.83

(B) 0.99

(C) 0.64

(D) 0.46

www.gatehelp.com

***************


UNIT 2

= 301 . ´ 10 -5 cm, CT =

-14

. ´ 8.85 ´ 10 ´ 10 eA 117 = 301 . ´ 10 -5 W

-4

= 35 . ´ 10 -12 F

ö ÷÷ ø

=

2 eEmax 2e

æ 1 1 ö ÷÷ çç + è Na Nd ø

(117 . ´ 8.85 ´ 10 -14 )( 3 ´ 10 5) 2 æ 1 1 ö V + ç 17 ÷ 16 ´ ´ 2 ´ 1.6 ´ 10 -19 4 10 4 10 è ø

= 8.008 V

1

ì 2 e ( V + VR ) é 1 1 ùü 2 W = í s bi + ê úý , e ë Na Na û þ î

VR = 8.008 - 0.826 = 7.18 V æN N ö 17. (B) Vbi = Vt ln çç a 2 d ÷÷ è ni ø

1

Þ

1

Vbi + VR =

N a = 4. 2 ´ 1015 cm -3, N d = 2.1 ´ 1017 cm -3

Nd > > Na

æ 4 ´ 1016 ´ 4 ´ 10 17 ö ÷÷ = 0.826 V = 0.0259 ln çç 2. 25 ´ 10 20 ø è é2 e( Vbi + VR ) Na Nd ù 2 | Emax | = ê e ( N a + N d ) úû ë

æN N ö 13. (A) Vbi = Vt ln çç a 2 d ÷÷ è ni ø æ 50 N a2 0.751 = 0.0259 ln çç 20 è 2.25 ´ 10

Electronics Devices

é2 e( Vbi + VR ) æ 1 öù 2 ÷÷ú çç W » ê e è N a øû ë 1 2

é2 ´ (117 . ´ 8.85 ´ 10 -4 )(10.752) ù -4 =ê ú = 1.8 ´ 10 cm 1.6 ´ 10 -19 ´ 4.2 ´ 1015 ë û = 1.8 mm

æ 1018 ´ 1015 = 0.0259 ln çç 20 è 2.25 ´ 10

ö ÷÷ = 0.754 V ø 1

é ù2 eeN a N d C¢ = ê ú ë2 ( Vi + VR )( N a + N d ) û 1

é ù ee N a N d 14. (D) C¢ = ê ú ë2( Vbi + VR )( N a + N d ) û é ee N a ù For N d >> N a , C¢ = ê ú ë2( Vbi + VR ) û é1.6 ´ 10 =ê ë

-19

. ´ 8.85 ´ 10 ´ 117 2 (10 + 0.754)

-4

é ù2 eeN d For N a >> N d , C¢ = ê ú ë2 ( Vbi + VR ) û

1 2

1

é1.6 ´ 10 -19 ´ 117 . ´ 8.85 ´ 10 -4 ´ 1015 ù 2 =ê ú 2 (10 + 0.754) ë û

1 2

´ 4.2 ´ 10 ù ú û 15

= 5.7 ´ 10 -9 F cm 2

1 2

= 2.77 ´ 10 -9 F cm 2 C = AC¢ = 6 ´ 10 -4 ´ 2.77 ´ 10 -9 = 1.66 ´ 10 -12 F 1 1 fo = = = 2.6 MHz. 2p LC 2 p 2. 2 ´ 10 -3 ´ 1.66 ´ 10 -12

1

ì 2 e ( V + VR ) é 1 1 ùü 2 15. (D) W = í s bi + êN úý e ë a Na û þ î 1

W A é( Vbia + V R ) ( N aA + N dA ) N aB N dB ù 2 = WB êë ( Vbib + R ) ( N aB + N dB ) N aA N dB úû æN N ö Vbi = Vt ln çç a 2 d ÷÷ è ni ø æ 1018 ´ 1015 = 0.0259 ln çç 20 è 2.25 ´ 10

ö ÷÷ = 0.754 V ø

æ 1018 ´ 1016 VbiB = 0.0259 ln çç 20 è 2.25 ´ 10

ö ÷÷ = 0.814 V ø

W A éæ 5.754 öæ 1018 + 1015 = êç ÷ç WB ëè 5.814 øçè 10 18 + 1016

öæ 1016 öù 2 . ÷÷çç 15 ÷÷ú = 313 øè 10 øû

VbiA

1

æN N ö 16. (C) Vbi = Vt ln çç a 2 d ÷÷ è ni ø Page 122

18. (B) | Emax | =

eN a xn e

é2e( Vbi + VR ) ù For a p+ n junction, xn » ê ú eN d ë û 1

é2 eN d ù2 So that | Emax | = ê ( Vbi + VR ) ú ë es û Assuming Vbi << VR , Nd =

2 eEmax (117 . ´ 8.85 ´ 10 -14 )(10 6 ) 2 = 2 eVR 2(1.6 ´ 10 -14 )(10)

= 3. 24 ´ 1017 cm -3 0 .9 æ VD ö 19. (B) I D = I s ç e hVt - 1 ÷ = 10 -10 ( e 2 ´( 0 .0259) - 1 ) = 35 mA ç ÷ è ø

æ VD ö 20. (B) I D = I s ç e hVt - 1 ÷ ç ÷ è ø

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Þ

æ I ö VD = hVt ln çç 1 + D ÷÷ Is ø è


The pn Junction

æ 70 ´ 10 -6 = (105 . )(0.0259) ln çç 1 + 10 -18 è

21. (B) I d » I s e

æ Vö ÷ çç V ÷ è tø

æI V1 - V2 = Vt ln çç d 2 è I d1

ö ÷÷ = 0.87 V ø -

V1 Vt

I d1 e = V2 = e I d2 e Vt

,

é çæç eVa ÷ö÷ ù 27. (A) N p = ALn np 0 êe è kT ø -1ú êë úû

Þ

ni2 (15 . ´ 1010 ) 2 = = 2.8 ´ 10 4 cm -3 Na 8 ´ 1015

np 0 =

( V1 - V2 ) Vt

Ln = Dn t no = 35 ´ 10 -6 = 5.9 ´ 10 -3 Cm

ö ÷÷ = 0.0259 ln 10 = 59.6 mV ø

æ VV ö 22. (A) I = I s ç e t - 1 ÷ ç ÷ è ø

Chap 2.2

é çæç 0 .5 ÷ö÷ ù N p = 10 -3 ´ 5.9 ´ 10 -3 ´ 2.8 ´ 10 4 êe è 0 .0259 ø - 1ú êë úû

ö æ I V = Vt ln çç + 1 ÷÷ I ø è s

= 4 ´ 10 7 cm -3 æ -Eg ö ç ÷ ç V ÷ t ø

æ Va ö ç ÷

I = -0.90 (–ive due to reverse current) Is

28. (A) I µ ni2 e çè Vt ÷ø µ e è

V = 0.0259 ln (1 - 0.9) = -59.6 mV

Þ

=

(10 -4 )(1.6 ´ 10 -19)(15 . ´ 10 10 ) 2 1016 Jn = 0.95, Jn + Jp

24. (D)

1 J n = en Na 2 i

Dn N Dn + a Nd Þ

Dn , t no

Dp

10 ´ 10 -3 =e 10 ´ 10 -6

t po

5 = 0.95 Na 5+ 10 Nd

ö ÷ ÷ ø

ö ÷÷ ø

t n 0 = t p 0 = 10 -6 s,

I p 0 + I n 0 = I dQ = 2 mA

Cd = Z=

Lp = Dp t p 0 = ( 8)(1 ´ 10 -8 ) = 2.83 ´ 10 -4 cm æ ö é æç 0 .61 ö÷ ùé ç- x ÷ ù dpn = 2.25 ´ 10 4 êe çè 0 .0259 ÷ø - 1ú êe çè 2 .83´10 -4 ÷ø ú ë ûë û

= 3.8 ´ 10 e

cm

2 ´ 10 ´ 10 2( 0.0259)

I dQt p 0 2 Vt

I dQ Vt

=

-6

=

= 3.86 ´ 10 -8 F

10 -3 = 3.86 ´ 10 -2 S 0.0259

10 -3 ´ 10 -7 = 193 . ´ 10 -9 F 2 ´ (0.0259)

1 1 = = 235 . - j7.5 W Y g d + jwCd

tp 0

æ 1 ö ÷÷( I pot po), Cd = çç è 2 Vt ø

-3

Þ

-4

)

=

2 Vt

= 2.5 ´ 10 -6

t p 0 = 2 ´ 0.0259 ´ 2.5 ´ 10 -6 = 1.3 ´ 10 -7 s

32. (C) RP =

x = 3 mm = 3 ´ 10 -4 cm J p = (1.6 ´ 10 -19)(18)( 3.8 ´ 1014 )( 3534) e - ( 3534 )( 3´10

-3

31. (A) For a p+ n diode I p 0 >> I n 0

¶ ( dpn ) 26. (A) J p = - eDp = eDp ( 3.8 ´ 1014 )( 3534) e -3534 x ¶x

= 0.6 A cm 2

æ E g2 - 0 .59 ö ç ÷ ç 0 .0259 ÷ è ø

30. (D) g d =

ù ú ú û

ni2 (15 . ´ 1010 ) 2 = = 2. 25 ´ 10 4 cm -3 Nd 1016

-3534 x

( 0 .0259)

æ I p 0 t p 0 + I n 0 tn 0 29. (B) Cd = çç 2 Vt è

Cd =

Na = 0.083 Nd

14

( 0 .255- 0 .32 - 0 .525+ E g2 )

E g 2 = 0.59 + 0.0259 ln 10 3 = 0.769 EV

Dp

æ é æçç eVa ö÷÷ ù é çç - Lx = pn 0 êe è kt ø - 1ú êe è p êë úû ê ë

1

ø t ( Va1 - Va2 - E g1 + E g2 ) I1 e è = æ V - E ö = e Vt a2 g2 ÷ ç I2 ç ÷ Vt ø eè

10 = e 1 J p = en Nd

= 0.95,

æ V -Eg ö ç a ÷ ç V ÷ ø t

3

2 i

25. (C) dpn = pn - pn 0

pn 0 =

12 = 394 . ´ 10 -15 A 10 -7

æ Va ö ç ÷ çV ÷ è tø

I µ eè

æ V a - E g1 ö ç ÷ ç ÷ V

D t po

1 23. (B) I s = Aen Nd 2 i

e

rp L A

=

L A( em p N a )

0.2 = 26 W (10 -2 )(1.6 ´ 10 -19)( 480)(10 16 )

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Page 123


UNIT 2

Rn = =

rn L L = A Ae(m n N d )

æ 5 ´ 1019 = 2(0.0259) ln çç . ´ 1010 è 15

0.10 = 46.3 W (10 )(1.6 ´ 10 -19)(1350)(1015)

1

R = Rp + Rn = 72.3 W

1

é2 ´ (117 . ´ 8.85 ´ 10 -14 )(114 . - 0.4) æ 2 öù 2 =ê ç -14 19 ÷ú 1.6 ´ 10 è 5 ´ 10 øû ë

æN N ö 33. (B) Vbi = Vt ln çç a 2 d ÷÷ è ni ø æ 10 = 2(0.0259) ln çç 6 è 1.8 ´ 10

= 6.19 ´ 10 -7 cm = 62 A°

ö ÷÷ =1.16 V ø

38. (B) Ln = Dn t n 0 , Lp = Dp t p 0

1

ì 2 e ( V + VR ) W = í s bi e î

Dn np 0

é 1 1 ùü 2 ê N + N úý ë a a ûþ

é2 ´ (131 . ´ 8.85 ´ 10 -14 )( 6.16) æ 2 öù -4 =ê ç 16 ÷ú = 1.34 ´ 10 cm -19 1 . 6 10 10 ´ è øû ë

=

eniW 2t o

1.6 ´ 10 -19 ´ 1.8 ´ 106 ´ 1.34 ´ 10 -4 = 193 . ´ 10 -9 A cm 2 2 ´ 10 -8

34. (A) VB = 25 =

Dn np 0 Ln

35. (D) Emax =

eN d xn e

= np 0

Lp

Dp Dn + pn 0 tn 0 tp 0

ni2 (1.8 ´ 106 ) 2 = = 324 . ´ 10 -4 cm -3 Na 1016

pn 0 =

ni2 (1.8 ´ 106 ) 2 = = 6.48 ´ 106 cm -3 Nd 5 ´ 1017

Dn = tn 0

25 = 5 ´ 10 4 , 10 ´ 10 -9

Dp

12 . ´ 10 4 = 35 10 ´ 10 -9

tp 0

=

(5 ´ 10 4 )( 3. 24 ´ 10 -4 ) (5 ´ 10 )( 3. 24 ´ 10 -4 ) + ( 35 . ´ 10 4 )( 6.48 ´ 10 -6 )

hinj =

N B = N d = 2 ´ 1016 cm -3

+

Dp pn 0

Dn tn 0

np 0 =

2 eEcrit 2 eN B

(117 . ´ 8.85 ´ 10 -4 )( 4 ´ 10 5) 2 2 ´ 1.6 ´ 10 -19 ´ N B

np 0

Ln

hinj = 1 2

J gen =

ö ÷÷ = 114 . V ø

ì 2 e ( V + VR ) é 1 1 ùü 2 W = í s bi + êN úý e ë a Na û þ î

-3

16

Electronics Devices

4

= 0.986 Þ

xn =

eEmax eN d

39. (B) The areal density at threshold is n2 D =

(117 . ´ 8.85 ´ 10 -14 )( 4 ´ 10 5) = = 5.18 ´ 10 -5 cm (1.6 ´ 10 -19)(5 ´ 1016 ) æN N ö æ 5 ´ 106 Vbi = Vt ln çç a 2 d ÷÷ = 2(0.0259) ln çç . ´ 1010 è 15 è ni ø

ö ÷÷ = 0.778 V ø

1

. ´ 10 -9) J th t r (500)(15 = = 4.69 ´ 1012 cm -3 1.6 ´ 10 -19 e

The carrier density is nth =

n2 D 4.69 ´ 1012 = = 4.69 ´ 1018 cm -3 dLas 10 -6

ì 2 e V æ N öæ öü 2 1 ÷ý xn = í s bi çç a ÷÷çç ÷ î e è N d øè N a + N d øþ

Once the threshold is reached, the carrier density does

é2(117 . ´ 8.85 ´ 10 -4 )( Vbi + VR ) ù (5.18 ´ 10 ) = ê -19 6 ú ë (1.6 ´ 10 )(2 ´ 5 ´ 10 ) û

15 . ´ 10 -9 J th = 3 ´ 10 -10 s t r ( J th ) = 5 J JA The optical power produced is p = hw e

not

Vbi + VR = 20.7,

VR = 19.9 V,

xn

J > J th

the

VR = VB

1

é2 eVB ù 2 é2(117 . ´ 8.85 ´ 10 -14 )(100) ù 2 » ê = -19 15 ú ê ú = 5.1 mm ë eN d û ë (1.6 ´ 10 )(5 ´ 10 ) û

=

(5 ´ 500)(2 ´ 10 -5)(1.43 ´ 1.6 ´ 10 -19) = 715 . MW 1.6 ´ 10 -19

****************

æN N ö 37. (D) Vbi = Vt ln çç a 2 d ÷÷ è ni ø Page 124

electron

tr ( J ) =

36. (A) For a p+ n diode, Neglecting Vi compared to VB , 1

When

recombination is

-5 2

Þ

change.

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hole


CHAPTER

2.3 THE BIPOLAR JUNCTION TRANSISTOR


5. A uniformly doped npn bipolar transistor is biased in

The parameters in the base region of an npn bipolar transistor are as follows Dn = 20 cm 2 s, nB0 = 10 4 cm -3, xB = 1 mm, ABE = 10 -4 cm 2 .

the

forward-active

region.

The

transistor

doping

concentration are N E = 5 ´ 1017 cm -3, N B = 10 16 cm -3 and N C = 1015 cm -3. The minority carrier concentration pE0 , nB0 and pC0 are

1. If VBE = 0.5 V, then collector current I C is

(A) 4.5 ´ 10 2 , 2. 25 ´ 10 4 , 2. 25 ´ 10 5 cm -3

(A) 7.75 mA

(B) 1.6 mA

(B) 2. 25 ´ 10 4 , 2. 25 ´ 10 5, 4.5 ´ 10 2 cm -3

(C) 0.16 mA

(D) 77.5 mA

(C) 2. 25 ´ 10 4 , 2. 25 ´ 10 5, 4.5 ´ 10 4 cm -3 (D) 4.5 ´ 10 4 , 2.25 ´ 10 4 , 2. 25 ´ 10 5 cm -3

2. If VBE = 0.7 V, then collector current I C is (A) 418 mA

(B) 210 mA

(C) 17.5 mA

(D) 98 mA

6. A uniformly doped silicon pnp transistor is biased in the forward-active mode. The doping profile is N E = 10 18 cm -3, N B = 5 ´ 1016 cm -3 and N C = 1015 cm -3. For VEB = 0.6 V, the pB at x = 0 is

(See fig. P2.3.7-8)

3. In bipolar transistor biased in the forward-active

(A) 5.2 ´ 10 cm

-3

(B) 5.2 ´ 10 13 cm -3

region the base current is I B = 50 mA and the collector

(C) 5.2 ´ 1016 cm -3

(D) 5.2 ´ 10 11 cm -3

currents is I C = 2.7 mA. The a is (A) 0.949

(B) 54

(C) 0.982

(D) 0.018

19

Statement for Q.7-8: An npn bipolar transistor having uniform doping of N E = 10 18 cm -3 N B = 1016 cm -3 and N C = 6 ´ 10 15 cm -3

4. A uniformly doped silicon npn bipolar transistor is to be biased in the forward active mode with the B-C junction reverse biased by 3 V. The transistor doping

is operating in the inverse-active mode with VBE = - 2 V and VBC = 0.6 V. The geometry of transistor is shown in fig P2.3.7-8. Emitter -n-

are N E = 1017 cm -3, N B = 1016 cm -3 and N C = 10 15 cm -3.

Base -p-

Collector -n-

The BE voltage, at which the minority carrier electron concentration at x = 0 is 10% of the majority carrier hole

xE

xB

xC

concentration, is (A) 0.94 V (C) 0.48 V

(B) 0.64 V

x' = xE

(D) 0.24 V

x'=0 x'

x=0 x

x = xB x'' = 0

x'' = xC x''

Fig. P2.3.7-8

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Page 125


UNIT 2

Electronics Devices

7. The minority carrier concentration at x = x B is

N E = 1018 cm -3, N B = 5 ´ 1016 cm -3,

(A) 4.5 ´ 1014 cm -3

(B) 2.6 ´ 10 12 cm -3

N C = 2 ´ 1019 cm -3,

(C) 2.6 ´ 1014 cm -3

(D) 39 . ´ 10 14 cm -3

DE = 8 cm 2 s , DB = 15 cm 2 s , DC = 14 cm 2 s xE = 0.8 mm, xB = 0.7 mm

8. The minority carrier concentration at x¢¢ = 0 is (A) 39 . ´ 1014 cm -3 (C) 2.7 ´ 10

14

cm

(B) 2.7 ´ 10 12 cm -3

-3

(D) 4.5 ´ 10

14

cm

The emitter injection efficiency g is

-3

(A) 0.999

(B) 0.977

(C) 0.982

(D) 0.934

9. An pnp bipolar transistor has uniform doping of N E = 6 ´ 1017 cm -3, N B = 2 ´ 10 16 cm -3 and N C = 5 ´ 1014 -3

15. A uniformly doped silicon epitaxial npn bipolar

cm . The transistor is operating is inverse-active mode.

transistor

The maximum V CB voltage, so that the low injection

N B = 3 ´ 10

condition applies, is

with N C = 5 ´ 10

fabricated

cm

-3

(A) 0.86 V

(B) 0.48 V

xB = 0.7 mm

(C) 0.32 V

(D) 0.60 V

punch-through is

Statement for Q.10-12: The

following

currents

are

measured

in

with

a

base

doping

of

and a heavily doped collector region cm -3. The neutral base width is

17

when

VBE = VBC = 0.

The

(A) 26.3 V

(B) 18.3 V

(C) 12.2 V

(D) 6.3 V

VBC

at

a 16. A silicon npn transistor has a doping concentration

uniformly doped npn bipolar transistor:

of

I nE = 120 . mA, I pE = 0.10 mA, I nC = 118 . mA

N B = 1017

cm -3

and

N C = 7 ´ 10 15

cm -3.

The

metallurgical base width is 0.5 mm. Let VBE = 0.6 V.

I R = 0.20 mA, I G = 1 mA, I pC0 = 1 mA

Neglecting the B–E junction depletion width the VCE at punch-through is

10. The a is (A) 0.667

(B) 0.733

(C) 0.787

(D) 0.8

(A) 146 V

(B) 70 V

(C) 295 V

(D) 204 V

17. A uniformly doped silicon pnp transistor is to

11. The b is (A) 3.69

(B) 0.44

(C) 2.27

(D) 8.39

designed with N E = 1019 cm -3 and N C = 10 16 cm -3. The metallurgical base width is to be 0.75 mm. The minimum

base

doping,

so

that

the

minimum

punch-through voltage is Vpt = 25 V, is

12. The g is (A) 0.816

(B) 0.923

(C) 1.083

(D) 0.440

(A) 4.46 ´ 1015 cm -3

(B) 4.46 ´ 10 16 cm -3

(C) 195 . ´ 1015 cm -3

(D) 195 . ´ 1016 cm -3

13. A silicon npn bipolar transistor has doping

18. For a silicon npn transistor assume the following

concentration of N E = 2 ´ 1018 cm -3, N B = 10 17 cm -3 and

parameters:

. ´ 1016 cm -3. The area is 10 -3 cm 2 and neutral N C = 15

I E = 0.5 mA, b = 48

base width is 1 mm. The transistor is biased in the active

xB = 0.7 mA, xdc = 2 mm

region at VBE = 0.5 V. The collector current is

Cs = Cm = 0.08 pF, C je = 0.8 pF

(DB = 20 cm 2 s)

Page 126

is 16

(A) 9 mA

(B) 17 mA

(C) 22 mA

(D) 11 mA

Dn = 25 cm 2 s, rc = 30 W The carrier cross the space charge region at a speed of 10 7 cm s. The total delay time t ec is

14. A uniformly doped npn bipolar transistor has

(A) 164.2 ps

(B) 234.4 ps

following parameters:

(C) 144.2 ps

(D) 298.4 ps

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The Bipolar Junction Transistor

19. In a bipolar transistor, the base transit time is 25%


of the total delay time. The base width is 0.5 mm and base diffusion coefficient is DB = 20 cm 2 s. The cut-off

Chap 2.3

For the transistor in circuit of fig. P2.3.24-26. The parameters are bR = 1 , bF = 100 , and I s = 1 fA .

frequency is (A) 637 MHz

(B) 436 MHz

(C) 12.8 GHz

(D) 46.3 GHz

5V

20. The base transit time of a bipolar transistor is 100 ps and carriers cross the 1.2 mm B–C space charge at a speed of 10 7 cm s . The emitter-base junction charging time is 25 ps and the collector capacitance and resistance are 0.10 pF and 10 W, respectively. The cutoff frequency is (A) 43.8 GHz

(B) 32.6 GHz

(C) 3.26 GHz

(D) 1.15 GHz

Fig. P2.3.24-26

24. The current I C is (A) 1 fA (C) 1.384 fA

(B) 2 fA (D) 0 A

25. The current I E is

Statement for Q.21-22: Consider the circuit shown in fig. P2.3.21-22. If voltage Vs = 0.63 V, the currents are I C = 275 mA and I B = 5 mA .

(A) 1 fA

(B) -1 fA

(C) 2 fA

(D) -2 fA

26. The current I B is (A) 2 fA

(B) -2 fA

(C) 1 fA

(D) -1 fA

27. For the transistor in fig. P2.3.27, I S = 10 -15 A, bF = 100 , bR = 1. The current I CBO is Vs 5V

Fig.P2.3.21-22

21. The forward common-emitter gain bF is (A) 56

(B) 55

(C) 0.9821

(D) 0.9818

Fig.P2.3.27

(A) 101 . ´ 10 -14 A (C) 101 . ´ 10 -15 A

22. The forward current gain a F is (A) 0.9821

(B) 0.9818

(C) 55

(D) 56

(B) 2 ´ 10 -14 A (D) 2 ´ 10 -15 A

Statement for Q.28-31: Determine

23. Consider the circuit shown in fig P2.3.23. If Vs = 0.63 V, I1 = 275 mA and I 2 = 125 mA, then the value of I 3 is

the

region

of

operation

for

the

transistor shown in circuit in question. 28.

I1

6V I2

I3

Vs

Fig. P2.3.23

(A) - 400 mA

(B) 400 mA

(C) - 600 mA

(D) 600 mA

Fig.P2.3.28

(A) Forward-Active (C) Saturation www.gatehelp.com

(B) Reverse-Active (D) Cutoff Page 127


UNIT 2

29.

Electronics Devices

33. The current I1 is (A) -12.75 mA

(B) 12.75 mA

(C) 12.5 mA

(D) - 12.5 mA


6V

The leakage current of a transistor are I CBO = 5 m A and I CEO = 0.4 mA, and I B = 30 mA.

Fig. P2.3.29

(A) Forward-Active

(B) Reverse-Active

(C) Saturation

(D) Cutoff

34. The value of b is

30.

(A) 79

(B) 81

(C) 80


35. The value of I C is (A) 2.4 mA

(B) 2.77 mA

(C) 2.34 mA

(D) 1.97 mA

6V

Statement for Q.36–37: Fig.P2.3.30

(A) Forward-Active

(B) Reverse-Active

(C) Saturation

(D) Cutoff

For a BJT, I C = 5 mA, I B = 50 mA and I CBO = 0.5 mA. 36. The value of b is

31. 3V

(A) 103

(B) 91

(C) 83

(D) 51

37. The value of I E is

6V

(A) 5.25 mA

(B) 5.4 mA

(C) 5.65 mA

(D) 5.1 mA

Fig.P2.3.31

(A) Forward-Active

(B) Reverse-Active

(C) Saturation

(D) Cutoff

********

Statement for Q.32-33: For the circuit shown in fig. P2.3.32-33, let the value of b R = 0.5 and bF = 50. The saturation current is 10 -16 A.

+3 V

250 mA

I1

Fig. P2.3.32-33

32. The base-emitter voltage is

Page 128

(A) 0.53 V

(B) 0.7 V

(C) 0.84 V

(D) 0.98 V www.gatehelp.com



SOLUTIONS 1. (A) I C = I s e

æ çç

ö ÷÷

0 .5

2. (C) I C = 32 . ´ 10

-14

e

æ 0 .7 ö çç ÷÷ è 0 .0259 ø

= 17.5 mA

pC (0) = 0.10 N C = 5 ´ 1013 cm -3, ni2 (15 . ´ 1010 ) 2 = = 4.5 ´ 10 5 NC 5 ´ 1014 ö æ ç VCB ÷ ç V ÷ t ø

pC (0) = pC 0 e è

10. (C) a =

ni2 (15 . ´ 1010 ) 2 = = 2. 25 ´ 10 4 cm -3 NB 1016 æ VBE ö ÷ ç ç V ÷ t ø

At x = 0, np (0) = np 0 e è

VBE

æ np (0) ö ÷ VBE = VT ln ç ç n ÷ è p0 ø

=

æ 1015 = 0.0259 ln çç 4 è 2.25 ´ 10

ö ÷÷ = 0.635 V ø

ni2 (15 . ´ 10 10 ) 2 = = 450 cm -3 NE 5 ´ 1017

nB 0 =

ni2 (15 . ´ 10 10 ) 2 = = 2. 25 ´ 10 4 cm -3 NB 5 ´ 1016

pC 0 =

ni2 (15 . ´ 10 10 ) 2 = = 2. 25 ´ 10 5 cm -3 NE 5 ´ 1015

6. (B) pB 0 = pB (0) = pB 0 e 7. (C) nB 0 =

ni2 (15 . ´ 1010 ) 2 = = 4.5 ´ 10 3 cm -3 NB 5 ´ 1016

æ VEB ç ç V è t

ö ÷ ÷ ø

æ 0 .6 ö ÷÷ çç 3 è 0 .0259 ø

= 4.5 ´ 10 e

= 5. 2 ´ 1013 cm -3

ni2 (15 . ´ 10 10 ) 2 = = 2. 25 ´ 10 4 cm -3 NB 1016

nB ( x = xB ) = nB 0 e

Þ

æ p (0) ö VCB = Vt ln çç C ÷÷ è pC 0 ø

ö æ ç VBC ÷ ç V ÷ è t ø

0 .6

= 2. 25 ´ 10 4 e 0 .0259 = 2.6 ´ 10 14 cm -3

J nE

J nC I nC = + J R + J pE I nE + I R + I pE

118 . = 0.787 12 . + 0.2 + 0.1

11. (A) b =

a 0.787 == = 3.69 1-a 1 - 0.787

12. (B) g =

J nE I nE 1. 2 = = = 0.923 J nE + J pE I nE + I pE 1. 2 + 0.1

10 1016 ´ NB = = 10 15 100 10

5. (A) pE 0 =

ö ÷÷

æ 5 ´ 1013 ö ÷ = 0.48 V = 0.0259 ln çç 5÷ è 4.5 ´ 10 ø

IC 2.7m = = 0.982 I C + I B 2.7m + 50m

np (0) =

0 .6

9. (B) Low injection limit is reached when

pC 0 =

I bF 3. (C) bF = C , a F = IB 1 + bF

Þ

æ VBC ö ÷ ç ç V ÷ t ø

æ çç

I C = 3. 2 ´ 10 -14 e è 0 .0259 ø = 7.75 mA

4. (B) np 0 =

ni2 (15 . ´ 1010 ) 2 = = 375 . ´ 10 4 cm -3 NC 6 ´ 1015

= 375 . ´ 10 4 e è 0 .0259 ø = 4.31 ´ 1014 cm -3

(1.6 ´ 10 -19)(20)(10 -4 )(10 4 ) = 32 . ´ 10 -14 A 10 -4

aF =

8. (D) pC 0 =

pC ( x¢¢ = 0) = pC 0 e è

æ VBE ö ÷ ç ç V ÷ è b ø

eDn ABE nB 0 Is = xB =

Chap 2.3

13. (B) nB 0 =

ni2 (15 . ´ 10 10 ) 2 = = 2.25 ´ 10 3 cm -3 NB 1017

æ VBE ö ÷ ç ç V ÷ t ø

nB (0) = nB 0 e è IC = =

0 .5

ö ÷÷

eDB AnB (0) xB

(1.6 ´ 10 -19)(20)(10 -3)(5.45 ´ 10 11 ) = 17.4 mA 10 -4

14. (B) g =

=

æ çç

= 2.25 ´ 10 3 e è 0 .0259 ø = 5.45 ´ 1011 cm -3

1 N B D E xB 1+ × × N E D B xE

1 = 0.977 5 ´ 1016 8 0.7 1+ 1018 15 0.8

æN N ö 15. (B) Vbi = Vt ln çç B 2 C ÷÷ è ni ø æ 3 ´ 1016 ´ 5 ´ 1017 = 0.0259 ln çç . ´ 1010 ) 2 è (15

ö ÷÷ = 0.824 V ø

At punch-through

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Page 129



ö æ æ VBE ö ç VBC ÷ ö æ çæç VBE ÷ö÷ ö ÷ ç I æ çç ÷÷ 27. (C) I E = I S ç e è Vt ø - e è Vt ø ÷ + S ç e è Vt ø - 1 ÷ = 0 ç ÷ bF ç ÷ è ø è ø

Þ

ö æ ç VBE ÷ ç V ÷ t ø

eè

=

33. (A) I E = (bF + 1) I B = 12.75 mA I1 = - I E = - 12.75 mA.

æ VBC ö ÷ Vt ÷ø

ç ç 1 bR + eè 1 + bF 1 + BF

34. (A) I CEO = (b + 1) I CBO b+1=

ö æ æ VBC ö ç VBC ÷ ù é æçç VBE ö÷÷ ù ç V ÷ I é çç ÷÷ I C = I S êe è Vt ø - e è t ø ú - S êe è Vt ø - 1ú ê ú bR ê úû ë ë û

I CBO =

IS é ê1 - e 1 + bF ê ë

æ VBC ö ÷ ç ç V ÷ è t ø

ù I é ú - S êe úû bR êë

æ VBC ç ç V è t

ö ÷ ÷ ø

0.4m = 80 5m

Þ

b = 79

35. (B) I C = bI B + I CEO = 79( 30m ) + 0.4m =2.77 mA

ù - 1ú úû

36. (A) I C = bI B + I CEO = b I B + (b + 1) I CBO b=

VBC = - 5 V, Vt = 0.0259 V I CBO =

Chap 2.3

Is I (1 - 0) - S (0 - 1) = 101 . I S = 101 . ´ 10 -15 A 101 1

I C - I CBO 5.2m - 0.5m = » 10396 . I B + I CBO 50m + 0.5m

37. (A) a =

28. (D)

IE =

b = 0.9904 b+1

I C - I CBO 5.2m - 0.5m = = 5.25 MA a 0.9904

B-C Junction VBC

B-E junction VBE

Reverse Bias

Forward bias

Forward bias

Forward-Active

Saturation

Reverse Bias

Cut-off

Reverse-Active

*******

VBE = 0 , VBC < 0, Thus both junction are in reverse bias. Hence cutoff region. 29.(A) VBE > 0 , VBC = 0, Base-Emitter junction forward bais,

Base-collector

junction

reverse

bias,

Hence

forward-active region. 30. (B) VBE = 0 , VBC > 0, Base-Emitter junction reverse bais , Base-collector junction forward bias, Hence reverse-active region. 31. (C) VBE = 6 V, VBC = 3 V, Both junction are forward biase, Hence saturation region. 32. (C) The current source will forward bias the base-emitter junction and the collector base junction will then be reverse biased. Therefore the transistor is in the forward active region IC = ISe

æ VBE ö ÷ ç ç V ÷ è t ø

I C = bF I B = 50 ´ 250 ´ 10 -6 = 12.5 ´ 10 -3 A æ 12.5 ´ 10 -3 ö æI ö VBE = Vt ln çç C ÷÷ = 0.0259 ln çç -16 ÷÷ = 0.84 V ø è 10 è IS ø

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Page 131


UNIT 2

electronics Devices

11. In n-well CMOS fabrication substrate is

18. Monolithic integrated circuit system offer greater

(A) lightly doped n - type

reliability than discrete-component systems because

(B) lightly doped p - type

(A) there are fewer interconnections

(C) heavily doped n - type

(B) high-temperature metalizing is used

(D) heavily doped p - type

(C) electric voltage are low (D) electric elements are closely matched

12. The chemical reaction involved in epitaxial growth in IC chips takes place at a temperature of about (A) 500° C

(B) 800° C

(C) 1200° C

(D) 2000° C

19. Silicon dioxide is used in integrated circuits (A) because of its high heat conduction (B) because it facilitates the penetration of diffusants

13. A single monolithic IC chip occupies area of about

(C) to control the location of diffusion and to protect and insulate the silicon surface.

(A) 20 mm 2

(B) 200 mm 2

(D) to control the concentration of diffusants.

(C) 2000 mm 2

(D) 20,000 mm 2 s

20. Increasing the yield of an IC

14. Silicon dioxide layer is used in IC chips for

(A) reduces individual circuit cost

(A) providing mechanical strength to the chip

(B) increases the cost of each good circuit (C) results in a lower number of good chips per wafer

(B) diffusing elements

(D) means that more transistor can be fabricated on the same size wafer.

(C) providing contacts (D) providing mask against diffusion

21. The main purpose of the metalization process is 15. The p-type substrate in a monolithic circuit should be connected to

(A) to act as a heat sink (B) to interconnect the various circuit elements

(A) any dc ground point

(C) to protect the chip from oxidation

(B) the most negative voltage available in the circuit

(D) to supply a bonding surface for mounting the chip

(C) the most positive voltage available in the circuit 22. In a monolithic-type IC

(D) no where, i.e. be floating 16. The collector-substrate junction in the epitaxial

(A) each transistor is diffused into a separate isolation region (B) all components are fabricated into a single crystal of silicon

collector structure is, approximately (A) a step-graded junction

(C) resistors and capacitors of any value may be made

(B) a linearly graded junction (C) an exponential junction

(D) all isolation problems are eliminated

(D) None of the above 23. Isolation in ICs is required 17. The sheet resistance of a semiconductor is

(A) to make it simpler to test circuits

(A) an important characteristic of a diffused region especially when used to form diffused resistors

(B) to protect the transistor from possible ``thermal run away’’

(B) an undesirable parasitic element

(C) to protect the components mechanical damage

(C) a characteristic whose value determines the required area for a given value of integrated capacitance

(D) to minimize electrical interaction between circuit components

(D) a parameter whose value is important in a thin-film resistance Page 140

24. Almost all resistor are made in a monolithic IC (A) during the base diffusion

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Integrated Circuits

Chapc2.5

(B) during the collector diffusion

30. For the circuit shown in fig. P2.5.30, the minimum

(C) during the emitter diffusion

number and the maximum number of isolation regions

(D) while growing the epitaxial layer

are respectively

25. The equation governing the diffusion of neutral

(A)

¶N ¶2 N =D ¶t ¶x 2

(C)

¶ N ¶N =D 2 ¶t ¶x

R1

R2

atom is (B)

¶N ¶2 N =D ¶x ¶t 2

(D)

¶ N ¶N =D 2 ¶x ¶t

2

Vo2

Q1 Q2

2

Fig. P2.5.32

26. The true statement is (A) thick film components are vacuum deposited (B) thin film component are made by screen-and- fire process

(A) 2, 6

(B) 3, 6

(C) 2, 4

(D) 3, 4

31. For the circuit shown in fig. P2.5.31, the minimum number of isolation regions are

(C) thin film resistor have greater precision and are more stable (D) thin film resistor are cheaper than the thin film resistor 27. The False statement is (A) Capacitor of thin film capacitor made with proper dielectric is not voltage dependent (B) Thin film resistors and capacitor need to be biased for isolation purpose (C) Thin film resistors and capacitor have smaller stray capacitances and leakage currents.

Fig. P2.5.31

(A) 2

(B) 3

(C) 4

(D) 7


*******

28. Consider the following two statements S1 : The dielectric isolation method is superior to junction isolation method. S2 : The beam lead isolation method is inferior to junction isolation method. The true statements is (are) (A) S1 , S2

(B) only

(C) only

(D) Neither nor S2

29. If P is passivation, Q is n-well implant, R is metallization and S is source/drain diffusion, then the order in which they are carried out in a standard n-well CMOS fabrication process is (A) S - R - Q - P

(B) R - P - S - Q

(C) Q - S - R - P

(D) P - Q - R - S

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Page 141


UNIT 2

SOLUTIONS 1. (D)

2. (D)

3. (B)

4. (B)

5. (C)

6. (B)

7. (C)

8. (B)

9. (C)

10. (D)

11. (B)

12. (C)

13. (C)

14. (D)

15. (B)

16. (A)

17. (A)

18. (A)

19. (C)

20. (A)

21. (B)

22. (B)

23. (D)

24. (A)

25. (A)

26. (C)

27. (B)

28. (B)

29. (C)

30. (D) The minimum number of isolation region is 3 one containing Q1 , one containing and one containing both and . The maximum number of isolation region is 4, or one per component. 31. (A) The minimum number of isolation region is two. One for transistor and one for resistor.

*******

Page 142

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electronics Devices


CHAPTER

3.1 DIODE CIRCUITS

vo


vo 22

18

In the question a circuit and a waveform for the t

input voltage is given. The diode in circuit has cutin (A)

voltage Vg = 0. Choose the option for the waveform of

(B)

vo

output voltage vo .

vo t

-3

(C)

vi

+ 20

2.2 kW

3.

vo 5V

(D)

2 kW

t

_

t -7

1.

vi

t

vi

+

16

-5

vo

vi

Fig.3.1.1

2 T

4V 20

_

15

t

t

-5

-10

(A)

(B)

Fig. P3.1.3

vo

vo

16

4

4

t

T

T 2

vo

20

(A)

20

(A) 2.

(D)

10 kW

T

T 2

-4

(D)

R

vi + D1

t -5

t

4.

vi

vo _

T

4

t

(C) 20

vi

T 2

t

16

12

2V +

T

vo

5

t

T 2

(B)

vo t

t

-16

12

vo

T

6V vo

vi 8V

Fig.3.1.2

10

D2

_

t -10

Fig.3.1.4

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Page 145


UNIT 3 vo

vo

Analog Electronics

vo

vo

10

8

9.42

9.42 5.7 t

4.3

t

5.7

-6

15

vi

(C)

-10

(A)

vi

(D)

7. For the circuit shown in fig. P3.1.7, each diode has

(B)

vo

15

4.3

Vg = 0.7 V. The vo for -10 £ vs +10 V is

vo

+10 V

8

6

t

10 kW

t

-6

D1

-8

(C)

D2

vs

(D)

vo

5. For the circuit in fig. P.3.1.5, let cutin voltage Vg = 0.7 V. The plot of vo verses vi for -10 £ vi £ 10 V is

10 kW

D4

D3

10 kW

+ 10 kW

20 kW

-10 V

vo

vi

Fig. P3.1.7.

10 V

10 V

vo

_

vo

8.43

7.48

Fig. P3.1.5 vo

vo

9.3

-10

10

9.3

3.33 3.33

-10

10

4.03

10

(B)

vo

vi

vo

5.68

vo 10

10

4.65

-10 -6.8

6.8 10

vs

-10

-5.68

4.65 -4.65

(C) 3.33

10

vi

4.33

-10

(C)

10

vi

(D)

clipping circuit shown in fig. P3.1.8. If diode has rf = 0

(D)

and rr = 2 MW and Vg = 0, the output waveform is

is Vg = 0.7 V. The plot of vo versus vi is

+ 1 MW

+

vo

vi

1 kW

2.5 V

vo

1 kW

_

vi

15 V

_

Fig. P3.1.8 vo

Fig. P3.1.6 vo

vo

10

vo 19.6

19.6

t

-5

5

5.7

(A) Page 146

15

(B) vo

vo

4.3 vi

15

4.3

5

vi

(B)

t

-5

(C) www.gatehelp.com

t

-10

(A) 5.7

vs

varies between +10 V and -10 V is impressed upon the

6. For the circuit in fig. P3.1.6 the cutin voltage of diode

2 kW

10

4.65

8. A symmetrical 5 kHz square wave whose output

3.33

3.33 -10

vs

-7.48

(B)

vo

10

(A)

-10

(A)

-10

-8.43

3.33 vi

vs

t

(D)


Diode Circuits

Chap 3.1

C

9. In the circuit of fig. P3.1.9, the three signals of fig are

vi

+

10

impressed on the input terminals. If diode are ideal then the voltage vo is

vi

D1

v D2

+

D3

+ v1

v2

+

vo

v3

10 kW

-

-

t

vo

R 5V

_

v3 v2

-20

Fig. P.3.1.11

v1

vo

t

35

vo

25

-

Fig. P.3.1.9

vo

5 t

vo

t

-5

(A)

(B)

vo t

t

(A)

15

(B)

vo


t

vo -15

(C) t

t

(D)

12. In the circuit of fig. P3.1.12, D1 and D2 are ideal diodes. The current i1 and i2 are

(C)

(D) D1

10. For the circuit shown in fig. P3.1.10 the input

i1

voltage vi is as shown in fig. Assume the RC time

D2

i2

500 W

constant large and cutin voltage of diode Vg = 0. The

3V

5V

output voltage vo is

5V

C vi

+

Fig. P3.1.12

10 vi

t

vo

R

-10 _

Fig. P.3.1.10

vo

vo

(A) 0, 4 mA

(B) 4 mA, 0

(C) 0, 8 mA

(D) 8 mA, 0

13. In the circuit of Fig. P3.1.13 diodes has cutin

20

voltage of 0.6 V. The diode in ON state are

10 t

t

(A)

D1

(B)

vo

12 W

6W

D2

vo t

5.4 V

t

18 W

5V

-10 -20

(C)

Fig. P3.1.13

(D)

11. For the circuit shown in fig. P.3.1.11, the input

(A) only D1

(B) only D2

voltage vi is as shown in fig. Assume the RC time

(C) both D1 and D2


constant large and cutin voltage Vg = 0. The output voltage vo is www.gatehelp.com

Page 147


Diode Circuits

Chap 3.1

22. The diodes in the circuit in fig. P3.1.22 has

26. If v2 = 0, then output voltage vo is

parameters Vg = 0.6 V and rf = 0. The current iD2 is

(A) 6.43 V

(B) 9.43 V

(C) 7.69 V

(D) 8.93 V

+10 V 9.5 kW

27. If v2 = 5 V, then vo is

D2

0.5 kW 0V

iD2

vo

+5 V D1

0.5 kW

D3

Fig. P3.1.22

(B) 10 mA

(C) 7.6 mA

(D) 0 mA

(B) 12.63 V

(C) 18.24 V

(D) 10.56 V

28. If v2 = 10 V, then vo is

+5 V

(A) 8.4 mA

(A) 8.93 V

(A) 10 V

(B) 9.16 V

(C) 8.43 V

(D) 12.13 V

Statement for Q.29–30: The diode in the circuit of fig. P3.1.29–30 has the


non linear terminal characteristic as shown in fig. Let

The diodes in the circuit in fig. P3.1.23-25 have linear parameter of Vg = 0.6 V and rf = 0.

the voltage be vs = cos wt V. 100 W vi

9.5 kW 0.5 kW

D2

iD 4

+ vD

100 W

-

2V

vo

v2

~

iD(mA)

a

+10 V

b

v1

vD(V)

Fig. P3.1.29–30

D1

0.5 kW

0.5 0.7

Fig. P3.1.23–25

29. The current iD is

23. If v1 = 10 V and v2 = 0 V, then vo is (A) 8.93 V

(B) 7.82 V

(C) 1.07 V

(D) 2.18 V

(A) 2.5(1 + cos wt) mA

(B) 5(0.5 + cos wt) mA

(C) 5(1 + cos wt) mA

(D) 5(1 + 0.5 cos wt) mA

30. The voltage vD is

24. If v1 = 10 V and v2 = 5 V, then vo is (A) 9.13 V

(B) 0.842 V

(A) 0.25( 3 + cos wt) V

(C) 5.82 V

(D) 1.07 V

(C) 0.5( 3 + 1 cos wt) V

(B) 0.25(1 + 3 cos wt) V (D) 0.5(2 + 3 cos wt) V

25. If v1 = v2 = 0, then output voltage vo is

31. The circuit inside the box in fig. P3.1.31. contains

(A) 0.964 V

(B) 1.07 V

only resistor and diodes. The terminal voltage vo is

(C) 10 V

(D) 0.842 V

connected to some point in the circuit inside the box. The largest and smallest possible value of vo most


nearly to is respectively +15 V

The diodes in the circuit of fig. P3.1.26–28 have linear parameters of Vg = 0.6 V and rf = 0. 500 W

D2

-9 V

v2

Circuit Containing Diode and Resistor

vo

vo

+10 V 500 W

D1

Fig. 3.1.31

9.5 kW

(A) 15 V, 6 V

(B) 24 V, 0 V

(C) 24 V, 6 V

(D) 15 V, -9 V

Fig. P3.1.26–28

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Page 149


UNIT 3

32. In the voltage regulator circuit in fig. P3.1.32 the


maximum load current iL that can be drawn is

In the voltage regulator circuit in fig. P3.1.36–38

15 kW

the Zener diode current is to be limited to the range iL

30 V

Analog Electronics

Vz = 9 V Rz = 0

5 £ iz £ 100 mA. 12 W

RL

iL

iz

Fig. 3.1.32

(A) 1.4 mA

(B) 2.3 mA

(C) 1.8 mA

(D) 2.5 mA

power dissipation in the Zener diode is 150 W

Vz = 15 V Rz = 0

Fig. P3.1.33

(B) 1.5 W

(C) 2 W

(D) 0.5 W

36. The range of possible load current is (A) 5 £ iL £ 130 mA

(B) 25 £ iL £ 120 mA

(C) 10 £ iL £ 110 mA


37. The range of possible load resistance is

75 W

(A) 1 W

RL

Fig. P3.1.36–38

33. In the voltage regulator shown in fig. P3.1.33 the

50 V

Vz = 4.8 V Rz = 0

6.3 V

(A) 60 £ RL £ 372 W

(B) 60 £ RL £ 200 W

(C) 40 £ RL £ 192 W

(D) 40 £ RL £ 360 W

38. The power rating required for the load resistor is

34. The Q-point for the Zener diode in fig. P3.1.34 is

(A) 576 mW

(B) 360 mW

(C) 480 mW

(D) 75 mW

11 kW

39. The secondary transformer voltage of the rectifier 20 V

Vz = 4 V Rz = 0

circuit shown in fig. P3.1.39 is vs = 60 sin 2 p60 t V. Each

3.6 kW

diode has a cut in voltage of Vg = 0.6 V. The ripple voltage is to be no more than Vrip = 2 V. The value of filter capacitor will be

Fig. P3.1.34

(A) (0.34 mA, 4 V)

(B) (0.34 mA, 4.93 V)

(C) (0.94 mA, 4 V)

(D) (0.94 mA, 4.93 V)

+

vi

35. In the voltage regulator circuit in fig. P3.1.35 the

+ vo 10 kW

+ vs -

power rating of Zener diode is 400 mW. The value of RL

C -

that will establish maximum power in Zener diode is

Fig. P3.1.39

222 W

20 V

Vz = 10 V Rz = 0

RL

(A) 48.8 mF

(B) 24.4 mF

(C) 32.2 mF

(D) 16.1 mF

40. The input to full-wave rectifier in fig. P3.1.40 is Fig. P3.1.35

Page 150

(A) 5 kW

(B) 2 kW

(C) 10 kW

(D) 8 kW

vi = 120 sin 2 p60 t V. The diode cutin voltage is 0.7 V. If the output voltage cannot drop below 100 V, the required value of the capacitor is www.gatehelp.com


Diode Circuits

+

1 : 1

vi

SOLUTIONS

+ + vs + vs -

2.5 kW

C

vo -

1. (D) Diode is off for vi < 5 V. Hence vo = 5 V. For vi > 5 V, vo = vi , Therefore (D) is correct option.

-

2. (C) Diode will be off if vi + 2 > 0.Thus vo = 0

Fig. P3.1.40

(A) 61.2 mF

(B) 41.2 mF

(C) 20.6 mF

(D) 30.6 mF

For vi + 2 < 0 V,

voltage is Vin = 0. The ripple voltage is to be no more than vrip = 4 V. The minimum load resistance, that can be connected to the output is +

~

50 mF

vo

vi < - 2, vo = vi + 2 = -3 V

Thus (C) is correct option.

41. For the circuit shown in fig. P3.1.41 diode cutin

75sin 2p60t V

Chap 3.1

3. (D) For vi < 4 V the diode is ON and output vo = 4 V. For vi > 4 V diode is off and output vo = vi . Thus (D) is correct option. 4. (C) During positive cycle when vs < 8 V, both diode are OFF vo = vs . For vs > 8 V , vo = 8 V, D1 is ON. During

RL

negative cycle when |vs | < 6 V, both diode are OFF,

-

vo = vs . For vs > 6 V, D2 is on vo = -6 V. Therefore (C) is

Fig. P3.1.41

correct.

(A) 6.25 kW

(B) 12.50 kW

(C) 30 kW


10 10 5. (B) For D off , vo = 20 20 = 3.33 V. 1 1 + 20 10 For vi £ 3.33 + 0.7 = 4.03 V, vo = 3.33 V

****************

For vi > 4.03 V , vo = vi - 0.7 For vi = 10 V, vo = 9.3 V 6. (C) Let v1 be the voltage at n-terminal of diode, v1 =

15 ´ 1 =5 V 2 +1

For vi £ 5.7 V, vo = vi v1 - 15 v1 v - vi + + o =0 2k 1k 1k

Þ

3v1 + 2 vo - 2 vi = 15

Þ

vo = 0.4 vi + 3.42

vo = v1 + 0.7 5 vo - 2 v1 = 15 + 2.1 = 17.9

7. (D) For vs > 0, when D1 is OFF, Current through D2 is i=

10 - 0.7 = 0.465 mA, vo = 10 ki = 4.65 V 10 + 10

vo = vs for 0 < vs < 4.65 V. For negative values of vs , the output is negative of positive part. Thus (D) is correct option. 8. (B) The diode conducts (zero resistance) when vi < 2.5 V and vo = vi . Diode is open (2 MW resistance) when v - 2.5 vi > 2.5 V and vo = 2.5 + i = 5 V. 3

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Page 151


UNIT 3

= 50 ´ 5 (1 + cos wt) ´ 10 -3 + 0.5

C=

= 0.75 + 0. 25 cos wt = 0. 25( 3 + cos wt) V 31. (D) The output voltage cannot exceed the positive power supply voltage and cannot be lower than the negative power supply voltage. 32. (A) At regulated power supply is =

vmax 2 fkVrip

RL = 30 - 9 = 1.4 mA iL 15 k

75(50) 50 V = 75 + 150 3

50 > VZ , RTH = 150 ||75 = 50 W 3 1 æ 50 ö iZ = - 15 ÷ = 33 mA, P = 15 iZ = 0.5 W ç 50 è 3 ø 3.6(20) = 4.93 V > VZ , 11 + 3.6 4.93 - 4 = 11 || 3.6 = 2.71 kW, iZ = = 0.34 mA 2.71k

34. (A) vTH = RTH

35. (B) iZ ( max ) = iL + iZ =

400m = 40 mA 10

20 - 10 = 45 mA 222

iL ( min ) = 45 - 40 = 5 mA, RL =

10 = 2 kW 5m

36. (B) Current through 12 W resistor is 6.3 - 4.8 i= = 125 mA 12 iL = i - iZ = 125 - iZ

Þ

25 £ iL £ 120 mA

37. (C) 25 £ iL £ 120 mA, iL RL = 4.8 V 25 £

4.8 £ 120 mA RL

Þ

40 £ RL £ 192 W

38. (A) PL = iL VZ = (120m)( 4.8) = 576 mV 39. (B) vs = 60 sin 2 p60 t V vmax = 60 - 1.4 = 58.6 V 58.6 vmax C= = = 24.4 mF 2 fRVrip 2( 60)10 ´ 10 3 ´ 2 40. (C) Full wave rectifier vs = vi = 120 sin 2 p60 t V vmax = 120 - 0.7 = 119.3 V Vrip = 119.3 - 100 = 19.3 V Page 154

=

41. (A) Vrip =

will remain less than 1.4 mA. 33. (D) vTH =

Analog Electronics

www.gatehelp.com

119.3 = 20.6 mF 2( 60)2.5 ´ 10 3 ´ 14.4

vmax fRL C

75 vmax = 6.25 kW = fCVrip 60 ´ 50 ´ 10 -5 ´ 4 ***************


CHAPTER

3.2 BASIC BJT CIRCUITS

VBE ( ON ) = 0.7

VCE ( Sat ) = 0.2

(A) 8.4 V

(B) 6.2 V

transistor if not given in problem.

(C) 4.1 V



3. I C , RC = ?

Use

V,

V

for

npn

+5 V

The common-emitter current gain of the transistor is b =75. The voltage VBE in ON state is 0.7 V. 1. I E , RC = ?

RC VC = 2 V

50 kW +12 V

10 kW

IQ = 1 mA

+ VEC = 6 V -

-5 V

Fig. P3.3.3

RC

(A) 0.987 mA, 3.04 kW (B) 1.013 mA, 2.96 kW

-12 V

(D) 0.946 mA, 4.18 kW

Fig. P3.3.1

(A) 1.46 mA, 6.74 kW

(B) 0.987 mA, 3.04 kW

(C) 1.13 mA,, 5.98 kW


(D) 1.057 mA, 3.96 kW 4. VC = ?

2. VEC = ?

+5 V

+8 V 10 kW 20 kW

10 kW

VC

10 kW -2 V 2 kW

3 kW

-8 V

Fig. P3.3.2

Fig. P3.3.4

(A) 1.49 V

(B) 2.9 V

(C) 1.78 V

(D) 2.3 V

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Page 155


UNIT 3


Analog Electronics

9. VB = 1 V

In the circuit of fig.P3.3.5-6 VB = - 1 V

(A) 4 V (C) 1 V

+3 V

(B) 3 V (D) 1.9 V

10. VB = 2 V 500 kW

4.8 kW

(B) 1.5 V

(C) 2.6 V



-3 V

The transistor in circuit shown in fig. P3.3.11-12 has b = 200. Determine the value of voltage Vo for given value of VBB .

Fig. P3.3.5-6

5. b = ? (A) 103.4 (C) 134.5

(A) -7 V

(B) 135.5 (D) 102.4

+5 V

5 kW

6. VCE = ? (A) 6.4 V (C) 1.3 V

Vo

50 kW

(B) 4.7 V (D) 4.2 V

10 kW VBB

7. In the circuit shown in fig. P3.3.7 voltage VE = 4 V. The value of a and b are respectively +5 V

Fig. P3.3.11-12

11. VBB = 0

2 kW

(A) 2.46 V (C) 3.33 V

100 kW VE

(B) 1.83 V (D) 4.04 V

12. VBB = 1 V 8 kW

(A) 4.11 V (C) 2.46 V

(B) 1.83 V (D) 3.44 V

-5 V

13. VBB = 2 V

Fig. P3.3.7

(A) 0.943, 17.54 (C) 0.914, 11.63

(B) 0.914, 17.54 (D) 0.914, 11.63

(A) 3.18 V (C) 0.2 V

(B) 1.46 V (D) None of the above


Statement for Q.8-10: For the transistor in circuit shown in fig. P3.3.8-10, b = 200. Determine the value of I E and I C for given value of VB in question.

The transistor shown in the circuit of fig. P3.3.14-16 has b = 150. Determine Vo for given value of I Q in question. +5 V

+6 V

5 kW 10 kW

Vo

VC VB

IQ 1 kW

-5 V

Fig. P3.3.14-16

Fig. P3.3.8-10

14. I Q = 0.1 mA

8. VB = 0 V (A) 6.43 mA, 2.4 V (C) 0 A, 6 V Page 156

(B) 2.18 mA, 3.4 V (D) None of the above

(A) 1.4 V

(B) 4.5 V

(C) 3.2 V


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Basic BJT Circuits

15. I Q = 0.5 mA (A) 3.16 V

(B) 2.52 V

(C) 2.14 V

(D) 394 . V

Chap 3.2

(A) 0.991

(B) 0.939

(C) 0.968

(D) 0.914

20. For the transistor in fig. P3.3.20 , b = 50. The value

16. I Q = 2 mA

of voltage VEC is

(A) 4.9 V

(B) -4.9 V

(C) 0.5 V

(D) -0.5 V

+9 V

1 mA

17. For the circuit in fig. P3.3.17 VB = VC and b = 50. The value of VB is

+6 V 50 kW

10 kW

4.7 kW

VC

-9 V

VB

Fig. P3.3.20 1 kW

Fig. P3.317

(A) 0.9 V

(B) 1.19 V

(C) 2.14 V

(D) 1.84 V

(A) 3.13 V

(B) 4.24 V

(C) 5.18 V

(D) 6.07 V

21. In the circuit shown in fig. P3.3.21 if b = 50, the power dissipated in the transistor is +9 V

18. For the circuit shown in fig. P3.3.18, VCB = 0.5 V and b = 100. The value of I Q is

0.5 mA

+5 V

5 kW Vo

50 kW

4.7 kW

-9 V

IQ

Fig. P3.3.21 -5 V

Fig. P3.3.18

(A) 1.68 mA

(B) 0.909 mA

(C) 0.134 mA


19. For the circuit shown in fig. P3.3.19 the emitter voltage is VE = 2 V. The value of a is

(A) 3.87 mW

(B) 10.46 mW

(C) 7.49 mW

(D) 18.74 mW

22. For the circuit shown in fig. P3.3.22 the Q-point is VCEQ = 12 V and I CQ = 2 A when b = 60. The value of resistor RC and RB are

+24 V

+10 V RC

RB

10 kW VE

50 kW

10 kW

-10 V

Fig. P3.3.19

Fig. P3.3.22

(A) 10 kW, 241 kW

(B) 10 kW, 699 kW

(C) 6 kW, 699 kW

(D) 6 kW, 241 kW

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Page 157


Basic BJT Circuits

Chap 3.2

29. For the transistor in the circuit of fig. P3.3.29,

32. The current gain of the transistor shown in the

b = 100. The voltage VB is

circuit of fig.P3.3.32 is b = 100. The values of Q-point ( I CQ , VCEQ) is

+10 V

+5 V

20 kW

1 kW 5 kW

12 kW

15 kW 0.5 kW

2 kW

Fig. P3.3.29 -5 V

(A) 3.6 V

(B) 4.29 V

(C) 3.9 V

(D) 4.69 V

Fig. P3.3.32

30. The current gain of the transistor shown in the

(A) (1.8 mA, 2.1 V)

(B) (1.4 mA, 2.3 V)

(C) (1.4 mA , 1.8 V)

(D) (1.8 mA, 1.4 V)

circuit of fig. P3.3.30 is b = 125. The Q-point values ( I CQ , VCEQ) are

33. For the circuit in fig. P3.3.33, let b = 60. The value of +24 V

VECQ is +5 V

+10 V

58 kW

42 kW

10 kW

2 kW

20 kW

2.2 kW

10 kW

Fig. P3.3.30

-5 V

(A) (0.418 mA, 20.4 V)

(B) (0.915 mA, 14.8 V)

(C) (0 .915 mA, 16.23 V)

(D) (0.418 mA, 18.43 V)

31. For the circuit shown in fig. P3.3.31, let b = 75. The

-10 V

Fig.P3.3.33

(A) 2.68 V

(B) 4.94 V

(C) 3.73 V

(D) 5.69 V

34. In the circuit of fig. P3.3.34 Zener voltage is VZ = 5

Q-point (I CQ , VCEQ) is +24 V

V and b = 100. The value of I CQ and VCEQ are +12 V

25 kW

3 kW 500 W

8 kW

1 kW

Fig. P3.3.34 Fig. P3.3.31

(A) ( 4.68 mA, 16.46 V)

(B) ( 312 . mA , 1.86 V)

(C) ( 312 . mA, 8.46 V)

(D) ( 4.68 mA , 5.22 V)

(A) 12.47 mA, 4.3 V

(B) 12.47 mA, 5.7 V

(C) 10.43 A, 5.7 V

(D) 10.43 A , 4.3 V

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Page 159


UNIT 3

35. The two transistor in fig. P3.2.35 are identical. If b = 25, the current I C2 is

Analog Electronics

(A) 36.63 mA (C) 49.32 mA

(B) 36.17 mA (D) 49.78 mA

+5 V IC2

39. In the bipolar current source of fig. P3.2.39 the diode voltage and transistor BE voltage are equal. If

25 mA

base current is neglected then collector current is

Fig. P3.2.35

(A) 28 mA

(B) 23.2 mA

(C) 26 mA

(D) 24 mA

10 kW

36. In the shunt regulator of fig. P3.2.26, the VZ = 8.2 V

4.7 kW

and VBE = 0.7 V. The regulated output voltage Vo is

10 kW

120 W

Vo

+22 V

-20 V

Fig. P3.2.39

(A) 6.43 mA (C) 1.48 mA

100 W

(B) 2.13 mA (D) 9.19 mA

40. In the current mirror circuit of fig. P3.2.40. the transistor parameters are VBE = 0.7 V, b = 50 and the Early voltage is infinite.

Fig. P3.2.36

(A) 11.8 V

(B) 7.5 V

(C) 12.5 V

(D) 8.9 V

Assume transistor are

matched. The output current I o is +5 V

37. In the series voltage regulator circuit of fig. P3.2.37

1 mA

VBE = 0.7 V, b = 50, VZ = 8.3 V. The output voltage Vo is +25 V

Io

Vo 220 W 20 kW

50 kW

Fig. P3.2.40 50 kW

(A) 1.04 mA (C) 962 mA

30 kW

(B) 1.68 mA (D) 432 mA

41. All transistor in the N output mirror in fig. P3.2.41 are matched with a finite gain b and early voltage

Fig. P3.2.37

(A) 25 V

(B) 25.7 V

(C) 15 V

(D) 15.7 V

V A = ¥. The expression for each load current is V

38. In the regulator circuit of fig. P3.2.38 VZ = 12 V, b = 50, VBE = 0.7 V. The Zener current is +20 V

+

Io1 Iref

Io2

Io3

R1

Vo

QS

220 W

QR

1 kW

Q1

V

Q2

-

Fig. P32.41 Fig. P3.2.38

Page 160

www.gatehelp.com

QN


Basic BJT Circuits

(A)

(C)

I ref

(B)

æ (1 + N ) ö çç 1 + ÷ b (b + 1) ÷ø è b I ref

(D)

æ (1 + N ) ö çç 1 + ÷ (b + 1) ÷ø è

I ref

Chap 3.2

SOLUTIONS

æ N ö çç 1 + ÷ ( b + 1) ÷ø è

1. (C) I E =

b I ref æ N ö çç 1 + ÷ + 1 ÷ø b è

12 - 0.7 10 k

Þ

. mA I E = 113

æ 75 ö I C = çç . ) = 112 . mA ÷÷(113 è 75 + 1 ø

42. Consider the basic three transistor current source in

VCE = 12 - 1.13 ´ 10 - 1.12 RC - ( -12) = 6 V

fig. P3.2.42. Assume all transistor are matched with

RC = 5.98 kW

finite gain and early voltage V A = ¥. The expression for I o is V

2. (C) 8 = 10 ´ (75 + 1) I B + 0.7 + 10 I B - 2

+

IB = Io

I C = bI B = 0.906 mA, I E = (b + 1) I B = 0.918 mA

R1

Iref

9.3 = 12.08 mA, 10 + 760

8 = 10(0.918) + VEC + 3(0.906) - 8 Þ

VEC = 4.1 V

75 æ 75 ö 3. (A) I C = ç (1m) = 0.987 mA ÷ IE = 76 è 75 + 1 ø V

-

RC =

Fig. P3.2.42

(A)

(C)

I ref

(B)

æ 2 ö çç 1 + ÷ (1 + b) ÷ø è I ref

(D)

æ ö 2 çç 1 + ÷÷ b ( 1 + b ) è ø

I ref æ 1 ö çç 1 + ÷ (2 + b) ÷ø è

4. (A) 5 = (1 + b)10 kI B + 20 kI B + 0.7 + b2 kI B 5 = (760 k + 20 k + 150 k) I B + 0.7

I ref

Þ

æ ö 1 çç 1 + ÷÷ b ( 2 + b ) è ø

VC = 5 - (b + 1) I B RC = 5 - 760 ´ 4.62 ´ 10 -3 = 1.49 V

Both of transistor are identical and b >> 1 and VBE1 = 0.7 V. The value of resistance R1 and RE to produce I ref = 1 mA and I o = 12 mA is ( Vt = 0.026)

5. (C) VB = - I B RB Þ

IB =

-VB 1 = = 2.0 mA RB 500k

VE = -1 - 0.7 = - 17 . V V - ( -3) -17 . +3 IE = E = = 0.271 mA 4.8 k 4.8 k IE 0.271m = (b + 1) = IB 2m

+5 V Iref Io Q1

I B = 4.62 mA,

I C = bI B = 0.347 mA

43. Consider the wilder current source of fig. P3.2.43.

R1

5.2 = 304 . kW 0.987m

Þ

Q2

b = 134.5

6. (B) VCE = 3 - VE = 3 - ( -17 . ) = 4.7 V

RE

-5 V

7. (C) I E =

Fig. P3.2.43

5.4 = 0.5 mA 2k

(A) 9.3 kW, 18.23 kW

(B) 9.3 kW , 9.58 kW

4 = 0.7 + I B RB + I C RC - 5, I C » I E ,

(C) 15.4 kW , 16.2 kW

(D) 15.4 kW , 32.4 kW

8.3 = 100 I B + 0.5 ´ 8

*****************

Þ

I B = 43 mA , IE 0.5m =1 + b = = 11.63 IB 43m

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Page 161


UNIT 3

b = 10.63, a =

b ® a = 0.914 1+b

Vo = VCE ( sat ) - VBE = 0.2 - 0.7 = - 0.5 V 17. (B) I E =

8. (C) VB = 0 Transistor is in cut-off region.

VB - 0.7 1k

æ b ö æ 50 ö I C = çç ÷÷I E = ç ÷ ( VB - 0.7) mA è 51 ø èb + 1ø

I E = 0, VC = 6 V 9. (B) VB = 1 V , I E =

Analog Electronics

1 - 0.7 = 0.3 mA 1k

IC =

I C » I E = 0.3 mA

6 - VC mA, VC = VB 10

50 6 - VB ( VB - 0.7) = 51 10

VC = 6 - I C RC = 6 - (0.3)(10) = 3 V

10.8 VB = 12.86 , VB = 119 . V

2 - 0.7 10. (B) VB = 2 V, I E = = 1.3 mA, 1

18. (B) VCB = 0.5 V , VC = 0.5 V

I C » I E = 1.3 mA VC = 6 - (1.3)(10) = - 7 V Transistor is in saturation. The saturation voltage

IC =

5 - 0.5 æ 101 ö =0.9 mA, I Q = ç ÷0.9 = 0.909 mA 5k è 100 ø

VCE = 0.2 V 19. (C) I E =

11. (C) VBB = 0, Transistor is in cutoff region

VB = VE - 0.7 = 1.3 V

RL 10(5) Vo = VCC = + 5 = 3.33 V RC + RL 10

IB =

12. (B) I B =

1 - 0.7 = 6 mA 50 k

I C = bI B = 75 ´ 6m = 0.45 mA 5 - Vo V = IC + o 5k 10 k Vo Vo Vo = 1.83 V (1 - 0.45) = + , Þ 5 10 13. (C) I B =

2 - 0.7 = 26 mA 50 k

I C = b I B = 75 ´ 26 mA =1.95 mA . = -4.75 V VC = 5 - I C RC = 5 - 5 ´ 195 Transistor is in saturation, VCE = 0.2 V = VC = Vo 14. (B) I E = 0.1 mA IC =

b 150 IE = (0.1) = 0.099 mA (b + 1) 151

VB 1.3 = = 26 mA RB 50 k

b+1= a=

I E 0.8m = = 30.77 I B 26m

Þ

b = 29.77

b 29.77 = = 0.968 b + 1 30.77

æ b ö 50 20. (D) I C = çç mA = 0.98 mA I E ÷÷ = b + 1 è ø 51 VC = I C RC - 9 = (0.98)( 4.7) - 9 = - 4.394 V IB =

IE 1 = mA = 19.6 mA (b + 1) 51

VE = I B RB + VEB = 50(0.0196) + 0.7 = 1.68 V VEC = 1.68 - ( -4.394) = 6.074 V æ b ö 50 21. (A) I C = çç (0.5) = 0.49 mA ÷÷I E = b + 1 51 è ø

Vo = 5 - RC I C = 5 - 5(0.099) = 4.50 V

IB =

15. (B) I E = I Q = 0.5 mA

VE = I B RB + VEB = (0.0098)(50) + 0.7 = 1.19 V

æ 150 ö IC = ç ÷ (0.5m) = 0.497 mA è 150 + 1 ø

VC = I C RC - 9 = (0.49)( 4.7) - 9 = - 6.7 V VEC = 119 . - ( -6.7) = 7.89 V

Vo = 5 - RC I C = 2.517 V

PQ = I C VEC + I B VEB

16. (D) Transistor is in saturation Page 162

10 - VE = 0.8 mA 10k

. V VE = (1.3)(1) = 1.3 V , VC = VCE + VE = 15

0.5 = 9.8 mA 51

= (0.49)(7.89) + (0.0098)(0.7) mW = 3.87 mW www.gatehelp.com


UNIT 3

æ R2 VTH = çç è R1 + R2

ö æ 15 ö ÷÷VCC = ç ÷(10) = 4.29 V è 20 + 15 ø ø

10 = I E (1k) + VEB + 10 = I E + 0.7 + Þ IB =

IE ( 8.57 k) + 4.29 b+1

Analog Electronics

5.82 - 0.7 = ( 6.06 k + 76 k) I BQ Þ

I BQ = 62.4 mA

I EQ = (b + 1) I BQ = 4.74 mA I CQ = bI BQ = 4.68 mA VCEQ = 24 - I CQ RC - I EQ RE

IE ( 8.57 k) + 4.29 101

= 24 - ( 4.68)( 3) - ( 4.74)(1) = 5.22 V

I E = 4.62 MA

32. (B) R1 = 12 kW, R2 = 2 kW

IE = 0.046 mA b+1

RTH = R1||R2 = 12 ||2 = 171 . kW æ 2 ö VTH = ç . V ÷(10) - 5 = - 357 è 12 + 2 ø

VB = ( 8.57)(0.046) + 4.29 = 4.69 V 30. (B) R1 = 58 kW, R2 = 42 kW

+5 V

+24 V 5 kW 1.71 kW -3.57 V

24.36 kW +10.1 V 0.5 kW

10 kW -5 V

Fig. S3.3.32 Fig. S3.3.30

-357 . = I BQ(171 . k) + VBE + (b + 1) I BQ(0.5 k) - 5

RTH = 58 ||42 = 24.36 kW VTH

5 - 357 . - 0.7 = (171 . + 50.5)I BQ

æ 42 ö =ç ÷(24) = 10.1 V è 42 + 58 ø

Þ

10.1 = I BQ(24.36 k) + VBE + (b + 1) I BQ(10 k) 10.1 - 0.7 = I BQ(24.36 k + 1260 k)

I BQ = 14 mA

I EQ = (100 + 1) I BQ = 1.412 mA I CQ = 100 I BQ = 1.4 mA VCEQ = 5 - RC I CQ - RE I EQ + 5

I BQ = 7.32 mA

= 5 - (5)(1.4) - (0.5)(1.412) + 5 = 2.3 V

I CQ = bI BQ = 0.915 mA I EQ = (b + 1) I BQ = 0.922 mA

33. (B) RTH = 20 ||10 = 6.67 kW

VCEQ = 24 - (0.922)(10) = 14.8 V

æ 20 ö VTH = ç ÷10 - 5 = 1.67 V è 10 + 20 ø

31. (D) R1 = 25 kW, R2 = 8 kW

+10 V +24 V 2 kW 3 kW 6.67 kW

6.06 kW

+1.67 V

+5.82 V 2.2 kW 1 kW -10 V

Fig. S3.3.33

Fig. S3.3.31

RTH = 25 ||8 = 6.06 kW, VTH

æ 8 ö =ç ÷(24) = 5.82 V è 25 + 8 ø

5.82 = ( 6.06 k)( I BQ) + VBE + (b + 1) I B (1k) Page 164

10 = (1 + b) I BQ(2) + VEB + I BQ( 6.67) + 1.67 10 - 1.67 - 0.7 = I BQ( 6.67 + 122) www.gatehelp.com


UNIT 3

41. (A) I ref = I CR + I BS = I CR +

æ I ref I o RE = Vt ln çç è Io

I ES 1+b

I ES = I BR + I B1 + I B 2 + ....... I BN I BR = I Bi , I CR = I Ci = I oi (1 + N ) I CR I ES = (1 + N ) I BR = b Then I ref = I CR +

æ 1 ´ 10 -3 0.026 ln çç -6 -6 12 ´ 10 è 12 ´ 10

R1 =

V + - VBE1 - V - 5 - 7 - ( -5) = = 9.3 kW I ref 1m

(1 + N ) I CR I ES = I CR + b (b + 1) b+1

+

V

Iref

Io = IC2 Q3

IC1

IE3 Q2

Q1

IB1 IB2

V

-

Fig. S3.2.42

I E 3 = (1 + b) I B 3 2 IB2 IE3 I ref = I C1 + = I C1 + (1 + b) (1 + b) I C1 = I C 2 = bI B 2 æ ö 2 IC2 2 = I C2 çç 1 + ÷÷ b(1 + b) b ( 1 + b ) è ø I ref

æ ö 2 çç 1 + ÷÷ 1 b ( + b ) è ø

43. (B) If b >> 1 and transistor are identical VBE1

VBE2

I ref » I C1 = I S e

Vt

, Io = IC2 = IS e

Vt

æ I ref VBE1 = Vt lnçç è IS

ö æI ÷÷ , VBE 2 = Vt lnçç o è IS ø

ö ÷÷ ø

æ I ref VBE1 - VBE 2 = Vt ln çç è Io

ö ÷÷ ø

From the circuit, VBE1 - VBE 2 = I E 2 RE » I o RE

Page 166

ö ÷÷ = 9.58 kW ø

***********

42. (C) I ref = I C1 + I B 3 , I B1 = I B 2 , I E 3 = 2 I B 2

IC2 = Io =

ö ÷÷ ø

RE =

æ (1 + N ) ö = I Oi çç 1 + ÷ b(b + 1) ÷ø è I ref I oi = æ (1 + N ) ö çç 1 + ÷ b(b + 1) ÷ø è

I ref = I C 2 +

Analog Electronics

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CHAPTER

3.3 BASIC FET CIRCUITS


Statement for Q.1–3: In the circuit shown in fig. P3.3.1–3 the transistor

parameter are as follows:

parameters are as follows: Threshold voltage

In the circuit shown in fig. P3.3.4–6 the transistor

VTN = 2 V

Conduction parameter K n = 0.5 mA / V

VTN = 2 V, k¢n = 60 mA / V 2 ,

2

W = 60 L

+10 V

+10 V

32 kW

4 kW

14 kW

1.2 kW

18 kW

2 kW

6 kW

0.5 kW

-10 V

Fig. P3.3.1–3

Fig. P3.3.4–6

4. VGS = ?

1. VGS = ? (A) 2.05 V

(B) 6.43 V

(A) -3.62 V

(B) 3.62 V

(C) 4.86 V

(D) 3.91 V

(C) -0.74 V

(D) 0.74 V

5. I D = ?

2. I D = ? (A) 1.863 mA

(B) 1.485 mA

(C) 0.775 mA


(A) 13.5 mA

(B) 10 mA

(C) 19.24 mA

(D) 4.76 mA

6. VDS = ?

3. VDS =? (A) 4.59 V

(B) 3.43 V

(A) 2.95 V

(B) 11.9 V

(C) 5.35 V

(D) 6.48 V

(C) 3 V

(D) 12.7 V

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Page 167


Basic FET Circuits

Chap 3.3

16. The parameter of the transistor in fig. P3.3.16 are

19. In the circuit of fig. P.3.3.19 the PMOS transistor

VTN = 1.2 V, K n = 0.5 mA / V 2 and l = 0. The voltage VDS

has parameter VTP = -1.5 V, kp¢ = 25 mA / V 2 , L = 4 mm

is

and l = 0. If I D = 0.1 mA and VSD = 2.5 V, then value of W will be

+5 V

+9 V

50 mA

R

Fig. P3.3.16

(A) 1.69 V

(B) 1.52 V

(C) 1.84 V

(D) 0

Fig. P3.3.19

17. The parameter of the transistor in fig. P3.3.17 are VTN = 0.6 V and K n = 0. 2 mA / V 2 . The voltage VS is

(A) 15 mm

(B) 1.6 mm

(C) 32 mm

(D) 3.2 mm

20. The PMOS transistor in fig. P3.3.20 has parameters

+9 V

VTP = -1.2 V,

24 kW

W = 20, and k¢p = 30 mA / V 2 . L +5 V

Rs

0.25 mA

RD

-9 V

Fig. P3.3.17 -5 V

(A) 1.72 V

(B) -1.72 V

(C) 7.28 V

(D) -7.28 V

Fig. P3.3.20

If I D = 0.5

mA and VD = -3 V, then value of RS

18. In the circuit of fig. P3.3.18 the transistor

and RD are

parameters are VTN = 17 . V and K n = 0.4 mA / V 2 .

(A) 4 kW, 5.8 kW

(B) 4 kW, 5 kW

(C) 5.8 kW, 4 kW

(D) 5 kW, 4 kW

+5 V

21. The parameters for the transistor in circuit of fig.

RD

P3.3.21 are VTN = 2 V and K n = 0.2 mA / V 2 . The power dissipated in the transistor is +10 V

RS

50 kW

-5 V

Fig. P3.3.18 10 kW

If I D = 0.8 mA and VD = 1 V, then value of resistor Fig. P3.3.21

RS and RD are respectively (A) 2.36 kW, 5 kW

(B) 5 kW, 2.36 kW

(C) 6.43 kW, 8.4 kW

(D) 8.4 kW, 6.43 kW

(A) 5.84 mW

(B) 2.34 mW

(C) 0.26 mW

(D) 58.4 mW

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Page 169


UNIT 3


Analog Electronics

25. The transistors in the circuit of fig. P3.3.25 have

Consider the circuit shown in fig. P3.2.22–33.

parameter VTN = 0.8 V, kn¢ = 40 mA / V 2 and l = 0. The width-to-length ratio of M 2

+5 V

when Vi = 5 V, then

( )

W L 1

is

( WL )2 = 1.

If Vo = 0.10 V

for M1 is +5 V

M1 Vo

M1

M2

Vo

The both transistor have parameter as follows VTN = 0.8 V,

M2

Vi

Fig. P3.3.22-23

Fig. P3.3.25

k¢n = 30 mA / V 2

22. If the width-to-length ratios of M1 and M 2 are æW ö æW ö ç ÷ = ç ÷ = 40 è L ø1 è L ø2

(A) 47.5

(B) 28.4

(C) 40.5

(D) 20.3


The output Vo is

All transistors in the circuit in fig. P3.3.26–27

(A) -2.5 V

(B) 2.5 V

(C) 5 V

(D) 0 V

have parameter VTN = 1 V and l = 0. +5 V

æW ö æW ö 23. If the ratio is ç ÷ = 40 and ç ÷ = 15, then Vo is è L ø1 è L ø2 (A) 2.91 V

(B) 2.09 V

(C) 3.41 V

(D) 1.59 V

RD M4

M1 ID1

24. In the circuit of fig. P3.324. the transistor

RG

ID4 M3

M2

parameters are VTN = 1 V and k¢n = 36 mA / V 2 . If I D = 0.5 mA, V1 = 5 V and V2 = 2 V then the width to-length

-5 V

Fig. P3.3.26–27

ratio required in each transistor is +5 V

The conduction parameter are as follows: M1

K n1 = 400 mA / V 2 V1

K n 2 = 200 mA / V 2

M2

K n 3 = 100 mA / V 2

V2

K n 4 = 80 mA / V 2

M3

Fig. P3.3.24

Page 170

26. I D1 = ? (A) 0.23 mA

(B) 0.62 mA

(C) 0.46 mA

(D) 0.31 mA

æW ö ç ÷ è L ø1



(A)

1.75

6.94

27.8

27. I D4 = ?

(B)

4.93

10.56

50.43

(A) 0.62 mA

(B) 0.31 mA

(C)

35.5

22.4

8.53

(C) 0.46 mA

(D) 0.92 mA

(D)

56.4

38.21

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Basic FET Circuits

Chap 3.3

28. For the circuit in fig. P3.3.28 the transistor

(A) 7.43 V

(B) 8.6 V

parameter are VTN = 0.8 V and kn¢ = 30 mA / V 2 . If output

(C) -1.17 V

(D) 1.17 V

voltage is Vo = 0.1 V, when input voltage is Vi = 4.2 V, the required transistor width-to length ratio is

32. A p-channel JFET biased in the saturation region with VSD = 5 V has a drain current of I D = 2.8 mA, and

+5 V

I D = 0.3 mA at VGS = 3 V. The value of I DSS is 10 kW Vo

(A) 10 mA

(B) 5 mA

(C) 7 mA

(D) 2 mA

Vi

Statement for Q.33–34: For the p-channel transistor in the circuit of fig.

Fig. P3.3.28

(A) 1.568

(B) 0.986

(C) 0.731

(D) 1.843

P3.3.33–34 the parameters are I DSS = 6 mA, VP = 4 V and l = 0. 1 kW

29. For the transistor in fig. P3.3.29 parameters are VTN = 1 V and K n = 12.5 mA / V 2 . The Q-point ( I D , VDS ) is +10 V

0.4 kW

20 kW -5 V

Fig. P3.3.33–34

10 kW

33. The value of I DQ is Fig. P3.3.29.

(A) 8.86 mA

(B) 6.39 mA

(C) 4.32 mA

(D) 1.81 mA

(A) (1 mA, 8 V)

(B) (0.2 mA, 4 V)

34. The value of VSD is

(C) (1.17 mA, 8 V)

(D) (0.23 mA, 3.1V)

(A) -4.28 V

(B) 2.47 V

30. For an n-channel JFET, the parameters are I DSS = 6

(C) 4.28 V

(D) 2.19 V

mA and VP = -3 V. If VDS > VDS ( sat ) and VGS = -2 V, then

35. The transistor in the circuit of fig. P3.3.35 has

I D is

parameters I DSS = 8 mA and VP = -4 V. The value of VDSQ

(A) 16.67 mA

(B) 0.67 mA

(C) 5.55 mA

(D) 1.67 mA

is

31. For the circuit in fig. P3.3.32 the transistor

+20 V

140 kW

2.7 kW

parameters are Vp = - 35 . V, I DSS = 18 mA, and l = 0. The value of VDS is

+15 V 60 kW

2 kW

0.8 kW

Fig. P3.3.35 IQ = 8 mA

(A) 2.7 V

(B) 2.85 V

(C) -1.30 V

(D) 1.30 V

-15 V

Fig. P3.3.32

******************

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Page 171


UNIT 3

10 - ( 4.67 + VSG) = (0.5)(1)( VGS )

SOLUTIONS 1. (A) R1 = 32 kW, æ R2 VG = çç è R1 + R2

R2 = 18 kW,

Þ

VDD = 10 V

ö æ 18 ö ÷÷VDD = ç ÷10 = 3.6 V è 18 + 32 ø ø

=

Þ

VGS = 2.05 V

10. (C) Assume transistor in saturation. I D = 0.4 mA, VG = 0,

Þ

VSG = 2 + 0.8 = 2.21 V

VSG = VS - VG = VS

11. (A) VD = I D RD - 5 = (0.4)(5) - 5 = -3 V

= 10 - 0.775( 4 + 2) = 5.35 V

VSD = VS - VD = 2. 21 - ( -3) = 5. 21 V

VDS ( sat ) = VGS - VTN = (2.05 - 0.8) = 1. 25 V VDS > VDS ( sat ) as assumed.

12. (C) R1 = 14.5 kW, R2 = 5.5 kW,

4. (B) R1 = 14 kW, R2 = 6 kW, RS = 0.5 kW, RD = 1.2 kW

RS = 0.6 kW, RD = 0.8 kW, æ RL VG = çç è R1 + R2

ö æ 6 ö ÷÷(20) - 10 = ç ÷(20) - 10 = -4 V è 14 + 6 ø ø

ö 5.5 æ ö ÷÷(10) - 5 = ç ÷(10) - 5 = -2.25 V è 14.5 + 5.5 ø ø

Assume transistor in saturation. V - ( -5) ID = S = K n ( VGS - VTN ) 2 RS

Assume transistor in saturation V - ( -10) VG - VGS + 10 ID = S = = K n ( VGS - VTN ) 2 RS RS

VS = VG - VGS

kn¢ W ( 60)( 60 ´ 10 -6 ) = = 1.8 mA / V 2 2 L 2

-2.25 - VGS + 5 = (0.6)(0.5) ( VGS - ( -1)) 2

Þ

- 4 - VGS + 10 = (0.5)(1.8)( VGS - 2) 2

Þ

Þ

VGS = 3.62, - 0.74 V, VGS will be positive.

VGS is positive. Thus (D) is correct option.

5. (D) I D =

VG - VGS + 10 -4 - 3.62 + 10 = = 4.76 mA RS 0.5 k

VGS = 124 . , - 6.58 V

13. (D) I D =

VS + 5 VG - VGS + 5 -2.25 - 124 . +5 = = RS RS 0.6 k

= 2.52 mA, Therefore (D) is correct option.

6. (B) 10 = I D( RS + RD) + VDS - 10 VDS = 20 - 4.76(12 . + 0.5) = 119 . V

14. (B) 5 = I D( RS + RD) + VDS - 5

VDS ( sat ) = VGS - VTN = 3.62 - 2 = 1.62 V

VDS = 10 - I D( RS + RD) = 10 - 2.52(0.8 + 0.6) = 6.47 V

. V > VDS ( sat ) , Assumption is correct. VDS = 119

VDS ( sat ) = VGS - VTH = 124 . - ( -1) =2.24

7. (B) R1 = 8 kW, RL = 22 kW, RS = 0.5 kW, RD = 2 kW æ R2 VG = çç è R1 + R2

ö æ 22 ö ÷÷(20) - 10 = ç ÷(20) - 10 = 4.67 V è 8 + 22 ø ø

Assume transistor in saturation 10 - VS ID = = K P ( VSG + VTP ) 2 RS VS = VG + VSG

VDS > VDS ( sat ) ,Assumption is correct. 15. (B) I S = 50 mA = I D ,I D = K n ( VGS - VTN ) 2 Þ

50 ´ 10 -6 = 0.5 ´ 10 -3( VGS - 12 . )2

V VG = 0, VS = VG - VGS = -1516 . VDS = VD - VS = 5 - ( -1516 . ) = 6.516 V 16. (B) I D = 50m = K n ( VGS - VTN ) 2 Þ

Page 172

0.4 = K P ( VGS + VTP ) 2

0.4 = (0.2)( VSG - 0.8) 2

3. (C) VDS = VDD - I D( RD + RS )

Kn =

10 - ( 4.67 + 377 . ) = 312 . mA 0.5

VSD = 20 - I D ( RS + RD)= 20 - 2.12(2 + 0.5) = 12.2 V

V - VGS 3.6 - 2.05 2. (C) I D = G = = 0.775 mA RS 2k

æ RL VG = çç è R1 + R2

10 - VS 10 - ( VG + VGS ) = RS RS

9. (C) 10 = I D( RS + RD) + VSD - 10

K n = 0.5 mA / V 2

3.6 - VGS = (2)(0.5)( VGS - 0.8) 2

VSG = 377 . V, - 177 . V, VSG is positive voltage.

8. (A) I D =

Assume that transistor in saturation region V V - VGS ID = S = G = K n ( VGS - VTN ) 2 RS RS RS = 2 kW,

Analog Electronics

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50 ´ 10 -6 = 0.5 ´ 10 -3( VGS - 12 . )2

Þ

V, VGS = 1516 .


Basic FET Circuits

VGS = 152 . V,

æW ö æW ö 23. (A) ç ÷ > ç ÷ thus VGS1 < VGS 2 è L ø1 è L ø2

VGS = VDS

17. (B) I D = K n ( VGS - VTN ) 2 Þ

0.25 = 0.2( VGS - 0.6)

VGS = VG - VS , 18. (A) I D =

Þ

2

VG = 0,

40( VGS1 - 0.8) 2 = 15( VGS 2 - 0.8) 2

V, VGS = 172 .

VGS 2 = 5 - VGS1

V VS = -172 .

5 - VD = 0.8 mA, RD

RD =

1.63( VGS1 - 0.8) = (5 - VGS1 - 0.8) VGS1 = 2.09, VGS 2 = 2.91 V, Vo = VGS 2 = 2.91 V

6 -1 = 5 kW 0.8m

24. (A) Each transistor is biased in saturation because

I D = K n ( VGS - VTN ) 2 Þ 0.8 = (0.4)( VGS - 17 . )2

Þ

VDS = VGS and VDS > VGS - VTN

VGS = 311 . V

For M 3 , V2 = 2 V = VGS 3

VGS = VG - VS , VG = 0, VS = -311 . V -311 . - ( -5) I D = 0.8 mA = Þ RS = 2.36 kW RS

æ 25 ö æ W 10 -4 = ç ÷ç è 2 øè 4

ö . )2 ÷ (2.5 - 15 ø

æ 30 ´ 10 -6 20. (D) K p = çç 2 è I D = K p ( VSG + VTP ) 2 Þ

æ 36 ´ 10 -3 öæ W ö ÷÷ç ÷ (2 - 1) 2 I D = 0.5 = çç 2 øè L ø3 è

Þ

æ 36 ´ 10 -3 öæ W ö ÷÷ç ÷ ( 3 - 1) 2 I D = 0.5 = çç 2 øè L ø2 è

25. (D) M 2 is in saturation because

0.5 = 0.3( VSG - 12 . )2

VGS 2 = VDS 2 > VGS 2 - VTN

VSG = 2.49 V, VG = 0

VD - ( -5) RD

Þ RD =

M1 is in non saturation because VGS1 = Vi = 5 V, VDS1 = VD = 0 V VDS1 < VGS1 - VTN , I D1 = I D2 æW ö æW ö 2 2 ç ÷ [2( VGS1 - VTN1 ) VDS1 - VDS 2 ]= ç ÷ ( VGS 2 - VTN 2 ) L L è ø1 è ø2

-3 + 5 = 4 kW 0.5m

æW ö Þ ç ÷ [2(5 - 0.8)(0.1) - (0.1) 2 ] = (1)(5 - 0.1 - 0.8) 2 è L ø1

21. (B) Assume transistor in saturation 10 - VGS ID = = K n ( VGS - VTN ) 2 10 k

æW ö ç ÷ (0.83) = 16.81 è L ø1

10 - VGS = (10)(0.2)( VGS - 2) 2 Þ

. V VGS = VDS = 377 10 - 377 . ID = = 0.623 mA 10 k

VGS1 = 5 - VGS 2 Þ

VDS > VGS - VTN assumption is correct.

transistor

I D1 = I D2

Þ

are

K n 1 ( VGS1 - VTN1 ) = K n 2 ( VGS 2 - VTN 2 )

K n 1 = K n 2 , VTN1 = VTN 2 5 VGS1 = VGS 2 = V 2 Vo = VGS 2 = 2.5 V

-6

(5 - VGS 2 - 1) 2 = 200 ( VGS 2 - 1) 2 VGS1 = 2.24 V

(2.24 - 1) 2 =0.62 mA

I D4 = K n 4 ( VGS 4 - VTN ) 2 = K n 3 ( VGS 3 - VTN ) 2 in

saturation. 2

æW ö ç ÷ =20.3 è L ø1

27. (B) VGS 2 = VGS 3 = 2.76 V

22. (B) For both transistor VDS = VGS , both

Þ

VGS 2 = 2.76 V,

I D1 = 400 ´ 10

Power = I DVDS = 2.35 mW

Therefore

Þ

26. (B) I D1 = K n 1 ( VGS1 - VTN ) 2 = K n 2 ( VGS 2 - VTN ) 2

. V, - 0.27 V, VGS will be 3.77 V VGS = 377

VDS > VGS - VTN

æW ö ç ÷ = 6.94 è L ø2

æ 36 öæ W ö æW ö I D = 0.5 = ç ´ 10 -5 ÷ç ÷ (5 - 1) 2 Þ ç ÷ = 174 . è 2 øè L ø1 è L ø1

VS = VSG = 2.49 V 5 - VS 5 - 2.49 ID = Þ RS = = 5.02 kW RS 0.5m ID =

Þ

For M1 , VGS1 = 10 - V1 = 10 - 5 = 5 V

W = 32 mm

ö ÷÷(20) = 0.3 mA / V 2 ø

Þ

æW ö ç ÷ = 27.8 è L ø3

Þ

For M 2 , VGS 2 = V1 - V2 = 5 - 2 = 3 V

k¢p W ( VGS + VTP ) 2 2 L

19. (C) VSD = VSG, I D =

Chap 3.3

= 100 ´ 10 -6 (2.76 - 1) 2 = 0.31 mA 28. (C) VGS = 4.2 V,

2

VDS = 0.1 V

VDS < VGS - VTN , Thus transistor is in non saturation. 5 - 0.1 ID = = 0.49 mA 10 k k¢ W ID = n {2 ( VGS - VTN ) VDS - VDS2 } 2 L

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Page 173


CHAPTER

3.4 AMPLIFIERS

1. If the transistor parameter are b = 180 and Early

3. The small signal votlage gain Av = vo vs is

voltage V A = 140 V and it is biased at I CQ = 2 mA, the

(A) -4.38

(B) 4.38

values of hybrid-p parameter g m , rp and ro are

(C) -1.88

(D) 1.88

respectively 4. The nominal quiescent collector current of a

(A) 14 A V, 2.33 kW, 90 kW

transistor is 1.2 mA. If the range of b for this transistor

(B) 14 A V, 90 kW , 2.33 kW

is 80 £ b £ 120 and if the quiescent collector current

(C) 77 mA V, 2.33 kW , 70 kW

changes by ±10 percent, the range in value for rp is

(D) 77.2 A V, 70 kW, 2.33 kW

(A) 1.73 kW < rp < 2.59 kW (B) 1.93 kW < rp < 2.59 kW

Statement for Q.2–3.

(C) 1.73 kW < rp < 2.59 kW

Consider the circuit of fig. P3.4.2–3. The transistor

(D) 1.56 kW < rp < 2.88 kW

parameters are b = 120 and V A = ¥. Statement for Q.5–6: +5 V

Consider the circuit in fig. P3.4.5.6. The transistor parameter are b = 100 and V A = ¥.

4 kW 250 kW vs

+10 V

vo

~

RC 50 kW

2V

vs

Fig. P3.4.2–3

~

vBB

2. The hybrid-p parameter values of g m , rp and ro are (A) 24 mA V, ¥, 5 kW

Fig. P3.4.5–6

(B) 24 mA V, 5 kW , ¥

5. If Q-point is in the center of the load line and I CQ = 0.5

(C) 48 mA V, 10 kW , 18.4 kW

mA, the values of VBB and RC are

(D) 48 mA V, 18.4 kW, 10 kW

(A) 10 kW , 0.95 V

(B) 10 kW , 1.45 V

(C) 48 kW , 0.95 V

(D) 48 kW , 1.45 V

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Page 175


Amplifiers


Chap 3.4

19. For an n-channel MOSFET biased in the saturation

Consider the common Base amplifier shown in fig. P3.4.14–15. The parameters are

g m = 2 mS

and

region, the parameters are VTN = 1 V, 12 m n Cox = 18 mA V 2 and l = 0.015 V -1 and I DQ = 2 mA. If transconductance is

ro = 250 kW. Find the Thevenin equivalent faced by load

g m = 3.4 mA V, the width-to-length ratio is

resistance RL .

(A) 80.6

(B) 43.2

(C) 190

(D) 110

Thevenin equivalent

20. In the circuit of fig. P3.4.20, the parameters are

270 W vi

~

g m = 1mA / V, ro = 50 kW. The gain Av = vo vs is

RL

VDD

Fig. P3.4.14–15

60 kW

14. The Thevenin voltage vTH is (A) 263vi

(B) 132vi

(C) 346vi

(D) 498vi

2 kW

vs

~

300 kW

15. The Thevenin equivalent resistance RTH is (A) 384 kW

(B) 697 kW

(C) 408 kW

(D) 915 kW

Fig. P3.4.20


(A) -8.01

(B) 8.01

(C) 14.16

(D) -14.16


The common-base amplifier is drawn as a two-port in fig. P3.4.16–17. The parameters are b = 100, g m = 3 mS, and ro = 800 kW. +

For the circuit shown in fig. P3.4.21–23 transistor parameters are VTN = 2 V, K n = 0.5 mA / V 2 and l = 0. The transistor is in saturation.

i2

i1

v1

+ 3.9 kW

10 kW

18 kW

_

+10 V

v2

10 kW vo

_

Fig. P3.4.16–17

vi

16. The h-parameter h21 is

~

VGG

(A) 2.46

(B) 0.9

(C) 0.5

(D) 0.67

Fig. P3.4.21–23

21. If I DQ is to be 0.4 mA, the value of VGSQ is

17. The h-parameter h12 is (A) 3.8 ´10 -4

(B) 4.83 ´10 -3

(C) 3.8 ´10 4

(D) 4.83 ´10 3

18. For an n-channel MOSFET biased in the saturation region, the parameters are K n = 0.5 mA V 2 , VTN = 0.8 V and l = 0.01 V -1 , and I DQ = 0.75 mA. The value of g m and ro are

(A) 5.14 V

(B) 4.36 V

(C) 2.89 V

(D) 1.83 V

22. The values of g m and ro are (A) 0.89 mS, ¥

(B) 0.89 mS, 0

(C) 1.48 mS, 0

(D) 1.48 mS, ¥

23. The small signal voltage gain Av is

(A) 0.68 mS, 603 kW

(B) 1.22 mS, 603 kW

(A) 14.3

(B) -14.3

(C) 1.22 mS, 133 kW

(D) 0.68 mS, 133 kW

(C) -8.9

(D) 8.9

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Page 177


Amplifiers

(A) 4.44

(B) -4.44

(C) 2.22

(D) -2.22

SOLUTIONS

the

source-follower

circuit

in

fig.

P3.4.33-34. The values of parameter are g m = 2 mS and ro = 100 kW. +5 V

Ro

vs

~

I CQ

1. (C) g m =

Statement for Q.33–34: Consider

Chap 3.4

IQ

500 kW

bVt b 180 = = = 2.33 kW I CQ g m 77.2m

ro =

V A 140 = = 70 kW I CQ 2m

gm =

Fig. P3.4.33-34

rp =

33. The voltage gain Av is (B) -0.89

(C) 2.79

(D) -2.79

2 - 0.7 = 5.2 m A 250 k

I CQ

=

Vt

0.624 = 24 mA V 0.0259

bVt b 120 = = = 5 kW, ro = ¥ I CQ g m 24m

ö æ rp bRC 3. (C) Av = - g m RC çç ÷÷ = r + R r B ø p + RB è p

34. The output resistance Ro is

æ ö 5k = - (24m)( 4 k) çç ÷÷ = -1.88 5 k + 250 k è ø

(A) 100 kW

(B) 0.498 kW

(C) 1.33 kW

(D) None of the above 4. (D) rp =

rp(min) =

bVT , I CQ

(120)(0.0259) = 2.88 kW, 108 . m

rp( max ) = *******************

2m = 77.2 mA V 0.0259

I CQ = bI B = (120)(5.2m ) = 0.642 mA

4 kW

-5 V

(A) 0.89

=

rp =

2. (B) I BQ =

vo

Vt

( 80)(0.0259) = 156 . kW 1.32m

5. (A) VECQ =

1 VCC = 5 V 2

VECQ = 10 - I CQ RC = 5 Þ

10 - (0.5m) RC = 5

RC = 10 kW, I BQ =

I CQ b

=

0.5 =5 mA 100

VEB ( ON ) + I BQ RB = VBB Þ

0.7 + (5m ) (50 k) = 0.95 V

6. (D) g m = rp =

I CQ Vt

=

0.5 = 19.3 mA V 0.0259

bVt (100)(0.0259) = = 5.18 kW , ro = ¥ 0.5m I CQ

æ 100 ö 7. (B) I CQ = ç ÷(0.35) = 0.347 mA è 1001 ø

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Page 179


UNIT 3

The small-signal equivalent circuit is as shown in fig. S3.4.7 500 W

vo +

~

vs

C

B

10 kW

rp

Vp _

gmVp

ro

7 kW

Analog Electronics

9. (A) DC Analysis: VTH =

RTH = 11||50 = 8.33 kW 12 - 0.7 - 10 I BQ = = 119 . mA 8.33k + (101)1k . mA, I EQ = 12 . mA I CQ = bI BQ = 119 . )2 = 8.42 V VECQ = 12 - (1. 20)1 - (119

E

AC Analysis: Ib

Fig. S3.4.7

rp vs

ö æ rp ||10 k vo ÷÷ ( ro ||7 k) = - g m çç vs 500 r | | 10 k + ø p è I CQ 0.347m gm = = = 1313 . mA V Vt 0.0259

rp =

288 ´ 7 7.6 ´ 10 = 6.83 kW, rp ||10k = = 4.32 kW 288 + 7 7.6 + 10

vo

10||50 kW

bIb E 2 kW (b+1)Ib

bVt (100)(0.0259 = = 2.18 kW 119 . m I CQ

vo = bI b (2k) , vs = - (b + 1) I b (1k) + I b ( rp) v -b(2 k) -(100)(2 k) Av = o = = = -196 . vs rp + (b + 1)1k 2.18 k + (100)(1k) 10. (B) VECQ = 8.42 V,

4.32 k æ ö Av = -1313 . mç ÷( 6.83k) = - 80 è 500 + 4.32 k ø

Þ Output voltage swing = 5.16 V peak to peak.

8. (C) DC Analysis: I CQ = I EQ

11. (B) Since the B–C junction is not reverse biased, the transistor continues to operate in the forward-active Þ

I CQ = 3.57 mA

-mode

AC Analysis: Ib

C

B

vo

+ rp vs

~

R1 || R2

Ie

+

+ vce _

vce

rp

gmvce

_

Vp bIb _ E

Fig. S 3.4.11 1.2 kW

bVt (0.0259) = (150) = 109 . kW , ro = ¥ 357 . m I CQ

æ 1 ö vce 1 , So rp || çç ÷÷ || ro = g m Vce g m è gm ø (100)(0.0259) rp = = 2.33 kW 2m I CQ 2m gm = = = 77.2 mA V Vt 0.0259

vo -(b I b ) RC , vs = I b rp + (b + 1) RE I b = vs vs

1 V 150 = 12.95 , ro = A = = 75 kW gm I CQ 2m

0.2 kW

Fig. S3.4.8

rp =

DvEC = 11 - 8.42 = 2.58 V

For 1 £ vEC £ 11 V,

VCEQ = 5 = 10 - I CQ ( RC + RE ) Þ 5 = 10 - I CQ(1. 2 k + 0. 2 k) 3.57 I BQ = = 23.8 m A 150

Av = Av =

Page 180

C

Fig. S3.4.9

VA 100 = = 288 kW I CQ 0.347m

ro ||7k =

~

Vp +

1 kW

b bV 100 = t = = 7.6 kW rp = . m g m I CQ 1313 ro =

Ic

B _

rp ||10 k ( vs ), vo = - g m Vp( ro ||7k) 500 + rp ||10 k

Vp =

50 (12) = 10 V 10 + 50

-bRC -(150)(12 . )k = = -5.75 rp + (1 + b) RE 109 . k + (151)(0.2 k)

r=

re = (2.33k)||(12.95)||(75 k) =12.87 W

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ro


Amplifiers

VA I CQ

12. (C) ro =

Þ

I CQ =

Vp Vp v + Vpi + + g m Vp + i =0 rp ro 270

VA 75 = = 0.375 mA ro 200 k

vi + Vpi Vp Vp + + 2mVp + =0 50 k 250 k 270

bV 75(0.0259) 13. (B) rp = t = = 194 . kW I CQ 1m E

bIb

Ie Ib

C

2.7 kW

B

Þ

16. (B) The equivalent small-signal circuit is shown in

+ vo _

fig. S3.4.16

1.5 kW

ro

Fig. S 3.4.13

i1

vi = I b ( rp + 15k . ), I in = I e = (b + 1) I b Vi ( rp + 15 . k) 194 . + 15 . k Rin = = = = 45 W Ie (b + 1) 76

E

~

vs

3.9 kW rp

~

rp = gmVp

Vp +

vi rp Removing the RL , -Vp = 270 + rp

rp =

vi rp(1 + g m ro) 270 + rp

i2 i1

, i2 = v2 = 0

Vp V V - p - p - g m Vp , 39 . k rp ro

h21 =

i2 = i1

. k - g m rp 39 - gm = = 0.91 1 1 39 . k . k r + + g p m rp 39 + + gm 39 . k rp

i1 = 0

E

3.9 kW rp

fig. S3.4.15

C

Vp +

gmVp 18 kW

i2

~

v2

B

ro

Fig. S 3.4.17

270 W _ rp

gmVp

Vp +

Fig. S 3.4.15

I sc = g m Vp +

v1 v v - v2 + 1 + 1 = g m Vp 39 . k rp ro

_

15. (A) The equivalent small-signal circuit is shown in

~

Vp can be neglected ro

ro

vi 50 k(1 + (2m)(250 k)) = 498 vi 270 + 50 k

vi

Vp + g m Vp r0

i1 = -

17. (A) v1 = -Vp ,

b 100 = = 50 kW g m 2m

vTH =

v2 = 0

b 100 = = 33.3 kW gm 3m

h21 =

Fig. S 3.4.14

vTH = -ro g m Vp - Vp =

18 kW

Vp +

Fig. S 3.4.16

270 W _

gmVp

B

ro

rp

C

_

14. (D) The equivalent circuit is shown below

vi

Vp = -0.647 vi ,

. mvi I sc = 1297 vTH 498 vi RTH = = = 384 kW I SC 1297 . mvi

rp

~

vi

Chap 3.4

Vp Vp = 2.004 mVp = 2mVp + ro 250 k

Isc

æ 1 1 1ö v + + ÷÷ - 2 = - g m v1 v1 çç 39 . k r ro ø ro è p 1 1 v1 ro 800 k = = 1 1 1 1 1 1 v2 + + + 3m + + + gm 39 . k 33.3k 800 k 39 . k rp ro = 3.8 ´ 10 -4 www.gatehelp.com

Page 181


Amplifiers

RS = 4 kW, v gs = 0.84 vi ,

v gs = vi ,

vo = - g m v gs ( ro ||RD)

Chap 3.4

vo = Av = - g m (7k) vi

g m = 2 K n ( VGS - VTN )

= -(1.41m)(0.84 vi )(100 k || 5 k) vo Þ = Av = - 5.6 vi

= 2 (1m)(151 . - 0.8) = 1.42 mS

28. (A) Ro = RD ||ro || 100k =4.76 kW

32. (A) The small-signal equivalent circuit is shown in

Av = -(1.42m) (7 k) = - 9.9

fig. S.3.4.34 29. (A) As shown in fig. S3.4.27, Ri = R1 || R2 = 20.6 kW

. V, I DQ = 0.5 mA VGSQ = 15

vi

g m = 2 K n ( VGS - VTN ) = 2(1m) (15 . - 0.8) = 1.4 mA V ro = [ lI DQ ]

vo

vgs

~

10 kW

RS

5 kW +

RD

RL

4 kW

G

=¥ Fig. S3.4.32

The resulting small-signal equivalent circuit is shown in fig. S5.4.30 G

D

+

~

vi

D

_

30. (C) From the DC analysis:

-1

gmvgs

S

vgs gmvgs

RTH

RD

vo = - g m v gs ( RD || RL ), vi = -v gs v Av = o = g m ( RD || RL ) = (2m)(5 k ||4 k) = 4.44 vi

vo 7 kW

_

33. (A) The small-signal equivalent circuit is shown in

S

fig. S3.4.33 RS

0.5 kW G

D

+

Fig. S 3.4.30

vi

vo = - g m v gs RD, vi = v gs + g m v gs RS vo - g m RD (7 k) Þ = = - (1.4m) = -5.76 vi 1 + g m RS 1 + (1.4m) (0.5 k)

~

ro

vgs gmvgs

500 kW

_ vo

S

4 kW

31. (B) Since the DC gate current is zero, VS = -VGSQ Fig. S3.4.33

I DQ = I Q = K n ( VGSQ - VTN ) 2 Þ

0.5 = 1( VGSQ - 0.8) 2

vo = g m v gs ( RL || ro)

VGSQ = 151 . V = -VS . ) = 301 . V VDSQ = 5 - (0.5m)(7 k) - ( -151 The transistor is therefore biased in the saturation region. The small-signal equivalent circuit is shown in fig.S3.4.31. D

G + vs

~

vgs _

gmvgs

vo

7 kW

vi = v gs + vo = v gs + g m v gs ( RL || ro) 1 v gs = 1 + g m ( RL || ro) vo g m ( RL || ro) = Av = vi 1 + g m ( RL || ro) RL ||ro = 4 k ||100 k = Av =

(2m)( 3.85 k) = 0.89 1 + (2m)( 3.85 k)

S

34. (B) Ro = Fig. S3.4.31

vo = - g m v gs (7k)

100 k = 3.86 kW 26

1 ||ro gm

1 = æç ö÷ ||(100) » 0.498 kW è2ø ********************

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Page 183


CHAPTER

3.5 OPERATIONAL AMPLIFIERS

1. Av =

vo =? vi

400 kW vi

(A) -2 sin wt m A

(B) -7 sin wt m A

(C) -5 sin wt m A

(D) 0

40 kW

4. In circuit shown in fig. P3.5.4, the input voltage vi is

vo

0.2 V. The output voltage vo is 50 kW

R

150 kW

10 kW vi

25 kW

Fig. P3.5.1

(A) -10

(B) 10

(C) -11

(D) 11

2. Av =

vo

Fig. P3.5.4

vo =? vi

400 kW 40 kW

vi

vo

(A) 6 V

(B) -6 V

(C) 8 V

(D) -8 V

5. For the circuit shown in fig. P3.5.5 gain is

60 kW

Av = vo vi = -10. The value of R is R

Fig. P3.5.2

(A) -10

(B) 10

(C) 13.46

(D) -13.46

100 kW vi

3.

The

input

to

the

circuit

in

fig.

100 kW

P3.5.3

100 kW vo

is

vi = 2 sin wt mV. The current io is 10 kW vi

Fig. P3.5.5

1 kW io vo

(A) 600 kW

(B) 450 kW

(C) 4.5 MW

(D) 6 MW

4 kW

Fig. P3.5.3

Page 184

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Operational Amplifiers

6. For the op-amp circuit shown in fig. P3.5.6 the

Chap 3.5

10. In the circuit of fig. P3.5.10 the output voltage vo is

voltage gain Av = vo vi is

20 kW

R

R

20 kW

+0.5 V

R

40 kW -1 V

R

R

vi

vo

60 kW +2 V

R vo

Fig. P3.5.10

Fig. P3.5.6

(A) -8

(B) 8

(C) -10

(D) 10

(B) -2.67 V

(C) -6.67 V

(D) 6.67 V

11. In the circuit of fig. P3.5.11 the voltage vi1 is

7. For the op-amp shown in fig. P3.5.7 open loop differential gain is Aod = 10 3. The output voltage vo for vi = 2 V is

(A) 2.67 V

(1 + 2 sin wt) mV and vi2 = -10 mV. The output voltage vo is 20 kW 20 kW

100 kW

vi1 vi

2 kW 1 kW

100 kW vo

vo

vi2 1 kW

Fig. P3.5.11

Fig. P3.5.7

(A) -1.996

(B) -1.998

(A) -0.4(1 + sin wt) mV

(B) 0.4(1 + sin wt) mV

(C) -2.004

(D) -2.006

(C) 0.4(1 + 2 sin wt) mV

(D) -0.4(1 + 2 sin wt) mV

8. The op-amp of fig. P3.5.8 has a very poor open-loop

12. For the circuit in fig. P3.5.12 the output voltage is

voltage gain of 45 but is otherwise ideal. The closed-loop

vo = 2.5 V in response to input voltage vi = 5 V. The finite

gain of amplifier is

open-loop differential gain of the op-amp is

100 kW

vi

2 kW

500 kW

vo

vo

1 kW

vi

Fig. P3.5.8

(A) 20

(B) 4.5

(C) 4

(D) 5

Fig. P3.5.12

(A) 5 ´ 10 4

(B) 250.5

(C) 2 ´ 10

(D) 501

4

9. For the circuit shown in fig. P3.5.9 the input voltage 13. vo = ?

vi is 1.5 V. The current io is 10 kW vi

100 kW

6 kW

100 kW

8 kW io

vo

vo

20 kW +18 V

5 kW

40 kW +15 V

Fig. P3.5.13 Fig. P3.5.9

(A) -1.5 mA

(B) 1.5 mA

(A) 34 V

(B) -17 V

(C) -0.75 mA

(D) 0.75 mA

(C) 32 V

(D) -32 V

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Page 185


UNIT 3

28. For the circuit shown in fig. P3.5.28 the input

Analog Electronics

31. io = ?

resistance is

6 kW

io

2 kW vo

6A 2 kW

4 kW is

Fig. S3.5.31 2 kW 10 kW

(A) -18 A

(B) 18 A

(C) -36 A

(D) 36 A

Fig. P3.5.28

(A) 38 kW

(B) 17 kW

(C) 25 kW

(D) 47 kW

Statement for Q.32–33: Consider the circuit shown below

29. In the circuit of fig. P3.5.29 the op-amp slew rate is SR = 0.5 V ms. If the amplitude of input signal is 0.02 V, then the maximum frequency that may be used is vi

3 kW

D1

6 kW

D2

2 kW vo

240 kW vi

10 kW vo

Fig. P3.5.32–33

32. If vi = 2 V, then output vo is

Fig. P3.5.29

(A) 0.55 ´ 106 rad/s

(B) 0.55 rad/s

(C) 1.1 ´ 106 rad/s

(D) 1.1 rad/s

(B) -4 V

(C) 3 V

(D) -3 V

33. If vi = -2 V, then output vo is

30. In the circuit of fig. P3.5.30 the input offset voltage and input offset current are Vio = 4 mV and I io = 150 nA. The total output offset voltage is

(A) 4 V

(A) -6 V

(B) 6 V

(C) -3 V

(D) 3 V

34. vo( t) = ?

500 kW vi

8 mF

5 kW vo 5u(t) mA

250 W

vo 50 W

1 kW

5 kW

Fig. P3.5.34 Fig. P3.5.30

(A) e (A) 479 mV (C) 168 mV

Page 188

(B) 234 mV (D) 116 mV

(C) e

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-

t 10

-

t 1 .6

-

u( t) V

(B) -e

u( t) V

(D) -e

t 10

-

t 1 .6

u( t) V u( t) V



Chap 3.5

35. The circuit shown in fig. P3.5.35 is at steady state

(A) vs vss

(B) -vs vss

before the switch opens at t = 0. The vC ( t) for t > 0 is t=0

v (C) - s vss

(D)

20 kW

39. If the input to the ideal comparator shown in fig.

vs vss

P3.5.39 is a sinusoidal signal of 8 V (peak to peak)

20 kW 20 kW

without any DC component, then the output of the +

4 mF

5V

comparator has a duty cycle of

vC -

Input Output

Fig. P3.5.35

(A) 10 - 5 e -12 .5t V (C) 5 + 5 e

-

t 12 .5

Vref = 2 V

(B) 5 + 5 e -12 .5t V (D) 10 - 5 e

V

-

t 12 .5

Fig. P3.5.39

V

36. The LED in the circuit of fig. P3.5.36 will be on if vi is 10 kW +10 V

470W 10 kW

vi

(A)

1 2

(B)

1 3

(C)

1 6

(D)

1 12

40. In the op-amp circuit given in fig. P3.5.40 the load current iL is

Fig. P3.5.36

(A) > 10 V

(B) < 10 V

(C) > 5 V

(D) < 5 V

R1 R1 vs

37. In the circuit of fig. P3.5.37 the CMRR of the op-amp is 60 dB. The magnitude of the vo is

R2

2V

IL

R2 RL

100 kW R

R

1 kW

Fig. P3.5.40 R

R

vo

1 kW

(A) -

vs R2

(B)

vs R2

(C) -

vs RL

(D)

vs RL

100 kW

Fig. P3.5.37

(A) 1 mV

(B) 100 mV

(C) 200 mV

(D) 2 mV

41. In the circuit of fig. P3.5.41 output voltage is |vo| = 1

V for a certain set of w, R, an C. The |vo| will be 2 V if R1

38. The analog multiplier X of fig. P.3.5.38 has the R1

characteristics vp = v1 v2 . The output of this circuit is vss

vi = sin wt V

vo

10 kW X vs

R

R

C vo

Fig. P3.5.41 Fig. P3.5.38

(A) w is doubled

(B) w is halved

(C) R is doubled


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Page 189


UNIT 3

Analog Electronics

42. In the circuit of fig. P3.5.42. the 3 dB cutoff

+15 V

frequency is 50 nF

47 kW

6 kW

3 kW

vo vi

vo

Fig. P3.5.42

(A) 10 kHz

(B) 1.59 kHz

(C) 354 Hz

(D) 689 Hz

Vz = 5 V

100 W

Fig. P3.5.45

43. The phase shift oscillator of fig. P3.5.43 operate at

46. In the circuit in fig. P3.5.46 both transistor Q1 and

f = 80 kHz. The value of resistance RF is

Q2 are identical. The output voltage at T = 300 K is

RF 100 pF

100 pF

100 pF

R

R

R1

R2

v1

R

v2

vo

333 kW 20 kW

Fig. P3.5.43

(A) 148 kW

(B) 236 kW

(C) 438 kW

(D) 814 kW

vo

20 kW

333 kW

44. The value of C required for sinusoidal oscillation of Fig. P3.5.46

frequency 1 kHz in the circuit of fig. P3.5.44 is 2.1 kW

1 kW

C 1 kW

æv R ö (A) 2 log10 çç 2 1 ÷÷ è v1 R2 ø

æv R ö (B) log10 çç 2 1 ÷÷ è v1 R2 ø

æv R ö (C) 2.303 log10 çç 2 1 ÷÷ è v1 R2 ø

æv R ö (D) 4.605 log10 çç 2 1 ÷÷ è v1 R2 ø

47. In the op-amp series regulator circuit of fig. P8.3.47 Vz = 6.2 V, VBE = 0.7 V and b = 60. The output voltage vo is

C

1 kW

vo

+36 V 1 kW

Fig. P3.5.44

(A) (C)

1 mF 2p 1 2p 6

30 kW

(B) 2p mF mF

(D) 2 p 6 mF 10 kW

45. In the circuit shown in fig. P3.5.45 the op-amp is ideal. If bF = 60, then the total current supplied by the 15 V source is (A) 123.1 mA

(B) 98.3 mA

(C) 49.4 mA

(D) 168 mA

Fig. P3.5.47

(A) 35.8 V

(B) 24.8 V

(C) 29.8 V


*******

Page 190

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26. (C) v2 + = v2 - = 0 V, current through 6 V source i=

6 = 2 mA, vo = -2m( 3k + 2 k) = -10 V 3k

33. (D) If vi < 0, then vo > 0, D2 blocks and D1 conduct Av = -

vo(1) vo v (2) vo(1) = , v- = i + 1+ 3 4 2+1 2 +1 2v v v v v+ = v- , o = o + i , o = -8 4 3 3 vi

Chap 3.5

3k . )=3 V = -15 . , vo = ( -2)( -15 2k

34. (A) Voltage follower vo = v- = v+

27. (D) v+ =

v+ (0 + ) = 5m(250 ||1000) = 1 V, v+ ( ¥) = 0 t = 8m(1000 + 250) = 10 s 35. (A) vc (0 - ) = 5 V = vc (0 + ) = 5 V

28. (B) Since op-amp is ideal

For t > 0 the equivalent circuit is shown in fig. S3.5.35 20 kW

i1 is

+ vC –

is 2 kW

i2

Fig. S3.5.35 10 kW

t = 20 k ´ 4m = 0.08 s vc = 10 + (5 - 10) e

Fig. S3.5.28

v- = v+ , 2 kis = 4 ki1

Þ

is = 2 i1

vs = 2 kis + 10 ki2

36. (C) v- =

i2 = is + i1 , vs = 2 kis + 10 k( is + i1 ), i1 = i ö æ vs = 2 kis + 10 kç is + s ÷ 2ø è

Þ

is 2

-

t 0 .08

= 10 - 5 e -12 .5t V for t > 0

(10)(10 k) =5 V 10 k + 10 k

When v+ > 5 V, output will be positive and LED will be on. Hence (C) is correct.

vs = 17k = Rin is

R R = 1 V, v- = (2) = 1 V, vd = 0 2R 2R v + vR VCM VCM = + = 1, vo = F 2 1 CMRR 100 1 CMRR = 60 dB = 10 3 , vo = = 100 mV 1 10 3

37. (B) v+ = (2)

½R ½ 240 k 29. (C) Closed loop gain A =½ F½ = = 24 ½ R1½ 10 k The maximum output voltage vom = 24 ´ 0.02 = 0.48 V w £

4 mF

10 V

4 kW

0.5 / m SR . ´ 106 rad/s = = 11 0.48 vom

38. (C) v+ = 0 = v- ,

æ R ö 30. (A) The offset due to Vio is vo = çç 1 + 1 ÷÷Vio R1 ø è

Let output of analog multiplier be vp . vp vs =Þ vs = -vp , vp = vss vo R R v vs = -vss vo , vo = - s vss

500 ö æ = ç1 + ÷ 4m = 404 mV 5 ø è Due to I io, vo = RF I io = (500 k)(150n) = 75 mV Total offset voltage vo = 404 + 75 = 479 mV

39. (B) When vi > 2 V, output is positive. When vi < 2 V,

-vo v , io = - 6 + o 6k 3k -6( 6 k) io = - 6 + = -18 A. 3k

output is negative.

31. (A) 6 =

V 4V 2V

32. (B) If vi > 0, then vo < 0, D1 blocks and D2 conducts Av = -

6k = -2 3k

p 6

Þ

2p 5p 6

t

vo = ( -2)(2) = -4 V Fig. S3.5.39

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Page 193


UNIT 3

5p p TON 1 Duty cycle = = 6 6 = 2p T 3 vs - v- v- - vo = R1 R1

40.(A)

v+ v v - vo + + + + =0 R2 RL R2

Þ

Analog Electronics

44. (A) This is Wien-bridge oscillator. The ratio R2 2.1k = = 2.1 is greater than 2. So there will be R1 1k oscillation

2 v1 = vs + vo

R2 R1

æ R ö vo = çç 2 + 2 ÷÷ v+ R è L ø

Þ

æ R ö 2 v- = vs + çç 2 + 2 ÷÷ v+ , v- = v+ R è L ø R Þ 0 = vs + 2 v+ RL v+ = -

RL vs , R2

iL =

v+ , RL

iL = -

R R

vs R2

Fig. S3.5.44

Frequency =

41. (D) This is a all pass circuit vo 1 - jwRC , = H ( jw) = vi 1 + jwRC

| H( jw)| =

1 + ( wR 2 C) 2 1 + ( wRC) 2

C

C

=1

C=

1 2pRC

Þ

1 ´ 10 3 =

1 mF 2p

Thus when w and R is changed, the transfer function is

45. (C) v+ = 5 V = v- = vE ,

unchanged.

The input current to the op-amp is zero. i+15V = iZ + iC = iZ + a F iE

42. (B) Let R1 = 3 kW , R2 = 6 kW , C = 50 nF vi v - vo + i =0 1 R2 æ ö R1 ||ç ÷ è sC ø

Þ

vi v v + i = o ö R2 R2 æ R1 ÷÷ çç 1 + sR C ø 1 è

éR ù vi ê 2 (1 + sR1 C) + 1ú = vo R ë 1 û vi [R2 + R1 + sR1 R2 C ] = vo R1 vo R + R1 = 2 vi R1 Þ f3dB =

é sR1 R2 C ù ê1 + R + R ú 1 2 û ë

vo æ R ö = ç 1 + 2 ÷÷(1 + s( R1 || R2 ) C) vi çè R1 ø 1 = 2 p( R1 || R2 ) C

1 1 = = 159 . kHz 2 p( 3k ||6 k)50n 2 p(2 k)50n

Þ R=

1 ( 80 k)(2 p 6 )(100 p)

RF = 29 R

Þ

=

15 - 5 60 æ 5 ö + ç ÷ = 49.4 mA 47 k 61 è 100 ø

46. (B) vo =

333 ( vo1 - vo2 ) 20

æi vo1 = -vBE1 - Vt ln çç c1 è is

43. (B) The oscillation frequency is 1 1 f = Þ 80 k = 2 p 6 RC 2 p 6 R(100 p) = 8.12 kW

RF = ( 8.12 k)(29) = 236 kW

æi vo1 - vo2 = -Vt ln çç c1 è ic 2 v1 , R1

ic1 =

ic 2 =

ö æi ÷÷, vo2 = -vBE 2 - Vt ln çç c 2 ø è is

æi ö ÷÷ = Vt ln çç c 2 è ic1 ø

ö ÷÷ ø

ö ÷÷ ø

v2 R2

æv R ö vo1 - vo2 = Vt ln çç 2 1 ÷÷, Vt = 0.0259 V è R2 v1 ø vo =

æv R ö 333 333 ( vo1 - vo2 ) = (0.0259) ln çç 2 1 ÷÷ 20 20 è v1 R2 ø

æv R ö æv R ö = 0.4329 lnçç 2 1 ÷÷ = 0.4329(2.3026) log10 çç 2 1 ÷÷ è v1 R2 ø è v1 R2 ø æv R ö = log10 çç 2 1 ÷÷ è v1 R2 ø 47. (B) v+ = v- , vZ =

10 vo v = o 10 + 30 4

vo = 4 vz = 6.2 ´ 4 = 24.8 V

************

Page 194

1 2 p(1k) C

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CHAPTER

4.1 NUMBER SYSTEMS & BOOLEAN ALGEBRA

1. The 100110 2 is numerically equivalent to 1. 2616

2. 3610

3. 468

5. A computer has the following negative numbers stored in binary form as shown. The wrongly stored

4. 212 4

number is

The correct answer are (A) 1, 2, and 3

(B) 2, 3, and 4

(C) 1, 2, and 4

(D) 1, 3, and 4

(B) 5

(C) 7

(D) 9

(B) -89 as 1010 0111

(C) -48 as 1110 1000

(D) -32 as 1110 0000

6. Consider the signed binary number A = 01010110

2. If (211) x = (152)8 , then the value of base x is (A) 6

(A) -37 as 1101 1011

and B = 1110 1100 where B is the 1’s complement and MSB is the sign bit. In list-I operation is given, and in list-II resultant binary number is given.

3. 11001, 1001 and 111001 correspond to the 2’s List–I

complement representation of the following set of

List-II

numbers (A) 25, 9 and 57 respectively

P. A + B

(B) -6, -6 and -6 respectively

Q. B - A

(C) -7, -7 and -7 respectively

R. A - B

(D) -25, -9 and -57 respectively

1. 0 1 0 0 2. 0 1 1 0 3. 0 1 0 0 4. 1 0 0 1 5. 1 0 1 1 6. 1 0 0 1 7. 1 0 1 1 8. 0 1 1 0

S. - A - B

4. A signed integer has been stored in a byte using 2’s

0011 1001 0010 0101 1100 0110 1101 1010

complement format. We wish to store the same integer The correct match is

in 16-bit word. We should copy the original byte to the less significant byte of the word and fill the more

P

Q

R

S

significant byte with (A) 0

(A)

3

4

2

5

(B) 1

(B)

3

6

8

7

(C) equal to the MSB of the original byte

(C)

1

4

8

7

(D) complement of the MSB of the original byte.

(D)

1

6

2

5

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Page 197


UNIT 4

28.

The

simplified

form

of

a

logic

Digital Electronics

A

function

A

B

Y = A( B + C( AB + AC)) is

Z

C

(A) A B

(B) AB

(C) AB

(D) AB

D

(B) A + B

(C) AB + AB

(D) A B + AB

B

(C) 0

(D) 1

Z

D

B

Z

C D

(C)

(D)

35. In fig. P.4.1.35 the input condition, needed to

30. If X Y + XY = Z then XZ + XZ is equal to (B) Y

A

C

(A) A + B

(B)

A

Y = ( AB ) × ( AB) is

Z

D

(A)

29. The reduced form of the Boolean expression of

(A) Y

B C

produce X = 1, is A B

31. If XY = 0 then X Å Y is equal to (A) X + Y

(B) X + Y

(C) XY

(D) X Y

Fig. P4.1.34

32. From a four-input OR gate the number of input condition, that will produce HIGH output are (A) 1

(B) 3

(C) 15

(D) 0

(A) A = 1, B = 1, C = 0

(B) A = 1, B = 1, C = 1

(C) A = 0, B = 1, C = 1

(D) A = 1, B = 0, C = 0

36. Consider the statements below:

33. A logic circuit control the passage of a signal according to the following requirements : 1. Output X will equal A when control input B and

1. If the output waveform from an OR gate is the same as the waveform at one of its inputs, the other input is being held permanently LOW. 2. If the output waveform from an OR gate is always HIGH, one of its input is being held permanently HIGH.

C are the same.

The statement, which is always true, is

2. X will remain HIGH when B and C are different. The logic circuit would be A X

C

X

C

(B)

A

(C) Only 2


is (A) 3

(B) 4

(C) 5

(D) 6

A

B

X

C

B

X

number of two-input NAND required to implement

(D)

34. The output of logic circuit is HIGH whenever A and B are both HIGH as long as C and D are either both LOW or both HIGH. The logic circuit is

38. If the X and Y logic inputs are available and their complements X and Y are not available, the minimum

C

(C)

Page 200

(B) Only 1

NAND gates, minimum number of requirement of gate

B

(A)

(A) Both 1 and 2

37. To implement y = ABCD using only two-input

A

B

X

C

X Å Y is (A) 4

(B) 5

(C) 6

(D) 7

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Number Systems & Boolean Algebra


Chap 4.1

Assuming complements of x and y are not

A Boolean function Z = ABC is to be implement

available, a minimum cost solution for realizing f using

using NAND and NOR gate. Each gate has unit cost.

2-input NOR gates and 2-input OR gates (each having

Only A, B and C are available.

unit cost) would have a total cost of

39. If both gate are available then minimum cost is (A) 2 units

(B) 3 units

(C) 4 units

(D) 6 units

(A) 1 units

(B) 2 units

(C) 3 units

(D) 4 units

44. The gates G1 and G2 in Fig. P.4.2.44 have propagation delays of 10 ns and 20 ns respectively.

40. If NAND gate are available then minimum cost is (A) 2 units

(B) 3 units

(C) 4 units

(D) 6 units

G1

1 Vi 0

G2

Vo

Vi

to

Fig. P4.1.44

If the input Vi makes an abrupt change from logic

41. In fig. P4.1.41 the LED emits light when

0 to 1 at t = t0 then the output waveform Vo is VCC = 5 V

1 kW

1 kW

[t1 = t0 + 10 ns, t2 = t1 + 10 ns, t3 = t2 + 10 ns] (B)

(A)

1 kW

t0

1 kW

t2

t1

t3

t1

t2

t3

t0

t1

t2

t3

(D)

(C) t0

Fig. P4.1.41

(A) both switch are closed

t0

t2

t1

t3

45. In the network of fig. P4.1.45 f can be written as

(B) both switch are open

X0

(C) only one switch is closed

1 2

X1

(D) LED does not emit light irrespective of the switch positions

3 X2

42. If the input to the digital circuit shown in fig.

n-1 X3

Xn-1

n

F

Xn

Fig. P4.1.45

P.4.1.42 consisting of a cascade of 20 XOR gates is X,

(A) X 0 X1 X 3 X 5 + X 2 X 4 X 5 .... X n -1 + .... X n -1 X n

then the output Y is equal to

(B) X 0 X1 X 3 X 5 + X 2 X 3 X 4 .... X n + .... X n -1 X n (C) X 0 X1 X 3 X 5 .... X n + X 2 X 3 X 5 K X n + .... + X n -1 X n

1

(D) X 0 X1 X 3 X 5... X n -1 + X 2 X 3 X 5 K X n +..+ X n -1 X n - 2 + X n Y X

Fig. P4.1.42

(A) X

(B) X

(C) 0

(D) 1

*******

43. A Boolean function of two variables x and y is defined as follows : f (0, 0) = f (0, 1) = f (1, 1) = 1; f (1, 0) = 0

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Page 201


UNIT 4

Digital Electronics

A - B = A + B,

SOLUTIONS

A

010 10110

B

+ 00010011 0110 1001

1. (D) 100110 2 = 2 5 + 2 2 + 21 = 3810

- A - B = A + B,

2616 = 2 ´ 16 + 6 = 3810 468 = 4 ´ 8 + 6 = 3810

A

1010 1001

B

+ 00010011 10111100

212 4 = 2 ´ 4 2 + 41 = 3810 7. (B) Here A , B are 2’s complement

So 3610 is not equivalent. 2. (C) 2 x 2 + x + 1 = 64 + 5 ´ 8 + 2

Þ

A + B,

x =7

A

0100 0110

B

+ 1101 0011 1 0001 1001

3. (C) All are 2’s complement of 7 Þ

11001

Discard the carry 1

00110 +

1

Þ

1001

Þ

B - A,

= 710

= 710

1101 0011

A

+ 1011 1010

Discard the carry 1 - A - B = A + B,

42 in a byte

00101010

42in a word

0 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0

-42 in a byte

11010110

-42 in a word

1 1 1 1 1 1 1 1 1 1 0 1 0 1 1 0

1011 1010

B

+ 0010 1101

8. (B) 1110 = 10112 2Fi-1

Bi

Fi

0.3

0.6

0

0.6

1.2

1

0.2

1

0.4

0

0.4

11010000

0.8

0

0.8

1.6

1

0.6

00110000 2 1100 1111 +

6. (D) Here A , B are 1’s complement A 01010110 B + 1110 1100 10100 0010 , + 1 0100 0011 B

1110 1100

A

+ 1010 1001 110010101 1 +

Repeat from the second line 0.310 = 0.01001

2

9. (C)

Received

b4

b3

b2

p3

b1

p2

p1

1

1

0

1

1

0

0

C1* = b4 Å b2 Å b1 Å p1 = 0 C2* = b4 Å b3 Å b1 Å p2 = 1 C3* = b4 Å b3 Å b2 Å p3 = 1

10010110 Page 202

A

1110 0111

Therefore (C) is correct.

B - A = B + A,

+ 0010 1101

1 1000 1101

4. (C) See a example

A + B,

B

B

1

000111

-4810 =

010 0 0110 0111 0011

000110 +

5. (C) 4810 =

A

0110 + 1 0111

111001

A - B = A + B,

= 710

00111

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Number Systems & Boolean Algebra

Chap 4.1

C3* C2* C1* = 110 which indicate position 6 in error

= ( AA + AB)( A + B + C) = A( A + B + C) = A

Transmitted code 1001100.

Therefore No gate is required to implement this function.

10. (D) X = MNQ + M NQ + M NQ

23. (A)

= MQ + M NQ = Q( M + M N ) = Q( M + N ) 11. (A) The logic circuit can be modified as shown in fig. S. 4.1.11 A B

Z

C+D E

Fig. S4.1.11

Now Z = AB + ( C + D) E

A

B

C

( A + BC)

( A + B)( A + C)

0

0

0

0

0

0

0

1

0

0

0

1

0

0

0

0

1

1

1

1

1

0

0

1

1

1

0

1

1

1

Fig. S 4.1.23

12. (D) You can see that input to last XNOR gate is same. So output will be HIGH.

24. (B) X = ABC + ABC + ABC = BC + ABC

13. (D) Z = A + ( AB + BC) + C

25. (B) ( A + B)( B + C ) = ( AB)( BC) = ABC

= A + ( A + B + B + C) + C = A + B + C

( A + B)( B + C ) = ( A + B) + ( B + C) = A + B + C

ABC = A + B + C

( A + B)( B + C) = ( A + B) + ( B + C)

AB + BC + AC = A + B + B + C + A + C = A + B + C

= AB + B + C = A + B + C

14. (C) ( X + Y )( X + Y ) = XY + X Y

From truth table Z = A + B + C

( X + Y )( X + Y )( X + Y ) = ( X + Y )( XY + X Y )

Thus (B) is correct.

= XY + XY = XY

26. (D) AC + BC = AC( B + B) + ( A + A) BC = ABC + ABC + ABC + ABC

15. (B) Using duality ( A + B)( A + C)( B + C) = ( A + B)( A + C)

27. (D) F = A + AB + A BC + A B C( D + DE)

Thus (B) is correct option.

= A + AB + A B( C + C( D + E)) = A + A( B + B( C + D + E)) = A + B + C + D + E

16. (B) Z = ( AB)( CD)( EF ) = AB + CD + EF 17. (A) X = ( A B + AB)( A + B) = ( AB + A B)( AB) = AB 18. (B) Y = ( A Å B) × C = ( AB + AC) × C = ( AB + AB) + C = A B + AB + C 19. (C) Z = A( A + A) BC = ABC 20. (A) Z = AB( B + C) = ABC 21. (A) Z = ( A + B ) × BC = ( AB) × BC = ABC

28. (B) A( B + C ( AB + AC)) = AB + AC ( AB × AC) = AB + AC[( A + B)( A + C)] = AB + AC ( A + AC + AB + B C) = AB 29. (C) ( AB ) × ( AB) = AB + AB = AB + AB 30. (B) X Z + XZ = X ( XY + XY ) + X ( X Y + XY ) = X ( XY + X Y ) + XY = XY + XY = Y 31. (A) X Å Y = X Y + XY = ( XY + XY ) = ( XY ) = X + Y

22. (A) A( A + B)( A + B + C) www.gatehelp.com

Page 203


UNIT 4

Digital Electronics

(A) ( w + y)( x + y + z)( w + x + z)

(A) AC + AB

(B) AC + AB + BC

(B) ( w + x)( w + z )( x + y)( y + z )

(C) AB + BC + ABC

(D) Above all

(C) ( x + z)( w + y)

12. In the logic circuit of fig. P4.2.12 the redundant gate

(D) ( x + z )( w + y)

is

7. A function with don’t care condition is as follows

x y z w x z w y z

f ( a, b, c, d) = Sm(0, 2, 3, 5, 7, 8, 9, 10, 11) + Sdc ( 4, 15) The minimized expression for this function is (A) ab + b d + cd + abc

(B) ab + b d + cd + abd

(C) ab + b d + b c + abd

(D) Above all

8. A function with don’t cares is as follows : g( X , Y , Z ) = Sm(5, 6) + Sdc (1, 2, 4)

1

2

4

X

3

Fig. P4.2.12

(A) 1

(B) 2

(C) 3

(D) 4

For above function consider following expression 1. XYZ + XYZ

2. XY + XZ

3. XZ + XZ + YZ

4. YZ + YZ

13. If function W, X, Y, and Z are as follow W = R + PQ + RS

The solution for g are (A) 1, 2, and 3

(B) 1, 2, and 4

(C) 1, and 4

(D) 1, and 3

X = PQRS + P Q RS + PQ R S Y = RS + PR + PQ + P Q Z = R + S + PQ + P Q R + PQ S

9. A logical function of four variable is given as

Then

f ( A, B, C, D) = ( A + B C)( B + CD) The function as a sum of product is (A) A + BC + ACD + BCD (B) A + BC + ACD + BCD

(A) W = Z , X = Z

(B) W = Z , X = Y

(C) W = Y

(D) W = Y = Z

14. In a certain application four inputs A, B, C, D are

(C) AB + BC + ACD + BCD

fed to logic circuit, producing an output which operates

(D) AB + AB + ACD + BCD

a relay. The relay turns on when f(A, B, C, D) =1 for the

10. A combinational circuit has input A, B, and C and its K-map is as shown in fig. P4.2.10. The output of the circuit is given by

00 00 01

0101, 0110, 1101 and 1110. States 1000 and 1001 do not occur, and for the remaining states the relay is off. The minimized Boolean expression f is

CD

A

following states of the inputs (ABCD) : 0000, 0010,

01

11

1

10 1

1

(A) ACD + BCD + BCD

(B) BCD + BCD + ACD

(C) ABD + BCD + BCD

(D) ABD + BCD + BCD

15. There are four Boolean variables x1 , x2 , x3 and x4 .

1

The following function are defined on sets of them f ( x3 , x2 , x1 ) = Sm( 3, 4, 5)

Fig. P4.2.1

Page 206

(A) ( AB + AB ) C

(B) ( AB + AB ) C

(C) ABC

(D) A Å B Å C

g( x4 , x3 , x2 ) = Sm(1, 6, 7) h( x4 , x3 , x2 , x1 ) = fg Then h( x4 , x3 , x2 , x1 ) is

11. The Boolean Expression Y = ( A + B)( A + C) is equal

(A) Sm(3, 12, 13)

(B) Sm(3, 6)

to

(C) Sm(3, 12)

(D) 0

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Combinational Logic Circuits


Chap 4.2

21. For a binary half subtractor having two input A and

A switching function of four variable, f ( w, x y, z) is to equal the product of two other function f1 and f2 , of

B, the correct set of logical expressions for the outputs D = ( A - B) and X (borrow) are

the same variable f = f1 f2 . The function f and f1 are as

(A) D = AB + AB , X = AB

follows :

(B) D = AB + AB , X = AB

f = Sm( 4, 7, 15)

(C) D = AB + AB , X = AB

f1 = Sm(0, 1, 2, 3, 4, 7, 8, 9, 10, 11, 15)

(D) D = AB + AB , X = AB 22. f1 f2 = ?

16. The number of full specified function, that will satisfy the given condition, is (A) 32

(B) 16

x0

I0

(C) 4

(D) 1

x1

I1

17. The simplest function for f2 is

x2

I2

(A) x

(B) x

(C) y

(D) y

D0 3-to-8 Decoder D1 D2 D3 D4 D5 D6 D7

and m9. If the literals in these minterms are complemented, the corresponding minterm numbers are (A) m3 and m0

(B) m9 and m6

(C) m2 and m0

(D) m6 and m9

(A) x0 x1 x2

(B) x0 Å x1 Å x2

(C) 1

(D) 0

23. The logic circuit shown in fig. P4.2.23 implements

A

I0

B

I1

C

I2

19. The minimum function that can detect a “divisible by 3’’ 8421 BCD code digit (representation D8 D4 D2 D1 ) is given by (A) D8 D1 + D4 D2 + D8 D2 D1 (B) D8 D1 + D4 D2 D1 + D4 D2 D1 + D8 D4 D2 D1

D0 3-to-8 D Decoder 1 D2 D3 D4 D5 D6 D7 EN

Fig. P4.2.23

(D) D4 D2 D1 + D4 D2 D1 + D8 D4 D2 D1

(A) D( A

I1

x2

I2

u C + AC )

(B) D( B Å C + AC )

(C) D( B Å C + AB )

20. f ( x2 , x1 , x0 ) = ?

x1

Z

D

(C) D4 D1 + D4 D2 + D8 D1 D2 D1

I0

f2

Fig. P4.2.22

18. A four-variable switching function has minterms m6

x0

f1

(D) D( B

u C + AB )


D0 3-to-8 D Decoder 1 D2 D3 D4 D5 D6 D7

The building block shown in fig. P4.2.24–25 is a f

active high output decoder.

Fig. P4.2.21

(A) P(1, 2, 4, 5, 7)

(B) S(1, 2, 4, 5, 7)

(C) S(0, 3, 6)

(D) None of Above

A

I0

B

I1

C

I2

D0 3-to-8 Decoder D1 D2 D3 D4 D5 D6 D7

X

Y

Fig. P4.2.24-25

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Page 207


UNIT 4

24. The output X is

Digital Electronics

28. Z 2 = ?

(A) AB + BC + CA

(B) A + B + C

(A) ab + bc + ca

(B) a + b + c

(C) ABC


(C) abc

(D) a u b u c

25. The output Y is

29. This circuit act as a

(A) A + B

(B) B + C

(A) Full adder

(B) Half adder

(C) C + A


(C) Full subtractor

(D) Half subtractor

26. A logic circuit consist of two 2 ´ 4 decoder as shown

30. The network shown in fig. P4.2.30 implements

in fig. P4.2.26. x

1 MUX

1

0

f

D0

A0

D1 D2 y

A

A1

D0

A1

D3

f

D1

B

1 MUX

0

0

S0

D2 z

A0

D3

S0 C

Fig. P4.2.26

Fig. P4.2.30

The output of decoder are as follow D0 = 1 when A0 = 0,

A1 = 0

D1 = 1 when A0 = 1,

A1 = 0

D2 = 1 when A0 = 0,

A1 = 1

D3 = 1 when A0 = 1,

A1 = 1

(A) NOR gate

(B) NAND gate

(C) XOR gate

(D) XNOR gate

31. The MUX shown in fig. P4.2.31 is 4 ´ 1 multiplexer. The output Z is C

The value of f ( x, y, z) is

I3 I2

(A) 0

(B) z

(C) z

(D) 1

MUX

I0


+5 V

S 1 S0 A

B) A Å B Å C

(A) ABC

c

1

0

MUX

Z1

B

Fig. P4.2.31

A MUX network is shown in fig. P4.2.27-29. c

Z

I1

(C) A u B

uC

(D) A + B + C

32. The output of the 4 ´ 1 multiplexer shown in fig.

S0

P4.2.32 is a

1

a

0

MUX Z0 S0 b

Y

c

b

1

S0 MUX

+5 V

Z2

X

0

Page 208

(A) a + b + c

(B) ab + ac + bc

(C) a u b u c

(D) a Å b Å c

MUX

I1 I0

Z

S 1 S0 Y

Fig. P4.2.32

Fig. P4.2.27-29

27. Z1 = ?

I3 I2

(A) X + Y

(B) X + Y

(C) XY + X

(D) XY

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Chap 4.2

33. The MUX shown in fig. P4.2.33 is a 4 ´ 1

36. The 4–to–1 multiplexer shown in fig. P4.2.36

multiplexer. The output Z is

implements the Boolean expression

I3

C

I3

I2

MUX

I1

C

I0

Z

uC (D) B u C

0 1 0 1 1

The input to I1 and I 3 will be (A) yz , y + z

(B) y + z, y u z

(C) y + z, y Å z

(D) x + y ,

yÅ z

37. The 8-to-1 multiplexer shown in fig. P4.2.37 realize 0 1 1

x

I0 I1 MUX I2 I3 EN S 1 S0 w

x

f ( w, x, y, z) = Sm( 4, 5, 7, 8, 10, 12, 15)

34. f = ?

w

S1 S0 w

(B) A

(C) B Å C

f

Fig. P4.2.36

Fig. P4.2.33

I0 I1 MUX I2 I3 EN S 1 S0

I0

0

B

(A) A Å C

0 1 1 0 1

MUX

I1

S1 S0 A

I2

z

I0 I1 MUX I2 I3 EN S1 S0 y

the following Boolean expression 1 z z

f

0 z 0 1 z

z

I0 I1 I2 I3 I4 I5 I6 I7

MUX

EN S2 S1 S0 0

x

w

x

y

Fig. P4.2.37

Fig. P4.2.34

(A) wxyz + wx yz + xy + yz

(A) wxz + w x z + wyz + xy z

(B) wxyz + wxyz + x y + yz

(B) wxz + wyz + wyz + w x y

(C) wx yz + w x yz + yz + zx

(C) w x z + wy z + w yz + wxz

(D) wxyz + wxyz + gz + zx

(D) MUX is not enable

35. For the logic circuit shown in fig. P4.2.35 the output


Y is

A PLA realization is shown in fig. P4.2.38–40 x0 1 0 I 0 I1 I 2 I 3 I 4 I 5 I 6 I 7 C B A

EN S2 S1 S0

x1 x2

MUX

Y

X

Fig. P4.2.35

(A) A Å B

(B) A Å B

(C) A Å B Å C

(D) A Å B Å C

X X

X X X

X

X

f1 f2 f3

Fig. P.4.2.38-40

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Page 209


UNIT 4

38. f1 ( x2 , x1 , x0 ) = ?

Digital Electronics

The terms P1 , P2 , P3 , P4 and P5 are

(A) x2 x0 + x1 x0

(B) x2 x0 + x1 x2

(C) x2 Å x0

(D) x2

ux

0

39. f2 ( x2 , x1 , x0 ) = ? (A) Sm(1, 2, 5, 6)

(B) Sm(1, 2, 6, 7)

(C) Sm(2, 3, 4)


(A) ab, ac, bc, bc, ab

(B) ab , bc, ac, ab , bc

(C) ac, ab, bc, ab , bc

(D) Above all

43. The circuit shown in fig. P.4.2.43 has 4 boxes each described by input P, Q, R and output Y , Z with Y = P Å Q Å R and Z = RQ + PR + QP . Q

40. f3( x2 , x1 , x0 ) = ? (A) P M(0, 4, 6, 7)

(B) P M(2, 4, 5,7)

(C) P M(1, 2, 3, 5)

(D) P M(2, 3, 4, 7)

P P

Q

P

Q

P

Q

P

Q

Z

R

Z

R

Z

R

Z

R

41. If the input X 3 X 2 X1 X 0 to the ROM in fig. P4.2.41 are 8–4–2–1 BCD numbers, then output Y3Y2 Y1 Y0 are X3

X2

X1

X0

Output

Fig. P4.2.43

The circuit act as a 4 bit

BCD to Decimal Decoder

(A) adder giving P + Q

D0 D1 D2 D3 D4 D5 D6 D7 D8 D9

X X X X Y3

(C) subtractor giving Q - P

X X X X X X Y2

(D) adder giving P + Q + R

X X X

(B) subtractor giving P - Q

X

X X Y1 X

X

44. The circuit shown in fig. P4.2.44 converts

X Y0

MSB

Fig. P4.2.41

(A) 2–4–2–1 BCD number

(B) gray code number

(C) excess 3 code converter

(D) none of the above +

42. It is desired to generate the following three Boolean

+

function f1 = ab c + abc + bc

MSB

Fig. P4.2.44

f2 = ab c + ab + abc, f3 = ab c + abc + ac

(A) BCD to binary code

by using an OR gate array as shown in fig. P4.2.42 where P1 and P5 are the product terms in one or more of the variable a, a , b, b , c and c. P1

X

P2

X

(B) Binary to excess (C) Excess–3 to Gray Code (D) Gray to Binary code

X X

P3

******* X

P4 P5

X F1

F2

F3

Fig. P4.2.42

Page 210

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+



16. (A) f = Sm(4, 7, 15),

RS 00

01

11

00 PQ

Chap 4.2

f1 = Sm(0, 1, 2, 3, 4, 7, 8, 9, 10, 11, 15)

10

f2 = Sm(4, 7, 15) + Sdc(5, 6, 12, 13, 14)

1

01

1

1

1

11

1

1

1

There are 5 don't care condition. So 2 5 = 32 different

1

functions f2 . 17. (A) f2 = Sm(4, 7, 15) + Sdc(5, 6, 12, 13, 14), f2 = x

1

10

yz 00

01

11

10

00

0

0

0

0

01

1

´

1

´

11

´

´

1

´

10

0

0

0

0

Fig. S 4.2.13c

Z = R + S + PQ + P Q R + PQ S= R + S + PQP Q R PQ S

wx

= R + S + ( P + Q )( P + Q + R)( P + Q + S) = R + S + ( PQ + PR + Q P + Q R)( P + Q + S) = R + S + PQ + PQS + PR + PRQ + PRS + Q PS + Q PR + Q RS

Fig. S 4.2.17

RS 00

01

11

10

1

1

1

1

1

1

11

1

1

1

10

1

1

1

00 PQ

01

1

18. (B) m6 = ABCD ,

m9 = ABCD

After complementing literal m6¢ = ABCD = m9 ,

m¢9 = ABCD = m6

19. (B) 0, 3, 6 and 9 are divisible by 3 D2 D1 00 00

Fig. S 4.2.13d

D8 D4 01 11

= R + S + PQ We can see that W = Z , X = Z

01

11

1

10

1 1

´

10

´

´

´

1

´

´

14. (D) ABD + BCD + BCD Fig. S 4.2.19

CD 00 00 AB

01

11

1

10 1

01

1

1

11

1

1

10

´

f = D8 D1 + D4 D2 D1 + D4 D2 D1 + D8 D4 D2 D1 20. (B) f = Sm(0, 3, 6) = Sm(1, 2, 4, 5, 7) 21. (C) D = AB + AB,

´

X = AB

A

B

D

X

0

0

0

0

Fig. S 4.2.14

0

1

1

1

15. (A) f = x3 x2 x1 + x3 x2 x1 + x3 x2 x1 = x3 x2 x1 + x3 x2

1

0

1

0

g = x4 x3 x2 + x4 x3 x2 + x4 x3 x2 = x4 x3 x2 + x4 x3

1

1

0

0

fg = x4 x3 x2 x1 + x4 x3 x2

Fig. S 4.2.20

= x4 x3 x2 x1 + x4 x3 x2 x1 + x4 x3 x2 x1 h = Sm(3, 12, 13) www.gatehelp.com

Page 213


UNIT 4

Digital Electronics

22. (D) f1 = Sm(0, 2, 4, 6),

29. (A) The equation of Z1 is the equation of sum of A

f2 = Sm(1, 3, 5, 7), f1 f2 = 0

and B with carry and equation of 2 is the resultant carry. Thus, it is a full adder.

23. (D) Z = D( ABC + ABC + ABC + ABC + ABC) = D( AB ( C + C) + BC( A + A) + ABC )

30. (B) f1 = CD + CB = CB ,

= D( AB + BC + ABC ) = D( B ( A + AC ) + BC)

f = f 1 + f1 A = CB + CBA = CB + A

= D( BA + BC + BC) = D( B

u C + AB )

24. (A) X = Sm(3, 5, 6, 7),

S = F1

= C + B + A = ABC

X = AB + BC + CA

31. (D) Z = ABC + AB + AB + AB = A ( BC + B) + A( B + B) = A ( B + C) + A = A + B + C

25. (D) Y = Sm(1, 3, 5, 7),

Y =C 32. (A) Z = XY + XY + XY ,

Z=X +Y

BC 00 A

01

11

00

33. (A) Z = ABC + ABC + ABC + ABC

10

= AC + AC = A Å C

1

01

1

1

1

34. (A) The output from the upper first level and from the lower first level

multiplexer is fa multiplexer is fb

Fig. S4.2.24

fa = wx + wx, 26. (D) D0 = A1 A0 , D1 = A1 A0 ,

fb = wx + wx = x

f = fa yz + fb yz + yz = ( wx + wx) yz + xyz + yz

D2 = A1 A0 ,

= wxyz + wx yz + xy + yz BC 00 A

01

11

00

1

1

01

1

1

10

35. (D) Output is 1 when even parity BA 00 00 C

Fig. S4.2.25

01

11

1

10

1 1

01

1

D3 = A1 A0 Fig. S4.2.35

For first decoder A0 = x , A1 = y, D2 = yx , D3 = xy For second decoder A1 = D2 D3 = yxxy = 0,

A0 = z

Therefore Y = A Å B Å C

f = D0 + D1 = A1 A0 + A1 A0 = A1 = 1

36. (B) I1 = y + z ,

27. (D) The output of first MUX is Z o = ab + ab = ( a Å b)

28. (A) Z 2 = bS0 + cS0 = b( ab + ab ) + c( ab + ab ) = ab + ab c + abc

z

yz 00

01

11

10

00

0

0

0

0

I0 = 0

01

1

1

1

0

I1 = y + z

11

1

0

1

0

10

1

0

0

1

This is input to select S0 of both second-level MUX Z1 = CS0 + CS0 = C Å S0 = a Å b Å c

I3 = y

S1 S0 bb wx

= a( b + b c) + abc = ab + ac + abc = ab + ac + bc

Fig. S 4.2.36

Page 214

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I 3 = yz + yz = y u z I2 = z



Chap 4.2

37. (C) Let z = 0, Then

42. (A) f1 = ab c + abc + bc = ac + ab

f = w x y + w xy + wx y + wxy = w x + w y

f2 = ab c + ab + abc = ac + b c

If we put z = 0 in given option then

f3 = ab c + abc + ac = ab + bc

(A) = w x + xy

(B) = wy + w x y

(C) = w x + wy

Since MUX is enable so option (C) is correct.

Thus P1 = ab, P2 = ac,

P3 = bc, P4 = bc, P5 = ab

43. (B) Let P = 1001 and Q = 1010 then

38. (C)f = x0 x2 + x0 x1 x2 + x0 x2 = x0 x2 (1 + x1 ) + x0 x2

Yn = Pn Å Qn Å Rn ,

= x0 x2 + x0 x2

Z n = Rn Qn + Pn Rn + Qn Pn

output is 1111 which is 2’s complement of –1. So it gives

39. (B) f2 = x0 x1 + x1 x2 + x0 x1 x2

P - Q . Let another example

= x0 x1 x2 + x0 x1 x2 + x1 x2 x0 + x1 x2 x0 + x0 x1 x2

then output is 00111. It gives P–Q.

P = 1101 and Q = 0110

So (B) is correct.

= x2 x1 x0 + x2 x1 x0 + x2 x1 x0 + x2 x1 x0 f2 ( x2 , x1 , x0 ) = Sm(1, 2, 6, 7)

Pn

Qn

Rn

Zn

Yn

40. (C) f3 = x0 x1 + x1 x2

n =1

1

0

0

0

1

= x2 x1 x0 + x2 x1 x0 + x2 x1 x0 + x2 x1 x0

n =2

0

1

0

1

1

f3( x2 , x1 , x0 ) = Sm(0, 4, 6, 7)

n=3

0

0

1

1

1

f3( x2 , x1 , x0 ) = PM(1, 2, 3, 5)

n=4

1

1

1

1

1 1

41. (A) Fig. S4.2.43a

Let X 3 X 2 X1 X 0 be 1001 then Y3Y2 Y1 Y0 will be 1111. Let X 3 X 2 X1 X 0 be 1000 then Y3Y2 Y1 Y0 will be 1110 Let X 3 X 2 X1 X 0 be 0110 then Y3Y2 Y1 Y0 will be 1100 bc 00

01

00 a

1

01

Pn

Qn

Rn

Zn

Yn

n =1

1

0

0

0

1

11

10

n =2

0

1

0

1

1

1

1

n=3

1

1

1

1

1

n=4

1

0

1

0

0

1

0 Fig. S4.2.43b

Fig. S4.2.42a

44. (D) Let input be

1010

10

output will be

1101

1

Let input be

0110

1

output will be 0100

bc 00

01

11

00 a

01

1

1

This convert gray to Binary code. Fig. S4.2.42b

bc 00 a

00

01

11

1

1

1

01

*******

10

1 Fig. S4.2.42b

So this converts 2–4–2–1 BCD numbers.

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Page 215


Sequential Logic Circuits

1

(B)

Q1

D1

D2

13. The counter shown in fig. P4.3.13 counts from

Y

Q2

Chap 4.3

J

C

X Q1

(C)

Q1

D1

D2

Q1

(D)

Y

Q2

1

D2

A CLR K

Fig.P4.3.13

Q2

Q1

D1

B CLR K

A B C

X

1

J

A

Q2 C CLR K

1

J

B

Y

Q2

X

(A) 0 0 0 to 1 1 1

(B) 1 1 1 to 0 0 0

(C) 1 0 0 to 0 0 0

(D) 0 0 0 to 1 0 0

14. The mod-number of the asynchronous counter Q1

Q2

shown in fig. P4.2.13 is J

Q0

J

Q1

J

Q2

Q3

J

J

Q4

11. The circuit shown in fig. P4.3.11 is K CLR

T

T

Q

K CLR

K CLR

K CLR

K CLR

Q All J.K. input are HIGH

CLK

CLK

A

B

Q

Q

Fig.P4.3.14

Fig.P4.3.11

(A) a MOD–2 counter

(A) 24

(B) 48

(C) 25

(D) 36

(B) a MOD–3 counter 15. The frequency of the pulse at z in the network

(C) generate sequence 00, 10, 01, 00.....

shown in fig. P4.3.15. is

(D) generate sequence 00, 10, 00, 10, 00 ......

160 kHZ

12. The counter shown in fig. P4.3.12 is a

10-Bit Ring Counter

w 4-Bit Parallel Counter

x

Mod-25 Ripple Counter

y

4-Bit Johnson Counter

z

Fig.P4.3.15

J

Q

Q

B

C Q

J

Q

K

Q

J

1

Q

K

1

(B) 160 Hz

(C) 40 Hz

(D) 5 Hz

16. The three-stage Johnson counter as shown in fig.

A K

(A) 10 Hz

P4.2.16 is clocked at a constant frequency of fc from the CLK

starting state of Q2Q1Q0 = 101. The frequency of output Q2Q1Q0 will be

Fig.P4.3.12

(A) MOD–8 up counter

J2

Q2

J1

Q1

J0

Q0

K2

Q2

K1

Q1

K0

Q0

(B) MOD–8 down counter (C) MOD–6 up counter

CLK

(D) MOD–6 down counter

Fig.P4.3.16

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Page 218


UNIT 4

(A)

fc 8

fc 6

21. In the circuit shown in fig. P4.3.21 is PIPO 4-bit

f (D) c 2

input lines are connected to a 4 bit bus. Its output acts

(B)

f (C) c 3

register, which loads at the rising edge of the clock. The as the input to a 16 ´ 4 ROM whose output is floating

17. The counter shown in the fig. P4.3.17 has initially Q2Q1Q0 = 000. The status of Q2Q1Q0 after the first pulse is

J2

Q1

J1

Q2

Digital Electronics

J0

when the enable input E is 0. A partial table of the contents of the ROM is as follows Address

0

2

4

6

8

10

12

Data

0011

1111

0100

1010

1011

1000

0010

The clock to the register is shown below, and the

Q0

data on the bus at time t1 is 0110. K2

Q2

Q1

K1

K0

MSB

Q0

CLK

Fig.P4.3.17

(A) 0 0 1

(B) 0 1 0

(C) 1 0 0

(D) 1 0 1

CLK

A

18. A 4 bit ripple counter and a 4 bit synchronous 1

counter are made by flips flops having a propagation

E

delay of 10 ns each. If the worst case delay in the ripple

ROM

counter and the synchronous counter be R and S respectively, then (A) R = 10 ns, S = 40 ns

(B) R = 40 ns, S = 10 ns

(C) R = 10 ns, S = 30 ns

(D) R = 30 ns, S = 10 ns

19. A 4 bit modulo–6 ripple counter uses JK flip-flop. If the propagation delay of each FF is 50 ns, the maximum clock frequency that can be used is equal to (A) 5 MHz

(B) 10 MHz

(C) 4 MHz

(D) 20 Mhz

t2

Fig. P4.3.21

right-shift, register shown in fig. P4.3.20 is 0 1 1 0. After three clock pulses are applied, the contents of the

CLK

t t1

20. The initial contents of the 4-bit serial-in-parallel-out

shift register will be

CLK

The data on the bus at time t2 is (A) 1 1 1 1

(B) 1 0 1 1

(C) 1 0 0 0

(D) 0 0 1 0

22. A 4-bit right shift register is initialized to value 1000 for (Q3 , Q2 , Q1 , Q0 ). The D input is derived from 0

1

1

0

Q0 , Q2 and Q3 through two XOR gates as shown in fig. P4.2.22. The pattern 1000 will appear at

Fig.P4.3.20

Page 219

(A) 0 0 0 0

(B) 0 1 0 1

(C) 1 1 1 1

(D) 1 0 1 0

D

Q0

Q1

Fig.P4.3.22

www.gatehelp.com

Q2

Q3



Chap 4.3

(A) 3rd pulse

(B) 7th pulse

28. To count from 0 to 1024 the number of required

(C) 6th pulse

(D) 4th pulse

flip-flop is

Statement for Q.23–24: The 8-bit left shift register and D-flip-flop shown in fig. P4.3.22–23 is synchronized with same clock. The

(A) 10

(B) 11

(C) 12

(D) 13

29. Four memory chips of 16 ´ 4 size have their address buses connected together. This system will be of size

b7

b 6 b 5 b4 b 3 b 2 b 1 b0

D

(A) 64 ´ 4

(B) 32 ´ 8

(C) 16 ´ 16

(D) 256 ´ 1

Q

30. The address bus width of a memory of size 1024 ´ 8

CLK

bits is Q

Fig.P4.3.23-24

(A) 10 bits

(B) 13 bits

(C) 8 bits

(D) 18 bits

31. For the circuit of Fig. P4.3.31 consider the

D flip-flop is initially cleared.

statement: 23. The circuit act as

Assertion (A) : The circuit is sequential

(A) Binary to 2’s complement converter

Reason (R) : There is a loop in circuit

(B) Binary to Gray code converter

d

(C) Binary to 1’s complement converter (D) Binary to Excess–3 code converter 24. If initially register contains byte B7, then after 4

c

(B) 72

(C) 7E

(D) 74

b e

z0

clock pulse contents of register will be (A) 73

z1

a b

Fig.P4.3.131

Choose correct option Statement for Q.25–26: A Mealy system produces a 1 output if the input has been 0 for at least two consecutive clocks followed immediately by two or more consecutive 1’s.

(A) Both A and R true and R is the correct explanation of A (B) Both A and R true but R is not a correct explanation on of A (C) A is true but R is false

25. The minimum state for this system is (A) 4

(B) 5

(C) 8

(D) 9

(D) A is false *****************

26. The flip-flop required to implement this system are (A) 2

(B) 3

(C) 4

(D) 5

27. The output of a Mealy system is 1 if there has been a pattern of 11000, otherwise 0. The minimum state for this system is (A) 4

(B) 5

(C) 6

(D) 7

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Page 220


UNIT 4

Digital Electronics

6. (D) Q + = LM + LMQ

SOLUTIONS

= L( M + MQ )

1. (C) Given FF is a negative edge triggered T flip-flop.

= L M + LQ

So at the negative edge of clock Vi FF will invert the

L

M

Q+

0

0

0

2. (A) At first rising edge of clock, D is HIGH. So Q will

0

1

0

be high till 2nd rising edge of clock. At 2nd rising edge,

1

0

1

D is low so Q will be LOW till 3rd rising edge of clock.

1

1

Q1

output if there is 1 at input.

At 3rd rising edge, D is HIGH, so Q will be HIGH till Fig. S4.3.6

4th rising edge. At 4th rising edge D is HIGH so Q will be HIGH till 5th rising. edge. At 5th rising edge, D is LOW, so Q will be LOW till 6th rising edge.

7. (D) Initially

3. (C)

J

Q

Q

1

0

1

Qn + 1

Qn +1

x

Q

S

R

Q+

Clock 1st

1

1

0

1

1

0

0

0

0

1

0

2nd

0

1

1

0

0

1

0

1

1

0

1

3rd

1

1

0

1

1

0

1

0

1

0

1

4th

0

1

1

0

0

1

1

1

0

1

0

5th

1

1

0

1

1

0

Fig. S4.3.3

Fig. S4.3.7

4. (D) Q + = x Å Q

Therefore sequence is 010101.

Q1+ = x1 Å Q0 = x1 0 + x1 0 = x1

8. (A)

Q2+ = x2 Å x1 ,

Q3+ = x3 Å x2 Å x1

Q4+ = x4 Å x3 Å x2 Å x1

5. (C)

A

B

X

Y

1

1

0

1

1

0

0

1

X and Y are fixed at 0 and 1.

So this generate the even parity and check odd parity.

9. (D) Z = XQ + YQ

A

B

S

R

Q

Q+

0

0

1

0

0

1

0

0

1

0

1

1

0

1

0

1

0

0

0

1

0

1

1

0

1

0

0

0

0

0

1

0

0

0

1

1

1

1

1

1

0

´

Comparing from the truth table of J–K FF

1

1

1

1

1

´

Y = J, X = K

X

Y

Z

0

0

Q

0

1

0

1

0

1

1

1

Q1

Fig. S4.3.9

Fig. S4.3.5

Q + = AB + AQ = AB + BQ Page 221

K

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Chap 4.3

14. (A) It is a 5 bit ripple counter. At 11000 the output

10. (C)

of NAND gate is LOW. This will clear all FF. So it is a t0

t2

t1

Mod–24 counter. Note that when 11000 occur, the

t3

CLR input is activated and all FF are immediately

Q1

cleared. So it is a MOD 24 counter not MOD 25.

0 1

15. (D) 10-bit ring counter is a MOD–10, so it divides

D2=Q1

the 160 kHz input by 10. therefore, w = 16 kHz. The Q2

four-bit parallel counter is a MOD–16. Thus, the t0

frequency at x = 1 kHz. The MOD–25 ripple counter

t1

produces a frequency at y = 40 Hz. (1 kHz/25 = 40 Hz).

Fig.S4.3.10

11. (B)

The four-bit Johnson Counter is a MOD-8. This, the

Present State

FF Input

Next State

Q A QB

TA TB

Q +A QB+

0

0

0

1

0

1

0

1

1

1

1

0

1

0

1

0

0

0

1

1

1

1

0

0

frequency at z = 5 Hz. 16. (D) Q0 Q0

Q2 Q2

Q1 Q1

J2 K2

J1 K 1

J0 K0

Q2+ Q1+ Q0+ 1 0 1

Fig. S4.3.11

From table it is clear that it is a MOD–3 counter. 12. (B) It is a down counter because 0 state of previous

0 1

1 0

0 1

0 1 0

1 0

0 1

1 0

1 0 1

0 1

1 0

0 1

0 1 0

1 0

0 1

1 0

1 0 1

FFs change the state of next FF. You may trace the following sequence, let initial state be 0 0 0

Fig. S4.3.16

We see that 1 0 1 repeat after every two cycles, hence frequency will be fc / 2 .

FF C

FF B

FF A

JK C

JK B

JK A

C+ B+ A+

111

111

111

111

J 2K 2 = 1 0

Þ

Q2 = 1,

000

000

110

110

J1 K 1 = 0 0

Þ

Q1 = 1,

000

110

111

101

J 0K 0 = 0 0

Þ

Q0 = 0

000

001

110

100

18. (B) In ripple counter delay 4Td = 40 ns.

111

111

111

011

The synchronous counter are clocked simultaneously,

001

000

110

010

then its worst delay will be equal to 10 ns.

001

110

111

001

19. (A) 4 bit uses 4 FF

000

001

110

000

Total delay Ntd = 4 ´ 50 ns = 200 ´ 10 -9 1 f = = 5 Mhz 200 ´ 10 -9

Fig. S4.3.12

13. (C) It is a down counter because the inverted FF

17. (C) At first cycle

output drive the clock inputs. The NAND gate will clear

20. (D) At pulse 1 input,

FFs A and B when the count tries to recycle to 111. This

So contents are 1 0 1 1,

will produce as result of 100. Thus the counting

At pules 2 input 1 Å 1 = 0‘

sequence will be 100, 011, 010, 001, 000, 100 etc.

So contents are 0 1 0 1,

1 Å 0 =1

At pules 3 input 0 Å 1 = 1, contents are 1 0 1 0

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Page 222


CHAPTER

4.4 DIGITAL LOGIC FAMILIES


4. In the circuit shown in fig. P.4.4.4. the output Z is +5 V

Consider the DL circuit of fig. P4.4.1–2.

+5 V

+5 V +5 V Z

A B

+

C

Fig. P4.4.4

+

V1

+ Vo -

V2 -

-

Fig. P4.4.1-2

1. For positive logic the circuit is a

(A) AB + C

(B) ABC

(C) ABC

(D) ABC


(A) AND

(B) OR

(C) NAND

(D) NOR

Consider the AND circuit shown in fig. P4.4.5–7. The binary input levels are V(0) = 0 V and V(1) = 25 V. Assume ideal diodes. If V1 = V (0) and V2 = V (1), then Vo

2. For negative logic the circuit is a

is to be at 5 V. However, if V1 = V2 = V (1), then Vo is to

(A) AND

(B) OR

(C) NAND

(D) NOR

rise above 5 V.

Vss

3. The diode logic circuit of fig. P4.4.3 is a

20 kW 1 kW

D2

D1

V1

V1

Vo

V2

V2

Vo

1 kW

D1

D2

D0 +5 V

Fig. P4.4.5-7 Fig. P4.4.3

5. If Vss = 20 V

and V1 = V2 = V (1), the diode current

I D1 , I D2 , and I D0 are

Page 224

(A) AND

(B) OR

(A) 1 mA, 1 mA, 4 mA

(B) 1 mA, 1 mA, 5 mA

(C) NAND

(D) NOR

(C) 5 mA, 5mA, 1 mA

(D) 0,

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0,

0


Digital Logic Families

Chap 4.4

6. If Vss = 40 V and both input are at HIGH level then,

+VCC

diode current I D1 , I D2 and I D0 are respectively (A) 0.4 mA, 0.4 mA, 0

RC

(B) 0, 0, 1 mA

(C) 0.4 mA, 0.4 mA, 1 mA

Vo RB1

(D) 0, 0, 0

RB2

V1

V2 Q2

Q1

7. The maximum value of Vss which may be used is (A) 30 V

(B) 25 V

(C) 125 V

(D) 20 V

Fig. P4.4.10-11

10. For positive logic the gate is

8. The ideal inverter in fig. P4.4.8 has a reference

(A) AND

(B) OR

voltage of 2.5 V. The forward voltage of the diode is 0.75

(C) NAND

(D) NOR

V. The maximum number of diode logic circuit, that may be cascaded ahead of the inverter without producing logic error, is +5 V

+5 V

11. For negative logic the gate is (A) AND

(B) OR

(C) NAND

(D) NOR

+5 V

Statement for Q.12–13: Consider the RTL circuit of fig. P4.4.12–13.

+5 V A

Z

+VCC

B C

RC

D

n Stages of Diode Logic

RB1

RB2

V1

Fig. P4.4.8

(B) 4

(C) 5

(D) 9

Vo2

RB3

V2 Q2

Q1

(A) 3

Vo1

Q3

Fig. P4.4.12-13

9. Consider the TTL circuit in fig. P4.4.9. The value of

12. If Vo1 is taken as the output, then circuit is a

V H and VL are respectively

(A) AND

(B) OR

(C) NAND

(D) NOR

+5 V

13. If Vo2 is taken as output, then circuit is a 2 kW

4 kW

Vo

(A) AND

(B) OR

(C) NAND

(D) NOR

Vi

Statement for Q.14–15: Consider the TTL circuit of fig. P4.4.14. If either or

Fig. P4.4.9

(A) 5 V, 0 V

(B) 4.8 V, 0 V

(C) 4.8 V, 0.2 V

(D) 5 V, 0.2 V

both V1

and V2

are logic LOW, Q1

saturation.

+VCC

Statement Q.10–11:

R1

R2

Consider the resistor transistor logic gate of fig.

Vo2

R3 V1 V2

P4.4.10-11.

is driven to

Q2

Q1 Vo1

Fig. P4.4.14-15

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Page 225



22. The circuit shown in fig. P4.4.22 is

25. The circuit shown in fig. P4.4.25. implements the function

+VDD

M2

B

M1

+VDD

C

Y A

Chap 4.4

A

B

D Y

Fig. P4.4.22

(A) NAND

(B) NOR

(C) AND

(D) OR

Fig. P4.4.25

+VDD

Y M2

(A) ( A + B) C + D

(B) ( AB + C) D

(C) ( A + B) C + D

(D) ( AB + C) D

26. Consider the CMOS circuit shown in fig. P4.4.26.

M3

M1

D

C

23. The circuit shown in fig. P4.4.23 acts as a

A

B

A

The output Y is

+VDD

B C

A

Fig. P4.4.23 B

(A) NAND

Y

(B) NOR C

(C) AND

(D) OR B

A

24. The circuit shown in fig. P4.4.24 implements the function

Fig. P4.4.26 +VDD A

B

C

A

B

C

+VDD

(A) ( A + C) B

(B) ( A + B) C

(C) AB + C

(D) AB + C

27. The CMOS circuit shown in fig. P4.4.27 implement Y

A

A

B

B

C

C

+5 V Logic Input A to E

PMOS Network Y

A

B

C

D

Fig. P4.4.24 E

(A) ABC + ABC

(B ABC + ( A + B + C)

(C) ABC + ( A + B + C)


Fig. P4.4.27

(A) AB + CD + E

(B) ( A + B)( C + D) E

(C) AB + CD + E

(D) ( A + B)( C + D) E

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Page 227


UNIT 4

V1

V2

Digital Electronics

For n attached gate I o = nI B ( sat ) .

Vo

To assure no logic error Vo = VCC - I o RC > V H = 35 . V V - 35 . 5 - 35 . n £ CC = = 15.6 Þ n £ 15 RC I B ( sat ) 640(0.15m)

Actual

Logic

Actual

Logic

Actual

Logic

VH

1

VH

1

VCE ( sat )

0

VL

0

VL

0

VCE

1

VH

1

VL

0

VCE ( sat )

0

19. (A) Let V1 = V2 = 0 V, then M 3 will be ON, M1 and

VL

0

VH

1

VCE ( sat )

0

M 2 OFF and M 4 ON, hence Vo = -VDD . Let V1 = 0 V and V2 = -VDD then M 3 will be ON, M1 OFF M 4 OFF, M 2

13. (B) The Q3 stage is simply an inverter (a NOT gate).

ON, hence Vo = -VDD. Let V1 = -VDD and V2 = 0 V, then

Thus output Vo2

M 3 OFF, M 4 ON, M 2 OFF hence Vo = -VDD. Finally if

is the logic complement of Vo1 .

Therefore this is a OR gate.

V1 = V2 = -VDD, M 3 and M 4 will be OFF and M1 , M 2

14. (A) When Q1

satisfies the function of a negative NAND gate.

will be ON, hence Vo = 0 V. Thus the given CMOS gate

is saturated, Vo1 is logic LOW

otherwise Vo1 is logic HIGH. The following truth table shows AND logic

20. (C) If V A = -VDD then M1

is ON and VY = 0 V. If

VB = VC = -VDD and V A = 0 V then M 3 and M 2 are ON

V1

V2

Vo1

but M1 is OFF hence VY = 0 V. If V A = 0 V and either or

1

1

1

both VB , VC are 0 V then M1 is OFF and either or both

0

1

0

1

0

0

0

0

0

M2

and M 3

will be OFF, which implies no current

flowing through M 4

hence VY = -VDD . Thus given

circuit satisfies the logic equation A + BC . 21. (A) Let V1 = V2 = 0 V = V (0) then M 4 and M 3 will be OFF hence Vo = VDD = V (1). Let

15. (C) The Q2 stage is simply an inverter. Thus output

ON and M 2 , M1

Vo2 is the logic complement of Vo1 .

V1 = 0 V, V2 = VDD then M 4 and M 2 will be ON but M 3 and

16. (C) If V1 = V2 £ VL , Vo1 » VCC . If V1 ( V2 ) > V H , while V2 ( V1 ) £ VL , Q1 (Q2 )

is ON and Q2 (Q1 )

is OFF and

Vo1 » VCC . If V1 = V2 > V H , both Q1 and Q2 are ON and Vo1 » 2 VCE ( sat ) . The truth table shows NAND logic

M1

will

be

OFF

hence

Vo = 0 = V (0).

Let

V1 = VDD , V2 = 0 V , then M 4 and M 3 will be OFF and M1 ON hence Vo = 0 V = V (0). Finally if V1 = V2 = VDD , M1

and M 2

will be ON but M 4 will be OFF hence

Vo = 0 V = V (0). Thus the given CMOS satisfy the function of a positive NOR gate.

V1

V2

Vo

Actual

Logic

Actual

Logic

Actual

Logic

VL

0

VL

0

VCC

1

VH

1

VL

0

VCC

1

VL

0

VH

1

VCC

1

VH

1

VH

1

2VCE ( sat )

0

22. (A) If either one or both the inputs are V(0) = 0 V the corresponding FET will be OFF, the voltage across the load FET will be 0 V, hence the output is VDD . If boths inputs are V (1) = VDD , both M1 and M 2 are ON and the output is V(0) = 0 V. It satisfy NAND gate. 23. (B) If both the inputs are at V(0) = 0 V,

17. (A) The Q3 stage is simple an inverter. Hence AND

transistor M1

and M 2 are OFF, hence the output is

logic.

V (1) = VDD . If either one or both of the inputs are at

18. (C) For each successive gate, that has a transistor in

output will be V (0) = 0 V. Hence it is a NOR gate.

V (1) = VDD , the corresponding FET will be ON and the

saturation, the current required is I B ( sat ) = Page 230

the

I C ( sat ) VCC - VCE ( sat ) 5 - 0. 2 = = = 0.15 mA b bRC 50( 640)

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Chap 4.4

24. (B) If all inputs A, B and C are HIGH, then input to

30. (A) When an output is LOW, it may be as high as

invertor is LOW and output Y is HIGH. If all inputs are

VOL ( max ) = 0.4 V. The maximum voltage that an input will

LOW, then input to inverter is also LOW and output Y

respond to as a LOW is V IL ( max ) = 0.8 V. A positive noise

is HIGH. In all other case the input to inverter is HIGH

spike can drive the actual voltage above the 0.8 V level

and output Y is LOW.

if its amplitude is greater than V NL = V IL ( max ) - VOL ( max ) = 0.8 - 0.4 = 0.4 V

Y = ABC + ABC = ABC + ( A + B + C)

Hence

31. (B) A positive noise spike can drive the voltage

25. (C) The operation of circuit is given below ABCD

PA

PB

´´´ 1

´

´

´ ´0 0

´

´

0010

PC

PD

above 1.0 V level if the amplitude is greater than Y

NA NB NC ND

´ OFF

V NL = V IL ( max ) - VOL ( max ) = 1 - 0.1 = 0.9 V, A negative noise spike can drive the voltage below 3.5 V

´

´

´

´

´

OFF OFF

HIGH

ON ON OFF ON

OFF OFF ON OFF

HIGH

0110

ON OFF OFF ON

OFF ON ON OFF

LOW

32. (B) V IH ( min ) = VOH ( min ) - V NH = - 0.8 - 0.5 = - 1.3 V

1010

OFF ON OFF ON

ON OFF ON OFF

LOW

V IL ( max ) = VOL ( max ) + V NL = 0.5 + ( -2) = -15 . V

1110

OFF OFF OFF ON

ON ON ON OFF

LOW

ON

ON

ON

LOW

if the amplitude is greater than V NH = VOH ( min ) - V IH ( min ) = 4.9 - 35 . = 1.4 V

33. (C) V NH = VOH ( min ) - V IH ( min ) , V NL = V IL ( max ) - VOL ( max ) V NH = 2.7 (for LS) -2.0 (for ALS) = 0.7 V

Y = ( A + B)C + D

26. (B) The operation of this circuit is given below :

V NL = 0.8 (for ALS) -0.5 (for LS) = 0.3 V 34. (B) V NH = 2.5 (for ALS) - 2.0 (for LS) = 0.5 V

A B C

PA

PB

PC

N A NB NC

´

´ 0

´

´

ON

´

0

0 1

´

1 1

1 ´

1

´

Y

V NL = 0.8 (for LS) - 0.4 (for ALS) = 0.4 V

OFF

HIGH

ON ON OFF

OFF OFF ON

HIGH

´

´

OFF OFF

OFF ´

OFF

ON

ON

LOW

´

ON

LOW

ON

35. (D) V NH ( min ) = 0.5 V ,

V NL ( min ) = 0.3 V

36. (B) fanout (LOW) =

I OL ( max ) 8m = = 80 I IL ( max ) 0.1m

fanout (HIGH) =

Y = ( A + B) C

I OH ( max ) 400m = = 20 I IH ( max ) 20m

The fanout is chosen the smaller of the two. 27. (B) If input E is LOW, output will not be LOW. It 37. (B) In HIGH state the loading on the output of gate

must be HIGH. Option (B) satisfy this condition.

1 is equivalent to six 74LS input load. 28. (A) In this circuit parallel combination are OR gate and series combination are AND gate.

Hence load = 6 ´ I IH = 6 ´ 20m = 120 mA 38. (C) The NAND gate represent only a single input

Hence Y = ( A + B)( C + D)( E + F )

load in the LOW state. Hence only five loads in the

29. (A) When an output is HIGH, it may be as low as VOH ( min ) = 2.4 V. The minimum voltage that an input will

LOW state. load = 5I IL = 5 ´ 0.4 = 2 mA

respond to as a HIGH is V IH ( min ) = 2.0 V. A negative noise spike that can drive the actual voltage below 2.0 V if its amplitude is greater than

*******

V NH = VOH ( min ) - V IH ( min ) = 2.4 - 2.0 = 0.4 V

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Page 231


CHAPTER

4.6 MICROPROCESSOR

1. After an arithmetic operation, the flag register of 8085 mP has the following contents D7

D6

D5

D4

D3

D2

D1

D0

1

0

´

1

´

0

´

1

HLT DSPLY : XRA OUT HLT

A PORT1

The output at PORT1 is (A) 00

(B) FEH

(C) 01H

(D) 11H

The contents of accumulator after operation may be (A) 75

(B) 6C

5. Consider the following 8085 assembly program

(C) DB

(D) B6

MVI A, DATA1 MOV B, A SUI 51H JC DLT MOV A, B SUI 82H JC DSPLY DLT : XRA A OUT PORT1 HLT DSPLY : MOV A, B OUT PORT2 HLT

2. In an 8085 microprocessor, the instruction CMP B has been executed while the contents of accumulator is less than that of register B. As a result carry flag and zero flag will be respectively (A) set, reset

(B) reset, set

(C) reset, reset

(D) set, set

3. Consider the following 8085 instruction MVI MVI ADD ORA

This program will display

A, A9H B, 57H B A

(A) the bytes from 51H to 82H at PORT2 (B) 00H AT PORT1

The flag status (S, Z, CY) after the instruction ORA A is executed, is

(C) all byte at PORT1 (D) the bytes from 52H to 81H at PORT 2

(A) (0, 1, 1)

(B) (0, 1, 0)

(C) (1, 0, 0)

(D) (1, 0, 1)

6. It is desired to mask is the high order bits ( D7 - D4 ) of the data bytes in register C. Consider the following set

4. Consider the following set of 8085 instructions MVI ADI JC OUT

A, 8EH 73H DSPLY PORT1

of instruction (a)

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MOV ANI MOV HLT

A, C F0H C, A Page 239


UNIT 4

(b)

(c)

(d)

Digital Electronics

MOV MVI ANA MOV HLT

A, C B, F0H B C, A

(C) 8529H are complemented and stored at location 529H

MOV MVI ANA MOV HLT

A, C B, 0FH B C, A

10. Consider the sequence of 8085 instruction

MOV ANI MOV HLT

A, C 0FH C, A

(D) 5829H are complemented and stored at location 85892H

MVI ADI MOV HLT

A, 5EH A2H C, A

The initial contents of resistor and flag are as follows

The instruction set, which execute the desired A ´´

operation are (A) a and b

(B) c and d

(C) only a

(D) only d

S 0

CY 0

After execution of the instructions the contents of

A

A B, 4AH 4FH B

The contents of register A and B are respectively

C

S

Z

CY

(A) 10H

10H

0

0

1

(B) 10H

10H

1

0

0

(C) 00H

00H

1

1

0

(D) 00H

00H

0

1

1

(A) 05, 4A

(B) 4F, 00

11. It is desired to multiply the number 0AH by 0BH

(C) B1, 4A


and store the result in the accumulator. The numbers

8. Consider the following 8085 assembly program : MVI MOV MOV MVI OUT HLT

B, 89H A, B C, A D, 37H PORT1

are available in register B and C respectively. A part of the 8085 program for this purpose is given below : A, 00H MVI LOOP : ----------------------------------------------------------------------HLT END

The output at PORT1 is (A) 89

(B) 37

(C) 00


The sequence of instruction to complete the program would be

9. Consider the sequence of 8085 instruction given

(A)

JNZ ADD DCR

LOOP B C

(B)

ADD JNZ DCR

B LOOP C

(C)

DCR JNZ ADD

C LOOP B

(D)

ADD DCR JNZ

B C LOOP

below LXI MOV CMA MOV

H, 9258H A, M M, A

By this sequence of instruction the contents of memory location (A) 9258H are moved to the accumulator (B) 9258H are compared with the contents of the accumulator Page 240

Z 0

register and flags are

7. Consider the following 8085 instruction XRA MVI SUI ANA HLT

C ´´

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Microprocessor

Chap 4.6

12. Consider the following assembly language program:

(A) A7H

(B) 98H

MVI MOV START : JMP MVI XRA OUT HLT NEXT : XRA JP OUT HLT

(C) 47H


B, 87H A, B NEXT B, 00H B PORT1

15. The memory requirement for this program is

B START1 PORT2

16. The instruction, that does not clear the accumulator

(A) 20 Byte

(B) 21 Byte

(C) 23 Byte

(D) 18 Byte

of 8085, is

The execution of the above program in an 8085 will result in

(A) XRA A

(B) ANI 00H

(C) MVI A, 00H


(A) an output of 87H at PORT1

17. The contents of some memory location of an 8085 mP

(B) an output of 87H at PORT2

based system are shown

(C) infinite looping of the program execution with accumulator data remaining at 00H

Address Hex.

Contents (Hex.)

(D) infinite looping of the program execution with accumulator data alternating between 00H and 87H.

3000

02

3001

30

13. Consider the following 8085 program

3002

00

MVI ORA JM OUT CMA DSPLY : ADI OUT HLT

3003

30

A, DATA1 A, DSPLY PORT1 01H PORT1

Fig. P4.6.17

The program is as follows

If DATA1 = A7H, the output at PORT1 is (A) 47H

(B) 58H

(C) 00


LHLD MOV INX MOV LDAX MOV INX LDAX MOV

Statement for Q.14–15: Consider the following program of 8085 assembly language: LXI LDA MOV LDA CMP JZ JC MOV JMP MOV FNSH : HLT

3000H E, M H D, M D L, A D D H, A

The contents if HL pair after the execution of the

H 4A02H 4A00H B, A 4A01H B FNSH GRT M, A FNSH M, B

program will be (A) 0030 H

(B) 3000 H

(C) 3002 H

(D) 0230H

18. Consider the following loop

14. If the contents of memory location 4A00H, 4A01H and 4A02H, are respectively A7H, 98H and 47H, then

XRA LXI LOOP : DCX JNZ

A B, 0007H B LOOP

This loop will be executed

after the execution of program contents of memory

(A) 1 times

(B) 8 times

location 4A02H will be respectively

(C) 7 times

(D) infinite times

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Page 241


UNIT 4

19. Consider the following loop LXI LOOP : DCX MOV ORA JNZ

24. Consider the following program MVI RRC RRC

H, 000AH B A, B C LOOP

(A) 1 time

(B) 10 times

(C) 11 times

(D) infinite times

execution of program will be (A) 08H

(B) 8CH

(C) 12H


25. Consider the following program

20. The contents of accumulator after the execution of following instruction will be A, A7H A

(A) CFH

(B) 4FH

(C) 4EH

(D) CEH

RJCT :

21. The contents of accumulator after the execution of following instructions will be MVI ORA RAL

A, BYTE1

If BYTE1 = 32H, the contents of A after the

This loop will be executed

MVI ORA RLC

Digital Electronics

MVI MVI MVI CMP JC CMP JNC OUT HLT SUB A OUT HLT

PORT1

If the following sequence of byte is loaded in accumulator,

A, B7H A

DATA (H)

(A) 6EH

(B) 6FH

(C) EEH

(D) EFH

58

64

73

C8

FA

(A) 00, 00, 73, B4, 00, FA (B) 58, 64, 00, 00, C8, FA (C) 58, 00, 00, 00, C8, FA

of the following program will be

(D) 00, 64, 73, B4, 00, FA

A, C5H A

26. Consider the following instruction to be executed by a 8085 mp. The input port has an address of 01H and has a data 05H to input:

(A) 45H

(B) C5H

(C) C4H


IN ANI

23. Consider the following set of instruction MVI RLC MOV RLC RLC ADD

B4

then sequence of output will be

22. The contents of the accumulator after the execution

MVI ORA RAL

01H 80H

After execution of the two instruction the contents

A, BYTE1

of flag register are

B, A

(A)

1

0

´

1

´

1

´

0

B

(B)

0

1

´

0

´

1

´

0

(C)

0

1

´

1

´

1

´

0

(D)

0

1

´

1

´

0

´

0

If BYTE1 = 07H, then content of A, after the execution of program will be

Page 242

A, DATA B, 64H C, C8H B RJCT C RJCT PORT1

(A) 46H

(B) 70H

(C) 38H

(D) 68H www.gatehelp.com


Microprocessor

ORA JM

A DSPLY

OUT DSPLY : CMA ADI OUT HLT

PORT1 01H PORT1

;Set flag ;If negative jump to ;DSPLY ;A ® PORT1 ;Complement A ;A+1 ® A ;A ® PORT1

Chap 4.6

18. (A) The instruction XRA will set the Z flag. LXI and DCX does not alter the flag. Hence this loop will be executed 1 times. 19. (B) LXI LOOP : DCX

This program displays the absolute value of DATA1. If

MOV ORA JNZ

DATA1 is negative, it determine the 2’s complements and display at PORT1.

B, 000AH B A, B C LOOP

;00 ® C, 0AH ® B ; CB – 1 ® B, ;flag not affected ;B ® A ;A OR C ® A, set flag

Hence this loop will be executed 0AH or ten times. 14. (A)

LXI

H, 4A02H

LDA

4A00H

MOV LDA

B, A 4A01H

CMA JZ

B FNSH

JC

GRT

MOV JMP GRT : MOV FNSH : HLT

M, A FNSH M, B

;Store destination address ;in HL pair ;Load A with contents of ;memory location A00H ;A ® B ;Load A with contents of ;memory location 4A01H ;Compare A and B ;Jump to FNSH if two ;number are equal ;If CY = 1, (A
MVI ORA RLC

A, B7H A

;B7H ® A ;Set Flags, CY = 1 ;Rotate accumulator left

The contents of bit D7 are placed in bit D0 . Accumulator Before RLC

10100111

After RLC

01001111

21. (A) RAL instruction rotate the accumulator left through carry. D7 ® CY ,

CY ® D0 ,

ORA reset the carry. Accumulator

This program find the larger of the two number stored in location 4A00H and 4A01H and store it in memory location 4A002. A7H > 98H

20. (B)

Thus A7H will be stored at 4A02H.

15. (C) Operand R, M or implied : 1–Byte instruction Operand 8–bit

: 2–Byte instruction

Operand 16–bit

: 3–Byte instruction

CY

Before RAL

10110111

0

After RAL

01101110

1

22. (A) RRC instruction rotate the accumulator right and D0 is placed in D7 . MVI ORA RAL

3–Byte instruction are: LXI, LDA, JZ, JC, JMP

A, C5H A

RRC

;C5H ® A ;Reset Carry flag ;Rotate A left through ;carry, A = 8AH ;Rotate A right, A = 45H

P–Byte instruction are : MOV, CMP, HLT Hence memory = 3 ´ 6 + 1 ´ 5 = 23

Byte.

23. (A) This program multiply BYTE1 by 10. Hence content of A will be 46H.

16. (D) All instruction clear the accumulator

17. (C)

XRA ANI MVI

A 00H A

LHLD MOV INX MOV LDAX MOV INX LDAX MOV

3000H E, M H D, M D L, A D D H, A

;A Å A ;A AND 00 ;00 ® A ;(3000A) ® HL = 3002H ;(3002H) ® E = 00 ;HL +1 ® HL = 3003H ;M ® D=(3003H) = 30H ;(DE) ® A=(3000H) = 02H ;A ® L = 02H ;DE +1 ® DE = 3001H ;(DE) ® A = (3001) = 30H ;A ® H = 30H

Hence HL pair contain 3002H.

07H = 0710 ,7 ´ 10 = 70, 7010 = 46H 24. (B) Contents of Accumulator A = 0011 0010 After First RRC

= 0001 1001

After second RRC

= 1000 1100

25. (D) This program will display the number between 64H to C8H including 64H. C8H will not be displayed. Thus (D) is correct option. 26. (C) 05H AND 80H =00

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Page 245


UNIT 4

After the ANI instruction S, Z and P are modified to reflect the result of operation. CY is reset and AC is set . Thus, S = 0, Z = 1, AC = 1, P =1, CY = 0 27. (B)

ACI

;A + 56H + CY ® A

56H

37H + 56H + 1 =8EH 28. (C) Instruction load the register pairs HL with 01FFH. SHLD instruction store the contents of L in the memory location 2050H and content of H in the memory location 2051H. Contents of HL are not altered. 29. (B) At a time 8085 can drive only a digit. In a second each digit is refreshed 500 times. Thus time given to 1 each digit = = 0.4 ms. (5 ´ 500) 30. (C) The stack pointer register SP point to the upper memory location of stack. When data is pushed on stack, it stores above this memory location. 31. (B) Line 5 push the content of HL register pair on stack. The contents of L will go to 03FFH and contents of H will go to 03FEH. Hence memory location 03FEH contain 22H. 32. (C) Contents of register pair B lie on the top of stack when POP H is executed, HL pair will be loaded with the contents of register pair B. 33. (C) The instruction PUSH B store the contents of BC at stack. The POP PSW instruction copy the contents of BC in to PSW. The contents of register C will be copied into flag register. D0 = 1 = carry flag,

D6 = 0 = zero flag.

Hence zero flag will be reset and carry will be set. 34. (A)

MVI A DATA1 ORA A JP DSPLY

XRA DSPLY OUT HLT

A PORT1

;DATA1 ® A ; Set flag ;If A is positive, then ;jump to DSPLY ; Clear A ; A ® PORT2

If DATA1 is positive, it will be displayed at port1 otherwise 00. ********************

Page 246

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Digital Electronics


UNIT 5

ì e2 t, t < 0 30. y( t) = u( t) * h( t) , where h( t) = í -3t îe , t >0 1 5 1 (A) e -2 t u( - t - 1) + - e -3t u( -t) 2 6 3

Signal & System

(A)

5 3 1 13 -4 t sin t + cos t + e - t e , t ³ 0 34 34 6 61

(B)

5 3 13 -4 t 1 - t sin t + cos t e + e , t ³ 0 34 34 51 6

(B)

1 2t 5 1 e u( -t - 1) + - e-3t u( -t) 2 6 3

(C)

3 5 13 -4 t 1 - t sin t + cos t e + e , t ³ 0 34 34 51 6

(C)

1 2t 1 e + [5 - 3e 2 t - 2 e -3t ]u( t) 2 6

(D)

3 5 1 13 -4 t sin t + cos t + e -4 t e , t ³ 0 34 34 6 51

(D)

1 2t 1 e + [5 - 3e 2 t - 2 e -3t ]u( -t) 2 6

37.

d 2 y ( t) dy ( t) +6 + 8 y( t) = 2 x( t), dt 2 dt


y (0 - ) = -1,

The impulse response of LTI system is given. Determine the step response.

(A)

2 - t 5 -2 t 5 -4 t e - e + e , t ³0 3 2 6

31. h( t) = e - |t |

(B)

2 5 -2 t 5 -4 t + e + e , t ³ 0 3 2 6

(A) 2 + e t - e - t

(B) e t u( -t + 1) + 2 - e - t

(C) e t u( -t + 1) + [2 - e - t ]u( t)

(D) e t + [2 - e - t - e t ]u( t)

(C) 4 + 5( 3e -2 t + e -4 t ) , t ³ 0 (D) 4 - 5( 3e -2 t + e -4 t ), t ³ 0

32. h( t) = d( 2 ) ( t) (A) 1

(B) u( t)

(C) d( 3) ( t)

(D) d( t)

38.

d 2 y( t) 3dx( t) , + y( t) = dt 2 dt y (0 - ) = -1,

33. h( t) = u( t) - u( t - 4) (A) tu( t) + (1 - t) u( t - 4)

(B) tu( t) + (1 - t) u( t - 4)

(C) 1 + t

(D) (1 + t) u( t)

(A) sin t + 4 cos t - 3te -3t + t, t ³ 0 (B) 4 sin t - cos t - 3te - t , t ³ 0 (D) 4 sin t + cos t - 3te - t , t ³ 0

(A) u( t)

(B) t

(C) 1

(D) tu( t)

39. The raised cosine pulse x( t) is defined as ì 1 ï (cos wt + 1) , x ( t) = í 2 ïî 0,

Statement for Q.35-38: The system described by the differential equations has been specified with initial condition. Determine the output of the system and choose correct option. dy( t) + 10 y( t) = 2 x( t), y(0 - ) = 1, x( t) = u( t) dx

(A) 15 (1 + 4 e -10 t ) u( t)

(B) 15 (1 + 4 e -10 t )

(C) - 15 (1 + 4 e -10 t ) u( t)

(D) - 15 (1 + 4 e -10 t )

-

p p £t£ w w otherwise

The total energy of x ( t) is 3p 3p (B) (A) 4w 8w (C)

3p w

(D)

3p 2w

40. The sinusoidal signal x( t) = 4 cos (200 t + p 6) is

d 2 y( t) dy( t) dx( t) , 36. +5 + 4 y( t) = 2 dt dt dt dy( t) = 1, x( t) = sin t u( t) y (0 - ) = 0, dt 0 -

Page 252

dy( t) = 1, x( t) = 2 te- t u( t) dt 0 -

(C) sin t - 4 cos t + 3te -3t + t, t ³ 0

34. h( t) = y( t)

35.

dy( t) = 1, x( t) = e - t u( t) dt 0 -

passed through a square law device defined by the input output relation y ( t) = x 2 ( t). The DC component in the signal is (A) 3.46

(B) 4

(C) 2.83

(D) 8

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Continuous-Time Systems

41. The impulse response of a system is h( t) = d( t - 0.5). If two such systems are cascaded, the impulse response

Chap 5.1

46. The y( t) = x( t) * h( t) is y(t)

y(t)

of the overall system will be (A) 0.58( t - 0.25)

(B) d( t - 0.25)

(C) d( t - 1)

(D) 0.5 d( t - 1)

a

t

1+a

a

t

y(t)

h(t)

a 1

t

5

3

t

1+a 2

1+a

t

a

1-a

(C)

Fig P5.1.42

The output of the system is zero every where except for the

value of a is (A) 1

(B) 2

(C) 1 < t < 5

(D) 1 < t < 8

(C) 3

(D) 0

S1 : h1 ( t) = e - (1 - 2 j ) t u( t)

48. Consider the signal x( t) = d( t + 2) - d( t - 2).The value of E¥ for the signal y( t) =

S2 : h2 ( t) = e cos 2 t u( t) The stable system is (A) S1

(B) S2

(C) Both S1 and S2

(D) None

ò x( t) dt

is

(A) 4

(B) 2

(C) 1

(D) ¥

49. The

response of a system S to a complex input

x( t) = e j 5t is specified as y( t) = te j 5t . The system

44. The non-invertible system is (B) y( t) =

t

ò

(A) is definitely LTI x( t) dt

(B) is definitely not LTI

-¥

dx( t) dt

t

-¥

-t

(A) y( t) = x( t - 4)

t

47. If dy( t) dt contains only three discontinuities, the

(B) 0 < t < 8

43. Consider the impulse response of two LTI system

1

(D)

(A) 0 < t < 5

(C) y( t) =

1

(B)

y(t)

impulse response h( t) of the system.

1-a

(A)

42. Fig. P5.1.40 show the input x( t) to a LTI system and x(t)

1


(C) may be LTI (D) information is insufficient

45. A continuous-time linear system with input x( t) and output y( t) yields the following input-output pairs: x( t) = e j 2 t Û y( t) = e j 5t

(B) is definitely not LTI

If x1 ( t) = cos (2 t - 1), the corresponding y1 ( t) is (B) e- j cos (5 t - 1) (D) e j cos (5 t - 1)

Suppose that 0 £ t £1 elsewhere

(C) may be LTI (D) information is insufficient. 51. The auto-correlation of the signal x( t) = e - t u( t) is


ì 1, x( t) = í î 0,

response of a system S to a complex input

x( t) = e j8 t is specified as y( t) = cos 8 t. The system (A) is definitely LTI

x( t) = e - j 2 t Û y( t) = e - j 5t (A) cos (5 t - 1) (C) cos 5( t - 1)

50. The

and

ætö h( t) = xç ÷, where 0 < a £ 1. èaø

(A)

1 t 1 e u( -t) + e - t u( t) 2 2

(B)

et 1 - t + ( e - e t ) u( t) 2 2

(C)

1 -t 1 e u( -t) + e - t u( t) 2 2

(D)

1 t 1 e u( -t) - e - t u( t) 2 2

*********************

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Page 253


UNIT 5

Signal & System

11. (C)

SOLUTIONS

x(10t)

x(10t-5)

1

1. (A)

2p = 60 p T

2. (C) T1 =

Þ

T=

p 30

1

-0.5 -0.4

0.4 0.5

t

0.1

0.9

1

t

Fig S5.1.11

12. (D) Multiplication by 5 will bring contraction on

2p 2p æ 2p 2p ö s, T2 = s, LCM ç , ÷ = 2p 5 7 7 ø è 5

time scale. It may be checked by x(5 ´ 0.8) = x( 4).

3. (D) Not periodic because of t. 4. (D) Not periodic because least common multiple is

13. (A) Division by 5 will bring expansion on time scale. æ 20 ö It may be checked by y( t) = xç ÷ = x( 4). è 5 ø

infinite. 5. (C) y( t) is not periodic although sin t and 6 cos 2 pt are independently periodic. The fundamental frequency can’t be determined.

¥

ò | x ( t)|dt < ¥

=

-¥

¥

¥

-¥

0

-4 t -4 t ò e u( t) dt = ò e dt =

¥

7. (A) | x( t)| = 1, E¥ =

ò | x( t)| dt 2

1 4

1 T® ¥ 2T

T

5

-5

4

otherwise

=8+ =¥

-¥

ò | x( t)| dt 2

5

4

5

0

0

4

15. (D) E = 2 ò x 2 ( t) dt = 2 ò (1)1 dt + 2 ò (5 - t) 2 dt 2 26 = 3 3

16. (B) Let x1 ( t) = v( t) then y1 ( t) = u{v( t)}

So this is a power signal not a energy. P¥ = lim

-4

for - 5 < t < - 4 for 4 < t <5

E = ò (1) 2 dt + ò ( -1) 2 dt = 2

6. (C) This is energy signal because E¥ =

ì 1, ï 14. (C) y( t) = í -1, ï 0, î

Let x2 ( t) = kv( t) then y2 ( t) = u{kv( t)} ¹ ky1 ( t)

=1

(Not homogeneous not linear)

-T

8. (D) v( t) is sum of 3 unit step signal starting from, 1, 2, and 3, all signal ends at 4.

y1 ( t) = u{v( t)},

y2 ( t) = u{v( t - to)} = y1 ( t - to)

(Time invariant)

The response at any time depends only on the

9. (A) The function 1 does not describe the given pulse.

excitation at time t = to and not on any future value. (Causal)

It can be shown as follows : u(a-t)

u(t-b)

17. (C) y1 ( t) = v( t - 5) - v( 3 - t)

u(a-t) - u(t-b)

y2 ( t) = kv( t - 5) - kv( 3 - t) = ky1 ( t) t

a

t

b

t

(Homogeneous)

Let x1 ( t) = v( t) then y1 ( t) = v( t - 5) - v( 3 - t) Let x2 ( t) = 2 w( t) then y2 ( t) = w( t - 5) - w( 3 - t)

Fig S5.1.3.9

Let x3( t) = x( t) + w( t) 10. (B)

Then

r(t-4)

r(t-6)

r(t-4) - r(t-6)

2

2

2

y3( t) = v( t - 5) + w( t - 5) - v( 3 - t) - w( 3 - t)

= y1 ( t) + y2 ( t)

(Additive)

Since it is both homogeneous and additive, it is also linear.

4

6

8

t

4

6

8

Fig S5.1.10

Page 254

t

4

6

8

t

y1 ( t) = v( t - 5) - v( 3 - t) y2 ( t) = v( t - to - 5) - v( 3 - t + to) = y1 ( t - to)

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(Time invariant)


Continuous-Time Systems

At

time,

t = 0, y(0) = x( -5) - x( 3).

Therefore

the

Chap 5.1

21. (C) All option are linear. So it is not required

response at time, t = 0 depends on the excitation at a

to check linearity.

later time t = 3.

d y1 ( t) - 8 y1 ( t) = v( t) dt d Let x2 ( t) = v( t - to) then t y2 ( t) - 8 y2 ( t) = v( t - to) dt

(Not causal)

If x( t) is bounded then x( t - 5) and x( 3 - t) are bounded and so is y( t).

(Stable)

ætö ætö 18. (D) y1 ( t) = vç ÷ , y2 ( t) = kvç ÷ = ky1 ( t) è2 ø è2 ø (Homogeneous)

Let x1 ( t) = v( t) then t

The first equation can be written as d ( t - to) y( t - to) - 8 y( t - to) = v( t - to) dt This equation is not satisfied if y2 ( t) = y1 ( t - to) therefore

x3 = v( t) + w( t) then ætö ætö y3( t) = vç ÷ + wç ÷ = y1 ( y) + y2 ( t) è2 ø è2 ø

(Additive)

Since it is both homogeneous and additive, it is also

y2 ( t) ¹ y1 ( t - to)

(Time Variant)

The system can be written as y( t) =

linear

x( l) y( l) dl + 8 ò dl l -¥ -¥ l t

t

ò

So the response at any time, t = to depends on the

ætö æt ö æ t - to ö y1 ( t) = vç ÷ , y2 ç - to ÷ ¹ y( t - to) = vç ÷ 2 2 è ø è ø è 2 ø

excitation at t £ to , and not on any future values.

(Time variant)

(Causal)

At time t = -2, y( -2) = x( -1), therefore, the response at

The Homogeneous solution to the differential equation

time t = -2, depends on the excitation at a later time,

is of the form y( t) = kt8 . If there is no excitation but the

t = -1.

zero excitation, response is not zero. The response will

(Not causal)

It x( t) is bounded then y( t) is bounded.

(Stable)

increases without bound as time increases. (Unstable)

19. (C) y1 ( t) = cos 2 pt v( t) y2 ( t) k cos 2 pt v( t) = ky1 ( t)

(Homogeneous)

22. (C) y1 ( t) =

x3( t) = v( t) + w( t)

t+ 3

ò v(l)dl

-¥

y3( t) = cos 2 pt [ v( t) + w( t)] = y1 ( t) + y2 ( t)

(Additive)

t+ 3

t+ 3

-¥

-¥

ò kv(l)dl = k ò v(l)dl = ky1 ( t)

y2 ( t) =

Since it is both homogeneous and additive. It is also linear.

x3( t) = v( t) + w( t)

y1 ( t) = cos 2 pt v( t)

y3( t) =

y2 ( t) = cos 2 pt ( t - to) ¹ y( t - to) = cos [2p( t - to)]v( t - to)

(Time Variant)

(Homogeneous)

t+ 3

t+ 3

t+ 3

-¥

-¥

-¥

ò [ v( l) + w( l)]dl =

ò v( l)dl +

ò w( l)dl

= y1 (t) + y2 (t)

(Additive)

The response at any time t = to depends only on the excitation at that time and not on the excitation at any

Since it is Homogeneous and additive, it is also linear.

later time.

(Causal)

y1 ( t) =

If x( t) is bounded then y( t) is bounded.

(Stable)

ò v(l)dl

-¥

y2 ( t) =

20. (C) y1 ( t) = |v( t)|, y2 ( t) = |kv( t)| = k y1 ( t)

t+ 3

t+ 3

t - to + 3

-¥

-¥

ò v(l - to )dl =

ò v(l)dl = y (t - t ) o

1

If k is negative k y1 ( t) ¹ ky1 ( t)

(Time invariant)

(Not Homogeneous Not linear). y1 ( t) = |v( t)|, y2 ( t) = | y( t - to)| = y1 ( t - to)

The response at any time, t = to , depends partially on the excitation at time to to < t < ( to + 3) which are in

(Time Invariant) The response at any time t = to depends only on the excitation at that time and not on the excitation at any later time. If x( t) is bounded then y( t) is bounded.

(Causal)

future. If x( t) is a constant k, then y ( t) =

(Not causal) t+ 3

t+ 3

-¥

-¥

ò kdl = k ò dl and as

t ® ¥ , y ( t) increases without bound.

(unstable)

(Stable) www.gatehelp.com

Page 255


UNIT 5

Signal & System

æ pn ö æ pn ö æ pn p ö 14. x[ n] = cos ç + ÷ ÷ - sin ç ÷ + 3 cos ç 3ø è 2 ø è 8 ø è 4

11. x[ n + 2 ] y[ n - 2 ] x[n] 3

(A) periodic with period 16

2

(A)

(B) periodic with period 4

1 -6

-5

-4

-3

-2

-1

0

1

2

3

4

5

6

n

(C) periodic with period 2 (D) Not periodic

x[n] 3

15. x[ n] = 2 e

2

(B)

1 -6

-5

-4

-3

-2

-1

-6

-5

-4

-3

-2

-1

0

1

2

3

4

5

6

1

2

3

4

5

6

n

(A) periodic with 12p

(B) periodic with 12

(C) periodic with 11p

(D) Not periodic

16. The sinusoidal signal has fundamental period n

N = 10 samples. The smallest angular frequency, for which x[ n] is periodic, is 1 (A) rad/cycle 10

-1

(C)

æn ö j çç - p÷÷ è6 ø

-2 -3

(C) 5 rad/cycle -6

-5

-4

-3

-2

-1

1

2

3

4

5

6

n

-1

(D)

(B) 10 rad/cycle (D)

p rad/cycle 5

17. Let x[ n], - 5 £ n £ 3 and h[ n], 2 £ n £ 6 be two finite duration signals. The range of their convolution is

-2 -3

(A) -7 £ n £ 9

(B) -3 £ n £ 9

(C) 2 £ n £ 3

(D) -5 £ n £ 6

Statement for Q.12–15: A discrete-time signal is given. Determine the


period of signal and choose correct option.

x[ n] and h[ n] are given in the question. Compute the convolution y[ n] = x[ n] * h[ n]

pn æ pn 1 ö 12. x[ n] = cos + sin ç + ÷ 9 2ø è 7 (A) periodic with period N = 126 (B) periodic with period N = 32

Page 260

and choose correct

option. 18. x[ n] = {1, 2, 4}, h[ n] = {1, 1, 1, 1, 1} (A) {1, 3, 7, 7, 7, 6, 4} (B) {1, 3, 3, 7, 7, 6, 4}

(C) periodic with period N = 252

(C) {1, 2, 4}

(D) Not periodic

(D) {1, 3, 7}

æ nö æ pn ö 13. x[ n] = cos ç ÷ cos ç ÷ è8ø è 8 ø

19. x[ n] = {1, 2, 3, 4, 5}, h [ n] = {1} (A) {1, 3, 6, 10, 15}

(B) {1, 2, 3, 4, 5}

(A) Periodic with period 16p

(C) {1, 4, 9, 16, 20}

(D) {1, 4, 6, 8, 10}

(B) periodic with period 16( p + 1)

20. x[ n] = {1, 2, -1}, h [ n] = x [ n]

(C) periodic with period 8

(A) {1, 4, 1}

(B) {1, 4, 2, -4, 1}

(D) Not periodic

(C) {1, 2, -1}

(D) {2, 4, -2}

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Discrete-Time Systems

21. x[ n] = {1, -2, 3}, h [ n] = {0, 0, 1, 1, 1, 1}

Chap 5.2

25. x[ n] = {1, 4, -3, 6, 4}, h[ n] = {2, -4, 3}

(A) {2, 4, -19, 36, -25, 2, 12}

(A) {1, -2, 4, 1, 1, 1}

(B) {4, -19, 36, -25}

(B) {0, 0, 3}

(C) {1, 4, -3, 6, 4}

(C) {0, 0, 3, 1, 1, 1, 1}

(D) {1, 4, -3, 6, 4}

(D) {0, 0, 1, -1, 2, 2, 1, 3}

22. x[ n] = {0, 0, 1, 1, 1, 1}, h[ n] = {1, -2, 3}

(A) {0, 0, 1, -1, 2, 2, 1, 3}

ì 1, ï 26. x[ n] = í 2, ï0 î

n = -2, 0, 1 n = -1 elsewhere

h (n) = d[ n ] - d[ n - 1] + d[ n - 4]

(A) d[ n ] - 2 d[ n - 1 ] + 4 d[ n - 4 ] + d[ n - 5 ]

(B) {0, 0, 1, -1, 2, 2, 1, 3}

(B) d[ n + 2 ] + d[ n + 1 ] - d[ n ] + 2 d[ n - 3 ] + d[ n - 4 ] + d[ n - 5 ]

(C) d[ n + 2 ] - d[ n + 1 ] + d[ n ] + 2 d[ n - 3 ] - d[ n - 4 ] + 2 d[ n - 5 ]

(C) {1, -2, 3, 1, 1, 2, 1, 1}

(D) d[ n ] + 2 d[ n - 1 ] + 4 d[ n - 5 ] + d[ n - 5 ]


(D) {1, -2, 3, 1, 1, 1, 1}

In question y[ n] is the convolution of two signal.

23. x[ n] = {1, 1, 0, 1, 1}, h[ n] = {1, -2, - 3, 4}

(A) {1, -1, - 2, 4, 1, 1}

(B) {1, -1, - 2, 4, 1, 1}

(C) {1, -1, - 5, 2, 3, -5, 1, 4}

(D) {1, -1, - 5, 2, 3, -5, 1, 4}

24. x[ n] = {1, 2, 0, 2, 1}, h[ n] = x[ n]

(A) {1, 4, 4, 4, 10, 4, 4, 4, 1}

(B) {1, 4, 4, 4, 10, 4, 4, 4, 1} (C) {1, 4, 4, 10, 4, 4, 4, 1}

(D) {1, 4, 4, 10, 4, 4, 4, 1}

Choose correct option for y[ n]. 27. y[ n] = ( -1) n * 2 n u[2 n + 2 ] (A)

4 6

(B)

4 u[ -n + 2 ] 6

(C)

8 ( -1) n u[ -n + 2 ] 3

(D)

8 ( -1) n 3

28. y[ n] =

1 u[ n] * u[ n + 2 ] 4n

æ1 1 ö (A) ç - n ÷u[ n] è3 4 ø

æ 1 12 ö (B) ç - n ÷u[ n + 2 ] è3 4 ø

æ 4 1 æ 1 ön ö ÷u[ n + 2 ] (C) ç ç 3 12 çè 4 ÷ø ÷ è ø

1 ö æ 16 (D) ç - n ÷u[ n + 2 ] è 3 4 ø

29. y[ n] = 3n u[ -n + 3] * u[ n - 2 ] ì 3n , ïï (A) í 2 ï 83 , ïî 2 ì 3n , ïï (C) í 2 ï 81 , ïî 2

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n £5 n³6 n £5 n³6

ì 3n , ï (B) í 83 ïî 2 , ì 3n , ïï (D) í 6 ï 81 , ïî 2

n £5 n³6

n £5 n³6 Page 261


UNIT 5

30. y[ n] = u[ n + 3] * u[ n - 3]

37. x[ n] as shown in fig. P5.2.37

(A) ( n + 1) u[ n]

(B) nu[ n]

(C) ( n - 1) u[ n]

(D) u[ n]

31.

The

convolution

of

x[ n] = cos ( 2p n) u[ n]

5

x[n]

and

+

10

y[n]

Fig. P5.2.37

h[ n] = u[ n - 1] is f [ n]u[ n - 1]. The function f [ n] is n = 4 m + 1, 4 m + 2 ì 1, (A) í n = 4 m, 4 m + 3 î 0, ì 0, (B) í î 1,

Signal & System

(A) P, Q, R, S

(B) Q, R, S

(C) P, R, S

(D) P, Q, S

n = 4 m + 1, 4 m + 2 n = 4 m, 4 m + 3

ì 1, (C) í î 0,

n = 4 m + 1, 4 m + 3

ì 0, (D) í î 1,

n = 4 m + 1, 4 m + 3 n = 4 m, 4 m + 2

38. x[ n] as shown in fig. P5.2.38 x[n]

n = 4 m, 4 m + 2

+

y[n] +

Fig. P5.2.38

Statement for Q.32–38: Let P be linearity, Q be time invariance, R be causality and S be stability. In question discrete time

(A) P, Q, R, S

(B) P, Q, R

(C) P, Q

(D) Q, R, S

Statement for Q.39–41: and S2

Two discrete time systems S1

input x[ n] and output y[ n] relationship has been given. In the option properties of system has been given. Choose the option which match the properties for system.

fig. P5.2.39–41.

32. y[ n] = rect ( x[ n])

x[n]

(A) P, Q, R

(B) Q, R, S

(C) R, S, P

(D) S , P, Q

S2

S1

y[n]

Fig. P5.2.39–41.

39. Consider the following statements

33. y[ n] = nx[ n] (A) P, Q, R, S

(B) Q, R, S

(a) If S1 and S2 are linear, the S is linear

(C) P, R

(D) Q, S

(b) If S1 and S2 are nonlinear, then S is nonlinear

34. y[ n] =

(c) If S1 and S2 are causal, then S is causal

n +1

å

u[ m ]

(d) If S1 and S2 are time invariant, then S is time invariant

m = -¥

(A) P, Q, R, S

(B) R, S

(C) P, Q

(D) Q, R

35. y[ n] = x[ n] (A) Q, R, S

(B) R, S, P

(C) S, P, Q

(D) P, Q, R

36. x[ n] as shown in fig. P5.2.36 x[n]

2

y[n]

True statements are : (A) a, b, c

(B) b, c, d

(C) a, c, d

(D) All

40. Consider the following statements (a) If S1 and S2 are linear and time invariant, then interchanging their order does not change the system. (b) If S1 and S2 are linear and time varying, then interchanging their order does not change the system. True statement are

Fig. P5.2.36

Page 262

are

connected in cascade to form a new system as shown in

(A) P, Q, R, S

(B) Q, R, S

(C) P, Q

(D) R, S

(A) Both a and b

(B) Only a

(C) Only b

(D) None

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41. Consider the statement

Chap 5.2

(D) Above all

(a) If S1 and S2 are noncausal, the S is non causal (b) If S1 and/or S2 are unstable, the S is unstable.

45. The system shown in fig. P5.2.45 is

(A) Both a and b

y[n]

+

1 4

x[n]

True statement are : (B) Only a

(C) Only b

D

y[n-2]

D

+

+

(D) None

1 4

+

-1 2

42. The following input output pairs have been

Fig. P5.2.45

observed during the operation of a time invariant (A) Stable and causal

system : x1 [ n] = {1, 0, 2}

S ¬¾ ®

y1 [ n] = {0, 1, 2}

(B) Stable but not causal

(C) Causal but unstable

x2 [ n] = {0, 0, 3}

¬¾® S

y2 [ n] = {0, 1, 0, 2}

x3[ n] = {0, 0, 0, 1}

¬¾® S

46. The impulse response of a LTI system is given as

y3[ n] = {1, 2, 1}

(D) unstable and not causal

n

æ 1ö h[ n] = ç - ÷ u[ n]. è 2ø

The conclusion regarding the linearity of the

The step response is

system is (A) System is linear

1 æç æ 1ö 2 -ç- ÷ ç 3è è 2ø

(C)

1 æç æ 1ö 2 + ç- ÷ 3 çè è 2ø

(B) System is not linear (C) One more observation is required.

n +1

(A)

(D) Conclusion cannot be drawn from observation.

n +1

ö ÷u[ n] ÷ ø

(B)

1 æç æ 1ö 2 -ç- ÷ ç 3è è 2ø

ö ÷u[ n] ÷ ø

(D)

1 æç æ 1ö 2 + ç- ÷ 3 çè è 2ø

n

n

ö ÷u[ n] ÷ ø ö ÷u[ n] ÷ ø

43. The following input output pair have been observed

47. The difference equation representation for a system

during the operation of a linear system:

is

x1 [ n] = { -1, 2, 1}

¬¾® S

y1 [ n] = {1, 2, - 1, 0, 1}

x2 [ n] = {1, - 1, - 1}

S ¬¾ ®

y2 [ n] = { - 1, 1, 0, 2}

x3[ n] = {0, 1, 1}

y[ n] -

S ¬¾ ®

(A)

3æ 1ö ç - ÷ u[ n] 2è 2ø

(C)

3æ 1ö ç ÷ u[ n] 2 è2 ø

(A) System is time-invariant

(A) y[ n] = x[ n] + 11 . y[ n - 1] (B) y[ n] = x[ n] -

1 ( y[ n - 1] + y[ n - 2 ]) 2

(D)

2æ1ö ç ÷ u[ n] 3è2 ø

n

y[ n] - 2 y[ n - 1] + y[ n - 2 ] = x[ n] - x[ n - 1] If y[ n] = 0 for n < 0 and x[ n] = d[ n], then y[2 ] will

(C) One more observation is required

44. The stable system is

2æ 1ö ç - ÷ u[ n] 3è 2 ø

48. The difference equation representation for a system is

(B) System is time variant

(D) Conclusion cannot be drawn from observation

n

(B)

n

The conclusion regarding the time invariance of the system is

y [ -1] = 3

The natural response of system is n

y3[ n] = {1, 2, 1}

1 y[ n - 1] = 2 x[ n], 2

be (A) 2

(B) -2

(C) -1

(D) 0

49. Consider a discrete-time system S whose response to a complex exponential input e jpn S : e jpn

(C) y[ n] = x[ n] - (15 . y[ n - 1] + 0.4 y[ n - 2 ]) www.gatehelp.com

2

Þ

e jp3n

2

is specified as

2

Page 263


UNIT 5

24. (B) y[ n] = {1, 4, 4, 10, 4, 4, 4, 1}

Signal & System

28. (C) For n + 2 < 0

for

n + 2 ³0

or

n < - 2, y [ n] = 0

n ³ - 2, y[ n] =

or

n+2

1

å4k

4 1 1 , 3 12 4 n

=

k=0

1

2

0

2

1

1

1

2

0

2

1

2

2

4

0

4

2

Þ

æ 4 1 æ 1 ön ö ÷u[ n + 2 ] y[ n] = ç ç 3 12 çè 4 ÷ø ÷ è ø n -2

0

0

0

0

0

k=¥

0

2

2

4

0

4

2

1

1

2

0

2

1

for n - 2 ³ 4 or n ³ 6, y[ n] =

Fig. S5.2.24

25. (A) y[ n] = {2, 4, -19, 36, -25, 2, 12}

-3

6

4

2

2

8

-6

12

8

-4

-4

-16

12

-24

-16

3

3

12

ì 3n , ïï y[ n] = í 6 ï 81 , ïî 2

-9

18

=

k

81 , 2

n³6 n < 0,

y[ n] = 0 n -3

n > 0, y[ n] =

for n - 3 ³ - 3 or

å 1 = n + 1,

k = -3

y[ n] = ( n + 1) u[ n] 31. (A) For n - 1 < 0

12

For

Fig. S5.2.25

26. (B) x[ n] = {1, 2, 1, 1}, h[ n] = {1, -1, 0, 0, 1}

å3

n £5

30. (A) For n - 3 < - 3 or

4

3

k = -¥

Þ

1

3n 6

å 3k =

29. (D) For n - 2 £ 3 or n £ 5 , y[ n] =

Þ

n -1 ³ 0 ì 1, y[ n] = í î 0,

or

or

n <1 ,

y[ n] = 0

æp ö n ³ 1, y[ n] = å cos ç k ÷ k=0 è2 ø n -1

n = 4 m + 1, 4 m + 2 n = 4 m, 4 m + 3

32. (B) y1 [ n] = rect ( v[ n]) ,

y2 [ n] = rect ( kv[ n])

1

2

1

1

y2 [ n] ¹ k y1 [ n]

1

1

2

1

-1

y1 [ n] = rect ( v[ n]), y2 [ n] = rect ( v[ n - no ])

-1

-1

-2

-1

-1

0

0

0

0

0

0

0

0

0

0

(Not Homogeneous not linear)

y1 [ n - no ] = rect ( v[ n - no ]) = y2 [ n]

(Time Invariant)

At any discrete time n = no , the response depends only on the excitation at that discrete time.

(Causal)

No matter what values the excitation may have the response can only have the values zero or one.

1

1

2

1

(Stable)

1

Fig. S5.2.26

33. (C) y1 [ n] = nv[ n] ,

y[ n] = {1, 1, -1, 0, 0, 2, 1, 1}

ky1 [ n] = y2 [ n]

y[ n ] = d[ n + 2 ] + d[ n + 1 ] - d[ n ] + 2 d[ n - 3 ] + d[ n - 4 ] + d[ n - 5 ] ¥

å ( -1)

27. (D) y[ n] =

k

k=n -2

æ 1ö 2n ç - ÷ è 2ø = 1 1+ 2 Page 266

n -2

=

8 ( -1) n 3

2n -k = 2n

¥

å

k=n -2

y2 [ n] = nkv[ n]

æ 1ö ç- ÷ è 2ø

k

(Homogeneous)

Let x1 [ n] = v[ n]

then

y1 [ n] = nv[ n]

Let x2 [ n] = w[ n]

then

y2 [ n] = nw[ n]

Let x3[ n] = v[ n] + w[ n] then y3[ n] = n( v[ n] + w[ n]) = nv[ n] + nw[ n] = y1 [ n] + y2 [ n]

(Additive)

Since the system is homogeneous and additive, it is also linear. y1 [ n - no ] = ( n - no) v[ n - no ] ¹ yn [ n] = nv[ n - no ]

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(Time variant)

Chap 5.2

If the excitation is bounded, the response is bounded.

At any discrete time, n = no the response depends only on the excitation at that same time.

(Causal)

(Stable).

If the excitation is a constant, the response is

37. (B) y1 [ n] = 10 v[ n] - 5,

unbounded as n approaches infinity.

y2 [ n] ¹ ky1 [ n]

n +1

34. (C) y1 [ n] =

å v[ m ] ,

y2 [ n ] =

m = -¥

(Unstable)

n +1

å

å kv[ m ]

n = -¥

y3[ n] =

y1 [ n - no ] = 10 v[ n - no ] - 5, = y2 [ n]

m = -¥

(Homogeneous)

v[ m ], y2 [ n] =

(Not Homogeneous so not linear)

y1 [ n] = 10 v[ n] - 5, y2 [ n] = 10 v[ n - no ] - 5

n +1

y2 [ n] = ky1 [ n] y1 [ n] =

y2 [ n] = 10 kv[ n] - 5

n +1

å w[ m ]

(Time Invariant)

At any discrete time, n = no the response depends only on the excitation at that discrete time and not on any future excitation.

n = -¥

(Causal)

If the excitation is bounded, the response is bounded.

n +1

å ( v[ n] + w[ m ])

(Stable).

m = -¥

=

n +1

n +1

m = -¥

m = -¥

å v[ m ] + å w[ n] = y [ n] + y [ n] 1

(Additive)

2

Since the system is homogeneous and additive it is also linear y1 [ n] =

n +1

å v[ n] , y2 [ n] =

m = -¥

y1 [ n - no ] =

y[ n] = x[ n] + x[ n - 1] + y[ n - 2 ], Then by induction ¥

y[ n] = x[ n - 1] + x[ n - 2 ] + K x[ n - k] + K = å x[ n - k] k=0

n +1

å v[ m - no ]

n +1

m = -¥

q = -¥

å v[ q - no ] = y2 [ n]

y1 [ n] =

n

å v[ m ] ,

n

m =n

m = -¥

n

å kv[ m ] = ky [ n]

y2 [ n ] =

m = -¥

-¥

å x[ m ] = å x[ m ]

Let m = n - k then y[ n] =

m = -¥

n -no +1

å v[ m ] =

38. (B) y[ n] = x[ n] + y[ n - 1], y[ n - 1] = x[ n - 1] + y[ n - 2 ]

1

m = -¥

(Time Invariant) At any discrete time, n = no , the response depends on the excitation at the next discrete time in future. (Anti causal)

(Homogeneous) y3[ n] =

n

¥

n

m = -¥

m = -¥

m = -¥

å (v[ m ] + w[ m ]) = å v[ m ] + å w[ m ]

= y1 [ n] + y2 [ n]

(Additive)

If the excitation is a constant, the response increases

System is Linear.

without bound.

y1 [ n] =

(Unstable)

¥

å v[ m ]

m = -¥

35. (A) y1 [ n] = v[ n] , y2 = kv[ n] = k v[ n] y1 [ n] = v[ n] , y2 [ n] = v[ n - no ] (Time Invariant)

At any discrete time n = no , the response depends only on the excitation at that time

o

m = -¥

y1 [ n] can be written as

ky1 [ n] = k v[ n] ¹ y2 [ n] (Not Homogeneous Not linear)

y1 [ n - no ] = v[ n - no ] = y2 [ n]

n

å v[ n - n ]

, y2 =

(Causal)

If the excitation is bounded, the response is bounded. (Stable). 36. (B) y[ n] = 2 x [ n]

y1 [ n - no ] =

n -no

n

m = -¥

q = -¥

å v[ m ] = å v[ q - n ] = y [ n] o

2

(Time Invariant) At any discrete time n = no the response depends only on the excitation at that discrete time and previous discrete time.

(Causal)

If the excitation is constant, the response increase without bound.

(Unstable)

2

Let x1 [ n] = v[ n]

Let x2 [ n] = kv[ n] then ky[ n] ¹ y2 [ n] Let x1 [ n] = v[ n]

39. (C) Only statement (b) is false. For example

then y1 [ n] = 2 v 2 [ n]

S1 : y[ n] = x[ n] + b, and S2 : y[ n] = x[ n] - b , where b ¹ 0

y2 [ n] = 2 k2 v 2 [ n]

S{x[ n]} = S2 {S1 {x[ n]}} = S2 {x[ n] + b} = x[ n]

(Not homogeneous Not linear) then

y1 [ n] = 2 v 2 [ n]

Hence S is linear.

Let x2 [ n] = v[ n - no ] then y2 [ n] = 2 v 2 [ n - no ]

40. (B) For example

y1 [ n - no ] = 2 v[ n - no ] = y2 [ n]

S1 : y[ n] = nx[ n]

(Time invariant)

At any discrete time, n = no , the response depends only on the excitation at that time.

(Causal)

and S2 : y[ n] = nx[ n + 1]

If x[ n] = d[ n] then S2 {S1 {d[ n]}} = S2 [0 ] = 0,

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Page 267


CHAPTER

5.3 THE LAPLACE TRANSFORM


6. x( t) = u( t) - u( t - 2)

Determine the Laplace transform of given signal. 1. x( t) = u( t - 2) -2 s

(A)

-e s

(B)

e

-2 s

e -2 s - 1 s

(B)

1 - e -2 s s

(C)

2 s

(D)

-2 s

s 7. x( t) =

-2 s

e (C) 1+ s

(A)

(D) 0

(A)

1 s( s + 1) 2

(B)

s ( s + 1) 2

(C)

e- s s+1

(D)

e- s ( s + 1) 2

2. x( t) = u( t + 2) (A)

1 s

(B) -

(C)

e -2 s s

(D)

1 s

-e - 2 s s

3. x( t) = e -2 t u( t + 1) (A) (C)

1 s+2

(B)

e- ( s + 2 ) s+2

(D)

e- s s+2 -e - s s+2

4. x( t) = e 2 t u( -t + 2) (A)

e

2 (s - 2 )

-1 s -2 -2 ( s - 2 )

(C)

1-e s -2

8. x( t) = tu( t) * cos 2pt u( t) (A)

1 s( s 2 + 4 p2 )

(B)

2p s 2 ( s 2 + 4 p2 )

(C)

1 s ( s + 4 p2 )

(D)

s3 s + 4 p2

-3 s4

e s+2 -2 s

(D)

2

2

2

9. x( t) = t 3u( t) (A)

3 s4

(B)

(C)

6 s4

(D) -

-2 s

(B)

d { te - t u( t)} dt

e s -2

6 s4

10. x( t) = u( t - 1) * e -2 t u( t - 1)

5. x( t) = sin 5 t 5 (A) 2 s +5

s (B) 2 s +5

(A)

e -2 ( s + 1 ) 2s + 1

(B)

e -2 ( s + 1 ) s+1

5 (C) 2 s + 25

s (D) 2 s + 25

(C)

e- ( s + 2 ) s+2

(D)

e -2 ( s + 1 ) s+2

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Page 269


UNIT 5

t

11. x( t) = ò e -3t cos 2 t dt

17. X ( s) =

0

s2 - 3 ( s + 2)( s 2 + 2 s + 1)

(A)

-( s + 3) s(( s + 3) 2 + 4)

(B)

( s + 3) s(( s + 3) 2 + 4)

(A) ( e -2 t - 2 te - t ) u( t)

(B) ( e -2 t + 2 te - t ) u( t)

(C) ( e - t - 2 te -2 t ) u( t)

(D) ( e - t + 2 te -2 t ) u( t)

(C)

s( s + 3) ( s + 3) 2 + 4

(D)

-s( s + 3) ( s + 3) 2 + 4

18. X ( s) =

12. x( t) = t

( s + 2 s + 2) ( s 2 + 4 s + 2) 2

(C) ( 3e - t cos 3t - e - t sin 3t) u( t)

-( s + 2 s + 2) ( s 2 + 4 s + 2) 2 2

(D)

(D) ( 3e - t sin 3t + 3e- t cos 3t) u( t) 19. X ( s) =

Statement for Q.13–24: Determine the time signal x( t) corresponding to given X ( s) and choose correct option.

(B) (2 e - t - e -2 t ) u( t)

(C) (2 e -2 t - e - t ) u( t)

(D) (2 e - t + e -2 t ) u( t)

(A) 2 d( t) + ( e -3t - e-2 t ) u( t)

(A) (2 e -2 t + 2 e - t sin 2 t - 2 e - t cos 2 t) u( t) (B) (2 e -2 t + 2 e - t cos 2 t - 2 e - t sin 2 t) u( t) (D) (2 e -2 t + 2 e - t sin 2 t - e - t cos 2 t) u( t)

(A) (2 e -2 t + e - t ) u( t)

2 s 2 + 10 s + 11 s2 + 5 s + 6

20. X ( s) =

(C) 2 d( t) + ( e

+ e ) u( t)

(D) 2 d( t) - ( e

-2 t

+ e -3t ) u( t)

15. X ( s) = -t

-3t

2s - 1 s2 + 2 s + 1 -t

(A) ( 3e - 2 te ) u( t) (B) ( 3e- t + 2 te- t ) u( t)

(B) ( e -2 t + 2 e -3t cos t - 6 e -3t sin t) u( t) (C) ( e -2 t + 2 e -3t cos t - 2 e -3t sin t) u( t) (D) (9 e -2 t - 6 e -3t cos t + 3e -3t sin t) u( t) 21. X ( s) =

16. X ( s) =

5s + 4 s 3 + 3s 2 + 2 s

2 s 2 + 11s + 16 + e -2 s ( s2 + 5 s + 6)

(A) 2 d( t) + ( 3e -2 t - 2 e-3t ) u( t - 2) (B) 2 d( t) + (2 e -2 t - e -3t + e -2 ( t - 2 ) + e -3( t - 2 ) ) u( t) (C) 2 d( t) + (2 e -2 t - e -3t ) u( t) + ( e -2 t - e -3t ) u( t - 2) (D) 2 d( t) + (2 e -2 t - e -3t ) u( t) + ( e -2 ( t - 2 ) - e -3( t - 2 ) ) u( t - 2) 22. X ( s) = s

(C) (2 e- t - 3te- t ) u( t) (D) (2 e- t + 3te- t ) u( t)

3s 2 + 10 s + 10 ( s + 2)( s 2 + 6 s + 10)

(A) ( e -2 t + 2 e -3t cos t + 2 e -3t sin t) u( t)

(B) 2 d( t) + ( e -2 t - e-3t ) u( t) -2 t

4 s 2 + 8 s + 10 ( s + 2)( s 2 + 2 s + 5)

(C) (2 e -2 t + 2 e - t cos 2 t - e - t sin 2 t) u( t)

s+3 13. X ( s) = 2 s + 3s + 2

14. X ( s) =

2

1 æ ö (B) ç 3e- t sin 3t - e- t cos 3t ÷u( t) 3 è ø

( s 2 + 4 s + 2) (B) 2 ( s + 2 s + 2) 2

2

(C)

3s + 2 s + 2 s + 10 1 æ -t ö (A) ç 3e cos 3t - e- t sin 3t ÷u( t) 3 è ø

d -t { e cos t u( t)} dt

-( s 2 + 4 s + 2) (A) 2 ( s + 2 s + 2) 2

d2 ds 2

æ 1 ö 1 çç 2 ÷÷ + 9 3 s + s + è ø

ö æ 2t t2 (A) çç e -3t + sin 3t + cos 3t ÷÷u( t) 3 9 ø è (B) ( e -3t + 2 t sin 3t + t 2 cos 3t) u( t)

(A) (2 + e - t + 3e -2 t ) u( t)

2t æ ö (C) ç e -3t + sin 3t + t 2 cos 3t ÷u( t) 3 è ø

(B) (2 + e - t - 3e -2 t ) u( t)

(D) ( e -3t + t 2 sin 3t + 2 t cos 3t) u( t)

(C) ( 3 + e - t - 3e -2 t ) u( t) (D) ( 3 + e - t + 3e -2 t ) u( t) Page 270

Signal & System

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The Laplace Transform

23. X ( s) = (A) e (C)

-0 .5t

Chap 5.3


1 (2 s + 1) 2 + 4

Given the transform pair

1 (B) e - t sin t u( t) 2

sin t u( t)

1 -0 .5t e sin t u( t) 4

x( t) u( t)

(D) e - t sin t u( t)

L ¬¾ ®

2s . s2 + 2

Determine the Laplace transform Y ( s) of the given time signal in question and choose correct option.

24. X ( s) = e

-2 s

1 d æ ç ds çè ( s + 1) 2

ö ÷÷ ø

30. y( t) = x( t - 2)

(A) - te - t u(1 - t)

(B) -te - t u( t - 1)

(C) -( t - 2) 2 e - ( t - 2 ) u( t - 2)

(D) te - t u( t - 1)


(A)

2 se -2 s s2 + 2

(B)

2 se 2 s s2 + 2

(C)

2( s - 2) ( s - 2) 2 + 1

(D)

2( s + 2) ( s + 2) 2 + 1

Given the transform pair below. Determine the 31. y( t) = x( t) *

time signal y( t) and choose correct option. cos 2t u( t)

L ¬¾ ®

X ( s).

dx( t) dt

(A)

4 s3 ( s 2 + 2) 2

(B)

4 ( s 2 + 2) 2

-4 s 3 ( s + 2) 2

(D)

4 ( s 2 + 2) 2

25. Y ( s) = ( s + 1) X ( s) (A) [cos 2 t - 2 sin 2 t ]u( t)

sin 2 t ö æ (B) ç cos 2 t + ÷u( t) 2 ø è

(C)

(C) [cos 2 t + 2 sin 2 t ]u( t)

sin 2 t ö æ (D) ç cos 2 t ÷u( t) 2 ø è

32. y( t) = e - t x( t) (A)

2( s + 1) ( s + 1) 2 + 2

(B)

2( s + 1) s2 + 2 s + 2

(C)

2( s + 1) s2 + 2 s + 4

(D)

2( s + 1) s2 + 2 s

26. Y ( s) = X ( 3s) æ2 ö (A) cos ç t ÷u( t) è3 ø

1 æ2 ö (B) cos ç t ÷u( t) 3 è3 ø

(C) cos 6t u( t)

(D)

1 cos 6 t u( t) 3

33. y( t) = 2 tx( t) (A)

8 - 4 s2 ( s 2 + 2) 2

(B)

4 s2 - 8 ( s 2 + 2) 2

(C)

4 s2 s2 + 1

(D)

s2 s +1

27. Y ( s) = X ( s + 2) (A) cos 2( t - 2) u( t)

(B) e2 t cos 2 t u( t)

(C) cos 2( t + 2) u( t)

(D) e-2 t cos 2 t u( t)

2

2

Statement for Q.34–43: X ( s) 28. Y ( s) = 2 s (A) 4 cos 2 t u( t) 2

(C) t cos 2 t u( t)

29. Y ( s) =

Determine the bilateral laplace transform and (B)

1 - cos 2 t u( t) 4

cos 2 t (D) u( t) t2

d -3s [ e X ( s)] ds

(A) t cos 2( t - 3) u( t - 3)

(B) t cos 2( t - 3) u( t)

(C) -t cos 2( t - 3) u( t - 3)

(D) -t cos 2( t - 3) u( t)

choose correct option. 34. x( t) = e - t u( t + 2) (A)

e2 ( s + 1 ) , s+1

Re ( s) > -1

(B)

1 , 1+ s

Re ( s) < - 1

(C)

e2 ( s + 1 ) , s+1

Re ( s) < - 1

(D)

1 , 1+ s

Re ( s) > -1

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Page 271


UNIT 5

35. x( t) = u( -t + 3) 1 - e -3s , (A) s (B)

1-e (C) s (D)

-3s

Re ( s) > 3

(B)

e 3s , s-3

Re ( s) < 3

(C)

e 3( s -1 ) , s-3

Re ( s) > 3

(D)

e 3( s -1 ) , s-3

Re ( s) < 3

Re ( s) < 0

,

-e -3s , s

Re ( s) > 0

(A) e , Re( s) > 0 s

(B) e , Re ( s) < 0

s

(D) None of above

s

(C) e , all s 37. x( t) = sin t u( t) 1 , (A) (1 + s 2 )

Re ( s) > 0

-1 (C) , (1 + s 2 )

Re ( s) < 0

-1 , (1 + s 2 )

Re ( s) > 0

-

t 2

41. x( t) = cos 3t u( -t) * e - t u( t) -s , (A) ( s + 1)( s2 + 9) -s , ( s + 1)( s2 + 9)

-1 < Re ( s) < 0

(C)

s , ( s + 1)( s 2 + 9)

-1 < Re ( s) < 0

(D)

s , ( s + 1)( s 2 + 9)

Re ( s) > 0

42. x( t) = e t sin (2 t + 4) u( t + 2)

-t

38. x( t) = e u( t) + e u( t) + e u( -t) t

(A)

6 s2 + 2 s - 2 , (2 s + 1)( s 2 - 1)

Re ( s) < - 0.5

(B)

6 s2 + 2 s - 2 , (2 s + 1)( s 2 - 1)

-1 > Re ( s) > 1

(C)

1 1 1 , + + s + 0.5 s + 1 s - 1

-1 < Re ( s) < 1

(D)

1 1 1 , + s + 0.5 s + 1 s - 1

-0.5 < Re ( s) < 1

(A)

e 2 ( s -1 ) , ( s - 1) 2 + 4

Re ( s) > 1

(B)

e 2 ( s -1 ) , ( s - 1) 2 + 4

Re ( s) < 1

(C)

e( s - 2 ) , ( s - 1) 2 + 4

Re ( s) > 1

(D)

e( s - 2 ) , ( s - 1) 2 + 4

43. x( t) = e t (A)

1-s , s+1

Re ( s) > -1

(C)

s -1 , s+1

Re ( s) < - 1

s -1 , s+1

Re ( s) > -1

(1 - s) 1 1 , 0.5 < Re ( s) < 1 + + 2 ( s - 1) + 4 s + 1 s - 0.5

(D)

(B)

(1 - s) 1 1 , -1 < Re ( s) < 1 + + ( s - 1) 2 + 4 s + 1 s - 0.5


(C)

( s - 1) 1 1 , 0.5 < Re ( s) < 1 + + ( s - 1) 2 + 4 s + 1 s - 0.5

bilateral Laplace transform.

40. x( t) = e ( 3t + 6 ) u( t + 3) Page 272

Re ( s) < - 1

(A)

( s - 1) 1 1 , -1 < Re ( s) < 1 + + ( s - 1) 2 + 4 s + 1 s - 0.5

Re ( s) < 1

d -2 t [ e u( -t)] dt

1-s , s+1

(B)

t

39. x( t) = e t cos 2 t u( -t) + e - t u( t) + e 2 u( t)

(D)

Re ( s) > 0

(B)

Re ( s) < 0

1 , (1 + s 2 )

(D)

e 3s , s-3

Re ( s) < 0

36. y( t) = d( t + 1)

(B)

(A) Re ( s) > 0

-e -3s , s

Signal & System

Determine the corresponding time signal for given

44. X ( s) =

e 5s with ROC: Re ( s) < -2 s+2

(A) e -2 ( t + 5) u( t + 5) www.gatehelp.com


UNIT 5

(A)

1 -t e sin t u( t) 2

(D)

58.

62. A stable system has input

(B) 2e- t cos t u( t)

(C) 2 e - t cos t u( t) +

is (A) d( t) - ( e -2 t cos t + e -2 t sin t) u( t)

1 -t e cos t u( t - 1) + 2 e - t sin t u( t - 1) 2

(B) d( t) - ( e -2 t cos t + e -2 t sin t) u( t - 2) (C) d( t) - ( e 2 t cos t + e 2 t sin t) u( t)

d 3 y( t) d 2 y( t) dy( t) + 4 +3 = x( t) 3 2 dt dt dt

(D) d( t) - ( e 2 t cos t + e 2 t sin t) u( t + 2)

All initial condition are zero, x( t) = 10 e -2 t

63. The relation ship between the input x( t) and output y( t) of a causal system is described by the differential

5 é5 ù (A) ê + 5 e - t - 5 e -2 t + e -3t ú u( t) 3 ë3 û

equation

5 é5 ù (B) ê - 5 e - t + 5 e -2 t + e -3t ú u( t) 3 ë3 û 5 5 u( t) - 5 u( t - 1) + 5 u( t - 2) + u( t - 3) 3 3

(D)

5 5 u( t) + 5 u( t - 1) - 5 u( t - 2) + u( t - 3) 3 3

dy( t) + 10 y( t) = 10 x( t) dt The impulse response of the system is

59. The transform function H ( s) of a causal system is

(A) -10 e -10 t u( -t + 10)

(B) 10 e -10 t u( t)

(C) 10 e -10 t u( -t + 10)

(D) -10 e -10 t u( t)

64. The relationship between the input x( t) and output y( t) of a causal system is defined as

2 s2 + 2 s - 2 H ( s) = s2 - 1

d 2 y( t) dy( t) dx( t) . - 2 y( t) = -4 x( t) + 5 2 dt dt dt

The impulse response is

The impulse response of system is

(A) 2d( t) - ( e- t + et ) u( -t) -t

(A) 3e - t u( t) + 2 e 2 t u( -t)

(B) 2d( t) - ( e + e ) u( t) t

-t

(B) ( 3e - t + 2 e 2 t ) u( t)

(C) 2d( t) + e u( t) - e u( -t) t

(C) 3e - t u( t) - 2 e 2 t u( -t)

(D) 2d( t) + ( e- t + et ) u( t)

(D) ( 3e - t - 2 e 2 t ) u( -t)

60. The transfer function H ( s) of a stable system is 2s - 1 H ( s) = 2 s + 2s + 1 The impulse response is (A) 2 u( -t + 1) - 3tu( -t + 1) (B) ( 3te- t - 2 e- t ) u( t) (C) 2 u( t + 1) - 3tu( t + 1) (D) (2 e- t - 3te- t ) u( t) 61. The transfer function H ( s) of a stable system is H ( s) =

s2 + 5 s - 9 ( s + 1)( s 2 - 2 s + 10)

The impulse response is (A) -e - t u( t) + ( e t sin 3t + 2 e t cos 3t) u( t) (B) -e - t u( t) - ( e t sin 3t + 2 e t cos 3t) u( -t) (C) -e - t u( t) - ( e t sin 3t + 2 e t cos 3t) u( t) (D) -e - t u( t) + ( e t sin 3t + 2 e t cos 3t) u( -t) Page 274

x( t) and output

y( t) = e -2 t cos t u( t). The impulse response of the system


(C)

Signal & System

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*******



SOLUTIONS ¥

¥

0

2

1. (B) X ( s) = ò x( t) e - st dt = ò e - st dt =

Chap 5.3

r ( t) = e -2 t u( t)

L ¬¾ ®

v( t) = e -2 t u( t - 1)

e -2 s s

L ¬¾ ®

x( t) = q( t) * v( t)

¥

¥

¥

0

0

0

2. (A) X ( s) = ò x( t) e -3t dt = ò u( t + 2) -3t dt = ò e -3t dt =

1 s

L ¬¾ ®

3. (A) X ( s) = ò e -2 t e - st dt = 0

1 s+2

¥

¥

0

0

- 1 1 - e -2 ( 2 - s ) = 2-s s -2

p( t) dt

1 P ( s) p( t) dt + ò s -¥ s

L ¬¾ ®

Þ

X ( s) =

5 ( e j 5t - e - j 5t ) - st e dt = 2 2 25 j s + 0

q( t) =

d p( t) dt

x( t) = tq( t) 6. (B) X ( s) = ò e - st dt =

1 - e -2 s s

7. (B) p( t) = te - t u( t)

L ¬¾ ®

2

0

d p( t) dt

L ¬¾ ®

Þ P ( s) =

1 ( s + 1) 2

s ( s + 1) 2

X ( s) =

8. (A) p( t) = tu( t)

L ¬¾ ®

P ( s) =

1 s2

q( t) = cos 2 pt u( t)

L ¬¾ ®

Q( s) =

s s + 4 p2

x( t) = p( t) * q( t) X ( s) =

L ¬¾ ®

q( t) = - tp( t)

t n u( t)

X ( s) = P ( s)Q( s)

L ¬¾ ®

P ( s) =

Q( s) = X ( s) =

1 s2

d -2 P ( s) = 3 ds s

d 6 Q( s) = 4 ds s

n! sn + 1

L ¬¾ ®

10. (D) p( t) = u( t) q( t) = u( t - 1)

L ¬¾ ®

L ¬¾ ®

Q( s) = X ( s) = -

P ( s) =

s+1 ( s + 1) 2 + 1

s( s + 1) ( s + 1) 2 + 1 d Q( s) ds

-( s 2 + 4 s + 2) ( s 2 + 2 s + 2) 2

13. (B) X ( s) = A=

L ¬¾ ®

s+3 A B = + ( s + 3s + 2) s + 1 s + 2 2

s+3 s+3 = 2, B = = -1 s + 2 s = -1 s + 1 s = -2

x( t) = [2 e - t - e -2 t ]u( t) 14. (A) X ( s) = 2 -

1 1 1 =2 + ( s + 2) ( s + 3) ( s + 2) ( s + 3)

x( t) = 2 d( t) + ( e-3t - e-3t ) u( t) 2s - 1 A B = + s 2 + 2 s + 1 ( s + 1) ( s + 1) 2

B = (2 s - 2) s = -1 = -3, A = 2 x( t) = x( t) = [2 e - t - 3te - t ]u( t) 16. (B) X ( s) =

5s + 4 A B C = + + 2 s + 3s + 2 s s s + 1 s + 2 3

A = sX ( s) s = 0 = 2, B = ( s + 1) X ( s) s = -1 = 1, C = ( s + 2) X ( s) s = -2 = -3 x( t) = [2 + e - t - 3e -2 t ]u( t)

L ¬¾ ®

L ¬¾ ®

L ¬¾ ®

15. (C) X ( s) =

2

9. (C) p( t) = tu( t)

X ( s) =

L ¬¾ ®

2

1 s( s + 4 p2 )

x( t) = - tq( t)

( s + 3) s[( s + 3) 2 + 4 ]

12. (A) p( t) = e - t cos t u( t)

¥

s+3 ( s + 3) 2 + 4

0

ò

5. (C) X ( s) = ò

Þ

X ( s) = Q( s) V ( s)

2 (2 - s )

0

x( t) =

e- ( s + 2 ) s2

e X ( s) = s+2

Þ

t

4. (C) X ( s) = ò x( t) e - st dt = ò e 2 t u( -t + 2) e - st dt e

V ( s) =

-2 ( s + 1 )

-¥

= ò e t ( 2 - s ) dt =

1 s+2

L 11. (B) p( t) = e -3t cos 2 t u( t) ¬¾ ® P ( s) =

¥

2

R( s) =

P ( s) =

Q( s) =

e- s s2

1 s

17. (C) X ( s) = =

s2 - 3 ( s + 2)( s 2 + 2 s + 1)

A B C + + ( s + 2) ( s + 1) ( s + 1) 2

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Page 275


UNIT 5

A = ( s + 2) X ( s) s = -2 = 1, C = ( s + 1) 2 X ( s) A + B =1 x( t) = [ e

-2 t

Þ

B =0

æsö L 23. (C) P ç ÷ ¬¾ ® ap( at) èaø 1 1 -t L ¬¾ ® e sin 2 t u( t) ( s + 1) 2 + 4 2

= -2

s = -1

-t

- te ]u( t) 3s + 2 3( s + 1) 1 = 2 2 ( s + 1) 2 + 32 s + 2 s + 10 ( s + 1) + 3

18. (A) X ( s) =

2

1 é ù x( t) = ê3e - t cos 3t - e - t sin 3t ú u( t) 3 ë û

Q( s) =

Þ

A = ( s + 2) X ( s) s = -2 = 2 Þ

Þ

C = -2

Þ

x( t) = [2 e -2 t + 2 e - t cos 2 t - e - t sin 2 t ]u( t)

A = ( s + 2) X ( s) s = -2 = 1, A + B = 3 x( t) = [ e

-2 t

+ 2e

-3t

21. (D) X ( s) =

Þ

Þ

-3t

sin t ]u( t)

( s + 2)(2 s 2 + 11s + 16) =2 ( s2 + 5 s + 6) s = -2

B=

( s + 3)(2 s + 11s + 16) = -1 ( s2 + 5 s + 6) s = -3

R( s) = sQ( s)

L ¬¾ ®

e-2 t x( t)

X ( s) s

L ¬¾ ®

p( t) =

ò x( t) dt

-¥ t

ò cos 2 t u( t) dt =

P ( s) s

1 sin 3t u( t) 3

sin 2 t 1 - cos 2 t dt = u( t), 2 4 0

ò

L ¬¾ ®

31. (A) p( t) =

q( t) = - p( t)

32. (A) e - t x ( t)

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L ¬¾ ®

d x( t) dt

y( t) = x( t) * p( t)

é2 t ù x( t) = v( t) + r ( t) = ê sin 3t u( t) + t 2 cos 3t u( t) + e -3t ú u( t) ë3 û

p( t) = x( t - 3)

= -t cos 2( t - 3) u( t - 3).

t2 sin 3t u( t) 3

2t sin 3t u( t) + t 2 cos 3t u( t) 3 1 L V ( s) = ¬¾ ® v( t) = e -3t u( t) s+3

L ¬¾ ®

= cos 2( t - 3) u( t - 3) d L Q( s) = P ( s) ¬¾ ® ds

30. (A) x( t - 2)

d q( t) - q(0 - ) dt

sin 2 t 2

t

29. (C) P ( s) = e -3s X ( s)

q( t) = ( -1) 2 t 2 p( t) =

r ( t) =

t

L ¬¾ ®

-¥

=

Page 276

L ¬¾ ®

L ¬¾ ®

x( t) = 2 d( t) + [2 e -2 t - e-3t ]u( t) + [ e -2 ( t - 2 ) - e -3( t - 2 ) ]u( t - 2)

L ¬¾ ®

ax( at)

1 æ2 ö cos ç t ÷ u( t) 3 è3 ø

L ¬¾ ®

28. (B) P ( s) =

2

d2 P ( s) ds 2

L ¬¾ ®

2 s 2 + 11s + 16 + e -2 s ( s2 + 5 s + 6)

A=

Q( s) =

dx( t) + x( t) dt

L ¬¾ ®

C = -6

cos t - 6 e

1 s2 + 9

x( t) = q( t - 2)

y( t) = ( -2 sin 2 t + cos 2 t ) u( t)

27. (D) X ( s + 2)

B =2

A B e -2 s e -2 s =2 + + + ( s + 2) ( s + 3) ( s + 2) ( s + 3)

22. (C) P ( s) =

L ¬¾ ®

x( t) = - ( t - 2) e ( t - 2 ) u( t - 2)

X ( 3s)

A B( s + 3) C = + + ( s + 2) ( s + 3) 2 + 1 ( s + 3) 2 + 1

q( t) = -tp( t) = -t 2 e - t u( t)

L ¬¾ ®

æsö 26. (B) X ç ÷ èaø

3s 2 + 10 s + 10 20. (B) X ( s) = ( s + 2)( s 2 + 6 s + 10)

10 A + 6 B + 2 C = 10

d P ( s) ds

p( t) = te - t u( t)

L ¬¾ ®

25. (A) sX ( s) + X ( s)

B =2

5 A + 2 B + 2 C = 10

1 ( s + 1) 2

X ( s) = e -2 sQ( s)

A B( s + 1) C = + + 2 2 ( s + 2) ( s + 1) + 2 ( s + 1) 2 + 2 2 A + B=4

1 -0 .5t e sin t u( t) 4

L ¬¾ ®

x( t)

24. (C) P ( s) =

4 s 2 + 8 s + 10 ( s + 2)( s 2 + 2 s + 5)

19. (C) X ( s) =

Signal & System

e -2 s X ( s), Y ( s) =

L ¬¾ ®

L ¬¾ ®

L ¬¾ ®

2 se -2 s s2 + 2

P ( s) = sX ( s)

Y ( s) = P ( s) X ( s) = s( X ( s)) 2 X ( s + 1) =

2( s + 1) ( s + 1) 2 + 2



33. (B) 2 tx( t)

L ¬¾ ®

¥

ò x( t) e

34. (A) X ( s) =

-2

- st

d 4 s2 - 8 X ( s) = 2 ds ( s + 2) 2

Chap 5.3

42. (A) x( t) = e -2 e t + 2 sin (2 t + 4) u( t + 2) p( t + 2) X ( s) =

dt

L ¬¾ ®

e 2 s P ( s),

e 2 s e -2 , ( s - 1) 2 + 4

Re ( s) > 1

-¥ ¥

¥

-2

-2

= ò e - t e - st dt = ò e - t ( s + 1 ) dt =

e2 ( s + 1 ) , Re( s) > - 1 s+1

¥

-3s

-e 35. (B) X ( s) = ò u( -t + 3) e - st dt = ò e - st dt = s -¥ -¥ 36. (C) Y ( s) =

3

¥

ò d( t + 1) e

- st

dt = e s ,

43. (A) p( t) = e -2 t u ( -t) q( t) = Re ( s) < 0

( e - e ) - st e dt 2j 0

37. (B) X ( s) = ò ¥

X ( s) = e 5s P ( s)

¥

1 1 1 , e t ( j - s ) dt e - t ( j + s ) dt = = 2 j ò0 2 j ò0 1 + s2 ¥

-

¥

t 2

38. (D) X ( s) = ò e e - st + ò e - t e - st dt + 0

=

0

Þ

Re ( s) > 0

òee

t - st

dt

-¥

òe

-¥

¥

X ( s) = -

0.5 < Re ( s) < 1

q( t) = p( t + 3) X ( s) =

e 3( s -1 ) , s-3

¬¾® L

L ¬¾ ®

Re ( s) > 3

41. (B) p( t) * q( t)

d2 P ( s) ds 2

1 R( s) s

u( -( t + 5))

p( t) = e 3t u( t)

L ¬¾ ®

x( t) = t 2 e 3t u( t)

Þ

L ¬¾ ®

Re ( s) > -1, Re ( s) < 0 -1 < Re ( s) < 0

P ( s)Q( s)

p( t) = u( t)

L ¬¾ ®

r ( t) = -tq( t) = -tu( t - 3) v( t) =

t

ò r( t) dt

-¥

t

3

e 3s s-3

L ¬¾ ®

q( t) = p( t - 3) = u( t - 3)

v( t) = -ò tdt = -

X ( s) =

1 s

L ¬¾ ®

L ¬¾ ®

1 v( s) s

1 2 ( t - 9) 2

L ¬¾ ®

x( t) = -

t

1 2 ò ( t - 9) 2 -¥

9 é -1 ù x( t) = ê ( t 3 - 27) + ( t - 3) ú u( t - 3) 6 2 ë û

48. (B) X ( s) =

-s æ 1 ö X ( s) = 2 ç ÷ s + 9 è s + 1ø Þ

x( t) = p( t + 5)

47. (C) Right-sided, P ( s) =

Þ

1 P ( s) = s-3 Q( s) = e 3s P ( s) =

-

V ( s) =

40. (C) x( t) = e -3e -3( t + 3) u( t + 3) p( t) = e u( t)

-2 ( t + 5)

Q( s) = e-3s P ( s) d R( s) = Q( s) ds

s -1 1 1 + + ( s - 1) 2 + 4 ( s + 1) s - 0.5

3t

L ¬¾ ®

t 2

( e - e ) - st e dt + ò e - t e- st dt + ò e e - st dt 2j 0 0

Re ( s) < 1, Re ( s) > -1, Re ( s) > 0.5 Therefore

p( t) = -e -2 t u( -t)

x( t) = -u( -t) + u ( -t + 1) + d( t + 2) ¥

- jt

jt

t

1-s 1+ s

46. (D) Left-sided

-0.5 < Re ( s) < 1

39. (A) X ( s) =

x( t) = - e

X ( s) =

Re ( s) > -0.5, Re ( s) > -1, Re ( s) < 1

0

X ( s) = Q( s - 1) =

45. (A) Right-sided 1 L P ( s) = ¬¾ ® ( s - 3)

0

1 1 1 + s + 0.5 s + 1 s - 1

Þ

L ¬¾ ®

-1 , Re ( s) < - 2 s+2

Q( s) = sP ( s)

44. (C) Left-sided 1 L P ( s) = ¬¾ ® s+2

- jt

jt

x( t) = e t q( t)

L ¬¾ ®

P ( s) =

Re ( s) < - 1 thus left-sided .

All s

-¥

¥

d p( t) dt

L ¬¾ ®

-s - 4 -3 2 = + s 2 + 3s + 2 ( s + 1) s + 2

Left-sided, x( t) = 3e - t u( -t) - 2 e -2 t u( -t) 49. (A) X ( s) = Left-sided, www.gatehelp.com

5 1 ( s + 1) ( s + 1) 2

x( t) = -5 u( -t) + te - t u( -t) Page 277


UNIT 5

50. (D) x(0 + ) = lim sX ( s) = x ®¥

h( t) = (2 e - t - 3te - t ) u( t).

s =0 s + 5s - 2 2

61. (A) H ( s) =

s2 + 2 s 51. (A) x(0 + ) = lim sX ( s) = 2 =1 s ®¥ s + 2s - 3 52. (D) x(0 + ) = lim sX ( s) = s ®¥

53. (A) x( ¥) = lim sX ( s) = s ®0

e

-2 s

h( t) = - e - t u( t) + (2 e t cos 3t + et sin 3t) u( -t)

(6s + s ) =0 s2 + 2 s - 2 3

s ®0

2

62. (A) X ( s) =

2 s3 + 3s =0 s2 + 5 s + 1

H ( s) = =1 -

1 , s+1

Y ( s) =

( s + 2) ( s + 2) 2 + 1

Y ( s) ( s + 1)( s + 2) = X ( s) ( s + 2) 2 + 1

( s + 2) 1 ( s + 2) 2 + 1 ( s + 2) 2 + 1

h( t) = d( t) - ( e-2 t cos t + e-2 t sin t) u( t)

e -3s (2 s2 + 1) 1 = s2 + 5 s + 4 4

63. (B) sY ( s) + 10 Y ( s) = 10 X ( s) H ( s) =

-

56. (C) sY ( s) - y(0 ) + 10 Y ( s) = 10( s)

Þ

1 y(0 ) = 1, X ( s) = s 10 1 1 Y ( s) = + = s( s + 1) ( s + 1) s -

Þ

-1 2( s - 1) 3 + + ( s + 1) ( s - 1) 2 + 32 ( s - 1) 2 + 32

System is stable

s+2 54. (C) x( ¥) = lim sX ( s) = 2 =2 s ®0 s + 3s + 1 55. (B) x( ¥) = lim sX ( s) =

Signal & System

Y ( s) 10 = X ( s) s + 10

h( t) = 10 e-10 t u( t)

64. (B) Y ( s)( s 2 - s - 2) = X ( s)(5 s - 4) H ( s) =

y( t) = u( t)

Y ( s) 5s - 4 3 2 = = + X ( s) s 2 - s - 2 s + 1 s - 2

h( t) = 3e - t u( t) + 2 e 2 t u( t).

57. (C) s Y ( s) - 2 s + 2 sY ( s) - 2 + 5 Y ( s) = 1 2

( s 2 + 2 s + 5) Y ( s) = 3 + 2 s 2s + 3 2( s + 1) 1 Y ( s) = 2 = + s + 2 s + 5 ( s + 1) 2 + 2 2 ( s + 1) 2 + 2 2 Þ

y( t) = 2 e - t cos t u( t) +


58. (B) s 3Y ( s) + 4 s 2 Y ( s) + 3sY ( s) = Y ( s) =

***********

10 ( s + 2)

10 A B C D = + + + s( s + 1)( s + 2)( s + 3) s ( s + 1) ( s + 2) s + 3

5 A = sY ( s) s = 0 = , B = ( s + 1) Y ( s) s = -1 = -5, 3 C = ( s + 2) Y ( s) s = -2 = 5, D = ( s + 3) Y ( s) s = 0 = Þ

5 é5 ù y( t) = ê - 5 e - t + 5 e -2 t + e -3t ú u( t) 3 3 ë û

59. (D) For a causal system H ( s) = 2 + Þ

h( t) = 0

for t < 0

1 1 + s + 1 s -1

h( t) = 2d( t) + ( e- t + e t ) u( t)

60. (D) H ( s) = Page 278

5 3

2 3 , System is stable s + 1 ( s + 1) 2 www.gatehelp.com


CHAPTER

5.4 THE Z-TRANSFORM

Statement forQ.1-12: Determine the z-transform and choose correct option. 1. x[ n] = d[ n - k] , k > 0 k

(A) z , k

(C) z ,

z >0 z ¹0

(C) (B) z

-k

,

z >0

(D) z

-k

,

z ¹0

(A) z - k , z ¹ 0

(B) z k , z ¹ 0

(C) z - k , all z

(D) z k , all z

(C)

z , |z|< 1 1 - z -1

(B) (D)

1 , |z|< 1 1 - z -1 z , |z|> 1 1 - z -1

(A)

z - 0.25 , z > 0.25 z 4 ( z - 0.25)

(B)

z - 0.25 , z >0 z 4 ( z - 0.25)

(C)

z 5 - 0.25 5 , z < 0.25 z 3( z - 0.25)

(D)

z 5 - 0.25 5 , all z z 4 ( z - 0.25)

5

(B)

-5 z 2 3 , <|z|< (2 z - 3)( 3z - 2) 3 2

(C)

5z 2 2 , <|z |< (2 z - 3) ( 3z - 2) 3 3

(D)

5z 3 2 , -
(A)

4z 1 , |z|> 4z - 1 4

(B)

4z 1 , |z|< 4z - 1 4

(C)

1 1 , |z|> 1 - 4z 4

(D)

1 1 , |z|< 1 - 4z 4

n

æ1ö æ1ö 8. x[ n] = ç ÷ u[ n] + ç ÷ u[ -n - 1] è2 ø è4ø (A)

(B)

4

æ1ö 5. x[ n] = ç ÷ u[ -n] è4ø

3 , |z|< 3 3-z

-5 z 3 2 , -
n

5

(D)

(A)

n

5

z , |z|< 3 3-z

|n|

æ1ö 4. x[ n] = ç ÷ ( u[ n] - u[ n - 5 ]) è4ø 5

3 , |z|> 3 3-z

(B)

æ2 ö 7. x[ n] = ç ÷ è 3ø

2. x[ n] = d[ n + k] , k > 0

3. x[ n] = u[ n] 1 (A) , |z|> 1 1 - z -1

6. x[ n] = 3n u[ -n - 1] z (A) , |z|> 3 3-z

(C)

1 1 , 1 -1 1 -1 1- z 1- z 2 4

1 1 <|z|< 4 2

1 1 + , 1 -1 1 -1 1- z 1- z 4 2

1 1 <|z|< 4 2

1 1 1 , |z|> 1 -1 1 -1 2 1- z 1- z 2 4

(D) None of the above www.gatehelp.com

Page 279


UNIT 5

33. X ( z) =

Signal & System

37. Consider three different signal

1 1 , |z|< -2 1 - 4z 4

n é æ1ö ù x1 [ n] = ê2 n - ç ÷ ú u[ n] è 2 ø úû êë

¥

(A) - å 2 2 ( k + 1 ) d[ -n - 2( k + 1)] k=0 ¥

(B) - å 2 2 ( k + 1 ) d[ -n + 2( k + 1)]

x2 [ n] = -2 n u[ -n - 1] +

1 u[ -n - 1] 2n

x3[ n] = - 2 n u[ -n - 1] -

1 u[ n] 2n

k=0 ¥

(C) - å 2 2 ( k + 1 ) d[ n + 2( k + 1)] k=0 ¥

(D) - å 2 2 ( k + 1 ) d[ n - 2( k + 1)]

Fig. P.5.4.37 shows the three different region.

k=0

Choose the correct option for the ROC of signal

34. X ( z) = ln (1 + z -1 ) , |z|> 0 ( -1) (A) k

k -1

Im

( -1) (B) k

d[ n - 1]

( -1) k (C) d[ n - 1] k

k -1

R1 z - plane

d[ n + 1]

R2 R3 2

( -1) k (D) d[ n + 1] k

Re 1 2

35. If z-transform is given by X ( z) = cos ( z -3), |z|> 0,

Fig. P5.4.37

R1

R2

R3

(A)

x1 [ n]

x2 [ n ]

x3[ n]

(B)

x2 [ n ]

x3[ n]

x1 [ n]

(C)

x1 [ n]

x3[ n]

x2 [ n ]

(D)

x3[ n]

x2 [ n ]

x1 [ n]

The value of x[12 ] is (A) -

1 24

(B)

(C) -

1 6

(D)

1 24 1 6

36. X ( z) of a system is specified by a pole zero pattern in fig. P.5.4.36.

38. Given 7 -1 z 6 X ( z) = 1 -1 öæ 1 -1 ö æ ç 1 - z ÷ç 1 + z ÷ 2 3 è øè ø

Im

1+

z - plane

1 3

Re 2

For three different ROC consider there different Fig. P.5.4.36

solution of signal x[ n] :

Consider three different solution of x[ n]

(a) |z|>

n é æ1ö ù x1 [ n] = ê2 n - ç ÷ ú u[ n] è 3 ø úû êë

n é 1 1 æ -1 ö ù , x[ n] = ê n -1 - ç ÷ ú u[ n] 2 è 3 ø úû êë2

(b) |z|<

é -1 æ -1 ö 1 , x[ n] = ê n -1 + ç ÷ 3 è 3 ø êë2

1 u[ n] 3n 1 x3[ n] = - 2 n u[ n - 1] + n u[ -n - 1] 3 Correct solution is x2 [ n] = - 2 n u[ n - 1] -

Page 282

n

ù ú u[ -n + 1] úû n

(c)

1 1 1 æ -1 ö <|z |< , x[ n] = - n -1 u[ -n - 1] - ç ÷ u[ n] 3 2 2 è 3 ø

Correct solutions are

(A) x1 [ n]

(B) x2 [ n]

(A) (a) and (b)

(B) (a) and (c)

(C) x3[ n]

(D) All three

(C) (b) and (c)

(D) (a), (b), (c)

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The z-Transform

39. X ( z) has poles at z = 1 2 and z = -1. If

x [1] = 1

Chap 5.4

44. The transfer function of a causal system is given as

x [ -1] = 1, and the ROC includes the point z = 3 4. The

H ( z) =

time signal x[ n] is (A)

1 u[ n] - ( -1) n u[ -n - 1] 2 n -1

The impulse response is (A) ( 3n + ( -1) n 2 n + 1 ) u[ n]

1 (B) n u[ n] - ( -1) n u[ -n - 1] 2 (C) (D)

1 2 n -1

(B) ( 3n + 1 + 2 ( -2) n ) u[ n] (C) ( 3n -1 + ( -1) n 2 n + 1 ) u[ n]

u[ n] + u[ -n + 1]

(D) ( 3n -1 - ( -2) n + 1 ) u[ n]

1 u[ n] + u[ -n + 1] 2n

45. A causal system has input

40. x[ n] is right-sided, X ( z) has a signal pole, and x[0 ] = 2, x[2 ] = 1 2. x[ n] is u[ -n] (A) n -1 2 (C)

u[ n] (B) n -1 2

u[ -n] 2n +1

as 3 -1 z 2

(1 - 2 z -1 ) (1 +

1 1 d[ n - 1] - d[ n - 2 ] and output 4 8

y[ n] = d[ n] -

3 d[ n - 1] . 4

(A)

n n 1 é æ -1 ö æ1ö ù 5 2 ÷ ç ÷ ú u[ n] ê ç 3 êë è 2 ø è 4 ø úû

(B)

1 é æ1ö æ -1 ö ÷ ê5ç ÷ + 2ç 3 êë è 2 ø è 4 ø

(C)

1 3

(D)

1 é æ1ö æ1ö ê5ç ÷ + 2ç ÷ 3 êë è 2 ø è4ø

41. The z-transform function of a stable system is given

2-

x[ n] = d[ n] +

The impulse response of this system is

u[ -n] 2n +1

(D)

H ( z) =

1 -1 z ) 2

n

n

æ1ö (A) 2 n u[ -n + 1] - ç ÷ u[ n] è2 ø

ù ú u[ n] úû

n

ù ú u[ n] úû

46. A causal system has input x[ n] = ( -3) n u[ n] and

n

æ -1 ö (B) 2 n u[ -n - 1] + ç ÷ u[ n] è 2 ø

output n é æ1ö ù y[ n] = ê4(2) n - ç ÷ ú u[ n]. è 2 ø úû êë

n

æ -1 ö (C) -2 n u[ -n - 1] + ç ÷ u[ n] è 2 ø

The impulse response of this system is

n

æ1ö (D) 2 n u[ n] - ç ÷ u[ n] è2 ø Let

n

n é æ 1 ön æ -1 ö ù 5 2 ç ÷ ç ÷ ê ú u[ n] è 4 ø úû êë è 2 ø n

The impuse response h[ n] is

42.

5z2 z2 - z - 6

x[ n] = d[ n - 2 ] + d[ n + 2 ].

The

unilateral

z-transform is (A) z -2

(B) z 2

(C) -2 z -2

(D) 2 z 2

n é æ 1 ön æ1ö ù (A) ê7ç ÷ - 10ç ÷ ú u[ n] è 2 ø úû êë è 2 ø

n é æ1ö ù (B) ê7(2 n ) - 10ç ÷ ú u[ n] è 2 ø úû êë

é æ 1 ö2 ù (C) ê10ç ÷ - 7(2) n ú u[ n] êë è 2 ø úû

n é æ1ö ù (D) ê10 (2 n ) - 7ç ÷ ú u[ n] è 2 ø úû êë

47. A system has impulse response h[ n] =

43. The unilateral z-transform of signal x[ n] = u[ n + 4 ] is (A) 1 + z + z 2 + 3z + z 4

(B)

1 1-z

(C) 1 + z -1 + z -2 + z -3 + z -4

(D)

1 1 - z -1

1 u[ n] 2n

The output y[ n] to the input x[ n] is given by y[ n] = 2 d[ n - 4 ]. The input x[ n] is (A) 2 d[ -n - 4 ] - d[ -n - 5 ]

(B) 2 d[ n + 4 ] - d[ n + 5 ]

(C) 2 d[ -n + 4 ] - d[ -n + 5 ]

(D) 2 d[ n - 4 ] - d[ n - 5 ]

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Page 283


UNIT 5

48. A system is described by the difference equation y[ n] -

1 2

n -2

53. The z-transform of a signal x[ n] is given by

1 y[ n - 1] = 2 x[ n - 1] 2

X ( z) =

The impulse response of the system is -1 1 (B) n - 2 u[ n + 1] (A) n - 2 u[ n - 1] 2 2 (C)

Signal & System

u[ n - 2 ]

(D)

If X ( z) converges on the unit circle, x[ n] is 1

(A) -

-1 u[ n - 2 ] 2n -2

(B) 49. A system is described by the difference equation (C)

y[ n] = x[ n] - x[ n - 2 ] + x[ n - 4 ] - x[ n - 6 ]

3n -1 8 1

3n -1 8 1 3n -1 8

The impulse response of system is (D) -

(A) d[ n] - 2 d[ n + 2 ] + 4 d[ n + 4 ] - 6 d[ n + 6 ] (B) d[ n] + 2 d[ n - 2 ] - 4 d[ n - 4 ] + 6 d[ n - 6 ]

u[ n] -

3n + 3 u[ -n - 1] 8

u[ n] -

3n + 3 u[ -n] 8

1 n -1

3

3n + 3 u[ -n - 1] 8

u[ n] -

8

u[ n] -

3n + 3 u[ -n] 8

54. The transfer function of a system is given as

(C) d[ n] - d[ n - 2 ] + d[ n - 4 ] - d[ n - 6 ] (D) d[ n] - d[ n + 2 ] + d[ n + 4 ] - d[ n + 6 ]

H ( z) =

50. The impulse response of a system is given by h[ n] =

3 u[ n - 1]. 4n

4 z -1 1 -1 ö æ ç1 - z ÷ 4 è ø

2

, |z|>

1 4

The h[ n] is

The difference equation representation for this system is (A) 4 y[ n] - y[ n - 1] = 3 x[ n - 1]

(A) Stable

(B) Causal

(C) Stable and Causal


55. The transfer function of a system is given as

(B) 4 y[ n] - y[ n + 1] = 3 x[ n + 1]

1ö æ 2ç z + ÷ 2ø è . H ( z) = 1 öæ 1ö æ ç z - ÷ç z - ÷ 2 øè 3ø è

(C) 4 y[ n] + y[ n - 1] = - 3 x[ n - 1] (D) 4 y[ n] + y[ n + 1] = 3 x[ n + 1]

Consider the two statements

51. The impulse response of a system is given by

Statement(i) : System is causal and stable.

h[ n] = d[ n] - d[ n - 5 ] The difference equation representation for this

Statement(ii) : Inverse system is causal and stable.

system is

The correct option is

(A) y[ n] = x[ n] - x[ n - 5 ]

(B) y[ n] = x[ n] - x[ n + 5 ]

(A) (i) is true

(C) y[ n] = x[ n] + 5 x[ n - 5 ]

(D) y[ n] = x[ n] - 5 x[ n + 5 ]

(B) (ii) is true

52. The transfer function of a system is given by H ( z) =

z( 3z - 2) 1 z2 - z 4

(C) Both (i) and (ii) are true (D) Both are false 56. The impulse response of a system is given by n

(A) Causal and Stable (B) Causal, Stable and minimum phase (C) Minimum phase (D) None of the above

n

æ -1 ö æ -1 ö h[ n] = 10ç ÷ u[ n] - 9ç ÷ u[ n] è 2 ø è 4 ø

The system is

Page 284

3 10 -1 1z + z -2 3

For this system two statement are Statement (i): System is causal and stable Statement (ii): Inverse system is causal and stable.

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The z-Transform

Chap 5.4

62. The impulse response of the system shown in fig.

The correct option is (A) (i) is true

(B) (ii) is true

(C) Both are true

(D) Both are false

P5.4.62 is X(z)

Y(z) z-1

+

z-1

57. The system

z-1

y[ n] = cy[ n - 1] - 0.12 y[ n - 2 ] + x[ n - 1] + x[ n - 2 ]

Fig. P5.4.62

is stable if (A) c < 112 .

(B) c > 112 .

(C) |c|< 112 .

(D) |c|> 112 .

æn ö çç - 2 ÷÷ ø

(A) 2 è 2 (B)

58. Consider the following three systems

y2 [ n] = x[ n] - 0.1 x[ n - 1]

(C) 2 è 2 (D)

y3[ n] = 0.5 y[ n - 1] + 0.4 x[ n] - 0.3 x[ n - 1]

(B) y2 [ n] and y3[ n]

(C) y3[ n] and y1 [ n]

(D) all

1 d[ n] 2

2n 1 [1 + ( -1) n ] u [ n] - d [ n] 2 2

H ( z) =

z z2 + z + 1

is shown in fig. P5.4.63. This system diagram is a X(z)

59. The z-transform of a causal system is given as X ( z) =

(1 + ( -1) n ) u[ n] -

63. The system diagram for the transfer function

The equivalent system are (A) y1 [ n] and y2 [ n]

1 d[ n] 2

2n 1 (1 + ( -1) n ) u[ n] + d[ n] 2 2 æn ö çç - 2 ÷÷ ø

y1 [ n] = 0.2 y[ n - 1] + x[ n] - 0.3 x[ n - 1] + 0.02 x[ n - 2 ]

(1 + ( -1) n ) u[ n] +

. z -1 2 - 15 -1 . z + 0.5 z -2 1 - 15

Y(z)

+

+

+

The x[0 ] is

z-1

z-1

(A) -15 .

(B) 2

(C) 1.5

(D) 0

Fig. P5.4.63

(A) Correct solution 60. The z-transform of a anti causal system is X ( z) =

12 - 21z 3 - 7 z + 12 z 2

The value of x[0 ] is 7 (A) 4

(B) Not correct solution (C) Correct and unique solution (D) Correct but not unique solution

(B) 0

(C) 4

*****************

(D) Does not exist

61. Given the z-transforms X ( z) =

z( 8 z - 7) 4z2 - 7z + 3

The limit of x[ ¥ ] is (A) 1

(B) 2

(C) ¥

(D) 0 www.gatehelp.com

Page 285


The z-Transform

z 2 - 3z 1 - 3z -1 = 3 3 z 2 + z -1 1 + z -1 - z -2 2 2 2 1 1 , ROC : <|z|< 2 = 1 + 2 z -1 1 - 1 z -1 2 2 1 Þ x[ n] = -2(2) n u[ -n - 1] - n u[ n] 2

Chap 5.4

ìæ 1 ö 2 ï , n even and n ³ 0 = í çè 4 ÷ø ï n odd î 0 , n

24. (A) X ( z) =

ì2-n , =í î 0,

n even and n ³ 0 n odd ¥

¥

k=0

k=0

33. (C) X ( z) = -4 z 2 å (2 z) 2 k = - å 2 2 ( k + 1 ) z 2 ( k + 1 )

25. (A) x[ n] is right sided 1 49 47 z - z -1 4 32 32 = + X ( z) = 1 - 16 z -1 1 + 4 z -1 1 - 4 z -1 47 n ù é 49 Þ x[ n] = ê ( -4) n + 4 ú u[ n] 32 ë 32 û

Þ

¥

x[ n] = - å 2 2 ( k + 1 ) d[ n + 2 ( k + 1)] k=0

34. (A) ln (1 + a) =

( -1) k -1 ( a) k k k =1 ¥

å

( -1) k -1 ( z -1 ) k k k =1 ¥

X ( z) = å

26. (C) x[ n] is right sided

( -1) k -1 d[ n - 1] k k =1 ¥

x [ n] = å

1 -1 ö 2 æ X ( z) = ç 2 + + ÷z -1 1+ z 1 - z -1 ø è

Þ

Þ

35. (B) cos a = å

x[ n] = 2 d[ n + 2 ] + (( -1) n - 1) u[ n + 2 ]

( -1) k 2 k a k = 0 (2 k) ! ¥

27. (A) d[ n] + 2 d[ n - 6 ] + 4 d[ n - 8 ]

( -1) k -3k 2 k (z ) k = 0 (2 k) ! ¥

X ( z) = å 10

1

å k d[ n - k]

28. (B) x[ n] is right sided, x[ n] =

Þ

k=5

x [ n] =

( -1) k

¥

å (2 k) ! d [ n - 6 k]

k=0

n = 12

29. (D) x[ n] is right sided signal X ( z) = 1 + 3z Þ

-1

+ 3z

-2

+z

x[ n] = d[ n] + 3d[ n - 1] + 3d[ n - 2 ] + d[ n - 3]

12 - 6 k = 0, k = 2,

36. (D) All gives the same z transform with different

30. (A) x[ n] = d[ n + 6 ] + d[ n + 2 ] + 3d[ n] + 2 d [ n - 3] + d[ n - 4 ] 31. (B) X ( z) = 1 + z +

z2 z3 + + ......... 2 ! 3! 1 1 1 ......... +1 + + + 2 z 2!z 3! z3

d[ n + 2 ] d[ n + 3] x[ n] = d[ n] + d[ n + 1] + + ...... 2! 3! d[ n - 2 ] d[ n - 3] +d[ n] + d[ n - 1] + + ......... 2! 3! 1 x [ n] = d [ n] + n! 32. (A) X ( z) = 1 + Þ

Þ

( -1) 2 1 x[12 ] = = 4! 24

-3

¥

k

-2

-2

æz z + çç 4 è 4

æ1ö x [ n] = å ç ÷ d[ n - 2 k] k=0 è 4 ø

2

ö æ1 ö ÷÷ = å ç z -2 ÷ k=0 è 4 ø ø ¥

k

ROC. So all are the solution. 37. (C) x1 [ n] is right-sided signal z1 > 2 , z1 >

1 gives z1 > 2 2

x2 [ n] is left-sided signal 1 1 z 2 < 2, z 2 < gives z 2 < 2 2 x3[ n] is double sided signal 1 1 z 3 > and z 3 < 2 gives < z3 < 2 2 2 38. (B) X ( z) =

2 -1 , + 1 -1 1 1- z 1 + z -1 2 z n

2 æ -1 ö u[ n] - ç ÷ u[ n] 2n è 3 ø

|z|>

1 (Right-sided) Þ 2

x[ n] =

|z|<

1 (Left-sided) Þ 3

é -2 æ -1 ön ù x[ n] = ê n + ç ÷ ú u[ -n - 1] è 3 ø úû êë 2

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Page 287


UNIT 5

n

1 1 2 æ -1 ö <|z|< (Two-sided) x[ n] = - n u[ -n - 1] - ç ÷ u[ n] 3 2 2 è 3 ø So (b) is wrong. 3 39. (A) Since the ROC includes the z = , ROC is 4 1 <|z|< 1, 2 A B X ( z) = + 1 -1 1 + z -1 1- z 2 A Þ x[ n] = n u[ n] × B ( -1) n u[ -n - 1] 2 A 1= Þ A =2 , 2

Signal & System

-2 5 Y ( z) 3 3 , H ( z) = = + X ( z ) 1 - 1 z -1 1 + 1 z -1 4 2 Þ

h[ n] =

n n 1 é æ -1 ö æ1ö ù 5 2 ç ÷ ç ÷ ê ú u[ n] 3 êë è 2 ø è 4 ø úû

Y ( z) =

H ( z) =

4 1 3 = -1 1 1 1 - 2z æ ö 1 - z -1 (1 - 2 z -1 )ç 1 - z -1 ÷ 2 2 è ø 10 Y ( z) -7 = + X ( z ) 1 - 2 z -1 1 - 1 z -1 2

x[ -1] = 1 = ( -1) B( -1) Þ B = 1 1 Þ x[ n] = n -1 u[ n] - ( -1) n u[ -n - 1] 2

Þ

40. (B) x[ n] = Cpn u[ n] , x[0 ] = 2 = C

47. (D) H ( z) =

x[2 ] =

1 = 2 p2 2

Þ

p=

X ( z) =

n

41. (B) H ( z) =

Þ

1 1 + -1 1 1 - 2z 1 + z -1 2

n

æ -1 ö h[ n] = (2) u[ -n - 1] + ç ÷ u [ n] è 2 ø ¥

-n

Y ( z) = 2 z -4 - z -5 H ( z)

x[ n] = 2[ d - 4 ] - d[ n - 5 ]

H ( z) =

Þ

n

å x[ n]z

= d[ n - 2 ]z - n = z -2

Y ( z) 2 z -1 = z -1 X ( z) 12

æ1ö h[ n] = 2ç ÷ è2 ø

49. (C) H ( z) =

n =0

+

43. (D) X ( z) =

¥

å x[ n]z

n =0

-n

=

¥

åz

n =0

Þ -n

1 = 1 - z -1

3 2 44. (B) H ( z) = + -1 1 - 3z 1 + 2 z -1 h[ n] is causal so ROC is |z|> 3, Þ

h[ n] = [ 3n + 1 + 2 ( -2) n ]u[ n]

45. (A) X ( z) = 1 +

1 , Y ( z ) = 2 z -4 1 -1 1- z 2

é z -1 ù -1 48. (A) Y ( z) ê1 ú = 2 z X ( z) 2 ë û

h[ n] is stable, so ROC includes |z|= 1 1 ROC : <|z|< 2 , 2

42. (A) X + ( z) =

n é æ1ö ù h[ n] = ê10(2) n - 7ç ÷ ú u( n) è 2 ø úû êë

1 , 2

æ1ö x[ n] = 2ç ÷ u( n) è2 ø

Page 288

1 1 + 3z -1

46. (D) X ( z) =

z -1 z -2 3z -1 , Y ( z) = 1 4 8 4

n -1

u[ n - 1] Y ( z) = (1 - z -2 + z -4 - z -6 ) X ( z)

h[ n] = d[ n] - d[ n - 2 ] + d[ n - 4 ] - d[ n - 6 ]

50. (A) h[ n] =

3æ1ö ç ÷ 4è4ø

n -1

u[ n - 1]

3 -1 z Y ( z) 4 H ( z) = = X ( z ) 1 - 1 z -1 4 1 3 Y ( z ) - z -1 Y ( z) = z -1 X ( z) 4 4 1 3 Þ y[ n] - y[ n - 1] = x[ n - 1] 4 4

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The z-Transform

51. (A) H ( z) = Þ

Y ( z) = 1 - z -5 X ( z)

So y1 and y2 are equivalent. 59. (B) Causal signal x[0 ] = lim X ( z) = 2

y[ n] = x[ n] - x[ n - 5 ]

52. (D) Zero at : z = 0,

Chap 5.4

z ®¥

1± 2 2 , poles at z = 2 3

60. (C) Anti causal signal, x[0 ] = lim X ( z) = 4 z ®¥

(i) Not all poles are inside |z|= 1, the system is not causal and stable.

3 61. (A) The function has poles at z = 1, . Thus final 4

(ii) Not are poles and zero are inside |z|= 1, the system

value theorem applies.

is not minimum phase.

7 ) 4 =1 lim x( n) = lim( z - 1) X ( z) = ( z - 1) n ®¥ z ®1 3ö æ ( z - 1) ç z - ÷ 4ø è z(2 z -

3 27 8 8 53. (A) X ( z) = + 1 -1 1 - 3z -1 1- z 3 -

62. (C) [2 Y ( z) + X ( z)] z -2 = Y ( z)

Since X ( z) converges on |z|= 1. So ROC must include this circle. 1 ROC : <|z|< 3, 3

H ( z) =

Þ

3n + 3 x[ n] = - n -1 u[ n] u[ -n - 1] 3 8 8 1

=-

z -2 1 - 2 z -2

1 1 1 4 4 h[ n] = - + + 2 1 - 2 z -1 1 + 2 z -1 1 1 d[ n] + {( 2 ) n + ( - 2 ) n } u[ n] 2 4

n

æ1ö 54. (C) h[ n] = 16 nç ÷ u[ n]. So system is both stable and è4ø

63. (D) Y ( z) = X ( z) z -1 - { Y ( z) z -1 + Y ( z) z -2 }

causal. ROC includes z = 1.

Y ( z) z -1 z = = 2 -1 -2 X ( z) 1 + z + z z + z +1

1 1 , 2 3 1 Pole of inverse system at : z = 2

So this is a solution but not unique. Many other correct

55. (C) Pole of system at : z = -

diagrams can be drawn.

For this system and inverse system all poles are inside

***********

|z|= 1. So both system are both causal and stable. 56. (A) H ( z) =

=

10 9 1 -1 1 1+ z 1 + z -1 4 2

1 - 2 z -1 1 -1 öæ 1 -1 ö æ ç 1 + z ÷ç 1 + z ÷ 2 4 è øè ø

Pole of this system are inside |z|= 1. So the system is stable and causal. For the inverse system not all pole are inside |z|= 1. So inverse system is not stable and causal. 57. (C) |a2|= 0.12 < 1, a1 =|-c|< 1 + 0.12, |c|< 112 . 58. (A) Y1 ( z) = 1 - 0.1z -1 , Y2 ( z) = 1 - 0.1z -1 Y3( z) =

0.4 - 0.3z -1 1 - 0.5 z -1

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Page 289


CHAPTER

5.6 THE DISCRETE-TIME FOURIER TRANSFORM

-n

Statement for Q.1–9: Determine the discrete-time Fourier Transform for the given signal and choose correct option. |n|£ 2

ì 1, 1. x[ n] = í î 0, sin 5W (A) sin W (C)

otherwise (B)

sin 2.5W sin W

sin 4W sin W

(C)

(

3 4

e - jW

)

4

(B)

1 - 43 e - jW 3 4

e - jW

1+

3 4

)

(

3 4

e jW

)

(B)

2 e jW 2 - e - jW

(C)

e jW 2 - e jW

(D)

2 e jW 2 - e jW

(A) 2 e - j 2 W

(B) 2 e j 2 W

(C) 1


(A) pd(W) -

1 1 + e - jW

(B)

1 1 - e - jW

(C) pd(W) +

1 1 - e - jW

(D)

1 1 + e - jW

4

1 - 43 e jW

4


e jW

8. x[ n] = {-2, - 1,

3. x[ n] = u[ n - 2 ] - u[ n - 6 ] (A) e 3 jW + e 3 jW + e 4 jW + e 5 jW (C) e -2 jW + e -3 jW + e -4 jW + e -5 jW

(B) (D)

e -2 jW(1 - e 3 jW) 1 - e jW e

-2 jW

-3 jW

(1 - e 1 - e - jW

)

0, 1, 2}

(A) 2 j(2 sin 2W + sin W)

(B) 2(2 cos 2W - cos W)

(C) -2 j(2 sin 2W + sin W)

(D) -2(2 cos 2W - cos W)

æp ö 9. x[ n] = sin ç n ÷ è2 ø

4. x[ n] = a|n| , |a|< 1

(A) p( d[W - p 2 ] - d[W + p 2 ])

(A)

1-a 1 - 2 a sin W + a 2

(B)

(C)

1 - a2 1 - 2 ja sin W + a 2


2

Page 300

e jW 2 - e - jW

7. x[ n] = u[ n]

n

(

(A)

6. x[ n] = 2 d[ 4 - 2 n]


æ 3ö 2. x[ n] = ç ÷ u[ n - 4 ] è4ø (A)

æ1ö 5. x[ n] = ç ÷ u[ -n - 1] è2 ø

1-a 1 - 2 a cos W + a 2 2

(B)

j ( d[W + p 2 ] - d[W - p 2 ]) 2

(C) 2 p( d[W - p 2 ] - d[W + p 2 ]) (D) jp( d[W + p 2 ] - d[W - p 2 ])

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The Discrete-Time Fourier Transform

16. X ( e jW) = j 4 sin 4W - 1

Statement for Q.10–21: Determine

the

signal

having

the

Fourier

(C) d[ n + 4 ] - d[ n - 4 ] - d[ n]

1 , |a|< 1 (1 - ae - jW) 2

(A) ( n - 1) a n u[ n]


(B) ( n + 1) a n u[ n]

n


(C) na u[ n]

(A) (d[ n + 2 ] + 2 d[ n] + d[ n - 2 ]) (C) -4(d[ n + 2 ] + d[ n] + d[ n - 2 ]) (d[ n + 2 ] + d[ n] + d[ n - 2 ])

(C)

ì 2 j, 0
2-n 5

2 + e - jW + 6

n +1 æ ö ç 1 + æç -2 ö÷ ÷u[ n] ç è 3 ø ÷ø è n +1

ö ÷u[ n] ÷ ø

n +1 æ ö ç ( -1) n + æç 2 ö÷ ÷u[ n] ç è 3 ø ÷ø è

(D) None of the above 4 æ pn ö (B) sin 2 ç ÷ pn è 2 ø

8 æ pn ö sin 2 ç ÷ pn è 2 ø

5 2-n

-e

- j2 W

æ æ -2 ö (B) 2 - n ç 1 - ç ÷ ç è 3 ø è

(B) 2(d[ n + 2 ] + 2 d[ n] + d[ n - 2 ])

1 2

17. X ( e jW) = (A)

11. X ( e jW) = 8 cos 2 w

(D)

(A) 4 pd[ n + 4 ] - 4 pd[ n - 4 ] - 2 pd[ n] (B) 2 d[ n + 4 ] - 2 d[ n - 4 ] - d[ n]

transform given in question. 10. X ( e jW) =

Chap 5.6

(D) -

18. X ( e jW) =

2+ - 18 e - j 2 W

e - jW + 14 e - jW + 1 1 4

(A) 2 - n + 1 [1 + ( -2) - n ]u[ n]

8 æ pn ö sin 2 ç ÷ pn è 2 ø

(B) 2 - n [1 + ( -2) - n ]u[ n] (C) 2 - n + 1 [( -1) n + 2 - n ]u[ n]

ì ï 1, jW 13. X ( e ) = í ï 0, î (A)

(B)

3p p £|W|< 4 4 p 3p 0 £|W|< , £|W|£ p 4 4

(D) 2 - n [( -1) n + 2 - n ]u[ n] 19. X ( e jW) =

2æ æ 3pn ö æ pn ö ö ç sin ç ÷ - sin ç ÷÷ nè è 4 ø è 4 øø

(A) 2 n -1 [1 + ( -1) n ]u[ n]

1 æ æ 3pn ö æ pn ö ö ç sin ç ÷ - sin ç ÷÷ pn è è 4 ø è 4 øø

(C) 21 - n [1 - ( -1) n ]u[ n]

(B) 21 - n [1 + ( -1) n ]u[ n] (D) 2 n -1 [1 - ( -1) n ]u[ n]

2æ æ 3pn ö æ pn ö ö (C) ç cos ç ÷ + cos ç ÷÷ nè è 4 ø è 4 øø (D)

20. X ( e jW) =

1 æ æ 3pn ö æ pn ö ö ç cos ç ÷ + cos ç ÷÷ pn è è 4 øø è 4 ø

14. X ( e jW) = e

-

æ2 (A) ç ç9 è

jW 2

for

-p £ W £ p

(A) pd[ n - 1 2 ] (C)

2 e - jW 1 - 14 e - j 2 W

( -1) n + 1 p(n - 12 )

(B) 2pd[ n - 1 2 ] (D) None of the above

1 - 13 e - jW 1 - 14 e - jW - 18 e -2 jW

n

7 æ 1ö æ1ö ç ÷ + ç- ÷ 2 9 è 4ø è ø

n

ö ÷u[ n] ÷ ø

æ 2 æ 1 ön 7 æ 1 ön ö (B) ç ç - ÷ + ç ÷ ÷u[ n] ç9 è 2 ø 9 è 4 ø ÷ø è æ 2 æ 1 ön 7 æ 1 ön ö (C) ç ç - ÷ - ç ÷ ÷u[ n] ç9 è 2 ø 9 è 4 ø ÷ø è

15. X ( e jW) = cos 2W + j sin 2W (A) 2 pd[ n + 2 ]

(B) d[ n + 2 ]

(C) 0


æ2 (D) ç ç9 è

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n

7 æ 1ö æ1ö ç ÷ - ç- ÷ 2 9 è 4ø è ø

n

ö ÷u[ n] ÷ ø Page 301


UNIT 5

n

n

æ 1ö æ1ö x[ n] = ç - ÷ u[ n] + ç ÷ u[ n] = 2 - n [( -1) n + 2 - n ]u[ n] 2 è ø è4ø

Signal & System

¥

å x[ n] , This condition is satisfied only

27. (D) X ( e j 0 ) =

n = -¥

if the samples of the signal add up to zero. This is true 2 e - jW 2 2 19. (C) X ( e ) = = 1 - j2 W 1 - jW 1 - jW 1- e 1- e 1+ e 4 2 2

for signal (b) and (h).

jW

n

28. (A) X ( e j 0 ) =

n

æ1ö æ 1ö x[ n] = 2ç ÷ u[ n] - 2ç - ÷ u[ n] è2 ø è 2ø 1 = n -1 [1 - ( -1) n ]u[ n] = 21 - n [1 - ( -1) n ]u[ n] 2

n = -¥

29. (A) y[ n] = x[ n + 2 ] Y ( e ) is real and even. Y ( e jW) = e j 2 W X ( e jW)

Thus arg{X ( e jW)} = -2W 30. (C)

=

p

ò X (e

jW

) dW = 2 px[0 ] = 4 p

-p

¥

å ( -1)

31. (A) X ( e jp) =

2æ1ö 7æ 1ö x[ n] = ç ÷ u[ n] + ç - ÷ u[ n] 9 è2 ø 9è 4ø

e j2 W

X ( e jW) = e - j 2 WY ( e jW),

Þ

Since Y ( e jW) is real. This imply arg{ Y ( e jW)} = 0

n

21. (C) X ( e jW) =

is an even signal. Therefore

jW

1 - jW 1- e jW 3 20. (A) X ( e ) = 1 1 1 - e - jW - e -2 jW 4 8 2 7 9 9 = + 1 - jW 1 - jW 1- e 1+ e 2 4 n

¥

å x[ n] = 6

n

x[ n] = 2

n = -¥

DTFT 32. (C) Ev{x[ n]} ¬ ¾¾ ® Re{X ( e jW)}

( b - a) e jW - ( a + b) e jW + ab

Ev{x[ n] =

- jW

( b - a) e 1 -1 = + 1 - ( a + b) e - jW + abe- j 2 W 1 - be - jW 1 - ae - jW

( x[ n] + x[ -n]) 2

1ü 1 1 ì 1 , 1, 0, 0, 1, 2, 1, 0, 0, 1, , 0, - ý = í - , 0, 2 2 2 2 î þ

x[ n] = bn u[ n] + a n u[ -n - 1] . 22. (D) The signal must be read and odd. Only signal ( h)

33. (D)

23. (A) The signal must be real and even. Only signal (c) and (e) are real and even. 24. (A) Y ( e jW) = e jaW X ( e jW),

y[ n] = x[ n + a ]

real.). Therefore x [ n + a ] is even and x [ n] has to be symmetric about a.This is true for signal (a), (c), (e), (f) and (g).

ò X (e

2

n = -¥

DTFT ¬ ¾¾ ®

j

) dW = 2 px[0 ] ,

35. (A) Y ( e jW) = e - j 4 W X ( e jW) |n - 4|

æ 3ö y[ n] = x[ n - 4 ] = ( n - 4) ç ÷ è4ø

imaginary. Thus y[ n] = 0.

x[0 ] = 0 is for signal (c), (f), (g) and (h).

37. (D) X 2 ( e jW) = X ( e j 2 W)

is always periodic with period 2p.

X ( e j 2 W)

Therefore all signals satisfy the condition.

DTFT ¬ ¾¾ ®

ì x[ n] , n even x 2 [ n] = í otherwise î 0,

|n| ì 2æ 3 ö jn ï ç ÷ , n even y[ n] = í è4ø ï0 , otherwise î

Page 306

d X ( e jW) dW

36. (C) Since x[ n] is real and odd, X ( e jW) is purely

-p

26. (D) X ( e jW)

34. (C) nx[ n]

¥

= 2 p å | x[ n]| = 28 p

¥ 2 ½dX ( e jW)½ ½ |n|2 x[ n] = 316 p ò- p½ dW ½½ = 2 pnå = -¥

If Y ( e ) is real, then y[ n] is real and even (if x[ n] is

25. (D)

2

p

jW

jW

jW

-p

is real and odd.

p

p

ò | X ( e )|

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The Discrete-Time Fourier Transform

38. (B) Y ( e jW) = X ( e jW ) * X ( e j ( W- p 2 ) ) y[ n] = 2 px[ n] x1 [ n], x1 [ n] = e y[ n] = 2 pn2 e jpn

Þ

39. (C) Y ( e jW) =

æ 3ö ç ÷ è4ø

2

jpn 2

This is possible only if b = -a.

x[ n] , 44. (A) For x[ n] = d[ n], X ( e jW) = 1,

2|n|

h[ n] =

d X ( e jW) dW

= |n|

æ 3ö y[ n] = - jnx[ n] = - jn2 ç ÷ è4ø

Þ

41. (C) For a real signal x[ n]

)

3=

2

dW =

å | x[ n]|

2

n = -¥

x[0 ] = ± 1,

å | x[ n]|

2

= ( x [0 ]) + 2 But

x[0 ] = 0,

Hence

n

n

n =0

=

-p

n

æ1ö h2 [ n] = ç ÷ u[ n] è 3ø

¬ ¾¾® DTFT

1 2 - jW 1- e 3 d j dW

æ ç 1 ç çç 1 - 2 e - jW è 3

2 - jW ö e ÷ ÷= 3 ÷÷ 1 - 2 e - jW ø 3

2 - jW e Y ( e jW ) 3 47. (B) H ( e ) = = X ( e jW) 1 - 2 e - jW 3 2 2 æ ö Þ ç 1 - e - jW ÷ Y ( e jW) = e- jW X ( e jW) 3 3 è ø jW

x[0 ] = 1

1 æ1ö ç ÷ 1 - jW è 4 ø 1- e 4

DTFT ¬ ¾¾ ®

n

1 - jW æ ö n e ÷ d ç 1 æ1ö DTFT 4 ç ÷= nç ÷ u[ n] ¬ ¾¾® j 2 dW ç 1 - 1 e - jW ÷ æ è2 ø ç ÷ ç 1 - 1 e - jW ö÷ è 4 ø è 4 ø 1

e jWn dW

2

æ1ö 42. (C) ç ÷ u[ n] è4ø

¥

- jW

2 - jW e 2 e - jW H( e ) = 3 = - jW 2 1 - e - jW 3 - 2 e 3

x[ n] = d[ n] + d[ n + 1] - d[ n + 2 ]

å næçè 2 ö÷ø

p

òe

jW

n =- ¥

-¥

-1

¥

DTFT ¬ ¾¾ ®

n

æ2 ö nç ÷ u[ n] è 3ø

Using Parseval’s relation jW

1 2p

Y ( e jW) , X ( e jW)

n

x[ n] = 2od{ x[ n] } = d[ n + 1] - d[ n + 2 ] For n < 0

ò | X ( e )|

1 , 1 - jW 1- e 3

æ2 ö ç ÷ u[ n] è 3ø

Since x[ n] = 0 for n > 0,

¥

-p

46. (D) H ( e jW) =

Therefore od{x[ n]} = F -1 { jIm{X ( e jW)}} 1 = ( d[ n + 1] - d[ n - 1] - d[ n + 2 ] + d[ n - 2 ]) 2 x[ n] - x[ -n] Od{ x[ n]} = 2

1 2p

) e jWn dW =

1 sin p( n - 1) e jW ( n -1 ) dW = 2 p -òp p( n - 1)

H 2 ( e jW) =

jIm{X ( e jW)}= j sin W - j sin 2W, 1 jW = e - e - j W - e 2 jW + e -2 jW 2

(

jW

p

jIm{X ( e jW)}

DTFT ¬ ¾¾ ®

p

ò Y (e

-12 + 5 e - jW 1 -2 = + 1 - jW 1 12 - 7 e - jW + e - j 2 W 1- e 1 - e - jW 3 4

y[ n] = x[ n] + x[ -n] = 0

od{x[ n]}

1 2p

dX ( e jW) =0 dW

45. (B) H ( e jW) = H1 ( e jW) + H 2 ( e jW)

40. (B) Y ( e jW) = X ( e jW ) + X ( e - jW) Þ

Chap 5.6

¥

å x[ n] = X ( e

j0

)=

n = -¥

y[ n] -

Þ

3 y[ n] - 2 y[ n - 1] = 2 x[ n - 1] .

4 9

|

2 2 y[ n - 1] = x[ n - 1] 3 3

Þ

*********

|

43. (A) For all pass system H ( e jW) = 1 for all W H ( e jW) =

- jW

b+ e , b + e - jW = 1 - a e - jW 1 - ae- jW

|

| |

|

1 + b2 + 2 b cos W = 1 + a 2 - 2 a cos W www.gatehelp.com

Page 307


CHAPTER

5.7 THE CONTINUOUS-TIME FOURIER SERIES

x(t)


A

Determine the Fourier series coefficient for given periodic signal x( t). -2p 3

1. x( t) as shown in fig. P5.7.1

Fig. P5.7.3

10

-5

0

5

10

(C) - j

t

Fig. P5.7.1

(A) 1

æp ö (B) cosç k ÷ è2 ø

æp ö (C) sinç k ÷ è2 ø

(D) 2

A æç - j ççè e 2 pk ç è

æ 4 pk ö ÷÷ 3 ø

(B) j

ö - 1÷ ÷ ø

(D)

A æç - j ççè e 2 pk ç è

æ 4 pk ö ÷÷ 3 ø

- A æç - j ççè e 2 pk ç è

æ 4 pk ö ÷÷ 3 ø

ö - 1÷ ÷ ø

ö - 1÷ ÷ ø

4. x( t) as shown in fig. P5.7.4 x(t) A 1 -1 -A


Fig. P5.7.4

x(t) A

-T

0

T 4

T 2

T

t

A (A) (1 - ( - 1 ) k ) kp A (C) (1 - ( - 1 ) k ) jkp

A (1 + ( - 1 ) k ) kp A (D) (1 + ( - 1 ) k ) jkp

(B)

5. x( t) = sin 2 t

Fig. P5.7.2

(A)

A æp ö sin ç k ÷ jpk è2 ø

(B)

A æp ö cos ç k ÷ jpk è2 ø

(A) -

1 1 1 d[ k - 1] + d[ k] - d[ k + 1] 4 2 4

(C)

A æp ö sin ç k ÷ pk è2 ø

(D)

A æp ö cos ç k ÷ pk è2 ø

(B) -

1 1 1 d [ k - 2 ] + d[ k] - d[ k + 2 ] 4 2 4

(C) -

1 1 d[ k - 1] + d[ k] - d[ k + 1] 2 2

(D) -

1 1 d[ k - 2 ] + d[ k] - d[ k + 2 ] 2 2


Page 308

t

4p

2p

æ 4 pk ö ö A æç - j ççè 3 ÷÷ø (A) - 1÷ e ÷ 2 pk ç è ø

x(t)

-10

-4p 3

0

www.gatehelp.com

t


The Continuous-Time Fourier Series


(A)

sin 9pt sin pt

(B)

(C)

sin 18 pt 2 sin pt


In the question, the FS coefficient of time-domain signal have been given. Determine the corresponding time domain signal and choose correct option. 7. X [ k] = jd[ k - 1] - jd[ k + 1] + d[ k + 3] + d[ k - 3], wo = 2 p (A) 2(cos 3pt - sin pt)

(B) -2(cos 3pt - sin pt)

(C) 2(cos 6 pt - sin 2 pt)

(D) -2(cos 6 pt - sin 2 pt)

Chap 5.7

sin 9pt p sin pt

11. X [ k] As depicted in fig. P5.7.11, wo = p X [k] 3 2

|k|

æ -1 ö 8. X [ k] = ç ÷ , wo = 1 3 ø 4è (A) 5 + 3 sin t

5 (B) 4 + 3 sin t

5 (C) 4 + 3 cos t

4 (D) 5 + 3 cos t

1 k -6

-5

-4

-3

-2

-1

0

1

2

3

4

5

6

Fig. P5.7.11

(A) 3 cos 3pt + 2 cos 2 pt + cos pt (B) 3 sin 3pt + 2 sin 2 pt + sin pt

9. X [ k] as shown in fig. P5.7.9 , wo = p |{X [k]}|

(C) 6 sin 3pt + 4 sin 2 pt + 2 sin pt

2

(D) 6 cos 3pt + 4 cos 2 pt + 2 cos pt

1 k -6

-5

-4

-3

-2

-1

1

0

2

3

4

5

6


Ð{X [k]} p 4

Consider a continuous time periodic signal x( t) k

-p 4

with

coefficients

(C) e

-1

1

0

Determine

the

Fourier

series

2

3

4

5

- to

X [ k] + e to X [ -k]

æ 2p ö (B) 2 sin ç kto ÷ X [ k] T è ø (D) e

- to

X [ -k] + e to X [ k]

(A)

X [ k] + X [ - k] 2

(B)

X [ k] - X [ - k] 2

6

(C)

X [ k] + X *[ - k] 2

(D)

X [ k] + X *[ - k] 2

Ð{X [k]} 8p 6p 4p 2p

14. y( t) =Re{x( t)} k

-2p -4p -6p -8p

series

13. y( t) = Ev{x( t)}

k -2

Fourier

æ 2p ö (A) 2 cos ç kto ÷ X [ k] T è ø

1 -3

and

12. y( t) = x( t - to ) + x ( t - to )

|{X [k]}|

-4

T

choose correct option.

10. X [ k] As shown in fig. P5.7.10 , wo = 2 p

-5

X [ k].

period

coefficient of the signal y( t) given in question and

Fig. P5.7.9

pö pö æ æ (A) 6 cos ç 2 pt + ÷ - 3 cos ç 3pt - ÷ 4 4 è ø è ø pö pö æ æ (B) 4 cos ç 4 pt - ÷ - 2 cos ç 3pt + ÷ 4ø 4ø è è pö pö æ æ (C) 2 cos ç 2 pt + ÷ - 2 cos ç 3pt - ÷ 4ø 4ø è è pö pö æ æ (D) 4 cos ç 4 pt + ÷ + 2 cos ç 3pt - ÷ 4 4 è ø è ø

-6

fundamental

(A)

X [ k] + X [ - k] 2

(B)

X [ k] - X [ - k] 2

(C)

X [ k] + X *[ - k] 2

(D)

X [ k] + X *[ - k] 2

Fig. P5.7.10

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Page 309


UNIT 5

d 2 x( t) dt 2

15. y( t) =

2

æ 2 pk ö (A) ç ÷ X [ k] è T ø 2

æ 2 pk ö (C) jç ÷ X [ k] è T ø

Signal & System

Statement for Q.20-21: Let x1 ( t) and x2 ( t) be continuous time periodic

2

æ 2 pk ö (B) - ç ÷ X [ k] è T ø

signal with fundamental frequency w1 and w2 , Fourier series coefficients X1 [ k] and X 2 [ k] respectively. Given

2

æ 2 pk ö (D) - jç ÷ X [ k] è T ø

that

x2 ( t) = x1 ( t - 1) + x1 (1 - t)

20. The relation between w1 and w2 is 16. y( t) = x( 4 t - 1) 8p (A) X [ k] T (C) e

æ 8 pö - jk çç ÷÷ èTø

(A) w2 =

4p (B) X [ k] T

X [ k]

(D) e

æ 8 pö jk çç ÷÷ èTø

w1 2

(B) w2 = w12

(C) w2 = w1

X [ k]

(D) w2 = w1

21. The Fourier coefficient X 2 [ k] will be

17. Consider a continuous-time signal

(A) ( X1 [ k] - jX1 [ -k]) e - jw 1 k

x( t) = 4 cos 100 pt sin 1000 pt 1 with fundamental period T = . The nonzero FS 50 coefficient for this function are

(B) ( X1 [ -k] - jX1 [ k]) e - jw 1 k

(A) X[ -4 ], X[ 4 ], X[ -7 ], X[7 ]

(C) ( X1 [ k] + jX1 [ -k]) e - jw 1 k (D) None of the above Statement for Q.22-23:

(B) X[ -1], X[1], X[ -10 ], X[10 ]

Consider three continuous-time periodic signals

(C) X[ -3], X[ 3], X[ -4 ], X[ 4 ]

whose Fourier series representation are as follows.

(D) X[ -9 ], X[9 ], X[ -11], X[11]

2p

k

100 æ 1 ö - jk t x1 ( t) = å ç ÷ e 50 k=0 è 3 ø

18. A real valued continuous-time signal x( t) has a fundamental period T = 8. The nonzero Fourier series

x2 ( t) =

coefficients for x( t) are

100

å cos kp e

- jk

2p t 50

k = -100

x3( t) =

X [1] = X [ -1] = 4, X [ 3] = X *[ -3] = 4 j

100

å

k = -100

2p

æ kp ö - jk 50 t j sinç ÷e è 2 ø

The signal x( t) would be æp ö æ 3p ö (A) 4 cos ç t ÷ + 4 j sin ç t÷ è4 ø è 4 ø

22. The even signals are (A) x2 ( t) only

(B) x2 ( t) and x3( t)

æp ö æ 3p ö (B) 4 cos ç t ÷ - 4 j cos ç t ÷ è4 ø è 4 ø

(C) x1 ( t) and x3( t)

(D) x1 ( t) only

pö æp ö æ 3p (C) 8 cos ç t ÷ + 8 cos ç t + ÷ 2ø è4 ø è 4

23. The real valued signals are (A) x1 ( t) and x2 ( t)

(B) x2 ( t) and x3( t)


(C) x3( t) and x1 ( t)

(D) x1 ( t) and x3( t)

19. The continuous-time periodic signal is given as æ 2p ö æ 5p ö x( t) = 4 + cos ç t ÷ + 6 sin ç t÷ è 3 ø è 3 ø The nonzero Fourier coefficients are (A) X [0 ], X [ -1], X [1], X [ -5 ], X [5 ]

24. Suppose the periodic signal x( t) has fundamental period T and Fourier coefficients X [ k]. Let Y [ k] be the Fourier coefficient of y( t) where Fourier coefficient X [ k] will be (A)

TY [ k] ,k¹0 j2 pk

(B)

TY [ k] j2pk

(C)

TY [ k] ,k¹0 jk

(D)

TY [ k] jk

(B) X [0 ], X [ -2 ], X [2 ], X [ -5 ], X [5 ] (C) X [0 ], X [ -4 ], X [ 4 ], X [ -10 ], X [10 ] (D) None of the above Page 310

y( t) = dx( t) dt . The

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25. Suppose we have given the following information

The signal will be æp ö æp ö (A) 4 cos ç t ÷ - 2 sin ç t ÷ è4 ø è4 ø

about a signal x( t) : 1. x( t) is real and odd.

æp ö æp ö (B) 2 cos ç t ÷ + 4 sin ç t ÷ è4 ø è4 ø

2. x( t) is periodic with T = 2 3. Fourier coefficients X [ k] = 0 for |k|> 1

æp ö æp ö (C) 2 cos ç t ÷ + 2 sin ç t ÷ è4 ø è4 ø

2

4.

Chap 5.7

2 1 x( t) dt = 1 ò 20


The signal, that satisfy these condition, is (A)

2 sin pt and unique

(B)

2 sin pt but not unique

Statement for Q.29-31: Consider the following three continuous-time

(C) 2 sin pt and unique

signals with a fundamental period of T = 1

(D) 2 sin pt but not unique

x( t) = cos 2pt , y( t) = sin 2pt , z( t) = x( t) y( t)

26. Consider a continuous-time LTI system whose

29. The Fourier series coefficient X [ k] of x( t) are

frequency response is

(A)

1 2

( d[ k + 1] + d[ k - 1])

(B)

1 2

( d[ k + 1] - d[ k - 1])

(C)

1 2

( d[ k - 1] - d[ k + 1])

H ( jw) =

¥

ò h( t) e

- jwt

dt =

-¥

sin 4 w w


The input to this system is a periodic signal ì 2, 0 £ t £ 4 x( t) = í î -2, 4 £ t £ 8

30. The Fourier series coefficient of y( t), Y [ k] will be

with period T = 8. The output y( t) will be æ pt ö (A) 1 + sin ç ÷ è4ø

æ pt ö (B) 1 + cos ç ÷ è4ø

æ pt ö æ pt ö (C) 1 + sin ç ÷ + cos ç ÷ è4ø è4ø

(D) 0

2

2

(A)

j 2

( d[ k + 1] + d[ k + 1])

(B)

j 2

( d[ k + 1] - d[ k - 1])

(C)

j 2

( d[ k - 1] - d[ k + 1])

(D)

1 2j

( d[ k + 1] + d[ k + 1])

31. The Fourier series coefficient of z( t) , Z [ k] will be

27. Consider a continuous-time ideal low pass filter having the frequency response ì 1, |w|£ 80 H ( jw) = í î 0, |w|> 80

(A)

1 4j

( d[ k - 2 ] - d[ k + 2 ])

(B)

1 2j

( d[ k - 2 ] - d[ k + 2 ])

(C)

1 2j

d[ k + 2 ] - d[ k - 2 ])


When the input to this filter is a signal x( t) with fundamental frequency wo = 12 and Fourier series

32. Consider a periodic signal x( t) whose Fourier series coefficients are ì 2, ï X [ k] = í æ 1 ö|k| ï jç 2 ÷ , î è ø

S coefficients X [ k], it is found that x( t) ¬¾ ® y( t) = x( t).

The largest value of|k, | for which X [ k] is nonzero, is (A) 6

(B) 80

(C) 7

(D) 12

28.

A

continuous-time

periodic

otherwise

Consider the statements signal

has

a

fundamental period T = 8. The nonzero Fourier series coefficients are as, X [1] = X [ -1] = j , X [5 ] = X [ -5 ] = 2, *

k =0

1. x( t) is real.

2. x( t) is even

3.

dx( t) is even dt

The true statements are (A) 1 and 2

(B) only 2

(C) only 1

(D) 1 and 3

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Page 311


UNIT 5


(A)

A 2A 1 1 + (sin t - sin 3t + sin 5 t....) 2 p 3 5

(B)

A 2A 1 1 + (cos t - cos 2 t + cos 3t....) 2 p 2 3

(C)

A 2A 1 1 + (cos t - cos 3t + cos 5 t....) 2 p 3 5

(D)

A 2A 1 1 + (sin t + cos t + sin 3t + cos 3t ....) 2 p 3 3

A waveform for one peroid is depicted in figure in question. Determine the trigonometric Fourier series and choose correct option. 33. x(t) 1 -p p

Signal & System

t

36. -1

x(t)

Fig. P5.7.33

2

2 1 1 1 (A) (cos t + cos 2 t + cos 3t + cos 4 t +....) p 2 3 4 2 1 1 1 (sin t - sin 2 t + sin 3t - sin 4 t +....) p 2 3 4

(C)

2 1 1 1 (sin t + cos t - sin 2 t - cos 2 t + sin 3t +....) p 2 2 3

Fig. P5.7.36

(A)

A

1 12 1 1 + (cos pt + cos 3pt + cos 5 pt +....) 2 p2 9 25

(B) 3 + (C)

x(t)

p

*****

Fig. P5.7.34

(A)

A 4A æ 1 1 ö + ç sin t + sin 2 t + sin 3t +.... ÷ 2 p è 2 3 ø

(B)

A 4A æ 1 1 ö + ç cos t + cos 3t + cos 5 t +.... ÷ 2 p è 3 5 ø

(C)

4A æ 1 1 ö ç sin t + sin 3t + sin 5 t + .... ÷ p è 3 5 ø

(D)

4A æ 1 1 ö ç cos t + cos 2 t + cos 3t +.... ÷ p è 2 3 ø

35. x(t) A

-p 2

p 2

p

t

Fig. P5.7.35

Page 312

12 1 1 (sin pt - sin 3pt + sin 5 pt -....) 2 p 9 25

t

-A

-p

12 1 1 (cos pt + cos 3pt + cos 5 pt +....) p2 9 25

1 12 1 1 + (sin pt - sin 3pt + sin 5 pt -....) 2 p2 9 25

(D) 3 +

-p

t

-1

2 1 1 1 (D) (sin t + cos t + sin 3t + cos 3t + sin 5 t + ....) p 3 3 5 34.

1

-1

(B)

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SOLUTIONS 1 T

T 2 - jkw t ò Ad( t) e o dt =

-T 2

- jkw ot

-T 2

1 dt = T

ò Ae

- jkw ot

- jkt ò x( t) e dt = 0

dt

1 2p

Ae- jkt dt =

0

jA é êe 2 pk ê ë

ù - 1ú úû

æ 4 pk ö ÷÷ j çç è 3 ø

å X [ k]e

k = -¥

å X [ k]e

=

sin 9 pt sin pt

jpkt

1 T

X1 [ k] =

ò x( t - t ) e o

-k

e-

T

j kw o to

T

ò x( t) e

- jkw ot

dt

T

Y [ k] = X1 [ k] + X 2 [ k] = e - jkw oto X [ k] + e jkw oto X [ k] = 2 cos ( wo kto ) X [ k] 13. (A) Ev{ x( t)} =

2p =2 , p

x( t) + x( - t) , 2

The FS coefficients of x( t) are 1 1 X1 [ k] = ò x( -t) e - jkw ot dt = ò x( t) e jkw ot dt = X [ -k] TT T T Therefore, the FS X [ k] + X [ -k] Y [ k] = 2

j 2 pt

æ -1 ö jkt = åç ÷ e + k = -¥ è 3 ø

-1 - jt e 4 1 = 3 + = 1 1 1 + e - jt 1 + e j t 5 + 3cos t 3 3

dt =

The FS coefficients of x( t - to ) + x( t + to ) are

2

-1

- jkw ot

X 2 [ k] = e jkw oto X [ k]

ö - jkpt ò0 Ae dt ÷÷ ø 1

= je j 2 p t - je -

series coefficients X1 [ k] of x( t - to ) are

+ e j6 pt + e -

k

æ -1 ö jkt ÷ e ç å k=0 è 3 ø ¥

coefficients

of

Ev{ x( t)}

are

j6 pt

14. (C) Re{ x( t)} =

jkt

- j 2 pk ( t -1 )

k = -4

Similarly, the FS coefficients of x( t + to ) are

ö -1 2 jt ÷÷ = ( e - 2 + e -2 jt ) 4 ø

j 2 pkt

4

åe

e jpkt =

= e - jkw oto X [ k]

= - 2 sin 2 pt + 2 cos 6 pt 8. (D) x( t) =

p

12. (A) x( t - to ) is also periodic with T. The Fourier

k = -¥

¥

j

e j ( 3) pt + 2 e 4 e j (4 )pt

= 6 cos 3pt + 4 cos 2 pt + 2 cos pt

-1 1 1 d[ k - 1] + d[ k] - d[ k + 1] 4 2 4 ¥

4

= 3e j ( -3) pt + 2 e j ( -2 ) pt + e j ( -1 ) pt + e j (1 ) pt + 2 e j ( 2 ) pt + 3e j ( 3) pt

4p 0
The fundamental period of sin 2 ( t) is p and wo =

å X [ k]e

jp

k = -¥

A æ 1 - e jkp e - jkp -1 ö A ÷÷ = (1 - ( -1) k ) = çç + 2 è jkp - jkp ø jkp

7. (C) x( t) =

- j 2 pk

¥

x( t) =

1 0 1 1æ X [ k] = ò x( t) e - jkt dt = çç ò - Ae - jkpt dt + 2 -1 2 è -1

X [ k] =

-

11. (D) X [ k] =|k|, - 3 £ k £ 3

4p 3

ò

4

k = -4

ì - A, - 1 < t < 0 2p 4. (C) T = 2, wo = = p, x( t) = í 0< t<1 2 î A,

æ e jt - e - jt 5. (A) sin 2 t = çç è 2j

p

-T 4

ì ï A, 2p 3. (B) T = 2p , wo = = 1, x( t) = í 2p ï0, î 2p

j

e j ( -4 ) pt + e 4 e j ( -3) pt + e

åe

x( t) =

T 4

4

1 2p

p 4

= 4 cos ( 4 pt + p 4) + 2 cos( 3pt - p 4)

é e - jkw o t ù 4 A æ pk ö ê - jkw ú - T = pk sinç 2 ÷ è ø o û ë

X [ k] =

- j

= 2( e - j ( 4 pt + p 4 ) + e j ( 4 pt + p 4 ) ) + ( e - j ( 3pt - p 4 ) + e j ( 3pt - p 4 ) )

A , T

T

A = T

jpkt

10. (A) X [ k] = e - j 2 pk , -4 £ k £ 4

T 2

ò x( t) e

å X [ k]e

k = -¥

A = 10 , T = 5, X [ k] = 2 1 2. (C) X [ k] = T

¥

9. (D) x( t) = = 2e

1. (D) X [ k] =

Chap 5.7

x ( t) + x *( t) , 2

The FS coefficient of x *( t) is 1 X1 [ k] = ò x *( t) e - jkw ot dt = X1*[ -k] TT X1*[ k] =

1 T

ò x( t) e

jkw ot

dt = X [ -k]

T

X1 [ k] = X *[ -k] Y [ k] = www.gatehelp.com

X [ k] + X *[ - k] 2 Page 313


UNIT 5

æ 10 p ö æ 4p ö 13. X [ k] = cos ç k ÷ + 2 j sin ç k÷ è 19 ø è 19 ø

18. y[ n] = x[ n] - x[ n - N 2 ] , (assume that N is even) (A) (1 - ( -1) k + 1 ) X [2 k]

(B) (1 - ( -1) k ) X [ k] (D) (1 - ( -1) k\ ) X [2 k]

(A)

19 ( d[ n + 5 ] + d[ n - 5 ]) + 19(d[ n + 2 ] - d[ n - 2 ]),|n|£ 9 2

(C) (1 - ( -1) k + 1 ) X [ k]

(B)

1 ( d[ n + 5 ] + d[ n - 5 ]) + (d[ n + 2 ] - d[ n - 2 ]),|n|£ 9 2

19. y[ n] = x *[ -n]

9 (C) ( d[ n + 5 ] + d[ n - 5 ]) + 9(d[ n + 2 ] - d[ n - 2 ]),|n|£ 9 2 (D)

1 ( d[ n + 5 ] + d[ n - 5 ]) + (d[ n + 2 ] - d[ n - 2 ]),|n|£ 9 2

æ pk ö 14. X [ k] = cos ç ÷ è 21 ø

(A) - X *[ k ]

(B) - X *[ -k ]

(C) X *[ k ]

(D) X *[ -k ]

20. y[ n] = ( -1) n x[ n], (assume that N is even) N ù é (A) X êk 2 úû ë

N ù é (B) X êk + 2 úû ë N é ù (D) X êk + - 1ú 2 ë û

(A)

21 (d[ n + 4 ] + d[ n - 4 ]),|n|£ 10 2

N é ù (C) X êk + 1ú 2 ë û

(B)

1 (d[ n + 4 ] + d[ n - 4 ]),|n|£ 10 2


(C)

21 (d[ n + 4 ] - d[ n - 4 ]),|n|£ 10 2

(D)

1 (d[ n + 4 ] - d[ n - 4 ]),|n|£ 10 2

Consider a discrete-time periodic signal ì 1, 0 £ n £ 7 x[ n] = í î 0, 8 £ n £ 9 with period N = 10. Also y[ n] = x[ n] - x[ n - 1 ]

Statement for Q.15-20: Consider a periodic signal x[ n] with period N and

21. The fundamental period of y[ n] is

FS coefficients X [ k]. Determine the FS coefficients Y [ k]

(A) 9

(B) 10

of the signal y[ n] given in question.

(C) 11


15. y[ n] = x[ n - no ] (A) e

æ 2 pö ÷÷ n ok j çç è Nø

(B) e

X [ k]

æ 2 pö ÷÷ n ok - j çç è Nø

22. The FS coefficients of y[ n] are (A)

æ 8 pö j çç ÷÷ k ö 1 æç 1-e è 5ø ÷ ÷ 10 ç è ø

(B)

æ 8 pö - j çç ÷÷ k ö 1 æç 1-e è 5ø ÷ ÷ 10 ç è ø

(C)

1 10

æ 4 pö ÷k ö æ jç ç 1 - e çè 5 ÷ø ÷ ç ÷ è ø

(D)

1 10

X [ k]

(D) -kX [ k]

(C) kX [ k] 16. y[ n] = x[ n] - x[ n - 2 ] æ 4p ö (A) sin ç k ÷ X [ k] è N ø

æ 4p ö (B) cos ç k ÷ X [ k] è N ø

æ 4 pö ÷÷ k ö æ - j çç (C) ç 1 - e è N ø ÷ X [ k] ç ÷ è ø

æ 4 pö ÷÷ k ö æ j çç (D) ç 1 - e è N ø ÷ X [ k] ç ÷ è ø

23. The FS coefficients of x[ n] are æ pk ö

(A) -

17. y[ n] = x[ n] + x[ n + N 2 ] , (assume that N is even) æN ö (A) 2 X [2 k - 1], for 0 £ k £ ç - 1÷ è 2 ø (B) 2 X [2 k - 1], for 0 £ k £

Page 318

Signal & System

N 2

(C) 2 X [2 k],

æN ö for 0 £ k £ ç - 1÷ è 2 ø

(D) 2 X [2 k],

for 0 £ k £

N 2

j - j ççè 10 ÷÷ø æ pk ö e cosec ç ÷ Y [ k], k ¹ 0 2 è 10 ø æ pk ö

(B)

j - j ççè 10 ÷÷ø æ pk ö e cosec ç ÷ Y [ k], k ¹ 0 2 è 10 ø æ pk ö

(C) -

1 - j ççè 10 ÷÷ø æ pk ö e sec ç ÷ Y [ k] 2 è 10 ø æ pk ö

(D)

1 - j ççè 10 ÷÷ø æ pk ö e sec ç ÷ Y [ k] 2 è 10 ø

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æ 4 pö ÷k ö æ - jç ç 1 - e çè 5 ÷ø ÷ ç ÷ è ø


The Discrete-Time Fourier Series


29. A real and odd periodic signal x[ n] has fundamental

Consider a discrete-time signal with Fourier

period N = 7 and FS coefficients X [ k]. Given that X [15 ] = j,

representation. x[ n]

¬ ¾ ¾¾®

X [ k]

Determine the corresponding signal y[ n] and choose correct option.

æp ö (B) 2 cos ç n ÷ x[ n] è5 ø

æp ö (C) 2 sin ç n ÷ x[ n] è2 ø

æp ö (D) 2 cos ç n ÷ x[ n] è2 ø

(A) 0, j, 2 j, 3 j

(B) 1, 1, 2, 3

(C) 1, -1, -2, -3

(D) 0, - j , -2 j, -3 j

(B)

4.

1 ( x[ n + 2 ] + x[ n - 2 ]) 2

9

1 10

å X [ k]

The signal x[ n] is æ p ö (A) 5 cos ç n÷ è 10 ø

æ p ö (B) 5 sin ç n÷ è 10 ø

æp ö (C) 10 cos ç n ÷ è5 ø

æp ö (D) 10 sin ç n ÷ è5 ø

31. Each of two sequence x[ n] and y[ n] has a period N = 4. The FS coefficient are

(B) j2 p( x[ n]) 2

The FS coefficient Z [ k] for the signal z[ n] = x[ n] y[ n] will be

x [ n] - x[ -n] 2

(B)

x [ n] + x[ -n] 2p

(D)

(A) 6

(B) 6|k|

(C) 6|k|

(D) e

æ 11p ö sin ç n÷ 20 ø è x[ n] = æ p ö sin ç n÷ è 20 ø

1. x[ n] is periodic with N = 6 5

å x[ n] = 2

n =0

n

with a fundamental period N = 20. The Fourier

x[ n] = 1

n -2

4. x[ n] has the minimum power per period among the set of signals satisfying the preceding three condition. The sequence would be.. ì 1 1 1 1 1 ü (A) í ... , , , , , ...ý (B) î 2 6 2 6 2 þ

ì 1 1 1 1 1 ü (C) í ... , , , , , ...ý î 3 6 3 6 3 þ

p j k 2

32. Consider a discrete-time periodic signal

28. Consider a sequence x[ n] with following facts :

7

1 1 X [1] = X [2 ] = 1 and 2 2

Y [0 ], Y [1], Y [2 ], Y [ 3] = 1

(D) 2 p( x[ n]) 2

x [ n] - x[ -n] 2p

å ( -1)

= 50

X [0 ] = X [ 3] =

27. Y [ k] = Re{ X [ k]} x [ n] + x[ -n] (A) 2

3.

2

n =0

2

(C) ( x[ n]) 2

2.

of

30. Consider a signal x[ n] with following facts

26. Y [ k] = X [ k] * X [ k]

(C)

values

3. X[11] = 5

1 ( x[ n + 10 ] + x[ n + 10 ]) (D) None of the above 2

( x[ n]) 2p

The

2. The period of x[ n] is N = 10

æp ö (A) 2 sin ç n ÷ x[ n] è5 ø

(A)

X [17 ] = 3 j.

1. x[ n] is a real and even signal

24. Y [ k ] = X [ k - 5 ] + X [ k + 5 ]

æ pk ö 25. Y [ k] = cos ç ÷ X [ k] è 5 ø 1 (A) ( x[ n + 5 ] + x[ n + 5 ]) 2

X [16 ] = 2 j,

X [0 ], X [ -1], X [ -2 ], and X [ -3] will be

p DTFS ; 10

In question the FS coefficient Y [ k] is given.

(C)

Chap 5.8

series coefficients of this function are 1 (A) ( u[ k + 5 ] - u[ k - 6 ]), |k|£ 10 20 (B)

1 1 1 ü ì í ...0, 1, , , , ...ý 2 3 4 þ î

1 ( u[ k + 5 ] - u[ k - 5 ]), |k|£ 10 20

(C) ( u[ k + 5 ] - u[ k + 6 ]), |k|£ 10 (D) ( u[ k + 5 ] - u[ k - 6 ]), |k|£ 10

(D) {...0, 1, 2, 3, 4, ...}

************

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Page 319


The Discrete-Time Fourier Series

11. (D) N = 7, W o = x[ n] =

3

å X [ k]e

2p , 7

æ 2 pö ÷÷ kn j çç è 7 ø

17. (C) Note that y[ n] = x[ n] + x [ n + N 2 ] has a period

= 2e

æ 2 pö ÷÷ n j ( -1 ) çç è 7 ø

- 1 + 2e

of N 2 and N has been assumed to be even,

æ 2 pö ÷÷ n j (1 ) çç è 7 ø

Y [ k] =

n = -3

æ 2p ö = 4 cos ç n÷ - 1 è 7 ø

6

æ pö - j çç ÷÷ k è6 ø

e

æ pö j çç ÷÷ kn è6 ø

e =

9p j ( n -1 ) 6

p j ( n -1 ) ö æ ÷ ç 1-e 6 ÷ ç ø è

13. (A) N = 19, W o =

ö ÷ sin æç 3p ( n - 1) ö÷ ÷ ø = è 4 ø æ p ö sin ç ( n - 1) ÷ è 12 ø

æ 4 pö ÷÷ kn - j çç è Nø

n =0

19. (C) y[ n] = x *[ -n] 1 Y [ k] = N

2p 19

N -1

å x [ -n]e *


= X *[ k]

n =0

20. (A) With N even

æ 10 p ö æ 10 p ö X [ k] = cos ç k ÷ + 2 j sin ç k÷ 19 è ø è 19 ø =

( x[ n] + x[ n + N 2 ]) e

k even ì 0, =í 2 X [ k ], k odd î

k = -6

æ ç 1-e ç è

å

æ 2 pö N ÷÷ k ö æ - j çç Y [ k] = ç 1 - e è N ø 2 ÷ X [ k] = (1 - e - jpk ) X [ k] ç ÷ è ø

æ pö j çç ÷÷ k ( n -1 ) è6 ø

6

åe

=

k = -6

p j ( -4 ) ( n -1 ) 6

N 2 -1

18. (B) y[ n] = x[ n] - x [ n - N 2 ]

æ pö

åe

2 N

= 2 X [2 k] for 0 £ k £ ( N 2 - 1)

- j çç ÷÷ k p 12. (C) N = 12, W o = , X [ k] = e è 6 ø 6

x[ n] =

Chap 5.8

y[ n] = ( -1) n x[ n] = e jpn x[ n] = e

2p 2p 2p 2p - j ( 5) kö k - j(2 ) kö æ - j ( -2 ) 19 1 æ - j ( -5) 19 k 19 ÷ 19 ÷ çe çe + e + + e ÷ ç ÷ 2 çè ø è ø

Y [ k] =

By inspection 19 x[ n] = ( d[ n + 5 ] + d[ n - 5 ]) + 19 ( d[ n + 2 ] - d [ n - 2 ]), 2 Where |n|£ 9

=

1 N

1 N

N -1

åe

æ 2 pö N ÷÷ j çç è Nø 2

x[ n]e

æ 2 pö N ÷÷ j çç è Nø 2

x[ n]


n =0

N -1

å x[ n]e

æ 2 pö æ Nö ÷÷ n çç k - ÷÷ - j çç 2 ø è Nø è

= X [k - N 2]

n =0

21. (B) y[ n] is shown is fig. S5.8.21. It has fundamental y[n]

14. (A) N = 21, W o =

2p 21

1 9 2p - j ( -4 ) k 21

2p - j(4 ) k 21

ö æ 8 p ö 1 æç ÷ X [ k] = cos ç k÷ = ç e +e ÷ 21 2 è ø è ø 1 Since X [ k] = å x[ n]e- jkWon , By inspection N n=N

-1

15. (B) Y [ k] = æ 2 pö

=

1 - j ççè N ÷÷ø kn o e N

1 N

N -1

å x[ n - n ]e o

n =0

å x[ n]e


æ 2 pö ÷÷ kn o - j çç è Nø

22. (B) Y [ k] = =

1 æç 1-e 10 ç è

16. (C) Y [ k] = X [ k] - e

æ X [ k] = ç 1 - e ç è

5

7

8

n 10 11

9

å y[ n]e

æ 2 pö ÷÷ kn - j çç è 10 ø

n =0

ö 1 ÷= ÷ 10 ø

æ ç1 - e ç è

æ 8 pö - j çç ÷÷ k è 5ø

ö ÷ ÷ ø

23. (A) y [ n] = x [ n] - x [ n - 1]

X [ k] æ 4 pö ÷÷ k - j çç è Nø

1 10

æ 2 pö ÷÷ k8 - j çç è 10 ø

n =0

æ 2 pö ÷÷ 2 k - j çç è Nø

4

period of 10.


=e

3

Fig. S5.8.21

ì 21 ï , n = ±4 x[ n] = í 2 ïî 0, otherwise n Î { -10, - 9, ......9. 10} N -1

2

1

ö ÷ X [ k] ÷ ø

Y [ k] = X [ k] - e

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æ 2 pö ÷÷ k - j çç è 10 ø

X [ k]

Þ

X [ k] =

Y [ k] 1-e

æ pö - j çç ÷÷ k è 5ø

Page 321


UNIT 5

Þ

e

X [ k] = e

=

-j e 2

æ p ö - j çç k ÷÷ è 10 ø

æ pö j çç ÷÷ k è 10 ø

Y [ k]

-e

æ pö - j çç ÷÷ k è 10 ø

æ p cosec ç è 10

24. (D) W o = Þ


=


29. (D) Since the FS coefficient repeat every N. Thus

Y [ k] æ pk ö 2 j sin ç ÷ è 10 ø e

Signal & System

X [1] = X [15 ], X [2 ] = X [16 ], X [ 3] = X [17 ] The signal real and odd, the FS coefficient X [ k] will be purely imaginary and odd. Therefore X[0 ] = 0

ö k ÷ Y [ k] ø

X [ -1] = - X [1], X [ -2 ] = - X [2 ], X [ -3] = - X [ 3] Therefore (D) is correct option.

p , Y [ k] = X [ k - 5 ] + X [ k + 5 ] 10

30. (C) Since N = 10, X [11] = X [1] = 5

p j ( -5) n ö æ j ( 5) p n æp ö y[ n] = çç e 10 + e 10 ÷÷ x[ n] = 2 cos ç n ÷ x[ n] è2 ø è ø

Since x[ n] is real and even X [ k] is also real and even. Therefore X [1] = X [ -1] = 5. Using Parseval’s relation

æ çe æp ö 25. (B) Y [ k] = cos ç k ÷ X [ k] = ç è5 ø ç è p

p j k 5

+e 2

p -j k 5

ö ÷ ÷ X [ k] ÷ ø

2

2

2

X [ -1] + X [1] + X [0 ] + 2

X [0 ] +

8

å X [ k]

2

27. (A) Y [ k] =Re{ X [ k]} Þ y[ n] =Ev{ x[ n]} =

x[ n] = å X [ k]e 2

x[ n] + x[ -n] 2

æ = 5ç e ç è

æ 2 pö ÷÷ kn j çç è Nø

=

2p , 6

åe

k = -1

1 Þ X [0 ] = , 3

Þ

1 6

åe

n =0

3

Z[ k ] = X [ 0 ] Y [ k ] + X [ 1 ] Y [ k - 1 ] + X [ 2 ] Y [ k - 2 ] + X [ 3 ] Y [ k - 3 ]

32. (A) N = 20 We know that

1 1 x[ n] = , X [ 3] = 6 6

P = å X [ k] ,

|n|£ 5 ì 1, í î 0, 5 <|n|£ 10

The value of P is minimized by choosing X [1] = X [2 ] = X [ 4 ] = X [5 ] = 0 Therefore x[ n] = X [0 ] + X [ 3]e

p 10

¬ ¾ ¾¾®

ö ÷÷ 3n ø

=

1 1 1 1 + ( -1) n = + ( -1) n 3 6 3 6

Using duality æ 11p ö sin ç n÷ p DTFS ; è 20 ø ¬ ¾ 10 ¾¾ ® æ p ö sin ç n÷ è 20 ø

ì 1 1 1 1 1 ü x[ n] = í ... , , , , , ...ý î 2 6 2 6 2 þ

æ 11p ö sin ç k÷ è 20 ø æ p ö sin ç k÷ è 20 ø

|k|£ 5 1 ì 1, í 20 î 0, 5 <|k|£ 10

*********

Page 322

DTFS ;

2

k=0

æ 2p çç è 6

å X [ l ]Y [ k - l ]

Thus Z [ k] = 6, for all k.

By Parseval’s relation, the average power in x[ n] is 5

DTFS ¬ ¾¾ ®

Since Y [ k] is 1 for all values of k.

å ( -1) n x[ n] = 1

æ 2 pö ÷÷ ( 3) k j çç è 6 ø

ö ÷ = 10 cos æç p n ö÷ ÷ è5 ø ø

= Y [ k] + 2 Y [ k - 1] + 2 Y [ k - 2 ] + Y [ k - 3]

1 x[ n] = 3

n =2 5

æ 2 pö ÷÷ n j çç è 10 ø

l= 0

Þ

7

From fact 3,

æ 2 pö ÷÷ kn j çç è 10 ø

Z [ k] = å X [ l ]Y [ k - l ]

Þ

å x[ n] = 2

n =0

+e

å X [ k]e

k =< N>

n =0

Þ

æ 2p ö - j çç ÷÷ n è 10 ø

8

31. (A) z[ n] = x[ n] y[ n]

5

1 6

= 50

Therefore X [ k] = 0 for k = 0, 2, 3, ..... 8.

26. (C) Y [ k] = X [ k] * X [ k] Þ y[ n] = x[ n] x[ n] = ( x[ n])

æ 2 pö ÷÷ ( 0 ) k j çç è 6 ø

2

2

k = -1

=0

N

5

8

å X [ k]

8

å X [ k]

k=2

1 y[ n] = ( x[ n - 2 ] + x[ n + 2 ]) 2

From fact 2,

= 50 =

k=2

p

28. (A) N = 6, W o =

2

N

j ( -2 ) kö 1 æ j(2 ) k 10 ÷ = çç e 10 + e ÷ X [ k] 2è ø

Þ

å X [ k]

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CHAPTER

6.1 TRANSFER FUNCTION

1. The equivalent transfer function of three parallel

R(s)

+

C(s)

G(s)

blocks G1 ( s) =

1 1 s+3 , G2 ( s) = and G3( s) = is s+1 s+4 s+5

H(s)

Fig. P.6.1.3

( s 3 + 10 s 2 + 34 s + 37) (A) ( s + 1)( s + 4)( s + 5)

(A)

( s + 3) ( s + 1)( s + 4)( s + 5)

s( s + 2)( s + 3) s + 7 s 2 + 12 s + 3

(B)

-( s 3 + 10 s 2 + 34 s + 37) ( s + 1)( s + 4)( s + 5)

s( s + 2)( s + 3) s + 5 s2 + 4 s - 3

(C)

-( s + 3) ( s + 1)( s + 4)( s + 5)

( s + 1)( s + 4) s + 7 s 2 + 12 s + 3

(D)

( s + 1)( s + 4) s + 5 s2 + 4 s - 3

(B) (C) (D)

2. The block having transfer function

3

3

3

3

4. A feedback control system is shown in fig. P.6.1.4.

1 1 s+1 , G2 ( s) = , G3( s) = G1 ( s) = s+2 s+5 s+3

The transfer function for this system is R

are cascaded. The equivalent transfer function is (A) (B)

( s 3 + 10 s 2 + 37 s 2 + 31) ( s + 2)( s + 3)( s + 5)

1 G1

G3

C

Fig. P.6.1.4

-( s 3 + 10 s 2 + 37 s 2 + 31) (C) ( s + 2)( s + 3)( s + 5)

(A)

G1 G2 1 + H1 G1 G2 G3

-( s + 1) (D) ( s + 2)( s + 3)( s + 5)

(B)

G2 G3 G1 (1 + H1 G2 G3)

(C)

G2 G3 1 + H1 G1 G2 G3

(D)

G2 G3 G1 (1 + H1 G2 G3)

3. For a negative feedback system shown in fig. P.6.1.3 s+1 s+3 and H ( s) = s( s + 2) s+4

G2

H2

s+1 ( s + 2) ( s + 3) ( s + 5)

G( s) =

+

The equivalent transfer function is www.gatehelp.com

Page 325


UNIT 6

5. Consider the system shown in fig. P.6.1.5. R(s)

+

G1

+

G2

+

(D)

Control Systems

G1 G2 1 + G1 H1 + G2 H 2 + G1 G2 H1 H 2

C(s) +

8. The closed loop gain of the system shown in fig. P6.1.8 is Fig. P.6.1.5

R

+

C

6

+

The input output relationship of this system is R(s)

C(s)

G1G2

R(s)

1 + G1 + G1G2

(A) R(s)

Fig.P6.1.8

(B) C(s)

G1 + G2

R(s)

1 + G2 + G1G2

(C)

1 3

C(s)

C(s)

(D)

(A) -2

(B) 6

(C) -6

(D) 2

9. The block diagrams shown in fig. P.6.1.9 are

6. A feedback control system shown in fig. P.6.1.6 is

equivalent if G is equal to R(s)

C(s)

s+2 s+1

subjected to noise N ( s). N(s) R(s)

R(s) +

G1

+

+

1 s+1

C(s)

G2

Fig. P.6.1.6

C ( s) The noise transfer function N is N ( s) G2 1 + G1 G2 H

(B)

G2 (C) 1 + G2 H

C(s) +

Fig. P.6.1.9

H2

(A)

+

G

(A) s + 1

(B) 2

(C) s + 2

(D) 1

10. Consider the systems shown in fig. P.6.1.10. If the

G2 1 + G1 H

forward path gain is reduced by 10% in each system, then the variation in C1 and C2 will be respectively


R1

R2

7. A system is shown in fig. P6.1.7. The transfer

16

C1

3

+

C2

10

function for this system is H1 R(s)

+

G1

Fig. P.6.1.10 +

G2

C(s)

(A) 10% and 1%

(B) 2% and 10%

(C) 10% and 0%

(D) 5% and 1%

H2

Fig. P.6.1.7

(A) (B) (C)

Page 326

G1 G2 1 + G1 G1 H 2 + G2 H1

11. The transfer function

C R

of the system shown in the

fig. P.6.1.11 is R

G1 G2 1 + G1 G2 + H1 H 2

1 H1

+

H2

G1

G2

G1 G2 1 - G1 H1 - G2 H 2 + G1 G2 H1 H 2

Fig. P.6.1.11

www.gatehelp.com

C


Transfer Function

(A)

(C)

G1 H 2 H1 (1 + G1 G2 H 2 )

(B)

G2 G1 1 + H1 H 2 G1 G2

(D)

Chap 6.1

15. A closed-loop system is shown in fig. P.6.1.15. The

G1 G2 H 2 H1 (1 + G1 G2 H 2 )

noise transfer function Cn ( s) N ( s) is approximately

G1 G2 H1 (1 + G1 G2 H 2 )

R(s) 1

G1 C(s) -H2

H1

12. In the signal flow graph shown in fig. P6.1.12 the

1

sum of loop gain of non-touching loops is t32 x2

x1

Fig. P.6.1.15 t56

t45

t34

t23

t12

N(s)

t44

t43

x5

x4

x3

x6

(A)

1 For |G1 ( s) H1 ( s) H 2 ( s)| << 1 G1 ( s) H1 ( s)

(B)

1 For |G1 ( s) H1 ( s) H 2 ( s)| >> 1 -H1 ( s)

(C)

1 For |G1 ( s) H1 ( s) H 2 ( s)| >> 1 H1 ( s) H 2 ( s)

(D)

1 For |G1 ( s) H1 ( s) H 2 ( s)| << 1 G1 ( s) H1 ( s) H 2 ( s)

t24 t25

Fig. P.6.1.12

(A) t32 t23 + t44

(B) t23t32 + t34 t43

(C) t24 t43t32 + t44

(D) t23t32 + t34 t43 + t44

13. For the SFG shown in fig. P.6.1.14 the graph determinant D is -c

16. The overall transfer function

of the system shown

in fig. P.6.1.16 will be

b

a

C R

d

1

-H1

-H2

1 R

i

e

G

1

C

h

j

Fig. P.6.1.16

f

(A) G

-g

(B)

G 1 + H2

(D)

G 1 + H1 + H 2

Fig. P.6.1.13

(A) 1 - bc - fg - bcfg + cigj

(C)

G (1 + H1 )(1 + H 2 )

(B) 1 - bc - fg - cigj + bcfg (C) 1 + bc + fg + cig j - bcfg

17. Consider the signal flow graphs shown in fig.

(D) 1 + bc + fg + bcfg - cigj

P6.1.17. The transfer 2 is of the graph 1

1

14. The sum of the gains of the feedback paths in the 1

signal flow graph shown in fig. P.6.1.13 is 1

a

b

c

f

e

d

1

1 1

1 1 2

1

Fig. P.6.1.14

(C) af + be + cd + abef + abcdef (D) af + be + cd + cbef + bcde + abcdef

1 1 2

1 2

(A) af + be + cd + abef + bcde (B) af + be + cd

1 2

1

Fig. P.6.1.17

(A) a

(B) b

(C) b and c

(D) a, b and c

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Page 327


UNIT 6

Control Systems

18. Consider the List I and List II List I

-H3

List II

(Signal Flow Graph)

R(s) 1

(A)

(Transfer Function)

1

1

2

-H2

3 G1

4

1 C(s)

4

1

C(s)

4

1

C(s)

4

1

C(s)

G2

-H1H2

P. Q.

a

xi

b

xi

xo

-H2

2 3

G1

G2

-H2

a xo

a 3. (1 - ab)

S. b

R.

1

-H2

1

xi 1

1

2. ab

b xi 1

R(s) 1

(B)

xo

a

-H3

1. a + b

1 xo

a

4.

b

R(s) 1

(C)

1

1

3

2

G2

G1

-H3

-H1H2

a 1-b

H3 R(s) 1

(D)

1

1

3

2

G2

G1

The correct match is

H2

P

Q

R

S

(A)

2

1

3

4

(B)

2

1

4

3

(C)

1

2

4

3

21. The block diagram of a system is shown in fig.

(D)

1

2

3

4

P.6.1.21. The closed loop transfer function of this system

-H1

19. For the signal flow graph shown in fig. P6.1.19 an

is H1

equivalent graph is ta

R(s) tc

e1

G1

e1

Fig. P.6.1.21 tc+ td

tatb e1

e4

(A)

e4

e3

(B)

ta + tb e1

tatb

tctd e2

e4

e1

(C)

1 +

G1 G2 G3 1 + G1 G2 G3 H1

(B)

G1 G2 G3 1 + G1 G2 G3 H1 H 2

(C)

G1 G2 G3 1 + G1 G2 H1 + G2 G3 H 2

(D)

G1 G2 G3 1 + G1 G2 H1 + G1 G3 H 2 + G2 G3 H1

tc+ td e4

e2

block diagram shown in figure

P.6.1.20 R(s)

(A)

(D)

20. Consider the

3

2 +

C(s)

H2

tctd e3

G3

e4

e3

Fig. P.6.1.19 ta + tb

G2

+

td

e2

tb

+

G2

+

4 G2

22. For the system shown in fig. P6.1.22 transfer C(s)

function C( s) R( s) is G3

H2

H1

R(s)

+

G1

+

+

G2

H3

H1

Fig. P.6.1.20 H2

For this system the signal flow graph is

Fig. P.6.1.22

Page 328

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+

C(s)


Transfer Function

(A)

26. The transfer function of the system shown in fig.

G3 1 - H1 G2 - H 2 G3 - G1 G2 H 2

(B)

G3 + G1 G2 1 + H1 G2 + H 2 G3 + G1 G2 H 2

(C)

G3 1 + H1 G2 + H 2 G3 + G1 G2 H 2

Chap 6.1

P.6.1.26 is R(s)

+

+ +

G1

C(s)

G2

H1

H2

Fig. P.6.1.26

G3 (D) 1 - H1 G2 - H 2 G3 - G1 G2 H 2

(A)

G1 G2 1 - G1 G2 H1 - G1 G2 H 2

(B)

G1 G2 1 - G2 H 2 - G1 G2 H1


(C)

G1 G2 1 - G2 H 2 + G1 G2 H1 H 2

(D)

G1 G2 1 - G1 G2 H1 H 2

transfer function is 3

2

C

27. For the block diagram shown in fig. P.6.1.27 transfer function C( s) R( s) is

-3

Fig. P.6.1.23

(B) -3

(C) 3

(D) -3.75

R(s)

+

G1

+ -1

-1

-1

2

3

4

1

1

G3

G5

G4


R

G2

C

+

‘(A) 3.75

gain C R is

C(s)

G8

+

5

G7

G6

+

R

Fig. P.6.1.27

5

(A)

G1 G2 1 + G1 G2 + G1 G7G3 + G1 G2 G8 G6 + G1 G2 G3G7G5

Fig. P.6.1.24

44 (A) 23

(B)

29 19

(B)

G1 G2 1 + G1 G4 + G1 G2 G8 + G1 G2 G5G7 + G1 G2 G3G6 G7

44 19

(D)

29 11

(C)

G1 + G2 1 + G1 G4 + G1 G2 G8 + G1 G2 G5G7 + G1 G2 G3G6 G7

(D)

G1 + G2 1 + G1 G2 + G3G6 G7 + G1 G3G4 G5 + G1 G2 G3G6 G7G8

(C)

25. The gain C( s) R( s) of the signal flow graph shown in fig. P.6.1.25 is

28. For the block diagram shown in fig. P.6.1.28 the

G4

G3 R(s) 1

1 C(s) G1

numerator of transfer function is

G2

G1

-H1

Fig. P.6.1.25

(A)

G1 G2 + G2 G3 1 + G1 G2 H1 + G2 G3 H1 + G4

(B)

G1 G2 + G2 G3 1 + G1 G3 H1 + G2 G3 H1 - G4

(C)

G1 G3 + G2 G3 1 + G1 G3 H1 + G2 G3 H1 - G4

(D)

G1 G3 + G2 G3 1 + G1 G3 H1 + G2 G3 H1 + G4

R(s)

+

G2

+

+

G5

+

+

G6

C(s)

G3

G4

+

+

Fig. P.6.1.28

(A) G6 [ G4 + G3 + G5( G3 + G2 )] (B) G6 [ G2 + G3 + G5( G3 + G4 )] www.gatehelp.com

Page 329


UNIT 6

Control Systems

33. The closed loop transfer function of the system

(C) G6 [ G1 + G2 + G3( G4 + G5)] (D) None of the above

G1

29. For the block diagram shown in fig. P.6.1.29 the

R

+

transfer function C( s) R( s) is 50( s - 2) 50( s - 2) (A) 3 (B) 3 2 s + s + 150 s - 100 s + s 2 + 150 s (C)

50 s s + s + 150 s - 100 3

(D)

2

+ +

G2

C

G3

H1

50 s + s + 150

Fig. P.6.1.32

2

shown in fig. P6.1.33 is 30. For the SFG shown in fig. P.6.1.30 the transfer function

C R

G2

is

R

G2

G1

-H1

G3

-H2

R

C

1

+

-H3

+ +

G1

+ +

G3

H3

G4

H2

Fig. P.6.1.30

H1

Fig. P.6.1.33

(A)

G1 + G2 + G3 1 + G1 H1 + G2 H 2 + G3 H 3

(A)

G1 G2 G3 + G2 G3G4 + G1 G4 1 + G1 G3G4 H1 H 2 H 3 + G2 H 4 H1 H 2 + G4 H1

(B)

G1 + G2 + G3 1 + G1 H1 + G2 H 2 + G3 H 3 + G1 G3 H1 H 3

(B)

G2 G4 + G1 G2 G3 1 + G1 G3 H1 H 2 H 3 + G4 H1 + G3G4 H1 H 2

(C)

G1 G2 G3 1 + G1 H1 + G2 H 2 + G3 H 3

(C)

G1 G3G4 + G2 G4 1 + G3G4 H1 H 2 + G4 H1 + G1 G3 H 3 H 2

(D)

G1 G2 G3 1 + G1 H1 + G2 H 2 + G3 H 3 + G1 G3 H1 H 3

(D)

G1 G3G4 + G2 G3G4 + G2 G4 1 + G1 G3G4 H1 H 2 H 3 + G3G4 H1 H 2 + G4 H1

31. Consider the SFG shown in fig. P6.1.31. The D for this graph is

Statement for Q.34-37: A block diagram of feedback control system is

G4

shown in fig. P6.1.34-37

-H3 R

1

G2

G1

G3

1

C

-H1

R1(s)

+ +

G

R2(s)

+ + +

G

C1(s)

-H2

Fig. P.6.1.31

(A) 1 + G1 H1 + G2 G3 H 3 + G1 G3 H 2 (B) 1 + G1 H1 - G2 G3 H 3 - G1 G3 H 3 + G2 G4 H 2 H 3 (C) 1 + G1 H1 + G2 G3 H 3 + G1 G3 H 3 - G2 G4 H 2 H 3

32. The transfer function of the system shown in fig.

(C)

Page 330

G2 G3 + G1 G3 1 + G3 H1 + G2 G3

C2(s)

Fig. P.6.1.34-37

(D) 1 + G1 H1 + G2 G3 H 3 + G1 G3 H 3 + G2 G4 H 2 H 3

P.6.1.32 is G2 G3 + G1 G3 (A) 1 - G3 H1 + G2 G3

C


C1 R1

is R2 = 0

(B)

G2 G3 + G1 G3 1 + G3 H1 - G2 G3

(A)

G 1 - 2G2

(B)

G(1 - G) 1 - 2G2

(D)

G2 G3 + G1 G3 1 - G3 H1 - G2 G3

(C)

G(1 - 2 G) 1 - G2

(D)

G 1 - G2

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Transfer Function

35. The transfer function (A) (C)

C1 R2

R1 = 0

(B)

G2 1 - 2G2

(D)

C2 R1

G (1 + G) (A) 1 - 2G2

5

G 1 - G2

G 1 - G2

+ +

2s

C2(s)

Fig. P.6.1.40-41

is R2 = 0

G 1 - G2

(D)

G(1 + G) (A) 1 - 2G2

1 s

+

2

G2 (B) 1 - 2G2

C 37. The transfer function 2 R2

(C)

R1(s)

G2 1 - G2

2

(C)

40. The transfer function for this system is

is

G 1 - 2G2


Chap 6.1

(A)

2 s(2 s + 1) 2 s 2 + 3s + 5

(B)

2 s(2 s + 1) 2 s 2 + 13s + 5

(C)

2 s(2 s + 1) 4 s 2 + 13s + 5

(D)

2 s(2 s + 1) 4 s 2 + 3s + 5

41. The pole of this system are

is R1 = 0

(A) -0.75 ± j1.39

(B) -0.41, - 6.09

(C) -0.5,

(D) -0.25 ± j0.88

- 1.67

G (B) 1 - 2G2

G 1+ G

(D)

G 1 - G2

********

Statement for Q.38–39: A signal flow graph is shown in fig. P.6.1.38–39. G4 Y1

1

Y2

G1

Y3

-H1

Y4 G3

G2

Y5

1

Y5

-H2 -H3

Fig. P.6.1.38-39


Y2 is Y1 1 + G2 H 2 D

(A)

1 D

(B)

(C)

G1 G2 G3 D



Y5 is Y2

(A)

G1 G2 G3 + G4 G3 D

(B) G1 G2 G3 + G4 G3

(C)

G1 G2 G3 + G4 G3 G1 G2 G3

(D)

G1 G2 G3 + G4 G3 1 + G2 H 2

Statement for Q.40–41: A block diagram is shown in fig. P6.1.40–41. www.gatehelp.com

Page 331


UNIT 6

SOLUTIONS

C2 =

C¢2 =

9 9 = , C2 is reduced by 1%. 9 + 1 10

11. (A) Apply the feedback formula and then multiply 1 by , H1

1. (A) Ge ( s) = G1 ( s) + G2 ( s) + G3( s) =

1 1 s+3 + + ( s + 1) ( s + 4) ( s + 5)

=

s 2 + 9s + 20 + s 2 + 6s + 5 + s 3 + 5s 2 + 4s + 3s 2 + 15s + 12 (s + 1)(s + 4)(s + 5)

=

s 3 + 10 s 2 + 34 s + 37 ( s + 1)( s + 4)( s + 5)

æ 1 ö ( H 2 G1 ) çç ÷÷ C H 2 G1 è H1 ø = = R H1 (1 + G1 G2 H 2 ) 1 + H 2 G1 G2 12. (A) There cannot be common subscript because

2. (B) Ge ( s) = G1 ( s) G2 ( s) G3( s) =

3. (C)

10 10 , = 10 + 1 11

Control Systems

( s + 1) ( s + 2)( s + 5)( s + 3)

subscript refers to node number. If subscript is common, that means that node is in both loop.

C( s) G( s) = R( s) 1 + H ( s) G( s)

13. (D) L1 = -bc, L2 = - fg, L3 = jgic, L1 L3 = bcfg D = 1 - ( -bc - fg + cigj) + bcfg = 1 + bc + fg - cig j + bcfg

s+1 ( s + 1)( s + 4) s( s + 2) = = 3 ( s + 3) ( s + 1) s + 7 s 2 + 12 s + 3 1+ ( s + 4) s( s + 2)

14. (A) In this graph there are three feedback loop. abef is not a feedback path because between path x2 is a summing node.

4. (B) Multiply G2 and G3 and apply feedback formula and then again multiply with T( s) =

1 . G1

P1 = - H 2 G1 , L1 = - G1 H 2 H1 , D1 = 1,

G2 G3 G1 (1 + G2 G3 H1 )

if |G1 H 2 H1 | >> 1, Tn ( s) =

5. (D) T( s) = G2 (1 + G1 ) + 1 = 1 + G1 + G1 G2

TN ( s) =

T( s) =

G2 1 + G1 G2 H

7. (D) Apply the feedback formula to both loop and then

=

öæ G2 ÷÷çç + 1 G2 H 2 øè

ö ÷÷ ø

- H 2 G1 -1 = G1 H 2 H1 H1

G G = 1 + H1 + H 2 + H1 H 2 (1 + H1 )(1 + H 2 )

17. (B) Ga = 1, Gb = 1 + 1 = 2, Gc =

1 1 1 1 + + + =1 4 4 4 4

18. (B) P. P1 = ab, D = 1, L = 0 , T = ab

G1 G2 1 + G1 H1 + G2 H 2 + G1 G2 H1 H 2

8. (C) For positive feedback

Q. P1 = a, P2 = b , D = 1, L = Dk = 0, T = a + b a R. P1 = a, L1 = b, D = 1 - b, D1 = 1, T = a-b a S. P1 = a, L1 = ab, D = 1 - ab, D1 = 1, T = 1 - ab

C 6 = =-6 R 1 - 6 ´31

9. (D) For system (b) closed loop transfer function G G + s+1 G + s+1 s+2 , , Hence G = 1 +1= = s+1 s+1 s+1 s+1

Page 332

- H 2 G1 1 + G1 H 2 H1

There are no loop in any graph. So option (B) is correct.

multiply æ G1 T( s) = çç 1 + G1 H1 è

Tn ( s) =

16. (C) P1 = G , L1 = - H1 , L2 = - H 2 , L1 L2 = H1 H 2 , D1 = 1

6. (A) Open-loop gain = G2 Feed back gain = HG1

15. (B) By putting R ( s) = 0

19. (A) Between e1 and e2 , there are two parallel path. Combining them gives ta + tb . Between e2 and e4 there is a path given by total gain tc td . So remove node e3 and

10. (A) In open loop system change will be 10% in C1

place gain tc td of the branch e2 e4 . Hence option (A) is

also but in closed loop system change will be less

correct.

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Transfer Function

20. (A) Option (A) is correct. Best method is to check the

Chap 6.1

28. (A) SFG:

signal flow graph. In block diagram there is feedback

G3

from 4 to 1 of gain -H1 H 2 . The signal flow graph of

G4 G3

option (A) has feedback from 4 to 1 of gain -H1 H 2 .

R

1

21. (C) Consider the block diagram as SFG. There are

G5

G2 -G1

22. (B) Consider the block diagram as a SFG. Two forward path G1 G2 and G3 and three loops

P1 = G2 G5G6 , P2 = G3G5G6 , P3 = G3G6 , P4 = G4 G6 If any path is deleted, there would not be any loop. Hence D1 = D2 = D3 = D4 = 1 C G4 G6 + G3G6 + G3G5G6 + G2 G5G6 = D R

-G1 G2 H 2 , - G2 H1 , - G3 H 2 . There are no nontouching loop. So (B) is correct. 23. (C) P1 = 5 ´ 3 ´ 2 = 30, D = 1 - ( 3 ´ - 3) = 10

29. (A)

C 30 D1 = 1 , = =3 R 10

R

1

-2 s

1 s2

L3 = -4, L4 = -5,

C

s

-1

L1 L3 = 8, D = 1 - ( -2 - 3 - 4 - 5) + 8 = 23,

Fig. S6.1.29

D1 = 1, D2 = 1 - ( -3) = 4, C 24 + 5 ´ 4 44 = = R 24 23

P1 =

1 50 50 × ×s= s2 ( s + 1) s( s + 1)

P2 =

1 50 -100 × × ( -2) = 2 s2 s + 1 s ( s + 1)

L1 =

50 -2 -100 × = s+1 s s( s + 1)

L2 =

1 50 -50 × × s × ( -1) = s2 s + 1 s( s + 1)

L3 =

1 50 100 × × ( -2) × ( -1) = 2 s2 s + 1 s ( s + 1)

P2 = G3G2

L1 = -G3G2 H1 , L2 = -G1 G2 H1 ,

-2 1

50 (s + 1)

24. (A) P1 = 2 ´ 3 ´ 4 = 24 , P2 = 1 ´ 5 ´ 1 = 5

25. (B) P1 = G1 G2 ,

C

Fig. S6.1.28

forward path G1 G2 G3 . So (D) is correct option.

L2 = -3,

1

-1

two feedback loop -G1 G2 H1 and -G2 G3 H 2 and one

L1 = -2 ,

G6

L3 = G4 , D1 = D2 = 1

There are no nontouching loop. P1 D1 + P2 D2 G1 G2 + G2 G3 T( s) = = 1 - ( L1 + L2 + L3) 1 + G1 G2 H1 + G2 G3 H1 - G4 26. (C) P1 = G1 G2 , L1 = -G1 G2 H1 H 2 , L2 = G2 H 2 C( s) G1 G2 = R( s) 1 + G1 G2 H1 H 2 - G2 H 2 27. (B) There is one forward path G1 G2 .

D =1 +

100 50 100 + s( s + 1) s( s + 1) s 2 ( s + 1)

D1 = D2 = 1 C P1 + P2 50( s - 2) = = 3 R D s + s2 + 150 s - 100

Four loops -G1 G4 , - G1 G2 G8 , - G1 G2 G5G7 30. (D) P1 = G1 G2 G3

and -G1 G2 G3G6 G7 .

L1 = - G1 H1 , L2 = - G2 H 2 , L3 = - G3 H 3 R(s) 1

G2

G1

1 C(s)

D = 1 - ( - G1 H1 - G2 H 2 - G3 H 3 ) + G1 G3 H1 H 3

1 -H1

L1 L3 = G1 G3 H1 H 3

H2

Fig. S6.1.27

There is no nontouching loop. So (B) is correct.

D = 1 + G1 H1 + G2 H 2 + G3 H 3 + G1 G3 H1 H 3 D1 = 1 C G1 G2 G3 = R 1 + G1 H! + G2 H 2 + G3 H 3 + G1 G3 H1 H 3

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Page 333


CHAPTER

6.2 STABILITY

1. Consider the system shown in fig. P6.2.1. The range of K for the stable system is R(s)

+

E(s)

C(s)

K(s2 - 2s + 2)

The range of K to ensure stability is 6 3 (A) K > (B) K < - 1 or K > 8 4 (C) K < - 1

(D) -1 < K <

3 4

1 s2 + 2s + 1

5. Consider a ufb system with forward-path transfer

Fig. P6.2.1

(A) -1 < K < -

1 2

function

(B) -

(C) -1 < K < 1

1 < K <1 2

G( s) =

(D) Unstable

The system is stable for the range of K

2. The forward transfer function of a ufb system is G( s) =

K ( s + 3) s 4 ( s + 2)

K ( s 2 + 1) ( s + 1)( s + 2)

(A) K > 0

(B) K < 0

(C) K > 1

(D) Always unstable

6. The open-loop transfer function of a ufb

The system is stable for

control

system is

(A) K < - 1

(B) K > -1

(C) K < - 2

(D) K > -2

G( s) =

3. The open-loop transfer function with ufb are given below for different systems. The unstable system is

For K > 6, the stability characteristic of the open-loop and closed-loop configurations of the system

(A)

2 s+2

(B)

2 s 2 ( s + 2)

are respectively

(C)

2 s( s + 2)

(D)

2( s + 1) s( s + 2)

(B) stable and stable

(A) stable and unstable

(C) unstable and stable

4. Consider a ufb system with forward-path transfer function

K ( s + 2) ( s + 1)( s - 7)

(D) unstable and unstable 7. The forward-path transfer function of a ufb system is

G( s) =

K ( s + 3)( s + 5) ( s - 2)( s - 4)

G( s) = www.gatehelp.com

K ( s 2 - 4) s2 + 3 Page 335


UNIT 6

For the system to be stable the range of K is 3 (A) K > -1 (B) K < 4 (C) -1 < K <

3 4

Control & System

13. The open-loop transfer function of a ufb system is R(s)

+

C(s)

1 s(s + 1)(s + 5)

K

(D) marginal stable Fig. P6.2.12

8. A ufb system have the forward-path transfer function G( s) =

K ( s + 6) s( s + 1)( s + 3)

G( s) =

K ( s + 10)( s + 20) s 2 ( s + 2)

The closed loop system will be stable if the value of

The system is stable for (A) K < 6

(B) -6 < K < 0

K is

(C) 0 < K < 6

(D) K > 6

(A) 2

(B) 3

(C) 4

(D) 5

9. The feedback control system shown in the fig. P6.2.8. R(s)

+

K(1 + Ts) s2(1 + s)


C(s)

A feedback system is shown in fig. P6.14-15. 2 s

Fig. P6.2.9 R(s) +

is stable for all positive value of K , if (A) T = 0

(B) T < 0

(C) T > 1

(D) 0 < T < 1

+

C(s)

+ s2

1 s+1

Fig. P6.2.14–15

10. Consider a ufb system with forward-path transfer function

14. The closed loop transfer function for this system is K G( s) = ( s + 15)( s + 27)( s + 38)

The system will oscillate for the value of K equal to (A) 23690

(B) 2369

(C) 144690

(D) 14469

(A)

s 5 + s 4 + 2 s 3 + ( K + 2) s 2 + ( K + 2) s + K s3 + s2 + 2 s + K

(B)

2 s 4 + ( K + 2) s 3 + Ks 2 s3 + s2 + 2 s + K

(C)

s3 + s2 + 2 s + K s 5 + s 4 + 2 s 3 + ( K + 2) s 2 + ( K + 2) s + K

(D)

s3 + s2 + 2 s + K 2 s + ( K + 2) s 3 + Ks 2

11. The forward-path transfer function of a ufb system is G( s) =

K ( s - 2)( s + 4)( s + 5) ( s 2 + 3)

1 3
4

15. The poles location for this system is shown in fig.

For system to be stable, the range of K is 1 3 (A) K > (B) K < 54 40 (C)

K s2

P6.2.15. The value of K is jw

(D) Unstable s

12. The closed loop system shown in fig. P6.2.12 become marginally stable if the constant K is chosen to be

Page 336

(A) 30

(B) -30

(C) 10

(D) -10

Fig. P6.2.15

(A) 4

(B) -4

(C) 2

(D) -2

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UNIT 6

(A) stable

Control & System

32. The closed loop transfer function of a system is

(B) unstable

(C) marginally stable

T( s) =

(D) More information is required.

( s + 8)( s + 6) s 5 - s 4 + 4 s 3 - 4 s 2 + 3s - 2

The number of poles in RHP and in LHP are 27. The forward path transfer of ufb system is 1 G( s) = 2 2 4 s ( s + 1)

(A) 4, 1

(B) 1, 4

(C) 3, 2

(D) 2, 3


The system is (A) stable

(B) unstable

T( s) =


s 3 + 3s 2 + 7 s + 24 s - 2 s 4 + 3s 3 - 6 s 2 + 2 s - 4 5

The number of poles in LHP, in RHP, and on jw

(D) More information is required

axis are 28. The forward-path transfer function of a ufb system is G( s) =

G( s) 2 s4 + 5 s3 + s2 + 2 s

(B) 0, 1, 4

(C) 1, 0, 4

(D) 1, 2, 2

34. For the system shown in fig. P6.2.34. the number

The system is (A) stable

(A) 2, 1, 2

of poles on RHP, LHP, and imaginary axis are

(B) unstable

R(s)


+

C(s)

507 s4 + 3s3 + 10s2 + 30s + 169

(D) more information is required. 1 s

29. The open loop transfer function of a system is as

Fig. P6.2.34

K ( s + 0.1) s( s - 0.2)( s 2 + s + 0.6)

G( s) H ( s) =

(A) 2, 3, 0

(B) 3, 2, 0

(C) 2, 1, 2

(D) 1, 2, 2

The range of K for stable system will be (A) K > 0.355

(B) 0.149 < K < 0.355

35. A Routh table is shown in fig. P6.2.36. The location

(C) 0.236 < K < 0.44

(D) K > 0.44

of pole on RHP, LHP and imaginary axis are

30. The open-loop transfer function of a ufb control system is given by G( s) =

K s( sT1 + 1)( sT2 + 1)

s7

1

2

s5

1

2

s5

3

4

s4

1

-1

For the system to be stable the range of K is æ 1 1 (A) 0 < K < çç + è T1 T2

ö ÷÷ ø

æ 1 1 (B) K > çç + è T1 T2

(C) 0 < K < T1 T2

ö ÷÷ ø

(D) K > T1 T2

(A) 1, 2, 4

(B) 1, 6, 0

(C) 1, 0, 6


36. For the open loop system of fig. P6.2.35 location of


poles on RHP, LHP, and an jw-axis are

s + 4 s + 8 s + 16 s + 3s 4 + 5 s 2 + s + 3 3

T( s) =

Fig. P6.2.35

2

R(s)

5

Page 338

(A) 3, 2

(B) 2, 3

(C) 1, 4

(D) 4, 1

C(s)

Fig. P6.2.35

The number of poles in right half-plane and in left half-plane are

-8 s6 + s5 - 6s4 + s2 + s - 6

(A) 3, 3, 0

(B) 1, 3, 2

(C) 1, 1, 4

(D) 3, 1, 2 ************

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UNIT 6

11. (C) T( s) = =

Control & System

K 2 æK 2ö + . Then apply feedback formula with ç 2 + ÷ and 2 s s sø ès 1 , and then multiply with s 2 . ( s + 1)

G( s) 1 + G( s)

K ( s - 2)( s + 4)( s + 5) Ks + (7 K + 1) s 2 + 2 Ks + ( 3 - 40 K ) 3

æK 2ö s2 ç 2 + ÷ 2 s 4 + ( K + 2) s 3 + Ks 2 sø ès T( s) = = 1 æK 2ö s3 + s2 + 2 s + K 1+ ç 2 + ÷ s + 1è s sø

Routh table is as shown in fig. S.6.211 s3

K

2K

s2

7K + 1

3 - 40 K

2

s1

54 K - K 7K + 1

s0

3 - 40 K

15. (C) Denominator = s 3 + s 2 + 2 s + K Routh table is as shown in fig. S.6.2.15

Fig. S.6.2.11

K > 0, 7K + 1 > 0

Þ

54 K 2 - K >0 7K + 1

Þ

3 - 40 K > 0

Þ

12. (A) T( s) =

1ü K >- ï 7 ï 1 ï K > ý 54 ï 3 ï K < 40 ïþ

Þ

1 3
1

5

s2

1

K

s1

2-K

s0

K Fig. S.6.2.15

Row of zeros when K = 2, s 2 + 2 = 0,

1 s + 6s + 5s + K 3

s3

Þ

s = -1, j 2 , - j 2

2

16. (D) Applying the feedback formula on the inner loop

Routh table is as shown in fig. S.6.212 s3

1

5

s2

6

K

s1

30 - K

s0

K

and multiplying by K yield K , Ge ( s) = s( s 2 + 5 s + 7) T( s) =

K s + 5s + 7s + K 3

2

17. (B) Routh table is as shown in fig. S.6.2.17

Fig. S.6.2.12

K ( s + 10)( s + 20) 13. (D) T( s) = 3 s + ( K + 2) s 2 + 30 Ks + 200 K

s3

1

7

s2

5

K

Routh table is as shown in fig. S.6.2.13

s1

35- K 5

s0

K

s3

1

30K

s2

K +2

200K

s1

30 K 2 - 140 K

s0

200K

Fig. S.6.2.17

K >0 ,

K < 35

Auxiliary equation 5 s 2 + 35 = 0,

200 K > 0 ® K > 0, 30 K 2 - 140 K > 0 14 Þ K > , 5 satisfy this condition. 3

Page 340

Þ

18. (C) At K = 35 system will oscillate.

Fig. S.6.2.13

14. (B) First combine the parallel loop

35 - K >0 5

K 2 and giving 2 s s

Þ

19. (B) For inner loop K K , Gi ( s) = = ( s - a)( s + 3a)( s + 4 a) P ( s) For outer loop, Go( s) = Ti ( s) =

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s=± j 7

Ti ( s) =

K , P ( s) + K

K P ( s) + K


Stability

To( s) =

K , P ( s) + 2 K

Therefore if inner loop is stable for X < K < Y , then outer loop will be stable for X < 2 K < Y X Y . Þ
2K - 1 >0 2 18 K - 109 >0 2K - 1

Chap 6.2

Þ Þ

1 ü 2 ï ý 109 ï K > 18 þ K >

K ( s + 2) s 4 + 3s 3 - 3s 2 + ( K + 3) s + (2 K - 4)


2K - 4

s4

1

20

s3

9

K K

1

-3

s3

3

K +3

s2

180 - K 9

s2

- ( K +312 )

2K - 4

s1

K ( K - 99) K -180

s1

K ( K + 33) K + 12

s0

K

s0

2K - 4

For stability 0 < K < 99

-( K + 12) > 0 Þ K < - 12, 2 K - 4 > 0 3

24. (C) T( s) =

Þ K > 2 and K > -33, These condition can not be met simultaneously. System is unstable for any value of K .

2

2

s4

1

-3

s3

3

K +3

s2

- K + 12 3

2K - 4

4

1

1

3

K

1

s1

K ( K + 33) K + 12

K -1 K

1

s0

2K - 4

s s

K ( s + 2) s + 3s - 3s + ( K + 3) s + (2 K - 4) 4


21. (D) Routh table is as shown in fig. S.6.2.21

1

1

K -1 - K 2 K -1

0

2K - 4

Fig. S.6.2.24

For K < - 33, 1 sign change

1

For -33 < K < - 12, 1 sign change

Fig. S.6.2.21

K > 0, K - 1 > 0 Þ

K

Fig. S.6.2.23

Fig. S.6.2.20

s2

109 18

s4 + 9 s3 + 20 s2 + Ks + K = 0

s4

s

K >

23. (B) Characteristic equation


s

Þ

K >1 ,

For -12 < K < 0, 1 sign change

K -1 - K2 > 0, K -1

For 0 < K < 2,

3 sign change

But for K > 1 third term is always -ive. Thus the three

For K > 2,

condition cannot be fulfilled simultaneously.

Therefore K > 2 yield two RHP pole.

22. (D) Routh table is as shown in fig. S.6.2.22


2 sign change

s4

1

8

9

s3

4

20

25

s2

3

15

18 K -109 2 K -1

s1

6

25

s0

15

s4

1

4+K

s3

2

s2

2 K -1 2

s1 s0

25

Fig. S.6.2.22

15

ROZ

Fig. S.6.2.25

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Page 341


UNIT 6

P ( s) = 3s 2 + 15,

d P ( s) = 6 s, No sign change from s 2 to s 0 ds

on jw-axis 2 roots, RHP 0,

Þ

s s

3

2

s1 s

0

Routh table is as shown in fig. S.6.2.29 2

0.4

s3

0.8

K - 0.12 0.1K

35

10

50

s2

0.55 - 125 . K

30

264

s1

-1 .25K 2 + 0 .63K - 0 .066 0 .55-1 .25K

s0

0.1K

-386

264

s4

1

ROZ

264

K > 0, 0.55 - 125 . K > 0 Þ K < 0.44

Two sign change. RHP-2 poles. System is not stable. 27. (C) Closed loop transfer function G( s) 1 T( s) = = 1 + G( s) 4 s 4 + 4 s 2 + 1

-125 . K 2 + 0.63K - 0.066 > 0 ( K - 0.149)( K - 0.355) < 0, 0.149 < K < 0.355 30. (A) Characteristic equation


s( sT1 + 1)( sT2 + 1) + K = 0

s4

4

4

1

s3

16

8

ROZ

s2

2

1

s1

46

s0

1

T1 T2 s 3 + ( T1 + T2 ) s2 + s + K = 0 Routh table is as shown in fig. S.6.2.30 s3

T1 T2

1

s2

T1 + T2

K

Fig. S.6.2.27

s1

( T1 + T2 ) - T1 T2K T1 + T2

dp( s) = 16 s 3 + 3s ds

s0

K

ROZ

Fig. S.6.2.30

There is no sign change. So all pole are on jw–axis. So

æ 1 1 ö ÷÷ 0 < K < çç + T T è 1 2 ø

system is marginally stable.

K > 0, ( T1 + T2 ) - T1 T2 K > 0

28. (B) Closed loop transfer function 1 G( s) T( s) = = 4 3 1 + G( s) 2 s + 5 s + s 2 + 2 s + 1



Þ

s5

1

5

1

s4

3

4

3

s4

2

1

s3

5

2

s3

3.67

0

s2

1 5

1

s2

4

3

s1

-23

s1

-2.75

s0

1

s0

3

Fig. S.6.2.28

Page 342

1

Fig. S.6.2.29

Fig. S.6.2.26

P ( s) = 4 s 4 + 4 s + 1,

s( s - 0.2)( s 2 + s + 0.6) + K ( s + 0.1) = 0

s 4 + 0.8 s 3 + 0.4 s 2 + ( K - 0.12) s + 0.1K = 0

Routh table is as shown in fig. S.6.2.26 s

2 RHP poles so unstable. 29. (B) The characteristic equation is 1 + G( s) H ( s) = 0

LHP 2.

26. (B) Closed-loop transfer function is G( s) 240 T( s) = = 4 3 s + 10 s + 35 s 2 + 50 s + 264 1 + G( s)

4

Control & System

1

In RHP -2 poles. In LHP -3 poles.

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Stability

32. (C) Routh table is as shown in fig. S.6.2.32 e

+

-

+

+

s5 s

- -

4

3

4

-1

-4

-2

3

e

1 -2

s

+

-

s2

1-4e e

+

+

s1

2e2 +1-4e 1-4e

-

-

s0

-2

From s 4 row down to s 0 there is one sign change. So LHP–1 + 1 = 2 pole. RHP–1 pole, jw-axis -2 pole. 35. (A) Notice that in s 5 row there would be zero. In this row coefficient of

Corresponding to this pole there is a pole on LHP. Rest 4 out of 6 poles are on imaginary axis. Rest 1 pole is on LHP. 36. (A) Routh table is as shown in fig. S.6.2.36

1

3

2

-2

-6

-4

-2

-3

ROZ

2

-3

-4

1

-

s

s s

3

s s

e

1 3

-4

0

Fig. S.6.2.33

dP ( s) = - 8 s 3 - 12 s , -2 , -3 P ( s) = - 2 s - 6 s , ds 4

2

4

0

No sign change exist from the s row down to the s row.

+ + + +

+

-

-

-

-

+

-

-

+

-

-

Thus, the even polynomial does not have RHP poles. Therefore 4 pole

RHP

1 pole

LHP

0 pole

s6

1

-6

s5

1

0

s4

-6

0

s3

-24

0

s2

e

s1

- 144 e

s0

-6

-6

ROZ

Fig. S.6.2.36

P ( s) = - 6 s 4 - 6,

because of symmetry all four poles must be on jw-axis. jw-axis

P ( s) = s6 + 2 s 4 - s 2 - 2

Corresponding to this pole there is a pole on LHP.


4

, where

there is one sign change. So there is one pole on RHP.

2 LHP poles.

s5

dP ( s ) ds

have been entered. From s6 to row down to the s 0 row,

Fig. S.6.2.32

3 RHP,

dP ( s) = 12 s 3 + 60 ds

P ( s) = 3s 4 + 30 s 2 + 507,

1

-

+

Chap 6.2

dP ( s) = -24 s 3 , ds

There is two sign change from the s 4 row down to the s 0 (1 sign change)

row. So two roots are on RHS. Because of symmetry rest two roots must be in LHP. From s6 to s 4 there is 1 sign change so 1 on RHP and 1 on LHP.

34. (D) Closed loop transfer function G( s) T( s) = 1 + G( s) + H ( s) =

Total

LHP 3 root,

RHP 3 root.

507 s s 5 + 3s 4 + 10 s 3 + 30 s 2 + 169 s + 507


***********

5

1

10

69

s4

3

30

57

s3

12

60

ROZ

s2

15

507

s1

-345.6

s0

507

s

Fig. S.6.2.33

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Page 343


Time Response

8. A system is shown in fig. P6.3.8. The rise time and

Chap 6.3

d. T( s) =

s+2 s2 + 9

e. T( s) =

( s + 5) ( s + 10) 2

settling time for this system is R(s)

1 s

10 (s + 10)

C(s)

Consider the following response

Fig. P6.3.8

(A) 0.22s, 0.4s

(B) 0.4s, 0.22s

(C) 0.12s, 0.4s

(D) 0.4s, 0.12s

1. Overdamped

2. Under damped

3. Undamped

4. Critically damped.

The correct match is 9. For a second order system settling time is Ts = 7 s and

1

2

3

4

peak time is Tp = 3 s. The location of poles are

(A)

a

c

d

e

(A) -0.97 ± j0.69

(B) -0.69 ± j0.97

(B)

b

a

d

e

(C) -1.047 ± j0.571

(D) -0.571 ± j1.047

(C)

c

a

e

d

(D)

c

b

e

d

10. For a second order system overshoot = 10% and peak time Tp = 5 s. The location of poles are (A) -0.46 ± j0.63

(B) -0.63 ± j0.46

(C) -0.74 ± j0.92

(D) -0.92 ± j0.74

15. The forward-path transfer of a ufb control system is G( s) =

11. For a second-order system overshoot = 12 % and settling time = 0.6 s. The location of poles are (A) -9.88 ± j6.67

(B) -6.67 ± j9.88

(C) -4.38 ± j6.46

(D) -6.46 ± j4.38

The step, ramp, and parabolic error constants are (A) 0, 1000, 0

(B) 1000, 0, 0

(C) 0, 0, 0

(D) 0, 0, 1000

16. The open-loop transfer function of a ufb control

Statement for Q.12–13: A system has a damping ratio of 1.25, a natural

system is G( s) =

frequency of 200 rad/s and DC gain of 1. 12. The response of the system to a unit step input is (A) 1 + (C) 1 +

1000 (1 + 0.1s)(1 + 10 s)

5 -50 t 2 -150 t e - e 3 3

(B) 1 -

1 -100 t 4 -400 t e - e 3 3

(D) 1 +

4 -100 t 1 -400 t e + e 3 3 2 -50 t 5 -150 t e - e 3 3

K (1 + 2 s)(1 + 4 s) s 2 ( s 2 + 2 s + 8)

The position, velocity and acceleration error constants are respectively K , 0 8

(A) 0, 0, 4K

(B) ¥ ,

(C) 0, 4 K , ¥

(D) ¥, ¥,

K 8

13. The system is (A) overdamped

(B) under damped

17. The open-loop transfer function of a unit feedback

(C) critically damped


system is G( s) =

14. Consider the following system a. T( s) =

5 ( s + 3)( s + 6)

10( s + 7) b. T( s) = ( s + 10)( s + 20) 20 c. T( s) = 2 s + 6 s + 144

50 (1 + 0.1s)(1 + 2 s)

The position, velocity and acceleration error constants are respectively (A) 0, 0, 250

(B) 50, 0, 0

(C) 0, 250, ¥

(D) ¥, 50, 0

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Page 345


UNIT 5

Control Systems


D(s)

The forward-path transfer function of a unity R(s)

feedback system is G( s) =

1 (s + 5)

+

100 s(s + 2)

+ +

C(s)

K s ( s + a) n

Fig. P6.3.22

The system has 10% overshoot and velocity error constant K v = 100.

(A) -

49 11

(B)

49 11

18. The value of K is

(C) -

63 11

(D)

63 11

(A) 237 ´ 10

3

(B) 144

(C) 14.4 ´ 10

(D) 237

3

23. The forward path transfer function of a ufb system

19. The value of a is (A) 237 . ´ 10

is

3

(C) 14.4 ´ 10

(B) 237

3

G( s) =

(D) 144

20. For the system shown in fig. P6.3.20 the steady state error component due to unit step disturbance is 0.000012 and steady state error component due to unit ramp input is 0.003. The values of K 1

and K 2

are

K s( s + 4)( s + 8)( s + 10)

If a unit ramp is applied, the minimum possible steady-state error is (A) 0.16

(B) 6.25

(C) 0.14

(D) 7.25

respectively 24. The forward-path transfer function of a ufb system D(s)

R(s)

K1(s + 2) (s + 3)

+

is K2 s(s + 4)

+ +

C(s)

G( s) =

1000( s 2 + 4 s + 20)( s 2 + 20 s + 15) s 3( s + 2)( s + 10)

The system has r ( t) = t 3 applied to its input. The steady state error is

Fig. P6.3.20

(A) 16.4, 1684

(B) 1250, 2.4

(C) 125 ´ 10 , 0.016

(D) 463, 3981

3

21. The transfer function for a single loop nonunity

(A) 4 ´ 10 -4

(B) 0

(C) ¥

(D) 2 ´ 10 -5

25. The transfer function of a ufb system is

feedback control system is G( s) =

G( s) =

1 1 , H ( s) = s2 + s + 2 ( s + 1)

10 5( s + 3)( s + 10)( s + 20) s( s + 25)( s + a)( s + 30)

The value of a to yield velocity error constant The steady state error due to unit step input is 6 6 (A) (B) 7 5 2 (C) 3

(B) 0

(C) 8

(D) 16

26. A system has position error constant K p = 3. The

error due to a unit step input and a unit step

Page 346

(A) 4

(D) 0

22. For the system of fig. P6.3.22 the total steady state disturbance is

K v = 10 4 is

steady state error for input of 8tu( t) is (A) 2.67

(B) 2

(C) ¥

(D) 0

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Time Response

27. The forward path transfer function of a unity

If

Chap 6.3

the

r ( t) = 1 + t +

feedback system is G( s) =

For input of 60u( t) steady state error is (B) 300

(C) ¥

(D) 10

E(s)

G(s)

subjected

(B) 0.1

(C) 10

(D) ¥

to

an

input

32. The system shown in fig. P6.3.32 has steady-state error 0.1 to unit step input. The value of K is R(s)

function is +

is

(A) 0

28. For ufb system shown in fig. P6.3.28 the transfer

R(s)

system

t 2 , t > 0 the steady state error of the

system will be

1000 ( s + 20)( s 2 + 4 s + 10)

(A) 0

1 2

+

C(s)

K (s + 1)(0.1s + 1)

C(s)

Fig. P6.3.32 Fig. P6.3.28

G( s) =

20( s + 3)( s + 4)( s + 8) s 2 ( s + 2)( s + 15)

(B) 0

(C) ¥

(D) 64

(B) 0.9

(C) 1.0

(D) 9.0


If input is 30 t 2 , then steady state error is (A) 0.9375

(A) 0.1

Block diagram of a position control system is shown in fig.P6.3.33–34.

29. The forward-path transfer function of a ufb control

R(s)

+

Ka

1 s(0.5s + 1)

+

C(s)

system is G( s) =

450( s + 8)( s + 12)( s + 15) s( s + 38)( s 2 + 2 s + 28)

sKt

Fig. P6.3.33–34

The steady state errors for the test input 37tu( t) is (A) 0

(B) 0.061

(C) ¥

(D) 609

unit ramp input is

30. In the system shown in fig. P6.3.30, r ( t) = 1 + 2 t, t > 0. The steady state error e( t) is equal to r(t)

+

10(s + 1) s2(s + 2)

e(t)

1 5

(B) 5

(C) 0

(D) ¥

G( s) =

10(1 + 4 s) s 2 (1 + s)

(B) 0.2

(C) ¥

(D) 0

0.7 without affecting the steady state error, then the

c(t)

value of K a and K t are (A) 86, 12.8

(B) 49, 9.3

(C) 24.5, 3.9

(D) 43, 6.4

35. A system has the following transfer function G( s) =

31. A ufb control system has a forward path transfer function

(A) 5

34. If the damping ratio of the system is increased to

Fig. P6.3.30

(A)

33. If K t = 0 and K a = 5, then the steady state error to

100( s + 15)( s + 50) s 4 ( s + 12)( s 2 + 3s + 10)

The type and order of the system are respectively (A) 7 and 5

(B) 4 and 5

(C) 4 and 7

(D) 7 and 4

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Page 347


Time Response

SOLUTIONS

Chap 6.3

10 1 1 = s( s + 10) s s + 10

8. (A) C( s) = Þ

1. (D) Characteristic equation is s + 9 s + 18.

c( t) = 1 - e -10 t

2

a = 10,

w = 18, 2 xwn = 9 2 n

Therefore x = 106 . ,

wn = 4.24 rad/s

Settling time Ts =

1 0.6 6 ( s + 0.8) 2 + (0.6) 2

2. (A) T( s) =

wn 1 - x2 = 0.6 , Hence wn = 1,

9. (D) xwn =

xwn = 0.8

Ds = { s - ( -3 + j 4)}{ s - ( -3 - j 4)} = ( s + 3) + 4 . 2

= s + 6 s + 25, w = 25 2 xwn = 6 ,

x=

w2n = 4

wn = 2 ,

5. (D) M p = e xp 1-x

2

T( s) =

=3

xp 1 - x2

Þ

=

x = 0.59

-

xp 1 - x2

Þ

x = 0.56, wn =

4 = 1192 . xTs

Note : 32 . , For 0 < x < 0.69 Ts = xwn

5 = 0.05, 100

Ts =

4.5 , For xwn

12. (B) T( s) =

1 = 1.45 0.69

=

Peak time, Tp =

Þ

Therefore Poles = - xwn ± jwn 1 - x2 = - 6.67 ± j9.88

x = 0.5

2 xwn = 2,

x = 0.69,

wn =

xp 1 - x2

1 - x2 = 0.779,

11. (B) 0.12 = e

1 K = 2 1 + G( s) s + 2 s + K

2 xwn = 2,

p Tp

-

Poles = - xwn ± jwn 1 - x2 = - 0.46 ± j0.63

16 4 = ( 4 s 2 + 8 s + 16) ( s 2 + 2 s + 4)

-

wn =

6 = 0.6 2´5

4. (A) T( s) = Þ

2

wn = 5 rad/s

Þ

4 = 0.4s a

4 p = 0.571, wn 1 - x2 = = 1047 . Ts Tp

10. (A) 0.1 = e

3. (A) Characteristic equation is 2 n

2.2 2.2 = = 0.22s a 10

Poles = - 0.571 ± j1.047

x = 0.8

2

Rise time Tr =

p p p = = = 3 sec wd wn - x2 1.45 (1 - 0.69 2 )

40000 w2n = 2 2 s + 2 xwn s + wn s + 500 s + 40000

40000 ( s + 100)( s + 400)

R( s) =

But the peak time Tp given is 1 sec. Hence these two

x > 0.69

40000 1 4 1 = + s( s + 100)( s + 400) s 3( s + 100) 3( s + 400)

r ( t) = 1 -

4 -100 t 1 -400 t e + e 3 3

specification cannot be met. 13. (A) System has two different poles on negative real

K1 , 6. (C) T( s) = 2 s + ( K 2 + s) + K 1 w2n = K 1 ,

14. (A) 1. Overdamped response (a, b)

2 xwn = 1 + K 2

wd = 0.10, wn = 12.5

axis. So response is over damped.

x = 0.6, Þ

wd = wn 1 - 0.6 2 = 10

2. Underdamped response (c)

K 1 = 156.25,

Poles : Two complex in left half plane

2 wn 3 = K 2 + 1 2 ´ 12.5 ´ 0.6 = K 2 + 1 7. (A) M p = e

-

Þ

K 2 = 14

xp 1 - x2

Poles : Two real and different on negative real axis.

3.Undamped response (d) Poles : Two imaginary. 4.Critically damped (e)

, At x = 0,

M p = 1 = 100%

Poles : Two real and same on negative real axis.

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Page 349


UNIT 5

15. (B) K p = lim G ( s) = 1000 s ®0

K v = lim sG( s) = 0 s ®0

ess = lim sE( s) = lim s ®0

R( s) =

K a = lim s 2 G( s) = 0 s ®0

16. (D) H ( s) = 1,

K p = lim G( s) × H ( s) = ¥ s ®0

s®

s ®0

17. (B) H ( s) = 1,

K 8

K p = lim G( s) × H ( s) = 50 s ®0

K v = lim sG( s) × H ( s) = 0

ess = lim s ®0

sR( s) , 1 + G ( s)

1 6 = K 1 K 2 ( s + 2) K 1K 2 s+ ( s + 3)( s + 4)

6 = 0 .003 125 ´ 10 3 K 2

Þ

K 2 = 0.016

21. (C) E( s) = R( s) - C( s) H ( s) = R( s) -

s ®0

s ®0

1 K 1 K 2 ( s + 2) , G( s) = s2 s( s + 3)( s + 4)

K v = lim sG( s) × H ( s) = ¥ K a = lim s 2 G ( s) × H ( s) =

Control Systems

K a = lim s 2 G( s) × H ( s) = 0.

R( s) G( s) H ( s) R( s) = 1 + G( s) H ( s) 1 + G( s) H ( s)

s ®0

18. (C) System type = 1, so n = 1 K v = lim sG( s) = s ®0

ess = lim sE( s) = lim s ®0

K = 100 a

s ®0

For 10% overshoot, 0.1 = = e T( s) =

-

22. (A) ess = lim

xp 1 - x2

Þ

s ®0

x = 0.6

G( s) K = 2 1 + G ( s) s + as + K

2 xwn = a ,

w2n = K

K ´ 0.6 K = 100 2

Þ Þ

2 ´ 0.6 K = a

R( s) = D( s) =

K = 14400

K = 100, K = 14400, a

14400 = 100 a

Þ

a = 144

Error in output due to disturbance E( s) = TD( s) D ( s), 1 , s

essD = lim sE( s) = lim s × 3 = 0.000012 2 K1

s ®0

Þ

1 3 × TD( s) = lim TD( s) = s ®0 s 2 K1

1 s+5

and

G2 ( s) =

100 s+2

1 s

23. (A) Using Routh-Hurwitz Criterion, system is stable 2000 = 6. 25 4 ´ 8 ´ 10 1 1 = = 0.16 K v 6.25

K v = lim sG( s) = s ®0

24. (A) R( s) =

6 , s4

E( s) =

R( s) 1 + G ( s)

ess = lim sE( s) s ®0

6s s4 = lim 2 s ®0 1000 ( s + 4 s + 20) ( s 2 + 20 s + 15) 1+ s 3( s + 2) ( s + 10) = lim s ®0

K 1 = 125 ´ 10 3

Error due to ramp input Page 350

sR( s) - sD( s) G2 ( s) 1 + G1 ( s) G2 ( s)

minimum possible error

K 2 ( s + 3) s( s + 3)( s + 4) + K 1 K 2 ( s + 2)

s ®0

2 3

100 -49 2 = = 1 100 11 1+ ´ 5 2

maximum

K2 s( s + 4) TD( s) = K 1 K 2 ( s + 2) 1+ s( s + 4)( s + 3)

If D( s) =

=

for 0 < K < 2000

20. (C) If R ( s) = 0

=

1 1 1+ 2 ( s + s + 2) ( s + 1)

1-

ess 19. (D)

G1 ( s) =

where

s s

=

6 1000 ( s 2 + 4 s + 20) ( s 2 + 20 s + 15) s + ( s + 2) ( s + 10) 3

6 0+

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1000 ´20 ´15 2 ´10

= 4 ´ 10 -4


Time Response

25. (A) K v = lim sG( s) 10 4 ´ 3 ´ 10 ´ 20 25 ´ a ´ 30

10 4 =

Þ

a=4 1 = ¥ K v = 0, ess = Kv

26. (C) System is zero type

34. (C) The equivalent open-loop transfer function Ka Ka s(0.5 s + 1) Ge = = sK t s(0.5 s + 1 + K t ) 1+ s(0.5 s + 1)

27. (D) K p = lim G( s) = 5 s ®0

60 = 10 1 + Kp

T( s) =

28. (A) K a = lim s 2 G( s) = 64 s ®0

ess =

30 ´ 2 = 0.9375 64

=

G( s) Ka = 1 + G( s) 0.5 s 2 + s(1 + K t ) + K a

2K a s 2 + 2 s(1 + K t ) + 2 K a

w2n = 2 K a Þ wn =

29. (B) K v = lim sG( s) = 609.02 s ®0

ess =

1 1 s2 = lim = = 0.2 s ®0 5 5 1+ s(0.5 s + 1 ) s×

ess

For input 60u( t), ess =

5 1 , H ( s) = 1, R( s) = 2 s(0.5 s + 1) s

G( s) =

s ®0

Chap 6.3

2K a

2 x wn = 2(1 + K t ) Kt x =1 + = 0.7 2K a

37 = 0.0607 Kv

30. (C) The system is type 2. Thus to step and ramp

ess = lim s ®0

input error will be zero. E( s) = R( s) - C( s) = R( s) -

G( s) R( s) 1 + G ( s)

R( s) 1 + G( s)

=

... (i)

sR( s) 1 , R ( s) = 2 1 + Ge ( s) s

1 1 + Kt = Ka ö æ Ka ÷÷ s çç 1 + s(0.5 s + 1 + K t ) ø è 1 + Kt = = 0. 2 Ka

ess = lim s ®0

1 2 s+2 + = s s2 s2 s+2 E( s) = 10( s + 1) s2 + ( s + 2) R( s) =

ess

...(ii)

Solving (i) and (ii) K a = 24.5 ,

ess ( t) = lim sE( s) = 0 s ®0

. K t = 39

35. (C) The s has power of 4 and denominator has order

31. (C) System is type 2. Therefore error due to 1 + t

of 7. So Type 4 and Order 7.

t2 1 would be . would be zero and due to 2 Ka

36. (D) For 8u( t),

K a = lim s 2 G( s) = 10,

ess ( t) =

s ®0

1 = 0.1 10

ess =

8 = 2. 1 + Kp

For 8tu( t), ess = ¥, since the system is type 0.

Note that you may calculate error from the formula ess ( t) = lim sE( s) = s ®0

37. (A) For equivalent unit feedback system the forward

sR( s) 1 + G( s)

transfer function is Ge =

32. (D) K p = lim G( s) = K s ®0

1 1 ess ( t) = = = 0.1 1 + Kp 1 + K

Þ

K = 9.

s ®0

When K t = 0 and K a = 5

10 ( s + 10 ) s (s + 2 ) 10 ( s + 10 )( s + 3) s (s + 2 )

10( s + 10) 11s + 132 s + 300 2

The system is of Type 0. Hence step input will produce a

33. (B) ess = lim e( t) = lim sE( s) = lim t ®¥

=

G( s) = 1 + G( s) H ( s) - G( s) 1 +

s ®0

sR( s) 1 + G( s) H ( s)

constant error constant.

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Page 351


CHAPTER

6.5 FREQUENCY-DOMAIN ANALYSIS


4. The gain-phase plots of open-loop transfer function of

An under damped second order system having a

correct sequence of the increasing order of stability of

transfer function of the form T( s) =

four different system are shown in fig. P6.5.4. The these four system will be

Kw2n 2 s + 2 xwn s + w2n

dB B

A

C

40 dB

has a frequency response plot shown in fig.

D

30 dB

P6.5.1–2.

20 dB 10 dB

|T( jw )|

-270o

2.5

-225o

o

-135

-90o

-45o

-20 dB -30 dB

1.0 w

wn

Fig. P6.5.4

Fig. P6.5.1-2

1. The system gain K is (A) 1 (C)

(B) 2 (D)

2

(A) D, C, B, A

(B) A, B, C, D

(C) B, C, A, D

(D) A, D, B, C

1

5. The open-loop frequency response of a

2

feedback system is shown in following table

2. The damping factor x is approximately

w

|G ( jw)|

ÐG ( jw)

(A) 0.6

(B) 0.2

2

8.5

-119°

(C) 1.8

(D) 2.4

3

6.4

-128°

4

4.8

-142°

5

2.56

-156°

6

1.4

-164°

8

1.00

-172°

10

0.63

-180°

3. For the transfer function G( s) H ( s) =

1 s( s + 1)( s + 0.5)

the phase cross-over frequency is (A) 0.5 rad/sec

(B) 0.707 rad/sec

(C) 1.732 rad/sec

(D) 2 rad/sec

Fig. P6.5.5

Page 362

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unity


Frequency-Domain Analysis

The gain margin and phase margin of the system

Chap 6.5

10. The gain margin of the ufb system

are (A) 2 dB, 8°

(B) 2 dB, -172 °

(C) 4 dB, 8°

(D) 4 dB, -172 °

G( s) =

Statement for Q.6–7: Consider the gain-phase plot shown in fig.

2 is ( s + 1)( s + 2)

(A) 1.76 dB

(B) 3.5 dB

(C) -3.5 dB

(D) -1.76 dB

11. The open-loop transfer function of a system is K s(1 + 2 s)(1 + 3s)

G( s) H ( s) =

P6.5.6–7. dB w=2

2 dB

0

The phase crossover frequency is (B) 2.46 rad/sec

(C) 0.41 rad/sec

(D) 3.23 rad/sec

ÐG( jw)

w = 10

-2 dB

(A) 6 rad/sec

12. The open-loop transfer function of a ufb system is

w = 100 o

-270

-180

o

G( s) = -140

o

-90

o

Fig. P6.5.6-7

1+ s s(1 + 0.5 s)

The corner frequencies are

6. The gain margin and phase margin are

(A) 0 and 2

(B) 0 and 1

(A) -2 dB, 40°

(B) 2 dB, 40°

(C) 0 and -1

(D) 1 and 2

(C) 2 dB, 140°

(D) -2 dB, 140°

13. In the Bode-plot of a unity feedback control system,

7. The gain crossover and phase crossover frequency are

the value of magnitude of G( jw) at the phase crossover

respectively

frequency is

(A) 10 rad/sec, 100 rad/sec

(A) 2

1 2

(B) 100 rad/sec, 10 rad/sec (C)

(C) 10 rad/sec, 2 rad/sec

. The gain margin is 1 (B) 2

1 3

(D) 3

(D) 100 rad/sec, 2 rad/sec 8. The phase margin of a system with the open loop transfer function

14. In the Bode-plot of a ufb control system, the value of phase of G( jw) at the gain crossover frequency is -120 °. The phase margin of the system is

G( s) H ( s) =

(1 - s) is (1 + s)( 3 + s)

(A) 68.3°

(B) 90°

(C) 0

(D) ¥

(A) -120 °

(B) 60°

(C) -60 °

(D) 120°

15. The transfer function of a system is given by G( s) =

9. Consider a ufb system having an open-loop transfer function

K 1 ; K < s( sT + 1) T

The Bode plot of this function is G( s) =

K s(0.2 s + 1)(0.05 s + 1)

dB -20 dB/dec

For K = 1, the gain margin is 28 dB. When gain margin is 20 dB, K will be equal to (A) 2

(B) 4

(C) 5

(D) 2.5

dB -40 dB/dec -40 dB/dec

0 dB

0.1 T

1 T

(A) www.gatehelp.com

w

0 dB

0.1 T

1 T

w

(B) Page 363


UNIT 6

dB

dB -20 dB/dec

(A)

100 s + 10

(B)

(C)

1 s + 10


-20 dB/dec 0 dB

w

1 T

0.1 T

0 dB

1 T

0.1 T

w

Control & System

10 s + 10

-40 dB/dec

19. Consider the asymptotic Bode plot of a minimum (C)

(D)

phase linear system given in fig. P6.5.19. The transfer function is dB

16. The asymptotic approximation of the log-magnitude versus frequency plot of a certain system is shown in

32

fig. P6.5.16. Its transfer function is

-20 dB/dec

dB 54 dB

6

-40 dB/dec

-20 dB/dec w1

0.1

-60 dB/dec

w2

w

10 -40 dB/dec

-40 dB/dec

-60 dB/dec

0.1

5

2

w

25

(C)

Fig. P6.5.16

50( s + 5) (A) 2 s ( s + 2)( s + 25) (C)

10 s 2 ( s + 5) ( s + 2)( s + 25)

Fig. P6.5.19

8 s( s + 2) (A) ( s + 5)( s + 10)

(B)

20( s + 5) s 2 ( s + 2)( s + 25)

(D)

20( s + 5) s( s + 2)( s + 25)

4( s + 2) s( s + 5)( s + 10)

(B)

4( s + 5) ( s + 2)( s + 10)

(D)

8 s( s + 5) ( s + 2)( s + 10)

20. The Bode plot shown in fig. P6.5.20 represent dB 100 dB

17. For the Bode plot shown in fig. P6.5.17 the transfer

-60 dB/dec

function is

40 dB/dec dB 4

-4

100 ( s + 4)( s + 10)

100 s 2 (1 + 0.1s) 3

(B)

1000 s 2 (1 + 0.1s) 3

100 s 2 (1 + 0.1s) 5

(D)

1000 s 2 (1 + 0.1s) 5

c

-2 0

de

B/

dB /

0d

(C)

w

Fig. P6.5.20

(A)

(B)

100( s + 4) s( s + 10) 2

(C)

(D)

100 s ( s + 4)( s + 10)


Fig. P6.5.17

100 s (A) ( s + 4)( s + 10) 2

w = 10

w

10

de c

0 dB

2

18. Bode plot of a stable system is shown in the fig. P6.3.18. The open-loop transfer function of the ufb

The Bode plot of the transfer function K (1 + sT) is given in the fig. P6.5.21–22. Phase

dB

system is dB -20 dB/dec 20 dB -20 dB/dec 1 T

w

Page 364

Fig. P6.5.21-22

www.gatehelp.com

-45

10 T

°/d

w

Fig. P6.5.18

0.1 T

ec

w


UNIT 6


Control & System

Im

Im 2

2

Consider the Bode plot of a ufb system shown in fig. P6.5.29–30.

w= 0

w=¥

Re

w= 0

w=¥

Re

dB 32 dB

0 dB

(A)

-20 dB/dec

18 dB

-40 dB/dec w1

0.1

(B)

Im

w

Im 2

2

Fig. P6.5.29-30

Re

w= 0

w=¥

w= 0

w=¥

Re

29. The steady state error corresponding to a ramp input is (A) 0.25

(B) 0.2

(C) 0

(D) ¥

(C)

(D)

35. Consider a ufb system whose open-loop transfer 30. The damping ratio is

function is

(A) 0.063

(B) 0.179

(C) 0.483

(D) 0.639

G( s) =

K s( s 2 + 2 s + 2)

The Nyquist plot for this system is

31. The Nyquist plot of a open-loop transfer function

Im

G( jw) H ( jw) of a system encloses the ( -1, j0) point. The

Im

gain margin of the system is (A) less than zero

(B) greater than zero

(C) zero

(D) infinity

w=¥

Re

Re

w=¥

w= 0

32. Consider a ufb system G( s) =

K s(1 + sT1 )(1 + sT2 )(1 + sT3)

w= 0

(A)

(B)

The angle of asymptote, which the Nyquist plot approaches as w ® 0, is

Im

(A) -90 °

(B) 90°

(C) 180°

(D) -45 °

Im

w=¥

33. If the gain margin of a certain feedback system is

Re

Re

w=¥

w= 0

given as 20 dB, the Nyquist plot will cross the negative real axis at the point

w= 0

(A) s = -0.05

(B) s = -0.2

(C) s = -0.1

(D) s = -0.01

(C)

34. The transfer function of an open-loop system is G( s) H ( s) =

36. The open loop transfer function of a system is

s+2 ( s + 1)( s - 1)

G( s) H ( s) =

K (1 + s) 2 s3

The Nyquist plot for this system is

The Nyquist plot will be of the form

Page 366

(D)

www.gatehelp.com



Chap 6.5

Im

Im

Im w= 0

w= 0

Re

w=¥

10.64

Re

w=¥

w=¥

Re

w= 0

(A)

(B) Im

Im

Fig. P6.5.38

The no. of poles of closed loop system in RHP are w=¥

w=¥

Re

Re

(C)

(B) 1

(C) 2

D) 4


w= 0

w= 0

(A) 0

The open-loop transfer function of a feedback

(D)

control system is 37. For the certain unity feedback system

G( s) H ( s) =

K G( s) = s( s + 1)(2 s + 1)( 3s + 1)

-1 2 s(1 - 20 s)

39. The Nyquist plot for this system is

The Nyquist plot is

Im Im

Im

Im w= 0

w= 0

w=¥ w=¥

Re

Re

w=¥

Re

Re

w=¥

w= 0

w= 0

(A)

(A)

(B)

Im

(B) Im

Im

Im

w=¥

w=¥

Re

Re

w=¥

Re

Re

w=¥

w= 0 w= 0

w= 0

w= 0

(C) (C)

(D)

(D)

38. The Nyquist plot of a system is shown in fig.

40. Regarding the system consider the statements

P6.5.38. The open-loop transfer function is

1. Open-loop system is stable

G( s) H ( s) =

4s + 1 s ( s + 1)(2 s + 1) 2

2. Closed-loop system is unstable 3. One closed-loop poles is lying on the RHP www.gatehelp.com

Page 367


UNIT 6

Control & System

Im

The correct statements are (A) 1 and 2

(B) 1 and 3

(C) only 2

-2 3

Im

w=¥

-3 2

Re

w=¥

Re

(D) All

41. The Nyquist plot shown in the fig. P6.5.41 is for (A) type–0 system

(B) type–1 system

(C) type–2 system

(D) type–3 system

w= 0

w= 0

(C)

Statement for Q.42–43: The open-loop transfer function of a feedback system is

(D)

45. The phase crossover and gain crossover frequencies are

K (1 + s) (1 - s)

(A) 1.414 rad/sec, 0.57 rad/sec

42. The Nyquist plot of this system is

(C) 0.707 rad/sec, 0.57 rad/sec

G( s) H ( s) =

(B) 1.414 rad/sec, 1.38 rad/sec

Im

(D) 0.707 rad/sec, 1.38 rad/sec

Im

46. The gain margin and phase margin are w=¥

w= 0

Re

w=¥

w= 0

(A)

Re

(A) -3.52 dB, -168.5 °

(B) -3.52 dB, 11.6 °

(C) 3.52 dB, -168.5 °

(D) 3.52 dB, 11.6 °

(B) **************** Im

Im

w=¥

w= 0

Re

w=¥

w= 0

(C)

Re

(D)

43. The system is stable for K (A) K > 1

(B) K < 1

(C) any value of K

(D) unstable

Statement for Q.44–46: A unity feedback system has open-loop transfer function G( s) =

1 s(2 s + 1)( s + 1)

44. The Nyquist plot for the system is Im

Im

w=¥

w= 0 2 3

w=¥

(A)

Page 368

Re

3 2

w= 0

Re

(B) www.gatehelp.com



At w = 10 rad/sec gain is 0 dB. Gain cross over frequency

SOLUTIONS 1. (A) T( s) =

w =10 rad/sec.

Kw2n s + 2 xwn s + w2n

8. (D) |GH( jw)|¹ 1, for any value of w. Thus phase

2

margin is ¥.

Kw2n T( jw) = -w2 + 2 jxwn w + w2n

|T( jw)|

2

=

9. (D) For 28 dB gain Nyquist plot intersect the real axis at a, 1 20 log = 28 a

K w ( w - w ) + 4 x2 w2n w2 2

2 n

4 n

2 2

From the fig. P6.5.1–2, |T( j0)| = 1

|T( j0)|

2

=

K 2 wn4 = K2 =1 wn4

Þ

denominator of function |T( jw)| is minimum i.e. when 2

Þ

w = wn

K w K2 |T( jwn )| = 2 = 2 4x w 4x 2

x=

2

4 n 4 n

Þ

. = 35 . dB = 15

ÐGH ( jw) = -180 ° GH ( jw) =

K jw(1 + 2 jw)(1 + 3 jw)

-90 °- tan -1 2 wp - tan -1 3wp = -180 °

- tan -1 2 w - tan -1 w - 90 ° = -180 °

tan -1 2 wp + tan -1 3wp = 90 ° 2 wp + 3wp = tan 90 ° 1 - (2 wp)( 3wp)

tan -1 2 w + tan -1 w = 90 ° 2w + w = tan 90 ° = ¥ 1 - (2 w)( w) 2

-1

11. (C) For phase crossover frequency

At phase cross over point f = -180 °

w=

1 2

-1

f = -90 °- tan -1 2 w - tan -1 w

1 - (2 w) w = 0 Þ

This is achieved if the system gain is increased by factor 0.1 = 2.5. Thus K = 2.5. 0.04

é KT1 T2 ù é(2)(0.5) ù Gain Margin = ê ú = ê 1 + 0.5 ú + T T ë û 2û ë 1

1 jw( jw + 1)(0.5 + jw)

1

a = 0.04

10. (B) Here K = 2, T1 = 1, T2 =

K |T( jwn )| = = 2.5 2x

K = 0.2 5

3. (B) G( jw) H ( jw) =

Þ

For 20 dB gain Nyquist plot should intersect at b, 1 20 log = 20 Þ b = 0.1. b

K =1

2. (B) The peak value of T( jw) occurs when the

w2n - w2 = 0

Chap 6.5

1 - 6 w2p = 0

= 0.707 rad/sec

Þ

12. (D) G( s) = 4. (B) For a stable system gain at 180° phase must be

wp = 0.41 rad/s

s+1 s+1 = s(1 + 0.5 s) æs ö sç + 1 ÷ è2 ø

negative in dB. More magnitude more stability.

The Bode plot of this function has break at w = 1 and

5. (C) At 180° gain is 0.63. Hence gain margin is 1 = 20 log = 4 dB 0.63

w = 2. These are the corner frequencies. 13. (A) G.M. =

At unity gain phase is -172 °, Phase margin = 180 °-172 ° = 8 °

14. (B) P.M. = 180 ° + ÐGH ( jw1 ) = 180 °-120 ° = 60 °

6. (A) At ÐG( jw) = 180 ° gain is -2 dB. Hence gain margin is 2 dB. At 0 dB gain phase is -140 °. Hence phase margin is 180 °-140 ° = 40 °. 7.

(A) At w = 100 rad/sec phase is 180°. Phase cross-

over frequency wp = 100 rad/sec.

1 1 = =2 GH ( jwp) 1 2

15. (D) Due to pole at origin initial plot has a slope of 1 -20 dB/decade. At s = jw = . Slope increases to -40 T 1 dB/decade. At w = , T

|G( jw)| » KT < 1 ,Gain www.gatehelp.com

in dB < 0. Page 369



27. (C) Initially slope is -20 dB/decade. Hence there is a

Chap 6.5

ÐGH ( jw) = - tan -1

pole at origin and system type is 1. For type–1 system position error coefficient is ¥. 20 log K = 6

Þ

GH ( jw) =

K = 2,

w (2 - w2 ) 2 + 4 w2

At w = 0, 28. (B) The system is type –0,

At w = ¥

1 1 . 20 log K p = 40, K p = 100, estep ( ¥) = = 1 + K p 101

At w = 1,

-20 dB/dec

0.5

0.1

1.4

K 2 18

Ð - 206.6 °,

correct option. 4

w

36. (B) GH ( jw) =

Fig.S6.5.29

K v = 4, eramp ( ¥) =

GH ( jw) = 0 Ð - 270 °, K GH ( jw) = Ð - 153.43°, 5

Due to s there will be a infinite semicircle. Hence (C) is

-40 dB/dec

18 dB 0 dB

GH ( jw) = ¥ Ð - 90 °,

At w = 2, GH ( jw) =

29. (A) The Bode plot is as shown in fig. S6.5.29 32 dB

2w - 90 ° 2 - w2 K

K (1 + w2 ) w3

|GH( jw)| =

1 1 = = 0.25 Kv 4

K (1 + jw) 2 ( jw) 3

ÐGH ( jw) = -270 °+ 2 tan -1 w For w = 0, GH ( jw) = ¥Ð - 270 °

0.5 w 30. (B) From fig. S6.5.29 x = 2 = = 0.179 2 w3 2(1.4)

For w = 1 , ÐGH ( jw) = -180 ° For w = ¥, GH ( jw) = 0 Ð - 90 °

31. (A) If Nyquist plot encloses the point ( -1, j0), the

As w increases from 0 to ¥, phase goes -270 ° to -90 °. Due to s 3 term there will be 3 infinite semicircle.

system is unstable and gain margin is negative. 37. (A) |GH ( jw)| =

K 32.(A) GH ( jw) = jw(1 + jwT1 )(1 + jwT2 ) (1 + jwT2 )

,

ÐGH ( jw) = -90 °- tan -1 w - tan -1 2 w - tan -1 3w ,

K K lim GH ( jw) = lim = lim Ð - 90 ° w ®0 w ®0 jw w ®0 w

For w = 0, GH ( jw) = ¥Ð - 90 °,

Hence, the asymptote of the Nyquist plot tends to an angle of -90 ° as w ® 0.

K 1 + w2 1 + 4 w2 1 + 9 w2

For w = ¥, GH ( jw) = 0 Ð - 360 °, Hence (A) is correct option. 38. (C) The open-loop poles in RHP are P = 0. Nyquist

33. (C) 20 log

1 = 20 GH ( jw)

path enclosed 2 times the point ( -1 + j0). Taking

1 = 10 GH ( jw)

Þ

N = P - Z,

GH ( jw) = 0.1

Since system is stable, it will cross at s = -0.1.

clockwise encirclements as negative N = -2. -2 = 0 - Z ,

Z = 2 which implies that two

poles of closed-loop system are on RHP. -1 , 2 s(1 - 20 s)

s+2 34. (B) GH ( s) = 2 ( s - 1)

39. (B) G( s) H ( s) =

jw + 2 GH ( jw) = ( -1 - w2 )

GH ( jw) =

At w = 0 ,

GH ( jw) = 2 Ð - 180 °

ÐGH ( jw) = 180 °-90 ° - tan -1

At w = ¥,

GH ( jw) = 0 Ð - 270 °

Hence (B) is correct option. 35. (C) GH ( jw) =

K jw( -w + 2 jw + 2) 2

1 2 w 1 + 400 w2 -20 w , 1

At w = 0 GH ( jw) = ¥Ð90 ° At w = ¥ GH ( jw) = 0 Ð180 ° At w = 0.1 GH ( jw) = 2.24 Ð153.43° At w = 0.01 GH ( jw) = 49 Ð9115 . ° www.gatehelp.com

Page 371


UNIT 6

40. (C) One open-loop pole is lying on the RHP. Thus

Control & System

The frequency at which magnitude unity is

open-loop system is unstable and P = 1. There is one

Im

clockwise encirclement. Hence N = -1. Z = P - N = 1 - ( -1) = 2, -2 3

Hence there are 2 closed-loop poles on the RHP and

Re

-1 wp

system is unstable.

w1

41. (B) There is one infinite semicircle. Which represent

Phase Margin

single pole at origin. So system type is1. K 1 + w2

42. (D) |GH ( jw)| =

1 + w2

ÐGH ( jw) = tan -1 w - tan -1 -

=K

Fig.S6.5.44

w (1 + w ) (1 + 4 w ) = 1 2 1

-w 1

At w = 2 GH ( jw) = KÐ127 °,

43. (A) RHP poles of open-loop system P = 1, Z = P - N . For closed loop system to be stable,

w ®0

At unit gain w1 = 0.57 rad/sec,

Phase margin = -168.42 °+180 ° = 11.6 ° Note that system is stable. So gain margin and phase margin are positive value. Hence only possible option is

H ( s) = 1

(D).

***************

1 jw(2 jw + 1)( jw + 1) w ®0

1 = ¥Ð - 90 ° jw

lim GH ( jw) = lim w w ®¥

w ®¥

1 = 0 Ð - 270 ° 2( jw) 3

The intersection with the real axis can be calculated as Im{ GH ( jw)} = 0, The condition gives w (2 w2 - 1) = 0 i.e. w = 0,

1 2

,

æ 1 ö -2 GH ç j ÷= 2ø 3 è

With the above information the plot in option (C) is correct. 45. (C) The Nyquist plot crosses the negative real axis 1 rad/sec. Hence phase crossover frequency is at w = 2 wp = Page 372

1 2

3 = 352 . dB. 2

ÐGH ( jw) = -90 °- tan -1 w - tan -1 2 w ,

1 s(2 s + 1)( s + 1)

lim GH ( jw) = lim

2 3

ÐGH ( jw1 ) = -90 °- tan -1 0.57 - tan -1 2(0.57) = -168.42 °

( -1 + j0). It is possible when K > 1.

GH ( jw) =

|GH( jwp)| =

Phase at this frequency is

N =1

There must be one anticlockwise rotation of point

GH ( s) =

,

|GH( jwp)|

Gain Margin = 20 log

At w = ¥ GH ( jw) = KÐ180 °,

1 , s(2 s + 1)( s + 1)

1

46. (D) G.M. = 20 log

At w = 1 GH ( jw) = KÐ90 °,

44. (C) G( s) =

2 1

w2 = 0.326, w1 = 0.57 rad/sec

At w = 0 GH ( jw) = KÐ0 °,

Z = 0, 0 = 1 - N Þ

2 1

= 0.707 rad/sec. www.gatehelp.com


CHAPTER

6.6 DESIGN OF CONTROL SYSTEMS

1. The term ‘reset control’ refers to

(C) is predictive in nature

(A) Integral control

(B) Derivative control

(D) increases the order of the system

(C) Proportional control


6. Consider the List I and List II

2. If stability error for step input and speed of response

List I

List II

be the criteria for design, the suitable controller will be

P. Derivative control

1. Improved overshoot response

(A) P controller

(B) PI controller

Q. Integral control

2. Less steady state errors

(C) PD controller

(D) PID controller

R. Rate feed back control

3. Less stable

S. Proportional control

4. More damping


1 + 0.5 s represent a 1+ s

The correct match is

(A) Lag network (B) Lead network (C) Lag–lead network (D) Proportional controller

P

Q

R

S

(A)

1

2

3

4

(B)

4

3

1

2

(C)

2

3

1

4

(D)

1

2

4

3

4. A lag compensation network

7. Consider the List–I (Transfer function) and List–II

(a) increases the gain of the original network without affecting stability.

(Controller) List I

List II

(b) reduces the steady state error.

P.

1. P–controller

(c) reduces the speed of response

Q.

(d) permits the increase of gain of phase margin is acceptable.

2. PI–controller

R.

K1s + K2 K 3s

3. PD–controller

S.

K1 K 2s

4. PID–controller

In the above statements, which are correct (A) a and b

(B) b and c

(C) b, c, and d

(D) all

The correct match is P

Q

R

S

(A)

3

4

2

1

5. Derivative control

(B)

4

3

1

2

(A) has the same effect as output rate control

(C)

3

2

1

4

(D)

4

1

2

3

(B) reduces damping

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Page 373


UNIT 6

Control Systems

8. The transfer function of a compensating network is of

(B) Two–position controller

form (1 + aTs) (1 + Ts). If this is a phase–Lag network,

(C) Floating controller

the value of a should be

(D) Proportional–position controller

(A) greater than 1 (B) between 0 and 1

14. In case of phase–lag compensation used is system,

(C) exactly equal to 1

gain crossover frequency, band width and undamped

(D) exactly equal to 0

frequency are respectively (A) decreased, decreased, decreased

9.

The

poll–zero

configuration

of

a

phase–lead

compensator is given by jw

(A)

(B) increased, increased, increased (C) increased, increased, decreased

jw

(B)

(D) increased, decreased, decreased s

s

15. A process with open–loop model (C)

jw

(D)

jw

s

G( s) = s

Ke - s TD ts + 1

is controlled by a PID controller. For this purpose 10. While designing controller, the advantage of pole– zero cancellation is

(A) the derivative mode improves transient performance

(A) The system order is increased

(B) the derivative mode improves steady state performance

(B) The system order is reduced

(C) the integral mode improves transient performance

(C) The cost of controller becomes low

(D) the integral mode improves steady state performance.

(D) System’s error reduced to optimum levels

The correct statements are 11. A proportional controller leads to

(A) (a) and (c)

(B) (b) and (c)

(A) infinite error for step input for type 1 system

(C) (a) and (d)

(D) (b) and (d)

(B) finite error for step input for type 1 system (C) zero steady state error for step input for type 1 system (D) zero steady state error for step input for type 0 system 12. The transfer function of a phase compensator is given by (1 + aTs) (1 + Ts) where a > 1 and T > 0. The

16. A lead compensating network (a) improves response time (b) stabilizes the system with low phase margin (c) enables moderate increase in gain without affecting stability. (d) increases resonant frequency In the above statements, correct are

maximum phase shift provided by a such compensator is æ a + 1ö (A) tan -1 çç ÷÷ è a -1ø (C) tan

-1

æ a -1ö çç ÷÷ è a + 1ø

(D) cos

-1

(B) (a) and (c)

(C) (a), (c) and (d)

(D) All

17. A Lag network for compensation normally consists

æ a -1ö çç ÷÷ è a + 1ø

of (A) R, L and C elements

13. For an electrically heated temperature controlled

(B) R and L elements

liquid heater, the best controller is

(C) R and C elements

(A) Single–position controller Page 374

æ a -1ö (B) sin -1 çç ÷÷ è a + 1ø

(A) (a) and (b)

(D) R only

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Design of Control Systems

18. The pole–zero plot given in fig.P6.6.18 is that of a

Chap 6.6

SOLUTIONS

jw s

1. (A)

2. (D)

3. (A)

4. (D)

5. (B)

6. (D)

7. (A)

8. (B)

9. (A)

10. (B)

11. (C)

12. (B)

13. (C)

14. (D)

15. (C)

16. (D)

17. (C)

18. (D)

19. (D)

20. (D)

Fig. P6.6.18

(A) PID controller (B) PD controller (C) Integrator

21. (D)

(D) Lag–lead compensating network 19. The correct sequence of steps needed to improve system stability is (A) reduce gain, use negative feedback, insert derivative action (B) reduce gain, insert derivative action, use negative feedback (C) insert derivative action, use negative feedback, reduce gain (D) use negative feedback, reduce gain, insert derivative action. 20. In a derivative error compensation (A) damping decreases and setting time decreases (B) damping increases and setting time increases (C) damping decreases and setting time increases (D) damping increases and setting time decreases 21. An ON–OFF controller is a (A) P controller (B) PID controller (C) integral controller (D) non linear controller **********

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Page 375


CHAPTER

6.7 THE STATE-VARIABLE ANALYSIS

1. Consider the SFG shown in fig. P6.7.1 u

1 s

1

1 s

x3

1 s

x2

x1

1

y

-1 -2 -3

Fig. P6.7.1

For this é x& 1 ù é3 (A) ê x& 2 ú = ê0 ê ú ê êë x& 3 úû êë3

system dynamic equation is 1 2 ù é x1 ù é0 ù 1 1ú ê x2 ú + ê0 ú u úê ú ê ú 2 1úû êë x3 úû êë1 úû

1 0ù é-2 ê (A) 0 -2 0 ú ê ú 0 -3úû êë 0

é2 -1 0 ù (B) ê0 2 0 ú ê ú êë0 0 3úû

1 -2 ù é 2 ê (C) 0 -2 0ú ê ú 0 úû êë-3 0

é0 -1 2 ù (D) ê0 2 0 ú ê ú êë3 0 0 úû

3. 5 1 s

1 0 ù é x1 ù é0 ù é x& 1 ù é 0 ê ú ê (B) x& 2 = 0 0 1ú ê x2 ú + ê0 ú u ê ú ê úê ú ê ú êë x& 3 úû êë-3 -2 -1ûú êë x3 úû êë1 ûú

u

-2 1 s

1

1 s

x2

1

x1

5

y

-2

1 x3

5

-3

é x& 1 ù é0 -1 0 ù é x1 ù é0 ù (C) ê x& 2 ú = ê0 0 -1ú ê x2 ú + ê0 ú u ê ú ê úê ú ê ú 1úû êë x3 ûú êë1 ûú êë x& 3 úû êë3 2

Fig. P6.7.3

(D) None of the above Statement for Q.2–4: Represent the given system in state-space equation x& = A ×x + B× u. Choose the correction option for

é 1 0 -2 ù (A) ê 0 -2 0ú ê ú 0 úû êë-3 0

é-1 0 2 ù (B) ê 0 2 0 ú ê ú êë 3 0 0 úû

0 1ù é-2 (C) ê 0 -2 0 ú ê ú 0 -3úû êë 0

é2 0 -1ù (D) ê0 2 0 ú ê ú 3úû êë0 0

matrix A. 4.

2.

1 s

x2

1 s

1

2

x1

5

y

1 u

-2 1 s

u

-2 x3

1 s

x3 1

1 s

x2

Fig. P6.7.4 Fig. P6.7.2

www.gatehelp.com

1

1 s -4

-3

5

-3

Page 376

1

x1 1

y


The State-Variable Analysis

Chap 6.7

é 0 1 -4 ù (A) ê 1 0 0ú ê ú 0 úû êë-3 0

é 0 -1 4 ù (B) ê-1 0 0 ú ê ú êë 3 0 0 úû

8. The F( t ) is é cos 2 t sin 2 t ù (A) ê ú ë- sin 2 t cos 2 t û

écos 2 t - sin 2 t ù (B) ê cos 2 t úû ësin 2 t

é- 4 1 0 ù (C) ê 0 0 1ú ê ú êë 0 0 -3úû

é4 - 1 0 ù (D) ê0 0 -1ú ê ú 3úû êë0 0

é sin 2 t cos 2 t ù (C) ê ú ë- cos 2 t sin 2 t û

ésin 2 t - cos 2 t ù (D) ê sin 2 t úû ëcos 2 t

9. The q( t ) is

5. The state equation of a LTI system is represented by 1ù é 0 é0 1ù x& = ê ú x + ê1 0 ú u 2 1 ë û ë û The Eigen values are (A) -1, + 1

(B) -0.5 ± j1.323

(C) -1, - 1


é-3 0 ù é0 ù x& = ê x + ê úu ú ë 0 -3û ë1 û The state-transition matrix F( t) is

é-e -3t (C) ê ë 0

0 ù ú e -3t û 0 ù ú -e -3t û

é-e -3t (B) ê ë 0

0 ù ú e -3t û

ée -3t (D) ê ë0

0 ù ú -e -3t û

ésin 2 t ù (B) ê ú ëcos 2 t û

é0.5(1 - cos 2 t) ù (C) ê ú ë 0.5 sin 2 t û

écos 2 t ù (D) ê ú ësin 2 t û

10. From the following matrices, the state-transition

6. The state equation of a LTI system is

ée -3t (A) ê ë0

é0.5(1 - sin 2 t) ù (A) ê ú ë 0.5 cos 2 t û

matrices can be é -e - t 0 ù (A) ê ú 0 1 e- t û ë

é1 - e - t (B) ê ë 0

0ù ú e- t û

é 1 (C) ê -t ë1 - e

é1 - e - t (D) ê ë 0

e- t ù ú e- t û

0ù e - t úû

Statement for Q.11–13: A system is described by the dynamic equations x& ( t) = A ×x ( t) + B ×u( t), y( t) = C × x( t) where 1 0ù é 0 é0 ù ê ú A= 0 0 1 , B = ê0 ú, C = [1 0 0 ] ê ú ê ú êë-1 -2 -3ûú êë1 ûú

7. The state equation of a LTI system is é 0 2ù é0 ù x& = ê x + ê úu ú ë-2 0 û ë1 û

11. The Eigen values of A are (A) 0.325, -1.662 ± j0.562

The state transition matrix is é cos 2 t sin 2 t ù (A) ê ú ë- sin 2 t cos 2 t û

écos 2 t - sin 2 t ù (B) ê cos 2 t úû ësin 2 t

(B) 2.325, 0.338 ± j0.562

é sin 2 t cos 2 t ù (C) ê ú ë- cos 2 t sin 2 t û

ésin 2 t - cos 2 t ù (D) ê sin 2 t úû ëcos 2 t

(D) -0.325, 1.662 ± j0.562

(C) -2.325, -0.338 ± j0.562

12. The transfer-function relation between X ( s) and

Statement for Q.8–9: The state-space representation of a system is given by x& ( t) = A ×x ( t) + B ×u( t), where

U ( s) is (A)

é 1ù 1 ê -s ú s 3 + 3s 2 + 2 s - 1 ê 2 ú êë s úû

(B)

(C)

é 1ù 1 ê -s ú s 3 + 3s 2 + 2 s + 1 ê 2 ú êë s úû


é0 ù é 0 2ù , B=ê ú A =ê ú ë1 û ë-2 0 û If x(0) is the initial state vector, and the component of the input vector u( t) are all unit step

é 1ù 1 ê sú s 3 + 3s 2 + 2 s + 1 ê 2 ú êës úû

function, then the state transition equation is given by x& ( t) = F( t )x (0) + q( t ), where F( t ) is a state transition

13. The output transfer function Y ( s) U ( s) is

matrix and q( t ) is a vector matrix.

(A) s( s 3 + 3s 2 + 2 s + 1) -1

(B) s( s 3 + 3s 2 + 2 s - 1) -1

(C) ( s 3 + 3s 2 + 2 s + 1) -1


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Page 377


UNIT 6

14. A system is described by the dynamic equation x& ( t) = A ×x ( t) + B ×u( t), y( t) = C × x( t) where é-1 0 ù é1 ù , B = ê ú and C = [1 1] A =ê ú ë 0 -2 û ë0 û

(C)

( s + 1) ( s + 2) 2

(B)

s+1 s+2

( s + 2) ( s + 1)


18.

15. The state-space representation of a system is given by

(C) s( s 2 + 3s + 2) -1

(D) ( s + 1) -1

1 0ù é 0 é10 ù x& = ê 0 0 1ú x + ê 0 ú u, y = [1 0 0 ]x ê ú ê ú êë-1 -2 -3úû êë 0 úû

(A)

10(2 s 2 + 3s + 1) s 3 + 3s 2 + 2 s + 1

(B)

10(2 s 2 + 3s + 1) s 3 + 2 s 2 + 3s + 1

(C)

10(2 s 2 + 3s + 2) s 3 + 3s 2 + 2 s + 1

(D)

10(2 s 2 + 3s + 2) s 3 + 2 s 2 + 3s + 1

1 é0 ê0 0 (C) x& = ê ê0 0 ê20 10 ë

0 ù é0 ù ú ê0 ú 0 úx + ê úr 0 1 ú ê0 ú ú ê1 ú 7 100 û ë û 0 1

1 0 0 ù é 0 é0 ù ú ê0 ú ê 0 0 1 0 úx + ê úr (D) x& = ê 0 0 1 ú ê0 ú ê 0 ê-20 -10 -7 -100 ú ê ú ë û ë1 û y = [100 0 0 0 ]x 19. A state-space representation of a system is given by

Statement for Q.17–18: Determine the state-space representation for the transfer function given in question. Choose the state variable as follows 2

3

d c d c dc c = c& , x3 = 2 = &&c , x4 = 2 = &&& dt dt dt

C( s) 24 = 3 R( s) s + 9 s 2 + 26 s + 24

1 0 ù é x1 ù é 0 ù é x& 1 ù é 0 ê ú ê (A) x& 2 = 0 0 1 ú ê x2 ú + ê 0 ú r ê ú ê úê ú ê ú êë x& 3 úû êë-24 -26 -9 ûú êë x3 úû êë24 úû 1 0 ù é x1 ù é 0 ù é x& 1 ù é 0 (B) ê x& 2 ú = ê 0 0 1 ú ê x2 ú + ê 0 ú r ê ú ê úê ú ê ú êë x& 3 úû êë24 26 9 úû êë x3 úû êë24 úû Page 378

0

y = [100 0 0 0 ]x

The transfer function Y ( s) U ( s) is

17.

0ù é 0 ù ú ê 0 ú 0 1 0 úx + ê ú r, 0 0 1ú ê 0 ú ê100 ú 7 10 20 úû ë û 1

y = [1 0 0 0 ]x

16. The state-space representation for a system is

x1 = c = y, x2 =

é 0 ê 0 (A) x& = ê ê 0 ê100 ë

1 0 0 ù é 0 é 0 ù ê 0 ú ê 0 ú 0 1 0 úx + ê úr (B) x& = ê 0 0 1 ú ê 0 ê 0 ú ê-100 -7 -10 -20 ú ê ú ë û ë100 û

The transfer function of this system is (B) ( s + 2) -1

C( s) 100 = 4 R( s) s + 20 s 3 + 10 s2 + 7 s + 100

y = [1 0 0 0 ]x

é-1 0 ù é1ù x& ( t) = ê x( t) + ê ú u( t), y( t) = [1 1]x( t) ú ë 0 -2 û ë1û

(A) ( s 2 + 3s + 2) -1

é x& 1 ù é0 1 0 ù é x1 ù é 0 ù (C) ê x& 2 ú = ê0 0 1 ú ê x2 ú + ê 0 ú r ê ú ê úê ú ê ú êë x& 3 úû êë9 26 24 úû êë x3 úû êë24 úû 1 0 ù é x1 ù é 0 ù é x& 1 ù é 0 (D) ê x& 2 ú = ê 0 0 1 ú ê x2 ú + ê 0 ú r ê ú ê úê ú ê ú êë x& 3 úû êë-9 -26 -24 úû êë x3 úû êë24 úû

The output transfer function Y ( s) U ( s) is (A)

Control & System

é 0 1ù é0 ù x& = ê ú x, y = [1 - 1]x, and x(0) = ê1 ú 2 0 ë û ë û The time response of this system will be 3 (A) sin 2t (B) sin 2 t 2 (C) -

1 2

(D)

sin 2 t

3 sin 2 t

20. For the transfer function Y ( s) s+3 = U ( s) ( s + 1)( s + 2) The state model is given by x& = A ×x + B× u, y = C × x. The A , B, C are www.gatehelp.com


UNIT 6

-4 -2 ù é-5 é1 ù ê ú 29. x& = -3 -10 0 x + ê1 ú r, y = [ -1 2 1]x ê ú ê ú 1 -5 úû êë -1 êë0 ûú estep ( ¥)

Statement for Q.34–36: Consider the system shown in fig. P6.7.34-36. 1 s

eramp ( ¥)

(A) 1.0976

1 u

0

(B) 1.0976

¥

(C) 0

1.0976

(D) ¥

1.0976

0.714

(B) ¥

0.714

(C) 0

4.86

(D) ¥

4.86

x2

1 s

1

-1 1 s

y

-10

x3

34. The controllability matrix for this system is 10 ù 1 -2 ù é 10 -10 é0 (A) ê-10 (B) ê1 -1 0 -20 ú 1ú ê ú ê ú 40 úû 4 úû êë 10 -20 êë1 -2 10 ù é 10 -10 (C) ê-10 0 20 ú ê ú êë 10 -20 -40 úû

1

-2

10

Fig. P6.7.34-36

Consider the system shown in fig. P6.7.31-33

u

x1

-2


1 s

1 s

1

10

eramp ( ¥)

(A) 0

x2

-1

1

1 0ù é 0 é0 ù ê ú 30. x& = -5 -9 7 x + ê0 ú r, y = [1 0 0 ]x ê ú ê ú êë -1 0 0 úû êë1 ûú estep ( ¥)

Control & System

x1

1

y

1 -1ù é0 (D) ê1 6 -1ú ê ú êë1 -4 -4 úû

35. The observability matrix is 10 ù 1 -2 ù é 10 -10 é0 (A) ê-10 (B) ê1 -1 0 -20 ú 1ú ê ú ê ú 40 úû 4 úû êë 10 -10 êë1 -2 10 ù é 10 -10 (C) ê-10 0 20 ú ê ú êë 10 -10 -40 úû

1 2ù é0 (D) ê1 -1 1ú ê ú êë1 -2 4 úû

-5

36. The system is

-6

Fig. P6.7.31-33

31. The controllability matrix is é1 0 ù é 1 -2 ù (A) ê (B) ê ú 4 úû ë0 1û ë-2 é1 0 ù (C) ê ú ë0 -1û

é 1 2ù (D) ê ú ë-2 -4 û

(A) Controllable and observable (B) Controllable only (C) Observable only (D) None of the above Statement for Q.37–38: A state flow graph is shown in fig. P6.7.37-38

32. The observability matrix is é1 0 ù é 1 -2 ù (A) ê (B) ê ú 0 1 4 úû ë û ë-2 é1 0 ù (C) ê ú ë0 -1û

(A) Controllable and observable

(C) Observable only

u

1 s x2

1 s

x1

5

y

-5

Fig. P6.7.37-38

37. The state and output equation for this system is éx ù é x& ù é0 -1ù é x ù é0 ù (A) ê 1 ú = ê 21 ú ê 1 ú + ê ú u, y = [5 4 ]ê 1 ú 5 & x x 1 ë 2 û êë ë x2 û 4 úû ë 2 û ë û

(D) None of the above Page 380

1

-21 4

é 1 2ù (D) ê ú ë-2 -4 û

33. The system is (B) Controllable only

4

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Chap 6.7

év& ù é-1 -1ù év1 ù é1 ù év ù (B) ê &1 ú = ê + ê ú vi , iR = [ 4 1]ê 1 ú ê ú ú ë i3 û ë-3 -1û ë i3 û ë0 û ë i3 û

1ù é x ù é0 ù é x& ù é 0 éx ù 21 ú ê 1 ú + ê ú u, y = [5 4 ]ê 1 ú (B) ê 1 ú = ê 5 & x 1 ë x2 û êë ë x2 û 4 úû ë 2 û ë û

é v& ù é1 -3ù é v1 ù é 1 ù év ù (C) ê 1 ú = ê + ê ú vi , iR = [1 4 ]ê 1 ú ê ú ú & ëv2 û ë1 6 û ëv2 û ë-1û ëv2 û

éx ù é x& ù é0 -1ù é x ù é1ù (C) ê 1 ú = ê 21 ú ê 1 ú + ê ú u, y = [ 4 5 ]ê 1 ú 5 x 1 ë x2 û ë x& 2 û êë 4 úû ë 2 û ë û

é v& ù é 1 3ù é v1 ù é 1 ù é v1 ù (D) ê 1 ú = ê ú êv ú + ê-1ú vi , iR = [1 4 ]êv ú & v 1 6 û ë 2û ë û ë 2û ë ë 2û

1ù é x ù é1ù é x& ù é 0 éx ù 21 ú ê 1 ú + ê ú u, y = [ 4 5 ]ê 1 ú (D) ê 1 ú = ê 5 & x x 1 ë 2 û êë ë x2 û 4 úû ë 2 û ë û


38. The system is

Consider the network shown in fig. P6.7.41-43. This system may be represented in state space representation x& = A × x + B × u

(A) Observable and controllable (B) Controllable only (C) Observable only

iC

(D) None of the above iR1

39. Consider the network shown in fig. P6.7.39. The state-space representation for this network is iL

iL

iR2

1 2F

1W

is

1W

4H

vs

2W

Fig. P6.7.41-43

iC

iR

41. The state variable may be

1F

Fig. P6.7.39

év& ù é-0.25 1ù évC ù é 1 ù évC ù (A) ê & C ú = ê ú ê i ú + ê0.25 ú vs , iR = [0.5 0 ]ê i ú i 0 5 0 . û ûë Lû ë ë Lû ë ë Lû

(A) iR1 , iR 2

(B) iL , iC

(C) vC , iL


42. If state variable are chosen as in previous question, then the matrix A is é 1 -1ù (A) ê ú ë-1 3û

é 1 -3ù (B) ê ú ë-1 1û

é-1 3ù (C) ê ú ë 1 -1û

é 3 -1ù (D) ê ú ë-1 1û

év& ù é-1 0.25 ù évC ù é0.25 ù év ù (D) ê & C ú = ê vs , iR = [0.5 0 ]ê C ú +ê ê ú ú ú ë iL û ë 0 -0.5 û ë iL û ë 0 û ë iL û

43. The matrix B is é 3ù (A) ê ú ë-1û

é-1ù (B) ê ú ë 3û

40. For the network shown in fig. P6.7.40. The output is

é-3ù (C) ê ú ë 1û

é 1ù (D) ê ú ë-3û

év& ù é -0.5 1ù évC ù é0.25 ù évC ù (B) ê & C ú = ê ú ê i ú + ê 1 ú vs , iR = [0.5 0 ]ê i ú i 0 25 0 . û ûë Lû ë ë Lû ë ë Lû év& ù é1 0.25 ù évC ù é0.25 ù év ù (C) ê & C ú = ê +ê vs , iR = [0.5 0 ]ê C ú ê ú ú ú ë iL û ë0 0.5 û ë iL û ë 0 û ë iL û

i1

1W

1H

v1

i3

v2

iR

i2 vi

1F

4vL

Statement for Q.44–47: 4v1

1W

Consider the network shown in fig. P6.7.44-47 i1

Fig. P6.7.40

iR ( t). The state space representation is

vi

év& ù é 1 -1ù év1 ù é1 ù év ù (A) ê &1 ú = ê + ê ú vi , iR = [ 4 1]ê 1 ú ê ú ú ë i3 û ë-3 1û ë i1 û ë0 û ë i3 û

1W

i3

1W

i5

i2

i4

1H

1H

1W

1F

+ vo -

Fig. P6.7.44-47

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Page 381


UNIT 6

Control & System

44. The state variable may be (A) i2 , i4

(B) i2 , i4 , vo

(C) i1 , i3

(D) i1 , i3 , i5

45. In state space representation matrix A is 1 1ù é 2 ê- 3 - 3 3ú ê 1 2 2ú (A) êú 3 3ú ê 3 ê- 1 - 2 - 1 ú êë 3 3 3 úû

1 2ù é 1 ê 3 - 3 - 3ú ê 2 2 1ú (B) ê - ú 3 3ú ê 3 ê- 1 - 2 - 1 ú êë 3 3 3 úû

1 1ù é 2 ê 3 - 3 - 3ú ê 1 2 2ú (C) ê ú 3 3ú ê 3 1ú ê- 1 - 2 êë 3 3 3 úû

1 2ù é 1 ê- 3 - 3 3ú ê 2 2 1ú (D) êú 3 3ú ê 3 ê- 1 - 2 - 1 ú êë 3 3 3 úû

46. The matrix B is é 2ù ê 3ú ê 1ú (A) ê- ú ê 3ú ê- 1 ú êë 3 úû

é2 ù ê3ú ê1 ú (B) ê ú ê3ú ê1 ú êë 3 úû

é 1ù ê- 3 ú ê 1ú (C) ê- ú ê 3ú ê 1ú êë 3 úû

é2 ù ê3ú ê1 ú (D) ê ú ê3ú ê2 ú êë 3 úû

47. If output is vo , then matrix C is (A) [-1 0 0]

(B) [1 0 0]

(C) [0 0 -1]

(D) [0 0 1]

SOLUTIONS 1. (B) From the SFG x& 3 = -3 x1 - 2 x2 x x2 = 3 Þ s x x1 = 2 Þ s

- x3 + u x& 2 = x3 x& 1 = x2

2. (A) x& 1 = - 2 x1 + x3 , x& 2 = - 2 x2 + u , x& 3 = -3 x3 + u 1 0 ù é x1 ù é0 ù é x& 1 ù é-2 ê x& ú = ê 0 -2 0 ú ê x ú + ê0 ú u ê 2ú ê ú ê 2ú ê ú & x 0 0 3 êë 3 úû êë ûú êë x3 úû êë1 ûú 3. (C) x& 1 = - 2 x1 + x3 , x& 2 = - 2 x2 + u , x& 3 = -3 x3 + u y = 5 x1 + 5 x2 + 5 x3 0 1ù é x1 ù é0 ù é x& 1 ù é-2 ê x& ú = ê 0 -2 0 ú ê x ú + ê1 ú u ê 2ú ê ú ê 2ú ê ú 0 -3úû êë x3 úû êë1 úû êë x& 3 úû êë 0 4. (C) x& 1 = -4 x1 + x2 , x& 2 = x3 + 2 u , x& 3 = - 3 x3 + u é x& 1 ù é-4 1 1ù é x1 ù é0 ù ê x& ú = ê 0 0 1ú ê x2 ú + ê2 ú u ê 2ú ê úê ú ê ú êë x& 3 úû êë 0 0 -3úû êë x3 úû êë1 úû 5. (B) Ds = |sI - A| = s 2 + s + 2 = 0 Þ s = - 0.5 ± j1.323 0 és + 3 6. (A) ( sI - A) = ê s+ ë 0 ( sI - A)

-1

é 1 ê = ês + 3 ê 0 ë

ù 0 ú 1 ú ú s + 3û

ée -3t F( t ) = L-1 {( sI - A )} == ê ë0 ************************

ù 1 , |sI - A|= 3úû ( s + 3) 2

0 ù ú e -3t û

é s -2 ù 2 7. (A) ( sI - A) = ê ú, |sI - A|= s + 4 ë2 s û ( sI - A)

-1

é s 1 é s 2 ù ê s2 + 4 = 2 = s + 4 êë-2 s úû ê -2 ê ë s2 + 4

2 ù s2 + 4 ú s ú ú 2 s + 4û

é cos 2 t sin 2 t ù F( t ) = L-1 {( sI - A )} = ê ú ë- sin 2 t cos 2 t û é s -2 ù 2 8. (A) ( sI - A) = ê ú, Ds =|sI - A|= s + 4 ë2 s û

Page 382

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( sI - A)

-1

é s 1 é s 2 ù ê s2 + 4 = 2 = s + 4 êë-2 s úû ê -2 ê ë s2 + 4

é1 ù é1ù 15. (D) T( s) = ê ú( sI - A) -1 ê ú ë0 û ë1û

2 ù s + 4ú s ú ú 2 s + 4û 2

( sI - A)

é cos 2 t sin 2 t ù F( t ) = L-1 {( sI - A )} = ê ú ë- sin 2 t cos 2 t û

ì 1 é s 2 ù é0 ù é1ù 1 ü é2 ù ü 1 -1 ì = L-1 í 2 ý =L í 2 ê ú ê ú ê ú ê úý î s + 4 ë-2 s û ë1 û ë1û s þ î s( s + 4) ë s û þ 2 é ù ê s( s 2 + 4) ú ê ú 1 ê 2 ú ë ( s + 4) û

10. (C) (A) is not a state-transition matrix, since F(0) ¹ I (C) is a state-transition matrix since F(0) = I and [ F( t)]-1 = F( -t) 0 ù é s -1 ê 11. (C) ( sI - A) = 0 s -1 ú ê ú êë1 2 s + 3úû

|sI - A| = s3 + 3s2 + 2 s + 1, Þ

s = - 2.325, - 0.338 ± j0.562 X ( s) = ( sI - A) -1 B U ( s)

12. (B)

és 3 + 3s + 2 s+3 1 ù é0 ù 1 ê ú s( s + 3) s ê0 ú = 3 -1 úê ú s + 3s 2 + 2 s + 1 ê 2 -s -2 s - 1 s úû êë1 úû êë é1 ù 1 êsú = 3 s + 3s 2 + 2 s + 1 ê 2 ú êës úû 13. (C)

Y ( s) CX ( s) = U ( s) U ( s)

1 ù é ê s 3 + 3s 2 + 2 s + 1 ú ú ê s 1 = [1 0 0 ]ê 3 ú= 3 2 2 ê s + 3s 2+ 2 s + 1 ú s + 3s + 2 s + 1 s ú ê êë s 3 + 3s 2 + 2 s + 1 úû 14. (D)

Y ( s) = C ( sI - A) -1 B U ( s)

B = [1 1]

1 ù é0 ù s+2 1 és + 1 = ê s + 1úû êë1 úû ( s + 1) 2 Ds ë 0

é 1 ê = ês + 1 ê 0 ë

ù 0 ú 1 ú ú s + 2û ù 0 ú é1 ù 1 = 1 ú êë0 úû s + 1 ú s + 2û

16. (C) x& = A ×x + B× u, y = C × x + Du

ü ïï é0.5(1 - cos 2 t) ù ý= ê ú ï ë 0.5 sin 2 t û ïþ

(B) is not a state-transition matrix since F(0) ¹ I

-1

é 1 é1ù ê s + 1 T( s) = ê ú ê ë1û ê 0 ë

9. (C) q( t) = L-1 {( sI - A) -1 BR( s)}

ì ïï = L-1 í ï ïî

Chap 6.7

1 0ù é 0 é10 ù A =ê 0 0 1ú, B = ê 0 ú, C = [1 0 0 ], D = 0 ê ú ê ú êë-1 -2 -3úû êë 0 úû Y ( s) T( s) = = C ( sI - A ) -1 B + D U ( s) ( sI - A)

-1

és 3 + 3s + 2 s+3 1ù 1 ê ú = 3 s ( s + ) s 1 3 ú s + 3s 2 + 2 s + 1 ê -s -2 s - 1 s 2 úû êë

Substituting the all values, T( s) =

10(2 s 2 + 3s + 2) s 3 + 3s 2 + 2 s + 1

17. (A)

C( s) b0 24 = = R( s) s 3 + a2 s 2 + a1 s + a0 ( s 3 + 9 s 2 + 26 s + 24)

( s 3 + a2 s 2 + a1 s + a0 ) C( s) = b0 R( s) Taking the inverse Laplace transform assuming zero initial conditions &&& c + a2 &&c + a1 c& + a0 c = b0 r x1 = c = y, x2 = c& , x3 = &&c x& 1 = c& = x2 , x& 2 = &&c = x3 x& 3 = &&& c = b0 r - a2 &&c - a1 c& - a0 c = - a0 x1 - a1 x2 - a2 x3 + b0 r , 1 0 ù é x1 ù é 0 ù é x& 1 ù é 0 ê x& ú = ê 0 0 1 ú ê x2 ú + ê 0 ú r ê 2ú ê úê ú ê ú êë x& 3 úû êë-a0 -a1 -a1 úû êë x3 úû êëb0 úû a0 = 24 , a1 = 26, a2 = 9, b0 = 24 1 0 ù é x1 ù é 0 ù é x& 1 ù é 0 ê x& ú = ê 0 0 1 ú ê x2 ú + ê 0 ú r ê 2ú ê úê ú ê ú êë x& 3 úû êë-24 -26 -9 úû êë x3 úû êë24 úû é x1 ù y = [1 0 0 ] ê x2 ú ê ú êë x3 úû 18. (B) Fourth order hence four state variable

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Page 383


UNIT 6

é 0 ê 0 x& = ê ê 0 ê -a ë 0

1 0

0 1

0

0

-a1

-a 2

0 ù é0 ù ú ê0 ú 0 ú x = ê ú r, y = [1 0 0 0 ]x 1 ú ê0 ú ú ê ú -a 3 û ëb0 û

a0 = 100, a1 = 7, a2 = 10,

a3 = 20 ,

=

b0 = 100

Y ( s) = [0 1] X ( s) =

é sin 2 t ù cos 2 t ú F( t ) = L-1 {( sI - A)} = ê 2 ú ê êë- 2 sin 2 t cos 2 t úû

Þ

20. (C) Find the transfer function of option Y ( s) 1 For (A) , , = U ( s) s - 2

y( t) =

0 ù é1ù és + 2 Y ( s) 1 = [0 1] s + 1úû êë1úû U ( s) ( s + 1)( s + 2) êë 2

|sI - A| = s

- 8 s - 11s + 8

Þ

-1

æ é2 ù é1ù 3 ö çç ê ú + ê ú 2 ÷÷ è ë1 û ë1û s + 9 ø

é2 s + 4 s + 21s + 45 ù 1 2 ( s + 5)( s + 9) êë s 3 - 7 s 2 + 12 s - 7 úû 3

1 1 -2 t - e 2 2

A

-1

0.05 -0.05 ù é-0.4 ê = -1 -0.25 -0.25 ú ê ú -0.5 ûú 15 . êë -2

1ù é -2 - ú 27. (A) estep ( ¥) = 1 + CA -1B, A -1 = ê 3 êë 1 0 úû

s = 9.11, 0.53, - 1.64

23. (D) X ( s) = ( sI - A) -1 (x(0) + B × u) és - 1 -2 ù =ê s + 1úû ë 3

1 s( s + 2)

= 1 - 0.2 = 0.8

s = -5.79, - 121 .

-2 -3 ù és 22. (B) ( sI - A) = ê 0 s - 6 -5 ú ê ú êë-1 -4 s - 2 úû 2

é0 ù é1 ù ö ê0 ú + ê0 ú 1 ÷ ê ú ê ú s÷ êë0 úû êë0 úû ÷ø

1 é ù ê ú s( s + 2) ê ú 1 ú =ê 2 ê s ( s + 2) ú 1 ê ú êë s 2 ( s + 1)( s + 2) úû

é-0.4 0.05 -0.05 ù é0 ù estep ( ¥) = 1 + [ -1 1 0 ]ê -1 0.25 -0.25 ú ê0 ú ê úê ú 15 . -0.5 úû êë1 úû êë -2

é-2 -1ù 21. (C) A = ê ú, ë -3 -5 û

3

æ ç ç ç è

X ( s) =

1 0ù é- 5 ê A= 0 -2 1ú, ê ú êë20 -10 1úû

So (C) is correct option.

Þ

-1

26. (D) For a unit step input estep ( ¥) = 1 + CA -1B

és + 2 ù s+3 1 = [0 1] = ( s + 1)( s + 2) êës + 3úû ( s + 1)( s + 3)

|sI - A| = s2 + 7 s + 7

1 1 - e - t + e -2 t 2 2

Y ( s) = [1 0 0 ],

Y ( s) 1 For (B) , = U ( s) s - 2 For (C),

y( t) =

1 s( s + 1)( s + 2)

0 ù és + 2 -1 ê = 0 -1 ú s ê ú 0 s + 1úû êë 0

sin 2 t

2

é1 ù é1ù 1 ö ê0 ú + ê1ú s ÷÷ ë û ë û ø

25. (B) X ( s) = ( sI - A) -1 (x(0) + B × u)

é sin 2 t ù cos 2 t + ê ú x( t) = F( t )x(0) = 2 ê ú êë- 2 sin 2 t + cos 2 t úû 3

0 ùæ és + 1 1 ç ( s + 1)( s + 2) êë -1 s + 2 úû çè

( s + 1) é ù ê s( s + 2) ú =ê ú 1 ê ú ë s( s + 1)( s + 2) û

é 0 1ù 1 é s 1ù 19. (B) A = ê , ( sI - A) -1 = 2 ú s + 2 êë-2 súû ë-2 0 û

y = x1 - x2 =

Control & System

2

1ù é 1 2 -2 - ú é0 ù estep ( ¥) = 1 + [1 1] ê 3 ê ú =1 - = 3 3 êë 1 0 úû ë1 û 28. (C) eramp ( ¥) = lim [(1 + CA -1B) t + C( A -1 ) 2 B] t ®¥

2 1 + CA -1B = , 3

é2 ù eramp ( ¥) = lim ê t + C( A -1 ) 2 Bú = ¥ t ®¥ ë 3 û

2

4 s 3 - 10 s 2 + 45 s - 105 Y ( s) = [1 2 ] X ( s) = ( s 2 + 5)( s 2 + 9) 24. (B) X ( s) = ( sI - A) -1 (x(0) + B × u) Page 384

29. (B) estep ( ¥) = 1 + CA -1B -4 -2 ù é-5 é1 ù A = ê -3 -10 0 ú, B = ê1 ú, C = [ -1 2 1] ê ú ê ú 1 -5 úû êë -1 êë0 úû

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A

-1

é-1 1 0 ù A = ê 0 -1 0 ú, C = [10 - 10 10 ], ê ú êë 0 0 -2 ûú

0.134 0.122 ù é-0.305 ê = 0.091 -0.140 -0.037 ú ê ú êë -0.079 -0.055 -0.232 úû

é-1 1 0 ù CA = [10 - 10 10 ] ê 0 -1 0 ú = [10 ê ú êë 0 0 -2 úû

. estep ( ¥) = 1 + 0.0976 = 10976 -1

-1 2

eramp ( ¥) = lim [(1 + CA B) t + C( A ) B] = ¥ t ®¥

-1 ù 0 é 0 30. (B) A -1 = ê 1 0 0 ú ê ú êë1.286 0.143 -0.714 úû

CA 2 = [10

0 -1 ù é0 ù é 0 ê 0 0 ú ê0 ú = 0 estep ( ¥) = 1 + [1 0 0 ] 1 ê úê ú êë1.286 0.143 -0.714 úû êë1 úû . -0.143 0.714 ù é -1286 ê 0 0 -1 ú (A ) = ú ê ëê-0.776 -0.102 -0.776 úû

nonsingular and system is controllable

1ù é 0 é 1ù A =ê , B=ê ú ú ë-6 -5 û ë-2 û

10 ù é 10 -10 det O M = detê-10 0 -10 ú = -3000 ê ú 40 úû êë 10 -20

1ù é 1ù ù é 1 -2 ù é 0 ê-6 -5 ú ê-2 ú ú = ê-2 4 úû ë ûë ûû ë

The rank of O M is 3. Hence system is observable.

32. (A) y = x1 , y = [1 0 ]x , 0 ], CA = [1

37. (B) x& 2 = - 5 x1 -

é C ù é1 0 ù 0 ] , OM = ê ú =ê ú ëCA û ë0 1 û

21 x2 + u , x& 1 = x2 , y = 5 x1 + 4 x2 4

1ù é x ù é0 ù é x& 1 ù é 0 é x1 ù 1 ê x& ú = ê-5 - 21 ú ê x ú + ê1 ú u, y = [5 4 ]ê x ú ë 2 û êë ë 2û 4 úû ë 2 û ë û

33. (C) det CM = 0. Hence system is not controllable. det O M = 1. Hence system is observable.

é C ù é 5 4ù 38. (B) O M = ê ú =ê ú ëCA û ë-20 1 û

34. (B) x& 1 = - x1 + x2 , x& 2 = - x2 + u , x& 3 = - 2 x3 + u

det O M = 0. Thus system is not observable

é-1 1 0 ù é0 ù é-1 1 0 ù é0 ù &x = ê 0 -1 0 ú x + ê1 ú u, A = ê 0 -1 0 ú , B = ê1 ú ê ú ê ú ê ú ê ú êë 0 0 -2 úû êë1 úû êë 0 0 -2 úû êë1 úû

1ù é0 21 ú CM = [B AB] = ê 1 êë 4 úû

é-1 1 0 ù é0 ù é 1ù AB = ê 0 -1 0 ú ê1 ú = ê -1ú ê úê ú ê ú êë 0 0 -2 úû êë1 úû êë-2 úû

det CM = -1. Thus system is controllable. 39. (B)

é-1 1 0 ù é 1ù é-2 ù A 2B = ê 0 -1 0 ú ê -1ú = ê 1ú ê úê ú ê ú êë 0 0 -2 úû êë-2 úû êë 4 úû 1 -2 ù é0 ê 1ú Cm = [B AB A B] = 1 -1 ê ú 4 úû êë1 -2 2

35. (A) y = 10 x1 - 10 x2 + 10 x3 , y = [10 - 10

40 ]

Since the determinant is not zero, the 3 ´ 3 matrix is

31. (B) x& 1 = x2 + u , x& 2 = - 5 x2 - 6 x1 - 2 u

C = [1

é-1 1 0 ù 0 - 20 ] ê 0 -1 0 ú = [10 - 10 ê ú êë 0 0 -2 úû

1 -2 ù é0 36. (A) det Cm = det ê1 -1 1ú = -1, ê ú 4 úû êë1 -2

C( A -1 ) 2 B = 0.714, eramp ( ¥) = 0.714

éé 1ù CM = [B AB] = êê ú ëë-2 û

0 - 20 ]

10 ù é C ù é 10 -10 ê ú ê O M = CA = -10 0 -20 ú ê ú ê ú 2 40 ûú êëCA úû êë 10 -10

-1 2

1ù é 0 é 1ù x& = ê x + ê ú u, ú ë-6 -5 û ë-2 û

Chap 6.7

10 ]x

dvc di v = ic , L = L = 0.25 vL dt dt 4

vC and iL are state variable. v iL = iC + iR , iC = iL - iR = iL - C , vL = vs - vC 2 dvL v Hence equations are = iL - C = - 0.5 vC + iL dt 2 diL = 0.25( vs - vC ) = - 0.25 vC + 0.25 vs dt év& C ù é -0.5 1ù évC ù é0.25 ù ê i& ú = ê-0.25 0 ú ê i ú + ê 1 ú vs , û ûë Lû ë ë Lû ë

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UNIT 6

iR =

év ù iR = [0.5 0 ] ê C ú ë iL û

vC = 0.5 vC , 2

45. (A)

Control & System

di2 di dvo = v2 , 4 = v4 , = i5 dt dt dt i1

40. (B)

dv1 di = i2 , 3 = vL dt dt

1W

vi

Hence v1 and i3 are state variable.

1W

v2 i3

v4 i5

i2

i4

1H

1H

1W

1F

+ vo -

i2 = i1 - i3 = ( vi - v1 ) - i3 , i2 = - v1 - i3 + vi vL = v1 - v2 = v1 - iR , = v1 - ( i3 + 4 v1 ) = -3v1 - i3 dv1 di = - v1 - i3 + vi , 3 = - 3v1 - i3, y = iR = 4 v1 + i3 dt dt év&1 ù é-1 -1ù év1 ù é1 ù év1 ù ê i& ú = ê-3 -1ú ê i ú + ê0 ú vi , iR = [ 4 1]ê i ú û ë 3û ë û ë 3û ë ë 3û 41. (C) Energy storage elements are capacitor and inductor. vC and iL are available in differential form and linearly independent. Hence vC and iL are suitable for state-variable. 1 dvC dvC = iC Þ = 2 iC 2 dt dt 1 diL diL = vL Þ = 2 vL 2 dt dt

42. (B)

iC iR1 is

1W

+

iL

+ vC 1 2F

vL

iR2 + vR2

-

1W

4vL

-

Fig. S6.7.42

vL = vC + vR 2 = vC + iR2 , iC + 4 vL = iR 2 vL = vC + iC + 4 vL , -3vL = vC + iC v iC = is - iR1 - iL , iC = is - L - iL 1

...(i) ...(ii)

Solving equation (i) and (ii) -3 ( is - iL - iC ) = vC + iC , 2 iC = vC - 3iL + 3is

Fig. S6.7.45

Now obtain v2 , v4 and i5 in terms of the state variable -vi + i1 + i3 + i5 + vo = 0 But i3 = i1 - i2 and i5 = i3 - i4 -vi + i1 + ( i1 - i2 ) + ( i3 - i4 ) + vo = 0 2 1 1 1 i1 = i2 + i4 - vo + vi 3 3 3 3 2 1 1 2 v2 = vi - i1 = - i2 - i4 + vo + vi 3 3 3 3 1 1 1 1 i3 = i1 - i2 = - i2 + i4 - vo + vi 3 3 3 3 1 2 1 1 i5 = i3 - i4 = - i2 - i4 - vo + vi 3 3 3 3 1 2 2 1 v4 = i5 + vo = - i2 - i4 + vo + vi 3 3 3 3 1 1ù é2 ù é 2 é i&2 ù ê 3 3 3 ú é i2 ù ê 3 ú ê& ú ê 1 2 2 ú ê ú ê1 ú ú i4 + ê ú vi ê i4 ú = ê- 3 - 3 3úê ú ê3ú êv& o ú ê 1 2 1 êëvo úû ê 1 ú ë û ê- ú êë 3 êë 3 úû 3 3 úû 1 1ù é 2 ê- 3 - 3 3ú ê 1 2 2ú A = êú, 3 3ú ê 3 ê- 1 - 2 - 1 ú êë 3 3 3 úû

47. (D) vo is state variable

-3vL = vC + is - vL - iL , 2vL = - vC + iL - is dvC di = vC - 3iL + 3is , L = - vC + iL - is dt dt év& C ù é 1 -3ù évC ù é 3 ù ê i& ú = ê-1 1ú ê i ú + ê-1ú is ûë Lû ë û ë Lû ë

y = vo ,

é i2 ù y = [0 0 1] = ê i4 ú ê ú êëvo úû

********

é 3ù 43. (A) B = ê ú ë-1û 44. (B) There are three energy storage elements, hence 3 variable. i2 , i4 and vo are available in differentiated form hence these are state variable.

Page 386

é2 ù ê3ú ê1 ú 46. (B) B = ê ú ê3ú ê1 ú êë 3 úû

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UNIT 7

10. If machine is not properly adjusted, the product

Statement for Question 15-16:

resistance change to the case where ax = 1050 W. Now the rejected fraction is

Communication System

The life time of a system expressed in weeks is a Rayleigh random variable X for which

(A) 5046% (C) 2.18%

(B) 10.57% (D) 6.43%

ì x -x e 400 ïï 200 f X ( x) = í ï ïî 0 2

11. Cannon shell impact position, as measured along the line of fire from the target point is described by a

0£x x <0

gaussian random variable X. It is found that 15.15% of

15. The probability that the system will not last a full

shell falls 11.2 m or farther from the target in a

week is

direction toward the cannon, while 5.05% fall farther

(A) 0.01%

(B) 0.25%

from the 95.6 m beyond the target. The value of ax and

(C) 0.40%

(D) 0.60%

sx

for X is

(Given that F(103 . ) = 0.8485 and 16. The probability that the system lifetime will exceed

F(1.64) = 0.9495) (A) T + 40 m and 50 m

(B) T + 40 m and 30 m

in year is

(C) T + 10 m and 50 m

(D) T + 30 m and 40 m

(A) 0.01%

(B) 0.05%

(C) 0.12%

(D) 0.22%

12. A gaussian random voltage X for which a X = 0 and s X = 4.2 V appears across a 100 W resistor with a power rating of 0.25 W. The probability, that the voltage will

17. The cauchy random variable has the following probability density function f X ( x) =

cause an instantaneous power that exceeds the resistor's rating, is æ 5 ö (A) 2Qç ÷ è 4.2 ø

æ 5 ö (B) Qç ÷ è 4.2 ø

æ 5 ö (C) 1 + Qç ÷ è 4.2 ø

æ 5 ö (D) 1 - Qç ÷ è 4.2 ø

For real numbers 0 < b and -¥ < a < ¥. The distribution function of X is 1 æ x -aö (A) tan -1 ç ÷ p è b ø

Statement for Question 13 -14 : Assume that the time of arrival of bird at Bharatpur

sanctuary

on

a

migratory

route,

first day), is approximated as a gaussian random variable X with a X = 200 and sx = 20 days. Given that : F(10 . ) = 0.8413.,

(B)

1 æ x -aö cot-1 ç ÷ p è b ø

(C)

1 1 æ x -aö + tan -1 ç ÷ 2 p è b ø

(D)

1 1 æ x -aö + cot-1 ç ÷ 2 p è b ø

as

measured in days from the first year (January 1 is the

F(0.5) = 0.6915,

b/ p b2 + ( x - a) 2

F(15 . ) = 0.8531,

Statement for Question 18 - 19 The number of cars arriving at ICICI bank

F(155 . ) = 0.9394 and F(2.0) = 0.9773.

drive-in window during 10-min period is Poisson 13. What is the probability that birds arrive after 160 days but on or before the 210

th

(A) 0.6687

(B) 0.8413

(C) 0.8531

(D) 0.9773

day ?

18. The probability that more than 3 cars will arrive during any 10 min period is

14. What is the probability that bird will arrive after 231st day ? (A) 0.0432

(B) 0.1123

(C) 0.0606

(D) 0.0732

Page 390

random variable X with b = 2.

(A) 0.249

(B) 0.143

(C) 0.346

(D) 0.543

19. The probability that no car will arrive is (A) 0.516 (C) 0.246

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(B) 0.459 (D) 0.135


Random Variables

Chap 7.1

20. The power reflected from an aircraft of complicated

26. The mean of random variable X is

shape that is received by a radar can be described by an

(A) 1/4

(B) 1/6

exponential random variable W . The density of W is

(C) 1/3

(D) 1/5

ì 1 - w W0 e ï fW ( w) = í W0 ïî 0

ü w > 0ï ý w < 0 ïþ

27. The variance of random variable X is

where W0 is the average amount of received power.

(A) 1/10

(B) 3/80

(C) 5/16

(D) 3/16

The probability that the received power is larger than the power received on the average is (A) e -2 (C) 1 - e

28. A Random variable X is uniformly distributed on X

(B) e -1 -1

(D) 1 - e

the interval (-5, 15). Another random variable Y = e -2

-

5

is formed. The value of E[ Y ] is


(A) 2

(B) 0.667

(C) 1.387

(D) 2.967

Delhi averages three murder per week and their 29. A random variable X has X = -3, x 2 = 11 and s2X = 2

occurrences follow a poission distribution.

2

For a new random variable Y = 2 x - 3, the Y , Y and s2Y 21. The probability that there will be five or more

are

murder in a given week is

(A) 0, 81, 8

(B) - 6, 8, 89

(C) - 9, 89, 8


(A) 0.1847

(B) 0.2461

(C) 0.3927

(D) 0.4167

22. On the average, how many weeks a year can Delhi

Statement for Question 31-32 : A joint sample space for two random variable X

expect to have no murders ? (A) 1.4

(B) 1.9

(C) 2.6

(D) 3.4

and Y has four elements (1, 1), (2, 2), ( 3, 3) and (4, 4). Probabilities of these elements are 0.1, 0.35, 0.05 and 0.5 respectively.

23. How many weeds per year (average) can the Delhi expect the number of murders per week to equal or exceed the average number per week ? (A) 15

(B) 20

(C) 25

(D) 30

30. The probability of the event{ X £ 2.5, Y £ 6} is (A) 0.45

(B) 0.50

(C) 0.55

(D) 0.60

31. The probability of the event { X £ 3} is

24. A discrete random variable X has possible values

(A) 0.45

(B) 0.50

xi = i , i = 1, 2, 3, 4 which occur with probabilities 0.4,

(C) 0.55

(D) 0.60

2

0.25, 0.15, 0.1,. The mean value X = E[ X ] of X is (A) 6.85 (C) 3.96

(B) 4.35 (D) 1.42

Statement for Question 32-34 : Random variable X and Y have the joint distribution

25. The random variable X is defined by the density x

1 f X ( x) = u( x) e 2 2

The expected value of g( X ) = X 3 is (A) 48

(B) 192

(C) 36

(D) 72

ì 5 æ x + e- ( x + 1 ) y2 2 ö - e - y ÷u( y), 0 £ x £ 4 ï çç ÷ x+1 ï4è ø ï FX , Y ( x, y) = í 0 x < 0 or y < 0 ï 1 -5 y 2 5 - y 2 - e , 4 £ x and any y ³ 0 ï1 + e 4 4 ï î www.gatehelp.com

Page 391


UNIT 7

32. The marginal distribution function FX ( x) is

37. The marginal distribution function FX ( x) is

x, > 0 ì 0, ïï 5 x (A) í , -4< x£4 ï 4( x + 1) ïî 1, x £ -4

x <0 ì 0, ïï 5 x (B) í , 0 £ x<4 ï 4( x + 1) ïî 1, x³4

x >0 ì 1, ïï 5 x (C) í , - 4 < x £0 ï 4( x + 1) ïî 0, x £ -4

x <0 ì 1, ïï 5 x (D) í , 0 £ x<4 ï 4( x + 1) ïî 0, x³4

33. The marginal distribution function FY ( y) is ì 5 - y2 ï- e , (A) í 4 2 ï 1 + 4 e -5 y , î 4

(A)

5 1 [ 4( x) - 4( x - 4)] 4 ( x + 1) 2

(B)

5 1 [ 4( x) - 4( x - 4)] 2 ( x + 1) 2

(C)

5 1 [ 4( x) - 4( x - 4)] 4 ( x + 1)

(D) None of the above 38. The marginal distribution function FY ( y) is

y >0

2

(A) [1 - e - y ]u( y)

y £0

(C)

y >0 ì 0, ï (B) í 1 -5 y 2 5 - y 2 - e , y £0 ïî 1 + 4 e 4

5 4

(A) 0.001

(B) 0.002

(C) 0.003

(D) 0.004

ì 5 é x + e- ( x + 1 ) y2 2ù - e - y ú, ï ê ï 8 x+1 û (C) í ë 2 1 -5 y 2 ï - 5 e - y ], ïî 1 + 4 [ e

Statement for Question 35-39 : Two random variable X and Y have a joint density 2 10 FX , Y ( x, y) = [ u( x) - u( x - 4)]u( y) y 3e - ( x + 1 ) y 4 35. The marginal density f X ( x) is u( x) - u( x - 4) u( x) - u( x - 4) (A) 5 (B) 5 ( x + 1) 2 ( x + 1) 5 u( x) - u( x - 4) (D) 4 ( x + 1)

2

2

2

2

(B)

5 2

y 2 [ e - y - e -5 y ]u( y)

(C)

5 4

y[ e - y - e -5 y ]u( y)

(D)

5 2

y[ e - y - e -5 y ]u( y)

Page 392

2

2

2

2

0 £ x £ 4 and y > 0 x > 4 and y > 0

0 £ x £ 4 and y > 0 x > 4 and y > 0

0 £ x £ 4 and y > 0 x > 4 and y > 0

ì 5 é x + e- ( x + 1 ) y2 2ù - e - y ú, ï ê ï4 x+1 û (D) í ë 2 1 -5 y 2 ï - 5 e - y ], ïî 1 + 2 [ e

0 £ x £ 4 and y > 0 x > 4 and y > 0

40. The function FX , Y ( x, y) =

36. The marginal density fY ( y) is y 2 [ e - y - e -5 y ]u( y)

2

[1 - e - y ]u( y)


ì 5 é x + e- ( x + 1 ) y2 2ù - e - y ú, ï ê ï 8 x+1 û (B) í ë 2 1 -5 y 2 ï - 5 e - y ], ïî 1 + 2 [ e

34. The probability P { 3 < X £ 5, 1 < y £ 2} is

5 4

2

[1 - e - y ]u( y)

ì 5 é x + e- ( x + 1 ) y2 2ù - e - y ú, ï ê ï4 x+1 û (A) í ë 2 1 -5 y 2 ï - 5 e - y ], ïî 1 + 4 [ e

y <0 ì 0, ï (D) í 1 -5 y 2 5 - y 2 - e , y ³0 ïî 1 + 4 e 4

(A)

5 2

(B)

39. The joint distribution function is

ì -5 - y 2 e , y <0 ï (C) í 4 2 2 ï 1 + 1 e -5 y - 5 e - y , y ³ 0 î 4 4

5 u( x) - u( x - 4) (C) 4 ( x + 1) 2


a ép æ x öù é p æ y öù + tan -1 ç ÷ú ê + tan -1 ç ÷ú 2 êë2 2 2 è øû ë è 3 øû

is a valid joint distribution function for random variables X and Y if the constant a is 1 2 (A) 2 (B) 2 p p (C)

4 p2

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(D)

8 p2


Random Variables

41. Random variable

and Y

X

have the joint

(A)

a2 1 - e- a

(B)

(C)

1 1 - e- a


distribution function ì 0, ï ï 27 ï 26 ï ï 27 FX , Y ( x, y) = í ï 26 ï 27 ï ï 26 ï 1, î The

x < 0 or y < 0 æ y ö ÷, xçç 1 27 ÷ø è æ y2 ö yçç 1 ÷, 27 ÷ø è

Chap 7.1

2

a 1 - e- a

0 £ x < 1 and 1 £ y Statement for Question 46-47 : 1 £ x and 0 £ y < 1

æ x2 y2 xyçç 1 27 è

ö ÷÷, ø

Random variable X and Y have the joint density x y - 1 f X , Y ( x, y) = u( x) u( y) e 4 3 2

0 £ x < 1 and 0 £ y < 1

46. The probability of the event {2 < X £ 4, - 1 < Y £ 5}

1 £ x and 1 £ y

probability

of

the

is event

{0 < X £ 0.5, 0 < Y £ 0.25} is (A) 0.13

(B) 0.24

(C) 0.69

(D) 1

(A) 0.1936

(B) 6.2964

(C) 0


47. The probability of the event {0 < X < ¥, < y £ -2} is

Statement for question 42-43 :

(A) 0.2349

(B) 0.3168

(C) 0.4946


The joint probability density function of random 48. Let X and Y be two statistically independent

variable X and Y is given by pXY ( x, y) = xye

-

random variables uniformly distributed in the ranges (

( x 2 + y2 ) 2

-1, 1) and (-2, 1) respectively. Let Z = X + Y . Then the

u( x) u( y)

probability that ( Z £ -2) is 42. The pX ( x) is 2

(A) 2 xe - x u( x)

(B) xe

-

x2 2 x

2

(C) xe - x u( x)

(A) zero (C) 1/3

u( x)

49. The probability density function of two statistically

2

(D) 2 xe 2 u( x)

independent random variable X and Y are f X ( x) = 5 u( x) e -5x

43. The pY / X ( y/x) is (A)

1 2

2

2

ye - y u( y)

(C) ye

-

(B) ye - y u( y)

y2 2

(D)

u( y)

(B) 1/6 (D) 1/12

1 2

ye

-

fY ( y) = 24( y) e -2 y The density of the sum W = X + Y is

y2 2

u( y)

(A)

10 6

[ e -2 w - e 5w ]u( w)

10 8

[ e -2 w - e 5w ]u( w)

44. The probability density function of a random

((B)

variable X is given as f X ( x) . A random variable Y is

(C)

10 13

[ e -2 w - e -5w ]u( w)

defined as y = ax + b where a < 0. The PDF of random

(D)

10 2

[ e -2 w - e -5w ]u( w)

variable Y is æ y - bö (A) bf X ç ÷ è a ø (C)

æ y - bö (B) af X ç ÷ è a ø

1 æ y - bö f Xç ÷ a è a ø

(D)

50. The density function of two random variable X and Y is ì 1 0 < x < 6 and 0 < y < 4 ï f X , Y ( x, y) = í 24 ïî 0 else where

1 æ y - bö f Xç ÷ b è a ø

45. The function ì be f X , Y ( x, y) = í î0

The expected value of the function g( x, y) = ( XY ) - ( x + y)

0 < x < a and 0 < y < ¥ else where

is a valid joint density function if b is

is (A) 64

(B) 96

(C) 32

(D) 48

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Page 393


UNIT 7

51. The density function of two random variable X and

56. The value of s2X , s2Y , R XY and r are respectively

Y is

f X , Y ( x, y) =

e

æ x 2 + y2 ö ÷ -ç ç 2 s2 ÷ ø è

2 ps2

with s2 a constant. The mean value of the function g( X , Y ) = X 2 + Y 2 is (A) s2

(B) s

(C) 2s2

(D) 2s

have

Y = E[ Y ] = Y .

mean They

X = E[ X ] = 2

values have

33 11 27 1 æ 1 ö , .ç 2 + ÷, and -3 2 4 2 2 è 3ø

(B)

33 11 11 1 æ 1 ö , ç2 + ÷, and -3 2 4 2 2è 3ø

(C)

1 2 9 11 1 æ 1 ö , ç2 ÷, and 3 33 4 2 2è 3ø

(D)

1 2 9 11 1 æ 1 ö , , ç2 ÷, and 3 33 4 2 2è 3ø

W = ( X + 3Y ) 2 + 2 X + 3 is

The statistically independent random variable X Y

(A)

57. The mean value of the random variable

Statement for Question 52-54 :

and


second

and

moments

(A) 98 + 3

(B) 98 - 3

(C) 49 - 3

(D) 49 + 3

X 2 = E[ X 2 ] = 8 and Y 2 = E[ Y 2 ] = 25. Consider a random ***********

variable W = 3 X - Y . 52. The mean value E[W ] is (A) 2

(B) 4

(C) 8

(D) 25

53. The second moment of W is (A) 145

(B) 49

(C) 97

(D) 0

54. The variance of the random variable is (A) 4

(B) 45

(C) 49

(D) 54

55. Two random variable X and Y have the density function ì xy , ï f X , Y ( x, y) = í 9 ïî 0

0 < x < 2 and 0 < y < 3 elsewhere

The X and Y are (A) Correlated but statistically independent (B) uncorrelated but statistically independent (C) Correlated but statistically dependent (D) Uncorrelated but statistically dependent Page 394

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Random Variables

= FX (900) + [1 - FX (1100)]

SOLUTION

æ 900 - ax = F çç sx è

¥

1

¥

1

x 2

e

-x

dx =

¥

1 x e - x dx = 0.368 2 ò1 0

2. (C) P { -1 < X £ 2} = ò -1

=1 -

= 2 - 2(0.9938) = 0.012 or 1.2 % æ 900 - 1050 ö 10. (B) P(resistor rejected) = F ç ÷+1 40 è ø

2

1 x 1 xe dx + ò xe - x dx 2 2 0

æ 1100 - 1050 ö - Fç . ) + 1 - F (125 . ) ÷ = F ( -375 40 è ø = 1 - F ( 375 . ) + 1 - F (125 . )

3. (A) Test 1: f X ( x) ³ 0 is true

= 2 - 0.9999 - 0.8944 = 0.1057 or 20.57 %

l 3x 1 é l 3b 1 ù Test 2: area must be 1 i.e. ò dx = ê - ú =1 4 4ë 3 3û 0 b

4. (C)

11. (D) P { x > T + 95.6} = 0.0505 æ T + 95.6 - ax ( T + 95.6 - ax ) = F çç sx sx è T + 95.6 - ax This occurs when = 1.64 sx

1 ln 13 3

¥

ò r( v) = 1 Þ

-¥

ö ÷÷ ø

= F ( -2.5) + 1 - F (2.5) = 1 - F (2.5) + 1 - F (2.5) = 2 - 2 F(2.5)

1 3 = 0.429 e 2 e2

Thus b =

æ 1100 - ax ö ÷÷ + 1 - F çç sx è ø

æ 900 - 1000 ö æ 11000 - 1000 ö = Fç ÷ + 1 - Fç ÷ 40 40 è ø è ø

1. (A) P { X > 1} = ò pX ( x) dx =ò

Chap 7.1

=1 - F

1 1 4k = 1 Þ k = 2 2

¥

4

2 2 ò x r( x) dx = ò x

-¥

0

x dx = 8 8

This occur when -

P ( A) = FX ( 3) - FX (1) = e

1 2

-e

6. (D) P ( B) = FX (2.5) = 1 - e

-

-

3 2

2 .5 2

12. (A) 0.25 exceeds when

P ( C) = FX (2.5) - FX (1) = e

-e

x

100

> 0.21 or x > 5 v

P(0.25 W exceeded) = P { x > 5}

= 0.7135

= P { x > 5} + P { x < -5} = 1 - P ( x £ 5) + P { x < -5} æ5 -0 ö æ -5 - 0 ö æ 5 ö æ -5 ö = 1 - Pç ÷ + Pç ÷ = 1 - Fç ÷ + Fç ÷ è 4.2 ø è 4.2 ø è 4.2 ø è 4.2 ø

2 .5 2

æ 5 ö æ 5 ö æ æ 5 öö æ 5 ö = 1 - Fç ÷ + 1 - Fç ÷ = 2ç 1 - F ç ÷ ÷ = 2Qç ÷ 4 . 2 4 . 2 4 . 2 è ø è ø è è øø è 4.2 ø

= 0.3200

13. (A) P {160 < X £ 120} = FX (210) - FX (160) æ 210 - 200 ö æ 160 - 200 ö = Fç ÷ - Fç ÷ 20 20 è ø è ø

8. (A) P ( X > 2) = P {2 < x} + P { x < -2} = 1 - P { x £ 2} + P { x < -2} = 1 - F (2) + F ( -2) We know that for gaussian function F ( - x) = 1 - F ( x)

= F (0.5) - F ( -2) = F (0.5) + F (2) - 1

Thus P ( X > 2)= 1 - F (2) + 1 - F (2)

= 0.6915 + 0.9773 - 1 = 0.6687

= 2 - 2 F(2) = 2 - 2(0.9772) = 0.0456 9. (C) Rejected resistor corresponds to { x < 900 W} and { x > 1100 W}.

Fraction

rejected

corresponds

...(ii)

2

= 0.3834

7. (D) C = A Ç B = {1 < X < 2.5} 1 2

. - ax T - 112 = 103 . sx

Solving (i) and (ii) we get ax = T + 30 and sx = 40

x x - ù é 1 5. (B) For x = 0, FX ( x) = ò e 2 dx = ê1 - e 2 ú u( x) 0 2 ë û x

-

...(i)

P { x £ T - 112 . } = 0.1515 ( T - 112 . - ax ) ( T - 112 . - ax ) =F =1 - F = 8485 sx sx

kv v Thus r( v) = = 4 8 Mean Square Value =

ö ÷÷ = 0.9495 ø

to

14. (C) P { X > 231}= 1 - P { X £ 231} = 1 - FX (231) æ 231 - 200 ö = 1 - Fç . ) = 1 - 0.9394 = 0.0606 ÷ = 1 - F(155 20 è ø

probability of rejection. P {resistor rejected} = P { X < 900} + P { X < 1100}

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Page 395


UNIT 7

15. (B) We use the Rayleigh distribution with a = 0 and b = 400 For probability P { X £ 1}= FX (1) = 1 - e

-

1 400

17 -3 ö æ = 52ç 1 e ÷ = 29.994 weeks 2 è ø 4

24. (B) E[ X ] = X = å xi P ( xi )

= 0.0025 or 0.25 %

i =1

= 10 . (0.4) + 4(0.25) + 9(0.15) + 16(0.1) = 4.35

16. (C) P { X ³ 52} = 1 - FX (52) 52 52 ù é = 1 - ê1 - e 400 ú = 1 - e 400 = 0.00116 or 0.12 % êë úû 2

2

x

òf

17. (C) FX =


X

( u) du =

-¥

x

òb

2

-¥

x 1 3 -2 1 é 6 ù xe = ê ú = 48 2 êë(1 2) 4 úû 0 2

¥

25. (A) E[ g( X )] = E[ X 3 ] = ò

( b p) du + ( u - a) 2

26. (A) Mean of X =

x

1

-¥

0

2 ò xf X ( x) dx = ò x 3(1 - x) dx =

1 4

Let v = u - a and dv = du to get b FX ( x) = p =

x -a

ò

-¥

27. (B) Variance of X is s2x = E[ X 2 ] - m 2x

x -a

dv b é1 æ v öù = tan -1 ç ÷ú b2 + v 2 p êë b è b øû -¥

E[ X 2 ] =

1 1 æ x -aö + tan -1 ç ÷ 2 p è b ø

òx

1

2

-¥

f X ( x) dx = ò x 2 3(1 - x) 2 dx = 0

s2x =

1 æ1ö 3 -ç ÷ = 10 è 4 ø 80

Hence B is correct option

P { x > 0} = 1 - P { x £ 3}

28. (B) Here Y = g( X ) = e

= 1 - P ( x = 0) - P ( x = 1) - P ( x = 2) - P ( x = 3)

So E[ Y ] = E[ g( Y )] =

ì 2 0 21 2 2 2 3 ü æ 19 ö =1 - e í + + + ý = 1 - e -2 ç ÷ = 0.1429 0 ! 2 ! 2 ! 3 ! è 3 ø î þ -2

-

=

0

2 = 0.135 0!

1 20

X 3

¥

ò g( X ) g

-¥

15

X

( x) dx = ò e -5

x 5

1 dx 15 - ( -5)

15

x - ù é 1 1 -3 5 ê-5 e ú = [ e - e ] = 0.667 5 ë û -5

E[ Y 2 ] = E[(2 X - 3) 2 ] = 4 X 2 - 12 X - 9

æ ö = 1 - ç 1 - e W0 ÷ = e -1 ç ÷ è ø W0

= 4(11) - 12( -3) - 9 = 89 2

s2Y = Y 2 - Y = 89 - 9 2 = 8

21. (A) P {5 or more}= 1 - P (0) - P (1) - P (2) - P ( 4) é 30 31 32 33 34 ù 131 -3 = 1 - e -3 ê + + + + =1 e = 0.1847 ú 8 ë 0 ! 11 2 ! 3 ! 4 ! û

30. (A) F

XY

(x, y) = 0.1u (x - 1)u ( y - 1) + 0.35u (x - 2)u ( y - 2) +0.05u (x - 3)u ( y - 3) + 0.5u (x - y)u ( y - 4)

P { X £ 2.5, Y £ 6.0} = f XY (2.5, 6.0) = 0.1 + 0.35 = 0.45

-3

22. (C) P(0) = e = 0.0498 average number of week, per year with no murder 52 e -3 = 2.5889 week.

31. (B) P { X £ 30 . }= FX ( 30 . ) = FXY ( 30 . , ¥) = 0.1 + 0.35 + 0.05 = 0.5 32. (B) FX ( x) = FX , Y ( x, ¥)

23. (D) P {3 or more}= 1 - P (0) - P (1) - P (2)

2 ö 5x 5 æ x + e- ( x + 1 ) y lim ç - e - y ÷u( y) = ÷ y ®¥ 4 ç x + 4 ( x + 1) 1 è ø 2

é 32 ù 17 -3 = 1 - e ê1 + 3 + ú = 1 - 2 e = 0.5768 2 ë û -3

Average number of weeks per year that number of murder exceeds the average

-

29. (C) E[ Y ] = E[2 X - 3] = 2 X - 3 = 2( -3) - 3 = -9

20. (B) P {W > W0 } = 1 - P {W £ W0 } = 1 - FW (W0 )

2 2 ö 1 5 æ limç 1 + e -5 y - e - y ÷ = 1 y ®¥ 4 4 è ø

33. (B) FY ( y)= FX , Y ( ¥, y) Page 396

1 10

2

¥ æ 3k ö 18. (B) Here f X ( x) = e -2 å çç ÷÷d( x - k) k=0 è k! ø

19. (D) P ( x = 0) = e -2

¥

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UNIT 7

¥

w

= ò 5 e - ( w- y) u( w - y) u( y)2 e -2 ydy = 10 ò e -5w+ 3 ydy, w > 0 -¥

0

10 = u( w)[ e -2 w - e -5w ] 3

=

6

2

y= 0 x = 0

òòx

2

y 2 f X , Y ( x, y) dx dy

2

51. (C) E[ g( X , Y ) ] =

=

¥

ò

-¥

x2 e

x2

-

2 s2

2 ps2

¥

dx ò

-¥

e

-

R XY = XY = C XY + X Y = -

¥ ¥

ò ò(x

2

+ y2 )

e

- y2

2 ps2

dy +

¥

ò

-¥

y2 e 2 s

¥

2

2 ps2

r=

æ x 2 + y2 ö ÷ -ç 2 ÷ ç è 2s ø

2 ps2

-¥ -¥

y2 2 s2

dy ò

-¥

dx dy e2s

2 ps2

dx

variance s2 . Thus E[ g( x, y)]= E[( x 2 + y 2 )]2s2 52. (A) E[W ] = E[ 3 X - Y ] = 3 X - Y = 6 - 4 = 2 53. (B) E[W 2 ] = E[( 3 X - Y ) 2 ] = E[9 X 2 - 6 XY + Y 2 ] = 9 X 2 - 6 XY + Y 2 = 9 X 2 - 6 X Y + Y 2 = 9( 8) - 6(2)( 4) + 25 = 49 2

54. (B) s2W = E[(W - W ) 2 ] = E[W 2 - 2W W + W ] 2

= W 2 - W = 49 - 4 = 45

-¥ -¥

E[ X ] = ò ò 0 0

3 2

E[ Y ] = ò ò 0 0

3 2

f X , Y ( x, y) dxdy = ò ò 0 0

x2 y2 8 dx dy = 9 3

2

4 x y dx dy = 9 3 x2 y dx dy = 2 9

Since R XY =

8 æ4ö 8 = E[ X ] E[ Y ] = 2ç ÷ = , we have X and Y 3 è 3ø 3

uncorrelated form 3

From marginal densities f X ( x) = ò 0

2

fY ( y) = ò 0

Page 398

xy x dy = , 0 < x < 2 9 2

xy 2y , 0< y<3 dy = 9 9

we have f X ( x) fY ( y) =

1 2 3

+

1 1æ 1 ö (2) = ç 2 ÷ 2 2è 3ø

1 ö æ1ö 5 æ 1 öæ æ 19 ö = 3 + 2ç ÷ + + 6ç ÷ç 2 ÷ + 9ç ÷ = 98 - 3 3ø è2 ø 2 è 2 øè è 2 ø

second moment of gaussian density with zero mean and

3 2

Y

2

= 3 + 2 X + X 2 + 6 XY + 9 Y 2

2

density. The first factor equal s2 because they are

ò ò xy

and

-1/2 3 -1 2 C XY = = s X sY ( 914 )( 11/2 ) 3 33

factors equal 1 because they are area of a gaussian

¥ ¥

X

57. (B) W = ( X + 3Y ) 2 + 2 X + 3

-x 2

Both double integral are of the same form. the second

55. (B) R XY =

and

5 æ1ö 9 -ç ÷ = 2 è2 ø 4 2 11 11 2 - (2) = s2Y = Y 2 - Y = 2 2

x y dx dy = 64 24

ò ò

= f X ( x) fY ( y)

2

-¥ -¥ 4

f X , Y ( x, y)

statistically independent. 56. (C) s2X = X 2 - X =

¥ ¥

50. (A) E [( XY ) 2 ] =

Thus


xy , 0 < x < 2 and 0 < y < 3 9

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are


CHAPTER

7.2 RANDOM PROCESS

Statement for Question 1 - 4 :

3. The value of E[ X 2 ( t)] is

A random process X ( t) has periodic sample

(A)

t0 A 2 T

(B)

(C)

2 t0 A 2 3T

(D) 0

functions as shown in figure where A, T and 4 t0 £ T are constant but Î is random variable uniformly distributed on the interval (0, T).

4. The value of s2X is

X (t)

t

0

Fig. P7.2.1-4

1. The first order density function is 2t ìï T - 2 t0 d( x) + 0 (A) í T AT ïî 0 (B)

0£x
t0 A 4T

(B)

t0 A 2 t0 T T

(C)

t0 A T

é2 t0 ù êë 3 + T úû

(D)

t0 A 2 T

é2 t0 ù êë 3 + T úû

5. An ergodic random power x( t) has an auto-correlation function R XX ( t) = 18 +

2 1 + 4 cos(12 t) 6 + t2

-

The X is (A) ± 18 (C) ± 17

(B) ± 13 (D) ± 18 ± 17

6. For random process X = 6 and

else where

R XX ( t, t + t) = 36 + 25 e -|t|. Consider following statements : 1. X ( t) is first order stationary. 2. X ( t) has total average power of 36 W.

2. The value of E[ X ( t)] is

(C)

t0 A é2 t0 ù T êë 3 T úû

else where

T - 2 t0 2 T0 d( x) + T AT

t A (A) 0 2T

(A)

0£x
T - 2 t0 2 t0 + [ u( x) - u( x - A)] T AT

2t ìï T + 2 t0 d( x) + 0 (C) í T AT ïî 0 (D)

t0 A 2 3T

3. X ( t) is a wide sense stationary. t A (B) 0 T (D) 0

4. X ( t) has a periodic component. The true statement is/are (A) 1, 2, and 4

(B) 2, 3, and 4

(C) 2 and 3

(D) only 3

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Page 399


UNIT 7


2

7. A random process is defined by X ( t) + A where A is

The variance of random variable Y = ò X ( t) dt will be

continuous random variable uniformly distributed on

(A) 1

(B) 2.31

(0,1). The auto correlation function and mean of the

(C) 4.54

0

(D) 0

process is (A) 1/2 & 1/3

(B) 1/3 & 1/2

13. A random process is defined by X ( t) = A cos( pt)

(C) 1 & 1/2

(D) 1/2 & 1

where A is a gaussian random variable with zero mean and variance s 2p . The density function of X(0)

Statement for Question 8 - 9 : A

random

process

(A) is

defined

by

Y ( t) = X ( t) cos( w0 t + q) where X ( t) is a wide sense

1 2 ps A

-

e

x2 2 s 2A

(B)

(C) 0

2 ps A e

-

x2 2 s 2A

(D) 1

stationary random process that amplitude modulates a carrier of constant angular frequency w0 with a random


phase q independent of X ( t) and uniformly distributed on ( -p / p).

The two-level semi-random binary process is defined by X ( t) = A or -A

8. The E[ Y ( t)] is

where ( n - 1) T < t < nt and the levels A and -A

(B) -E[ X ( t)] (D) 0

(A) E[ X ( t)] (C) 1

occur with equal probability. T is a positive constant

9. The autocorrelation function of Y ( t) is 1 (B) R XX ( t) cos( w0 t) (A) R XX ( t) cos( w0 t) 2 (D) None of the above (C) 2 R XX ( t) cos( w0 t) Statement for Question 10 - 11 : Consider a low-pass random process with a white-noise power spectral density S X ( w) = N/2 as shown in fig.P7.2.10-11.

and n = 0, ± 1, ± 2 14. The mean value E[ X ( t)] is (A) 1/2

(B) 1/4

(C) 1

(D) 0

15. The auto correlation R XX ( t1 = 0.5 T, t2 = 0.7 T) will be (A) 1

(B) 0

(C) A

2

(D) A 2/2

16. A random process consists of three samples function X ( t, s1 ) = 2, X ( t, s2 ) = 2 cos t1 and X ( t, s3) = 3 sin t- each occurring with equal probability. The process is (A) First order stationary (B) Second order stationary

Fig.P7.2.10-11

(C) Wide-sense stationary 10. The auto correlation function R X ( t) is

(D) Not stationary in any sense

(A) 2 NB sinc (2pbt)

(B) pNB sinc (2pbt)

(C) NB sinc (2pbt)



(A) 2 NB

(B) pNB

ergodic random process is shown in fig.P.7.2.17-19

(C) NB

D) D

The auto correlation function of a stationary

11. The power PX is NB 2p

50

12. If X ( t) is a stationary process having a mean value E[ X ( t)] = 3 R XX ( t) = 9 + 2 e Page 400

and

autocorrelation

function

-|t|

.

20 -10

10

Fig. P7.2.17-19

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Random Process

17. The mean value E[ X ( t)] is

Chap 7.2

24. A complex random process Z ( t) = X ( t) + jY ( t) is

(A) 50

(B)

50

defined by jointly stationary real process X ( t) and Y ( t).

(C) 20

(D)

20

The E[|Z ( t)|2 ] will be (A) 2 R XY (0) + R XX (0) + RYY (0) (B) R XX (0) + RYY (0) (C) R XX (0) - RYY (0)

18. The E[ X 2 ( t)] is (A) 10

(B)

10

(C) 50

(D)

50

(D) RYY (0) - R XX (0)

25. Consider random process X ( t) = A0 cos( w0 t + q) where A0 and w0 are constant and q is a random variable uniformly distributed on the interval (0, p). The

19. The variance s2X is

power in X ( t) is

(A) 20

(B) 50

(A) A 2

(C) 70

(D) 30

(C)

20. Two zero mean jointly wide sense stationary random process

X ( t) and Y ( t) have no periodic

components. It is know that s 2X = 5 and s2Y = 10. The function, that can apply to the process is 2

(A) R XX ( t) = 6 u( t) e -3t (C) R XY ( t) = 9(1 + 2 e 2 ) -1

ésin( 3t) ù (B) RYY ( t) = 5 ê ë 3t úû (D) None of the above

A2

1 4

(D) 1

real random process is (A) d( w + w0 ) - d( w - w0 ) (C) d( w) +

w2 w + 16 2

w2 w2 + 25 w2 (D) 2 w + 16 (B)

27. The valid power density spectrum is

21. A stationary zero mean random process X ( t) is

w2 1 + w2 + jw2

(C) e - ( w -1 )

2

(B)

w2 - d( w) w +1

(D)

w2 w + 3w2 + 3

component. The valid auto correlation function is (A) 16 + 18 cos( 3t)

(B) 24 da 2 (2 p)

e ( -6 t ) (C) (1 + 3t 2 )

(D) 24d( t - t)

A2

1 2

26. The non valid power spectral density function of a

(A) ergodic has average power of 24 W and has no periodic

(B)

4

6

28. A power spectrum is given as P é r XX ( w) = ê1 + ( w / W ) 2 ê 0 ë

22. Air craft of Jet Airways at Ahmedabad airport

w < KW w > KW

where P , W , and K are real positive constants. The

arrive according to a poisson process at a rate of 12 per hour. All aircraft are handled by one air traffic

sums bandwidth of power spectrum is

controller. If the controller takes a 2 - minute coffee

(A) W

tan + k -1 k

(B) W

(C) W

tan -1 k +1 k

(D) ¥

break, what is the probability that he will miss one or more arriving aircraft ? (A) 0.33

(B) 0.44

(C) 0.55

(D) 0.66

29. Consider the power spectrum given by ìP r XX ( w) = í î0

23. Delhi airport has two check-out lanes that develop waiting lines if more than two passengers arrives in any one minute interval. Assume that a poission process describes the number of passengers that arrive

(A) 0.16 (C) 0.32

(B) 0.29 (D) 0.49

w W

where P and W are real positive constants. The rms bandwidth of the power spectrum is

for check-out. The probability of a waiting line if the average rate of passengers is 2 per minute, is

k -1 tan -1 k

(A)

W

(C)

W

2 3

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(B)

W2 3

(D)

W 2 Page 401


UNIT 7


30. For a random process R XX ( t) = P cos 4 ( w0 t) where P

37. A random process X ( t) has an autocorrelation

and w0 are constants. The power in process is

function R XX ( t) = A 2 + Be -|t| Where A and B are

(A) P

(B) 2P

constants. A system have an input response

(C) 3P

(D) 4P

ì e - Wt 0 < t h( t) = í î 0 t <0

31. A random process has the power density spectrum r XX ( w) =

6w2 [1 + w 2 ] 3

where W is a real positive constant, which X ( t) is

. The average power in process is

(A) 1/4

(B) 3/8

(C) 5/8

(D) 1/2

its input. The mean value of the response is A A (B) (A) W 2W

32. A deterministic signal A cos( w0 t), where A and w0

(C)

are real constants is added to a noise process N ( t) for which r NN ( w) =

W2 W2+ w2

and W > 0 is a constant. The ratio

2A W

38. In previous question if impulse response of system is

of average signal power to average noise power is

(C)

ì e - Wt sin( w0 t) 0 < t h( t) = í 0 t <0 î

A (B) W A2 (D) W

(A) 1 2A W

where W and w0 are real positive constants, the mean value of response is

33. The autocorrelation function of a random process

(A)

Aw0 w + W2

(C)

2A æ 1 çç 2 w0 è w0 + W 2

X ( t) is 2

R XX ( t, t + t) = 12 e Yt cos 2 (24 t) The R XX ( t) is (A) 6 e -4 t (C) 48 e

2

(B) 12 e -4 t

-4 t 2

power density spectrum f XX ( w) is (B)

(C) 3 + jw2 35.

The

4 e -3|t| 1 + w2

of jointly

wide

sense - Wt

where A > 0 and W > 0 are constants. The r XX ( w) is A W - w2

(B)

A W + w2

(C)

A W + jw

(D)

A W - jw

2

E[ X ( t)] = 2, the mean value of the system's response Y ( t) is (A)

1 128

(B)

1 64

(C)

3 128

(D)

1 32

A random process X ( t) is applied to a network with

stationary process X ( t) and Y ( t) is R XY ( t) = Au( t) e

(A)

impulse response h( t) = u( t) te - at

Y ( t) is known to have the same form R XY ( t) = u( t) te - at 40. The auto correlation of Y ( t) is 4 + at - a|t| 1 + at - a|t| (A) (B) e e 3 4a 3a 2

2

36. A random process X ( t) is applied to a linear time system.

A response

Y ( t) = X ( t) - X ( t - t)

occurs when t is a real constant. The system's transfer function is (A) 1 - e jwt (C) 2 je - jwt/ 2 cos Page 402

wt 2

(B) 2 je - jwt/ 2 sin (D) 1 + e - jwt

where a > 0 is a

constant. The cross correlation of X ( t) with the output

(C) invariant

ö ÷÷ ø ö 1 ÷ 2 ÷ +W ø

1 + W2


(D) 18d( w) cross correlation

ö ÷÷ ø

A 2 w0

input of a system for which h( t) = 3u( t) t 2 e -8 t . If


6 6 + 7 w3

æ çç 2 è w0 A æ (D) ç 2 w0 çè w02

(B)

2 0

39. A stationary random process X ( t) is applied to the

2

34. If X ( t) and Y ( t) are real random process, the valid

(A)

(D) 0

wt 2

4 + at - a|t| e 8a2

(D)

1 + at - a|t| e 4 a3

41. The average power in Y ( t) is 1 1 (A) (B) 3 a 4a3 (C)

1 3a 2

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Random Process

Chap 7.2


SOLUTION

A random noise X ( t) having a power spectrum r XX ( w) =

3 49+ w 2

is applied to a differentiator that has a

transfer function H ( w) = jw. The output is applied to a network for which h( t) = u( t) t 2 e -7t

1.

(A)

Î

Let

have

value

Now

e.

P { X £ x Î= e} = FX ( X|Î= e) and for any Î must be zero for x < 0 because x( t) is never negative. The event { X £ 0} is satisfied whenever x( t) is zero. This happens

42. The average power in X ( t) is (A) 5/21

(B) 5/24

(C) 5/42

(D) 3/14

during

the

fraction

of

time ( T - 2 t0 ) / T.

FX ( x|Î= e) = [( T - 2 T0 ) / T} u( x).

For

Hence

0£x
the

additional time interval or fraction of time where X £ x becomes 2 to 2 t0 x / AT.

43. The power spectrum of Y ( t) is

2t x æ T - 2 t0 ö Thus FX ( x Î= e) = ç ÷u( x) + 0 , 0 £ x < A T AT è ø

(A)

4 w2 ( 49 + w2 ) 3

(B)

12 w2 ( 49 + w2 ) 4

(C)

42 w3 ( 49 + w2 ) 2


= 1,A £ x = 0,x < 0 By differentiation

44. White noise with power density N0/2 is applied to a lowpass network for which H(0) = 2. It has a noise

2t æ T - 2 t0 ö f X ( x Î= e) = ç ÷d( x) + 0 , 0 £ x < A AT è T ø = 0 else where

bandwidth of 2 MHz. If the average output noise power

f X , e ( x, e) = f X ( x Î= e) fÎ( e)

is 0.1 W in a 1 - W resistor, the value of N0 is (A) 12.5 nW/Hz

(B) 12.5 mW/Hz

(C) 25 nW/Hz

(D) 25 mW/Hz

2 T0 æ T - 2 t0 ö ,0 £ x < A and 0 < e < T =ç ÷d( x) + 2 AT 2 è T ø

45. An ideal filter with a mid-band power gain of 8 and

f X ( x) =

power spectrum r XX ( w) =

e

- w 2 /8

the network's output is ( F (2) = 0.9773) (B) 90.3

(C) 20.2

(D) 100.4

( x, e) de

X ,Î

2t æ T - 2 t0 ö =ç ÷d( x) + 0 .0 £ x < A AT è T ø

. The noise power at

(A) 60.8

òf

-¥

bandwidth of 4 rad/s has noise X ( t) at its input with 50 8p

¥

= 0 elsewhere. ¥

ò xf

2. (B) E[ X ( t)] =

( x) dx

X

-¥

46. White noise with power density N0/2 = 6 mW/Hz is applied to an ideal filter of gain 1 and bandwidth W

=

¥

2t x t A æ T - 2t ö ò-¥ xçè T 0 ÷ød( x) dx + ò0 AT0 dx = 0T A

rad/s. If the output's average noise power is 15 watts, the bandwidth W is

3. (C) E[ X 2 ( t)] =

(A) 2.5 ´ 10 -6

(B) 2.5 p ´ 10 -6

(C) 5 ´ 10 -6

(D) p5 ´ 10 -6

òx

-¥

A

2

2 t0 x 2 2 t0 A 2 = 3T AT 0

f X ( x) dx = ò

4. (D) s2X = E[ X 2 ( t)] - { E[ X ( t)]} 2 2

47. A system have the transfer function H ( w) = 1 + ( w1/W ) 4 where W is a real positive constant. The noise bandwidth of the system is

=

2 t0 A 2 t02 A 2 t0 A 2 = 3T T T2

é2 t0 ù êë 3 - T úû

5. (A) We know that ( i) if X ( t) has a periodic component

(A)

1 3

pW 2

(B)

(C)

1 6

pW 2


1 4

¥

pW 2

then R XX ( t) will have a periodic component with the -

same period. (ii) if E[ X ( t)] = X m 0 and X ( t) is ergodic with no periodic components then lim R XX ( t) = X

2

|t|®¥

2

************

Thus we get X = 18 or X= ± 18 www.gatehelp.com

Page 403


UNIT 7

6. (C) X = Constant and R XX ( t) is not a function of t, so X ( t) is a wide sense stationary. So 1 is false & 3 is true.


13. (A) For t = 0,X(0)= A,

1

So f X ( x) =

2 ps A

-

e

x2 2 2 sA

PXX = R XX (0) = 36 + 25 = 61. Thus 2 is false if X ( t) has a periodic component, then

14. (D) E[ X ( t)] = AP ( A) + ( - A) P ( - A) =

R XX( t) will have a periodic component with the same period. Thus 4 is false.

15. (C) Here R XX ( t1 , t2 )= A 2 1

7. (B) R XX ( t, t + t) = E[ X ( t) X ( t + t)] = E[ A 2 ] = ò a 2 da = 0

1

X= E[ X ( t)] = E[ A ] = ò a da = 0

1 3

= E X [ X ( t)] ò cos( w0 t + q) -X

If

both

and

t1

are

t2

in

the

same

interval

( n - 1) T < t, t2 < nT, n = 0, ±, ±2... and R XX ( t1 , t2 ) = 0 otherwise

1 2

Hence R XX (0.5 T, 0.7 T) = A 2 16. (D) Let x1 = 2, x2 = 2 cos t and x3 = 3 sin( t) 1 1 1 Then f X ( x) = d( x - x1 ) + d( x - x2 ) + d( x - x3) 2 3 3

8. (D) E[ Y ( t)} = E[ X ( t) cos( w0 t + q)] X

A A - =0 2 2

1 dq = 0 2p

where E X [×] represent expectation with respect to X

¥

and E[ X ( t)] =

ò xf

X

( x) dx

-¥

only =

9. (B) RYY ( t, t + t)

é1

1

ù

1

ò x êë 3 d( x - x ) + 3 d( x - x ) + 3 d( x - x ) úû 1

2

2

-¥

= E[ X ( t) cos( w0 t + q) X ( t + t) cos( w0 t + q + w0 t)] 1 = R XX ( t) [cos( w0 t) + cos(2 w0 t + 2 q + w0 t)] 2 1 = R XX ( t) cos( w0 t) 2 10. (C) S X ( w) =

¥

=

1 [2 + 2 cos t + 3 sin t ] 3

The mean value is time dependent so X ( t) is not stationary in any sense. 17. (D) We know that for ergodic with no periodic

N æ w ö rectç ÷ 2 è 4 pb ø

component 2

lim R XX ( t) = X ,

We know that R X ( t)¬¾® S X ( w) W æ w ö sin(Wt)¬¾® rect ç ÷ p è 2W ø

|t|®¥

2

Thus X = 20

or

X = 20

18. (C) R XX (0) = E[ X 2 ( t)] = R XX (0) = 50 = X 2

Here W = 2pB Hence R X ( t) =

2

2 pB N sin (2pBt) = NB sinc (2pBt) p 2

19. (D) s2X = X 2 - X = 50 - 20 = 30 20. Here X = 0, Y = 0, R XX (0) = 5, sY2 = RYY (0) = 10

11. (C) PX = X = R X (0) = NB since sinc (0) = 1 2

2

2

For (A) : Function does not have even symmetry For (B) : Function does not satisfy RYY (0) = 10

2

12. (C) E[ Y ] = E[ ò X ( t) dt = ò E[ X ( t)]dt = 3ò dt = 6 0

0

2

2

0

0

For

0

2 2

E[ Y 2 ] = E[ ò X ( t) dt ò X ( u) du]] = ò ò E[ X ( t) X ( u) du dt ]

(C)

:

Function

not

satisfy

|R XY ( t)|£ R XX (0) RYY (0) = 50

0 0

2 2

2 2

21. (D) For (A) : It has a periodic component.

0 0

0 0

For (B) ; It is not even in t, total power is also incorrect.

= ò ò R XX ( t - u) dt du = ò ò [9 + 2 e -|t - u| ]dt du 2 2

= 36 + 2 ò ò e -|t - u|dt du = 4(10 + e -2 ) 0 0

s2Y = E[ Y 2 ] - ( E[ Y ]) 2 = 4(1 + e -2 ) = 4.541

For (C) It depends on t not even in t and average power is ¥. 22. (A) P (miss/or more aircraft)= 1 - P(miss 0) = 1 - P (0 arrive) = 1 -

Page 404

does

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( lt) 0 e - lt 0!


UNIT 7

¥

¥

-¥

0

E[ Y ( t)] = A ò h( x)dx = A ò e - Wt dt =

So W = 2.5 p ´ 106

A W

¥

38. (A) X = A ¥

¥

-¥

0

E[ Y ( t)] = Y = X ò h( t) dt = A ò e

47. (B) Noise bandwidth Wn = ¥

ò H( w)

2

H (0) ¥

¥

òR

XY

2 -8 t

*********

( t + x)h( x) dx

-¥ ¥

ò u( x)u( x + t)( tx + x ) e

= e - at

2

-2 ax

dx

-¥

There are two cases of interest t ³ 0 and t < 0 Since RYY ( t) is an even function we solve only the ease t ³ 0 ¥

RYY ( t) = e - at ò ( tx + x2 )e -2 ax dx = 0

1 + at - ax e 4 a3

41. (B) Power in y( t) = RYY (0) =

1 4a3

¥

¥

dw 1 3 3 ò r XX ( w) dw = 2 p -¥ò 49 + w2 = 14 2p -¥

42. (D) PXX =

F 43. (B)h2 = 49 t) t 2 e -7t ¬¾ ®

2 = H 2 ( w) (7 + jw) 3 2

sYY ( w) = s XX ( w) = H1 ( w) H 2 ( w) =

12 w2 ( 49 + w2 ) 4

2

44. (A) PYY = So N0 =

N0 H (0) Wn 2p

2 p(0.1) 2

H (0) Wn

=

= 0.1

2 p(0.1) (2) 2 2 p ´ 2 ´ 106

. ´ 10 -8 W/Hz = 12.5 nW/Hz = 125 45. (A) PYY = 4

¥

2 1 r XX ( w) H ( w) dw ò 2 p -¥ w2

1 50 - 8 = e ò 2p 4 8p =

Page 406

200 = p

4

ò 4

e

-

w2 2 (u)

2 p( 4)

200 200 [ F (2) - F ( -2)] = [2 F (2) - 1)] = 60.8 p p

46. (B) PYY = =

(8 ) dw

¥

2 1 ò r XX ( w) H( w) dw 2 p -¥

1 6 ´ 10 -6 W 6 ´ 10 -6 W -6 d w ´ = = 6 10 = 15 2 p -òW p p W

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dw

2

dw pW = 4 ( ) + 1 w /W 2 2 0

3 39. (C)Y = X ò h( t) dt = 2 ò 3t e dt = 128 0 -¥ 40. (D) RYY ( t)=

2

0

Wn = ò H ( w) dwsince H(0) = 1 = ò 0

¥

¥

Aw sin( w0 t) dt = 2 0 2 w0 + W

- Wt



CHAPTER

7.3 NOISE

1. The power spectral density of a bandpass white noise

5. A mixer stage has a noise figure of 20 dB. This mixer

n( t) is N / 2 as shown in fig.P.7.3.1. the value of n is

stage is preceded by an amplifier which has a noise

2

figure of 9 dB and an available power gain of 15 dB. The overall noise figure referred to the input is

Fig. P7.3.1

(A) 11.07

B) 18.23

(C) 56.48

(D) 97.38

6. A system has three stage cascaded amplifier each

(A) NB

(B 2 NB

(C) 2pNB

(D)

stage having a power gain of 10 dB and noise figure of 6

NB p

dB. the overall noise figure is

2. In a receiver the input signal is 100 mV, while the

(A) 1.38

(B) 6.8

(C) 4.33

(D) 10.43

internal noise at the input is 10 mV. With amplification

7. A signal process m( t) is mixed with a channel noise

the output signal is 2 V, while the output noise is 0.4 V.

n( t). The power spectral density are as follows

The noise figure of receiver is Sm ( w) =

(A) 2

(B) 0.5

(C) 0.2


3. A receiver is operated at a temperature of 300 K. The transistor used in the receiver have an average output resistance of 1 kW. The Johnson noise voltage for a receiver with a bandwidth of 200 kHz is (A) 1.8 mV

(B) 8.4 mV

(C) 4.3 mV

(D) 12.6 mV

(C) 4 m V

(D) 16 ´ 10 -18 V

(C)

w2 + 10 w2 + 9

(B)

1 w + 10 2


A sonar echo system on a sub marine transmits a

noise voltage generated in a bandwidth of 10 kHz is (B) 0.4 m V

The optimum Wiener-Hopf filter is w2 + 9 (A) 2 w + 10

Statement for Question 8-9

4. A resistor R = 1 kW is maintained at 17 o C. The rms (A) 16 ´ 10 -14 V

6 , Sn ( w) = 6 9 + w2

random noise n( t) to determine the distance to another targeted submarine. Distance R is given by vt R / 2 where v is the speed of the sound wave in water and t R is the time it takes the reflected version of n( t) to return. Assume that n( t) is a sample function of an ergodic random process N ( t) and T is very large.

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Page 407


UNIT 7

8. The V will be


14. The standard spot noise figure of amplifier is

(A) 2 R NN ( t R - t T )

(B) R NN ( t R - 2 t T )

(A) 4 dB

(B) 5 dB

(C) R NN ( t R - t T )

(D)

R NN ( t R - t T )

(C) 7 dB

(D) 9 dB

1 2

9. What value of the delay t T will cause v to be

15. If a matched attenuator with a loss of 3.2 dB is

maximum ?

placed between the source and the amplifier's input, what is the operating spot noise figure of the attenuator

(A) t R

(B) 2t R

(C) 3t R


amplifier

cascade

if

the

attenuator's

temperature is 290 K ?

10. Two resistor with resistance R1 and R2 are

(A) 9 dB

(B) 10.4 dB

connected in parallel and have Physical temperatures

(C) 11.3 dB

(D) 13.3 dB

T1 and T2 respectively. The effective noise temperature Ts of an equivalent resistor is

physical

16. In previous question what is the standard spot noise figure of the cascade ?

(A)

T1 R1 + T2 R2 R1 + R2

(B)

T1 R1 + T2 R1 R1 + R2

(C)

T1 T2 ( R1 + R2 ) 2 ( T1 + T2 ) R1 R2

(D)

( T1 + T2 ) R1 R2 T1 + T2 ( R1 + R2 ) 2

(A) 10.3 dB (C) 14.9 dB

(B) 12.2 dB (D) 17.6 dB

17. Omega Electronics sells a microwave receiver (A) having an operating spot noise figure of 10 dB when


driven by a source with effective noise temperature 130

An amplifier has a standard spot noise figure

K Digilink (B) sells a receiver with a standard spot

F0 = 6.31 (8.0 dB). The amplifier, that is used to amplify

noise figure of 6 dB. Microtronics (C) sells a receiver

the output of an antenna have antenna temperature of

with standard spot noise figure of 8 dB when driven by

Ta = 180 K

a source with effective noise temperature 190 K. The best receiver to purchase is

11. The effective input noise temperature of this

(A) A

(B) B

amplifier is

(C) C

(D) all are equal

(A) 2520 K

(B) 2120 K

(C) 2710 K

(D) 1540 K

Statement for Question 18-20 : An amplifier has three stages for which Te1 = 150 K

12. The operating spot noise figure is

(first stage), Te 2 = 350 K, and Te 3 = 600 K (output stage).

(A) 3.2 dB

(B) 6.4 dB

Available power gain of the first stage is 10 and overall

(C) 9.8 dB

(D) 11.9 dB

input effective noise temperature is 190 K

13. An amplifier has three stages for which Te1 = 200 K

18. The available power gain of the second stage is

(first stage), Te 2 = 450 K, and Te 3 = 1000K (last stage). If

(A) 12

(B) 14

the available power gain of the second stage is 5, what

(C) 16

(D) 18

gain must the first stage have to guarantee an effective input noise temperature of 250 K ?

19. The cascade's standard spot noise figure is

(A) 10 (C) 16

(A) 1.3 dB

(B) 2.2 dB

(C) 4.3 dB

(D) 5.3 dB

(B) 13 (D) 19


20. What is the cascade's operating spot noise figure

An amplifier has an operating spot noise figure of

Page 408

when used with a source of noise temperature Ts = 50 K

10 dB when driven by a source of effective noise

(A) 1.34 dB

(B) 3.96 dB

temperature 225 K.

(C) 6.81 dB

(D) None of the above.

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Noise

Chap 7.3

21. Three network are cascaded. Available power gains

elements

are G1 = 8, G2 = 6 and G3 = 20. Respective input effective

matched attenuator with an overall loss of 2.4 dB and a

spot noise temperature are Te1 = 40 K, Te 2 = 100 K and

physical temperatures of 275 K. The overall system

Te 3 = 180 K.

noise temperature of the receiver T sys = 820 K.

(A) 58.33 K

(B) 69.41 K

(C) 83.90 K

(D) 98.39 K

that can be modeled as an impedance

26. The average effective input noise temperature of the receiver is

22. Three identical amplifier, each having a spot

(A) 420.5 K

(B) 320.5 K

effective input noise temperature of 125 K and available

(C) 220.5 K

(D) 10.5 K

power G are cascaded. The overall spot effective input noise temperature of the cascade is 155 K. The G is

27.

(A) 3

(B) 5

attenuator-receiver cascade is

(C) 7

(D) 9

(A) 13.67 d

(B) 11.4 dB

(C) 1.4 dB

(D) 1.367 dB

23. Three amplifier that may be connected in any order in a cascade are defined as follows:

The

average

operating

noise

figure

of

the

28. If receiver has an available power gain of 110 dB and a noise bandwidth of 10 MHz, the available output

Amplifier

Effective Input Noise Temperature

Available Power Gain

A

110 K

4

B

120 K

6

C

150 K

12

noise power of receiver is (A) 6.5 mW

(B) 8.9 mW

(C) 10.3 mV

(D) 11.4 mV

29. If antenna attenuator cascade is considered as a noise source, its average effective noise temperature is

The sequence of connection that will give the lowest overall effective input noise temperature for the cascade is

(A) 63 K

(B) 149 K

(C) 263 K

(D) 249 K


(A) ABC

(B) CBA

(C) ACB

(D) BAC

An amplifier when used with a source of average noise temperature 60 K, has an average operating noise

24. What is the maximum average effective input noise temperature that an amplifier can have if its average standard noise figure is to not exceed 1.7 ? (A) 203 K

(B) 215 K

(C) 235 K

(D) 255 K

figure of 5. 30. The T e is (A) 70 K

(B) 110 K

(C) 149 K

(D) 240 K

25. An amplifier has an average standard noise figure

31. If the amplifier is sold to engineering public, the

of 2.0 dB and an average operating noise figure of 6.5

noise figure that would be quoted in a catalog is

dB when used with a source of average effective source

(A) 0.46

(B) 0.94

temperature Ts . The Ts is

(C) 1.83

(D) 2.93

(A) 156.32 K

(B) 100.81 K

(C) 48.93 K


32. What average operating noise figure results when the amplifier is used with an antenna of temperature

Statement for Question

30 K ?

An antenna with average noise temperature 60 K connects to a receiver through various microwave

(A) 9.54 dB

(B) 10.96 dB

(C) 11.23 dB

(D) 12.96 dB

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Page 409


UNIT 7


33. An engineer of RS communication purchase an

(A) 1 / W

(B) 2.5 / W

amplifier with average operating noise figure of 1.8

(C) 4.5 / W

(D) 6 / W

when used with a 60 W broadband source having average source temperature of 80 K. When used with a different 60 W source the average operating noise figure

39. White noise, for which R XX ( t) = 10 -2 8( t) is applied to a network with impulse response h( t) = 4( t) 3 - te-4 t The

is 1.25. The average noise temperature of the source is

network's output noise power in a 1 W resistor is

(A) 125 K

(B) 156 K

(A) 0.15 mW

(B) 0.35 mW

(C) 256 K

(D) 292 K

(C) 0.55 mW

(D) 0.95 mW

34. The Te for unit 1 and 2 unit are, respectively (A) 126.4 K and 256.9 K

40. White noise with power density N0/2 = 6(10 -6 ) W/Hz is applied to an ideal fitter (gain= 1) with bandwidth W

(B) 256.9 K and 126.4 K

(rad/sec). For output's average noise power to be 15 W,

(C) 527.8 K and 864.2 K

the W must be

(D) 864.2 K and 527.8 K

(A) 2.5 p(10 -6 )

(B) -2.5 p(106 )

(C) 4.5 p(10 -2 )

(D) 4.5 p(106 )

35. The excess noise power of unit 1 and unit 2 are respectively

41. An ideal filter with a mid-band power gain of 8 and

(A) 15.4 nW and 27.1 nW

bandwidth of 4 rad/s has noise X ( t) at its input with

(B) 23.8 nW and 21.1 nW

power spectrum ( F (2) = 0.9773)

(C) 23.8 nW and 27.1 nW

w2

æ 50 ö - 8 r XX ( w) = ç ÷e è 8p ø

(D) 15.4 nW and 21.1 nW 36. Consider following statement S1 : If the source noise temperature T s is very small,

The noise power at the network's output is 164 343 (B) (A) p p

unit-2 is the best to purchase S2 : If the source noise temperature T s is very small

(C)

211 p

(D)

191 p

unit - 1 is the best to purchase. 42. A system has the power transfer function

correct statement is (A) S1

(B) S2

(C) both S1 and S2

(D) None

37. A source has an effective noise temperature of and feeds an amplifier that has an Ts ( w) = 100800 + w2

available power gain of Ga ( w) =

(

8 10 + jw

1

2

H( w) =

) . The T 2

s

for this

source is (A) 10 K

(B) 20 K

(C) 30 K

(D) 40 K

4

where W is a real positive constant. The noise bandwidth of the system is pW (A) 2 2 (C)

38. A system have an impulse response

æ wö 1+ç ÷ èW ø

pW 2

(B)

pW 2


43. White noise with power density N0/2 is applied to a low pass network for which H(0) = 2. It has a noise

- Wt

ìe e
bandwidth of 2 MHz. If the average output noise power

where W is a real positive constant. White noise

is 8.1 W in a 1 W resistor, the N0 is

with power density 5w/Hz is applied to this system. The

(A) 6.25 ´ 108 W/Hz

(B) 6.25 ´ 10 -8 W/Hz

mean-squared value of response is

(C) 125 . ´ 108 W/Hz

(D) 125 . ´ 10 -8 W/Hz

Page 410

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Noise

Chap 7.3


SOLUTION

An amplifier has a narrow bandwidth of 1 kHz and standard spot noise figure of 3.8 at its frequency of operation. The amplifier's available output noise power

fc + B

ò

1. (B) n2 = 2

fc + B

N df = 2 NB 2

is 0.1 mW when its input is connected to a radio receiving antenna having an antenna temperature of 80 K.

2. (A) NF =

100/10 S/N i i = =2 2/0.4 So/N o

44. The amplifier's input effective noise temperature Te

3. (A) vn2 = 4kTBR

is

= 4 ´ 1.38 ´ 10 -23 ´ 300 ´ 200 ´ 10 3 ´ 10 3 = 3.3 ´ 10 -12

(A) 812 K

(B) 600 K

(C) 421 K

(D) 321 K

vnrms = 1.8m V 4. (B) vn2 = 4kTBR,

45. Its operating spot noise figure Fop is (A) 5.16

(B) 7.98

(C) 11.15

(D) 16.23

46. Its available power gain Ga is (A) 2 ´ 1012

(B) 4 ´ 1012

(C) 8 ´ 1012

(D) 11 ´ 1012

T = (273 + 17) K = 290 K,

R = 1000W, B = 10 Hz, k = 1.38 ´ 10 -23 J/K 4

vn2 = 4 ´ 1.38 ´ 10 -23 ´ 290 ´ 10 3 ´ 10 4 = 16 ´ 10 -14 V 2 vnrms = 0.4m V 5. (A) F1 = 9 dB = 7.94,

F2 = 20 dB = 100

A1 = 15 dB = 31.62, 100 - 1 F -1 . = 7.94 + = 1107 F = F1 + 2 31.62 A 6. (C) Gain of each stage A1 = A2 = A3 = 10 dB Noise figure of each stage F1 = F2 = F3 = 6 dB or F1 = F2 = F3 = 4 db F - 1 F3 - 1 4 -1 4 -1 + =4+ + = 4.33 F = F1 + 2 A1 A1 A2 10 100 7. (B) H op ( w) =

8. (C) V =

1 2T

Sm ( w) = Sm ( w) + Sm ( w)

6 9+ w 2 6 9+ w 2

+6

=

1 10 + w2

T

ò n( t - t

T

)n( t - t R ) dt

-T

Since T is very large 1 T ®¥ 2 T

V = lim

T

ò n( t - t

T

)n( t - t R ) dt = A[ n( t - t T ) n( t - t R )]

-T

Since N ( t) is ergodic,

V » R NN ( t R - t T )

9. (A) Because R NN ( t) £ R NN (0) for any auto correlation function, V will be maximum if t R = t T 10. (B) Use the current form of equivalent circuit 2 kT1 dw 2 kT2 dw 2kTs dw where in2 = , in2 = i12 + i22 = + pR1 pR2 pR æT T Thus Ts = çç 1 + 2 è R1 R2

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ö T R + T2 R1 ÷÷ R = 1 2 R1 + R2 ø Page 411


UNIT 7

11. (D) Te = T0 ( F0 - 1) = 290( 6.31 - 1) = 1539.9 K

13. (B) Te = 250 = Te1 +

6 G 2 - 25 G - 25 = 0 or G = 5 23. (A) Sequence Te 120 150 ABC110 + + = 146.25 4 4( 6)

Te 2 Te 3 + G1 G2

450 1000 250 = 200 + + G1 5 G2 14. (D) F0 = 1 +

(125 - 155) G 2 + 125 G + 125 = 0

Te 1540 =1 + = 9.56 or 9.8 dB Ta 180

12. (C) Fop = 1 +

or


ACB110 +

G1 = 13

150 120 + = 150.00 4 4(12)

BAC 120 +

Ts 225 ( Fop - 1) = 1 + (10 - 1) T0 290

CBA150 +

110 150 + = 144.583 ¬ Best 6 6( 4)

120 110 + = 161528 . 12 (12)( 6)

= 7.98 or 9.0 dB 24. (A) Te = T0 ( F - 1) £ 290(17 . - 1) = 203 K

15. (D) Here L = 2.089 or 3.2 dB, TL = 290 K Te = Te1 +

Te 2 T ( F - 1) = TL ( L - 1) + 0 0 G1 1/ L

(or 2.0 dB) and F OP » 4.467 (or 25. (D) Here F0 » 1585 . 6.5 dB)

= 290[(2.089 - 1) + (2.089)(7.98 - 1)] = 4544.4 K

Ts =

4544.4 = 212 . or 13.3 dB Fop = 1 + 225 16. (B) F0 = 1 +

T0 ( F0 - 1) 290(1585 - 1) . = = 48.93 K 4.467 - 1 F op - 1

(or 2.4 dB), TL = 275 K 26. (B) Here Ta = 60 K, L = 1738 .

4544.4 = 16.67 or 12.2 dB 290

and T sys = 820 K.

17. (B) For A: Fop = 10(or 10 dB) when Ts = 130 K

TR =

[ T sys - Ta - TL ( L - 1)] 820 - 60 - 275(1738 . - 1) = L 1738 .

= 320.5 K

TeA = 130(10 - 1) = 1170 K For B: Fo = 398 . (or 6 dB) when Ts = 290 K

27. (B) F op = 1 +

. - 1) = 364.2 K TeB = 290( 398 For C: Fo = 6.3(or 8 dB) when Ts = 190 K TeC = 190( 6.3 - 1) = 1007 K, (B) is better as TeB is less. 18. (A) Te = Te1 +

Te1 Te 3 + G1 G1 G2

Te 3 G2 = G1 ( Te - Te1 -

600 = Te 2 ) 10(190 - 150 G1

19. (B) F0 = 1 +

350 10

)

N clo =

= 12

1 1 ù Te 2 Te 3 é = Te1 ê1 + + 2ú + G G G1 G1 G2 ë û

Page 412

kT sys GR ( w)Wn 1.38(10 -23)( 820)(1011 )(10 7) = 2 pL 1738 .

-5 or 6.51 mW . = 651110

29. (C) dN ao = k[ Ta + TL ( L - 1)]

dw dw = kTs 2p 2p

Thus Ts = Ta + TL ( L - 1) = 60 + 275(1738 . - 1) = 263 K

T Te 3 100 280 21. (A) Te = Te1 + e 2 + = 40 + + = 58.33 K G1 G1 G2 8 8( 6)

or ( Te1 - Te ) G + Te1 G + Te1 = 0

= 13.67 or 11.4 dB

and WPV = 2 p(10 7) Hz

T 190 20. (C) Fop = 1 + e = 1 + = 4.8 or 6.81 dB Ts 50

2

Te T sys - Ta 820 - 60 =1 + =1 + 60 Ts Ts

28. (A) Here GR ( w0 ) = 1011 (or 110 dB)

Te 190 =1 + = 1.655 or 2.19 dB T0 290

22. (B) Te = Te1 +

We know that

30. (D) T e = T s ( F op - 1) = 60(5 - 1) = 240 K 31. (C) F o = 1 +

Te 240 =1 + = 1.8276 290 290

32. (A) F op = 1 + www.gatehelp.com

Te 240 =1 + = 9 or 9.54 dB 30 Ts


CHAPTER

7.4 AMPLITUDE MODULATION

7. An AM broadcast station operates at its maximum allowed total output of 50 kW with 80% modulation. The power in the intelligence part is

Statement for Question 1 - 3 An AM signal is represented by x( t) = (20 + 4 sin 500 pt) cos(2 p ´ 10 5 t) V 1. The modulation index is (A) 20 (C) 0.2

(B) 4 (D) 10

(B) 204 W (D) 416 W

3. The total sideband power is (A) 4 W (C) 16 W

AM

signal

has

(C) 6.42 kW


(A) 0.68

(B) 0.73

(C) 0.89

(D) None fo the above

9. A modulating signal is amplified by a 80% efficiency amplifier before being combined with a 20 kW carrier to generate an AM signal. The required DC input power to the amplifier, for the system to operate at 100% modulation, would be

(B) 8 W (D) 2 W

Statement for Question 4 - 5 : An

(B) 31.12 kW

8. The aerial current of an AM transmitter is 18 A when unmodulated but increases to 20 A when modulated.The modulation index is

2. The total signal power is (A) 208 W (C) 408 W

(A) 12.12 kW

the

form

(A) 5 kW

(B) 8.46 kW

fc = 10 5 Hz.

(C) 12.5 kW

(D) 6.25 kW

4. The modulation index is

10. A 2 MHz carrier is amplitude modulated by a 500 Hz modulating signal to a depth of 70%. If the unmodulated carrier power is 2 kW, the power of the modulated signal is

x( t) = [20 + 2 cos 3000 pt + 10 cos 6000 pt ]cos 2 pfc t

(A)

201 400

(B) - 201 400

(C)

199 400

(D) - 199 400

where

5. The ratio of the sidebands power to the total power is (A)

43 226

(C)

26 226

(B)

26 226

(D)

43 224

(A) 2.23 kW

(B) 2.36 kW

(C) 1.18 kW

(D) 1.26 kW

11. A carrier is simultaneously modulated by two sine

6. A 2 kW carrier is to be modulated to a 90% level. The total transmitted power would be

waves with modulation indices of 0.4 and 0.3. The

(A) 3.62 kW (C) 1.4 kW

(A) 1.0 (C) 0.5

Page 414

(B) 2.81 kW (D) None of the above

resultant modulation index will be

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(B) 0.7 (D) 0.35


Amplitude Modulation

Chap 7.4

12. In a DSB-SC system with 100% modulation, the power saving is

Which of the following frequencies will NOT be present in the modulated signal?

(A) 50%

(B) 66%

(A) 990 KHz

(B) 1010 KHz

(C) 75%

(D) 100%

(C) 1020 KHz

(D) 1030 KHz

13. A 10 kW carrier is sinusoidally modulated by two carriers corresponding to a modulation index of 30% and 40% respectively. The total radiated power(is (A) 11.25 kW

(B) 12.5 kW

(C) 15 kW

(D) 17 kW

14. In amplitude modulation, the modulation envelope has a peak value which is double the unmodulated carrier value. What is the value of the modulation index ? (A) 25% (C) 75%

(A) 695 kHz

(B) 700 kHz

(C) 705 kHz

(D) 710 kHz

21. A message signal m( t) = sinc t + sinc 2 ( t) modulates the carrier signal ( t) = A cos 2pfc t. The bandwidth of the modulated signal is (A) 2 fc

(B) 50% (D) 100%

(C) 2

15. If the modulation index of an AM wave is changed from 0 to 1, the transmitted power (A) increases by 50% (C) increases by 100%

20. For an AM signal, the bandwidth is 10 kHz and the highest frequency component present is 705 kHz. The carrier frequency used for this AM signal is

(B) increases by 75% (D) remains unaffected

(B)

1 2

(D)

1 4

fc

22. The signal m( t) = cos 2000 pt + 2 cos 4000 t is multiplied by the carrier c( t) = 100 cos 2pfc t where fc = 1 MHz to produce the DSB signal. The expression for the upper side band (USB) signal is (A) 100 cos(2 p( fc + 1000) t) + 200 cos(2 p( fc + 200) t)

16. A diode detector has a load of 1 kW shunted by a 10000 pF capacitor. The diode has a forward resistance

(B) 100 cos(2 p( fc - 1000) t) + 200 cos(2 p( fc - 2000) t)

of 1 W. The maximum permissible depth of modulation,

(C) 50 cos(2 p( fc + 1000) t) + 100 cos(2 p( fc + 2000) t)

so as to avoid diagonal clipping, with modulating signal

(D) 50 cos(2 p( fc - 1000) t) + 100 cos(2 p( fc - 100) t)

frequency fo 10 kHz will be (A) 0.847 (C) 0.734


Statement for Question 23-26 : The Fourier transform

M ( f ) of a signal m( t) is

17. An AM signal is detected using an envelop detector.

shown in figure. It is to be transmitted from a source to

The carrier frequency and modulating signal frequency

destination. It is known that the signal is normalized,

are 1 MHz and 2 kHz respectively. An appropriate value

meaning that -1 £ m( t) £ 1 M( f)

for the time constant of the envelope detector is. (A) 500 m sec

(B) 20 m sec

(C) 0.2 m sec

(D) 1 m sec -10000

18. An AM voltage signal s( t), with a carrier frequency of 1.15 GHz has a complex envelope g( t) = AC [1 + m( t)], where Ac = 500 V, and the modulation is a 1 kHz sinusoidal test tone described by m( t) = 0.8 sin(2 p ´ 10 3 t) appears across a 50 W resistive load. What is the actual power dissipated in the load ? (A) 165 kW

(B) 82.5 kW

(C) 3.3 kW

(D) 6.6 kW

10000

f

Fig.P7.4.23-26

23. If USSB is employed, the bandwidth of the modulated signal is (A) 5 kHz

(B) 20 kHz, 10 kHz

(C) 20 kHz


24. If DSB is employed, the bandwidth of the modulated signal is

19. A 1 MHz sinusoidal carrier is amplitude modulated by a symmetrical square wave of period 100 m sec.

(A) 5 kHz

(B) 10 kHz

(C) 20 kHz


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Page 415


UNIT 7

25. If an AM modulation scheme with a = 0.8 is used, the bandwidth of the modulated signal is. (A) 5 kHz

(B) 10 kHz

(C) 20 kHz


m(t)


S

x(t) Square- Law Device

y(t) Filter

AM Signal

cos wct

Fig.P7.4.30-31

26. If an FM signal with kf = 60 kHz is used, then the bandwidth of the modulated signal is

30. The filter should be a

(A) 5 kHz

(B) 10 kHz

(A) LPP with bandwidth W

(C) 20 kHz


(B) LPF with bandwidth 2W

27. A DSB modulated signal x( t) = Am( t) cos 2pfc t is mixed

(multiplied)

with

a

local

carrier

xL ( t) = cos(2pfc t + q) and the output is passed through a LPF with a bandwidth equal to the bandwidth of the message m( t). If the power of the signal at the output of the low pass filter is pout and the power of the modulated signal by pu , the

p out pu

is

(A) 0.5 cos q

(B) cos q

(C) 0.5 cos q

(D)

2

cos 2 q

28. A DSB-SC signal is to be generated with a carrier frequency fc = 1 MHz using a non-linear device with the input-output characteristic vo = a0 vi + a1 vi3 where a0 and a1 are constants. The output of the non-linear device can be filtered by an appropriate band-pass filter. Let vi = Ac¢ cos(2pfc¢t) + m( t) where m( t) is the message signal. Then the value of fc¢(in MHz) is (A) 1.0

(B) 0.333

(C) 0.5

(D) 3.0

31. The modulation index is 2b 2a (B) (A) Am Am a b a Am b

b Am a

(D)

32. A message signal is periodic with period T, as shown in figure. This message signals is applied to an AM modulator with modulation index a = 0.4. The modulation efficiency would be m(t) K1 T t

-K1

Fig.P7.4.32

29. A non-linear device with a transfer characteristic given by i = (10 + 2 vi + 0.2 vi2 ) mA is supplied with a carrier of 1 V amplitude and a sinusoidal signal of 0.5 V amplitude in series. If at the output the frequency component of AM signal is considered, the depth of modulation is (A) 18 %

(D) a BPF with center frequency f0 and BW = W such that f0 - Wm > f0 - W2 > Wm

(C)

2

1 2

(C) a BPF with center frequency f0 and BW = W such that f0 - Wm > f0 - W2 > 2Wm

(A) 51 %

(B) 11.8 %

(C) 5.1 %


Statement for Question 33-36 The figure 6.54-57 shows the positive portion of the envelope of the output of an AM modulator. The message signal is a waveform having zero DC value. m(t)

(B) 10 %

45

(C) 20 %

(D) 33.33 %

30 15


t

Consider the system shown in figP7.4.30-31. The modulating signal m( t) has zero mean and its maximum (absolute) value is Am = max m( t) . It has bandwidth Wm . The nonlinear device has a input-output characteristic y( t) = ax( t) + bx 2 ( t). Page 416

Fig.P7.4.33-36

33. The modulation index is (A) 0.5

(B) 0.6

(C) 0.4

(D) 0.8

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Chap 7.4

34. The modulation efficiency is

41. The lower sideband of the SSB AM signal is

(A) 8.3 %

(B) 14.28 %

(A) -100 cos(2 p( fc - 1000) t) + 200 sin(2 p( fc - 1000) t)

(C) 7.69 %


(B) -100 cos(2 p( fc - 1000) t) - 200 sin(2 p( fc - 1000) t)

35. The carrier power is

(C) 100 cos(2 p( fc - 1000) t) - 200 sin(2 p( fc - 1000) t)

(A) 60 W

(B) 450 W

(C) 30 W

(D) 900 W

(D) 100 cos(2 p( fc - 1000) t) + 200 sin(2 p( fc - 1000) t) Statement for Question 42-43

36. The power in sidebands is (A) 85 W

Consider the system shown in figure 6.69-70. The

(B) 42.5 W

(C) 56 W

average value of m( t) is zero and maximum value of

(D) 37.5 W

m( t) is M. The square-law device is defined by 37. In a broadcast transmitter, the RF output is represented as e( t) = 50[1 + 0.89 cos 5000 t + 0.30 sin 9000 t ]cos( 6 ´ 106 t)V

y( t) = 4 x( t) + 10 x( t). S

m(t)

x(t) Square- Law Device

y(t) Filter

AM Signal

What are the sidebands of the signals in radians ? cos wct

(A) 5 ´ 10 3 and 9 ´ 10 3 (B) 5.991 ´ 106 , 5.995 ´ 106 , 6.005 ´ 106 and 6.009 ´ 106 (C) 4 ´ 10 3, 1.4 ´ 10 4 (D) 1 ´ 10 , 11 . ´ 10 , 3 ´ 10 , and 15 . ´ 10 6

7

6

42. The value of M, required to produce modulation index of 0.8, is

7

38. An AM modulator has output x( t) = 40 cos 400 pt + 4 cos 360 pt + 4 cos 440 pt The modulation efficiency is (A) 0.01

(B) 0.02

(C) 0.03

(D) 0.04

Fig. P7.4.42-43

(A) 0.32

(B) 0.26

(C) 0.52

(D) 0.16

43. Let W be the bandwidth of message signal m( t). AM signal would be recovered if

39. An AM modulator has output

(A) fc > W

(B) fc > 2W

(C) fc ³ 3W

(D) fc > 4W

x( t) = A cos 400 pt + B cos 380 pt + B cos 420 pt The carrier power is 100 W and the efficiency is 40%. The value of A and B are (A) 14.14, 8.16

(B) 50, 10

(C) 22.36, 13.46


44. A super heterodyne receiver is designed to receive transmitted signals between 5 and 10 MHz. High-side tuning is to be used. The tuning range of the local oscillator for IF frequency 500 kHz would be (A) 4.5 MHz - 9.5 MHz


(B) 5.5 MHz - 10.5 MHz

A single side band signal is generated by modulating signal of 900-kHz carrier by the signal

(C) 4.5 MHz - 10.5 MHz

m( t) = cos 200 pt + 2 sin 2000 pt. The amplitude of the


carrier is Ac = 100.

(B) - sin(2 p1000 t) + 2 cos(2000 pt)

45. A super heterodyne receiver uses an IF frequency of 455 kHz. The receiver is tuned to a transmitter having a carrier frequency of 2400 kHz. High-side tuning is to be used. The image frequency will be

(C) sin(2 p1000 t) + 2 cos(1000 t)

(A) 2855 kHz

(B) 3310 kHz

(C) 1845 kHz

(D) 1490 kHz

$ ( t) is 40. The signal m (A) - sin(2 p1000 t) - 2 cos(2000 pt)

(D) sin(2 p1000 t) - 2 cos(2 p1000 t)

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Page 417


UNIT 7

46. In the circuit shown in fig.P7.4.46, the transformers are center tapped and the diodes are connected as shown in a bridge. Between the terminals 1 and 2 an a.c. voltage source of frequency 400 Hz is connected. Another a.c. voltage of 1.0 MHz is connected between 3 and 4. The output between 5 and 6 contains components at


49. In fig.P7.4.49 m( t) =

2 sin 2 pt sin 199pt , s( t) = cos 200 pt and n( t) = t t Multiplier

3

LPF 1 Hz

y(t)

s(t)

n(t)

Fig.P7.4.49

The output y( t) will be sin 2pt (A) t

4

6

2

Multiplier

|H(jw)|=1 s(t)

5

1

Adder

S

m(t)

Fig.P7.4.46

(A) 400 Hz, 1.0 MHz, 1000.4 kHz, 999.6 kHz

(B)

sin 2 pt sin pt + cos 3pt t t

(C)

sin 2 pt sin 0.5 pt + cos 15 . pt t t

(D)

sin 2 pt sin pt + cos 0.75 pt t t

(B) 400 Hz, 1000.4 kHz, 999.6 kHz (C) 1 MHz, 1000.4 kHz, 999.6 kHz (D) 1000.4 kHz, 999.6 kHz 47. A superheterodyne receiver is to operate in the frequency range 550 kHz-1650 kHz, with the denote intermediate frequency of 450 kHz. Let R = CCmax min the required capacitance ratio of the local oscillator and I denote the image frequency (in kHz) of the incoming signal. If the receiver is tuned to 700 kHz, then

50. 12 signals each band-limited to 5 kHz are to be transmitted over a single channel by frequency division multiplexing. If AM -SSNB modulation guard band of 1 kHz is used, then the bandwidth of the multiplexed signal will be (A) 51 kHz

(B) 61 kHz (D) 81 kHz

(A) R = 4.41, I = 1600

(B) R = 2.10, I = 1150

(C) 71 kHz

(C) R = 3, I = 1600

(D) R = 9.0, I = 1150

51. Let x( t) be a signal band-limited to 1 kHz. Amplitude modulation is performed to produce signal A proposed demodulation g( t) = x( t) sin 2000pt. technique is illustrated in figure 6.83. The ideal low pass filter has cutoff frequency 1 kHz and pass band gain 2. The y( t) would be

48. Consider a system shown in Figure . Let X ( f ) and Y ( f ) denote the Fourier transforms of x( t) and y( t) respectively. The ideal HPF has the cutoff frequency 10 kHz. The positive frequencies where Y ( f ) has spectral peaks are

(A) 2 y( t) x(t)

HPF 10 kHz

Balanced Modulator

Balanced Modulator

~

~

10 kHz

13 kHz X(f )

-3

-1

1

Fig.P7.4.48

(A) 1 kHz and 24 kHz (B) 2 kHz and 24 kHz (C) 1 kHz and 14 kHz (D) 2 kHz and 14 kHz Page 418

3

f (kHz)

y(t)

(C)

1 2

y( t)

(B) y( t) (D) 0

52. Suppose we wish to transmit the signal x( t) = sin 200 pt + 2 sin 400 pt using a modulation that create the signal g( t) = x( t) sin 400pt. If the product g( t) sin 400pt is passed through an ideal LPF with cutoff frequency 400p and pass band gain of 2, the signal obtained at the output of the LPF is 1 2

(A) sin 200pt

(B)

(C) 2 sin 200pt

(D) 0

sin 200 pt

53. In a AM signal the received signal power is 10 -10 W with a maximum modulating signal of 5 kHz. The noise spectral density at the receiver input is 10 -18 W/Hz. If the noise power is restricted to the message signal

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bandwidth only, the signals-to-noise ratio at the input to the receiver is (A) 43 dB

(B) 66 dB

(C) 56 dB

(D) 33 dB

Chap 7.4

SOLUTION 1. (C) u( t) = (20 + 4 sin 500 pt) cos(2 p ´ 10 5 t) V = 20(1 + 0.2 sin 500 pt) cos(2 p ´ 10 5 t) V,

a = 0.2

Statement for Question 54-55 Consider the following Amplitude Modulated (AM)

2. (B) Pc =

signal, where fm < B x AM ( t) = 10(1 + 0.5 sin 2 pfm t) cos 2 pfc t.

(B) 12.5

(C) 6.25

(D) 3.125

25 2 N0 B

(D)

4. (B) x( t) = [20 + 2 cos(2 p1500 t) + 10 cos(2 p3000 t)]cos(2 pfc t) 1 1 æ ö = 20ç 1 + cos(2 p1500 t) + cos(2 p3000 t) cos(2 pfc t) ÷ 10 2 è ø

55. The AM signal gets added to a noise with Power Spectra Density Sn ( f ) given in the figure below. The ration of average sideband power to mean noise power would be 25 25 (B) (A) 8 N0 B 4 N0 B (C)

25 N0 B

Statement for Question 56-57 A certain communication channel is characterized by 80 dB attenuation and noise power-spectral density of 10 -10 W/Hz. The transmitter power is 40 kW and the message signal has a bandwidth of 10 kHz. 56. In the case of conventional AM modulation, the predetecion SNR is

This is the form of a conventional AM signal with message signal 1 1 m( t) = cos(2 p1500 t) + cos(2 p3000 t) 10 2 1 1 = cos 2 (2 p1500 t) + cos(2 p1500 t) 10 2 1 1 The minimum of g( z) = z 2 + is achieved for z10 2 1 201 1 and it is min( g( z)) = . Since z = is in z =20 400 20 the range of cos (2 p1500 t), we conclude that the 201 minimum value of m( t) is . Hence, the modulation 400 201 index is a = 400 5. (B) x( t) = 20 cos(2 pfc t) + cos(2 p( fc - 1500) t) + cos(2 p( fc - 1500) t)

(A) 108

(B) 2 ´ 108

= 5 cos(2 p( fc 3000) t) + 5 cos(2 p( fc + 3000) t)

(C) 10 2

(D) 2 ´ 10 2

The power in the sidebands is 1 1 25 25 + = 26 Psidebands = + + 2 2 2 2

57. In case of SSB, the predetecion SNR is (A) 2 ´ 10 2

(B) 4 ´ 10 2

(C) 2 ´ 10 3

(D) 4 ´ 10 3

*************

ö ÷÷ = 204 W ø

3. (A) Psb = Pt - Pc = 204 - 200 = 4 W

54. The average side-band power for the AM signal given above is (A) 25

æ (0.2) 2 Pt = Pc çç 1 + 2 è

20 2 = 200 W, 2

The total power is Ptotal = Pcarrier + Psidebands = 200 + 26 = 226 The ratio of the sidebands power to the total power is Psidebands 26 = Ptotal 226 æ æ 0.9 2 ö a2 ö 6. (B) Pt = Pc çç 1 + ÷÷ = 2000çç 1 + ÷ = 2810 W 2 ÷ø 2 ø è è æ a2 ö 7. (A) Pt = Pc çç 1 + ÷ 2 ÷ø è Pc = 37.88 kW, www.gatehelp.com

or

æ 0.8 2 ö ÷ 50 ´ 10 3 = Pc çç 1 + 2 ÷ø è

Pi = ( Pt - Pc ) = (50 - 37.88) = 12.12 kW Page 419


UNIT 7

1

1

æ æ a 2 ö2 a 2 ö2 8. (A) I t = I c çç 1 + ÷÷ or 20 = 18çç 1 + ÷ or a = 0.68 2 ÷ø 2 ø è è

18. (A) Pt =

Pt =

500 2 2

Pi = 30 - 20 = 10 kW The DC input power =

10 = 12.5 kW. 0.8

m( t) =

1

é 4 ¥ ( -1) n -1 ù x( t) = ê å cos[2 pfm (2 n - 1)]ú sin(2 p1000 ´ 10 3 t) ë p n =1 2 n - 1 û So frequency component (106 ± fm (2 n - 1) will be present

2 P. If carrier is 3

2 suppressed then P or 66% power will be saved. 3

where n = 1, 2, 3, .... For fm = 10 kHz and n = 1 & 2 frequency present is 990, 970, 1030 kHz.

Thus 1020 kHz will be absent.

20. (B) fc + fm = 705 kHz,

ö ÷÷ ø

BW = 2 fm = 10 kHz or fm = 5 kHz fc = 705 - 5 = 700 kHz

= 1125 . kW

21. (C) x( t) = m( t) c( t) = A(sinc ( t) + sinc 2 ( t) cos(2pfc t)

14. (D) x( t) = Ac (1 + a cos 2 pfm t) cos 2 pfc t Thus a =1, therefor modulation

Here Ac (1 + a) = 2 Ac ,

4 ¥ ( -1) n -1 cos[2 pfm (2 n - 1) ] å p n =1 2 n - 1

The modulated output

11. (C) a 2 = a1a + a 22 = 0.32 + 0.4 2 = 0.5 2 or a = 0.5

ö æ 0.32 0.4 2 ÷÷ = 10çç 1 + + 2 2 ø è

fc = 1000 kHz, x( t) = c( t) m( t)

Expressing square wave as modulating signal m( t)

æ æ 0.7 2 ö a 2 ö2 ÷÷ = 2çç 1 + ÷ = 2.23 kW Pt = Pc çç 1 + 2 ÷ø 2 ø è è

æ a2 a2 13. (A) Pt = Pc çç 1 + 1 + 2 2 2 è

é 0.8 2 ù = 165 kW 1 + ê 2 úû ë

19. (C) c( t) = sin 2p fc t,

a = 70% = 0.7

12. (B) In previous solution Pc =

2 a 2 m( t) ù Ac2 é ê1 + ú 2 ê 2 úû ë

Here modulation index a =1. Thus

1ö æ 9. (C) Pt = 20000ç 1 + ÷, Pt = 30 kW, 2ø è

10. (A) Pc = 2 kW,


Taking the Fourier transform of both sides, we obtain

index is 1 or 100% modulation.

X( f ) =

15. (A) If modulation index a is 0, then

=

A [ P( f ) + L( f )] * ( d( f - fc ) + d( f + fc )) 2

A [ P( f - fc ) + L( f - fc ) + P( f + fc ) + L( f - fc )] 2

æ 0 2 ö Ac2 ÷= çç 1 + 2 ÷ø 2 è

1 Since P( f - fc ) ¹ 0 for f - fc < , whereas L( f - fc ) ¹ 0 2

If modulation index is 1 then

for f - fc < 1. Hence, the bandwidth of the bandpass

Pt1 =

Pt2 =

Ac2 2

Ac2 2

æ 12 ö 3 2 ÷ = Ac , çç 1 + 2 ÷ø 4 è

filter is 2.

Pt 2 3 = Pt1 2

22. (C) x( t) = m( t) c( t)

Thus Pt2 = 15 . Pt1 and Pt2 is increases by 50%

= 100[cos(2 p1000 t) + 2 cos(2 p2000 t)]cos(2 pfc t) = 100 cos(2 p1000 t) cos(2 pfc t) + 200 cos(2 p2000 t) cos(2 pfc t) 100 = [cos(2 p( fc + 1000) t) + cos(2 p( fc - 1000) t)] 2

16. (A) fm = 10 kHz, R = 1000 W, C = 10000 pF Hence 2pfm RC = 2 p ´ 10 4 ´ 10 3 ´ 10 -8 = 0.628 a max = (1 + (0.628) 2 ) 17. (B)

-

1 2

= 0.847

1 1 , £ RC £ fc BWm

+

200 [cos(2 p( fc + 2000) t) + cos(2 p( fc - 2000) t)] 2

Thus, the upper sideband (USB) signal is

Here fc = 1 MHz

xu ( t) = 50 cos[2 p( fc + 1000) t ] + 100(2 p( fc + 2000) t)

Signal Bandwidth BWm = 2 fm = 2 ´ 2 ´ 10 3 = 4 kHz Thus

1 1 or 10 -6 £ RC £ 250 ms £ RC £ 106 4 ´ 10 3

Thus appropriate value is 20 m sec Page 420

23. (B) When USSB is employed the bandwidth of the modulated signal is the same with the bandwidth of the message signal. Hence WUSSB = W = 10 4 Hz www.gatehelp.com


UNIT 7

37.

(B)

Sidebands

are

( 6 ´ 106 ± 5000)

and


43. (C) The filter characteristic is shown in fig.S7.4.43

( 6 ´ 10 ± 9000) 6

H(f )

Thus 6.005 ´ 10 , 5.995 ´ 10 , 5.991 ´ 10 or 5.991 ´ 10 , 6

6

6

6

m(t)

6.005 ´ 106 and 6.009 ´ 106 38. (B) x( t) can be written as

f W

x( t) = ( 40 + 8 cos 40 pt) cos 400 pt 8 modulation index a = = 0.2 40 1 Pc = ( 40) 2 = 800 W 2

fc + W < 2 f or fc > W

1 1 Psb = B 2 + B 2 = 66.67 2 2

the

Hilbert

44. (B) Since High-side tuning is used fLO = fm + f IF = 500 kHz, fLOL = 5 + 0.5 = 5.5 MHz,

45. (B) fimage = fL + 2 f IF = 2400 = 3310 kHz 46. (D) The given circuit is a ring modulator. The output

or B = 8.161

is DSB-SC signal. So it will contain m( t) cos( nwc t) where n = 1, 2, 3...... Therefore there will be only (1 MHz ± 400

transform

of

sin (2 p1000 t) is cos (2 p1000 t)

Hz) frequency component. 47. (A) fmax = 1650 + 450 = 2100 kHz 1

$ ( t)= sin (2 p1000 t) - cos (2 p1000 t) Thus m

fmin = 550 + 450 = 1000 kHz.

41. (D) The expression for the LSSB AM signal is.

frequency is minimum, capacitance will be maximum

x l ( t) = Ac m( t) cos(2 pfc t) + Ac m( t) sin(2 pfc t)

R=

Substituting $ ( t) = sin(2 p1000 t) - 2 cos(2 p1000 t) and m

= 100[cos(2 p1000 t) cos(2 pfc t) + sin(2 p1000 t) sin(2 pfc t)] +200[cos(2 pfc t) sin(2 p1000 t) - sin(2 pfc t) cos(2 p1000 t)] = 100 cos(2 p( fc - 1000) t) - 200 sin(2 p( fc - 1000) t)

= 4m(t) + 4 cos wc t + 10m2 (t) + 20m(t) cos wc t + 5 + 5 cos 2wc t

xc ( t) = 4[1 + 5 Mmn ( t)]cos wc t 5M = 0.8

or

M = 0.16

component of 11 kHz will remain. The output of 2nd modulator will be (13 ± 11) kHz. So Y ( f ) has spectral peak at 2 kHz and 24 kHz. 49. (C) m( t) s( t) = y1 ( t)

42. (D) y( t) = 4( m( t) + cos wc t) + 10( m( t) + cos wc t) 2

m( t) = Mmn ( t)

or R = 4.41

kHz and (10 - 1) kHz. After passing the HPF frequency

+100[sin(2 p1000 t) - 2 cos(2 p1000 t) sin(2 pfc t)]

xc ( t) = 4[1 + 5 m( t)]cos wc t

2p LC

output of first modulator spectral peak will be at (10 + 1)

xl ( t) = 100[cos(2 p1000 t) + 2 sin(2 p1000 t) cos(2 pfc t)]

The AM signal is,

f =

48. (B) Since X ( f ) has spectral peak at 1 kHz so at the

we obtain

= 5 + 4m(t) + 10m2 (t) + 4[1 + 5m(t)] cos wc t + 5 cos 2wc t

2 Cmax fmax = 2 = (2.1) 2 Cmin fmin

or

fi = fc + 2 f IF = 700 + 2( 455) = 1600 kHz

Ac = 100, m( t) = cos(2 p1000 t) + 2 sin(2 p1000 t)

Page 422

Therefore fc > 3W

fLOU = 10 + 0.5 = 10.5 MHz

40. (D) The Hilbert transform of cos (2 p1000 t) is whereas

2fc

fc+W

fc - W > 2W or fc > 3W ,

A2 39. (A) Carrier power Pc = = 100 W, A = 14.14 2 40 Psb Psb or Eeff = = 0.4 = 100 + Psb Pc + Psb 100

sin (2 p1000 t),

fc-W fc

Fig.S7.4.43

The components at 180 Hz and 220 Hz are side band 1 1 Psb = ( 4) 2 + ( 4) 2 = 16 W, 2 2 Psb 16 Eeff = = Pc + Psb 800 + 16

Psb = 66.67 W,

2W

=

2 sin(2 pt) cos(200 pt) sin(202 pt) - sin(198 pt) = t t

y1 ( t) + n( t) = y2 ( t) =

sin 202 pt - sin 198 pt sin 198 pt + t t

y2 ( t) s( t) = y( t) =

[sin 202 pt - sin 198 pt + sin 199 pt ]cos 200 pt t

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=

Chap 7.4

55. (D) Noise power = Area rendered by the spectrum

1 [sin( 402 pt) + sin(2 pt) - {sin( 398 pt) - sin(2 pt)} 2 + sin( 399pt) - sin( pt)]

After filtering sin(2 pt) + sin(2 pt) - sin( pt) 2t sin(2 pt) + 2 sin(0.5 t) cos(15 . pt) = 2t sin 2 pt sin 0.5 pt = + cos 15 . pt 2t t y( t) =

= N0 B Ratio of average sideband power to mean noise 6.25 25 Power = = N0 B 4 N0 B 56. (C) Since the channel attenuation is 80 db, then P 10 log T = 80 PR or PR = 10 -8 PT = 10 -8 ´ 40 ´ 10 3 = 4 ´ 10 -4 Watts If the noise limiting filter has bandwidth B, then the

50. (D) The total signal bandwidth = 5 ´ 12 = 60 kHz

pre-detection noise power is

There would be 11 guard band between 12 signal. So

fc +

Pn = 2

guard band width = 11 kHz Total band width = 60 + 11 = 71 kHz

ò

fc -

B 2

B 2

N0 df = N 0 B = 2 ´ 10 -10 B Watts 2

In the case of DSB or conventional AM modulation,

51. (D) x1 ( t) = g( t) cos(2000pt)

B = 2W = 2 ´ 10 4

1 = x( t) sin(2000 pt) cos(2000 pt) = x( t) sin( 4000 pt) 2 1 X1 ( jw) = X ( j( w - 4000 p)) - X ( j( w + 4000 p)) 4j

Hz, whereas in SSB modulation

B = W = 10 . Thus, the pre-detection signal to noise 4

ratio in DSB and conventional AM is SNR DSB,AM =

This implies that X1 ( jw) is zero for w £ 2000 p because w < 2 pfm = 2 p1000. When x1 ( t) is passed through a LPF with cutoff frequency 2000p, the output will be zero.

PR 4 ´ 10 -4 = = 10 2 Pn 2 ´ 10 -10 ´ 2 ´ 10 4

57. (A) In SSB modulation B = W = 10 4 SNR SSB =

52. (A) y( t) = g( t) sin( 400 pt) = x( t) sin 2 ( 400 pt) (1 - cos)( 800 pt) = (sin(200 pt) + 2 sin( 400 pt) 2 1 = [sin(200 pt) - sin(200 pt) cos( 800 pt) + 2 sin( 400 pt) 2

4 ´ 10 -4 = 2 ´ 10 2 2 ´ 10 -10 ´ 10 4

***********

-sin( 400 pt) cos( 800 pt) 1 1 = sin(200 pt) - [sin(1000 pt) - sin( 6000 pt)] 2 4 1 + sin( 400 pt) - [sin(1200 pt) - sin( 400 pt)] 4 If this signal is passed through LPF with frequency 400p and gain 2, the output will be sin(200pt) 53. (A) Message signal BW fm = 5 kHz Noise power density is 10 -18 W/Hz Total noise power is 10 -18 ´ 5 ´ 10 3 = 5 ´ 10 -15 W Input signal-to-noise ratio SNR=

10 -10 = 2 ´ 10 4 or 43 dB 5 ´ 10 -15

54. (C) Average side band power is Ac2 a 2 10 2 (0.5) 2 = = 6.25 W 4 4

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CHAPTER

7.6 DIGITAL TRANSMISSION



Fig. P7.6.1-5 shows fourier spectra of signal x( t) and y(t). Determine the Nyquist sampling rate for the

A signal x( t) is multiplied by rectangular pulse train c( t) shown in fig.P7.6.6-7.. c(t)

given function in question. X( jw)

0.25 ms

Y( jw)

-2´10-3

-10-3

w

w

Fig.P7.6.1-5

10-3

2´10-3

t

Fig.P7.6.6-7

6. x( t) would be recovered form the product. x( t) c( t) by using an ideal LPF if X ( jw) = 0 for

1. x( t) (A) 100 kHz

(B) 200 kHz

(A) w > 2000 p

(B) w > 1000 p

(C) 300 kHz

(D) 50 kHz

(C) w < 1000 p

(D) w < 2000 p

7. If X ( jw) satisfies the constraints required, then the

2. y( t)

pass band gain A of the ideal lowpass filter needed to

(A) 50 kHz

(B) 75 kHz

recover x( t) from e( t) x( t) is

(C) 150 kHz

(D) 300 kHz

(A) 1

(B) 2

(C) 4

(D) 8

3. x 2 ( t) (A) 100 kHz

(B) 150 kHz

(C) 250 kHz

(D) 400 kHz

4. y 3( t) (A) 100 kHz

(B) 300 kHz

(C) 900 kHz

(D) 120 kHz

8. Consider a set of 10 signals xi ( t), i = 1, 2, 3, ...10.. Each signal is band limited to 1 kHz. All 10 signals are to be time-division multiplexed after each is multiplied by a carrier e( t) shown in Figure. If the period T of e( t) is chosen the have the maximum allowable value, the largest value of D would be c(t)

D

5. x( t) y( t) (A) 250 kHz

(B) 500 kHz

(C) 50 kHz

(D) 100 kHz

Page 434

-2T

-T

0

Fig.P7.6.8

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T

2T

t


Digital Transmission

(A) 5 ´ 10 -3 sec -5

(C) 5 ´ 10 sec

(B) 5 ´ 10 -4 sec

Chap 7.6

15. A CD record audio signals digitally using PCM. The audio signal bandwidth is 15 kHz. The Nyquist samples

-6

(D) 5 ´ 10 sec

are quantized into 32678 levels and then binary coded. 9. A compact disc recording system samples a signals

The minimum number of binary digits required to

with a 16-bit analog-to-digital convertor at 44.1 kbits/s.

encode the audio signal

The CD can record an hours worth of music. The

(A) 450 k bits/sec

(B) 900 k bits/sec

approximate capacity of CD is

(C) 980 340 k bits/sec

(D) 490 170, k bits/sec

(A) 705.6 M Bytes

(B) 317.5 M Bytes

(C) 2.54 M Bytes

(D) 5.43 M Bytes

16. The American Standard Code for Information Interchange has 128 characters, which are binary

10. An analog signal is sampled at 36 kHz and

coded. If a certain computer generates 1,000,000

quantized into 256 levels. The time duration of a bit of

character per second, the minimum bandwidth required

the binary coded signal is

to transmit this signal will be

(A) 5.78 ms

(B) 3.47 ms

(A) 1.4 M bits/sec

(B) 14 M bits/sec

(C) 6.43 ms

(D) 7.86 ms

(C) 7 M bits/sec

(D) 0.7 M bits/sec

11. An analog signal is quantized and transmitted using

17. A binary channel with capacity 36 k bits/sec is

a PCM system. The tolerable error in sample amplitude

available for PCM voic transmission. If signal is band

is 0.5% of the peak-to-peak full scale value. The

limited to 3.2 kHz, then the appropriate values of

minimum binary digits required to encode a sample is

quantizing level L and the sampling frequency will be

(A) 5

(B) 6

(A) 32, 3.6 kHz

(B) 64, 7.2 kHz

(C) 7

(D) 8

(C) 64, 3.6 kHz

(D) 32, 7.2 kHz

18. Fig.P7.4.18 shows a PCM signals in which

Statement for Question 12-13. Ten telemetry signals, each of bandwidth 2kHz, are to be transmitted simultaneously by binary PCM. The maximum tolerable error in sample amplitudes is 0.2% of the peak signal amplitude. The signals must be

amplitude level of + 1 volt and - 1 volt are used to represent binary symbol 1 and 0 respectively. The code word used consists of three bits. The sampled version of analog signal from which this PCM signal is derived is

sampled at least 20% above the Nyquist rate. Framing and synchronizing requires an additional 1% extra bits. Fig.P7.4.18

12. The minimum possible data rate must be (A) 272.64 k bits/sec

(B) 436.32 k bits/sec

(A) 4 5 1 2 1 3

(B) 8 4 3 1 2

(C) 936.32 k bits/sec


(C) 6 4 3 1 7

(D) 1 2 3 4 5

13. The minimum transmission bandwidth is

19. A PCM system uses a uniform quantizer followed by

(A) 218.16 kHz

(B) 468.32 kHz

a 8-bit encoder. The bit rate of the system is equal to 108

(C) 136.32 kHz


bits/s. The maximum message bandwidth for which the system operates satisfactorily is

14. A Television signal is sampled at a rate of 20% above the Nyquist rate. The signal has a bandwidth of 6 MHz. The samples are quantized into 1024 levels. The

(A) 25 MHz

(B) 6.25 MHz

(C) 12.5 MHz

(D) 50 MHz

minimum bandwidth required to transmit this signal

20. Twenty-four voice signals are sampled uniformly at

would be

a rate of 8 kHz and then time-division multiplexed. The

(A) 72 M bits/sec

(B) 144 M bits/sec

(C) 72 k bits/sec

(D) 144 k bits/sec

sampling process uses flat-top samples with 1 ms duration. The multiplexing operating includes provision

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UNIT 7


for synchronization by adding and extra pulse of 1 ms

quantized into 256 level using a m-low quantizer with

duration. The spacing between successive pulses of the

m = 225.

multiplexed signal is (A) 4 ms

(B) 6 ms

(C) 7.2 ms

(D) 8.4 ms

26. The signal-to-quantization-noise ratio is

21. A linear delta modulator is designed to operate on speech signals limited to 3.4 kHz. The sampling rate is 10 time the Nyquist rate of the speech signal. The step size d is 100 m V. The modulator is tested with a this test signal required to avoid slope overload is (A) 2.04 V

(B) 1.08 V

(C) 4.08 V

(D) 2.16 V

a

linear

DM

(B) 38.06 dB

(C) 42.05 dB

(D) 48.76 dB

27. It was found that a sampling rate 20% above the rate wou7ld be adequate. So the maximum SNR, that can be realized without increasing the transmission bandwidth, would be (A) 60.4 dB

(B) 70.3 dB

(C) 50.1 dB


28. For a PCM signal the compression parameter

Statement fo Question 22-23 : Consider

(A) 34.91 dB

m = 100 and the minimum signal to quantization-noise system

designed

to

ratio is 50 dB. The number of bits per sample would be.

accommodate analog message signals limited to bandwidth

(A) 8

(B) 10

of 3.5 kHz. A sinusoidal test signals of amplitude Amax = 1

(C) 12

(D) 14

V and frequency fm = 800 Hz is applied to system. The

29. A sinusoid massage signal m( t) is transmitted by

sampling rate of the system is 64 kHz. 22. The minimum value of the step size to

binary avoid

overload is

PCM

without

compression.

the

signal

to-quantization-noise ratio is required to be at least 48 dB, the minimum number of bits per sample will be

(A) 240 mV

(B) 120 mV

(A) 8

(B) 10

(C) 670 mV

(D) 78.5 mV

(C) 12

(D) 14

23. The granular-noise power would be (A) 1.68 ´ 10 -3 W (C) 2.48 ´ 10

If

-3

W

30. A speech signal has a total duration of 20 sec. It is

(B) 2.86 ´ 10 -4 W (D) 112 . ´ 10

-4

sampled at the rate of 8 kHz and then PCM encoded. The signal-to-quantization noise ratio is required to be

W

40 dB. The minimum storage capacity needed to 24. The SNR will be (A) 298 (C) 4.46 ´ 10

3

accommodate this signal is (B) 1.75´10 -3

(A) 1.12 KBytes

(B) 140 KBytes

(D) 201

(C) 168 KBytes


25. The output signal-to-quantization-noise ratio of a 10-bit

31. The input to a linear delta modulator having fa

PCM was found to be 30 dB. The desired SNR is 42 dB. It

step-size D = 0.628 is a sine wave with frequency fm and

can be increased by increasing the number of quantization

peak amplitude Em . If the sampling frequency fs = 40

level.In this way the fractional increase in the transmission

kHz, the combination of the sinc-wave frequency and

bandwidth would be (assume log 2 10 = 0.3)

the peak amplitude, where slope overload will take

(A) 20%

(B) 30%

piace is

(C) 40%

(D) 50%

Em (A) 0.3 V

fm 8 kHz

(B) 1.5 V

4 kHz

A signal has a bandwidth of 1 MHz. It is sampled

(C) 1.5 V

2 kHz

at a rate 50% higher than the Nyquist rate and

(D) 3.0 V

1 kHz

Statement for Question 26-27.

Page 436

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Chap 7.6

32. A sinusoidal signal with peak-to-peak amplitude of

38. Four signals g1 ( t), g 2 ( t), g s ( t) and g 4 ( t) are to be

1.536 V is quantized into 128 levels using a mid-rise

multiplexed and transmitted. g1 ( t) and g 4 ( t) have a

uniform quantizer. The quantization-noise power is

bandwidth of 4 kHz, and the remaining two signals

-6

(A) 0.768 V

(B) 48 ´ 10 V

(C) 12 ´ 10 -6 V 2

(D) 3.072 V

have bandwidth of 8 kHz,. Each sample requires 8 bit

2

for encoding. What is the minimum transmission bit rate of the system.

33. A signal is sampled at 8 kHz and is quantized using

(A) 512 kbps

(B) 16 kbps

for a

(C) 192 kbps

(D) 384 kbps

8 bit uniform quantizer. Assuming SNR q

sinusoidal signal, the correct statement for PCM signal 39. Three analog signals, having bandwidths 1200 Hz,

with a bit rate of R is

600 Hz and 600 Hz, are sampled at their respective

(A) R = 32 kbps, SNR q = 25.8 dB

Nyquist rates, encoded with 12 bit words, and time

(B) R = 64 kbps, SNR q = 49.8 dB

division multiplexed. The bit rate for the multiplexed

(C) R = 64 kbps, SNR q = 55.8 dB

signal is

(D) R = 32 kbps, SNR q = 49.8 dB

(A) 115.2 kbps

(B) 28.8 kbps

(C) 57.6 kbps

(D) 38.4 kbps

34. A 1.0 kHz signal is flat-top sampled at the rate of 180 samples sec and the samples are applied to an ideal rectangular LPF with cat-off frequency of 1100 Hz, then the output of the filter contains

40. The minimum sampling frequency (in samples/sec) required to reconstruct the following signal form its samples without distortion would be

(A) only 800 Hz component

3

2

(B) 800 and 900 Hz components

æ sin 2 p1000 t ö æ sin 2 p1000 t ö x( t) = 5ç ÷ + 7ç ÷ pt pt è ø è ø

(C) 800 Hz and 1000 Hz components

(A) 2 ´ 10 3

B) 4 ´ 10 3

(D) 800 Hz, 900 and 1000 Hz components

(C) 6 ´ 10 3

(D) 8 ´ 10 3

35. The Nyquist sampling interval, for the signal sinc (700 t) + sinc (500 t) is 1 (A) sec 350

p (B) sec 350

1 (C) sec 700

p (D) sec 175

41.

The

minimum

step-size

required

for

a

Delta-Modulator operating at32 K samples/sec to track the signal (here u( t) is the nuit function) x( t) = 125 t( u( t) - u( t - 1)) + (250 - 125 t)( u( t - 1) - u( t - 2)s so that slope overload is avoided, would be

36. A signal x( t) = 100 cos(24 p ´ 10 ) t is ideally sampled 3

(A) 2 -10

(B) 2 -8

(C) 2 -6

(D) 2 -4

with a sampling period of 50 m sec and then passed through an ideal lowpass filter with cutoff frequency of

42. Four signals each band limited to 5 kHz are

15 KHz. Which of the following frequencies is/are

sampled at twice the Nyquist rate. The resulting PAM

present at the filter output

samples are transmitted over a single channel after

(A) 12 KHz only

(B) 8 KHz only

time division multiplexing. The theoretical minimum

(C) 12 KHz and 9 KHz

(D) 12 KHz and 8 KHz

transmissions bandwidth of the channel should be equal to.

37. In a PCM system, if the code word length is

(A) 5 kHz

(B) 20 kHz

increased form 6 to 8 bits, the signal to quantization

(C) 40 kHz

(D) 80 kHz

noise ratio improves by the factor. (A) 8/6

(B) 12

(C) 16

(D) 8

43. Four independent messages have bandwidths of 100 Hz, 100 Hz, 200 Hz and 400 Hz respectively. Each is

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Page 437


UNIT 7


sampled at the Nyquist rate, time division multiplexed

50. A speech signal occupying the bandwidth of 300 Hz

and transmitted. The transmitted sample rate, in Hz,

to 3 kHz is converted into PCM format for use in digital

is given by

communication. If the sampling frequency is8 kHz and

(A) 200

(B) 400

each sample is quantized into 256 levels, then the

(C) 800

(D) 1600

output bit the rate will be

44.

The

Nyquist

sampling

rate

for

the

signal

g( t) = 10 cos(50 pt) cos 2 (150 pt). Where ' t ' is in seconds, is

(A) 3 kb/s

(B) 8 kb/s

(C) 64 kb/s

(D) 256 kb/s

(A) 150 samples per second

51. If the number of bits in a PCM system is increased

(B) 200 samples per second

from n to n + 1, the signal-to-quantization noise ratio

(C) 300 samples per second

will increase by a factor.

(D) 350 samples per second

(A)

45. A TDM link has 20 signal channels and each

(C) 2

channel is sampled 8000 times/sec. Each sample is represented by seven binary bits and contains an additional bit for synchronization. The total bit rate for the TDM link is (A) 1180 K bits/sec

(B) 1280 K bits/sec

(C) 1180 M bits/sec

(D) 1280 M bits/sec

46. In a CD player, the sampling rate is 44.1 kHz and the samples are quantized using a 16-bit/sample quantizer. The resulting number of bits for a piece of music with a duration of 50 minutes is (A) 1.39 ´ 10 9

(B) 4.23 ´ 10 9

(C) 8.46 ´ 10 9

(D) 12.23 ´ 10 9

( n + 1) n

(B)

(D) 4

52. In PCM system, if the quantization levels are increased

form

2

to

8,

the

256

quantization

levels.

(A) remain same

(B) be doubled

(C) be tripled

(D) become four times

53. Assuming that the signal is quantized to satisfy the condition of previous question and assuming the approximate bandwidth of the signal is W. The minimum required bandwidth for transmission of a will be. (A) 5 W

(B) 10 W

(C) 20 W


The

bit

transmission rate for the time-division multiplexed signal will be (A) 8 kbps (C) 256 kbps

************ (B) 64 kbps (D) 512 kbps

48. Analog data having highest harmonic at 30 kHz generated by a sensor has been digitized using 6 level PCM. What will be the rate of digital signal generated? (A) 120 kbps (C) 240 kbps

(B) 200 kbps (D) 180 kbps

49. In a PCM system, the number of quantization levels is 16 and the maximum lsignal frequency is 4 kHz.; the bit transmission rate is (A) 32 bits/s (C) 32 kbits/s Page 438

(B) 16 bits/s (D) 64 dbits/s

bandwidth

requirement will.

sampled at Nyquist rate are converted into binary PCM using

relative

binary PCM signal based on this quantization scheme

47. Four voice signals. each limited to 4 kHz and signal

( n + 1) 2 n2

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UNIT 7

16. (C) 128= 2 7,. We need 7 bits/character. For 1,000,000 character we need 7 Mbits/second. Thus minimum bandwidth = 7 Mbits/sec.

26. (B)


3 L2 So = = 6394 = 38.06 dB N o [ln(m + 1)]2

27. (C) Nyquist Rate = 2 MHz

17. (D) fs > 2 fm = 6400 Hz, nfs £ 63000 36000 36000 n£ = 5.63, n = 5, L = 2 n = 32, fs = = 7.2 kHz. 6400 5

50% higher rate = 3 MHz,

Thus transmission bandwidth is 3 MHz ´ 8 = 24 Mbits/s. New sampling rate is at 20% above the Nyquist rate. Sampling rate= 12 . ´ 2 = 2.4 MHz.

18. (D) The transmitted code word are

bits per second= 0

0

1

1

0

0

0

1

1

1

0

0

1

0

bits 24 M sec = 10 bits 2.4 MHz

1

So 3(1024) 2 = = 102300 = 50.1 dB N o (ln 256) 2

Level = 210 = 1024, Fig.S.7.6.18

In 1st word 001(1) 28. (B)

In 2nd word 010(2) In 3rd word 011(3)

3 L2 So = ³ 50 dB, m = 100 N o [ln(m + 1)]2

3 L2 = 100 000 or [ln 101) 2

In 4th word 100(4) In 5th word 101 (5)

L = 842.6

Because L is power of 2, we select L = 1024 = 210 .

19. (B) Message bandwidth= W, Nyquist rate = 2W

Thus 10 bits are required.

Bandwidth = 2W ´ 8 = 16W bit/s 29. (A) So = m 2 ( t), N o =

108 W= = 6.25 MHz 16

8

16W= 10 , or

1 20. (A) Sampling interval T, = = 125 ms. There are 24 8k channels and 1 sync pulse, so the time allotted to each channel is Tc =

T 25

= 5 ms. The pulse duration is 1 ms. So

the time between pulse is 4 ms. 21. (B) Amax =

22. (D) Amax =

23. (D) N o =

24. (C)

L = 256 = 28

So 3 L2 m 2 ( t) , = 3 L2 No mp2

since signal is sinusoidal

m 2 ( t) 1 = , mp2 2

3 L2 = 48 dB = 63096, L = 205.09 2 Since L is power of 2, so we select L = 256 Hence 256 = 28 ,

dfs 0.1 ´ 68 k = 108 . V = wm 2 p ´ 10 3

3mp2

So 8 bits per sample is required .

30. (B) ( SNR) q = 176 . + 6.02( n) = 40 dB, n = 6.35 We take the n = 7.

df s A w 1 ´ 2 p ´ 800 or d = max m = = 78.5 mV wm fs 50 ´ 10 3

d2 B (0.0785) 2 ´ 3500 = = 1122 . ´ 10 -4 W 3 fs 3 ´ 64000

So 0.5 = 4.46 ´ 10 3 = N o 112 . ´ 10 -4

Capacity = 20 ´ 8 k ´ 7 = 112 . Mbits = 140 Kbytes 31. (B) For slope overload to take place Em ³

df s 2 pfm

This is satisfied with Em = 15 . V and fm = 4 kHz. 32. (C) Step size d=

2 mp L

=

1536 . = 0.012 V 128

quantization noise power S 25. (A) o µ L2 , No

æ S ö L = 2 n çç o ÷÷ = 10 log( C2 2 n ) è N o ødB

= log C + 20 n log 2 = a + 6 n dB.

=

This equation shows

that increasing n by one bits increase the by 6 dB. Hence an increase in the SNR by 12 dB can be accomplished by increasing 9is form 10 to 12, the transmission bandwidth would be increased by 20% Page 440

d2 (0.012) 2 = 12 ´ 10 -6 V 2 = 12 12

33. (B) Bit Rate = 8 k ´ 8 = 64 kbps (SNR) q = 176 . + 6.02 n dB = 176 . + 6.02 ´ 8 = 49.8 dB 34. (B) fs = 1800 samples/sec,

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fm =

1800 = 900 Hz 2



Since the sampling rate is 1800 samples/sec the highest frequency that can be recovered is 900 Hz.

The

maximum

frequency

component

will

be

150 + 25 = 175 Hz.

x( t) is band limited with fm = 350 Hz,

Thus Nyquist 1 rate is 2 fm = 700 Hz, Sampling interval = sec 700 1 1 = = 20 kHz, T 50 ´ 10 -6

æ 1 + cos 300 pt ö 44. (D) g( t) = 10 cos 50 ptç ÷ 2 è ø = 5 cos 50 pt + 5 cos 50 pt cos 300 pt

35. (C) x( t) = sinc 700t + sinc 500t 1 = [sin 700 pt + sin 500 pt ] pt

36. (D) fs =

Chap 7.6

Thus fs = 2 ´ 175 = 350 . sample per second. 45. (B) Total sample = 8000 ´ 20 = 160 k sample/sec Bit for each sample = 7 + 1 = 8 Bit Rate = 160 k ´ 8 = 1280 ´ 10 3 bits/sec

fc = 12 kHz

The frequency passed through LPF are fc , fs - fm or 12 kHz, 8 kHz

46. (B) The sampling rate is fs = 44100 meaning that we take 44100 samples per second. Each sample is quantized using 16 bits so the total number of bits per

37. (C) P =

( SNR)1 2 , Here n = code word length, = (SNR)2 2 2 n 1 2n 2

16

n1 = 61 n2 = 8,

Thus rate =

second is 44100´16. For a music piece of duration 50 min = 3000- sec the resulting number of bits per channel

2 = 16 212

(left

= 2.1168 ´ 10

9

and

right)

is 44100 ´ 16 ´ 3000

and the overall number of bits is

2.1168 ´ 10 ´ 2 = 4.2336 ´ 10 9 9

38. (D) Signal g1 ( t), g 2 ( t), g s ( t) and g 4 ( t) will have 8 k, 8 k, 16 k and 16 k sample/sec at Nyquist rate. Thus

47. (C) Nyquist Rate = 2 ´ 4k = 8 kHz

48000 sample/sec bit rate 48000 ´ 8 = 384 kbps

Total sample = 4 ´ 8 = 32 k sample/sec

39. (C) Analog signals, having bandwidth 1200 Hz, 600 Hz and 600 Hz have 2400, 1200 samples/sec at Nyquist

256 = 28 , so that 8 bits are required Bit Rate = 32 k ´ 8 = 256 kbps

rate. Hence 48000 sample/sec

48. (D) Nyquist Rate = 2 ´ 30 k = 60 kHz

bit rate = 48000 sample/sec ´12 = 57.6 kbps

2 n ³ 6 Thus n = 3,

3

æ sin 2 p1000 t ö æ sin 2 p1000 t ö 40. (C) x( t) = 5ç ÷ + 7ç ÷ pt pt è ø è ø

2

49. (C) Nyquist rate= 2 ´ 4 = 8 kHz 2 n = 16 or n = 4,

Maximum frequency component = 3 ´ 1000 = 3 kHz Sampling rate = 2 fm = 6 kHz

50. (C) fs = 8 kHz,

41. (B) Here fs = 32 k sample/sec 1 1 fm = = Em = 125, T 2

51. (D)

For slope-overload to be averted Em ³ D£

Em fm fs

or D £

Bit Rate = 4 ´ 8 = 32 kbits/sec 2 n = 256 Þ n = 8

Bit Rate = 8 ´ 8k = 64 kb/x

1 2

So a 2 2 n , If PCM is increased form n to n + 1, No

the ratio will increase by a factor 4. Which is

fs fm

independent of n. 1 2

125( ) (128)( ) or D £ 2 -8 or D £ 3 32 ´ 10 32 ´ 1024

42. (D) fm = 5 kHz,

Bit Rate = 60 ´ 3 = 18 kHz

Nyquist Rate = 2 ´ 5 = 10 kHz

Since signal are sampled at twice the Nyquist rate so

52. (C) If L = 2, then 2 = 2 n or n = 1 ND If L = 8, then 8 = 2 n or n = 3. So relative bandwidth will be tripled. 53. (B) The minimum bandwidth requirement for transmission of a binary PCM signal is BW= vW. Since

sampling rate = 2 ´ 10 = 20 kHz.

v = 10, we have BW = 10 W

Total transmission bandwidth = 4 ´ 20 = 80 kHz 43. (D) Signal will be sampled 200, 200, 400 and 800 sample/sec thus 1600 sample per second,

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CHAPTER

7.8 SPREAD SPECTRUM


5. The Antijam margin is

A pseudo-noise (PN) sequance is generated using a

(A) 47.5 dB

(B) 93.8 dB

feedback shift register of length m = 4. The chip rate is

(C) 86.9 dB

(D) 12.6 dB

7

10 chips per second 6. 1. The PN sequance length is

A slow

FH/MFSK

system

has

the

following

parameters.

(A) 10

(B) 12

Number of bits per MFSK symbol = 4

(C) 15

(D) 18

Number of MFSK symbol per hop = 5 The processing gain of the system is

2. The chip duration is (A) 1ms

(B) 0.1 ms

(C) 0.1 ms

(D) 1 ms

(B) 15 ms

(C) 6.67 ns

(D) 0.67 ns

(B) 37.8 dB

(C) 6 dB

(D) 26 dB

7.

A fast

FH/MFSK

system

has

the

following

parameters.

3. The period of PN sequance is (A) 15 . ms

(A) 13.4 dB

Number of bits per MFSK symbol = 4 Number of pops per MFSK symbol = 4 The processing gain of the system is


(A) 0 dB

(B) 7 dB

(C) 9 dB

(D) 12 dB

A direct sequence spread binary phase-shiftkeying system uses a feedback shift register of Length


19 for the generation of PN sequence . The system is required to have an average probability of symbol error due to externally generated interfering signals that does not exceed 10 -5

A rate 1/2 convolution code with dfrec = 10 is used to encode a data requeence occurring at a rate of 1 kbps. The modulation is binary PSK. The DS spread spectrum sequence has a chip rate of 10 MHz

4. The processing gain of system is

8. The coding gain is

(A) 37 dB

(B) 43 dB

(A) 7 dB

(B) 12 dB

(C) 57 dB

(D) 93 dB

(C) 14 dB

(D) 24 dB

Page 450

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Spread Spectrum

9. The processing gain is

Chap 7.8


(A) 14 dB

(B) 37 dB

(C) 58 dB

(D) 104 dB

A CDMA system consist of 15 equal power user that transmit information at a rate of 10 kbps, each using a DS spread spectrum signal operating at chip

10. A total of 30 equal-power users are to share a

rate of 1 MHz. The modulation scheme is BPSK.

common communication channel by CDM. Each user transmit information at a rate of 10 kbps via DS spread

16. The Processing gain is

spectrum and binary PSK. The minimum chip rate to

(A) 0.01

(B) 100

(C) 0.1

(D) 10

obtain a bit error probability of 10

-5

(A) 1.3 ´ 106 chips/sec

(B) 2.9 ´ 10 5 chips/sec

(C) 19 . ´ 106 chips/sec

(D) 1.3 ´ 10 5 chips/sec

11. A CDMA system is designed based on DS spread

17. The value of e b/J 0 is (A) 8.54 dB

(B) 7.14 dB

(C) 17.08 dB

(D) 14.28 dB

spectrum with a processing gain of 1000 and BPSK modulation scheme. If user has equal power and the

18. How much should the processing gain be increased

desired level of performance of an error probability of

to allow for doubling the number of users with affecting

10 -6 , the number of user will be

the autopad SNR

(A) 89

(B) 117

(A) 1.46 MHz

(B) 2.07 MHz

(C) 147

(D) 216

(C) 4.93 MHz

(D) 2.92 MHz

12. In previous question if processing gain is changed to

19.

500, then number of users will be

processing gain of 500. If the desired error probability is

(A) 27 users

(B) 38 users

10 -5 and ( e b / J 0 ) required to obtain an error probability

(C) 42 users

(D) 45 users

of 10 -5 for binary PSK is 9.5 dB, then the Jamming

A DS/BPSK

spread

spectrum

signal

has

a

margin against a containers tone jammer is Statement for Question 13-15 : A DS spread spectrum system transmit at a rate of

(A) 23.6 dB

(B) 17.5 dB

(C) 117.4 dB

(D) 109.0 dB

1 kbps in the presets of a tone jammer. The jammer power is 20 dB greater then the desired signal, and the required Îb / J 0 to achieve satisfactory performance is 10 dB.

Statement for Question 20-21 : An m = 10 ML shift register is used to generate the pre hdarandlm sequence in a DS spread spectrum

13. The spreading bandwidth required to meet the specifications is

system. The chip duration is Tc = l ms and the bit duration is Tb = NTc , where N is the length (period of the m sequence).

(A) 10 7 Hz

(B) 10 3 Hz

(C) 10 5 Hz

(D) 106 Hz

20. The processing gain of the system is

14. If the jammer is a pulse jammer, then pulse duty cycle that results in worst case jamming is (A) 0.14

(B) 0.05

(C) 0.07

(D) 0.10

(A) 10 dB

(B) 20 dB

(C) 30 dB

(D) 40 dB

21. If the required e b/J 0 is 10 and the jammer is a tone jammer with an average power J av, then jamming

15. The correspond probability of error is

margin is.

(A) 4.9 ´ 10 -3

(B) 6.3 ´ 10 -3

(A) 10 dB

(B) 20 dB

(C) 9.4 ´ 10 -4

(D) 8.3 ´ 10 -3

(C) 30 dB

(D) 40 dB

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Page 451


UNIT 7

Statement for Question 22-23 : An FH binary orthogonal FSK system employs an m = 15 stage liner feedback shift register that generates an ML sequence. Each state of the shift register selects one of L non over lapping frequency bands in the hopping pattern. The bit rate is 100 bits/s. The demodulator employ non coherent detection. 22. If the hop rate is one per bit, the hopping bandwidth for this channel is


(A) 0.4 GHz

(B) 0.6 GHz

(C) 0.7 GHz

(D) 0.9 GHz

28. The probability of error for the worst-case partial band jammer is (A) 0.2996

(B) 0.1496

(C) 0.0368

(D) 0.0298

29. The minimum hop rate for a FH spread spectrum system that will prevent a jammer from operating five onives away from the receiver is

(A) 6.5534 MHz

(B) 9.4369 MHz

(C) 2.6943 MHz


(A) 3.2 bHz

(B) 3.2 MHz

(C) 18.6 MHz

(D) 18.6 kHz

23. Suppose the hop rate is increased to 2 hops/bit and the receiver uses square law combining the signal over two hops. The hopping bandwidth for this channel is (A) 3.2767 MHz

(B) 13.1068 MHz

(C) 26.2136 MHz

(D) 1.6384 MHz

***********

Statement for Qquestion 24-25 : In a fast FH spread

spectrum system, the

information is transmitted via FSK with non coherent detection. Suppose there are N = 3 hops/bit with hard decision decoding of the signal in each hop. The channel is AWGN with power spectral density

1 2

N0 and an SNR

20-13 dB (total SNR over the three hops) 24. The probability of error for this system is (A) 0.013

(B) 0.0013

(C) 0.049

(D) 0.0049

25. In case of one hop per bit the probability of error is (A) 196 . ´ 10 -5

(B) 196 . ´ 10 -7

(C) 2.27 ´ 10 -5

(D) 2.27 ´ 10 -7

Statement for Question 26-29 : A slow FH binary FSK system with non coherent detection

operates

at

e b / J 0 = 10,

with

hopping

bandwidth of 2 GHz, and a bit rate of 10 kbps. 26. The processing gain of this system is (A) 23 dB

(B) 43 dB

(C) 43 dB

(D) 53 dB

27. If the jammer operates as a partial band jammar, the bandwidth occupancy for worst case jamming is Page 452

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Spread Spectrum

Chap 7.8

W/R W/R Î = = b J av/Pav N u - 1 J 0

SOLUTION

æÎ W/R = çç b è J0

1. (C) The PN sequence length is N= 2 m - 1 = 2 4 - 1 = 15

æÎ W = Rçç b è J0

2. (B) The chip duration is 1 TC = 7 s = 0.1 ms 10

ö ÷÷( N u - 1) ø

ö ÷÷( N u - 1) ø

where R = 10 4 bps, N u = 30 and Îb / J 0 = 10 Therefore, W = 2.9 ´ 106 Hz The minimum chip rate is 1 / Tc = W = 2.9 ´ 106 chips/sec

3. (A) The period of the PN sequence is T= NTC = 15 ´ 0.1 = 15 . ms

11. (D) To achieve an error probability of 10 -6 , we

4. (C) m= 19

æÎ required çç b è J0

n= 2 m - 1 = 219 - 1 = 219

ö ÷÷ = 10.5 dB ødB

Then, the number of users of the CDMA system is

The processing gain is 10 log10 N= 10 log10 219

W/R 1000 +1= + 1 = 89 users Îb /J 0 11.3

= 190 ´ 0.3 or 57 dB

Nu =

æ Eb ö 5. (A) Antijam margin = (Processing gain) - 10 log10 çç ÷÷ è N0 ø

12. (D) If the processing gain is reduced to W/R = 500,

The probability of error is 1 Pe = erfc 2

æ ç ç è

Eb N0

N u=

ö ÷ ÷ ø

500 + 1 = 45 users 11.3

13. (D) We have a system where ( J av/Pav) dB = 20 dB,

With Pe = 10 -5, we have Eb / N 0 = 9.

R = 1000 bps and (Îb /J 0 ) dB = 10 dB

Hence, Antijam margin = 57 - 10 log10 9 = 57 - 9.5

æÎ æJ ö æW ö Hence, we obtain ç ÷ = çç av ÷÷ + çç b è R ødB è Pav ødB è J 0

or =47.5 dB 6. (D) The precessing gain (PG) is FH Bandwidth W c PG = = 5 ´ 4 = 20 = Symbol Rate Rs

ö ÷÷ = 30 dB ødB

W = 1000 R W = 1000 R = 106 Hz

Hence, expressed in decibels, PG= 10 log10 20 = 26 db 7. (D) The processing gain is PG = 4 ´ 4 = 16 Hence, in decibels,

14. (C) The duty cycle of a pulse jammer of worst-case 0.71 0.7 jamming is a = = = 0.07 Îb /J 0 10 15. (D) The corresponding probability of error for this

PG = 10 log10 16 = 12 dB 8. (A) The coding gain is Rcd min =

then

worst-case jamming is 1 ´ 10 = 5 or 7 dB 2

9. (B) The processing gain

P2 =

0.083 0.083 = = 8.3 ´ 10 -3 e b/J 0 10

16. (B) Precessing gain is

10 7 W = = 5 ´ 10 3 or 37 dB R 2 ´ 10 3

W 106 = = 100 R 10 4

17. (A) We have N u = 15 users transmitting at a rate of

10. (C) We assume that the interference is characterized

10,000 bps each, in a bandwidth of W = 1MHz.

as a zero-mean AWGN process with power spectral

The e b/J 0 is.

-5

density J 0 . To achieve an error probability of 10 , the required Îb /J 0 = 10 we have

e b W/R 106 / 10 4 100 = = = 14 14 J 0 Nu - 1

= 7.14 or 8.54 dB

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Page 453


UNIT 7

18. (B) With N u = 30 and e b/J 0 = 7.14, the processing


ec

gain should be increased to

1 P= e 2 N 0 2

W/R= (7.14)(29) = 207

where e c / N 0 = 20 / 3. The probability of a bit error is

W= 207 ´ 104 = 2.07 MHz

Pb = 1 - (1 - p) 2 = 1 - (1 - 2 p + p2 ) = 2 p - p2

Hence the bandwidth must be increased to 2.07 MHz 19. (B) The processing gain is given as W = 500 or 27 dB R

=e

-

ec 2 N0

ec

-

1 - 2 N0 e = 0.0013 2

25. (C) In the case of one hop per bit, the SNR per bit is ec

The ( e b/J 0 ) required to obtain an error probability of

20, Hence,

Pb =

1 - 2 N 0 1 -10 e = e = 2.27 ´ 10 -5 2 2

10 -5 for binary PSK is 9.5 dB. Hence, the jamming 26. (D) We are given a hopping bandwidth of 2 GHz and

margin is

a bit rate of 10 kbs.

æe ö æ J av ö æW ö ÷÷ = ç ÷ - çç b ÷÷ = 27.95 or 17.5 dB çç R P è ø è J 0 ødB è av ødB dB

Hence,

20. (C) The period of the maximum length shift register sequence is

a* W = 2/( e b/J 0 ) = 0.2

Since Tb = NTc then the processing gain is T N b = 1023 or 30 dB Tc

Hence a* W = 0.4 GHz 28. (C) The probability of error with worst-case partial-band jamming is

21. (B) A Jamming margin is æe ö ö ÷÷ - çç b ÷÷ = 30 - 10 = 20 dB ødB è J 0 ødB

where J av = J 0W » J 0/Tc = J 0 ´ 10

22. (A) The length of the shift-register sequence is L = 2 m - 1215 - 1 = 32767 bits For binary FSK modulation, the minimum frequency separation is 2/T, where 1/T is the symbol (bit) rate. The hop rate is 100 hops/sec. Since the shift register

e -1 e -1 = 3.68 ´ 10 -2 = ( e b/J 0 ) 10

Dd= 2 ´ 8050 = 16100 Dd Dd= x ´ t or t = t Dd 16100 Þ t= = = 5.367 ´ 10 5 x 3 ´ 108 1 1 f= = = 18.63 kHz t 5.367 ´ 10 -5

L = 32767 states and each state utilizes a

bandwidth of 2/T = 200 Hz, then the total bandwidth for the FH signal is 6.5534 MHz. 23. If the hopping rate is 2 hops/bit and the bit rate is 100 bits/sec, then, the hop rate is 200 hops/sec. The minimum

frequency

separation

for

orthogonality

2/T = 400 Hz. Since there are N = 32767 states of the shift register and for each state we select one of two frequencies

separated

by

400

Hz,

the

hopping

bandwidth is 13.1068 MHz. 24. (B) The total SNR for three hops is 20 ~ 13 dB. Therefore the SNR per hop is 20/3. The probability of a chip error with non-coherent detection is Page 454

P2 =

29. (D) d= 5 miles = 8050 meters

6

has

27. (A) The bandwidth of the worst partial-band jammer is a* W, where

N = 210 - 1 = 1023

æW æ J av ö ÷÷ = çç çç P è av ødB è Rb

W 2 ´ 10 9 = = 2 ´ 10 5or 53 dB 10 4 R

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***********


UNIT 8

10. A field is given as G=

Electromagnetics

16. A field is given as

13 ( yu x + 3u y + xu z ) x2 + y2

G =

25 ( xu x + yu y ) x2 + y2

The unit vector in the direction of G at P(3, 4, -2)

The field at point (-2, 3, 4) is (A) 13( -2 u x + 3u y + 4 u z )

(B) -2 u x + 3u y + 4 u z

(C) 13( 3u x + 4 u y - 2 u z )

(D) 3u x + 4 u y - 2 u z

11. A field is given as F = yu x + zu y + xu z The angle between G and u x at point (2, 2, 0) is (A) 45°

(B) 30°

(C) 60°

(D) 90°

is (A) 0.6 u x + 0. 8 u y

(B) 0. 8 u x + 0.6 u y

(C) 0.6 u y + 0. 8 u z

(D) 0.6 u z + 0.6 u x

17. A field is given as F = xyu x + yzu y + zxu x The value of 4 2

the double integral I = ò ò F × u ydzdx in the plane y = 7 is 0 0

12. A vector field is given as G = 12 xyu x + 6( x 2 + 2) u y + 18 z 2 u z

(A) 128

(B) 56

(C) 190

(D) 0

18. Two vector extending from the origin are given as

The equation of the surface M on which |G | = 60 is (A) 4 x 2 y 2 + 4 x 4 + 9 z 4 + 2 x 2 = 96 (B) 2 x 2 y 2 + x 4 + 9 z 4 + 2 x 2 = 96 (C) 2 x 2 y 2 + 4 x 4 + 9 z 4 + 2 x 2 = 96 (D) 4 x 2 y 2 + x 4 + 9 z 4 + 2 x 2 = 96

R1 = 4 u x + 3u y - 2 u z and R 2 = 3u x - 4 u y - 6 u z . The area of the triangle defined by R1 and R 2 is (A) 12.47

(B) 20.15

(C) 10.87

(D) 15.46

19. The four vertices of a regular tetrahedron are located at O (0, 0, 0), A(0, 1, 0), B(0.5

13. A vector field is given by

(

E = 4 zy u z + 2 y sin 2 x u y + y sin 2 x u z 2

(C) Plane x =

3p 2

, 0.5,

2 3

). The unit vector perpendicular (outward) to

(A) 0.41u x + 0.71u y + 0.29 u z (B) 0.47 u x + 0.82 u y + 0.33u z

(B) Plane x = 0

(C) -0.47 u x - 0.82 u y - 0.33u z

(D) all

(D) -0.41u x - 0.71u y - 0.29 u z 20. The two vector are R AM = 20 u x + 18 u y - 18 u z and

14. The vector field E is given by E = 6 zy 2 cos 2 x u x + 4 xy sin 2 x u y + y 2 sin 2 x u z The region in which E = 0 is

R AN = - 10 u x + 8 u y + 15 u z . The unit vector in the plane of the triangle that bisects the interior angle at A is (A) 0.168 u x + 0.915 u y + 0.367 u z

(A) y = 0

(B) x = 0

(B) 0.729 u x + 0.134 u y - 0.672 u z

(C) z = 0

np (D) x = 2

(C) 0.729 u x + 0.134 u y + 0.672 u z (D) 0.168 u x + 0.915 u y - 0.367 u z

15. Two vector fields are F = -10 u x + 20 x( y - 1) u y and G = 2 x 2 yu x - 4 u y + 2 u z . At point A(2, 3, -4) a unit vector in the direction of F - G is

21. Two points in cylindrical coordinates are A( r = 5, f = 70 ° , z = -3) and B(r = 2, f = 30 ° , z = 1). A unit vector at A towards B is

(A) 0.18 u x + 0.98 u y - 0.05 u z

(A) 0.03u x - 0.82 u y + 0.57 u z

(B) -0.18 u x - 0.98 u y + 0.05 u z

(B) 0.03u x + 0.82 u y + 0.57 u z

(C) -0.37 u x + 0.92 u y + 0.02 u z

(C) -0.82 u x + 0.003u y + 0.57 u z

(D) 0.37 u x - 0.92 u y - 0.02 u z Page 458

3, 0.5, 0) and C

the face ABC is

2

The surface on which E y = 0 is (A) Plane y = 0

0 .5 3

(D) 0.003u x - 0.82 u y + 0.57 u z www.gatehelp.com


Vector Analysis

22. A field in cartesian form is given as D = xu x +

29. The vector

yu y x +y 2

B=

2

10 u r + r cos q u q + u f r

in cartesian coordinates at (-3, 4, 0) is

In cylindrical form it will be u u uf (B) D = r + (A) D = r r r cos f (C) D = ru r

Chap 8.1

(D) D = ru r + cos f u f

(A) u x - 2 u y

(B) - 2u x + u y

(C) 1.36 u x + 2.72 u y

(D) -2.72 u x + 1.36 u x

30. The two point have been given A (20, 30 ° , 45 ° ) and

23. A vector extends from A(r = 4, f = 40 ° , z = -2) to B(

B ( 30, 115 ° , 160 ° ). The |R AB | is

r = 5, f = -110 ° , z = 1). The vector R AB is

(A) 22.2

(B) 44.4

(A) 4.77 u x + 7.30 u y + 4 u z

(C) 11.1

(D) 33.3

(B) -4.77 u x - 7.30 u y + 4 u z (C) -7.30 u x - 4.77 u y + 4 u z

31. The surface r = 2 and 4, q = 30 ° and 60°, f = 20 ° and

(D) 7.30 u x + 4.77 u y + 4 u z

80° identify a closed surface. The enclosed volume is

24. The surface r = 3, r = 5, f = 100 °, f = 130 °, z = 3 and

(A) 11.45

(B) 7.15

(C) 6.14

(D) 8.26

z = 4.5 define a closed surface. The enclosed volume is (A) 480

(B) 5.46

32. The surface r = 2 and 4, q = 30 ° and 50° and f = 20 °

(C) 360

(D) 6.28

and 60° identify a closed surface. The total area of the enclosing surface is

25. The surface r = 2, r = 4, f = 45 °, f = 135 °, z = 3 and z = 4 define a closed surface. The total area of the enclosing surface is

(A) 6.31

(B) 18.91

(C) 25.22

(D) 12.61

(A) 34.29

(B) 20.7

33. At point P(r = 4, q = 0.2 p , f = 0.8 p), u r in cartesian

(C) 32.27

(D) 16.4

component is

26. The surface r = 3, r = 5, f = 100 °, f = 130 °, z = 3 and

(A) 0.48 u x + 0.35 u y + 0.81u z

z = 4.5 define a closed volume. The length of the longest

(B) 0.48 u x - 0.35 u y - 0.81u z

straight line that lies entirely within the volume is

(C) -0.48 u x + 0.35 u y + 0.81u z

(A) 3.21

(B) 3.13

(D) 0.48 u x - 0.35 u y - 0.81u z

(C) 4.26

(D) 4.21 34. The expression for u y in spherical coordinates at P(

27. A vector field H is

r = 4, q = 0.2 p, f = 0.8 p) is

æ fö H = rz 2 sin f u r + e - z sin ç ÷ u f + r 3u z è2 ø

(A) 0.48 u r + 0.35 u q - 0.81u f (B) 0.35 u r + 0.48 u q - 0.81u f

p æ ö At point ç 2, , 0 ÷ the value of H × u x is 3 è ø (A) 0.25

(B) 0.433

(C) -0.433

(D) -0.25

(C) -0.48 u r + 0.35 u q - 0.81u f (D) -0.35 u r + 0.48 u q - 0.81u f 35. Given a vector field

28. A vector is A = yu x + ( x + z) u y. At point P(-2, 6, 3)

D = r sin f u r -

A in cylindrical coordinate is (A) -0.949 u r - 6.008 u f

(B) 0.949 u r - 6.008 u f

(C) -6.008 u r - 0.949 u f

(D) 6.008 u r + 0.949 u f

1 sin q cos f u q + r 2 u f r

The component of D tangential to the spherical surface r = 10 at P(10, 150°, 330°) is

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Page 459


UNIT 8

Electromagnetics

(A) 0 .043u q + 100 u f

40. The gradient of the functionG = r 3 sin 2 q sin 2 f sin q

(B) -0 .043u q - 100 u f

at point P ( 12 ,

(C) 110 u q + 0.043u f

(A) 1.41u r + 3u z

(B) u x + u y + u z

(D) 0 .043u q - 100 u f

(C) 3.46 u r + 9.3u q

(D) All

36. The circulation of F = x 2 u x - xzu y - y 2 u z around the

41.

path shown in fig. P8.1.36 is

The

1 2

1 2

,

) is

directional

derivative

of

function

F = xy + yz + zx at point P(3, -3, - 3) in the direction toward point Q(4, -1, -1) is

z 1

(A) -3

(B) 1

(C) -2

(D) 0

42. The temperature in a auditorium is given by T = 2 x 2 + y 2 - 2 z 2 . A mosquito located at (2, 2, 1) in the y

auditorium desires to fly in such a direction that it will

1

get warm as soon as possible. The direction, in that it

1

x

must fly is (A) 8 u x + 8 u y - 4 u z

Fig. P8.1.36

(A) -

1 3

(B)

1 6

(B) 2 u x + 2 u y - u z

(C) -

1 6

(D)

1 3

(D) -(2 u x + 2 u y - u z )

(C) 4 u x + 4 u y - 4 u z

37. The circulation of A = r cos f u r + z sin f u z around the edge L of the wedge shown in Fig. P8.1.37 is

43. The angle between the normal to the surface x2 y + z = 3

x ln z - y 2 = -4

and

at

the

point

of

intersection (-1, 2, 1) is

y

(A) 73.4°

(B) 36.3°

(C) 16.6°

(D) 53.7°

L

44. The divergence of vector A = yzu x + 4 xyu y + yu z at 60

point P(1, -2, 3) is

o

x

0

2

Fig. P8.1.37

(A) 2

(B) -2 (D) 4

(A) 1

(B) -1

(C) 0

(C) 0

(D) 3

45.

The

divergence

of

the

38. The gradient of field f = y 2 x + xyz is (A) y( y + z) u x + x(2 y + z) u y + xyu z (B) y(2 x + z) u x + x( x + z) u y + xyu z (C) y 2 u x + 2 yxu y + xyu z

(A) 2.6

(B) 1.5

(C) 4.5

(D) -4.5

46. The divergence of the vector

(D) y(2 y + z) u x + x(2 y + z) u y + xyu z

A = rz 2 cos f u r + z sin 2 f u z is

39. The gradient of the field f = r z cos 2 f at point 2

(1, 45 ° , 2) is (A) 4u f

(B) 4 2u f

(C) -4u f

(D) -4 2u f

Page 460

vector

A = 2 r cos q cos f u r + r u f at point P(1, 30°, 60°) is 1 2

(A) 2rz 2 cos f u r + sin 2 f u z (B) 2rz 2 cos f u r + sin 2 f u z (C) 2 z 2 cos f u r + sin 2 f u z (D) z sin 2

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f u r + 2rz cos f u f + z 2 sin f u z r


Vector Analysis

Chap 8.1

47. The flux of D = r 2 cos 2 f u r + 3 sin f u f over the

54. If V = x 2 y 2 z 2 , the laplacian of the field V is

closed surface of the cylinder 0 £ z < 3,

(A) 2( xy 2 + yz 2 + zx 2 )

r = 3 is

(A) 324

(B) 81p

(B) 2( x 2 y 2 + y 2 z 2 + z 2 x 2 )

(C) 81

(D) 64p

(C) ( x 2 y 2 + y 2 z 2 + z 2 x 2 )

48. The curl of vector A = e xyu x + sin xy u y + cos 2 xz u z is

(D) 0 55. The value of Ñ 2 V at point P(3, 60°, -2) is if

(A) y e xyu x + x cos xy u y - 2 x sin 2 xz u z

V = r 2 z(cos f + sin f)

(B) z sin 2 xy u y + ( y cos xy - xe xy) u z (C) z sin 2 xy u y + ( x cos xy - xe xy) u z (D) xy e xyu x + xy cos xy u y - 2 xz sin 2 xz u z

(A) -8.2

(B) 12.3

(C) -12.3

(D) 0

56. If the scalar field V = r 2 (1 + cos q sin f) then Ñ 2 V is

49. The curl of vector field A = rz sin f u r + 3rz 2 cos f u r at point (5, 90°, 1) is (A) 0

(B) 12u q

(C) 6u r

(D) 5u f

(A) 1 + 2(1 - r 2 ) cos q sin f (B) 6 + 4 cos q sin f - cot q cosec q sin f (C) 2 + 2(1 - r 2 ) cos q sin f (D) 0

50. The curl of vector field 1 A = r cos q u r - sin q u q + 2 r 2 sin q u f is r 1 (A) cos q u r - cos q u q r

57. Ñ ln r is equal to

æ 1 ö (B) 2 r 2 cos q u r - 4 r sin q u q + ç 2 sin q - r sin q ÷u f r è ø

(A) Ñ ´ ( fu z )

(B) Ñ ´ ( zu f)

(C) Ñ ´ (ru f)

(D) Ñ ´ (ru z )

58. If r = xu x + yu y + zu x then ( r × Ñ ) r 2 is equal to (A) 2 r 2

(B) 3r 2

(D) 0

(C) 4 r 2

(D) 0

51. If A = ( 3 y 2 - 2 z) u x - 2 x 2 z u y + ( x + 2 y) u z , the value

59. If r = xu x + yu y + zu x is the position vector of point

of Ñ ´ Ñ ´ A at P(-2, 3, -1) is

P( x, y, z) and r =|r| then Ñ × r n r is equal to

(C) 4 r cos q u r - 6 r sin q u q + sin q u f

(A) -( 6 u x + 4 u y)

(B) 8 ( u x + u y)

(A) nr n

(B) ( n + 3) r n

(C) -8( u x + u y)

(D) 0

(C) ( n + 2) r n

(D) 0

52. The grad × Ñ ´ A of a vector field

60. If F = x 2 yu x - yu y, the circulation of vector field F

A = x 2 yu x + y 2 zu y - 2 xzu z is

around closed path shown in fig. P8.1.60 is,

(A) 2 xy + 2 yz - 2 x

y

(B) x 2 y + y 2 z - 2 xz

1

(C) 2 x 2 y + 2 y 2 z - 2 xz S

(D) 0 53. If V = xy - x 2 y + y 2 z 2 , the value of the div grad V

0

is

L

2

1

x

Fig. P8.1.60

(A) 0 (B) z + x 2 + 2 y 2 z (C) 2 y( z 2 - yz - x) (D) 2( z - y - y) 2

2

(A)

7 3

(B) -

7 6

(C)

7 6

(D) -

7 3

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Page 461


UNIT 8

Electromagnetics

61. If A = r sin f u r + r 2 u r , and L is the contour of fig.

SOLUTIONS

ò A. dL is

P8.1.61, then circulation

L

1. (D) d = ( x1 - x2 ) 2 + ( y1 - y2 ) 2 + ( z1 - z 2 ) 2

y

2

= ( 4 - 2) 2 + ( -6 - 3) 2 + ( 3 - ( -1)) 2 = 4 + 81 + 16 = 101 L

1

2. (A) R AB = R B - R A = ( 3u x + 0 u y + 2 u z ) - (5 u x - u y + 0 u z ) -2

-1

1

2

= -2 u x + u y + 2 u z

x

|R AB | =

Fig. P8.1.61

(A) 7 p + 2

(B) 7 p - 2

(C) 7p

(D) 0

uR = -

22 + 1 + 22 = 3 2 1 2 ux + u y + uz 3 3 3

3. (C) The component of F parallel to G is

62. The surface integral of vector F = 2r 2 z 2 u r + r cos 2 f u z

=

over the region defined by 2 £ r £ 5, -1 < z < 1,

F ×G (10, - 6, 5) × (0.1, 0.2, 0.3) G = (0.1, 0.2, 0.3) 2 G 0.12 + 0.2 2 + 0.32

= 9.3(0.1, 0.2, 0.3) = (0.93, 1.86, 2.79)

0 < f < 2 p is (A) 44

(B) 176

(C) 88

(D) 352

63. If D = xyu x + yzu y + zxu z , then the value of òò A × dS

is, where S is the surface of the cube defined by 0 £ x £ 1, 0 £ y £ 1, 0 £ z £ 1 (A) 0.5

(B) 3

(C) 0

(D) 1.5

4. (C) The vector component of F perpendicular to G is F ×G ( 3, 2, 1) × ( 4, 4, - 2) =F G = ( 3, 2, 1) ( 4, 4, - 2) G2 42 + 42 + 22 = (3, 2, 1)- (2, 2, - 1) = (1, 0, 2) = u x + 2 u z 5. (C) R = 3u x + 4 M - N = 3u x + 4(2 u x + 3u y - 4 u z ) -( -4 u x + 4 u y + 3u z ) = 15 u x + 8 u y - 19 u z

64. If D = 2rzu r + 3z sin f u r - 4r cos f u z and S is the

|R| =

15 2 + 8 2 + 19 2 = 25.5 = 25.5

surface of the wedge 0 < r < 2, q < f < 45 °, 0 < z < 5, then 6. (B) R = - M + 2N

the surface integral of D is (A) 24.89

(B) 131.57

= - ( 8 u x + 4 u y - 8 u z ) + 2( 8 u x + 6 u y - 2 u z )

(C) 63.26

(D) 0

= 8u x + 8u y + 4u z 8u x + 8u y + 4u z uR = 82 + 82 + 42

65. If the vector field F = ( axy + bz 3) u x + ( 3 x 2 - gz) u y + ( 3 xz 2 - y) u z is irrotational, the value of a , b and g is

=

2 2 1 æ2 2 1ö ux + u y + uz = ç , , ÷ 3 3 3 è 3 3 3ø

(A) a = b = g = 1

(B) a = b = 1, g = 0

-M + 2N = -( 8, 4, - 8) + 2( 8, 6, - 2) = (8, 8, 4)

(C) a = 0,

(D) a = b = g = 0

uR =

b = g =1

**************

uR =

æ2 2 1ö =ç , , ÷ è 3 3 3ø 8 +8 +4 ( 8, 8, 4)

2

2

2

2 2 1 ux + u y + uz 3 3 3

æ 1 + 7 -6 - 2 4 + 0 ö 7. (C) Mid point is ç , , ÷ = (4, -4, 2) 2 2 ø è 2` uR = Page 462

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( 4, - 4, 2) 4 + 4 +2 2

2

2

2 1ö æ2 =ç , - , ÷ 3 3 3ø è


Vector Analysis

=

2 2 1 ux - u y + uz 3 3 3

8. (A) G = 24(1)(2) u x + 12(1 + 2) u y + 18( -1) 2 u z = 48 u x + 36 u y + 18 u z 2 1ö æ2 9. (A) A = ( 6, - 2, - 4), B = k ç , - , ÷ 3 3ø è3 |B - A |= 10 2

2

2

2 ö 2 ö 1 ö æ æ æ ç 6 - k ÷ + ç -2 + k ÷ + ç -4 - k ÷ = 100 3 3 3 ø è ø è ø è k2 - 8 k - 44 = 0

Þ

k = 1175 . ,

Chap 8.1

= -34 u x + 84 u y - 2 u z -34 u x + 84 u y - u z uR = 34 2 + 84 2 + 2 2 = - 0.37 u x + 0.92 u y - 0.02 u z 16. (A) At P(3, 4, -2) 25 G= 2 ( 3u x + 4 u y) = 3u x + 4 u y 3 + 42 3u x + 4 u y = 0.6 u x + 0.8 u y uG = 32 + 4 2 17. (B) F × u y = Fy = yz

2 1ö æ2 B = 1175 . ç ,- , ÷ 3 3ø è3

4 2

I=

ò ò yzdzdx = 0 0

= (7.83, -7.83, -3.92)

æ2 ò0 çç ò0 yzdz è 4

4 ö ÷dx = ò 2 ydx = 2( 4) y = 8 y ÷ 0 ø

At y = 7, I = 8(7) = 56

13 10. (D) G = ( 3u x + 4 u y - 2 u z ) 2 ( -2) + ( 3) 2 = 3u x + 4 u y - 2 u z

éu x u y u z ù 1 1ê 18. (B) Area = R1 ´ R 2 = 4 3 -2 ú ú 2 2ê êë 3 -4 -6 úû

11. (A) Let q be the angle between F and u x ,

= u x ( -18 - 8) - u y( -24 + 6) + u z ( -16 - 9)

Magnitude of F is |F|= y + z + x

= -26 u x + 18 u y - 25 u z

2

F × u x = (F)(1) cos q = y y cos q = = 2 y + z 2 + x2

2

2

R1 ´ R 2 = 26 2 + 18 2 + 25 2 = 40.31 2 2 +2 2

2

=

1

area =

2

q = 45 °

40.31 = 20.15 2

19. (B) R BA = (0, 1, 0) - (0.5 3 , 0.5, 0) = ( -0.5 3 , 0.5, 0) æ 0.5 R BC = çç , 0.5, è 3

12. (D) |G|= 60 (12 xy) 2 + ( 6( x 2 + 2)) 2 + (18 z 2 ) 2 = 60 Þ

144 x 2 y 2 + 36( x 4 + 4 x 2 + 4) + 324 z 4 = 3600

Þ

4 x y + ( x + 4 x + 4) + 9 z = 100

Þ

4 x 2 y 2 + x 4 + 9 z 4 + 4 x 2 = 96

2

2

4

2

4

R BA ´ R BC

13. (D) For E y = 0, 2 y sin 2 x = 0

Þ

sin 2 x = 0, Þ

3p x = 0, 2

2 x = 0, p , 3p , Þ

y =0

Hence (D) is correct. 14. (A) E = y( 6 zy cos 2 x u x + 4 x sin 2 x u y + y sin 2 x u z ) Hence in plane y = 0,

E = 0.

15. (C) R = F - G = ( -10 u x + 20 x( y - 1) u y) - (2 x 2 yu x - 4 u y + 2 u z ) At P(2, 3, - 4) , R = F - G = ( -10 u x + 80 u y) - (24 u x - 4 u y + 2 u z )

= u x 0.5

2 3

ö ÷ - (0.5 3 , 0.5, 0) ÷ ø

é uy ê ux = ê0.5 3 0.5 ê ê 1 0 êë 3

æ 1 = çç , 0, 3 è

2 3

ö ÷ ÷ ø

ù uz ú 0 ú ú 2ú 3 úû

2 0.5 - u y (0.5 2 ) + u z 3 3

= 0.41u x - 0.71u y + 0.29 u z The required unit vector is 0.41u x + 0.71u y + 0.29 u z = 0.412 + 0.712 + 0.29 2 = 0.47 u x + 0.81u y + 0.33u z 20. (A) The non-unit vector in the required direction is 1 = ( u AN + u AM ) 2 ( -10, 8, 15) u AN = = ( -0.507, 0.406, 0.761) 100 + 64 + 225

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Page 463


UNIT 8

(20, 18, - 10)

u AM =

400 + 324 + 100

Electromagnetics

26. (A) A(r = 3, f = 100 ° , z = 3) = A( -0.52, 2.95, 3)

= (0.697, 0.627, -0.348)

B(r = 5, f = 130 ° , z = 4.5) = B( -321 . , 3.83, 4.5)

1 ( u AM + u AN ) 2 1 = [(0.697, 0.627, - 0.348) + ( -0.507, 0.406, 0.761) ] 2

length = |B - A| B - A = ( -321 . , 3.83, 4.5) - ( -0.52, 2.95, 3) = ( -2.69, 0.88, 15 . )

= (0.095, 0.516, 0.207) (0.095, 0.516, 0.207) = (0.168, 0.915, 0.367) u bis = 0.095 2 + 0.516 2 + 0.2612 Hence (A) is correct.

|B - A| = |( -2.69,

0.88, 15 . )| = 2.69 2 + 0.88 2 + 15 . 2 = 3.21

æ p ö 27. (C) At P ç 2, , 0 ÷, H = 0.5 u f + 8 u z è 3 ø u x = cos f u r - sin f u f =

21. (D) In cartesian coordinates A (5 cos 70 ° , 5 sin 70 ° , - 3) = A(171 . , 4.70, - 3)

1 ( u r - 3u f) 2

æ 3 ö ÷ = -0.433 H × u x = (0.5)çç ÷ 2 è ø

B (2 cos ( -30 ° ), 2 sin ( -30 ° ), - 3) = B(173 . , - 1, 1) R AB = R B - R A = (173 . , - 1, 1) - (171 . , 4.70, - 3)

é Ar ù é cos f sin f 0 ù é Ax ù 28. (A) ê Af ú = ê- sin f cos f 0 ú ê A y ú ê ú ê úê ú 0 1 úû êë Az úû êë Az úû êë 0

= (0.02, - 5.70, 4) (0.02, - 5.70, 4) = (0.003, -0.82, 0.57) u AB = 0.02 2 + 5.70 2 + 4 2 22. (A) x = r cos f ,

At P (-2, 6, 3)

y = r sin f

æ 6 ö A = 6 u x + u y , f = tan -1 ç ÷ = 108.43° è -2 ø

r cos f u x + r sin f u y 1 D= 2 = (cos f u x + sin f u y) r cos 2 f + r 2 sin 2 f r

cos f = - 0.316, sin f = 0.948

1 Dr = D × u r = [cos f ( u x × u r) + sin f ( u y × u r)] r =

1 1 [cos 2 f + sin 2 f] = r r

Df = D × u f = =

0.948 0 ù é6 ù é Ar ù é-0.316 ê A ú = ê -0.948 -0.316 0 ú ê1 ú ê fú ê úê ú 0 1 úû êë0 úû êë Az úû êë 0 Ar = 6 ( -0.316) + 0.948 = -0.949,

1 [cos f ( u x × u f) + sin f ( u y × uf)] r

Af = 6( -0.948) - 0.316 = -6.008, Az = 0 Hence (A) is correct option.

1 [cos f ( - sin f) + sin f (cos f)] = 0 r

Therefore D =

29.(B) At P (-3, 4, 0)

1 ur r

r = x 2 + y 2 + z 2 = 32 + 4 2 + 0 2 = 5

23. (B) A( 4 cos 40 ° , 4 sin 40 ° , - 2) = A( 306 . , 2.57, - 2) B (5 cos ( -110 ° ), 5 sin ( -110 ° ), 2) = B ( -171 . , - 4.7, 2) R AB = R B - R A = ( -171 . , - 4.7, 2) - ( 306 . , 2.57, - 2)

4 .5 130 ° 5

ò ò ò rdrdfdz

éBx ù ésin q cos f cos q cos f - sin q ù éBr ù êB ú = êsin q sin f cos q cos f cos f ú êB ú ê yú ê ú ê qú - sin q 1 úû êëBf úû êë Bz úû êë cos q

= 6.28

3 100 ° 3

25. (C) Area is =2

135° 4

ò

ò rdrdf +

45° 2

4

4 135°

ò

ò 2 dfdz +

3 45°

4 135°

4 4

3 45°

3 2

ò ò 4 dfdz + 2 ò ò drdz

ér ù æ p ö æ pö = 2 ê ú ç ÷ + (2)(1)ç ÷ = 32.27 è2 ø ë 2 û2è 2 ø 2

Page 464

f = tan -1

x2 + y2 p = z 2 y 4 = tan -1 = 126.87 ° x -3

B = 2u r + u f

= ( -4.77, - 7.3, 4) 24. (D) Vol =

q = tan -1

é-0.6 0 -0.8 ù é2 ù = ê 0.8 0 -0.6 ú ê0 ú ê úê ú -1 0 úû êë1 úû êë 0 Bx = 2( -0.6) + 0.8 = -2 B y = 2(0.2) - 0.6 = 1 Bz = 0

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UNIT 8

2

C3 = ò r cos f dr

Along 3,

0

ò A × dL

= f=60 °

r2 = -1 2

Electromagnetics

44. (D) Ñ × A =

¶ Ax ¶ A y ¶ Az =0 + 4x + 0 + + ¶x ¶y ¶z

At P (1, -2, 3), Ñ × A = 4

= C1 + C2 + C3 = 1

L

45. (A) 38. (A) Ñf = u x

¶f ¶f ¶f + uy + uz ¶x ¶y ¶z

Ñ× A =

= y( y + z) u x + x(2 y + z) u y + xyu z 39. (C) Ñf = u r

=

¶f 1 ¶f ¶f + uf + uz ¶x r ¶y ¶z

1 ( 6 r 2 cos q cos f) + 0 + 0 r2

At P (1, 30°, 60°), Ñ × A = 6(1)(cos 30 ° )(cos 60 ° ) = 2.6

= 2 r 2 z cos 2 f u r - 2rz sin 2 f u f + r 2 cos 2 f u z

46. (C) Ñ × A =

At P (1, 45°, 2), Ñf = - 4u f 40. (B) r sin q cos f = x , r sin q sin f = y , r cos q = z G = r 3 sin 2 q sin 2 f sin q

=

1 ¶(rrz 2 cos f) ¶( z sin 2 f) = 2 z 2 cos f + sin 2 f + r ¶r ¶z

ò D × dS

¶( 4 xyz) ¶( 4 xyz) ¶( 4 xyz) ÑG = u x + uy + uz ¶x ¶x ¶z

S

Ñ ×D =

= 4 yzu x + 4 xzu y + 4 xyu z æ1 1 1ö At P ç , , ÷ , ÑG = u x + u y + u z è2 2 2 ø

ÑF = -6u x , R PQ = ( 4, - 1, - 1) - ( 3, - 3, - 3) = (1, 2, 2)

¶T ¶T ¶T + uy + uz ¶x ¶y ¶z

= 2 xu x + 2 yu z - 4 z u z

v

1 ¶(rr 2 cos 2 f) 1 ¶( z sin f) 3 + = 3r cos 2 f + sin f r r r ¶r ¶r 2

f+

v

z sin r

ö f ÷÷rdfdzdr ø

3

3

2p

3

3

2p

0

0

0

0

0

0

= ò dz ò zr 2 dr ò cos 2 f df + ò zdz ò dr ò sin f df = 81p

At P(3, -3, - 3),

= -2

ò Ñ × D dv

æ

41. (C) ÑF = ( y + z) u x + ( x + z) u y + ( x + y) u z

( -6 u x ) × ( u x + 2 u y + 2 u z )

=

ò Ñ × D dv = ò ççè 3r cos v

42. (C) ÑT = u x

By divergence theorem

S

= 4( r sin q cos f)( r sin q sin f)( r cos q) = 4 xyz

3

1 ¶(rAr) 1 ¶( Af) ¶( Az ) + + r ¶r r ¶q ¶z

47. (B) The flux is ò D × dS ,

= r 3(2 sin q cos q)(2 sin f cos f) sin q

ÑF × u R =

1 ¶(r 2 Ar ) 1 ¶(sin q Aq) 1 ¶(sin q Af) + + r2 r sin q r sin q ¶f ¶r ¶q

éu x ê¶ 48. (B) Ñ ´ A = ê ê ¶x êë Ax

uy ¶ ¶y Ay

u z ù éu x uy uz ù ¶ú ê¶ ¶ ¶ ú ú =ê ú ¶z ú ê ¶x ¶y ¶z ú Az úû êë e xy sin xy cos 2 xz úû

u x (0 - 0) - u y(2 cos xz ( - sin xz) z) + u z ( y cos xy - xexy) = z sin 2 xy u y + ( y cos xy - xe xy) u z

At P(2, 2, 1), ÑT = 4 u x + 4 u y - 4 u z 43. (A) Let f = x 2 y + z - 3, g = x ln z - y 2 + 4 Ñf = 2 xyu x + x 2 u y + u z Ñg = ln z u x - 2 yu y +

x uz z

At point P (-1, 2, 1) uf =

-4 u x + u y + u z 18

cos q = ± u f × u g = ± -1

q = cos 0.28 = 73.4 ° Page 466

, ug = -5 18 ´ 17

-4 u y - u z 17 = 0.286

éu r 1ê ¶ 49. (D) Ñ ´ A = ê r ê ¶r ëê Ar

uf uz ù ¶ ¶ú ú ¶f ¶z ú Af Az úû

é ur uf uz ù ê ¶ ¶ ¶ú 1 = ê ú r ê ¶r ¶f ¶z ú êërz sin f 2r 2 z 2 cos f 0 úû =

1 1 u r (-6r 2 z cos f) - u f (-r sin f) u z (6rz 2 cos f - rz cos f) r r

At point P(5, 90°, 1), Ñ ´ A = 5 u f www.gatehelp.com


Vector Analysis

éu r ê¶ 1 50. (C) Ñ ´ A = 2 ê r sin q ê ¶r êë Ar

56.(B)

r sin qu f ù ú ¶ ú ¶f ú r sin q Af úû

ru q ¶ ¶q rAq

Ñ2 V =

1 r 2 sin q

u r ( 2r 3 sin q - 0) -

¶2 V 1 ¶ æ 2 ¶V ö 1 ¶ æ ¶V ö 1 çr ÷+ 2 ç sin q ÷+ 2 2 2 r ¶r è ¶r ø r sin q ¶q è ¶q ø r sin q ¶f2

= 6 + 4 cos q sin f - cot q cosec q sin f 57. (A) r = x 2 + y 2

1 1 u q ( 6r 2 sin 2 q - 0) + u f ( 0 + r sin q) r sin q r

Ñ ln r = u x

= 4 r cos q u r - 6 r sin q u q + sin q u f ux uy uz ù é ê ú ¶ ¶ ¶ 51. (A) Ñ ´ A = ê ú ¶x ¶y ¶z ú ê 2 2 êë( 3 y - 2 z) ( -2 x z) ( x + 2 y) úû ux uy ¶ ¶ ¶x ¶y + 2 x2 ) 3

=

uz ù ú ¶ ú ¶z ú -( 4 xz + 6 y) úû

Ñ ´ Ñ ´ A = -6 u x - 4 u y = -( 6 u x + 4 u y) uz ù ¶ ú ú ¶z ú -2 xz úû

n

-1 æ nö Ñ × r n r = 2 x 2 ç ÷( x 2 + y 2 + z 2 ) 2 è2 ø n

2

n

= n( x 2 + y 2 + z 2 )( x 2 + y 2 + z 2 ) 2 2

2

2

L

2

¶ 55. (A) Ñ V = ¶r 2

2

æ ¶V çç r è ¶r

2

2

S

Ñ ´ F = -x uz dS = dxdy ( -u z )

ò (Ñ ´ F) × dS = - òò ( - x ) dxdy 2

= 2( y z + x z + x y ) = 2( x y + y z + z x ) 2

ò (Ñ ´ F) × dS

2

¶2V ¶2V ¶2V 54. (B) Ñ 2 V = + + ¶x 2 ¶y 2 ¶z 2 2

+ 3r n

60. (C) By Stokes theorem ò F × dL =

= -2 y + 2 z - 2 y = 2( z - y - y) 2

-1

= nr n + 3r n = ( n + 3) r n

¶( z - 2 xy) ¶(2 yz - x ) ¶( x - 2 y z) + + ¶x ¶y ¶z 2

2

n

+ rn + rn + rn

= ( z - 2 xy) u x + (2 yz - x ) u y + ( x - 2 y z) u z

2

¶( xr n ) ¶( yr n ) ¶( zr n ) + + ¶x ¶y ¶z

-1 -1 æ nö æ nö + 2 y 2 ç ÷( x 2 + y 2 + z 2 ) 2 + 2 z 2 ç ÷( x 2 + y 2 + z 2 ) 2 è2 ø è2 ø

2

2

x y x y ux + 2 u y = ux + u y 2 2 r r x +y x +y

n

¶V ¶V ¶V 53. (D) ÑV = u x + uy + uz ¶x ¶y ¶z

2

tan -1

where r n = ( x 2 + y 2 + z 2 ) 2

Ñ (Ñ ´ A) = 0

2

0

ù ú ú ú ú yú x úû

uz ¶ ¶z

2

59. (B) Ñ × r n r =

= - y 2 u x + 2 zu y - x 2 u z

Ñ × (ÑV ) =

uy ¶ ¶y

= x(2 x) + y(2 y) + z(2 z) = 2( x 2 + y 2 + z 2 ) = 2 r 2

At P(-2, 3, -1),

uy ¶ ¶y y2 z

é êu x ê¶ y uz = ê x ê ¶x ê 0 êë

æ ¶ ¶ ¶ ö 2 58. (A) ( r × Ñ ) r 2 = çç x +y +z ÷÷( x + y 2 + z 2 ) x y z ¶ ¶ ¶ è ø

= u x ( -6) - u y( -4 z) + u z (0) = -6 u x + 4 zu y

é ux ê ¶ 52. (D) Ñ ´ A = ê ê ¶x 2 ëê x y

¶ ln r ¶ ln r ¶ ln r x y + uy + uz = 2 ux + 2 u y r r ¶x ¶y ¶z

Ñ ´ f u z = Ñ ´ tan -1

= u x (2 + 2 x 2 ) - u y(1 - ( -2)) + u z ( -4 xz - 6 y) é ê Ñ ´Ñ ´ A =ê ê êë(2

2 sin q cos q sin f cos q sin f sin q sin 2 q

= 6(1 + cos q sin f) -

é ur ru q r sin q u f ù ú ê ¶ ¶ ¶ 1 = 2 ê ú r sin q ê ¶r ¶q ¶f ú 3 2 ëêr cos q - sin q 2 r sin q úû =

Chap 8.1

2

2

2

S

2

ö 1 ¶2V ¶2V + ÷÷ + 2 2 ¶z 2 ø r ¶f

ò ò x dydx + ò ò x dydx 2

0 0

2

1

1

= 4 z(cos f + sin f) - z(cos f + sin f) + 0 = 3z(cos f + sin f) æ1 3ö ÷= -8.2 At P(3, 60°, -2), Ñ 2 V = 3( -2)çç + 2 ÷ø è2

2 -x + 2

1 x

=

0

1

=

2

ò x dx + ò x (2 - x) dx 3

0

2

1

2

é x4 ù 1 ù é2 = ê ú + ê x3 - x4 ú 4 3 4 û1 ë ë û0 =

1 é16 1 7 ù é2 1 ù 1 14 + - 4ú - ê - ú = + -4+ = 4 êë 3 3 4 6 û ë3 4 û 4

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Page 467


UNIT 8

æ

ò A × dL = ççè ò

61. (C)

C

ab

+

ò

+

bc

ò

ò

+

cd

da

ò D × dS = ò (Ñ × D)dv

ö ÷÷A × dL ø

S

2

2

= 4 ò rdr

L

1

0

a

b

d

-2

-1

1

2

=0

p

c

ò A × dL

ò r df

=

3

b

A × dL = r 3df

0

= r 3 ò df = (1) 3( -p) = - p p

62. (B) Using divergence theorem

Ñ ×F =

ò

=

Ñ × Fdv

v

1 ¶ (2r 2 z 2 ) = 4 z 2 = 4 z 2 r ¶r 1

ò Ñ × F dv = òòò 4 z rdrdfdz = 4 ò z 2

2

0

¶( xy) ¶( yz) ¶( zx) + + x =y+z+ x ¶x ¶y ¶z

S

æ1 = 3çç ò xdx è0

1

1

1

0

0

0

( y + z + x) dxdydz

1 ö æ1ö dy ò0 ò0 dz ÷÷ = 3çè 2 ÷ø = 15. ø 1

64. (B) Ñ × D =

Page 468

2p

v

ò A × dS = ò ò ò

= 4z +

5

dz ò rdr ò df = 176

ò A × dS = ò (Ñ × A)dv S

Ñ ×A =

2

-1

v

63. (D)

0

0

0

uy ¶ ¶y 3 x 2 - gz

*******

=0

ò A × dL = 0 + 8 p + 0 - p = 7 p ò F × dS

0

f = p, A × dL = 0,

Along da, dr = 0,

d

0

0

c

ò A × dL

45°

i.e. a = 1 = b = g.

= (2) 3 p = 8 p

d

a

5

If F is irrotational, Ñ ´ F = 0

A × dL = r 3df

Along cd, df = 0,

ò A × dL

2

uz ¶ ¶z 3 xz 2 -

= ( -1 + g) u x + ( 3bz 2 - 3z 2 ) u y + ( 6 x - ax) u z

a

Along bc, dr = 0,

45°

é ux ê ¶ 65. (A) Ñ ´ F = ê ¶ x ê a x bz 2 + êë

f = 0,

ò A × dL

5

ò zdz ò df + 3ò dr ò zdz ò cos f df

x

b

A × dL = 0,

æ ö 3z cos f ÷÷ rdrdfdz çç 4 z + r è ø

æ 4 öæ 25 öæ p ö æ 25 öæ 1 ö = 4ç ÷ç . ÷ = 13157 ÷ç ÷ + 3(2)ç ÷ç è 2 øè 2 øè 4 ø è 2 øè 2 ø

Fig. S8.1.61

Along ab, df = 0,

v

ò D × dS = ò ò ò

y

c

Electromagnetics

1 ¶(rDr) 1 ¶( Df) ¶( Dz ) + + r ¶r r ¶q ¶z

3 z cos f r www.gatehelp.com

ù ú ú ú y úû


CHAPTER

8.2 ELECTROSTATICS

1. Let Q1 = 4 mC

be located at P1 (3, 11, 8) while

Q2 = -5 mC is at P2 (6, 15, 8). The force F2 on Q2 will be

6. A 2 mC point charge is located at A (4, 3, 5) in free space. The electric field at P(8, 12, 2) is

(A) -( 4.32 u x + 5.76 u y) N

(B) 4.32 u x + 5.76 u y N

(A) 131.1u r + 159.7 u f - 49.4 u z

(C) -( 4.32 u x + 5.76 u y) mN

(D) 4.32 u x + 5.76 u y mN

(B) 159.7 u r + 27.4 u f - 49.4 u z

2. Four 5 nC positive charge are located in the z = 0 plane at the corners of a square 8 mm on a side. A fifth

(C) 1311 . u r + 27.4 u f - 49.4 u z (D) 159.7 u r + 137.1u f - 49.4 u z

5 nC positive charge is located at a point 8 mm distant from each of the other charge. The magnitude of the

7. A point charge of -10 nC is located at P1 (0, 0, 0.5)

total force on this fifth charge is

and a charge of 2 mC at the origin. The E at P(0, 2, 1) is

(A) 2 ´ 10 -4 N (C) 0.014 N

(B) 4 ´ 10 -4 N (D) 0.01 N

3. Four 40 nC are located at A(1, 0, 0), B(-1, 0, 0), C(0,

(A) 68.83u r + 14.85 u f

(B) 68.83u r + 64.01u f

(C) 68.83u r - 14.85 u f

(D) 68.83u r - 64.01u f

8. Charges of 20 nC and -20nC are located at (3, 0, 0)

1, 0) and D(0, -1, 0) in free space. The total force on the

and (-3, 0, 0) and (-3, 0, 0), respectively. The magnitude

charge at A is

of E at y axis is

(A) 24.6u x mN (C) -13.6u x mN

(B) -24.6u x mN (D) 13.76u x mN

4. Let a point charge 41 nC be located at P1 (4, -2, 7) and a charge 45 nC be at P2 (–3, 4, –2). The electric field E at P3(1, 2, 3) will be

1080 (9 + y 2 ) 3 2

(B)

1080 (9 + y 2 ) 3

(C)

108 (9 + y 2 ) 3 2

(D)

108 (9 + y 2 ) 3

9. A charge Q0 located at the origin in free space

(A) 0.13u x + 0.33u y + 0.12 u z

produces a field for which E2 = 1 kV m at point P(–2, 2,

(B) -0.13u x - 0.33u y - 0.12 u z

–1). The charge Q0 is

(C) 115 . u x + 2.93u y + 109 . uz (D) -115 . u x - 2.93u y - 109 . uz 5. Let a point charge 25 nC be located at P1 (4, -2, 7) and a charge 60 nC be at P2 (-3, 4, -2). The point, at which on the y axis, is Ex = 0, is (A) -7.46 (C) -6.89

(A)

(B) -22.11 (D) (B) and (C)

(A) 2 mC

(B) -3 mC

(C) 3 mC

(D) -2 mC

10. The volume charge density r n = r oe -|x|-| y|-|z| exist over all free space. The total charge present is (A) 2r o

(B) 4r o

(C) 8r o

(D) 3r o

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Page 469


UNIT 8

Electromagnetics

11. A uniform volume charge density of 0.2 mC m 2 is

18. Two identical uniform charges with r l = 80 nC m are

present throughout the spherical shell extending from

located in free space at x = 0, y = ± 3 m. The force per

r = 3 cm to r = 5 cm. If r = 0 elsewhere, the total charge

unit length acting on the line at positive y arising from

present throughout the shell will be

the charge at negative y is

(A) 41.05 pC

(B) 257.92 pC

(A) 9.375u y mN

(B) 37.5u y mN

(C) 82.1 pC

(D) 129.0 pC

(C) 19.17u y mN

(D) 75u y mN

12.

If r v =

1 z 2 + 10

5 e -0 .1 r( p-|f|) mC m 3 in

the region

19. A uniform surface charge density of 10 nC m 2 is

0 £ r £ 10, -p < f < p and all z, and r v = 0 elsewhere,

present in the region x = 0, - 2 < y < 2 and all z if e = e o

the total charge present is

, the electric field at P(3, 0, 0) has

(A) 1.29 mC

(B) 2.58 mC

(C) 0.645 mC

(D) 0

(A) x component only (B) y component only

13. The region in which 4 < r < 5, 0 < q < 25 °, and

(C) x and y component

0.9 p < f < 11 . p contains the volume charge density of

(D) x, y and z component

f 2

r v = 10 ( r - 4) ( r - 5) sin q sin . Outside the region, r v = 0. The charge within the region is

20. The surface charge density is r s = 5 nC m 2 , in the

(A) 0.57 C

(B) 0.68 C

region r < 0.2 , z = 0, and is zero elsewhere. The electric

(C) 0.46 C

(D) 0.23 C

field E at A(r = 0, z = 0.5) is

14. A uniform line charge of 5 nC m is located along the

(A) 5.4 V m

(B) 10.1 V m

(C) 10.5 V m

(D) 20.2 V m

line defined by y = -2, z = 5. The electric field E at P(1, 21. Three infinite charge sheet are positioned as

2, 3) is (A) -9 u y + 4.5 u z

(B) 9 u y - 4.5 u z

follows: 10 nC m 2 at x = -3, - 40 nC m 2 at y = 4 and 50

(C) -18 u y + 9 u z

(D) 18 u y - 9 u z

nC m 2 at z = 2. The E at (4, 3, -2) is

15. A uniform line charge of 6.25 nC m is located along the line defined by y = -2, z = 5. The E at that point in the z = 0 plane where the direction of E is given by

(

1 3

u y - 23 u z ), is

(A) 4.5 u y + 9 u z

(B) 4.5 u y - 9 u z

(C) 9 u y - 18 u z

(D) 18 u y - 36 u z

(C) -48u y V m

(D) 24u y V m

(D) -0.56 u x - 2.23u y + 2.8 u z kV m Let E = 5 x 3u x - 15 x 2 yu y . The equation of the

(A) y =

m

respectively. The E at P(6, 0, 6) will be (B) 48u y V m

(C) 0.56 u x + 2.23u y + 2.8 u z kV m

stream line that passes through P(4, 2, 1) is

and y = -3

(A) -24u y V m

(B) 0.56 u x - 2.23u y + 2.8 u z kV m

22.

16. Uniform line charge of 20 nC m and -20 nC m are located in the x = 0 plane at y = 3

(A) 0.56 u x + 2.23u y - 2.8 u z kV m

128 x3

(B) x =

128 y3

64 x2

(D) x =

64 y2

(C) y =

23. A point charge 10 nC is located at origin. Four

17. Uniform line charges of 100 nC m lie along the

uniform line charge are located in the x = 0 plane as

entire extent of the three coordinate axes. The E at

follows : 40 nC m

at y = 1 and -5 m, -60 nC m

P(-3, 2, -1) is

y = -2 and - 4 m. The D at P(0, -3, 4) is

(A) -192 . u x + 2 u y - 108 . u z kV m

(A) -19.1u y + 25.5 u z pC m 2

(B) -0.96 u x + u y - 0.54 u z kV m

(B) 19.1u y - 25.5 u z pC m 2

(C) 0.96 u x - u y + 0.54 u z kV m

(C) -16.4 u y + 219 . u z pC m 2

(D) 192 . u x - 2 u y + 108 . u z kV m

(D) 16.4 u y - 219 . u z pC m 2

Page 470

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at


Electrostatics

24. A point charge 20 nC

Chap 8.2

is located at origin. Four

31. A spherical surface of radius of 3 mm is centered at

uniform line charge are located as follows 40 nC m at

P(4, 1, 5) in free space. If D = xu x C m 2 the net electric

y = ± 1 and 50 nC m at y = ± 2. The electric flux that

flux leaving the spherical surface is

leaves the surface of a sphere, 4 m in radius, centered at origin is (A) 1.33 nC

(B) 1.89 mC

(C) 1.33 mC

(D) 1.89 mC

(A) 113.1 mC

(B) 339.3 nC

(C) 113.1 nC

(D) 452.4 nC

32. The electric flux density is

25. The cylindrical surface r = 8 C contains the surface charge density, r s = 5 e

-20|z|

D=

2

nC m . The flux that leaves

the surface r = 8 cm, 1 cm < z < 5 cm

30 ° < f < 90 ° is

(A) 270.7 nC

(B) 9.45 nC

(C) 270.7 pC

(D) 9.45 pC

1 [10 xyzu x + 5 x 2 zu y + (2 z 3 - 5 x 2 y) u z ] z2

The volume charge density r v at (-2, 3, 5) is (A) 6.43

(B) 8.96

(C) 10.4

(D) 7.86

26. Let D = 4 xy u x + 2( x 2 + z 2 ) u y + 4 yzu z C m 2 . The total charge enclosed in the rectangular parallelepiped

33. If D = 2ru r C m 2 , the total electric flux leaving the

0 < x < 2, 0 < y < 3, 0 < z < 5 m is

surface of the cube, 0 < x , y, z < 0.4 is

(A) 360 C

(B) 180 C

(A) 0.32

(B) 0.34

(C) 100 C

(D) 560 C

(C) 0.38

(D) 0.36

27. Volume charge density is located in free space as rv = 2e

-1000 r

nC m

for 0 < r < 1 mm

3

and

rv = 0

34. If E = 4 u x - 3u y + 5 u z in the neighborhood of point P(6, 2, -3). The incremental work done in moving 5 C

elsewhere. The value of Dr on the surface r = 1 mm is

charge a distance of 2 m in the direction u x + u y + u z is

(A) 1.28 pC m 2

(B) 0.28 pC m 2

(A) -60 J

(B) 34.64 J

(C) 0.78 pC m 2

(D) 0.32 pC m 2

(C) -34.64 J

(D) 60 JJ

and 4 carry uniform

35. If E = 100 u r V m , the incremental amount of work

charge densities of 20 nC m and -4 nC m . The Dr at

done in moving a 60 mC charge a distance of 2 mm from

2 < r < 4 is

P(1, 2, 3) toward Q(2, 1, 4) is

28. Spherical surfaces at r = 2 2

(A) -

16 nC m 2 r2

2

(B)

80 (C) 2 nC m 2 r

16 nC m 2 r2

80 (D) - 2 nC m 2 r

+ 6 z 3u z C m 2 . The total charge enclosed in the volume 0 < x, y , z < a is 5 4 a 3

(B) a 5 + 6 a 4

(C) 6 a 5 + a 4

(D)

(B) 3.1 mJ

(C) -31 . mJ

(D) 0

36. A 10 C charge is moved from the origin to P(3, 1, -1)

29. Given the electric flux density, D = 2 xyu x + x 2 u y

(A) 6 a 5 +

(A) -5.4 mJ

5 5 a + 6a4 3

in the field E = 2 xu x - 3 y 2 u y + 4 u z V m along the straight line path x = -3 y, y = x + 2 z. The amount of energy required is (A) -40 J

(B) 20 J

(C) -20 J

(D) 40 J

37. A uniform surface charge density of 30 nC m 2 is present on the spherical surface r = 6 mm in free space.

30. Let D = 5 x y z u y . The flux enclosed by volume

The V AB between A ( r = 2 cm, q = 35 ° , f = 55 ° )

x = 3 and 3.1, y = 1 and 1.1, and z = 2 and 2.1 is

( r = 3 cm, q = 40 ° , f = 90 ° ) is

(A) 49.6

(B) 24.8

(A) 2.03 V

(B) 10.17 V

(C) 35.4

(D) 36.4

(C) 4.07 mV

(D) -10.17 V

4

4

4

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and B

Page 471


UNIT 8

Electromagnetics

38. A point charge is located at the origin in free space.

(A) -4.94 nC

(B) -4.86 nC

The work done in carrying a charge 10 C from point A

(C) -5.56 nC

(D) -3.68 nC

( r = 4, q = p 6 , f = p 4) to B( r = 4, q = p 3 , p 6) is (A) 0.45 mJ

(B) 0.32 mJ

(C) -0.45 mJ

(D) 0

46. A dipole having Qd = 100 V × m 2 4 pe o

39. Let a uniform surface charge density of 5 nC m 2 be present at the z = 0 plane, a uniform line charge density of 8 nC m be located at x = 0, z = 4 and a point charge of 2 mC be present at P(2, 0, 0). If V = 0 at A(0, 0, 5), the V at B(1, 2, 3) is (A) 10.46 kV

(B) 1.98 kV

(C) 0.96 kV

(D) 3.78 kV

is located at the origin in free space and aligned so that its moment is in the u z direction. The E at point ( r = 1, 45 ° , f = 0) is (A) 158.11 V m

(B) 194.21 V m

(C) 146.21 V m

(D) 167.37 V m

47. A dipole located at the origin in free space has a 40. A non uniform linear charge density, r L = 6 ( z + 1)

moment p = 2 ´ 10 -9 u z C.m. The points at which | E|q = 1

nC m lies along the z axis. The potential at P(r = 1, 0, 0)

mV m on line y = z, x = 0 are

in free space is ( V¥ = 0)

(A) y = z = ± 23.35

(B) y = z = ± 16.5, x = 0

(C) y = z = 16.5

(D) y = 0, z = 23.35, x = 0

2

(A) 0 V

(B) 216 V

(C) 144 V

(D) 108 V 48. A dipole having a moment p = 3u x - 5 u y + 10 u z

41. The annular surface, 1 cm < r < 3 cm carries the

nC.m is located at P(1, 2, -4) in free space. The V at Q

nonuniform surface charge density r s = 5r nC m . The V

(2, 3, 4) is

2

at P(0, 0, 2 cm) is (A) 81 mV

(B) 90 mV

(C) 63 mV

(D) 76 mV

42. If V = 2 xy 2 z 3 + 3 ln( x 2 + 2 y 2 + 3z 2 ) in free space the magnitude of electric field E at P (3, 2, -1) is (A) 72.6 V/m

(B) 79.6 V/m

(C) 75 V/m

(D) 70.4 V/m

V. The volume charge density at r = 0.6 m is (B) -1.79 nC m 3

(C) 1.22 nC m 3

(D) -1.22 nC m 3

(B) 1.26 V

(C) 2.62 V

(D) 2.52 V

49. A potential field in free space is expressed as V = 40 xyz . The total energy stored within the cube 1 < x, y, z < 2 is

43. It is known that the potential is given by V = 70 r 0 .6 (A) 1.79 nC m 3

(A) 1.31 V

(A) 1548 pJ

(B) 0

(C) 774 pJ

(D) 387 pJ

50. Four 1.2 nC point charge are located in free space at the corners of a square 4 cm on a side. The total potential energy stored is

44. The potential field V = 80 r 2 cos q V. The volume

(A) 1.75 mJ

(B) 2 mJ

(C) 3.5 mJ

(D) 0

charge density at point P(2.5, q = 30 °, f = 60 °) in free 51. Given the current density

space is (A) -2.45 nC m 3 (C) -1.42 nC m

3

(B) 1.42 nC m 3 (D) 2.45 nC m

J = 10 5[sin (2 x) e -2 yu x + cos (2 x) e -2 yu y ] kA m 2

3

The total current crossing the plane y = 1 in the

45. Within the cylinder r = 2, 0 < z < 1 the potential is

u y direction in the region 0 < x < 1, 0 < z < 2 is

given by V = 100 + 50r + 150r sin f V. The charge lies

(A) 0

(B) 12.3 mA

within the cylinder is

(C) 24.6 mA

(D) 6.15 mA

Page 472

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UNIT 8

Electromagnetics

64. In a certain region where the relative permitivity is

69. A potential field exists in a region where e = f ( x). If

2.4, D = 2 u x - 4 u y + 5 u z nC m 2 . The polarization is

r v = 0, the Ñ 2 V is 1 dF ¶V (A) f ( x) dx ¶x

(A) 2.8 u x - 5.6 u y + 7 u z nC m 2 (B) 3.4 u x - 6.9 u y + 8.6 u z nC m 2 (C) 1.2 u x - 2.3u y + 2.9 u z nC m 2

(C)

(D) 3.89 u x - 6.43u y + 8.96 u z nC m 2 65. Medium 1 has the electrical permitivity e1 = 15 . eo and occupies the region to the left of x = 0 plane. Medium 2 has the electrical permitivity e 2 = 2.5 e o and occupies the region to the right of x = 0 plane. If E1 in medium 1 is E1 = (2 u x - 3u y + 1u z ) V m

then E2

in

(B) f ( x)

1 df ¶V f ( x) dx ¶x

(D) -f ( x)

where r v = 0. It is know that both Ex and V are zero at the origin. The V ( x, y) is (A) 3( x 2 - y 2 )

(B) 3( y 2 - x 2 )

(C) 4 x 2 - 3 y 2

(D) 4 y 2 - 3 x 2

(A) (2.0 u x - 1.8 u x + 0.6 u z ) V m (B) (1.67 u x - 3u y + u z ) V m (C) (1.2 u x - 3u y + u z ) V m (D) (1.2 u x - 1.8 u y + 0.6 u z ) V m

*********

66. Two perfect dielectrics have relative permitivities e r1 = 2 and e r 2 = 8. The planner interface between them is the surface x - y + 2 z = 5. The origin lies in region 1. If E1 = 100 u x + 200 u y - 50 u z V m then E2 is (A) 400 u x + 800 u y - 200 u z V m (B) 400 u x + 200 u y - 50 u z V m (C) 25 u x + 200 u y - 50 u z V m (D) 125 u x + 175 u z V m 67. The two spherical surfaces r = 4 cm and r = 9 cm are for

u r then E2 4 < r < 6 and e r 2 = 5 for 6 < r < 9. If E1 = 1000 r2 is (A)

5000 ur V m r2

(B)

400 ur V m r2

(C)

2500 ur V m r2

(D)

2000 ur V m r2

68. The surface x = 0 separate two perfect dielectric. For x > 0,

let

e r1 = 3,

while

er 2 = 5

where

x < 0.

If

E1 = 80 u x - 60 u y - 40 u z V m then E2 is (A) (133.3u x - 100 u z - 66.7 u z ) V m (B) (133.3u x - 60 u z - 40 u z ) V m (C) ( 48 u x - 36 u y - 24 u z ) V m (D) ( 48 u x - 60 u y - 40 u z ) V m Page 474

df ¶V dx ¶x

70. If V ( x, y) = 4 e 2 x + f ( x) - 3 y 2 in a region of free space

medium 2 is

separated by two perfect dielectric shells, e r1 = 2

df ¶V dx ¶x

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Electrostatics

SOLUTIONS 1. (C) F2 = =

=

f = tan -1

= 65.9 cos 56.3° + 148.3 sin 56.3 = 159.7

(4, 4), (-4, 4), (4, -4), (-4, 4), the fifth charge will be on the z-axis at location z = 4 2. By symmetry, the force on the fifth charge will be z

directed, and will be four

times the z component

=

2

q2 4 pe o d 2

(5 ´ 10 -9) 2 ´ = 10 -2 N 4 p ´ ( 8.85 ´ 10 -12 )( 8 ´ 10 -3) 2 2

= -65.9 sin 56.3° + 148.3 cos 56.3° = 27.4 , Ez = -49.4 7. (C) Ep =

2 ´ 10 -8 4 pe o

é R1 2R 2 ù ê- |R |3 + |R |3 ú ë 1 2 û

R 2 = (0, 2, 1) - (0, 0, 0) =(0, 2, 1) é - (2 u y + 0.5 u z ) 2(2 u y + u z ) ù Ep = 9 ´ 10 9 ´ 10 -8 ê + ú ( 4.25) 3 2 ( 5)3 2 û ë E p = 54.9 u y + 44.1u z

3. (D) The force will be ( 40 ´ 10 -9) 2 é R CA R DA R BA ù F= + + 3 3 3ú ê 4 pe o ë|R CA| |R DA| |R BA| û where R CA = u x - u y , R DA = u x + u y , R BA = 2 u x

F=

Ef = E r × u f = 65.9( u x × u f) + 148.3 ( u y × u f)

R1 = (0, 2, 1) -(0, 0, 0.5) =(0, 2, 0.5)

4

|R CA|=|R DA|= 2,

12 = 56.3°, and z = 2 8

Er = Ep × u p = 65.9( u x × u r) + 148.3 ( u y × u r)

2. (D) Arranging the charge in the xy plane at location

´

é 4 u x + 9 u y - 3u z ù ê ú (106) 3 / 2 ë û

Then at point P, r = 8 2 + 12 2 = 14.4,

= ( 4.32 u x + 5.76 u y) mN

4

2 ´ 10 -6 4 pe o

= 65.9 u x + 148.3u y - 49.4 u z

Q1Q2 R12 4 pe o |R12|3

( 4 ´ 10 -6 )( -5 ´ 10 -6 ) ( 3u x + 4 u y) ´ 4 pe o 53

F=

Chap 8.2

|R BA|= 2

éu x - u y u x + u y 2u x ù ( 40 ´ 10 -9) 2 + + -9 ê 4 p ´ ( 8.85 ´ 10 ) ë 2 2 8 úû 2 2

At P, r = 5 , q = cos -1

1 5

= 63.4 ° and f = 90 °

So Er = E p × u r = 54.9[ u y × u r ] + 44.1[ u z × u r ] = 54.9 sin q sin f + 44.1 cos q = 68.83 Eq = E r × u q = 54.9[ u y × u q ] + 44.1[ u z × u q ] = 54.9 cos q sin f + 44.1 ( - sin q) = -14.85 Ef = Er × u f = 54.9( u y × u f) + 44.1( u z × u f) =54.9 cos f = 0

= 1376 . u x mN

8. (A) Let a point on y axis be P(0, y, 0)

10 -9 é 41R13 45R 23 ù 4. (C) E = + 2 ê |R 23|2 úû 4 pe o ë |R13|

Ep =

R13 = -3u x + 4 u y - 4 u z ,

R 23 = 4 u x - 2 u y + 5 u z

é 41( -3u x + 4u y - 4u z 45( 4u x + -2u y + 5u z ù E = 9 ´ 10 9 ´ 10 -9 ´ ê + ú ( 41) 3 / 2 ( 45) 3 / 2 ë û

= 1152 . u x + 2.93u y + 1089 . uz 5. (D) The point is P3(0, y, 0) R13 = -4 u x + ( y + 2) u y - 7 u z , R12 = 3u x + ( y - 4) u y + 2 u z Ex =

ù 10 -9 é 25 ´ ( -4) 60 ´ 3 + 4 pe o êë[ 65 + ( y + 2) 2 ]3 2 [13 + ( y - 4) 2 ]3 2 úû

To obtain Ex = 0,

0.48 y 2 + 1392 . y + 7312 . =0

which yields the two value y = -6.89, - 22.11 6. (B) Ep =

2 ´ 10 -6 R AP 4 pe o |R AP|3

20 ´ 10 -8 4 pe o

é R1 R2 ù ê-|R |3 - |R |3 ú ë 1 2 û

R1 = (0, y, 0) - (3, 0, 0) =(-3, y, 0) R 2 = (0, y 0) - (-3, 0, 0) = (3, y, 0),

|R1 | = |R 2 | =

9 + y2

é -3u x + yu y 3u x + yu y ù + Ep = 20 ´ 10 -9 ´ 9 ´ 10 9 ê ú 2 3 ( 9 + y 2 ) 3 úû êë ( 9 + y ) -1080 u x 1080 , |E| = = = 2 32 (9 + y ) (9 + y 2 ) 3 2 9. (B) The field at P will be Q é -2 u x + 2 u y - u z ù Ep = 0 ê ú, Ez = 1 kV m 4 pe o ë 93 2 û Q0 = -4 pe o ´ 9 3 2 ´ 10 3 = -3 m C 10. (C) This will be 8 times the integral of r n over the first octant

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Page 475


UNIT 8

¥

Q = 8ò

¥

¥

0

0

ò ò

0

é 6 u x - 3u y 6 u x + 3u y ù EP = 20nC ´ 2 ´ 9 ´ 10 9 ê 36 + 9 úû ë 36 + 9

r oe - x - y- z dxdydz = 8ro

11. (C) Q =

2p

p

0

0

= -48u y V m

0 .05

ò ò ò

0.2 r sin q drd qdf 2

0 .03

é r3 ù = ê4 p(0.2) ú = 82.1 pC 3 û 0 .03 ë 12. (A) Q =

òòò ¥ -p 0

ée =5 ê ë

R xp = ( -3, 0, - 1, ) - ( -3, 0, 0, ) = (0, 2, -1) Similarly R yp = (-3, 0, -1), R zp = (-3, 2, 0)

5 e -0 .1 r( p-|f|) z 2 + 10 10

-0 .1 r

é R xp R yp R zp ù + + ê 3 3 3ú ë|R xp| |R yp| |R zp| û

rL 2 pe o

17. (B) EP =

0 .05

-¥ p 10

Electromagnetics

ù ( -0.1 - 1) ( -0.101) ú ´ (0.1) 2 û0

Ep = 100 ´ 10 -9 ´ 2 ´ 9 ´ 10 -9 ¥

é 2u y - u z -3u x - u z -3ux + 2u y ù + + E p = 100 ´ 10 -9 ´ 2 ´ 9 ´ 10 -9 ´ ê ú 5 10 13 ë û

2p

( p - f) dz 2 + 10 0 z

ò 2ò

-¥

= -0.96 u x + u y - 0.54 u z kV m

¥

p2 dz -¥ z + 10

Q = 5 ´ 26.4 ò

18. (C) At y = 4, E =

2

¥

z ù é 1 = 5 ´ 26.4 ´ p2 ê tan -1 = 129 . mC 10 úû -¥ ë 10

dF = dqE = r L dzE r 2L dzu y

1

F=

ò

2 pe o

0

13. (A) f = 10

1 .1 p 25° 5

f

ò ò ò ( r - 4)( r - 5) sin q × sin 2 r

0 .9p 0 4

25°

5

1 .1 p

é r 5 9 r 4 20 r 3 ù é 1 1 fù é ù = 10 ê + ú ê2 q - 4 sin 2 q ú ´ ê-2 cos 2 ú 5 4 3 ë û 0 .9p ë û ë û4 0 = 10 [ -3.39 ][0.0266 ][0.626 ] = 0.57 C

( 6) = 18.75 u y mN r s dS R - R¢ 4 pe o |R - R¢|3

ò ò

19. (A) E =

sin q drdqdf

2

rL uy , 2 pe o

where R = 3u x and R¢ = yu y + zu z , ¥ 2 3u x - yu y - zuz r dydz EPA = L ò ò 4 pe o -¥ -2 (9 + y 2 + z 2 ) 3 2 Due to odd function

Rp r , 14. (D) Ep = L 2 pe o |R p|2

EPA =

R p = (1, 2, 3) - (1, -2, 5) = (0, 4, -2)

So there is only x component.

|R | p

2

= 20,

5 ´ 10 -9 Ep = 2 pe o

é4u y - 2u z ù ê ú = 18 u y - 9 u z 20 ë û

15. (C) With z = 0, the general field is Ez = 0 =

r L é( y + 2) u y - 5 u z ù 2 pe o êë ( y + 2) 2 + 25 úû

we require |E2 | = |2E y| So 2 ( y + 2) = 5 E=

Þ

y=

1 2

6.25 é2.5 u y - 5 u z ù = 9 u y - 18 u z 2 pe o êë 6.25 + 25 úû

rL 4 pe o

¥

2

òò

-¥ -2

3u x dydz (9 + y 2 + z 2 ) 3 2

20. (D) There will be z component of E only R = zu z , R¢ = ru r , Ez , Pa =

rs 4 pe o

2 p 0 .2

R - R¢ = zu z - ru z

zrdrdf 2 + r 2)3 2

ò ò (z 0

0

0 .2

ù ù 2pr s z é rzé 1 1 1 = ú = s ê 2 ê 2 ú 2 2 eo ê z 4 pe o ê z + r ú z + 0.04 úû û0 ë ë Ez =

rs eo

é z ê1 2 z + 0.04 êë

ù ú, at z = 0.5, Ez = 20.2 V m úû

21. (A) Since charge sheet are infinite, the field magnitude associated with each one will be r s 2 e o , which is position independent. The field direction will

r 16. (C) Ep = L 2 pe o

é R+ Q R- Q ù ê 2 2ú ë|R + Q| |R - Q| û

R+ Q = (6, 0, 6) - (0, 3, 6) = (6, -3, 0) R- Q = (6, 0, 6) - (0, -3, 6) = (6, 3, 0) Page 476

depend on which side of a given sheet one is positioned. é10 ´ 10 -9 40 ´ 10 -9 50 ´ 10 -9 ù EA = ê ux uy uzú 2 eo 2 eo ë 2 eo û = 0.56 u x + 2.23u y - 2.8 u z www.gatehelp.com


Electrostatics

22. (A)

dy E y -15 x 2 y -3 y = = = 5 x3 dx Ex x

dy 3dx =y 3

Þ

At P, 2 =

C1 43

Þ

a a

ln y = -3 ln x ln C y = C1 = 128

Þ

Chap 8.2

Þ

y =

C1 , x3

+

128 x3

ò ò -x

a a

= a4 + 0 -

a a + + 0 + 6a5 = a4 + 6a5 3 3

of

¶D y ¶y

= 20 x 4 y 3z 4

cube

=(3.05

1.05

2.05)

and

Volume

Ñ V = (0.1) 3 = 0.001 f = 20 ( 305 . ) 4 (105 . ) 3 (2.05) 3 (2.05) 4 0.001 =35.4

24. (C) h1 = 2 4 2 - 1 = 7.75,

31. (C) F = (Ñ × D ) Dv

h2 = 2 4 2 - 1 = 6.93

=

QT = 2 ´ 7.75 ´ 40n + 2 ´ 6.93 ´ 50 + 20n =1.33 mC

ò ò

Top

4

30. (C) Ñ × D = Center

Botoom

Right

4

D = -19.1u y + 25.5 u z pC m 2

a a

dxdz + ò ò x 2 dxdz + ò ò -6(0) 3 dxdy + ò ò 6a 3dxdy 0 0 0 0 0 0 0 0 1 4 4244 3 1 4243 1 442443 1 4 4244 3 Left

at that point. Thus D arises from the point charge alone 10 ´ 10 -9 ( -3u y + 4 u z ) , D= ( 32 + 4 2 )1 .5 4p

a a

2

arrangement of line charges, whose field will all cancel

0 .05 90 °

Back

Front

a a

23. (A) This point lies in the center of a symmetric

25. (D) Q =

a a

29. (C) Q = ò D × n dS = ò ò 2 aydydz + ò ò -2(0) dydz S 0 0 0 0 1 4 24 3 1 44244 3

¶( x) æ 4 3ö . nC ç p(0.003) ÷ = 1131 ¶x è 3 ø

é10 y æ 2 + 10 x 2 y öù =8.96 32. (B) r v = Ñ × D = ê ÷÷ú + 0 + çç z3 øû ( -2 , 3, 5) è ë z

5 e -20 z (0.08) dfdz nC

0 .01 30 0 .05

æp pö æ 1 ö -20 z = ç - ÷(5)(0.08)ç ÷e è2 6 ø è 20 ø 0 .01

33. (C) Ñ × D =

= 9.45 ´ 10 -3 nC = 9.45 pC

ò Ñ × D dv = 6 ´ (0.4)

26. (A) Out of the 6 surface only 2 will contribute to the net outward flux. The y component of D will penetrate the surface y = 0 and y = z and net flux will be zero. At x = 0 plane Dx = 0 and at z = 0 plane Dz = 0. This leaves the 2 remaining surfaces at x = 2 and z = 5. The net outward flux become 5 3

f = ò ò D x = 2 × u x dydz + 0 0

3

3

0

0

2p p

ò òD

x =2

× u z dzdy

0 0

é -r e Q = 8 pê êë 1000

2 -1000 r

-1000 r

0

2 e -1000 r ( -1000 r - 1) 1000(1000) 2 0

Q = 4 ´ 10 -9nC, Dr =

4.0 ´ 10 -9 Q = 0.32 C m 2 = 2 4 pr 4 p ´ (0.001) 2

28. (C) 4 pr Dr = 4 p(2 ) 20 ´ 10 2

Dr =

80 nC m 2 r2

2

= -5( 4 u x - 3u y + 5 u z ) × =-

10 3

( u x + u y + u z )(2) 3

( 4 - 3 + 5) = -34.64 J

æ ( u x - u y + u z )(2 ´ 10 -3) ö ÷ = -( 60 ´ 10 -6 )ç 100 u r × ç ÷ 3 è ø

r 2 sin q drdqdf 0 .001

+

34. (C) dW = - qE × dL

(2, 1, 4) - (7, 2, 3) = (1, -1, 1) ux - u y + uz , dW = - qE × dL u PQ = 3

0

0 .001

= 0.38

V

35. (B) The vector in this direction is

0 0

0 .001

ò ò ò 2e

3

3 2

= 5 ò 4(2) ydy + 2 ò 4(5) ydy = 360 C.

27. (D) Q =

1 d 2 ( r ´ 2 r) = 6 , r 2 dr

-9

2

Cm ,

ù ú úû

= - 12 ´

10 -6 3

( u r × u x - u r × u y)

æ2 ö At r , f = tan -1 ç ÷ = 63.4 ° è1ø u r × u x = cos 63.4 ° = 0.447, u r × u y = sin 63.4 ° = 0.894 dW = 31 . mJ 36. (A) W = - q ò E × dL

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Page 477


UNIT 8

Electromagnetics

= -10 ò (2 xu x - 3 y 2 u y + 4 u z ) ( dxu x + dyu y + dzu z ) 3

1

-1

0

0

0

é ù 18 z -ê6 xy 2 z 2 + 2 uz 2 2 ú x + y + z 2 3 ë û

= -10 ò 2 xdx + 10 ò 3 y 2 dy - 10 ò 4 dz = -40 J

. u z V m, Ep = 7.1u x + 22.8 u y - 711 |E|= 75 V m

37. (D) rA = 2 cm, rB = 3 cm V ( r) =

43. (D) E = -ÑV = -

Q 4 pa 2r s ( 6 ´ 10 -3) 2 ´ 30 ´ 10 -9 = = 4 pe o r 4 pe o r 8.85 ´ 10 -12 ´ r

= -42 r -0 .4 u r V m,

0.122 0.122 0.122 , V AB = V A - VB = V ( r) = = = 2.03 r 0.02 0.03 38. (D) W = -

Q1Q2 4 pe o

rB

ò

rA

dr Q1Q2 = r 2 4 pe o

D = e oE = -42 r -0 .4 e o u r C m 2 1 d 2 1 d e rv = Ñ × D = 2 ( r Dr ) = 2 ( -42 e o r 1 .6 )= - 67.2 1o.4 r dr r dr r

é1 1ù êr - r ú ë B Aû

At r = 0.6 m, rv = -

rA = rB , W = 0 39. (B) VP ( s) =

Q , 4pe o r

r L dr r + C1 = - L ln r + C1 2 pe or 2 pe o

Vs ( z) = - ò

rs rz dz + C2 = s + C2 2 eo 2 eo

D = e oE = -80 e o (2 r cos q u r - r sin q u q) é1 2 2 ù 1 ¶ rv = Ñ × D = ê 2 ( r Dr ) + ( Dr sin q) ú r sin q ¶q ë r ¶r û

Here r , r , z are the scalar distance from the charge. r = 2 2 + 5 2 = 29 , r = (5 - 4) 2 = 1 , z = 5 By putting these value. C = -193 . ´ 10 3 2

r = 12 ` + 12 = 2 , z = 3, 40. (D) Vr =

¥

-¥

2

é1 ¶ ù 1 ¶ r v = eo ê (rEr) + Ef ú r ¶f û ër ¶r é (50 + 150 sin f) 150 sin f ù 50e o = e o ê+ =C m2 ú r r r ë û 1 2p 2

Q =ò

R = zu z , R¢ = ru r, dS = rdrdf, Vr =

ò ò 0

0 .01

(5 ´ 10 -9)r 2 drdf 4 pe o r 2 + z 2

0

ù z2 5 ´ 10 -9 ér 2 2 r + z ln (r + r 2 + z 2 ) ú ê 2 eo 2 ë2 û 0 .01

At z = 0.02, Vp = 0.081 V

0 0

50 e o rdrdfdz = - 2 p(50) e o 2 = -5.56 nC r

46. (A) V =

42. (C) E = -ÑV é ù 6x = - ê2 y 2 z 2 + 2 ux 2 2ú x + 2 y + 3z û ë é ù 12 y - ê4 xyz 3 + 2 uy 2 2 ú x + 2 y + 3z û ë

Qd cos q 100 cos q , = 4 pe o r 2 r2

1 ¶V æ ¶V ö E = -ÑV = -ç ur + uq ÷ r r ¶ ¶q è ø =

Page 478

òò

, 0 .03

Vp =

1 ¶V ¶V ur uf ¶r r ¶f

D = e o E, r v = Ñ × D = e oÑ × E

¥

r s dS 41. (A) Vr = òò , 4 pe o|R - R¢| 2 p 0 .03

r V = -320 e o cos q = -2.45 nC m 3

= -(50 + 150 sin f) u r - (150 cos f) u f ,

VB = 198 . kV

6 ´ 10 -9 dz r L dz = ò = 108 V 4 ( 2 1) 3 2 oR -¥ pe o z +

ò 4 pe

é1 12 sin q cos q ù r v = -80 e o ê 2 ´ 3r 2 cos q ú r sin q ër û

45. (C) E = -ÑV = -

At point N, r = (2 - 1) + 2 + z = 14 2

¶V 1 ¶V ur uq ¶r r ¶q

= -160 r cos q u r + 80 r sin q u q V m

Q r r - L ln r - s z + C 4 pe o r 2 pe o 2 pe o

V =

67.2 ´ 8.85 ´ 10 -12 = -1.22 nC m 3 (0.6)1 .4

44. (A) E = -ÑV = -

Vl (r) = - ò

dV u r = -(0.6)(70) r -0 .4 u r dr

100 (2 cos q u r + sin q u q), r3

|E| = 100( 4 cos 2 q + sin 2 q)1 2 47. (A) E = www.gatehelp.com

= 100 ´

5 = 158.11 V m 2

2 ´ 10 -9 [2 cos q u r + sin q u q ], 4 pe o r 3


Electrostatics

55. (A) I = òò J × n z = 0 .2 dS

y = z lies at q = 45° 2 ´ 10 -9 1 = 10 -3 Eq = 4 pe o r 3 2

S

(required) =

r 3 = 12.73 ´ 10 3, r = 23.35

0

56. (B) So Eal = Est =

Solving

4 pe o (1 + 1 + 8 2 )1 .5

é 1 ù 1 1 49. (A) E = -ÑV = 40 ê 2 u x + uy + uzú 2 2 x yz xy z xyz ë û

òòò E × Edv 2

2

2

ò ò

1

1

1

Þ

J al =

sac J st sst

J st = 3.2 ´ 10 5 A m 2

57. (B) J =

We = 800 e o ò

J al J st = sal sst

I = p(2 ´ 10 -3) J st + p[( 4 ´ 10 -3) 2 - (2 ´ 10 -3) 2 ]J al = 80

( 3u x - 5 u y + 10 u z ) × ( u x + u y + 8 u z ) ´ 10 -9

=1.31 V

eo 2

ò ò

0 .4

40 40 rdrdf = log[r 2 + (0.1) 2 ] (2 p) 2 2 r + (0.1) 2 0

= 40 p log 17 = 356 A

where R - R¢ = Q - P = (1, 1, 8)

We =

2 p 0 .4

0

P × (R - R¢) 48. (A) V = 4 pe o|R - R¢|3

So VP =

Chap 8.2

E=

4 u r A m2, 2prl

4 12.73 J ur V m = = s 2 prls rl 3

5

12.73 12.73 5 6.51 V u r × u rdr = ln = l r l 3 l 3

V = - ò E × dL = ò

é 1 1 1 ù ê x 4 y 2 z 2 + x 2 y 4 z 2 + x 2 y 2 z 4 ú dxdydz ë û

5

V 6.51 1.63 R= = = W I 4l r

= 1548 pJ æ ¶V ö 1 ¶V ¶V 58. (D) E = -ÑV = -çç ur + uf + u z ÷÷ r ¶f ¶z è ¶r ø (r + 1) 2 = -z 2 cos f u r + z sin f u f - 2(r + 1) z cos f u z r

1 4 50. (A) W = å qn Vn 2 n =1 V1 = V21 + V31 + V41 =

q é 1 1 1 ù + + ê 4 pe o ë0.04 0.04 0.04 2 úû

E = -1.82 u f + 14.5 u f - 2.67 u z V m E ×E r s = eo E × n s = eo |E|

V1 = V2 = V3 = V4 W =

2 ´ (1.2 ´ 10 -9) 2 1 ( 4) q1 V1 = 4 pe o (0.04) 2

51. (B) I = òò J × n =

ò ò 10

5

r s = e o 1.82 2 + 14.5 2 + 2.67 2 = 1315 . pC m 2

2 1

y =1

S

2 1

1 ù é . mJ êë2 + 2 úû = 175

dS = ò ò J × u y 0 0

y =1

dxdz 59. (C) V =

cos (2 x) 2 -2 y dxdz = 12307 kA = 12.3 MA

p p sin 3 2 = 2.5 V 23

40 cos

So the equation of the surface is 40 cos q sin f = 2.5, 16 cos q sin f = r 3 r2

0 0

52. (C) I = òò J × n dS S

=

60. (A) E = -ÑV, D = eE = - e oÑV

2 p 0 .3p

ò 0

800 sin q (0.80) 2 sin q dqdf =154.8 A ò 2 + ( 0 . 80 ) 4 0 .1 p

53. (D) F = ma = qE, a=

qE ( -1.602 ´ 10 -19)( -4 ´ 106 ) = u z = 7.0 ´ 1017 u z m s 2 , m 9.11 ´ 10 -3

v = at = 7.0 ´ 10 tu z m s 17

54. (D) =

¶r V 1 ¶ ¶J z = -Ñ × J = (rJ r) + ¶t r ¶r ¶z

1 ¶ ¶ æ -20 ö (25) + ç ÷ =0 r ¶r ¶z çè r 2 + 0.01 ÷ø

é ù x 200 x ¶ ux + 2 u z ú C m2 = -e o ê200 z 2 ( ) x x + 4 x + 4 ¶ ë û 200 e o x D (z= 0 ) = - 2 u z C m2, x +4 r s = D × u z z= 0 = Q=

0

2

-3

0

ò ò

- 200 e o x C m2, x2 + 4 2

1 - 200 e o x dxdy = -( 3)(200) e o ln[ x 2 + 4 ] 2 x2 + 4 0

= - 300 ln 2 = - 1.84 nC

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Page 479


UNIT 8

61. (B) E = -ÑV ,

-3 xy 2 z 3 satisfy this equation.

62. (D) The plane can be replaced by -60 nC at Q (2, 5, -6).

Electromagnetics

Since E is normal to the surface e 2 1000 400 En 2 = r1 En 1 = ´ ur = 2 ur V m r2 r er 2 5

R = (5, 3, 1) - (2, 4, 6) = (3, -1, -5)

68. (D) D n 1 = D n 2

R¢ = (5, 3, 1)-(2, 4, -6) = (3, -1, 7),

D n 1 = 80 ´ 3e o = D n 2 = 5 e oEn 1 ,

|R |= 35 , |R¢|= 59

E2 = 48 u x - 60 u y - 40 u z

VP =

q q q é 1 1 ù = ê 4 pe o R 4 pe o R¢ 4 pe o ë 35 59 úû -9

Vp = 60 ´ 10 ´ 9 ´ 10

-9

rL 2p eo

0 = Ñ × ( - f ( x)ÑV ) é dF ¶V ¶2V ¶2V ¶2V ù =-ê + f ( x) 2 + f ( x) 2 + f ( x) 2 ú ¶x ¶y ¶z û ë dx ¶x

é Final Distance from the charge ù rL ln ê 2 pe o ëInitial Distance from the charge úû

é 1 12 + (2 - 1) 2 12 + (2 - (-2)) 2 12 + 32 ù - ln - ln êln + ln ú 1 1 1 êë 2 úû

= 2.40 kV 64. (C) P = D - e oE = D -

Þ

. e o = 2.5 e o En 2 = 2 ´ 15 Et1 = Et 2 ,

Þ

D D = ( e r - 1) er er

e1E N1 = e 2E N 2

En 1 = - 817 .

1 6

6

f ( x1 ) = -4 e 2 x + 3 x 2 + C, f (0, 1) = -4 + C2 V (0, 0) = 0 = 4 + f (0) f ( x) = -4 e

C2 = 0. f (0) = -4

+ 3x , 2x

- 4 e2 x + 3 x 2 - 3 y 2 = 3 ( x 2 - y 2 )

[u x - u y + 2u z ] *********

Et1 = Et 2 and D n 1 = D n 2 e r1 e oEn 1 = e r 2 e oEn 2 e 1 1 = r1 En 1 = E n 1 , E2 = Et 2 + En 1 er 2 4 4

= 133.3 u x + 166.7 u y + 16.67 u z - 8.33 u x + 8.33 u y - 16.67 u z

= 125 u x + 175 u y V m 67. (B) D n 1 = D n 2 , Page 480

Þ

2

[100 - 200 - 100 ] = -817 . Vm

Et1 = E1 - En 1 = 133.3u x + 166.7 u z + 16.67 u z

En 2

2x

V ( x, y) = 4 e

= -33.33u x + 33.33u y - 66.67 u z V m

Þ

Ñ 2 V = 0,

Integrating again

66. (D) The unit vector that is normal to the surface is u x - u y + 2u z ÑF , uN = = |ÑF| 6 1

Þ

It follow that C1 = 0

En 2 = 12 .

E2 = 1.2 u x - 3u y + 1u z

En 1 = E1 × u N =

70. (A) r v = 0

¶2 f - 6 =0 ¶x 2 ¶f ¶f = -16 e 2 x + 6 Þ = - 8 e 2 x + 6 x + C1 ¶x ¶x ¶V ¶f Ex = = 8 e2 x + ¶x ¶x ¶f ¶f Ex (0) = 8 + =0 Þ = -8 ¶x x = 0 ¶x x = 0

P = 12 . u x - 2.3u y + 2.9 u z nC m 2 65. (C) D n 1 = D n 2

é dF ¶V ù =-ê + f ( x)Ñ 2 V ú ë dx ¶x û 1 dF ¶V Þ Ñ 2V = f ( x) dx ¶x

Ñ 2 V = 16 e2 x +

(2.4 - 1) ´ 10 -9 2.4

P = (2 u x - 4 u y + 5 u z )

e r1 e oEn 1 = e r 2 e oEn 2

En1 = 48

Ñ × D = r v = 0 = Ñ × ( - f ( x)ÑV )

1 ù é 1 êë 35 - 59 úû = 21 V

V0 = 0, Vp = -

D = eE

69. (A) D = eE = - f ( x)ÑV ,

63. (A) Using method of images Vp - V0 = -

and Et1 = Et 2 ,

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UNIT 8

Electromagnetics


Statement for Q.9–11: An infinite filament on the z-axis carries 10 mA in

In the cylindrical region

the u z direction. Three uniform cylindrical current

2 r + for r £ 0.6 r 2 3 for r > 0.6 Hf = r Hf =

sheets are also present at 400 mA m at r = 1 cm, -250 mA m at r = 2 cm and 300 mA m at r = 3 cm. 9. The magnetic field H f at r = 0.5 cm is (A) 0.32 A m

(B) 0.64 A m

15. The current density J for r < 0.6 mm is

(C) 1.36 mA m

(D) 0

(A) 2u z A m

(B) -2u z A m

(C) u z A m

(D) 0

10. The magnetic field H f at r = 15 . cm is (A) 1.63 A m

(B) 0.37 A m

16. The current density J for r > 0.6 mm is

(C) 2.64 A m

(D) 0

(A) 2u z A m

(B) -3u z A m

(C) 3u z A m

(D) 0

11. The magnetic field H f at r = 3.5 cm is (A) 0.14 A m (C) 0.27 A m

(B) 0.56 A m

17.

(D) 0.96 A m

v = ( 3u x + 12 u y - 4 u z ) ´ 10

An

electron 5

with

velocity

m s experiences no net

forces at a point in a magnetic field B = u x + 2 u y + 3u z Statement for Q.12–14:

mWb m 2 . The electric field E at that point is

In the fig. P8.3.12–14 The region 0 £ z £ 2 is filled

(A) -4.4 u x + 1.3u y + 0.6 u z kV m

with an infinite slab of magnetic material (m r = 2.5). The

(B) 4.4 u x - 1.3u y - 0.6 u z kV m

surface of the slab at z = 0 and z = 2, respectively, carry

(C) -4.4 u x + 1.3u y + 0.6 u z kV m

surface current 30u x A m and -40u x as shown in fig.

(D) 4.4 u x - 1.3u y - 0.6 u z kV m

z

18. A point charge of 2 ´ 10 -16 C and 5 ´ 10 -26 kg is

mo

z=2

-40ux A/m

moving in the combined fields B = - 3u x + 2 u y - u z mT and E = 100 u x - 200 u y + 300 u z V m. If the charge velocity at t = 0 is v(0) = (2 u x - 3u y - 4 u z ) 10 5 m s, the

mr = 2.5 mo

z=0

30ux A/m

Fig. P8.3.12–14

12. In the region 0 < z < 2 the H is (A) -35u y A m

(B) 35u y A m

(C) -5u y A m

(D) 5u y A m

x

acceleration of charge at t = 0 is (A) 600[ 3u x + 2 u y - 3u z ]10 9 m s 2 (B) 400[ 6 u x + 6 u y - 3u z ]10 9 m s 2 (C) 400[ 6 u x - 6 u y + 3u z ]10 9 m s 2 (D) 800[ 6 u x + 6 u y - u z ]10 9 m s 2 19. An electron is moving at velocity v = 4.5 ´ 10 7 u y

13. In the region z < 0 the H is (A) 5u y A m

(B) -5u y A m

(C) 10u y A m

(D) -10u y A m

14. In the region z > 2 the H is (A) 5u y A m

(B) -5u y A m

(C) 35u y A m

(D) -35u y A m

Page 482

m s along the negative y-axis. At the origin, it encounters the uniform magnetic field B = 2 .5 u z mT, and remains in it up to y = 2.5 cm. If we assume that the electron remains on the y –axis while it is in the magnetic field, at y = 50 cm the x and z coordinate are respectively (A) 1.23 m, 0.23 m

(B) -1.23 m, -0.23 m

(C) -11.7 cm, 0

(D) 11.7 cm, 0

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Magnetostatics


Chap 8.3

25. If the current filament is located at y = 0.5, z = 0,

A rectangular loop of wire in free space joins points

and u L = u x , then F is

A(1, 0, 1) to B(3, 0, 1) to C(3, 0, 4) to D(1, 0, 4) to A. The

(A) 35 .2 u y nN m

(B) 68.3u x nN m

wire carries a current of 6 mA flowing in the u z

(C) 105.6 u z nN m

(D) 0

direction from B to C. A filamentary current of 15 A flows along the entire z, axis in the u z directions. 20. The force on side BC is (A) -18u x nN

(B) 18u x nN

(C) 3.6 u x nN

(D) -3.6 u x nN

21. The force on side AB is (A) 23.4 u z mN

(B) 16.4 u z mN

(C) 19.8 u z nN

(D) 26.3u z nN

(B) -36u x nN

(C) 54u x nN

(D) -54u x nN

fig.

P8.3.23.

The

magnetic

B = 6 xu x - 9 yu y + 3zu z Wb m 2 .

flux

The

(C) 4u x mN m

(D) -4u x mN m

27. The force exerted on the filament by the current

density total

(B) -0.4 u x N m

A conducting current strip carrying K = 6 u z A m lies in the x = 0 plane between y = 0.5 and y = 15 . m. There is also a current filament of I = 5 A in the u z direction on the z –axis.

23. Consider the rectangular loop on z = 0 plane shown in

(A) 0.4 u x N m


22. The total force on the loop is (A) 36u x nN

26. Two infinitely long parallel filaments each carry 100 A in the u z direction. If the filaments lie in the plane y = 0 at x = 0 and x = 5 mm, the force on the filament passing through the origin is

is

force

experienced by the rectangular loop is

strip is (A) 12.2 u y mN m

(B) 6.6 u y mN m

(C) -12.2 u y mN m

(D) -6.6 u y mN m

28. The force exerted on the strip by the filament is

y 2 5A

(A) -6.6 u y mN m

(B) 6.6 u y mN m

(C) 2.4 u x mN m

(D) -2.4 u x mN m


1

In a certain material for which m r = 6.5, 0

1

2

3

x

H = 10 u x + 25 u y - 40 u z A m

Fig. P8.3.23

(A) 30u z N

(B) -30u z N

(C) 36u z N

(D) -36u z N

29. The magnetic susceptibility c m of the material is (A) 5.5 (C) 7.5


30. The magnetic flux density B is

Statement for Q.24–25: Three uniform current sheets are located in free

(A) 82 u x + 204 u y - 327 u z mWb m 2

space as follows: 8u z A m at y = 0, -4u z A m at y = 1

(B) 82 u x + 204 u y - 327 u z mA m

and -4u z A m at y = -1. Let F be the vector force per

(C) 82 u x + 204 u y - 327 u z mT

meter length exerted on a current filament carrying 7

(D) 82 u x + 204 u y - 327 u z mA m

mA in the u L direction.

31. The magnetization M is

24. If the current filament is located at x = 0, y = 0.5

(A) 75 u x + 187.5 u y - 300 u z A m 2

and u L = u z , then F is

(B) 75 u x + 187.5 u y - 300 u z A m 2

(A) 35 .2 u y nN m

(B) -35 .2 u y nN m

(C) 55 u x + 137.5 u y - 220 u z A m 2

(C) 105.6 u y nN m

(D) 0

(D) 55 u x + 137.5 u y - 220 u z A m 2 www.gatehelp.com

Page 483


UNIT 8

32. The magnetic energy density is (A) 19 mJ m

2

If H increases from 0 to 210 A m, the energy

(B) 9.5 mJ m

(C) 16.3 mJ m

2

Electromagnetics

stored per unit volume in the alloy is

2

(D) 32.6 mJ m 2

(A) 6.2 MJ m 3

(B) 1.3 MJ m 3

(C) 2.3 kJ m 3

(D) 2.9 kJ m 3

Statement for Q.33–34: For a given material magnetic susceptibility c m = 31 . and within which B = 0.4 yu z T.

(B) 151.6 yu z kA m

(C) 102.7 yu z kA m

(D) 77.6 yu z kA m

cube of size a, the magnetization volume current density is 12 (A) uz a

33. The magnetic field H is (A) 986.8 yu z kA m

40. If magnetization is given by H = 6a ( - yu x + xu y) in a

(C)

34. The magnetization M is

6 uz a

(B)

6 ( x - y) a

(D)

3 ( x - y) a

41. The point P(2, 3, 1) lies on the planner boundary

(A) 241yu z kA m

(B) 318.2 yu z kA m

separating region 1 from region 2. The unit vector

(C) 163yu z kA m


u N12 = 0.6 u x + 0.48 u y + 0.64 u z is directed from region 1 to

35. In a material the magnetic field intensity is H = 1200 A m when B = 2 Wb m 2 . When H is reduced to 400

A m,

B = 1.4

Wb m 2 .

The

change

in

the

magnetization M is (A) 164 kA m

(B) 326 kA m

(C) 476 kA m

(D) 238 kA m

region

2.

If

m r1 = 2,

H1 = 100 u x - 300 u y + 200 u z A m, then H 2 is (A) 40.3u x + 48.3u y - 178.9 u z A m (B) 80.2 u x - 315.8 u y + 178.9 u z A m (D) 80.2 u x + 48.3u y + 178.9 u z A m

each atom has a dipole moment of 2.6 ´ 10 -30 u y A × m 2 . The H in material is (m r = 4.2) (A) 2.94 u y A m

(B) 0.22 u y A m

(C) 0.17 u y A m

(D) 2.24 u y A m

42. The plane separates air ( z > 0, m r = 1) from iron ( z £ 0, m r = 20). In air magnetic field intensity is H = 10 u x + 15 u y - 3u z A m. The magnetic flux density in iron is (A) 5.02 u x + 7.5 u y - 0.076 u z mWb m 2 (B) 12.6 u x + 18.9 u y - 75.4 u z mWb m 2

37. In a material magnetic flux density is 0.02 Wb m 2

(C) 251u x + 377 u y - 377 . u z mWb m 2

and the magnetic susceptibility is 0.003. The magnitude

(D) 251u x + 377 u y - 1508 u z mWb m 2

of the magnetization is (B) 23.4 A m

(C) 16.3 A m

(D) 8.4 A m

43. The plane 2 x + 3 y - 4 z = 1 separates two regions. Let m r1 = 2 in region 1 defined by 2 x + 3 y - 4 z > 1, while m r 2 = 5 in region 2 where 2 x + 3 y - 4 z < 1. In region

38. A uniform field H = - 600 u y A m exist in free space.

H1 = 50 u x - 30 u y + 20 u z A m. In region 2, H 2 will be

The total energy stored in spherical region 1 cm in

(A) 63.4 u x + 4318 . u y - 19.4 u z A m

radius centered at the origin in free space is

(B) 52.9 u x - 25.66 u y + 14.2 u z A m

(A) 0.226 J m

3

(B) 1.452 J m

(C) 1.68 J m 3

(C) 48.6 u x - 16.4 u y - 46.3u z A m

3


(D) 0.84 J m 3

39. The magnetization curve for an iron alloy is approximately given by B= Page 484

1 H + H 2 mWb m 2 3

and

(C) 40.3u x - 315.8 u y - 178.9 u z A m

36. A particular material has 2.7 ´ 10 29 atoms m 3 and

(A) 47.6 A m

m r2 = 8

********

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Magnetostatics

SOLUTIONS 1. (C) H = =

I 4p

IdL ´ u R 4 pR 2 -¥

ò

[2 + ( 3 - y) ] 2

0

2 32

R

R = (1 + x) + 32 + 2 2 = x 2 + 2 x + 14 2

H=

-u ydy[2 u x + ( 3 - y) u y ]

¥

-(1 + x) u x + 3u y + 2 u z

uR =

¥

ò

Chap 8.3

4 dxu x ´ [ -(1 + x) u x + 3u y + 2 u z ]

¥

ò

4 p ( x 2 + 2 x + 14) 3 2

-¥

=

(12 u z - 8 u y) dx

¥

ò 4 p( x

=

+ 2 x + 14)

2

-¥

¥

3/2

=

2 u z dy I = ò 2 4 p 0 [2 + ( 3 - y) 2 ]3 2

= 0.147 u z - 0.098 u y A m

3 - y = 2 tan q, -dy = 2 sec 2 q,

5. (B) H =

q1 = 56.31°, q 2 = -90 ° H=

I 4p

56 .31 °

ò

-90 °

2 u z dq 2 sec q

=

I u z [sin q ] -5690.31° °= 145.8 u z mA m = 4p 2. (A) H = H y + H z , H z =

Iz uf 2pr

r = ( -3) 2 + ( 4) 2 = 5 3u y + 4 u x = 5 5 24 ( 4 u x + 3u y) Hz = = 0.611u x + 0.458 u y mA m 2 p(5) 5 uf =

Hy =

- u z ´ ( -3u x + 4 u y)

Iy 2pr

2

I uf 2 pr

( -3u x + 5 u z ) 34

=

3u z - 5 u x 34

( -5 u x + 3u z )

12 2 p 34

-a

3. (A) H =

I uf 2 pr

4

ò

-4

Idzu z ´ (ru r - zu z ) 4 p(r 2 + z 2 ) 3 2

I æç a 12 2 pr ç (r + a 2 ) è I At r = 1, H = 2pr

Þ

a

1Þ

1+ a 1

a=

3

=

2p

ò

-a

Iau f 2 pr(r 2 + z 2 ) 3 2

ö ÷u A m ÷ f ø

1 2

= 0.577 m

IdL ´ u R 4 pR 2 Idfu f ´ ( - ur ) I uz A m = 2a 4 pa

K ´ u R dxdy 4 pR 2 ¥ 4 u x ´ ( - xu x - yu y - 3u z ) dx dy

7. (D) H = òò 2

=

ò ò

4 p( x 2 + y 2 + 9) 3 2

-2 -¥

=

ö I æç 4 ÷u 1f 2 pr ç (r 2 + 16) ÷ø è

=-

IdL ´ u R , IdL = 4 dxu x 4 pR 2

=

=

I = 3 A, a = 0.5 m, H = 3u z A m

I I 8 uf uf 2 pr 4 p r (r 2 + 16)

4. (B) H = ò

2

a

6. (A) H = ò

=

H = 0.1u f A m

rIdzu f 4 p(r 2 + z 2 ) 3 2

=

4

f = 60 °, I = 3p,

-a

Idzu z ´ (ru r - zu z ) 4 p(r 2 + z 2 ) 3 2

rIdzu f rIu f z ò- a 4 p(r 2 + z 2 ) 3 2 = 4 p r 2 (r 2 + z 2 ) 3 2

rdz I I uf uf ò 2 2 pr 4 p -4 4 p(r + z 2 ) 3 2

At r = - 3,

a

ò

4 p13

a

0

34

H = H y + H z = 0.331u x + 0.458 u y + 0.168 u z mA m

=

ò

2

= - 0.281u x + 0.168 u z mA m

=

a

u f , r = ( -3) + (5) = 34

uf = u y ´ Hy =

I uf 2 pr

2(12 u z - 8 u y)

2

¥

ò ò

4 ( - yu z - 3u y) dx dy

-2 -¥ 2

¥

4 p( x 2 + y 2 + 9) 3 2 12 u ydx dy

ò ò 4 p( x

-2 -¥

3 uy p

2

ò

-2

2

+ y 2 + 9) 3 2 2 dy y2 + 9 2

-

6 1 æ yö u y tan -1 ç ÷ = -0.75 u y A/m p 3 è 3 ø -2

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Page 485


UNIT 8

IdL ´ u R 4 pR 2 -Idzu z ´ ( -zu z + u y)

8. (D) H = ò ¥

=ò 0

¥

=ò 0

4 p(1 + z )

2 32

Idzu x + 4 p(1 + z 2 ) 3 2

æ I ç zu x = 4 p çç (1 + z 2 ) è =

0

¥

+

Idxu x ´ ( - xu x + u y)

¥

ò

4 p(1 + x )

2 32

0

¥

ò 0

Idxu z 4 p(1 + x 2 ) 3 2

uy 2 12

uzù 3 ú ´ 10 5 ´ 10 -3 ú -4 úû

= [ u x ( -8 - 36) - u y( -4 - 9) + u z (12 - 6)] ´ 10 2 V m = -4.4 u x + 1.3u y + 0.6 u z kV m 18. (D) v(0) ´ B = (2 u x - 3u y - 4 u z )10 5

ö ÷ + ÷ (1 + x 2 ) ¥ ÷ ø xu z

éu x ê1 ê êë 3

Electromagnetics

0

´ ( -3u x + 2 u y - u z )10 -3

I ( u x + u z ) = 0.8 ( u x + u z ) mA m 4p

= 1100 u x + 1400 u y - 500 u z F(0) = Q [E + v ´ B] = 2 ´ 10 -16 [1200 u x + 1200 u y - 200 u z ]

9. (A) Using Ampere’s circuital law

= 4 ´ 10 -14 [ 6 u x + 6 u y - u z ]

ò H × dL = 2prHf = I encl

F = ma

At r = 0.5 cm,

I encl = 10 mA

-3

2 p(5 ´ 10 ) H f = 10, H f = 0.32 A m 10. (B) At r = 15 . cm enclosed current I encl = 10 + 2 p(0.01)( 400) = 35.13 mA 2 p(0.015) H f = 35.13 ´ 10

-3

Þ

H f = 0.37 A m

11. (C) The enclosed current is I encl = 10 + 2 p(0.01) 400 - 2 p(0.02)250 + 2 p(0.03) 300 = 60.3 mA m 2 p(0.035) H f = 60.3 M 12. (A) H =

Þ

H f = 0.27 A m

1 ( -30 - 40) u x ´ ( -u z ) = -35 u y A m 2

Þ

a=

F 4 ´ 10 -14 = [6u x + 6u y - u z ] m 5 ´ 10 -26

= 800[ 6 u x + 6 u y - u z ] 10 9 m s 2 19. (C) F = e v ´ B = - (1.6 ´ 10 -19)( 4.5 ´ 10 7 u y)(2.5 ´ 10 -3 u z ) = -1.8 ´ 10 -14 u x N This force will be constant during the time the electron travels the field. It establishes a negative x –directed velocity as it leaves the field, given by the acceleration times the transit time tt , Ftt æ -1.8 ´ 10 -14 öæ 2.5 ´ 10 -2 ÷ç =ç m çè 9.1 ´ 10 -31 ÷øçè 4.5 ´ 10 7 0.5 - 0.025 = = 106 . ´ 10 -8 s 4.5 ´ 10 7

vx = t50

ö ÷÷ = -11 . ´ 10 7 m s ø

In that time, the electron moves to an x coordinate 1 13. (B) H = K ´ u n 2 1 = ( 30 - 40) u x ´ ( -u z ) = -5 u y A m 2

given by . ´ 10 7)(106 . ´ 10 -8 ) = -0.117 m x = vx t50 = -(11 x = -117 . cm, z = 0 C

1 14. (A) H = ( -30 + 40) u x + ( -u z ) = 5 u y A m 2 15. (C) J = Ñ ´ H = =

1 ¶ æ r2 çç 2 + 2 r ¶r è

1 ¶(rH f) uz r ¶r

ö ÷÷u z = u z A m ø

20. (A) FBC = ò I loop dL ´ Bfrom 4

= ò ( 6 ´ 10 -3) dzu z ´ 1

1 ¶ æ 3ö çr ÷ = 0 r ¶r çè r ÷ø

position along the loop segment. 3

ò ( 6 ´ 10 1

=

-3

) dxu x ´

15m o uy 2 px

45 ´ 10 -3 m o ln 3u z = 19.8 u z nN p

17. (A) F = e(E + v ´ B) If F = 0, E = -v ´ B = B ´ v Page 486

15m o u y = -1.8 ´ 10 -8 u x = -18 u x nN 2 p( 3)

21. (C) The field from the long wire now varies with

FAB = 16. (D) J =

wire at BC

B

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Magnetostatics

22. (A) This will be the vector sum of the forces on the four sides. By symmetry, the forces on sides AB and CD will be equal and opposite, and so will cancel. This leaves the sum of forces on side BC and DA 4

FDA =

ò -( 6 ´ 10

-3

) dxu z ´

1

15m o u y = 54 u x nN 2 p(1)

Chap 8.3

28. (A) F =

ò K ´ BdS =

area

1 1 .5

ò ò 6u

z

´

0 0 .5

-5m o u x dy 2 py

15m o æ 15 . ö =ln ç ÷u y = -6.6 u y mN m p è 0.5 ø 29. (A) c m + 1 = m r , c m + 1 = 6.5 , c m = 5.5

Ftotal = FDA + FBC = (54 - 18) u x = 36 u x nN 30. (A) B = mH = m om rH

23. (A) F = ò IdL ´ R

= 4 p ´ 10 -7 ´ 6.5(10 u x + 25 u y - 40 u z )

2

2

3

1

1

1

1

2

= I ò dxu x ´ B + I ò dyu y ´ B + I ò dxu x ´ B + I ò dyu y ´ B

= 82 u x + 204 u y - 327 u z mWb m 2

u x ´ B = u x [ 6 xu x - 9 yu y + 3zu z ] = 3zu y - 9 yu z

31. (C) M = c m H = 55 u x + 137.5 u y - 220 u z A m

u y ´ B = u y[ 6 xu x - 9 yu y + 3zu z ] = 3zu x - 6 xu z z = 0 for all element 3

2

1

1

1

1

3

2

32. (B) W =

F = I ò dx (-9 yu z) y=1 + I ò dy(-6 xu z) x=3 + I ò dx (-9 yu z) y=2 + I ò dy(-6 xu z) x=1

=

= I ( -18 - 18 + 36 + 6) u z = 5 ´ 6 u z = 30 u z N 24. (B) Within the region -1 < y < 1, the magnetic fields from the two outer sheets (carrying -4u z A m) cancel, leaving only the field from the center sheet. Therefore

1 1 H × B = mH 2 2 2

1 ´ 6.5 ´ 4 p ´ 10 -7 (100 + 625 + 1600) = 9.5 mJ m 2 2

33. (D) m r = c m + 1 = 31 . + 1 = 4.1, m = m om r = 4.1m o H=

0.4 yu z B = = 77.6 yu z kA m m 4.1 ´ 4 p ´ 10 -7

H = -4 u x A m (0 < y < 1) and H = 4 u x A m ( -1 < y < 0). Outside (y > 1 and y < - 1) the fields from all three sheet cancel, leaving H = 0 ( y > 1, y < - 1). So at x = 0, y = 0.5 F = Iu z ´ B = (7 ´ 10 -3) u z ´ - 4m o u x = -35.2 u y nN m m

0

0

-100 m o u y 2 p (5 ´ 10 -3)

= 0.4 u x N m

m r1 =

27. (B) The field from the current strip at the filament location 3m o . ö 6m o u x æ 15 dy = ln ç ÷u x y p 0 2 p è .5 ø 0 .5

1 .5

B=

ò

= 1.32 ´ 10 -6 u x Wb m 2 1

F = ò IdL ´ B 0

1

1 1 m = ´ = 1326.3 m o 600 4 p ´ 10 -7

M1 = c m H1 = 1590 . ´ 106 A m 1.4 B For case 2, m = 2 = H 2 400

ò IdL ´ B

= ò 100 dzu z ´

2 B1 = H 1200

c m = m r - 1 = 1325.3

1

1

35. (C) For case 1, m = m r1 =

F 25. (D) = Iu x ´ ( -4m o u x ) = 0 m 26. (A) F =

34. (A) M = c m H = ( 31 . )(77.6) yu z = 241 yu z kA m

= ò 5 dzu z ´ 1.32 ´ 10 -6 u x dz = 6.6 u y mN m

1.4 = 2785.2 400 ´ 4 p ´ 10 -7

c m = 2784.2 2784.2 M 2 = c m H 2 = 1114 . ´ 106 A m DM = (1590 . - 1114 . ) ´ 106 = 476 kA m 36. (B) M = Nm = (2.7 ´ 10 29)(2.6 ´ 10 -30 u y) = 0.7 u y A m H=

0.7 u y M = = 0.22 u y A m m r - 1 4.2 - 1

0

37. (A) M = www.gatehelp.com

B mo

ö æ 1 çç + 1 ÷÷ ø è cm

-1

Page 487


UNIT 8

-1

=

Electromagnetics

The normal component of H1 is

0.02 æ 1 ö + 1 ÷ = 47.6 A m ç 4 p ´ 10 -7 è 0.003 ø

H N1 = (H1 × u N 21 ) u N 21 H1 × u N 21 = (50 u x - 30 u y + 20 u z ) × (0.37 u x + 0.56 u y - 0.74u z )

= 18.5 - 16.8 - 14.8 = -131 .

1 1 38. (A) W = H × B = m o H 2 2 2 1 = ( 4 p ´ 10 -7)( 600) 2 = 0.226 J m 3 2

=

Ho

0

0

= -4.83u x - 7.24 u y + 9.66 u z A m Tangential component of H1 at the boundary

æ1

HT1 = H1 - H N1

ö

ò H. dB = ò Hçè 3 + 2 H ÷ødH

39. (A) W = 2 o

Ho

(H1 × u N 21 ) u N 21 = ( -131 . )(0.37 u x + 0.56 u y - 0.74 u z )

= (50 u x - 30 u y + 20 u z ) - ( -4.83u x - 7.24 u y + 9.66 u z ) = 54.83u x - 22.76 u y + 10.34 u z A m

3 o

2H H + = 6.2 MJ m 3 6 3

40. (A) J b = Ñ ´ M =

H T 2 = H T1 m 2 H N 2 = r1 H N1 = ( -4.83u x - 7.24 u y + 9.66 u z ) m r2 5

12 uz a

= -193 . u x - 2.90 u y + 3.86 u z A m H 2 = H T 2 + H N 2 = (54.83u x - 22.76 u y + 10.34 u z )

41. (B) B1 = 200m o u x - 600m o u y + 400m o u z

+ ( -193 . u x - 2.9 u y + 3.86 u z ) = 52.9 u x - 25.66 u y + 14.2 u z

Its normal component at the boundary is B1 N = (B1 × u N12 ) u N12

********

= (52.8 u x + 42.24 u y + 56.32 u z ) m o = B2 N B Þ H 2 N = 2 N = 6.60 u x + 5.28 u y + 7.04 u z 8m o H1 N =

B1 N = 26.40 u x + 2112 . u y + 28.16 u z 12m o

H1 T = H1 - H1 N = (100 u x - 300 u y + 200 u z ) -(26.4 u x + 2112 . u y + 28.16 u z ) = 73.6 u x - 32112 . u y + 171.84 u z H1 T = H 2 T H 2 = H 2 N + H 2 T = 80.2 u x - 315.8 u y + 178.9 u z A m 42. (C) H N1 = - 3u z , H T1 = 10 u x + 15 u y H T 2 = H T1 = 10 u x + 15 u y m 1 ( -3u z ) = 0.15 u z H N 2 = 1 H N1 = m2 20 H 2 = H N 2 + H T 2 = 10 u x + 15 u y - 0.15 u z B2 = m 2H 2 = 20 ´ 4 p ´ 10 -7(10 u x + 15 u y - 0.15 u z ) = 251u x + 377 u y - 377 . u z mWb m 2 43. (B) At the boundary normal unit vector 2 u x + 3u y - 4 u z Ñ (2 x + 3 y - 4 z) un = = |Ñ (2 x + 3 y - 4 z)| 29 = 0.37 u x + 0.56 u y - 0.74 u z Since this vector is found through the gradient, it will point

in

the

direction

of

increasing

values

of

2 x + 3 y - 4 z, and so will be directed into region 1. Thus u n = u n 21 . Page 488

www.gatehelp.com


UNIT 8


Electromagnetics


The location of the sliding bar in fig. P8.4.8–9 is

Consider the fig. P8.4.12–13. The rails have a

given by x = 5 t + 4 t . The separation of the two rails is

resistance of 2 W m. The bar moves to the right at a

30 cm. Let B = x u z T.

constant speed of 9 m s in a uniform magnetic field of

3

2

y

0.8 T. The bar is at x = 2 m at t = 0.

B

B

z

B

a

VM

v

0.2 cm I

v

x

b

Fig. P8.4.8–9.

8. The voltmeter reading at t = 0.5 s is

16 cm

(A) -21.6 V

(B) 21.6 V

Fig. P8.4–12–14

(C) -6.3 V

(D) 6.3 V

12. If 6 W resistor is present across the left-end with the

9. The voltmeter reading at x = 0.6 m is

right end open-circuited, then at t = 0.5 sec the current

(A) -1.68 V

(B) 1.68 V

I is

(C) -0.933 V

(D) 0.933 V

(A) -45 mA

(B) 45 mA

(C) -60 mA

(D)60 mA

Statement for Q.10–11: A perfectly conducting filament containing a 250W

13. If 6 W resistor is present across each end, then I at

resistor is formed into a square as shown in fig.

0.5 sec is

P8.4.10-11.

(A) -12.3 mA

(B) 12.3 mA

(C) -7.77 mA

(D) 77.7 mA

y I(t)

Statement for Q.14–15: 0.5 cm

B

The internal dimension of a coaxial capacitor is

250 W

a = 1.2 cm, b = 4 cm and c = 40 cm. The homogeneous material inside the capacitor has the parameter

x

e = 10 -11 F m, m = 10 -5 H m and s = 10 -5 S m.The electric

Fig. P8.4.10–11

field intensity is E = 10r cos (10 5 t) u p V m. 7

10. If B = 6 cos (120 pt - 30 ° ) u z T, then the value of I ( t) is

14. The current density J is

(A) 2.26 sin (120 pt - 30 ° ) A (B) 2.26 cos (120 pt - 30 ° ) A (C) -2.26 sin (120 pt - 30 ° ) A (D) -2.26 cos (120 pt - 30 ° ) A 11. If B = 2 cos p( ct - y) u z mT, where c is the velocity of light, then I ( t) is

200 sin (10 5 t) u r A m 2 r

(B)

400 sin (10 5 t) u r A m 2 r

(C)

100 cos (10 5 t) u r A m 2 r

(A) 1.2(cos pct - sin pct) mA


(B) 1.2(cos pct - sin pct) mA

15. The quality factor of the capacitor is

(C) 1.2(sin pct - sin pct) mA

(A) 0.1

(B) 10

(C) 0.2

(D) 20

(D) 1.2(sin pct - sin pct) mA Page 490

(A)

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Maxwell’s Equations

16. The following fields exist in charge free regions

(B) -4 a sin (15 . ´ 108 t - ax) u z mA m

P = 60 sin ( wt + 10 x) u z

(C) 4 a sin (15 . ´ 108 t - ax) u z m A m

Q = 10r cos ( wt - 2r) u f

(D) 4 a sin (15 . ´ 108 t - ax) u z mA m

R = 3r 2 cot f u r + 1r cos fu f

21. The value of a is

S=

(A) 4.3

(B) 2.25

(C) 5

(D) 6

1 r

sin q sin ( wt - 6 r) u q

The possible electromagnetic fields are (A) P, Q

(B) R, S

(C) P, R

(D) Q, S

Statement for Q.22–23: Let H = 2 cos (1010 t - bx) u z A m, m = 3 ´ 10 -5 H m,

17. A parallel-plate capacitor with plate area of 5 cm 2 and plate separation of 3 mm has a voltage 50 sin (10 3 t) V

Chap 8.4

applied to its plates. If e r = 2,

the displacement

current is

e = 1.2 ´ 10 -10 F m and s = 0 everywhere. 22. The electric flux density D is (A) 120 cos (1010 t - bx) nC m 2

(A) 148 cos (1010 t) nA

(B) 261 cos (1010 t) mA

(B) -120 cos (1010 t - bx) nC m 2

(C) 261 cos (1010 t) nA

(D) 148 cos (1010 t) mA

(C) 120 cos (1010 t + bx) nC m 2 (D) None of the above

18. In a coaxial transmission line ( e r = 1), the electric 23. The magnetic flux density B is

field intensity is given by E=

(A) 6.67 ´ 10 4 cos (1010 t + bx) T

100 cos (10 9 t - 6 z) u r V m. r

(B) 6.67 ´ 10 4 cos (1010 t - bx) (C) 6 ´ 10 -5 cos (1010 t + bx) T

The displacement current density is 100 (A) sin (10 9 t - 6 z) u r A m 2 r (B)

(D) 6 ´ 10 -5 cos (1010 t - bx) T Statement for Q.24–25:

116 sin (10 9 t - 6 z) u r A m 2 r

(C) -

0.9 sin (10 9 t - 6 z) u r A m 2 r

(D) -

216 cos (10 9 t - 6 z) u r A m 2 r

A material has s = 0 and e r = 1. The magnetic field intensity is H = 4 cos (106 t - 0.01z) u y A m. 24. The electric field intensity E is (A) 4.52 sin (106 t - 0.01z) kV m (B) 4.52 sin (106 t - 0.01z) V m


(C) 4.52 cos (106 t - 0.01z) V m

Consider the region defined by |x|,|y| and |z|< 1. Let e = 5 e o , m = 4m o , and s = 0 the displacement current

(D) 4.52 cos (106 t - 0.01z) kV m

density J d = 20 cos (15 . ´ 108 t - ax) u y m A m 2 . Assume

25. The value of m r is

no DC fields are present.

(A) 2

(B) 3

(C) 4

(D) 16

19. The electric field intensity E is (A) 6 sin (15 . ´ 108 t - ax) u y mV m

26. The surface r = 3 and 10 mm, and z = 0 and 25 cm

(B) 6 cos (15 . ´ 10 t - ax) u y mV m

are perfect conductors. The region enclosed by these

(C) 3 cos (15 . ´ 108 t - ax) u y mV m

surface has m = 2.5 ´ 10 -6 H m, e = 4 ´ 10 -11 F m and

8

(D) 3 sin (15 . ´ 108 t - ax) u y mV m

s = 0. If H = 2r cos 8 pz cos wt u f A m, then the value of w is

20. The magnetic field intensity is

(A) 2 p ´ 106 rad s

(B) 8 p ´ 106 rad s

(A) -4 a sin (15 . ´ 108 t - ax) u z m A m

(C) 2 p ´ 108 rad s

(D) 8 p ´ 108 rad s

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Page 491


UNIT 8

Electromagnetics

27. For distilled water m = m o , e = 81e o , and s = 2 ´ 10 -3

SOLUTIONS

S m, the ratio of conduction current density to displacement current density at 1 GHz is (A) 111 . ´ 10 -5

(B) 4.44 ´ 10 -4

(C) 2.68 ´ 10 -6

(D) 1.68 ´ 10 -7

1. (B) emf = -

= 2 p(0.2) 2 (20)( 377) sin 377 t mV = 0.95 cos 377 t V

28. A conductor with cross-sectional area of 10 cm 2 carrier

a

conductor

current

2 sin (10 9 t)

mA.

If

2. (A) emf =

s = 2.5 ´ 106 S m and e r = 4.6, the magnitude of the (A) 48.4 m A m

(B) 8.11 nA m

2

F = p(0.1) 2 B = 0.31 cos (120 pt)

(D) 16.4 m A m 2

(C) 32.6 nA m 2

emf = Vba ( t) = -

29. In a certain region

dF dt

= 0.31(120 p) sin (120 pt) = 118.43 sin(120 pt)

J = ( 4 yu x + 2 xzu y + z u z ) sin (10 t) A m 3

4

Vab = - 118.43 sin (120 pt)

If volume charge density r v in z = 0 plane is zero, 4. (D) I =

then r v is (A) 3z 2 cos (10 4 t) mC m 3 (B) 0.3z 2 cos (10 4 t) mC m 3 (C) -3z cos (10 t) mC m 2

1 1 Bo wL2 = ( 4)(2)(2) 2 = 16 V 2 2

3. (C) Since B is constant over the loop area, the flux is

displacement current density is 2

dF d B × dS =dt dt ò

4

Vab 118.43 sin (120 pt) = = 0.47 sin (120 pt) R 250

5. (A) emf = -

3

(D) -0.3z 2 cos (10 4 t) mC m 3

=

30. In a charge-free region (s = 0, e = e o e r , m = m o ) magnetic field intensity is

H = 10 cos (10 t - 4 y) u z 11

dF d =dt dt

òò B × u dz z

loop area

d ( 3)( 4)( 6) cos 5000 t = -360000 sin 5000 t dt

I=

emf 360000 sin 5000 t = - 0.4 sin 5000 t A =R 900 ´ 10 3

A m. The displacement current density is 1 1

(A) -40 sin (10 9 t - 4 y) u y A m

6. (C) F =

(B) 40 sin (10 9 t - 4 y) u y A m

ò ò 20m

o

cos ( 3 ´ 108 t - y) dx dy

0 0

= [20m o sin ( 3 ´ 108 t - y)]10

(C) -40 sin(10 3 t - 4 y) u x A m

= 20 m o [sin ( 3 ´ 108 t - 1) - sin ( 3 ´ 108 t)] Wb

(D) 40 sin (10 9 t - 4 y) u x A m 31. In a nonmagnetic medium ( e r = 6.25) the magnetic field of an EM wave is H = 6 cos bx cos (108 t) u z A m. The corresponding electric field is (A) 903 sin (0.83 x) sin (108 t) V m

Emf = -

dF dt

= - 20 ´ (4p ´ 10 -7)(3 ´ 108 ) ´ [cos (3 ´ 108 t - 1) - cos (3 ´ 108 t)]

= 7540[cos ( 3 ´ 108 t) - cos ( 3 ´ 108 t - 1)] V 7. (D) In this case F = [20m o (2 p) sin ( 3 ´ 108 t - y)]20 p = 0

8

(B) 903 sin (12 . x) sin (10 t) V m (C) 903 sin (0.83 x) cos (108 t) V m

8. (A) F =

(D) 903 sin (12 . x) cos (108 t) V m

òò B × dS =

area

E = 5 cos(10 9 t - 8 x) u x + 4 sin(10 9 t - 8 x) u z V m.

(A) 3.39

(B) 1.84

(C) 5.76

(D) 2.4 ***********

Page 492

ò ò t dtdy 2

0

0

= 0.1 x 3 = 0.1(5 t + 4 t 3) 3 Wb

32. In a nonmagnetic medium

The dielectric constant of the medium is

0 .3 x

emf = -

dF = -0.1( 3)(5 t + 4 t 3) 2 (5 + 12 t 2 ) dt

At t = 0.5 s, emf = -0.1( 3)(2.5 + 0.5) 2 (5 + 3) = -21.6 V 9. (C) At x = 0.6 m, 0.6 = 5 t + 4 t 3 At t = 0.119 s, www.gatehelp.com

emf = -0.933 V

Þ

t = 0.119 s


Maxwell’s Equations

10. (A) F =

òò B × dS = 6(0.5)

2

Chap 8.4

I d = 2 prlJ d = - 2 pl(10) sin (10 5 t) = - 8 psin (10 5 t) A

cos (120 pt - 30 ° ) Wb

area

Quality factor

dF emf = = 6(0.5) 2 (120 p) sin (120 pt - 30 ° ) dt The current is

emf 6(0.5) 2 (120 p) = sin (120 pt - 30 ° ) A R 250

= 2.26 sin (120 pt - 30 ° ) A

òò B × dS = (0.5)(2) ò cos( py - pct) dy

area

0

1é æ pö ù 1 sinç pct - ÷ - sin pct ú = [ - cos pct - sin pct ] mWb p êë è 2ø û p dF emf = = c [cos pct - sin pct ] mV dt =

I ( t) =

=

Ic

8 = 0.1 80

16. (A) Ñ × P = 0, Ñ ´ P = -

emf 3 ´ 108 = [cos pct - sin pct ] mA 250 R

Q

is a possible EM field sin f 1 ¶ ( 3r 2 cot f) × Ñ ×R = ¹ 0, R is not an EM field. r ¶r r Ñ ×S =

¶(sin 2 f) 1 sin ( wt - 6 r) ¹ 0 r sin q ¶r 2

S is not an EM field. Hence (A) is correct.

= 12 . [cos pct - sin pct ] A

17. (A) D = eE = e

12. (A) The flux in the left-hand closed loop is Id = J × S =

Fl = B ´ area = (0.8)(0.2)(2 + 9t) dF L = - (0.16)(9) = -1.44 V emfl = dt

V d

Þ

increasing with time, Rl = 6 + 2[2 (2 + 9 t)]W, At t = 0.5, Rl = 32 W 1.44 Il = = - 45 mA 32

18. (C) ¶D ¶E 100 =e Jd = = eo [ - sin (10 9 t - 6 z)] 10 9 u r A m 2 ¶t ¶t r =-

current from the right loop, which is now closed. The flux in the right loop, whose area decreases with time, is Fr = (0.8)(0.2)(16 - 2 - 9 t) dF R = 1.44 V emfR = dt

0.9 sin (10 9 t - 6 z) u r A m 2 r

19. (D) D = ò J d dt + C1 =

20 ´ 10 -6 sin (15 . ´ 108 - ax) u y 15 . ´ 108

. ´ 108 t - ax) u y C m 3 = 1.33 ´ 10 -13 sin (15 C1 is set to zero since no DC fields are present. E=

Rr = 44 W

The contribution to the current from the right loop -144 Ir = = 032.7 mA 44 The total current = -32.7 - 45 = -77.7 mA 14. (C) J = s E =

dD e dV = dt d dt

eS dV 2 e o 5 ´ 10 -4 = 10 3 ´ 50 cos (10 3 t) 3 ´ 10 -3 d dt

13. (C) In this case, there will be contribution to the

At 0.5 s,

Jd =

= 148 cos (1010 t) nA

While the bar in motion, the loop resistance is

Rr = 6 + 2 (2 (14 - 9 t)),

¶Pz uy ¹ 0 ¶x

P is a possible EM field 1 ¶ Ñ × Q = 0, Ñ ´ Q = [10 cos ( wt - 2r)]u z ¹ 0 r ¶r

0 .5

11. (D) F =

Id

100 cos (10 5 t) u r A m 2 r

15. (A) Total conduction current 100 cos (10 5 t) u r A m 2 I C = òò J × dS = 2prlJ = 2 prl r = 80 pcos (10 5 t) A ¶D ¶eE 10 = sin (10 5 t) A m 2 Jd = =¶t ¶t r

D 1.33 ´ 10 -13 = sin (15 . ´ 108 - ax) u y e 5 eo

= 3 ´ 10 -3 sin (15 . ´ 108 t - ax) V m 20. (D) Ñ ´ E =

¶E y ¶x

uz = -

¶B ¶t

= -a( 3 ´ 10 -3) cos (15 . ´ 108 t - ax) u z = -

¶B ¶t

B=

a( 3 ´ 10 -3) sin (15 . ´ 108 t - ax) u z 15 . ´ 108

H=

2 ´ 10 -11 B = sin (15 . ´ 108 t - ax) u z A m m 4 ´ 4 p ´ 10 -7

. ´ 108 t - ax) u z mA m = 4 ´ 10 -6 a sin (15 21. (B) Ñ ´ H = www.gatehelp.com

¶H z u y = Jd ¶x Page 493


UNIT 8

= a 2 ( 4 ´ 10 -6 ) cos (15 . ´ 108 t - ax) = J D

Electromagnetics

27.(B) At high frequency

Comparing the result a 2 4 ´ 10 -6 = 20 ´ 10 -6 , a = 5 = 2.25 22. (B) Ñ ´ H = -

=

¶H z ¶D uy = ¶x ¶t

Þ

2b cos (1010 t - bx) u y C m 2 D =1010

Þ

b=

w . ´ 10 -10 = 600 = 1010 me = 1010 3 ´ 10 -5 ´ 12 v

Þ

¶z

Jd =

E=

Ic sS

4.6 e o (10 9) 2 cos (10 9 t) 10 -3 2.5 ´ 106 ´ 10 ´ 10 -4

29. (B) Ñ × J = (0 + 0 + 3z 2 ) sin (10 4 t) = -

23. (D) B = mH = 6 ´ 10 -5 cos (1010 t - bx) u z T ¶H y

Ic = sE Þ S ¶E e ¶I c Jd = e = ¶t sS ¶t

|J d | = 32.6 nA m 2

D = -120 cos (1010 t - bx) u y nC m 2

24. (A) Ñ ´ H = -

2 ´ 10 -3 = 4.44 ´ 10 -4 2 p ´ 10 9 ´ 81 ´ e o

28. (C) J c =

¶D = 2b sin (1010 t - bx) u y ¶t

Jc sE s = = J d weE we

3z 2 cos (10 4 t) + C1 10 4

rv =

ux

8r ¶t

At z = 0, r v = 0, C1 = 0 r v = 0.3z 2 cos (10 4 t) mC m 3

¶E Ñ ´ H = 0.04 cos (10 t - 0.01z) u x = e o ¶t 6

30. (D) J d = Ñ ´ H =

0.04 sin (106 t - 0.01z) u x E= 106 e o

¶H z ux ¶y

= 40 sin (10 9 t - 4 y) u y A m

= 4.52 sin (10 t - 0.01z) u x kV m 6

31. (A) Ñ ´ H = -

¶Ex ¶H 25. (B) Ñ ´ E = uy = -m ¶z ¶t

= 6b sin (bx) cos (108 t) u y 1 E = ò 6b sin (bx) cos (108 t) u y dt e 6b sin (bx) sin (108 t) u y = e1010

-0.04(0.01) ¶H cos (106 t - 0.01z) u y = - m rm o 106 e o ¶t H=

0.04(0.01) sin (106 t - 0.013) u y (106 )(106 )m rm o e o

0.04(0.01) =4 1012 m rm o e o

Þ

mr =

(0.04)(0.01) ´ ( 3 ´ 108 ) 2 = 9 4(1012 )

H=

¶Er (16 p)( 8 p) ¶H uf = cos ( 8 pz) sin ( wt) u f = -m ¶z rew ¶t

6(0.833) sin (bx) sin (108 t) u y V m 6.25 e o ´ 108

b=

w w = er , v c

8=

10 9 ´ 108 e r 3

128 p2 cos ( 8 pz) cos ( wt) u f rew2

w=

8p me

=

8p 4 ´ 10

-11

´ 2.5 ´ 10 -6

= 8 p ´ 108 rad s Page 494

w = 10 9, b = 8, Þ

e r = 5.76

**********

This result must be equal to the given H field. Thus Þ

6.25 = 0.833

32. (C) For nonmagnetic medium m r = 1

16 p sin ( 8 pz) sin ( wt) u r rew

128 p2 2 = remw2 r

w w 108 = er = v c 3 ´ 108

= 903 sin (0.83 x) sin (108 t) u y V m

16 p ¶E = sin ( 8 pz) cos ( wt) u r = e r ¶t

Ñ É=

e = 6.25 e o , b = E=

¶H f 26. (D) Ñ ´ H = ur ¶z

E=

¶H z ¶E uz = e ¶y ¶t

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CHAPTER

8.5 ELECTROMAGNETIC WAVE PROPAGATION

Statement for Q.1–3: A y-polarized uniform plane wave with a frequency

(A) 6.43 ´ 106 m s

(B) 2.2 ´ 10 7 m s

(C) 1.4 ´ 108 m s


of 100 MHz propagates in air in the + x direction and impinges normally on a perfectly conducting plane at


x = 0. The amplitude of incident E-field is 6 mV m. 1. The phasor H s of the incident wave in air is (A) 16 e (C) 16 e

-j

2p x 3

-j

2p x 3

uz m A m

(B) -16 e

ux m A m

-j

2p x 3

-j

2p x 3

(D) -16 e

A uniform plane wave in free space has electric field Es = (2 u z + 3u y) e - jbx V m.

uz m A m

5. The magnetic field phasor H s is

ux m A m

(B) (5.3u y - 8 u z ) e - jbx m A m

(A) ( -5.3u y - 8 u z ) e - jbx m A m (C) ( -5.3u y + 8 u z ) e - jbx m A m

2. The E-field of total wave in air is

(D) (5.3u y + 8 u z ) e - jbx m A m

æ 2p ö (A) j12 sin ç x ÷ u y mV m è 3 ø

6. The average power density in the wave is

æ 2p ö (B) - j12 sin ç x ÷ u y mV m è 3 ø

(A) 34 mW m 2

(B) 17 mW m 2

(C) 22 mW m 2

(D) 44 mW m 2

æ 2p ö (C) 12 cos ç x ÷ u y mV m è 3 ø

7. The electric field of a uniform plane wave in free

æ 2p ö (D) -12 cos ç x ÷ u y mV m è 3 ø

field phasor H s is

space is given by Es = 12 p( u y + ju z ) e - j15x . The magnetic

3. The location in air nearest to the conducting plane, where total E-field is zero, is (A) x = 15 . m

(B) x = -15 . m

(C) x = 3 m

(D) x = - 3 m

vp is

( -u z + ju y) e - j15x

(B)

12 ho

( u z + ju y) e - j15x

(C)

12 ho

( -u z - ju y) e - j15x

(D)

12 ho

( u z - ju y) e - j15x

A lossy material has m = 5m o , e = 2 e o . The phase

uniform plane wave propagating in a certain lossless material is ( 6 u y - j5 u z ) e

12 ho


4. The phasor magnetic field intensity for a 400 MHz - j18 x

(A)

A m . The phase velocity

constant is 10 rad m at 5 MHz. 8. The loss tangent is (A) 2913

(B) 1823

(C) 2468

(D) 1374

www.gatehelp.com

Page 495


UNIT 8

Electromagnetics

9. The attenuation constant a is

16.

(A) 4.43

(B) 9.99

m r = 1, e 2 = 1) pipe with inner and outer radii 9 mm and

(C) 5.57


12 mm carries a total current of 16 sin (106 pt) A. The

A

60

m

long

aluminium

( s = 35 . ´ 10 7 S m,

effective resistance of the pipe is Statement for Q.10–11: At

50

MHz

a

lossy

dielectric

material

is

characterized by m = 2.1m o , e = 3.6 e o and s = 0.08 S m.

(A) 0.19 W

(B) 3.48 W

(C) 1.46 W

(D) 2.43 W

The electric field is Es = 6 e - jgx u z V m.

17. Silver plated brass wave guide is operating at 12

10. The propagation constant g is

m r = e r = 1) is 5d, the minimum thickness required for

(A) 7.43 + j2.46 per meter

wave-guide is

(B) 2.46 + j7.43 per meter

(A) 6.41 mm

(B) 3.86 mm

(C) 6.13 + j5.41 per meter

(C) 5.21 mm

(D) 2.94 mm

GHz. If at least the thickness of silver ( s = 6.1 ´ 10 7 S m,

(D) 5.41 + j 6.13 per meter


11. The impedance h is

A uniform plane wave in a lossy nonmagnetic

(A) 101.4 W

(B) 167.4 W

(C) 98.3 W

(D) 67.3 W

media has Es = (5 u x + 12 u y) e - gz , g = 0.2 + j 3.4 m -1

Statement for Q.12–13: A

non

magnetic

18. The magnitude of the wave at z = 4 m and t = T 8 medium

has

an

intrinsic

impedance 360 Ð30 ° W.

is (A) 10.34

(B) 5.66

(C) 4.36

(D) 12.60

12. The loss tangent is (A) 0.866

(B) 0.5

19. The loss suffered by the wave in the interval

(C) 1.732

(D) 0.577

0 < z < 3 m is

13. The Dielectric constant is (A) 1.634

(B) 1.234

(C) 0.936

(D) 0.548

(A) 4.12 dB

(B) 8.24 dB

(C) 10.42 dB

(D) 5.21 dB

Statement for Q.20–22: The plane wave E = 42 cos ( wt - z) u x V m in air


normally hits a lossless medium (m r = 1, e r = 4) at z = 0.

The amplitude of a wave traveling through a lossy nonmagnetic medium reduces by 18% every meter. The

20. The SWR s is

wave operates at 10 MHz and the electric field leads the

(A) 2

magnetic field by 24°.

(C)

14. The propagation constant is (A) 0.198 + j0.448 per meter (B) 0.346 + j0.713 per meter (C) 0.448 + j0.198 per meter

1 3

(D) 3

22. The reflected electric field is

15. The skin depth is (A) 2.52 m

(B) 5.05 m

(C) 8.46 m

(D) 4.23 m

Page 496


21. The transmission coefficient t is 2 4 (A) (B) 3 3 (C)

(D) 0.713 + j0.346 per meter

1 2

(B) 1

(A) -14 cos ( wt - z) u x V m (B) -14 cos ( wt + z) u x V m www.gatehelp.com


UNIT 8

Electromagnetics

33. The region z < 0 is characterized by e r = m r = 1 and

Region 2 (0 < z < 6 cm): m 2 = 2 mH m, e 2 = 25 pF m

s = 0.

Region 3 (z > 6 cm):

The

total

electric

field

E s = 150 e - j10 z u x + 50 Ð20 ° e j10 z u x

here

is

given

39. The lowest frequency, at which a uniform plane

impedance of the region z > 0 is (A) 692 + j176 W (C) 176 + j 692 W

(B) 193 - j 49 W

wave incident from region 1 onto the boundary at z = 0

(D) 49 - j193 W

will have no reflection, is

Statement for Q.34–35: Region 1, z < 0 and region 2, z > 0, are both perfect dielectrics. A uniform plane wave traveling in the u z direction

has

a

frequency

m 3 = 4 mH m, e 3 = 10 pF m

V m. The intrinsic

of

3 ´ 1010

rad s.

Its

wavelength in the two region are l1 = 5 cm and l2 = 3 cm.

(A) 2.96 GHz

(B) 4.38 GHz

(C) 1.18 GHz

(D) 590 MHz

40. If frequency is 50 MHz, the SWR in region 1 is (A) 0.64

(B) 1.27

(C) 2.38

(D) 4.16

41. A uniform plane wave in air is normally incident

34. On the boundary the reflected energy is

onto a lossless dielectric plate of thickness l 8 , and of

(A) 6.25%

(B) 12.5%

intrinsic impedance h = 260 W. The SWR in front of the

(C) 25%

(D) 50%

plate is

35. The SWR is (A) 1.67

(B) 0.6

(C) 2

(D) 1.16

(A) 1.12

(B) 1.34

(C) 1.70

(D) 1.93

42. The E-field of a uniform plane wave propagating in a dielectric medium is given by

36. A uniform plane wave is incident from region 1 (m r = 1, s = 0) to free space. If the amplitude of incident

z ö z ö æ æ 8 E = 2 cos ç 108 t ÷ u x - sin ç 10 t ÷ uy V m 3ø 3ø è è

wave is one-half that of reflected wave in region, then the value of e r is (A) 4

(B) 3

(C) 16

(D) 9

37. A 150 MHz uniform plane wave is normally incident from air onto a material. Measurements yield a SWR of 3 and the appearance of an electric field minimum at 0.3l in front of the interface. The impedance of material is (A) 502 - j 641 W

(B) 641 - j502 W

(C) 641 + j502 W

(D) 502 + j 641 W

The dielectric constant of medium is (A) 3

(B) 9

(C) 6

(D)

43. An electromagnetic wave from an under water source with perpendicular polarization is incident on a water-air interface at angle 20° with normal to surface. For water assume e r = 81, m r = 1. The critical angle q c is (A) 83.62°

(B) 6.38°

(C) 42.6°

(D) None of the above ***********

38. A plane wave is normally incident from air onto a semi-infinite slab of perfect dielectric ( e r = 3.45). The fraction of transmitted power is (A) 0.91

(B) 0.3

(C) 0.7

(D) 0.49

Statement for Q.39–40: Consider three lossless region : Region 1 (z < 0): Page 498

6

m 1 = 4 mH m, e1 = 10 pF m www.gatehelp.com


Electromagnetic Wave Propagation

SOLUTIONS 1. (A) w = 2 p ´ 10

8

8. (B) Loss tangent

rad s

w 2 p ´ 10 2p = = rad m c 3 ´ 108 3

Es = 6 e

2p -j x 3

u y mV m

uE ´ u H = ux , Hs =

6 e 120 p

-j

2p x 3

u y ´ u H = ux , uz

= 16 e

-j

2p x 3

u H = uz

Þ

10 =

Þ

x=

uz m A m

Er = -6 e

j

2p x 3

5´2 2

[ 1+ x

s = 1823 we 1 + x2 - 1 1 + x2 + 1

2 ù me é æ s ö ê 1+ç ÷ - 1ú 2 ê è we ø úû ë

æ 2p ö = - j12 sin ç x ÷ u y mV m è 3 ø

10. (D) a = w

3. (B) The electric field vanish at the surface of the

s 0.08 = =8 we 3.6 ´ 50 ´ 106 ´ 2 pe o

conducting plane at x = 0. In air the first null occur at 3 l p x =- 1 ==- m 2 2 b1 w 2 p ´ 400 ´ 10 = = 1.4 ´ 108 m s b 18

5. (C) The wave is propagating in forward x direction.

2 p ´ 50 ´ 106 3 ´ 108

=

(2.1)( 3.6) ( 65 - 1) = 5.41 2

2 ù me é æ s ö ê 1+ç ÷ + 1ú 2 ê è we ø úû ë

2 p ´ 50 ´ 106 3 ´ 108

(2.1)( 3.6) ( 65 + 1) = 6.13 2

g = a + jb = 5.41 + j 6.13 per meter.

Therefore u E ´ u H = u x . Þ

uH = - uy

For u E = u y , u y ´ u H = u x Þ u H = u z 1 Hs = ( -2 u y + 3u z ) e - jbx = ( -5.3u y + 8 u z ) e - jbx mA m 120 p 6. (B) Pavg =

a=

b=w

6

For u E = u z , u z ´ u H = u x

]

+1

a = 10 ´ 0.999 = 9.99

2p -j x æ - j 2 px ö E = Ei + Er = çç 6 e 3 u y - 6 e 3 u y ÷÷ mV m è ø

4. (C) vp =

2

a 1822 = b 1824

Þ

u y mV m,

2 p ´ 5 ´ 106 3 ´ 108

a = b

9. (B)

2. (B) For conducting plane G = -1,

s =x we

2 ù me é æ s ö ê 1+ç ÷ + 1ú 2 ê è we ø úû ë

b=w

8

b=

Chap 8.5

11. (A) |h| =

1 Re {Es ´ H *s } 2 12. (C)

1 {(5.3) u x + 3( 8) u x } ´ 10 -3 = 17.3u x mW m 2 2 7. (D) Since Pointing vector is in the positive x direction, therefore u E ´ u H = u x . For u E = u y , u y ´ u H = u x

Þ

u H = uz

For u E = u z , u z ´ u H = u x 12 Hs = ( u z - ju y) e - j 15x ho

Þ

u H = -u y ,

m e æ ç 1 + æç s ö÷ ç è we ø è

ö ÷ ÷ ø

=

1 4

2.1 3.6 = 101.4 1

64 4

s = tan 2 q n = tan 60 ° = 1732 . we

13. (D) |h| =

Þ

2

120 p

360 =

m e

1

2 4 æ ö ç 1 + æç s ö÷ ÷ ç è we ø ÷ø è 120p er Þ 1

e r = 0.548

2 4 (1 + 1732 . )

www.gatehelp.com

Page 499


UNIT 8

14. (A) |E| = Eo e - az

ho - ho 1 h2 - h1 G= = 2 =ho 3 h2 + h1 + ho 2 1 1 + |G| 1 + 3 s= = =2 1 - |G| 1 - 1 3

Eo e - a1 = (1 - 0.18) Eo e - a1 = 0.82 q n = 24 °

Þ Þ

1 = 0.198 0.82 s tan 2 q n = = 1111 . we a = ln

2

æ s ö 1+ç ÷ -1 è we ø

a = b

h 2 o 2h2 2 =2 21. (A) t = = h2 + h1 ho + 2 3 2

2

æ s ö 1+ç ÷ +1 è we ø

0.198 = b

234 - 1

Þ

234 + 1

b = 0.448

22. (A) Eor = GEoi = -

g = a + jb = 0.198 + j0.448 15. (B) d =

d=

1 pfsm

Rac =

=

Þ

23. (C) h1 = ho , h2 =

1 p ´ 5 ´ 10 ´ 35 . ´ 10 7 ´ m o

= 120 mm

l sdw

Since d is very small, w = 2pr outer 60 = 0.19 W Rac = 7 . ´ 10 ´ 120 ´ 10 -6 ´ 2 p ´ 12 ´ 10 -3 35

=

5 p ´ 12 ´ 10 ´ m o ´ 6.1 ´ 10 7

= 2.94 mm

h1 = ho

18. (B) E = Re{E s e jwt } = (5 u x + 12 u y) e -0 .2 z cos ( wt - 3.4 z) T 8

Þ

377 h - 377 = 2 3000 h 2 + 377 Þ

e r = 12.5,

Þ

æp ö |E|= 13e -0 .8 cos ç - 13.6 ÷ = 5.66 è4 ø

G=

19. (D) Loss = aDz = 0.2 ´ 3 = 0.6 Np

Eor = -

20. (A) h1 = ho , h2 = ho

Page 500

mr h = o 2 er

æ h - h1 ö 18 ÷÷ h1 = çç 2 -3 è h2 + h1 ø 6 ´ 10

h2 = 485.37 = ho

mr er

m r = 20.75

25. (A) h1 = ho , h2 = ho

0.6 Np = 5. 21 dB.

Þ

æ h - ho ö ÷÷3000 ho = çç 2 è h2 + ho ø

æp ö E = (5 u x + 12 u y) e -0 .8 cos ç - 13.6 ÷ è4 ø

1 Np = 8.686 DB,

mr mr = ho 12.5 er

Eor h - h1 =G = 2 Eoi h2 + h1 æ h - h1 ö ÷÷ Eoi h1 H or = çç 2 è h2 + h1 ø

pfms

At z = 4 m, t =

24. (D) h1 = ho , h2 = ho

But Eor = h1 H or = GEoi

5

9

m ho ho = = 2 e er

ho - ho 1 h2 - h1 G= = 2 =ho 3 h2 + h1 1 + ho 2

f = 5 ´ 10 5 Hz, 5

17. (D) t = 5 d =

1 ( 42) = -14 3

Er = - 14 cos ( wt - z) u x V m

1 1 = = 5.05 a 0.198

16. (A) w = p106

Electromagnetics

m r ho = 2 er

1 h2 - h1 =3 h2 + h1

1 10 (10) = 3 3 10 Eor H or = = = 8.8 ´ 10 -3 ho 3 ´ 377 uE ´ u H = uk ,

-u y ´ u H = -u z

H r = -8.8 cos ( wt - z) u x mA m w w b =1 = = m r er v c

www.gatehelp.com

Þ

u H = -u x


Electromagnetic Wave Propagation

w=

3 ´ 108

= 0.5 ´ 108 rad s.

12 ´ 3

26. (D) h1 = ho , h2 = ho

33.(A) G = G=

m r ho = = 0.58ho er 3

h - h1 0.58ho - ho G= 2 = = -0. 266 h2 + h1 0.58h o + h o

27. (B) ETotal = Ei + Er ,

Eor = GEoi = -2.66

ho G=

Þ

1 sin 45 ° = 0.333 4.5

sin q t1 =

q t1 = 19.47 °

q t2 = sin -1 0.47 = 28 °

But cos q t =

Þ

G=

h1 h cos q B = o cos 58.2 ° = 2.6 cos 58.2 ° ho h2 2.6

q t = 31.8 ° mr h = o = 0.447ho er 5

Ei Er

=

ho -

Þ

h - h1 G= 2 = - 0.38, t = 1 + G = 0.62 h2 + h1

G=

Þ Þ

h2 - h1 = h2 + h1

m r2 m r1 + ho er 2 e r1

ho

=

m r 2 1 + 0.447 = = 0.382, 2.62 m r1 1 ± 0.447 e r1 æ m r 2 =ç e r 2 çè m r1

3

ö ÷÷ = 0.056, 17.9 ø

m r1 m r2 m r32 m r31 m r2 m r1 + m r32 m r31

2

e r1 -1 er 2 e r1 er 2

l2 -1 l = 1 l2 +1 +1 l1

mr h = o er er

1 h2 - h1 = 2 h2 + h1 ho

er 1 = ho 2 ho + er

Þ

37. (C) At minimum

G = ± 0.447

m r2 m r1 - ho er 2 e r1

ho

ö ÷÷ ø

1+

Et = tEi = 92.7 cos ( wt - 8 y) u z V m 32. (B) |G| = 0.2 ,

e r1 æ l2 =ç e r 2 çè l1

1 l2 - l1 3 - 5 = =4 l2 + l1 3 + 5

36. (D) h2 = ho , h1 = ho

G=

2

Þ

1 4 = 5 35. (A) s = = 1 1 - |G| 3 14

2.6 e o = 2.6 eo

31. (A) h1 = ho , h2 = ho

ho

er 2 e r1 h2 - h1 = = ho ho h2 + h1 + er 2 e r1

1 + |G|

30. (A) Since both media are non magnetic e1 = e2

Þ

-

2

The fraction of the incident energy that is reflected is 1 = 6.25%. G2 = 16

e 4.5 29. (B) sin q t 2 = 1 sin q t1 = (0.333) = 0.47 e2 2.25

tan q B =

ö ÷ ÷ = 692 + j176 W ÷ ÷ ø

e j 20 3 e j 20 3

2

28. (B) m o = m 1 = m 2 eo sin q i e1

æ ç1 + ö ÷÷ = 377ç ç ø ç1è

æ 2 pc ö æ 2 pc ö 34. (A) e r1 = çç ÷÷ , e r 2 = çç ÷÷ è l1 w ø è l2 w ø

ETotal = 10 cos ( wt - z) u y - 2.66 cos ( wt + z) u y V m

Þ

h2 - h1 , h1 = ho , h2 + h1

æ1 + G h2 = ho çç è1-G

Et = 7.34 cos ( wt - z) u y V m

Þ

Er 50 Ð20 ° e j 20 = = Ei 150 3

Eot = tEoi = 7.34

t = 1 + G = 0.734,

sin q t1 =

Chap 8.5

er = 9

( f + p) = 0.3l , 2b

2p Þ f = 0.2 p l s -1 3 -1 1 = = |G| = s+1 3+1 2 h - ho G = 0.5 e j 0 .2 p = 2 h2 + ho b=

=

m r1 - m r 2 m r1 + m r 2

Þ

æ 1 + 0.5 e j 0 .2 p ö ÷ = 641 + j502 W h2 = ho çç j 0 .2 p ÷ è 1 - 0.5 e ø

www.gatehelp.com

Page 501


UNIT 8

ho 3.45 ho 3.45

- ho + ho

= -0.3

43. (B) q c = sin -1

39. (C) This frequency gives the condition b2 d = p Where d = 6 cm, b2 = w m 2 e 2

Þ

f =

p 0.06 1

2 ´ 0.06 2 ´ 10 -6 ´ 25 ´ 10 -12

= 118 . GHz

40. (B) At 50 MHz, b2 = w m 2 e2 = 2 p ´ 50 ´ 10 6 2 ´ 10 -6 ´ 25 ´ 10 -12 = 2. 2 b2 d = 2.22(0.06) = 0.133 h1 =

m1 4 ´ 10 -6 = = 632 W e1 10 -11

h3 = 632 W h2 =

m2 2 ´ 10 -6 = = 283 W e2 25 ´ 10 -12

The input impedance at the first interface is æ h + jh2 tan (b2 d) ö æ 632 + j283(0.134) ö ÷÷ = 283çç hin = h2 çç 3 ÷÷ è 283 + j 632(0.134) ø è h2 + jh3 tan (b2 d) ø = 590 - j138 h - h1 590 - j138 - 632 G = in = = 0.12 Ð - 100.5° h in + h1 590 - j138 + 632 s=

1 + |G| 1 - |G|

=

1 + 0.12 = 1.27 1 - 0.12

41. (C) bd =

2p l p p × = , tan =1 l 8 4 4

h2 = 260, h1 = h3 = ho æ h + jh2 tan (b2 d) ö æ 377 + j260 ö ÷÷ = 260çç hin = h2 çç 3 ÷÷ + j tan ( d ) h h b è 260 + j 377 ø ø è 2 3 2 = 243 - j92 W h - ho 243 - j92 - 377 G = in = = 0.26 Ð - 137 ° hin + ho 243 - j92 + 377 s=

1 + |G| 1 - |G|

=

126 . = 170 . 0.74

42. (A) w = 108 rad s, b = Page 502

er

=

1

3

Þ

e r = 3.

er 2 = sin -1 e r1

1 = 6.38 ° 81

******** 2

w m 2 e2 =

108

Þ

The transmitted fraction is 1 - |G| = 1 - 0.09 = 0.91.

Þ

3 ´ 108

mr ho = er 3.45

38. (A) h1 = ho , h2 = ho h - h1 G= 2 = h2 + h1

Electromagnetics

1 3

rad m, v =

c er

=

w b www.gatehelp.com


CHAPTER

8.7 WAVEGUIDES



A 2 cm by 3 cm rectangular waveguide is filled

In an air-filled rectangular waveguide the cutoff

with a dielectric material with e r = 6. The waveguide is

frequencies for TM11 and TE03 modes are both equal to

operating at 20 GHz with TM11 mode.

12 GHz.

1. The cutoff frequency is

6. The dominant mode is

(A) 3.68 GHz

(B) 22.09 GHz

(A) TM10

(B) TM 01

(C) 9.02 GHz

(D) 16.04 GHz

(C) TE01

(D) TE10

2. The phase constant is (A) 816 rad m

(B) 412 rad m

(C) 1009 rad m

(D) 168 rad m

7. At dominant mode the cutoff frequency is (A) 11.4 GHz

(B) 4 GHz

(C) 5 GHz

(D) 8 GHz

3. The phase velocity is (A) 1.24 ´ 108 m s

(B) 154 . ´ 106 m s

(C) 305 . ´ 10 m s

(D) 7.48 ´ 10 m s

8

8. For an air-filled rectangular waveguide given that

8

4. In an an-filled rectangular wave guide, the cutoff frequency of a TE10 mode is 5 GHz where as that of TE01

æ 2 px ö æ 3py ö 12 Ez = 10 sin ç ÷ sin ç ÷ cos (10 t - bz) V m è a ø è b ø If the waveguide has cross-sectional dimension

mode is 12 GHz. The dimensions of the guide is

a = 6 cm and b = 3 cm, then the intrinsic impedance of

(A) 3 cm by 1.25 cm

(B) 1.25 cm by 3 cm

this mode is

(C) 6 cm by 2.5 cm

(D) 2.5 cm by 6 cm

(A) 373.2 W

(B) 378.9 W

(C) 375.1 W

(D) 380.0 W

5. Consider a 150 m long air-filled hollow rectangular waveguide with cutoff frequency 6.5 GHz. If a short pulse of 7.2 GHz is introduced into the input end of the guide, the time taken by the pulse to return the input end is

Statement for Q.9–10: In an air-filled waveguide, a TE mode operating at 6 GHz has

(A) 920 ns

(B) 460 ns

(C) 230 ns

(D) 430 ns

www.gatehelp.com

æ 2 px ö æ py ö E y = 15 sin ç ÷ cos ç ÷ sin ( wt - 12 z) V m è a ø è b ø Page 511


UNIT 8

9. The cutoff frequency is

Electromagnetics

16. The cross section of a waveguide is shown in fig.

(A) 4.189 GHz

(B) 5.973 GHz

P8.7.16. It has dielectric discontinuity as shown in fig.

(C) 8.438 GHz

(D) 7.946 GHz

P8.7.16. If the guide operate at 8 GHz in the dominant mode, the standing wave ratio is y

10. The intrinsic impedance is (A) 35.72 W

(B) 3978 W

(C) 1989 W

(D) 71.44 W

2.5 cm mo, eo

Statement for Q.11–12.

x

mo, 2.25eo z

5 cm

Fig. P8.7.16

Consider an air-filled rectangular wave guide with cm. The y-component of the a = 2.286 cm and b = 1016 .

(A) -3.911

(B) 2.468

TE mode is

(C) 1.564

(D) 4.389

æ 2 px ö æ 3py ö 10 E y = 12 sin ç ÷ cos ç ÷ sin (10 p ´ 10 t - bz) V m è a ø è b ø

Statement for Q.17–19: Consider the rectangular cavity as shown in fig.

11. The propagation constant g is

P8.7.17–19.

(A) j4094.2

(B) j400.7

(C) j2733.3

(D) j276.4

y

z

12. The intrinsic impedance is (A) 743 W

(B) 168 W

(C) 986 W

(D) 144 W

b

x

Consider a air-filled waveguide operating in the TE12 mode at a frequency 20% higher than the cutoff frequency. 13. The phase velocity is (A) 1.66 ´ 108 m s

(B) 5.42 ´ 108 m s

(C) 2.46 ´ 108 m s

(D) 9.43 ´ 108 m s

(A) 1.66 ´ 108 m s

(B) 4.42 ´ 108 m s

(C) 2.46 ´ 108 m s

(D) 9.43 ´ 108 m s

rectangular

waveguide

is

filled

17. If a < b < c, the dominant mode is (A) TE011

(B) TM110

(C) TE101

(D) TM101

18. If a > b > c, then the dominant mode is

14. The group velocity is

A

0

Fig. P8.7.17–19


15.

a

c

(A) TE011

(B) TM110

(C) TE101

(D) TM101

19. If a = c > b, then the dominant mode is

with

(A) TE011

(B) TM110

(C) TE101

(D) TM101

a

polyethylene ( e r = 2.25) and operates at 24 GHz. The

20. The air filled cavity resonator has dimension a = 3

cutoff frequency of a certain mode is 16 GHz. The

cm, b = 2 cm, c = 4 cm. The resonant frequency for the

intrinsic impedance of this mode is

TM110 mode is

(A) 2248 W

(B) 337.2 W

(A) 5 GHz

(B) 6.4 GHz

(C) 421.4 W

(D) 632.2 W

(C) 16.2 GHz

(D) 9 GHz

Page 512

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UNIT 8

Electromagnetics

SOLUTIONS

frequency, the TM1 mode propagates through the guide without suffering any reflective loss at the dielectric interface. This frequency is

1. (A) fc = er2 = 2.1

er1 = 4 Incident wave

1 cm z

=

Fig. P8.7.34

(A) 8.6 GHz

(B) 12.8 GHz

(C) 4.3 GHz

(D) 7.5 GHz

2

2

2

2

v æmö æ nö ç ÷ +ç ÷ 2 è aø è bø

3 ´ 108 2 6 ´ 10 -2

æ1ö æ1ö ç ÷ + ç ÷ = 3.68 GHz è2 ø è 3ø 2

æf ö æf ö w 2. (C) bp = b 1 - çç c ÷÷ = 1 - çç c ÷÷ f v è f ø è ø


2

2

A 6 cm ´ 4 cm rectangular wave guide is filled with

Þ

bp =

2 p ´ 20 ´ 10 9 6 æ 3.68 ö 1 -ç ÷ = 1009 rad m 3 ´ 108 è 20 ø

dielectric of refractive index 1.25. 35. The range of frequencies over which single mode

3. (A) vp =

w 2 p ´ 20 ´ 10 9 = = 1.24 ´ 108 m s bp 1009

operation will occur is (A) 2.24 GHz < f < 3.33 GHz

4. (A) For TE10 mode fc =

(B) 2 GHz < f < 3 GHz (C) 4.48 GHz < f <7.70 GHz

v 3 ´ 108 = = 3 cm 2 fc 2 ´ 5 ´ 10 9

a=

(D) 4 GHz < f < 6 GHz

For TE01 mode fc = 36. The range of frequencies, over which guide support b=

both TE10 and TE01 modes and no other, is (A) 3.35 GHz < f < GHz (B) 2.5 GHz < f < 3.6 GHz

c æf ö 1 - çç c ÷÷ è f ø

(C) 3 GHz < f < 3.6 GHz

37. Two identical rectangular waveguide are joined end

t=

v , 2b

v 3 ´ 108 = = 1.25 cm 2 fc 2 ´ 12 ´ 10 9

5. (D) v =

(D) 2.5 GHz < f < 4.02 GHz

v , 2a

2

3 ´ 108

=

æ 6.5 ö 1 -ç ÷ è 7.2 ø

2

= 6.975 ´ 108 ms

2l 2 ´ 150 = = 430 ns v 6.975 ´ 108

to end where a = 2 b. One guide is air filled and other is filled with a lossless dielectric of e r . it is found that up

6. (C) 12 ´ 10 9 =

to a certain frequency single mode operation can be simultaneously

ensured

in

both

guide.

For

this

frequency range, the maximum allowable value of e r is (A) 4

(B) 2

(C) 1

(D) 6

3 ´ 108 æ 3ö 0+ç ÷ 2 è bø 2

12 . ´ 10 9 =

2

Þ

b = 375 . cm 2

3 ´ 108 æ 1 ö 1 æ ö ç ÷ +ç ÷ Þ a = 1.32 cm 2 . ´ 10 -2 ø èaø è 375

Since a < b, the dominant mode is TE01 .

38. A parallel-plate guide operates in the TEM mode

7. (B) fc 01 =

3 ´ 108 v = = 4 GHz 2 b 2 ´ 375 . ´ 10 -2

only over the frequency range 0 < f < 3 GHz. The dielectric between the plates is teflon ( e r = 2.1). The

8. (C) Ez ¹ 0, this must be TM 23 mode ( m = 2, n = 3)

maximum allowable plate separation b is (A) 3.4 cm

(B) 6.8 cm

(C) 4.3 cm

(D) 8.6 cm *************

Page 514

2

2

fc =

3 ´ 108 2 ´ 10 -2

f =

w 1012 = = 159.2 GHz 2p 2p

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æ2 ö æ 3ö ç ÷ + ç ÷ = 15.81 GHz è6ø è 3ø


Waveguides

2

2

Chap 8.7

ho

377

æf ö æ 15.81 ö hTM = 377 1 - çç c ÷÷ = 377 1 - ç ÷ = 375.1 W f è 159.2 ø è ø

h1 =

9. (B) m = 2, n = 1, bp = 12, f = 6 GHz

In dielectric medium

æf ö w 1 - çç c ÷÷ bp = v è f ø Þ

2

2 p ´ 6 ´ 10 9 æf ö 12 = 1 -ç c ÷ 8 3 ´ 10 è6ø

Þ

2

fc = h=

fc = 5.973 GHz 377

10. (B) hTE =

æf ö 1 - çç c ÷÷ è f ø

377

=

2

æ 5.973 ö 1 -ç ÷ è 6 ø

= 3978 W

2

11. (B) m = 2, n = 3, 2

fc =

2

2

æ 2 ö æ 3 ö çç ÷÷ + ç ÷ 2 . 286 . è 1016 ø è ø

3 ´ 108 c æmö æ nö ç ÷ +ç ÷ = 2 è aø 2 ´ 10 -2 è bø

2


2

er

=

3 ´ 108

377 2.25

= 251.33 W, h2 =


259.23 - 406.7 h2 - h1 = = -0.22 259.23 + 406.7 h2 + h1

s=

1 + |G| 1 + 0.22 = = 1564 . 1 -|G| 1 - 0.22

æ 46.2 ö 1 -ç ÷ è 50 ø

377

=

æ 46.2 ö 1 -ç ÷ è 50 ø

2

hTE =

h æf ö 1 - çç c ÷÷ è f ø

2

æ f ö 1 - çç c ÷÷ è 1.2 f c ø

=

n = 1, 2, 3.....

= 986 W

2

p = 1, 2, 3...... , 1 1 1 > > a b c

if a < b < c, then

fr1 =

2

2

377 = 251.33 W . 15 251.33 æ 16 ö 1 -ç ÷ è 24 ø

2

v æ1ö æ1ö ç ÷ +ç ÷ 2 èaø è bø

2

The lowest TE mode is TE011 with

= 337.2 W

16. (C) Since a > b, the dominant mode is TE10 . In free space fc =

2

For TE mode to z

= 5. 42 ´ 108 m s

2

er

2

v æmö æ nö æ pö ç ÷ +ç ÷ +ç ÷ 2 è aø b è ø ècø

The lowest TM mode is TM110 with

3 ´ 10

=

= 259.23 W

p = 0, 1, 2 ......

æ f ö æf ö 14. (A) v g = v 1 - çç c ÷÷ = c 1 - çç c ÷÷ = 1.66 ´ 108 m s . fc ø è f ø è 12 377

2

n = 1, 2, 3...... ,

2

fr 2 =

15. (B) h =

æ2 ö 1 -ç ÷ è8ø

m = 1, 2, 3.....,

8

=

2

251.33

G=

2

v

= 2 GHz

2 ´ 0.05 2.25

m = 1, 2, 3...... ,

13. (A) v = c, f = 1.2 fc vp =

= 406.7 W

2

where for TM mode to z

2 p ´ 50 ´ 10 9 = 3 ´ 108

377

ho

æ 3ö 1 -ç ÷ è8ø

2

= 400.7 m -1 , g = jbp = j 400.7 12. (C) hTE =

2 a er

=

17. (A) fr =

10 p ´ 1010 = 50 GHz 2p

æf ö w 1 - çç c ÷÷ bp = v è f ø

c

=

2

= 46.2 GHz f =


2

c 3 ´ 108 = = 3 GHz 2 a 2 ´ 0.05

v æ1ö æ1ö ç ÷ +ç ÷ 2 è bø è cø

2

fr 2 > fr1 , Hence the dominant mode is TE011 18. (B) If a > b > c then

1 1 1 < < a b c

The lowest TM mode is TM110 with 2

fr1 =


2

The lowest TE mode is TE101 with 2

fr 2 =

v æ1ö æ1ö ç ÷ +ç ÷ 2 èaø è cø

fr 2 > fr1 www.gatehelp.com

2

Hence the dominant mode is TM110 . Page 515


UNIT 8

1 1 1 = < a c b

19. (C) If a = c > b, then

The lowest TM mode is TM110 with 2

fr1 =


fr 2 =

v æ1ö æ1ö ç ÷ +ç ÷ 2 èaø è cø

2

2

2

v æmö æ nö æ pö ç ÷ +ç ÷ +ç ÷ 2 è aø è bø ècø 3 ´ 108 2 2 a

m<

mc

Þ

=

2 b er

f >

,

Þ

a = 10.6 cm

2 ´ 3 ´ 108 2 ´ 0.01 e r

mc 2 b er

Þ

15 . ´ 108 a

12 . ´ 15 . ´ 108 a

Þ

3 ´ 10 9 <

9

Þ

er = 9

m<

15 . ´ 108 , b

0.8 ´ 15 . ´ 108 b

b < 4 cm, Thus (C) is correct option.

27. (C) fc =

= 10 ´ 10

Þ

c 3 ´ 108 = = 2.3 GHz 2 a 2 ´ 0.065

c

vp =


2

=

3 ´ 108 æ 2.3 ö 1 -ç ÷ è 3 ø

2

= 4.7 ´ 108 m s

28. (B) For TE10 mode

2 fb e r c

Rs =

m < 316 .

p ´ 9 ´ 10 9 ´ 4 p ´ 10 -7 = 0.0568 . ´ 10 7 11

1 pfm = = sc d sc

2 2 æ æ 2 b æ fc ö ö÷ 2 ´ 15 . æ 3.876 ö ö÷ Rs ç 1 + çç ÷÷ 0.0568ç 1 + ç ÷ ç ç a è f ø ÷ 2.4 è 9 ø ÷ø ø= è ac = è 2 2 æf ö æ 3.876 ö 15 . ´ 10 -2 ´ 233.8 1 - ç bh 1 - çç c ÷÷ ÷ è 9 ø è f ø

= 0.022 ,

fc 2 = 2 fc1 = 15 GHz 29. (B)

f =

c 3 ´ 108 = = 20 GHz l 0.015

vg2

æf ö æ 15 ö 8 = c 1 - çç c ÷÷ = 3 ´ 108 1 - ç ÷ = 2 ´ 10 m s è 20 ø è f ø

2

2

s

2

Page 516

3 ´ 10 9 >

m 2 + k2 n2

a > 6 cm

Þ

Thus total 7 modes.

25. (A) fc =

Þ

3 GHz < 0.8 fc 01

The maximum allowed m is 3. The propagating mode

2 b er

2

c æmö 15 . ´ 108 æ nö ç ÷ +ç ÷ = 2 è aø a è bø

The next higher mode is TE01 , fc 01 =

will be TM1 , TE1 , TM 2 , TE2 , TM 3 , TE3 and TEM

24. (B) fcm =

7.21

2

2 ´ 30 ´ 10 9 ´ 0.01 2.5 3 ´ 108

mc

6.15

3 GHz > 12 . fc

23. (A) For a propagating mode f > fcm , fcm =

6.8

2

2

2 b er

14.4

Dominant mode is TE10 , fc10 =

21. (A) m = n = 1, p = 0, a = b = c, fr = 2 Ghz,

22. (A) fc =

lc (cm)

2

v æmö æ nö æ pö ç ÷ +ç ÷ +ç ÷ 2 è aø è bø ècø

mc

TE20

fcmn =

2

2 ´ 10 9 =

TE11

26. (C) Let a = kb , 1 < k < 2

3 ´ 108 æ 1 ö æ1ö ç ÷ + ç ÷ = 9 GHz 2 ´ 0.01 è 3 ø è2 ø

fr =

TE01

2

2

=

TE10

l > lc . Hence TE10 mode can be used.

fr2 < fr1 Hence the dominant mode is TE101 . 20. (D) fr =

Mode

2

The lowest TE mode is TE101 with 2

Electromagnetics

sd 10 -15 10 -15 = = we 2 p ´ 9 ´ 10 9 ´ 2.6 ´ 8.85 ´ 10 -12 1.3

sd << 1, hence v » we fc =

1 me

3 ´ 108 2 ´ 2.4 ´ 10 -2 2.6

=

c 2.6

, h»

377 2.6

= 233.8

= 3.876 GHz

2

2 c æmö æ nö ç ÷ + ç ÷ , lc = 2 2 2 è aø b è ø æmö æ nö ç ÷ +ç ÷ è aø è bø

ad =

www.gatehelp.com

sdh æf ö 1 - çç c ÷÷ è f ø

2

=

10 -15 ´ 233.8 æ 3.876 ö 2 1 -ç ÷ è 9 ø

2

= 1.3 ´ 10 -13 Np m


Waveguides

30. (D) Dominant mode is TE10 mode fc =

2

9

-7

= 1.429 ´ 10 -2 W

2 æ 2 b æ fc ö ö÷ 2 ´ 3.4 æ 2.08 ö ö -2 æ Rs ç 1 + çç ÷÷ ç a è f ø ÷ 1.429 ´ 10 ç 1 + 7.2 çè 3 ÷ø ÷ ø= è ø ac = è 2 2 æf ö æ 2.08 ö 377 ´ 0.034 1 - ç bh 1 - çç c ÷÷ ÷ è 3 ø è f ø

e- a c z

1 = 2

-3

Np m

Þ

z=

1 ln 2 = 308 m ac

2

2

2

c æmö 3 ´ 108 æ m ö æ n ö æ nö ÷ ç ÷ +ç ÷ = ç ÷ +ç 2 è aø 2 ´ 0.01 è 8 ø è 10 ø è bø 2

34. (B) The ray angle is such that the wave is interface

fc1 = f =

c

2

=

2 b e r1

3 ´ 1010 = 7.5 GHz 2 ´ 1´ 2

7.5 fc1 = = 12.8 GHz cos q cos 54.1°

fc10 =

c a2 e r

fc 01 =

2

æmö æ n ö = 15 ç ÷ + ç ÷ GHz è 8ø è 10 ø

c b2 e r

2

c 2 er

æmö æ nö ç ÷ +ç ÷ è aø è bø

3 ´ 108 = 2 GHz . ´ 0.06 2 ´ 125

=

3 ´ 108 = 3 GHz . ´ 0.04 2 ´ 125

2 GHz < f < 3 GHz

fc 01 = 15 . GHz,

fc11 = 2.4 GHz

fc 20 = 375 . GHz ,

fc 02 = 3 GHz,

fc 21 = 4.04 GHz,

fc12 = 354 . GHz,

36. (C) fc11 =

2

3 ´ 108 . ´ 10 -2 2 ´ 125

If fc < f , then mode will be transmit. Hence six mode

37. (A) fc =

will be transmitted.

2

c 2 er

2

æmö æ nö ç ÷ + ç ÷ , In guide 1 e r = 1 è 2b ø è bø

lowest cutoff frequency fc10 = 32. (C) For dominant mode (m = 1, n = 0)

Since given frequency is below the cutoff frequency, 3 GHz will not be propagated and get attenuated 2

æ mp ö æ np ö æ w ö g = a + jb = ç ÷ -ç ÷ ÷ +ç è a ø è b ø ènø

2

æ mp ö æ w ö æ p ö æ 2 p ´ 3 ´ 10 a= ç ÷ -ç ÷ = ç ÷ - çç 8 a c è ø è ø è 0.04 ø è 3 ´ 10 2

33. (B) fc = fc10 =

c æmö æ nö ç ÷ +ç ÷ 2 è aø è bø

c 2b

In guide 2 lowest cutoff frequency fc¢10 = Next lowest cutoff frequency fc¢20 =

2

c 2 e r 2 (2 b)

c 2 e r 2 ( b)

For single mode fc¢10 < f < fc¢10

b = 0, Since wave is attenuated, 2

c 2(2 b)

Next lowest cutoff frequency fc 20 =

2

c æmö 3 ´ 108 æ nö GHz . fc = = 375 ç ÷ +ç ÷ = 2 è aø 2 ´ 0.04 è bø

2

2

æ1ö æ1ö ç ÷ + ç ÷ = 3.6 GHz 6 è ø è4ø

3 GHz < f < 3.6 GHz

fc 30 = 5.625 GHz , fc 03 = 4.5 GHz

2

2

=

fc10 = 1.875 GHz

2

2.1 = 35.9 °. 4

The ray angle q = 90 °-35.9 ° = 54.1°

35. (A) fc =

31. (B) fcmn =

2

æ 2p ´ 1.5 ´ 1.875 ö æ p ö ÷ ç ÷ -ç 3 ´ 108 è ø è 0.08 ø

at Brewster’s angle q B = tan -1

For TE10 mode

= 2.25 ´ 10

2

= j439 .

p ´ 3 ´ 10 ´ 4 p ´ 10 5.8 ´ 10 7

pfm o = sc

2

æ mp ö æ np ö æwö ç ÷ + ç ÷ -ç ÷ = è a ø è b ø ènø

g=

c 3 ´ 108 = = 2.08 GHz 2 a 2 ´ 0.072

Rs =

Chap 8.7

Þ 9

2

ö ÷÷ = 47.1 ø

c c
38. (A) f < fc

Þ

f <

2

Þ

3 ´ 10 9 <

3 ´ 108 c = = 1.875 GHz 2 a 2 ´ 0.08

Þ

er < 4

v 3 ´ 108 = 2 b 2 ´ b ´ 2.1

3 ´ 108 2 ´ b ´ 2.1

Þ

b < 3.4 cm

*******************

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Page 517


CHAPTER

8.8 ANTENNAS

1. A Hertizian dipole at the origin in free space has

5. The time-average poynting vector at 50 km is

dl = 10 cm and I = 20 cos (2 p ´ 10 t) A. The | E| at the

(A) 6.36u r mW m 2

(B) 4.78u r mW m 2

distant point (100, 0, 0) is

(C) 9.55u r mW m 2

(D) 12.73u r mW m 2

7

(A) 0.252 V m

(B) 0.126 V m

(C) 0.04 V m

(D) 0.08 V m

6. The maximum electric field at that location is

Statement for Q.2–3: A 25 A source operating at 300 MHz feeds a Hertizian dipole of length 4 mm situated at the origin. Consider the point P(10, 30°, 90°). 2. The H at point P is (A) j0.25 mA m

(B) 94.25 mA m

(C) j0.5 mA m

(D) 188.5 mA m

(A) 24 mV m

(B) 85 mV m

(C) 109 mV m

(D) 12 mV m

7. In free space, an antenna has a far-zone field given by E =

1 r

10 sin 2 q e - jbr u q V m. The radiated power is

(A) 0.23 W

(B) 0.89 W

(C) 1.68 W

(D) 1.23 W

8.

At

Pave =

the

far

field,

an

antenna

cos q cos f u r W m , where

1 r2 p 2

2

0
and

0 < f < . The directive gain of the antenna is

3. The E at point P is (A) j0.25 mV m

(B) j0.5 mV m

(A) cos q cos f

(B) 2 sin q cos f

(C) j94.25 mV m

(D) j188.5 mV m

(C) 8 cos q sin f

(D) 8 sin q cos f

4. An antenna can be modeled as an electric dipole of


length 4 m at 3 MHz. If current is uniform over its length, then radiation resistance of the antenna is (A) 1.974 W

(B) 1.263 W

(C) 2.186 W

(D) 2.693 W

Statement for Q.5–6: A antenna located on the surface of a flat earth

Page 518

produces

The radiation intensity of antennas has been given. Determine the directivity of antenna. 9. U( q, f) = sin 2 q, 0 < q < p ,

0 < f < 2p

(A) 1.875

(B) 2.468

(C) 3.943

(D) 6.743

transmit an average power of 150 kW. Assume that all

10. U( q, f) = 4 sin 2 q sin 2 f ,

the power is radiated uniformly over the surface of

(A) 15

(B) 12

hemisphere with the antenna at the center.

(C) 3

(D) 6

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0 < q < p,

0

Antennas

11. The radiation intensity of a antenna is given by U( q, f) = 8 sin q cos f , where 0 < q < p and 0 < f < p. The 2

2

directive gain is (A) 6 sin 2 q cos 2 f (C) 3 sin 2 f cos 2 q

(B) 3 sin 2 q cos 2 f (D) 6 sin 2 f cos 2 q

Chap 8.8

19. Two identical antenna separated by 12 m are oriented for maximum directive gain. At a frequency of 5 GHz, the power received by one is 30 dB down from the transmitted by the other. The gain of antenna is (A) 22 dB (C) 19 dB


(B) 16 dB (D) 13 dB


At the far field, an antenna radiates a field

12. The total radiated power is

An L-band pulse radar has common transmitting and receiving antenna. The antenna having directive gain of 36 dB operates at 1.5 GHz and transmits 200 kW. The object is 120 km from the radar and its scattering cross section is 8 m 2 .

(A) 1.36 W (C) 0.844 W

20. The magnitude of the incident electric field

0.4 cos 2 q - jbr Ef = e kV m 4 pr

(B) 2.14 W (D) 3.38 W

intensity of the object is 13. The directive gain at q = p 3 is (A) 0.3125 (C) 1.963

14. An antenna has directivity of 100 and operates at 150 MHz. The maximum effective aperture is (A) 31.8 m 2 (C) 26.4 m 2

21. The magnitude of the scattered electric field at the radar is (A) 18 mW (C) 17 mW

(B) 62.4 m 2 (D) 13.2 m 2

15. Two half wave dipole antenna are operated at 100 MHz and separated by 1 km. If 100 W is transmitted by one, the power received by the other is (D = 1.68) (A) 12 mW (C) 18 mW

(B) 2.46 V m (D) 0.17 V m

(A) 1.82 V m (C) 0.34 V m

(B) 0.625 (D) 3.927

(B) 12 mW (D) 126 mW

22. A transmitting antenna with a 300 MHz carrier frequency produces 2 kW of power. If both antennas has unity power gain, the power received by another antenna at a distance of 1 km is (A) 11.8 mW (C) 18.4 mW

(B) 10 mW (D) 16 mW

16. The electric field strength impressed on a half wave

(B) 18.4 mW (D) 12.7 mW

23. A bistatic radar system shown in fig. P8.7.23 has

dipole is 6 mV m at 60 MHz. The maximum power

following parameters: f = 5 GHz,

received by the antenna is (D = 1.68)

22 dB. To obtain a return power of 8 pW the minimum

(A) 159 nW (C) 196 mW

necessary radiated power is

(B) 230 nW (D) 318 mW

Gdt = 34 dB,

Gdr =

Target s = 2.4m2

Scattered wave

17. The power transmitted by a synchronous orbit

Incident wave

satellite antenna is 480 W. The antenna has a gain of

4 km

40 dB at 15 GHz. The earth station is located at distance of 24, 567 km. If the antenna of earth station has a gain of 32 dB, the power received is (A) 32 pW (C) 10.2 pW

Receiving antenna

Transmitting antenna

3 km

Fig. P8.7.23

(B) 3.2 fW (D) 1.3 fW

(A) 1.394 kW (C) 1.038 kW

18. The directive gain of an antenna is 36 dB. If the

(B) 2.046 kW (D) 3.46 kW

time average power density at that distance is

24. The radiation resistance of an antenna is 63 W and loss resistance 7 W. If antenna has power gain of 16, then directivity is

(A) 9.42 mW m 2 (C) 1.32 mW m 2

(A) 48.26 dB (C) 38.96 dB

antenna radiates 15 kW at a distance of 60 km, the (B) 6.83 mW m 2 (D) 10.46 mW m 2

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(B) 12.5 dB (D) 24.7 dB Page 519


UNIT 8

Electromagnetics

25. An antenna is desired to operate on a frequency of

SOLUTIONS

40 MHz whose quality factor is 50. The bandwidth of antenna is (A) 5.03 MHz (C) 127 kHz

(B) 800 kHz (D) None of the above

26. A thin dipole antenna is l 15 long. If its loss (B) 59%

(C) 74.5%

(D) 25.5%

Statement for Q.27–29: An array comprises of two dipoles that are separated by the wavelength. The dipoles are fed by currents of the same magnitude and phase. 27. The array factor is (A) 2 cos ( p cos q + 45 ° ) (C) 2 cos ( p sin q + 45 ° )

(B) 2 cos ( p sin q ) (D) 2 cos ( p cos q )

(B) 60°, 120°

(C) 45°, 135°

(D) 0, 180°

At far field | Eq| =

2. (A) b =

Hf =

29. The maximum of the pattern occur at (A) q = 45 ° , 135 °

(B) q = 0, 90 ° , 180 °

(C) q = 30 ° , 150 °

(D) q = 60 ° , 150 °

) (C) cos ( 2p cos q + 2p )

(B) cos ( 4p cos q +

) (D) sin ( 2p cos q + 2p ) p 2

(B) 4 cos ( 4p cos q) cos (8p cos q) (C) 4 cos ( 4p cos q) sin ( 2p cos q) (D) 4 cos ( 4p cos q) sin (8p cos q)

***********

Page 520

c 3 ´ 108 = = 100 f 3 ´ 106

dl 4 1 1 = = < l 100 25 10 80 p2 æ dl ö Rrad = 80 p2 ç ÷ = = 1263 . W 625 è lø 5. (C) Prad = ò Pave × dS = Pave × 2 pr 2 Pave =

Prad 150 ´ 10 3 = = 9.55 mW m 2 2 pr 2 2 p(50 ´ 10 3) 2

P = 9.55 u r mW m 2

31. An antenna consists of 4 identical Hertizian dipoles uniformly located along the z–axis and polarized in the z-direction. The spacing between the dipole is l / 4 . The group pattern function is (A) 4 cos ( 4p cos q) cos ( 2p cos q)

j(2.5)(2 p) ( 4 ´ 10 -3) - j 2 p10 e = j0.25 mA m 4 p(10)

2

30. An array comprises two dipoles that are separated by half wavelength. If the dipoles are fed by currents, that are 180° out of phase with each other, then array factor is p 4

w 2 p ´ 300 ´ 106 = = 2p c 3 ´ 108

3. (C) E = Eq = hH f = 377 H f = j94.25 mV m 4. (B) l =

(A) sin ( 4p cos q +

hI obdl sin q 4 pr

r = 10 m, q = 30 ° , f = 90 ° jI bdl At far field H = H f = o sin q e - jbr 4 pr Þ

28. The nulls of the pattern occur when q is (A) 30°, 150°

w 2 p ´ 10 7 2 p = = c 3 ´ 108 30

h = 120 p = 377, I o = 20, dl = 10 cm p At (100 cm, 0, 0), q = 2 120 p ´ 20 ´ 0.1 2 p ´ = 0.126 V m | Eq| = 4 p ´ 100 30

resistance is 1.2 W, the efficiency is (A) 41.1%

1. (B) b =

6. (B) Pave =

( Emax ) 2 2h

Þ

Emax = 2hPave

Emax = 2 ´ 377 ´ 9.55 ´ 10 -6 = 85 mV m

Þ

7. (B) Pave = Prad = òò

|E|2 2h

100 sin 2 2 q sin q dq df 2h p

=

100 (2 p)(2 sin q cos q) 2 sin q dq 2 ´ 120 p ò0

=

10 sin 3 q cos 2 qdq = 0.89 W 3 ò0

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p


UNIT 8

Gd =

2p l = 2p , a = 0 l æ bd cos q + a ö AF = 2 cos ç ÷ = 2 cos ( p cos q)= 2 cos ( p cos q) 2 è ø

4 p(12) 1 = 79.48 = 19 dB 0.06 10 3

27. (D) bd =

20. (A) Gd = 36 dB = 3981 Gd =

4 pr 2 Pi Prad

Pi =

Þ

Gd Prad |E|2 = 4 pr 2 2h

28. (B) cos ( p cos q) = 0

(240 p) Gd Prad Þ |E|= 4 pr 2 ( 60)( 3981)(200 ´ 10 3) (120 ´ 10 3) 2

=

= 1.82 V m 21. (B) |Es|= 1.82 = 120 ´ 10 3 22. (D) l =

|Er|2 s |Ei| s = 4 pr 2 4p r 8 4p

2

æ l ö æ 1 ö Pr = Gdr Gdt ç 2000 = 12.7 mW ÷ P = (1)(1) ç 3÷ 4 p r è ø è 4 p10 ø Gdr Gdt æ l ç 4 p çè 4 pr1 r2

p 3p , ± 2 2

cos q = ±

1 2

q = 60 ° , 120 °

Þ

29. (B) Maxima occur when

d( AF ) =0 dq

sin ( p cos q) p sin q = 0

q = 0, 90 ° , 180 °

Þ

31. (A) ( AF ) N

æ Nf ö sinç ÷ è 2 ø = æyö sinç ÷ è2ø

y = bd cos q + a , N = 4 sin 4 x 2 sin 2 x cos 2 x = = 4 cos x cos 2 x sin x sin x y p 2p l p = cos q bd = = , a = 0, 2 4 l 4 2 æp ö æp ö AF = 4 cos ç cos q ÷ cos ç cos q ÷ . 4 2 è ø è ø

2

ö ÷÷ sPrad ø

Gdt = 34 dB = 2512, Gdr = 22 dB = 158.5 r1 = 3 km, r2 = 32 + 4 2 = 5 km l=

p cos q = ±

2p l =p , a =p l 2 pö æ bd cos q + a ö æp AF = 2 cos ç ÷ = cos ç cos q + ÷ 2 2ø è ø è4

c 3 ´ 108 = =1 m f 300 ´ 106

23. (C) Pr =

Þ

30. (B) bd =

= 12 mW

2

Electromagnetics

3 ´ 108 = 0.06 m, Pr = 8 pW 5 ´ 10 9 2

8 ´ 10 Þ

-12

0.06 (2512)(158.5) æ ö = ç ÷ (2.4) Prad 4p 4 3 5 p ( k )( k ) è ø

kW Prad = 1038 .

24. (B) Efficiency = D=

************

63 = 0.9 63 + 7

Gain 16 = = 17.78 = 12.5 dB Efficiency 0.9

25. (B) BW =

f 40 ´ 106 = = 800 kHz Q 50

æ dl ö 26. (C) Radiation resistance Rr = 80 p2 ç ÷ è l ø

2

2

æ 1 ö = 80 ´ p2 ´ ç . W ÷ = 351 è 15 ø Efficiency = Page 522

351 . Rr = = 74.5 % . + 12 . Rr + RL 351 www.gatehelp.com


CHAPTER

9.1 LINEAR ALGEBRA

é 0 1. If A = ê - 1 ê êë 2

1 0 -2

-2 3 l

ù ú is a singular matrix, then l is ú úû

(A) 0

(B) -2

(C) 2

(D) -1

7. Every diagonal elements of a Hermitian matrix is (A) Purely real

(B) 0

(C) Purely imaginary

(D) 1

8. Every diagonal element of a Skew–Hermitian matrix is

2. If A and B are square matrices of order 4 ´ 4 such

(A) Purely real

(B) 0

that A = 5B and A = a × B , then a is

(C) Purely imaginary

(D) 1

(A) 5

(B) 25

(C) 625

(D) None of these

9. If A is Hermitian, then iA is

3. If A and B are square matrices of the same order such that AB = A and BA = A , then A and B are both

(A) Symmetric

(B) Skew–symmetric

(C) Hermitian

(D) Skew–Hermitian

(A) Singular

(B) Idempotent

10. If A is Skew–Hermitian, then iA is

(C) Involutory

(D) None of these

(A) Symmetric

(B) Skew–symmetric

(C) Hermitian

(D) Skew–Hermitian.

é-5 -8 0 ù 4. The matrix, A = ê 3 5 0 ú is ê ú êë 1 2 -1ûú (A) Idempotent (C) Singular

é- 1 - 2 11. If A = ê 2 1 ê êë 2 - 2

(B) Involutory (D) None of these

5. Every diagonal element of a skew–symmetric matrix

-2ù - 2 ú, then adj. A is equal to ú 1úû

(A) A

(B) c t

(C) 3A t

(D) 3A

is (A) 1 (C) Purely real é 6. The matrix, A = ê êë-

(B) 0 (D) None of these 1 2 i 2

i 2

-

1 2

ù ú is úû

(A) Orthogonal

(B) Idempotent

(C) Unitary

(D) None of these

é-1 2 ù 12. The inverse of the matrix ê úis ë 3 -5 û é5 (A) ê ë3

2ù 1 úû

é-5 -2 ù (C) ê ú ë -3 -1û

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é5 (B) ê ë2

3ù 1 úû

(D) None of these Page 525


UNIT 9

é1 13. Let A = ê5 ê êë3

0

é 4 0 1 (A) ê-10 2 4ê êë -1 -1

0ù 0ú ú 2 úû

é 1 (C) ê-10 ê êë -1

0 2 -1

0ù 0 ú, then A -1 is equal to ú 2 úû

2 1

0 é 2 1 (B) ê-5 1 2ê êë -1 -1

0ù 0ú ú 2 úû

Engineering Mathematics

19. The system equationsx + y + z = 6, x + 2 y + 3z = 10, x + 2 y + lz = 12 is inconsistent, if l is 0ù 0ú ú 2 úû

(A) 3

(B) -3

(C) 0

(D) None of these.

20.

The

system

of

equations

5 x + 3 y + 7 z = 4,

3 x + 26 y + 2 z = 9, 7 x + 2 y + 10 z = 5 has (A) a unique solution

(D) None of these

(B) no solution (C) an infinite number of solutions

é 2 -1 14. If the rank of the matrix, A = ê 4 7 ê êë 1 4

3ù lú is 2, then ú 5 úû

the value of l is

(D) none of these 21. If A is an n–row square matrix of rank (n - 1), then (A) adj A = 0

(B) adj A ¹ 0 (D) None of these

(A) -13

(B) 13

(C) adj A = I n

(C) 3

(D) None of these

22.

The

system

of

equations

x - 4 y + 7 z = 14,

15. Let A and B be non–singular square matrices of the

3 x + 8 y - 2 z = 13, 7 x - 8 y + 26 z = 5 has

same order. Consider the following statements.

(A) a unique solution

(I) ( AB) T = A TBT

(II) ( AB) -1 = B-1 A -1

(B) no solution

(III) adj( AB) = (adj. A)(adj. B)

(IV) r( AB) = r( A)r(B)

(C) an infinite number of solution (D) none of these

(V) AB = A × B

(A) I, III & IV

(B) IV & V

4ù é3 23. The eigen values of A = ê ú are 9 5 ë û

(C) I & II

(D) All the above

(A) ± 1

(B) 1, 1

(C) -1, - 1

(D) None of these

Which of the above statements are false ?

1 -1ù 3 -2 ú is ú 4 -3úû

é 2 16. The rank of the matrix A = ê 0 ê êë 2 (A) 3

(B) 2

2ù é 8 -6 ê 24. The eigen values of A = -6 7 - 4 ú are ê ú 3 úû êë 2 - 4

(C) 1

(D) None of these

(A) 0, 3, -15

(B) 0, - 3 , - 15

(C) 0, 3, 15

(D) 0, - 3, 15

17.

The

system

15 x - 6 y + 5 z = 0,

of

equations

lx - 2 y + 2 z = 0

has

3 x - y + z = 0, a

non–zero

solution, if l is (A) 6

(B) -6

(C) 2

(D) -2

18.

The

system

of

then the eigen values of the matrix 2A are 1 3 (B) 2 , - 4 , 6 (A) , - 1 , 2 2 (C) 1 , - 2, 3

equation

x - 2 y + z = 0,

2 x - y + 3z = 0, lx + y - z = 0 has the trivial solution as the only solution, if l is (A) l ¹ (C) l ¹ 2 Page 526

4 5

(B) l =

25. If the eigen values of a square matrix be 1, - 2 and 3,

4 3

(D) None of these

(D) None of these.

26. If A is a non–singular matrix and the eigen values of A are 2 , 3 , - 3 then the eigen values of A -1 are 1 1 -1 (A) 2 , 3 , - 3 (B) , , 2 3 3 (C) 2 A , 3 A , - 3 A

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(D) None of these


Linear Algebra

é cos 2 f cos f sin fù B=ê ú cos f sin f sin 2 f û ë

27. If -1 , 2 , 3 are the eigen values of a square matrix A then the eigen values of A 2 are (A) -1 , 2 , 3

(B) 1, 4, 9

(C) 1, 2, 3

(D) None of these

is a null matrix, then q and f differ by (A) an odd multiple of p

28. If 2 , - 4 are the eigen values of a non–singular matrix A and A = 4, then the eigen values of adj A are (A)

1 2

, -1

(C) 2 , - 4

,

1 4

(C) 4, 16

(C) an odd multiple of

p 2

p 2

(D) an even multiple

(D) 8 , - 16

35. If A and B are two matrices such that A + B and AB

eigenvalues of A T are 1 2

(B) an even multiple of p

(B) 2 , - 1

29. If 2 and 4 are the eigen values of A then the (A)

Chap 9.1

are both defined, then A and B are (A) both null matrices

(B) 2, 4

(B) both identity matrices

(D) None of these

(C) both square matrices of the same order (D) None of these

30. If 1 and 3 are the eigenvalues of a square matrix A é 0 36. If A = ê ëtan

then A 3 is equal to (A) 13( A - I 2 )

(B) 13A - 12I 2

(C) 12( A - I 2 )

(D) None of these

31. If A is a square matrix of order 3 and A = 2 then A (adj A) is equal to é2 (A) ê0 ê êë0

0 2 0

0ù 0ú ú 2 úû

é 12 ê (B) ê0 êë0

é1 (C) ê0 ê êë0

0 1

0ù 0ú ú 1 úû

(D) None of these

0

0 1 2

0

é8 32. The sum of the eigenvalues of A = ê4 ê êë2

0ù ú 0ú 1 ú 2 û

écos a - sin a2 ù then (I - A ) × ê ú is equal to cos a û ësin a (A) I + A

(B) I - A

(C) I + 2 A

(D) I - 2 A - 4ù , then for every positive integer - 1 úû

é3 37. If A = ê ë1

2 5 0

- tan a2 ù ú 0 û

a 2

n, A n is equal to

3ù 9 ú is ú 5 úû

é1 + 2 n (A) ê ë n

4n ù 1 + 2 núû

é1 + 2 n (B) ê ë n

é1 - 2 n (C) ê ë n

4n ù 1 + 2 núû

(D) None of these

(A) 18

(B) 15

é cos a 38. If A a = ê ë- sin a

(C) 10

(D) None of these

statements :

equal to

sin a ù , then consider the following cos a úû

I. A a × A b = A ab 33. If 1, 2 and 5 are the eigen values of the matrix A then A is equal to (A) 8

(B) 10

(C) 9

(D) None of these

34. If the product of matrices écos q cos q sin q ù A =ê ú and 2 ëcos q sin q sin q û 2

- 4n ù 1 - 2 núû

II. A a × A b = A ( a + b)

é cos n a III. ( A a ) n = ê n ë- sin a

sin n a ù ú cos n a û

é cos na IV. ( A a ) n = ê ë- sin na

sin na ù cos na úû

Which of the above statements are true ? (A) I and II

(B) I and IV

(C) II and III

(D) II and IV

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Page 527


UNIT 9

39. If A is a 3-rowed square matrix such that A = 3, then adj (adj A) is equal to : (A) 3A

(B) 9A

(C) 27A

(D) none of these

40. If A is a 3-rowed square matrix, then adj (adj A) is


é1 2 0 ù T 45. If A = ê ú, then AA is ë3 -1 4 û é 1 3ù (A) ê ú ë-1 4 û

é 1 0 1ù (B) ê ú ë-1 2 3û

1ù é2 (C) ê ú ë1 26 û

(D) Undefined

equal to (A) A

6

(B) A

3

46. The matrix, that has an inverse is

(C) A

4

(D) A

2

é 3 1ù (A) ê ú ë6 2 û

é5 2 ù (B) ê ú ë2 1 û

é6 2 ù (C) ê ú ë9 3û

é8 2 ù (D) ê ú ë4 1 û

41. If A is a 3-rowed square matrix such that A = 2, then adj (adj A 2 ) is equal to (A) 2 4

(B) 28

47. The skew symmetric matrix is

(C) 216

(D) None of these

é 0 -2 5 ù (A) ê 2 0 6ú ê ú êë-5 -6 0 úû

é1 5 2ù (B) ê6 3 1ú ê ú êë2 4 0 úû

é0 1 3ù (C) ê1 0 5 ú ê ú êë3 5 0 úû

é0 3 3ù (D) ê2 0 2 ú ê ú êë1 1 0 úû

é2 x 42. If A = ê ë x

0ù é 1 and A -1 = ê x úû ë-1

0ù , then the value 2 úû

of x is (A) 1 (C)

1 2

(B) 2 (D) None of these

é1 ù é1 1 0 ù ê ú 48. If A = ê ú and B = ê0 ú, the product of A and B ë1 0 1û êë1 úû

é1 2 ù 43. If A = ê2 1ú then A -1 is ê ú êë1 1úû

is

é1 4 ù (A) ê3 2 ú ê ú êë2 5 úû

é 1 -2 ù (B) ê-2 1ú ê ú êë 1 2 úû

é2 3ù (C) ê3 1ú ê ú êë2 7 úû

(D) Undefined

é-1 -8 -10 ù (C) ê 1 -2 -5 ú ê ú 15 úû êë 9 22 Page 528

é1 0 ù (B) ê ú ë0 1 û

é1 ù (C) ê ú ë2 û

é1 0 ù (D) ê ú ë0 2 û

éA 49. Matrix D is an orthogonal matrix D = ê ëC

é 2 -1ù é1 -2 - 5 ù 44. If A = ê 1 0 ú and B = ê then AB is ê ú 0 úû ë3 4 3 4 êë úû é-1 -8 -10 ù (A) ê-1 -2 5ú ê ú 15 úû êë 9 22

é1 ù (A) ê ú ë0 û

0 -10 ù é 0 ê (B) -1 -2 -5 ú ê ú êë 0 21 -15 úû é0 -8 -10 ù (D) ê1 -2 -5 ú ê ú êë9 21 15 úû

value of B is 1 (A) 2 (C) 1

(B)

Bù . The 0 úû

1 2

(D) 0

50. If A n ń is a triangular matrix then det A is n

(A)

Õ ( -1) a i =1

n

ii

(B)

ii

(D)

i =1

n

(C)

å ( -1) a i =1

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Õa

ii

n

åa i =1

ii


Linear Algebra

ét 2 cos t ù dA 51. If A = ê t ú, then dt will be sin e t ë û

Chap 9.1

SOLUTIONS

ét 2 sin t ù (A)ê t ú ë e sin t û

é2t cos t ù (B) ê t ú ë e sin t û

1. (B) A is singular if A = 0

é2t - sin t ù (C) ê t cos t úû ëe

(D) Undefined

é 0 Þ ê -1 ê êë 2

1 0 -2

-2 ù 3 ú =0 ú l úû -2 ½ ½ + 2½ ½1 l½ ½ 0

½ 1 - ( -1)½ ½-2

- 2½ ½

½ 0 + 0½ 3½ ½ - 2

3½ ½= 0 l½

52. If A Î R n ń , det A ¹ 0, then

Þ

(A) A is non singular and the rows and columns of A are linearly independent.

Þ ( l - 4) + 2( 3) = 0

(B) A is non singular and the rows A are linearly dependent.

2. (C) If k is a constant and A is a square matrix of

(C) A is non singular and the A has one zero rows.

order n ´ n then kA = kn A .

(D) A is singular.

A = 5B Þ A = 5B = 5 4 B = 625 B

Þ l - 4 + 6 =0 Þ l = -2

Þ a = 625 ************

3. (B) A is singular, if A = 0, A is Idempotent, if A 2 = A A is Involutory, if A 2 = I Now, A 2 = AA = ( AB) A = A( BA) = AB = A and B2 = BB = (BA)B = B( AB) = BA = B Þ

A 2 = A and B2 = B,

Thus A & B both are Idempotent. é-5 -8 4. (B) Since, A = ê 3 5 ê êë 1 2

0 ù é-5 -8 0úê 3 5 úê -1úû êë 1 2

2

é1 0 0 ù = ê0 1 0 ú = I, ê ú êë0 0 1úû

A2 = I

Þ

0ù 0ú ú -1úû

A is involutory.

5. (B) Let A = [ aij ] be a skew–symmetric matrix, then AT = - A ,

Þ

aij = - aij ,

if i = j then aii = - aii

Þ

2 aii = 0

Þ

aii = 0

Thus diagonal elements are zero. 6. (C) A is orthogonal if AA T = I A is unitary if AA Q = I , where A Q is the conjugate transpose of A i.e., A Q = ( A) T . Here, é ê AA Q = ê êêë

1 2 i 2

ùé 2 úê úê 1 úê 2 ûú ëê i

1 2 i 2

ù 2 ú é1 ú= 1 ú êë0 2 ûú i

0ù = I2 1úû

Thus A is unitary.

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Page 529


UNIT 9

7. (A) A square matrix A is said to be Hermitian if A = A. So aij = a ji . If i = j then aii = aii i.e. conjugate of Q

an element is the element itself and aii is purely real. 8. (C) A square matrix A is said to be Skew-Hermitian if A = - A.

If A is Skew–Hermitian then A = - A

Q

Þ

Q

a ji = - aij ,

if i = j then aii = - aii Þ

Now, ( iA) Q = i A Q = - iA Q = - iA, Þ

Now, ( iA) Q = i A Q = - ( -A ) = iA then iA is Hermitian. 11. (C) If A = [ aij ]n ń then det A = [ cij ]nT ń Where cij is the cofactor of aij Also cij = ( -1) i + j M ij , where M ij is the minor of aij , row

and

the

the remaining matrix. ½ 1 Now, M11 = minor of a11 i.e. -1 =½ ½-2

-2 ½ ½ 1½

= -3

Similarly 1½ ½ ½2 - 2 ½ ½2 ½= 6 ; M13 =½ =-6 M12 = ½ 1½ ½2 - 2 ½ ½2 -2½ ½ ½-1 = - 6 ; M 22 =½ 1½ ½ 2

½-1 M 23 =½ ½ 2

- 2½ ½= 6 ; M 31 - 2½

½-2 =½ ½ 1

½-1 - 2 ½ ½= 6 ; M 33 M 32 =½ ½ 2 - 2½

-2 ½ ½ =3 ; 1½ - 2½ ½= 6 ; -2½

½-1 - 2 =½ ½ 2

½ ½= 3 1½

C11 = ( -1)1 + 1 M11 = -3; C12 = ( -1)1 + 2 M12 = -6 ; 1+ 3

C13 = ( -1)

C22 = ( -1)

2+ 2

M13 = -6; C21 = ( -1) M 22 = 3; C23 = ( -1)

2+1

2+ 3

M 21 = 6; M 23 = -6;

C31 = ( -1) 3+ 1 M 31 = 6; C32 = ( -1) 3+ 2 M 32 = -6 ; C33 = ( -1) 3+ 3 M 33 = 3 éC11 det A = êC21 ê êëC31

C12 C22 C32 T

C13 ù C23 ú ú C33 úû

T

Þ

é-5 adj A = ê ë-3

0 0 = 4 ¹ 0,

3

1

2

2ù 1 úû

T

10 -10 ù é 4 0 ú 2 -1 = ê10 2 ú ê 0 2 úû êë-1 -1

4 0 0

é 4 0 1ê 10 2 4ê êë-1 -1

0ù 0ú ú 2 úû

0ù 0ú ú 2 úû

14. (B) A matrix A ( m ń ) is said to be of rank r if (i) it has at least one non–zero minor of order r, and (ii) all other minors of order greater than r, if any; are zero. The rank of A is denoted by r( A). Now, given that r( A) = 2 ® minor of order greater than 2 i.e., 3 is zero. ½2 - 1 Thus A =½4 7 ½ 1 4 ½

3½ l½ = 0 ½ 5½

Þ

2( 35 - 4 l) + 1(20 - l) + 3(16 - 7) = 0,

Þ

70 - 8 l + 20 - l + 27 = 0,

Þ

9 l = 117

Þ

l = 13

15. (A) The correct statements are ( AB) T = BT A T , ( AB) -1 = B-1 A -1 , adj ( AB) = adj (B) adj ( A) r( AB) ¹ r( A) r(B), AB = A × B Thus statements I, II, and IV are wrong.

A = 2( -9 + 8) + 2( -2 + 3) = - 2 + 2 = 0 Þ T

- 2ù - 1 úû

1 adj A A

2

A -1 =

T

- 2 ù é5 = - 1 úû êë3

0

é adj A == ê ê êë

= -1

16. (B) Since

é -3 -6 -6 ù é -1 -2 -2 ù =ê 6 3 -6 ú = 3ê 2 1 -2 ú = 3A T ú ê ú ê 3úû 1ûú êë 6 -6 êë 2 -2 Page 530

é-5 ê-3 ë

1

column

corresponding to aij and then take the determinant of

½-2 M 21 =½ ½-2

1 -1

- 3ù - 1 úû

A =5

10. (C) A is Skew–Hermitian then A Q = - A

the

3 -5

é-5 Also, adj A = ê ë-2

( iA ) Q = - ( iA )

Thus iA is Skew–Hermitian.

leaving

2

13. (A) Since, A -1 =

9. (D) A is Hermitian then A Q = A

by

-1

Now, Here A =

it is only possible when aii is purely imaginary.

obtained

1 adj A A

12. (A) Since A -1 =

A -1 =

aii + aii = 0


r( A ) < 3

½2 1½ ½= 6 ¹ 0 Again, one minor of order 2 is ½ ½0 3½ Þ www.gatehelp.com

r( A ) = 2


UNIT 9

5 ½ ½3-l ½= 0 Þ ½ - 5 - l½ ½-4

31. (A) Since A(adj A) = A I 3

Þ

( 3 - l)( - 5 - l) + 16 = 0 Þ

Þ

l + 2l + 1 = 0 2


Þ

- 15 + l + 2 l + 16 = 0 2

( l + 1) = 0 2

Þ

Þ

l = - 1, - 1

é1 A(adj A) = 2 ê0 ê êë0

0 1 0

0 ù é2 0 ú = ê0 ú ê 1 úû êë0

0 2 0

0ù 0ú ú 2 úû

Thus eigen values are -1 , - 1 32. (A) Since the sum of the eigenvalues of an n–square

24. (C) Characteristic equation is A - lI = 0

matrix is equal to the trace of the matrix (i.e. sum of the

-6 2½ ½8 -l Þ ½-6 7-l - 4½= 0 ½ ½ - 4 3 - l½ ½ 2 Þ

l2 - 18 l2 + 45 l = 0

Þ

l( l - 3)( l - 15) = 0

diagonal elements) so, required sum = 8 + 5 + 5 = 18 33. (B) Since the product of the eigenvalues is equal to

Þ

the determinant of the matrix so A = 1 ´ 2 ´ 5 = 10

l = 0 , 3 , 15

25. (B) If eigen values of A are l1 , l2 , l3 then the eigen values of kA are kl1 , kl2 , kl3. So the eigen values of 2A are 2 , - 4 and 6

34. (C) écos q cos f cos ( q - f) cos q sin f cos ( q - f) ù AB = ê ú =A ëcos f sin q cos ( q - f) sin q sin f cos ( q - f) û null matrix when cos ( q - f) = 0

26. (B) If l1 , l2 ,........, l

n

are the eigen values of a

non–singular matrix A, then A -1 has the eigen values 1 1 1 1 1 , , ........, . Thus eigen values of A -1 are , , l1 l2 ln 2 3 -1 . 3

This happens when ( q - f) is an odd multiple of

p . 2

35. (C) Since A + B is defined, A and B are matrices of the same type, say m ´ n. Also, AB is defined. So, the number of columns in A must be equal to the number of rows in B i.e. n = m. Hence, A and B are square matrices

27. (B) If l1 , l2 , ......, l

n

are the eigen values of a matrix

A, then A has the eigen values l12 , l22 , ........, l 2n . So,

of the same order.

2

eigen values of A 2 are 1, 4, 9. 28. (B) If l1 , l2 ,...., l

n

are the eigen values of A then

the eigen values adj A are eigenvalues of adj A are

A l1

,

A l2

,......,

A l

; A ¹ 0. Thus

n

4 -4 , i.e. 2 and-1. 2 4

29. (B) Since, the eigenvalues of A and A are square so T

the eigenvalues of A T are 2 and 4. 30. (B) Since 1 and 3 are the eigenvalues of A so the characteristic equation of A is ( l - 1) ( l - 3) = 0 Þ l2 - 4 l + 3 = 0 Also, by Cayley–Hamilton theorem, every square matrix satisfies its own characteristic equation so A 2 - 4 A + 3I 2 = 0 Þ

A 2 = 4 A - 3I 2

Þ A 3 = 4 A 2 - 3A = 4( 4 A - 3I) - 3A Þ A 3 = 13A - 12I 2 Page 532

a 1 - tan 2 2 a 2 = 1-t 36. (A) Let tan = t, then, cos a = a t + t2 2 1 + tan 2 2 2 tan and sin a =

a 1 + tan 2 2

=

2t 1 + t2

- sin a ù cos a úû

écos a (I - A ) × ê ësin a é ê 1 =ê ê- tan a ë 2

a 2

aù 2ú´ ú 1 ú û

tan

é1 - t2 t ù ê1 + t 2 é 1 =ê ú´ê ë -t 1 û ê 2 t 2 ëê(1 + t )

écos a êsin a ë

- sin a ù cos a úû

-2 t ù (1 + t 2 ) ú ú 1 - t2 ú 1 + t 2 úû

aù é 1 - tan ú é 1 - tù ê 2 = (I + A ) =ê ú ú =ê t 1 a ë û êtan 1 ú ë û 2

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Linear Algebra

- 4 ù é3 - 1 úû êë1

é3 37. (B) A 2 = ê ë1 é1 + 2 n =ê ë n

- 4 ù é5 = - 1 úû êë2

- 8ù - 3 úû

é1 3ù é1 2 0 ù ê ú 45. (C) AA = ê ú 2 -1 ë3 -1 4 û ê0 4 ú êë úû T

- 4n ù , where n = 2. 1 - 2 núû

é cos a 38. (D) A a × A b = ê ë- sin a é cos ( a + b) =ê ë- sin ( a + b)

sin a ù é cos b cos a úû êë- sin b

(1)( 3) + (2)( -1) + (0)( 4) ù é (1)(1) + (2)(2) + (0)(0) =ê ( 3 )( 1 ) + ( + 1 )( 2 ) ( 4 )( 0 ) ( 3 )( 3) + ( -1)( -1) + ( 4)( 4) úû ë

sin bù cos b úû

1ù é5 =ê ú 1 26 ë û

sin ( a + b) ù = Aa+ b cos ( a + b) úû

46. (B) if A is zero, A -1 does not exist and the matrix A

Also, it is easy to prove by induction that sin na ù é cos na (A a )n = ê ú ë- sin na cos na û 39. (A) We know that adj (adj A) = A

n -2

is said to be singular. Only (B) satisfy this condition. 5 2 A = = (5)(1) - (2)(2) = 1 2 1

× A.

47. (A) A skew symmetric matrix A n ń is a matrix with

Here n = 3 and A = 3.

A T = -A . The matrix of (A) satisfy this condition.

So, adj (adj A) = 3( 3- 2 ) × A = 3A. 40. (C) We have adj (adj A) = A

é1ù é1 1 0 ù ê ú é(1)(1) + (1)(0) + (0)(1) ù é1 ù 48. (C) AB = ê ú 0 =ê ú =ê ú ë1 0 1û ê1ú ë(1)(1) + (0)(0) + (1)(1) û ë2 û êë úû

( n -1 ) 2

4

Putting n = 3, we get adj (adj A) = A .

49. (C) For orthogonal matrix

41. (C) Let B = adj (adj A 2 ). Then, B is also a 3 ´ 3

2

[K

A2 = A

é2 x 42. (C) ê ë x Þ

é2 x ê0 ë

2

( 3-1 ) 2

]

0ù é 1 x úû êë-1

0 ù é1 = 2 x úû êë0

det M = 1 And M -1 = M T , therefore Hence D -1 = D T é A Cù 1 é 0 -B ù DT = ê = D -1 = ú A úû -BC êë-C ëB 0 û -C 1 This implies B = Þ B= Þ B = ±1 -BC B

matrix.

adj {adj (adj A 2 )} = adj B = B3 = adj (adj A 2 ) = é A 2 ëê

Chap 9.1

3-1

2

2

ù = A 16 = 216 ûú

0 ù é1 = 2 úû êë0 0ù , 1úû

=B

Hence B = 1 50. (B) From linear algebra for A n ń triangular matrix

0ù 1úû

So, 2 x = 1

n

det A = Õ aii , The product of the diagonal entries of A i =1

Þ

1 x= . 2

43. (D) Inverse matrix is defined for square matrix only. é 2 -1ù é1 -2 -5 ù 44. (C) AB = ê 1 0 ú ê ê ú ë3 4 0 úû êë-3 4 úû

é d( t 2 ) dA ê dt 51. (C ) =ê t dt ê d( e ) ë dt

d(cos t) ù dt ú = é2 t - sin t ù d(sin t) ú êë e t cos t úû ú dt û

52. (A) If det A ¹ 0, then A n ń is non-singular, but if A n ń is non-singular, then no row can be expressed as a

é(2)(1) + ( -1)( 3) (2)( -2) + ( -1)( 4) (2)( -5) + ( -1)(0) ù = ê (1)(1) + (0)( 3) (1)( -2) + (0)( 4) (1)( -5) + (0)(0) ú ê ú êë( -3)(1) + ( 4)( 3) ( -3)( -2) + ( 4)( 4) ( -3)( -5) + ( 4)(0) úû

linear combination of any other. Otherwise det A = 0

é-1 -8 -10 ù = ê 1 -2 -5 ú ê ú 15 úû êë 9 -22

************

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Page 533


CHAPTER

9.2 DIFFERENTIAL CALCULUS

1. If f ( x) = x 3 - 6 x 2 + 11 x - 6 is on [1, 3], then the point

6. A point on the curve y = x - 2 on [2, 3], where the

c « ] 1, 3 [ such that f ¢ ( c) = 0 is given by 1 1 (A) c = 2 ± (B) c = 2 ± 2 3

tangent is parallel to the chord joining the end points of

(C) c = 2 ±

1 2

(D) None of these

2. Let f ( x) = sin 2 x, 0 £ x £ Then, c is equal to p (A) 4 (C)

the curve is

p 2

and f ¢ ( c) = 0 for c « ] 0, 2p [.

x 2

3. Let f ( x) = x( x + 3) e , -3 £ x £ 0. Let c « ] - 3, 0 [ such that f ¢ ( c) = 0. Then, the value of c is (B) -3

(C) -2

1 (D) 2

æ7 1ö (C) ç , ÷ è4 2ø

æ9 1 ö (D) ç , ÷ è2 4 ø

value of c of the mean value theorem is

(D) None

(A) 3

æ7 1 ö (B) ç , ÷ è2 4 ø

7. Let f ( x) = x( x - 1)( x - 2) be defined in [0, 12 ]. Then, the

p (B) 3

p 6

æ9 1ö (A) ç , ÷ è4 2ø

(A) 0.16

(B) 0.20

(C) 0.24

(D) None

8. Let f ( x) = x 2 - 4 be defined in [2, 4]. Then, the value of c of the mean value theorem is (A) - 6

(B)

(C)

(D) 2 3

3

6

4. If Rolle’s theorem holds for f ( x) = x 3 - 6 x 2 + kx + 5 on 1 , the value of k is [1, 3] with c = 2 + 3 (A) -3 (B) 3

mean-value theorem is (A) 0.5

(B) ( e - 1)

(C) 7

(C) log ( e - 1)

(D) None

(D) 11

5. A point on the parabola y = ( x - 3) 2 , where the

Page 534

9. Let f ( x) = e x in [0, 1]. Then, the value of c of the

tangent is parallel to the chord joining A (3, 0) and B (4,

10. At what point on the curve y = (cos x - 1) in ]0, 2p[ ,

1) is

is the tangent parallel to x –axis ?

(A) (7, 1)

æ3 1ö (B) ç , ÷ è2 4 ø

æp ö (A) ç , - 1 ÷ è2 ø

(B) ( p, - 2)

æ7 1 ö (C) ç , ÷ è2 4 ø

æ 1 1ö (D) ç - , ÷ è 2 2ø

æ 2 p -3 ö (C) ç , ÷ è 3 2 ø

(D) None of these

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Differential Calculus

11. log sin ( x + h) when expanded in Taylor’s series, is equal to

Chap 9.2

æ 16. If u = tan -1 ç ç è

(A) log sin x + h cot x -

1 2 h cosec2 x + K 2

(A) 2 cos 2 u

(B) log sin x + h cot x +

1 2 h sec 2 x + K 2

(C)

(C) log sin x - h cot x +

1 2 h cosec2 x + K 2

(B)

1 tan u 4

(D) None of these x

2

1 sin 2 u 2

(B) sin 2u

(C) sin u

2

3

x 3 + y 3 + x 2 y - xy 2 , then the value of x 2 - xy + y 2

¶u ¶u is +y ¶x ¶y

(A)

pö pö æ æ çx- ÷ çx- ÷ 2 2 è ø è ø (B) 1 + -K 2! 4!

1 sin 2 u 4

(D) 2 tan 2 u

17. If u = tan -1

pö æ 12. sin x when expanded in powers of ç x - ÷ is 2 è ø 2 3 2 pö pö pö æ æ æ çx- ÷ çx- ÷ çx- ÷ 2ø 2ø 2ø è è è (A) 1 + + + +K 2! 3! 4!

x + y ö÷ ¶u ¶u equals , then x +y ÷ ¶ x ¶y x+ yø

(D) 0

æ yö æ yö 18. If u = fç ÷ + xyç ÷, then the value of è xø è xø x2

5

pö pö æ æ çx- ÷ çx- ÷ 2ø 2ø pö è è æ (C) ç x - ÷ + + +K ! ! 2 3 5 è ø

¶2u ¶2u ¶2u + 2 xy + y 2 2 , is 2 dx dx dy ¶y

2

(A) 0

(B) u

(C) 2u

(D) -u

(D) None of these æp ö 13. tanç + x ÷ when expanded in Taylor’s series, gives è4 ø 4 2 (A) 1 + x + x + x 3 + K 3 (B) 1 + 2 x + 2 x 2 + (C) 1 +

8 3 x +... 3

x2 x4 + +K 2! 4!

19. If z = e x sin y, x = log e t and y = t 2 , then by the expression

dz is given dt

(A)

ex (sin y - 2 t 2 cos y) t

(B)

ex (sin y + 2 t 2 cos y) t

(C)

ex (cos y + 2 t 2 sin y) t

(D)

ex (cos y - 2 t 2 sin y) t

20. If z = z( u, v) , u = x 2 - 2 xy - y 2 , v = a, then (A) ( x + y)

¶z ¶z = ( x - y) ¶x ¶y

(B) ( x - y)

¶z ¶z = ( x + y) ¶x ¶y

¶ 3u is equal to 14. If u = e , then ¶ x¶ y¶ z

(C) ( x + y)

¶z ¶z = ( y - x) ¶x ¶y

(D) ( y - x)

¶z ¶z = ( x + y) ¶x ¶y

(A) e xyz [1 + xyz + 3 x 2 y 2 z 2 ]

21. If f ( x, y) = 0, f( y, z) = 0, then

(D) None of these xyz

(B) e

xyz

[1 + xyz + x y z ]

(C) e

xyz

[1 + 3 xyz + x y z ]

3

2

2

¶f ¶f ¶f ¶f dz × = × × ¶y ¶z ¶x ¶y dx

(B)

(C)

¶f ¶f dz ¶f ¶f × × = × ¶y ¶z dx ¶x ¶y

(D) None of these

2

(D) e xyz [1 + 3 xyz + x 3 y 3z 3 ] 15. If z = f ( x + ay) + f( x - ay), then (A)

¶ z ¶ z = a2 2 ¶x 2 ¶y

(B)

¶ z ¶ z = a2 2 ¶y 2 ¶x

(C)

¶2z 1 ¶2z =- 2 2 ¶y a ¶x 2

(D)

¶2z ¶2z = - a2 2 2 ¶x ¶y

2

(A)

3 3

2

2

¶f ¶f ¶f ¶f dz × × = × ¶y ¶z ¶x ¶x dx

22. If z = x 2 + y 2 and x 3 + y 3 + 3 axy = 5 a 2 , then at

2

x = a, y = a,

dz is equal to dx

(A) 2a

(B) 0

(C) 2 a 2

(D) a 3

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Page 535


UNIT 9

23. If x = r cos q, y = r sin q where r and q are the functions of x, then

dx is equal to dt


¶2u ¶2 y æ yö 30. If u = x n -1 yf ç ÷, then x 2 + y is equal to ¶x ¶y ¶x è xø (B) n( n - 1) u

(A) nu

(A) r cos q

dr dq - r sin q dt dt

(B) cos q

dr dq - r sin q dt dt

(C) r cos q

dr dq + sin q dt dt

(D) r cos q

dr dq - sin q dt dt

¶u ¶x

(C) ( n - 1)

(D) ( n - 1)

¶u ¶y

31. Match the List–I with List–II. List–I

¶ r ¶ r is equal to + dx 2 ¶y 2 2

24. If r 2 = x 2 + y 2 , then 2 2 æ ¶r ö ïü ïì æ ¶r ö (A) r 2 í ç ÷ + çç ÷÷ ý ïî è ¶x ø è ¶y ø ïþ

æ ¶r ö 1 ìï æ ¶r ö ÷ + çç ÷÷ 2 íç r ï è ¶x ø è ¶y ø î 2

(C)

2

ïü ý ïþ

2

2 2 æ ¶r ö ïü ïì æ ¶r ö (B) 2 r 2 í ç ÷ + çç ÷÷ ý ïî è ¶x ø è ¶y ø ïþ

(D) None of these

25. If x = r cos q, y = r sin q, then the value of is (A) 0 (C)

(i) If u =

(D)

¶2q ¶2q + ¶x 2 ¶y 2

¶r ¶y

1

1

then x 2

x4 + y4 1

3 u 16

(C) udu = mxdx + nydy

(D)

2

27. If y 3 - 3ax 2 + x 3 = 0, then the value of to (B)

(C) -

2 a2 x4 y5

(D) -

d y is equal dx 2

2 a2 x2 y5

(4) -

(B) dz =

xdy + ydx x2 + y2

(C) dz =

xdx - ydy x2 + y2

(D) dz =

xdx - ydy x2 + y2

(B) 1

(C) u

(D) eu

(II)

(III)

(IV)

(A)

1

2

3

4

(B)

2

1

4

3

(C)

2

1

3

4

(D)

1

2

4

3

in the area is approximately equal to (A) 1%

(B) 2%

(C) p%

(D) 4%

33. Consider the Assertion (A) and Reason (R) given below:

x2 + y2 ¶u ¶u , then x is equal to +y x+ y ¶x ¶y

(A) 0

(I)

and minor axes of an ellipse, then the percentage error

y , then x

xdy - ydx x2 + y2

1 u 4

32. If an error of 1% is made in measuring the major

2 a2 x2 y5

(A) dz =

¶u ¶x

Correct match is—

du dx dy =m +n u x y

a2 x2 y5

¶2u ¶2u ¶2u + 2 xy + y2 2 ¶x ¶x ¶y ¶y 2

(2)

(3) 0

(A) -

2 ¶2u ¶2u 2 ¶ u + 2 xy + y ¶x 2 ¶x ¶y ¶y 2

¶u ¶u æ yö (iv) If u = f ç ÷ then x +y ¶x ¶y è xø

(1) -

(B) du = mdx + ndy

Page 536

1

List–II

(A) du = mx m -1 y n + nx m y n -1

29. If u = log

(ii) If u =

1

26. If u = x m y n , then

28. z = tan -1

1

x2 - y2

(iii) If u = x 2 + y 2 then x 2

(B) 1

¶r ¶x

x2 y ¶2u ¶2u then x 2 + y x+ y ¶x ¶x ¶y

¶u ¶u æ yö Assertion (A): If u = xyf ç ÷, then x = 2u +y ¶x ¶y è xø Reason (R): Given function u is homogeneous of degree 2 in x and y. Of these statements (A) Both A and R are true and R is the correct explanation of A

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Chap 9.2

(B) Both A and R are true and R is not a correct explanation of A

41. If a < 0, then f ( x) = e ax + e - ax is decreasing for (A) x > 0

(B) x < 0

(C) A is true but R is false

(C) x > 1

(D) x < 1

(D) A is false but R is true

42. f ( x) = x 2 e - x is increasing in the interval

34. If u = x log xy, where x 3 + y 3 + 3 xy = 1, then equal to (A) (1 + log xy) -

x æ x2 + y ö ÷ ç y çè y 2 + x ÷ø

(B) (1 + log xy) -

yæ y + xö ÷ ç x çè x 2 + y ÷ø

(C) (1 - log xy) -

xæ x + yö ÷ ç y çè y 2 + x ÷ø

du is dx

(A) ] -¥, ¥ [

(B) ] -2, 0 [

(C) ] 2, ¥ [

(D) ] 0, 2 [

43. The least value of a for which f ( x) = x 2 + ax + 1 is increasing on ] 1, 2, [ is

2

(A) 2

(B) -2

(C) 1

(D) -1

2

44. The minimum distance from the point (4, 2) to the parabola y 2 = 8 x, is

y æ y2 + x ö (D) (1 - log xy) - çç 2 ÷ x è x + y ÷ø

(A)

(B) 2 2

2

(C) 2

(D) 3 2

¶z ¶z æ yö 35. If z = xyf ç ÷, then x is equal to +y ¶x ¶y è xø (A) z (B) 2z

45. The co-ordinates of the point on the parabola

(C) xz

y = 3 x - 3, are

y = x 2 + 7 x + 2 which is closest to the straight line

(D) yz

36. f ( x) = 2 x 3 - 15 x 2 + 36 x + 1 is increasing in the interval

(A) (-2, -8)

(B) (2, -8)

(C) (-2, 0)

(D) None of these

(A) ] 2, 3 [

(B) ] -¥, 3 [

46. The shortest distance of the point (0, c), where

(C) ] -¥, 2 [È ] 3, ¥

(D) None of these

0 £ c < 5, from the parabola y = x 2 is

37. f ( x) =

x is increasing in the interval ( x 2 + 1)

(A) ] -¥, - 1 [ È ] 1, ¥ [

(B) ] -1, 1 [

(C) ] -1, ¥ [

(D) None of these

4c + 1 2

(A)

4c + 1

(B)

(C)

4c - 1 2

(D) None of these x

æ1ö 47. The maximum value of ç ÷ is è xø

38. f ( x) = x 4 - 2 x 2 is decreasing in the interval (A) ] -¥, -1 [ È ] 0, 1 [

(B) ] -1, 1 [

(C) ] -¥, -1 [ È ] 1, ¥ [

(D) None of these

(A) e æ1ö (C) ç ÷ è eø

39. f ( x) = x 9 + 3 x 7 + 6 is increasing for (A) all positive real values of x

(B) e

-

1 e

e

(D) None of these

(B) all negative real values of x

250 ö æ 48. The minimum value of ç x 2 + ÷ is x ø è

(C) all non-zero real values of x

(A) 75

(B) 50

(C) 25

(D) 0

(D) None of these 40. If f ( x) = kx 3 - 9 x 2 + 9 x + 3 is increasing in each

49. The maximum value of f ( x) = (1 + cos x) sin x is

interval, then

(A) 3

(B) 3 3

(C) 4

(D)

(A) k < 3

(B) k £ 3

(C) k > 3

(D) k ³ 3

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3 3 4 Page 537


UNIT 9

50. The greatest value of f ( x) =

(A)

SOLUTIONS

sin 2 x pö æ sinç x + ÷ 4ø è

on the interval [0, 2p ] is 1 2

1. (B) A polynomial function is continuous as well as differentiable. So, the given function is continuous and differentiable.

(B)

2

f (1) = 0 and f ( 3) = 0. So, f (1) = f ( 3). By Rolle’s theorem Ec such that f ¢( c) = 0.

(D) - 2

(C) 1


Now, f ¢ ( x) = 3 x 2 - 12 x + 11

51. If y = a log x + bx 2 + x has its extremum values at

Þ

x = -1 and x = 2, then 1 (A) a = - , b = 2 2

Now, f ¢ ( c) = 0

(C) a = 2, b = 52.

The

(B) a = 2, b = -1

1 2

Þ

f ¢ ( c) = 3c 2 - 12 c + 11. Þ

3c 2 - 12 c + 11 = 0

1 ö æ c = ç2 ± ÷. 3 ø è

(D) None of these 2. (A) Since the sine function is continuous at each

co-ordinates

of

the

point

on

the

curve

4 x 2 + 5 y 2 = 20 that is farthest from the point (0, -2) are (A) ( 5 , 0)

(B) ( 6 , 0)

(C) (0, 2)

(D) None of these

pö æ 53. For what value of xç 0 £ x £ ÷, the function 2 è ø x has a maxima ? y= (1 + tan x) (A) tan x

(B) 0

(C) cot x

(D) cos x

é pù x « R, so f ( x) = sin 2 x is continuous in ê0, ú. ë 2û f ¢ ( x) = 2 cos 2 x, which clearly exists for all p p x « ]0, [ .So, f ( x) is differentiable in x « ]0, [. 2 2

Also,

æ pö Also, f (0) = f ç ÷ = 0. By Rolle’s theorem, there exists è2 ø p c « ]0, [ such that f ¢( c) = 0. 2 2 cos 2 c = 0

2c =

Þ

p 2

Þ

c=

p . 4

3. (C) Since a polynomial function as well as an exponential function is continuous and the product of two continuous functions is continuous, so f ( x) is *************

continuous in [-3, 0]. f ¢ ( x) = (2 x + 3) × e

-

x 2

- é x + 6 - x2 ù 1 -2 2 e ( x + 3 x) = e 2 ê ú 2 2 ë û x

-

x

which clearly exists for all x « ] - 3, 0 [. f ( x) is differentiable in ] -3, 0 [. Also, f ( -3) = f (0) = 0. By Rolle’s theorem c « ] -3, 0 [ such that f ¢ ( c) = 0. - é c + 6 - c2 ù e 2ê ú =0 2 ë û c

Now, f ¢ ( c) = 0

Þ

c + 6 - c 2 = 0 i.e. c 2 - c - 6 = 0 Þ

( c + 2) ( 3 - c) = 0

Þ

c = -2, c = 3.

Hence, c = -2 « ] -3, 0 [ . 4. (D) f ¢ ( x) = 3c 2 - 12 x + k f ¢ ( c) = 0 Page 538

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Þ

3c 2 - 12 c + k = 0


UNIT 9

¶u ¶u 1 + y sec 2 u = tan u ¶x ¶y 2

æ pö æ pö æ pö f ç ÷ = 1, f ¢ç ÷ = 0, f ¢¢ç ÷ = -1, è2 ø è2 ø è2 ø

x sec 2 u

æ pö æ pö f ¢¢¢ç ÷ = 0, f ¢¢¢¢ç ÷ = 1, .... 2 è ø è2 ø

Þ

13. (B) Let f ( x) = tan x Then,

17. (A) Here tan u =

2 3 æp ö æ pö æ pö x æ p ö x ¢¢¢æ p ö f ç + x ÷ = f ç ÷ + xf ¢ç ÷ + × f ¢¢ç ÷ + f ç ÷+... è4 ø è4ø è 4 ø 2! è 4 ø 3! è 4 ø

f ¢( x) = sec , f ¢¢( x) = 2sec x tan x, 2

2

f ¢¢¢( x) = 2sec 4 x + 4sec 2 x tan 2 x etc. æ pö æ pö æ pö æ pö f ç ÷ = 1, f ¢ç ÷ = 2, f ¢¢ç ÷ = 4, f ¢¢¢ç ÷ = 16, ... è4ø è4ø è4ø è4ø x2 x3 æp ö Thus tanç + x ÷ = 1 + 2 x + ×4 + × 16 + K 2 6 è4 ø 8 = 1 + 2 x + 2 x + x3 + K 3 2

14. (C) Here u = e xyz

Þ

¶u = exyz × yz ¶x

= e xyz (1 + 3 xyz + x 2 y 2 z 2 )

Which is homogeneous of degree 1 ¶f ¶f Thus x +y = f ¶x ¶y

¶z = f ¢ ( x + ay) + f¢ ( x - ay) ¶x

¶ z = a 2 f ¢¢( x + ay) + a 2 f¢¢( x - ay)....(2) ¶y 2 2

Hence from (1) and (2), we get

tan u =

x+

x+ y x+

y

Þ

x2

and x 2

2 ¶2v ¶2v 2 ¶ v + 2 xy + y = 0....(1) ¶x 2 ¶x ¶y ¶y 2

¶2w ¶2w ¶2w + 2 xy + y2 = 0....(2) 2 ¶x ¶x ¶y ¶y 2

Adding (1) and (2), we get x2

2 ¶2 ¶2 2 ¶ ( v + w ) + 2 xy ( v + w ) + y ( v + w) = 0 ¶x 2 ¶x ¶y ¶y 2

x2

¶2u ¶2u ¶2u + 2 xy + y2 2 = 0 2 ¶x ¶x ¶y ¶y

¶z = e x sin y ¶x ¶z dx 1 And = e x cos y, x = log e t Þ = ¶y dt t

¶ z = f ¢¢( x + ay) + f¢¢( x - ay)....(1) dx 2 ¶z = af ¢( x + ay) - af¢ ( x - ay) ¶y

x+ y

Then u = v + w

19. (B) z = e x sin y

2

æ 16. (B) u = tan -1 ç ç è

¶f ¶u 1 +y = sin 2 u ¶x ¶y 2

æ yö æ yö 18. (A) Let v = fç ÷ and w = xYç ÷ è xø è xø

Þ

15. (B) z = f ( x + ay) + f( x - ay)

¶2z ¶2z = a2 2 2 ¶y ¶x

ö ÷ y ÷ø

= f (say)

Which is a homogeneous equation of degree 1/2 ¶f ¶f 1 By Euler’s theorem. x +y = f ¶x ¶y 2 x

x 3 + y 3 + x 2 y - xy 2 = f (say) x 2 - xy + y 2

homogeneous of degree one

¶ 3u = e xyz × (1 + 2 xyz) + ( z + xyz 2 ) e xyz × xy ¶ x¶ y¶ z

Þ

¶u ¶u 1 1 +y = sin u cos u = sin 2 u ¶x ¶y 2 4

Now v is homogeneous of degree zero and w is

¶2u = ze xyz + yzexyz × xz = e xyz ( z + xyz 2 ) ¶x¶y

Þ

x

As above question number 16 x

Now,

Page 540


¶(tan u) ¶(tan u) 1 +y = tan u ¶x ¶y 2

Þ

dy = 2t dt dz ¶z dx ¶z dy = × + × dt ¶x dt ¶y dt

And y = t 2

= e x sin y ×

Þ

ex 1 + e x cos y × 2 t = (sin y + 2 t 2 cos y) t t

20. (C) Given that z = z( u, v), u = x 2 - 2 xy - y 2 , v = a....(i) ¶z ¶z ¶u ¶z ¶v = × + × ....(ii) ¶x ¶u ¶x ¶v ¶x ¶z ¶z ¶u ¶z ¶v and = × + × ....(iii) ¶y ¶u ¶y ¶v ¶y From (i), ¶u ¶u ¶v ¶v =2x -2y , = -2 x - 2 y, = 0, =0 ¶x ¶y ¶x ¶y www.gatehelp.com



Substituting these values in (ii) and (iii)

24. (C) r 2 = x 2 + y 2

¶z ¶z ¶z = (2 x - 2 y) + × 0....(iv) ¶x ¶u ¶v and

and

¶z ¶z ¶z = × ( -2 x - 2 y) + × 0....(v) ¶y ¶u ¶v

Þ

¶2r ¶2r + =2 + 2 + 4 ¶x 2 ¶y 2

2

2 2 æ ¶r ö üï ¶ 2 r ¶ 2 y 1 ìï æ ¶r ö + = + ç ÷ ç ÷ í ç ¶y ÷ ý ¶x 2 ¶ y 2 r 2 ï è ¶x ø è ø þï î

Þ

21. (C) Given that f ( x, y) = 0, f( y, z) = 0

25. (A) x = r cos q ,

These are implicit functions ¶f dz ¶y =¶f dy ¶z

æ ¶f ö æ ¶f ö ç÷ ç÷ dy dz ç ¶x ÷ ç ¶y ÷ × = ´ dx dy ç ¶f ÷ ç ¶f ÷ ç ¶y ÷ ç ¶z ÷ è ø è ø or,

¶r ¶r =2y = 2 x and ¶y ¶x

¶2r ¶2r = 2 and =2 2 ¶x ¶y 2 2

¶z ¶z = ( y - x) ¶x ¶y

¶f dy = - ¶x , ¶f dx ¶y

Þ

æ ¶r ö æ ¶r ö and ç ÷ + çç ÷÷ = 4 x 2 + 4 y 2 = 4 r 2 ¶ x è ø è ¶y ø

From (iv) and (v), we get ( x + y)

Chap 9.2

Þ

æ yö q = tan -1 ç ÷ è xø

Þ

tan q =

Þ

¶q 1 -y æ -y ö = = 2 2 ç 2 ÷ ¶x 1 + ( y x) è x ø x + y 2

and

¶2q -2 xy = ¶x 2 ( x 2 + y 2 ) 2

Similarly

¶f ¶f dz ¶f ¶f × × = × ¶y ¶z dx ¶x ¶y

y x

y = r sin q

¶2q 2 xy ¶2q ¶2q and = 2 + =0 2 2 2 ¶y (x + y ) ¶x 2 ¶y 2

26. (D) Given that u = x m y n Taking logarithm of both sides, we get

22. (B) Given that z = x + y 2

2

log u = m log x + n log y

and x + y + 3axy = 5 a ...(i) dz ¶z ¶z dy = + × ....(ii) dx ¶x ¶y dx 3

3

from (i),

Differentiating with respect to x, we get 1 du 1 1 dy du dx dy or, = m× + n× =m + n× u dx x y dx u x y

2

¶z 1 ¶z = ×2x , = 2 2 ¶x 2 x + y ¶y 2

1 x + y2 2

×2y

dy dy and 3 x + 3 y + 3ax + 3ay.1 = 0 dx dx 2

Þ

2

æ x 2 + ay ö dy ÷÷ = - çç 2 dx è y + ax ø

Substituting these value in (ii), we get dz = dx

x x2 + y2

+

æ x 2 + ay ö ÷÷ çç - 2 x 2 + y 2 è y + ax ø y

a æ dz ö = + ç ÷ 2 dx è ø( a , a ) a + a2

æ a 2 + aa ö ÷÷ = 0 çç - 2 a 2 + a 2 è a + a. a ø a

27. (D) Given that f ( x, y) = y 3 - 3ax 2 + x 3 = 0 fx = - 6 ax + 3 x 2 , f y = 3 y 2 , fxx = - 6 a + 6 x , f yy = 6 y , fxy = 0 é fxx ( f y) 2 - 2 fx f y fxy + f yy( fx ) 2 ù d2 y =-ê ú 2 ( f y) 3 dx ë û é( 6 x - 6 a( 3 y 2 ) 2 - 0 + 6 y( 3 x 2 - 6 ax) 2 ù = -ê ú (3 y2 )3 ë û 2 = - 5 ( -ax 3 - ay 3 + 4 a 2 x 2 ) y =-

2 [ -a( a 3 + y 3) + 4 a 2 x 2 ] y5

23. (B) Given that x = r cos q, y = r sin q....(i)

=-

2 [ -a( 3ax 2 ) + 4 a 2 x 2 ] [ \ x 3 + y 3 - 3ax 2 = 0] y5

dx ¶x dr ¶x dq = × + × ....(ii) dt ¶r dt ¶q dt

=-

2 a2 x2 y5

From (i),

¶x ¶x = cos q, = - r sin q ¶r ¶q

Substituting these values in (ii), we get dx dr dq = cos q - r sin q × dt dt dt

28. (A) Given that z = tan -1

www.gatehelp.com

y ....(i) x

Page 541


UNIT 9

dz ¶z ¶z dy = + × ....(ii) dx ¶x ¶y dx From (i)

¶z = ¶y

¶z = ¶x

In (c) u = x1 2 + y1 2 It is a homogeneous function of 1 degree . 2

-y æ -y ö ×ç 2 ÷ = 2 + y2 x x ø æ yö è 1+ç ÷ è xø 1

2

x2

x æ1ö ×ç ÷ = 2 2 æ yö è xø x + y 1+ç ÷ è xø 1

=

2

Hence correct match is

f is a homogeneous function of degree one Þ

¶e u ¶e u x +y = eu ¶x ¶y

¶u ¶u +y =1 ¶x ¶y

d

2

1

3

4

Þ

log A = log p + log a + log b

Þ

¶(log A) = ¶(log p) + ¶(log a) + ¶(log b) ¶A ¶a ¶b Þ =0 + + A a b 100 100 100 Þ ¶A = ¶a + × ¶b A a b 100 100 But it is given that ¶a = 1, and ¶b = 1 a b 100 ¶A = 1 + 1 = 2 A

Differentiating partially w.r.t. x, we get ¶ 2 u ¶u ¶2u n ¶u + + y = 2 ¶x ¶x ¶y ¶x ¶x x

c

Area A = pab

It is a homogeneous function of degree n ¶u ¶u Euler’s theorem x +y = nu ¶x ¶y

Þ

b

ellipse

æ yö 30. (C) Given that u = x n -1 yf ç ÷. è xø

x

a

32. (B) Let 2a and 2b be the major and minor axes of the

¶u ¶u or xe u + ye u = eu ¶x ¶y or, x

1 æ1 1 ö ç - 1 ÷u = - u 2 è2 4 ø

zero. ¶u ¶u x +y = 0. u = 0 ¶x ¶y

x2 + y2 x2 + y2 , eu = = f (say) x+ y x+ y

¶f ¶f x +y = f ¶x ¶y

¶2u ¶2u ¶2u + 2 xy + y2 = n( n - 1) u 2 ¶x ¶x dy ¶y 2

æ yö In (d)u = f ç ÷ It is a homogeneous function of degree è xø

Substituting these in (ii), we get dz -y x dy xdy - ydx , dz = = + × dx x 2 + y 2 x 2 + y 2 dx x2 + y2 29. (B) u = log

Thus percentage error in A =2%

¶2u ¶2u ¶u +y = ( n - 1) 2 ¶x ¶y ¶x ¶x

æ yö 33. (A) Given that u = xyf ç ÷. Since it is a homogeneous è xø

x2 y It is a homogeneous function of 31. (B) In (a) u = x+ y

function of degree 2. By Euler’s theorem x

degree 2. ¶2u ¶2u ¶u ¶u (as in question 30) x 2 +y = ( n - 1) = ¶x ¶x ¶y ¶x ¶x In (b) u =

x1 2 - y1 2 . It is a homogeneous function of x1 4 + y1 4

æ1 1ö 1 degree ç - ÷ = è2 4 ø 4 x2 = Page 542

¶2u ¶2u ¶2u + 2 xy + y 2 2 = n( n - 1) u 2 ¶x ¶x ¶y ¶y

1 æ1 3 ö u ç - 1 ÷u = 4 è4 16 ø


Thus x

¶u ¶u +y = nu (where n = 2) ¶x ¶y

¶u ¶u +y = 2u ¶x ¶y

34. (A) Given that u = x log xy....(i) x 3 + y 3 + 3 xy = 1....(ii) ¶u ¶u ¶u dy we know that ....(ii) = + ¶x ¶x ¶y dx From (i) and

www.gatehelp.com

¶u 1 = x× × y + log xy = 1 + log xy ¶x xy

¶u 1 x = x× ×x= ¶y xy y



Now, D = (2 t 2 - 4) 2 + ( 4 t - 2) 2 is minimum when

From (ii), we get 3 x2 + 3 y2

Chap 9.2

dy æ dy ö + 3ç x + y × 1÷ = 0 dx è dx ø

æ x + yö dy ÷÷ = - çç 2 dx è y + xø 2

Þ

Substituting these in (A), we get du x ì æ x 2 + y öü ÷ý = (1 + log xy) + í -çç 2 dx y î è y + x ÷øþ 35. (B) The given function is homogeneous of degree 2. ¶z ¶z Euler’s theorem x +y = 2z ¶x ¶y

E = (2 t 2 - 4) 2 + ( 4 t - 2) 2 is minimum. Now, E = 4 t 4 - 16 t + 20 dE Þ = 16 t 3 - 16 = 16 ( t - 1) ( t 2 + t + 1) dt dE =0 Þ t =1 dt éd2 E ù d2 E 2 . So, = 48 t = 48 > 0. ê dt 2 ú dt 2 ë û ( t =1 ) So, t = 1 is a point of minima. Thus Minimum distance = (2 - 4) 2 + ( 4 - 2) 2 = 2 2 .

36. (C) f ¢( x) = 6 x - 30 x + 36 = 6( x - 2)( x - 3) 2

Clearly, f ¢( x) > 0 when x < 2 and also when x > 3.

45. (A) Let the required point be P ( x, y). Then,

f ( x) is increasing in ] -¥, 2 [ È ] 3, ¥ [.

perpendicular distance of P ( x, y) from y - 3 x + 3 = 0 is

37. (B) f ¢ ( x) =

p=

( x + 1) - 2 x 1- x = 2 ( x 2 + 1) 2 ( x + 1) 2 2

2

2

Clearly, ( x 2 + 1) 2 > 0 for all x. So, f ¢ ( x) > 0 Þ

Þ

=

y - 3x + 3 10

x2 + 4 x + 5

(1 - x 2 ) > 0

10

= =

x2 + 7 x + 2 - 3 x + 3 10 ( x + 2) 2 + 1 10

or p =

( x + 2) 2 + 1 10

dp 2 ( x + 2) d p 2 and = = 2 dx dx 10 10 2

(1 - x) (1 + x) > 0

So,

This happens when -1 < x < 1.

æ d2 p ö ÷ > 0. x = -2, Also, çç 2 ÷ è dx øx = -2

So, f ( x) is increasing in ] -1, 1 [.

dp =0 dx

38. (A) f ¢ ( x) = 4 x 3 - 4 x = 4 x( x - 1)( x + 1).

So, x = -2 is a point of minima.

Clearly, f ¢ ( x) < 0 when x < - 1 and also when x > 1.

When x = -2, we get y = ( -2) 2 + 7 ´ ( -2) + 2 = -8.

Sol. f ( x) is decreasing in ] -¥, -1 [ È ] 1, ¥ [.

The required point is ( -2, - 8).

39.(C) f ¢ ( x) = 9 x8 + 21 x6 > 0 for all non-zero real values

46. (C) Let A (0, c) be the given point and P ( x, y) be any

of x.

point on y = x 2 .

Þ

D = x 2 + ( y - c) 2 is shortest when E = x 2 + ( y - c) 2 is 40. (C) f ¢ ( x) = 3kx - 18 x + 9 = 3 [ kx - 6 x + 3]

shortest.

This is positive when k > 0 and 36 - 12 k < 0 or k > 3.

Now,

2

2

41. (A) f ( x) = ( e ax + e - ax ) = 2 cosh ax. f ¢( x) = 2 a sinh ax < 0 When x > 0 because a < 0 42. (D) f ¢( x) = - x 2 e - x + 2 xe - x = e - x x(2 - x). Clearly, f ¢( x) > 0 when x > 0 and x < 2. 43. (B) f ¢( x) = (2 x + a) 1 < x <2 Þ Þ

2 <2x < 4 Þ 2 + a <2x + a < 4 + a

(2 + a) < f ¢( x) < ( 4 + a).

For f ( x) increasing, we have f ¢( x) > 0.

E = x 2 + ( y - c) 2 = y + ( y - c) 2 Þ E = y 2 + y - 2 cy + c2 dE d2 E = 2 > 0. = 2 y + 1 - 2 c and dy dy 2 dE =0 dy

Þ

1ö æ y =çc - ÷ 2 è ø

1ö æ Thus E minimum, when y = ç c - ÷ 2ø è 1ö æ 1 ö æ Also, D = ç c - ÷ + ç c - - c ÷ 2 2 è ø è ø = c-

\2 + a ³ 0 or a ³ - 2. So, least value of a is -2.

2

é 1 öù æ 2 ê . .. x = y = ç c - 2 ÷ú è øû ë

1 4c - 1 = 4 2

44. (B) Let the point closest to (4, 2) be (2 t 2 , 4). www.gatehelp.com

Page 543


UNIT 9

x

æ1ö 47. (B) Let y = ç ÷ then, y = x - x è xø

51. (C)

1 d2 y = x - x (1 + log x) 2 - x - x × 2 dx x dy =0 Þ 1 + log x = 0 Þ dx éd y ù æ1ö ê dx 2 ú æ 1 ö = -ç e ÷ è ø ë û çç x = ÷÷ 2

è

So, x =

-

1 -1 e

é dy ù =0 êë dx úû (x = 2 ) x=

1 e

Þ

48. (A) f ¢( x) = 2 x -

500 ö 250 æ and f ¢¢( x) = ç 2 + 3 ÷ x ø x2 è

f ¢( x) = 0

250 =0 x2

Þ

a + 4b + 1 = 0 2

a + 8 b = -2....(ii)

52. (C) The given curve is ellipse.

dz =0 df Þ x = p 3 or

x = p. æ p ö -3 3 f ¢¢ç ÷ = < 0. So, x = p 3 is a point of maxima. 2 è 3ø p öæ pö 3 3 æ . Maximum value = ç sin ÷ç 1 + cos ÷ = 3 øè 3ø 4 è 2 sin x cos x sin x + cos x

Þ

cos f = 0

2 cos f ( 4 - sin f) = 0 Þ

when f =

d 2z = -2 cos 2 f - 8 sin f df2

p d 2z , < 0. 2 df2

p z is maximum when f = . So, the required point is 2 p pö æ ç 5 cos , sin ÷ i.e. (0, 2). 2 2ø è 53. (D) Let z =

1 + tan x 1 tan x = + x x x

dz 1 d 2z 2 = - 2 + sec 2 x and = + 2sec2 x tan x dx x dx 2 x 3 dz 1 Þ x = cos x. =0 Þ - 2 + sec 2 x = 0 x dx

2 2 2 2 (say), = (sec x + cosec x) z

Then,

where z = (sec x + cosec x). dz cos x = sec x tan x - cosec x cot x = (tan 3 x - 1). dx sin 2 x

éd 2z ù = 2 cos 3 x + 2sec2 x tan x > 0. ê dx 2 ú ë û x = cos x

p é pù in ê0, ú. 4 ë 2û

dz =0 dx

Þ

Sign of

dz changes from -ve to +ve when x passes dx

x=

p f= . 2

dz = - sin 2 f + 8 cos f Þ df

2

Þ

x2 y2 + = 1 which is an 5 4

z = 5 cos 2 f + 4 (1 + sin f) 2 dz Þ = -10 cos f sin f + 8(1 + sin f) cos f df

49. (D) f ¢( x) = (2 cos x - 1)(cos x + 1) and

tan x = 1

1 and a = 2. 2

when z = D 2 is maximum

250 ö æ Thus minimum value = ç 25 + ÷ = 75. 5 ø è

=

Thus z has a minima and therefore y has a maxima at x = cos x.

through the point p 4. So, z is minimum at x = p 4 and ************

therefore, f ( x) is maximum at x = p 4. Maximum value = Page 544

a + 2 b = 1....(i)

D = ( 5 cos f - 0) 2 + (2 sin f + 2) 2 is maximum

x = 5.

f ¢¢( x) = - sin x(1 + 4 cos x). 1 or cos x = -1 Þ f ¢( x) = 0 Þ cos x = 2

Þ

Let the required point be ( 5 cos f , 2 sin f). Then,

f ¢¢(5) = 6 > 0. So, x = 5 is a point of minima.

50. (C) f ( x) =

Þ

- a - 2b + 1 = 0

< 0.

1 is a point of maxima. Maximum value = e1 e . e

2x -

Þ

Solving (i) and (ii) we get b = -

eø

Þ

dy a = + 2 bx + 1 dx x

é dy ù =0 êë dx úû ( x = -1 )

dy = - x - x (1 + log x) dx

Þ


2 2 = 1. [sec( p 4) + cosec ( p 4)] www.gatehelp.com


CHAPTER

9.3 INTEGRAL CALCULUS

1.

òx

(A)

2

x dx is equal to +1

1 log ( x 2 + 1) 2

(C) tan -1

(B) log ( x 2 + 1)

x 2

2. If F ( a) =

5.

(D) 2 tan -1 x 1 , log a

to 1 (A) ( a x - a a + 1) log a (C)

3.

1 ( a x + a a + 1) log a

1 (B) (ax - aa) log a (D)

1 ( a x + a a - 1) log a

(B) cot x + cosec x + c

(C) tan x - sec x + c

(D) tan x + sec x + c

4.

( 3 x + 1) dx is equal to 2 -2x + 3

ò 2x

æ2x -1ö 3 5 ÷ (A) log (2 x 2 - 2 x + 3) + tan -1 çç ÷ 4 2 è 5 ø (B)

æ2x -1ö 4 ÷ log (2 x 2 - 2 x + 3) + 5 tan -1 çç ÷ 3 è 5 ø

æ2x -1ö 4 2 ÷ (C) log (2 x 2 - 2 x + 3) + tan -1 çç ÷ 3 5 è 5 ø æ2x -1ö 3 2 ÷ (D) log (2 x 2 - 2 x + 3) + tan -1 çç ÷ 4 5 è 5 ø

is equal to

tan -1 (tan x)

(B) 2 tan -1 (tan x)

(C)

1 2

tan -1 (2 tan x)

(D) 2 tan -1 ( 12 tan x)

2 sin x + 3 cos x

ò 3 sin x + 4 cos x dx is equal to

(A)

9 1 x+ log( 3 sin x + 4 cos x) 25 25

(B)

18 2 x+ log( 3 sin x + 4 cos x) 25 25

(C)

18 1 x+ log( 3 sin x + 4 cos x) 25 25

(D) None of these

ò

(A) (A) - cot x + cosec x + c

x

1 2

7.

dx ò 1 + sin x is equal to

2

(A)

6.

a > 1 and F ( x) = ò a 2 dx + K is equal

dx

ò 1 + 3 sin

(B) (C)

3 + 8 x - 3 x 2 dx is equal to 3x - 4 3 3

3 + 8 x - 3 x2 -

æ 3x - 4 ö sin -1 ç ÷ 18 3 è 5 ø 25

3x - 4 25 3 æ 3x - 4 ö sin -1 ç 3 + 8 x - 3 x2 + ÷ 6 18 è 5 ø 3x - 4 6 3

3 + 8 x - 3 x2 -

æ 3x - 4 ö sin -1 ç ÷ 18 3 è 5 ø 25

(D) None of these 8.

dx

ò

(A) (C)

2 x + 3x + 4 2

1 2 1 2

www.gatehelp.com

sin -1

is equal to

4x + 3

cosh -1

23 4x + 3 23

(B)

1 2

sinh -1

4x + 3 23

(D) None of these

Page 545


UNIT 9

9.

2x + 3

ò

x + x+1 2

dx is equal to

(A) 2 x 2 + x + 1 + 2 sinh -1

(C) 2 x 2 + x + 1 + sinh -1 (D) 2 x 2 + x + 1 - sinh -1

10. (A)

ò1+ x + x

(A)

1 2

é ( x + 1) 2 -1 êlog x 2 + 1 + tan ë

(B)

1 4

é ( x + 1) 2 log + 2 tan -1 ê x2 + 1 ë

ù xú û

2x + 1

(C)

1 2

é ( x + 1) 2 -1 êlog x 2 + 1 - 2 tan ë

3

ù xú û

2x + 1

+ x3

is equal to ù xú û

(D) None of these

is equal to

x - x2

2

3 16.

dx

ò

3

dx

15.

3

2x + 1

x 2 + x + 1 + 2 sinh -1

(B)

2x + 1


x - x2 + c

(C) log (2 x - 1) + c

sin x

ò 1 - sin x dx

is equal to

(A) - x + sec x + tan x + k

(B) - x + sec x + tan x

(B) sin -1 (2 x - 1) + c

(C) - x + sec x - tan x

(D) - x - sec x - tan x

(D) tan -1 (2 x - 1) + c

17.

ò e { f ( x) + f ¢( x)} dx is equal to x

(A) e f ¢( x)

(B) e x f ( x)

(C) e x + f ( x)

(D) None of these

x

1

11.

ò ( x + 1)

(A)

æ 2 ö ÷ 2 cosh -1 çç ÷ è1 + xø

1 - 2 x - x2

dx is equal to

(B)

æ 2 ö ÷ (C) - 2 cosh -1 çç ÷ è1 + xø 12. (A)

(C)

13.

dx

ò sin x + cos x 1 2

æ 2 ö ÷ cosh -1 çç ÷ 2 è1 + xø

1

æ 2 ö ÷ cosh -1 çç ÷ 2 è1 + xø

1

(D) -

is equal to

pö æ log tanç x + ÷ 4 è ø

æ x pö log tanç + ÷ 2 è2 8ø dx

ò sin( x - a) sin( x - b)

(D)

1 2

æ x pö log tanç + ÷ è2 6ø

æ x pö log tanç + ÷ 2 è4 4ø

1

is equal to

(A) sin( x - a) log sin( x - b)

ì sin( x - a) ü (C) sin( a - b) log í ý î sin( x - b) þ

dx ò ex - 1 is equal to

(C) log ( e Page 546

-x

- 1)

x

x +c 2

(B) e x cot

(C) e x tan x + c

òx

x +c 2

(D) e x cot x + c

x3 dx is equal to +1

2

(A) x 2 + log ( x 2 + 1) + c (B) log ( x 2 + 1) - x 2 + c (C)

1 2 1 x - log ( x 2 + 1) + c 2 2

(D)

1 2 1 x + log( x 2 + 1) + c 2 2

(A) x sin -1 x + 1 - x 2 + c

(B) x sin -1 x - 1 - x 2 + c

(C) x sin -1 x + 1 + x 2 + c

(D) x sin -1 x - 1 - x 2 + c

21.

ì sin( x - a) ü 1 (D) log í ý sin( a - b) î sin( x - b) þ

(A) log ( e x - 1)

æ 1 + sin x ö

ò e ççè 1 + cos x ÷÷ødx is

20. ò sin -1 x dx is equal to

æ x -aö (B) log sinç ÷ è x -bø

14.

(A) e x tan

19. (B)

1

18. The value of

ò

sin x + cos x 1 + sin 2 x

dx is equal to

(A) sin x

(B) x

(C) cos x

(D) tan x 1

22. The value of (B) log (1 - e x ) (D) log (1 - e ) x

(A) -1/2 (C) 1/2

www.gatehelp.com

ò 5 x - 3 dx is 0

(B) 13/10 (D) 23/10


UNIT 9

1

39.

x

ò ò(x

46. The area bounded by the curve r = q cos q and the p lines q = 0 and q = is given by 2 ö ö p æ p2 p æ p2 (B) (A) çç çç - 1 ÷÷ - 1 ÷÷ 4 è 16 16 è 6 ø ø

+ y 2 ) dydx is equal to

2

0 x

(A)

7 60

(B)

(C)

4 49

(D) None of these

3 35

(C) 1

1 + x2

0

0

ò ò dy dx is

40. The value of p log ( 2 + 1) 4

(B)

(C)

p log ( 2 + 1) 2

(D) None of these

p log ( 2 - 1) 4

(C)

A

(B)

36 5

(D) None of these

x + y = a and x + y = a in the first quadrant is given 2

2

by a2 - x 2

a

(A)

(C) 4 ò

p2

ò ò

a cos 2 q

0

rdrdq

rdrdq

(B) 2 ò

p2

ò

0

(D) 2 ò

cos 2 q

a

0

p a cos 2 q

ò

0 0

rdrdq

rdrdq

48. The area of the region bounded by the curve 1

x2 4

0

y= 0

(A)

ò ò

(C)

ò ò

2

0

dxdy

3x ( x 2 + 2)

y= x 2 4

dydx

x2 4

y= 0

1

ò

ò ò

(D)

ò

y= 0

dydx

3x ( x 2 + 2)

y= x 2 4

dxdy

below by z = 0 and bounded above by z = h is given by (B) pa 2 h

1 pa 3h 3

(D) None of these

50.

òòò

0

1

0

(B)

49. The volume of the cylinder x 2 + y 2 = a 2 bounded

(C)

0

cos 2 q

a

0

ò ò dxdy

(B)

a-x

0

p4

(A) pah

a2 - x 2

a

ò ò dxdy

(D) None of these

y ( x 2 + 2) = 3 x and 4 y = x 2 is given by

42. The area of the region bounded by the curves 2

(A) 4 ò

0

òò ydxdy is equal to

32 5

ö æ p2 çç - 1 ÷÷ ø è 16

0

41. If A is the region bounded by the parabolas y 2 = 4 x and x = 4 y, then 48 (A) 5

p 16

47. The area of the lemniscate r 2 = a 2 cos 2 q is given by

(A)

2


(D) None of these

a 2 - y2 a

ò ò dxdy

(C)

a-x

0

43. The area bounded by the curves y = 2 x , y = - x, x = 1 and x = 4 is given by 33 (B) 2

(A) 25 47 (C) 4

44. The area bounded by the curves y 2 = 9 x, x - y + 2 = 0

e x + y+ z dxdydz is equal to (B)

(C) ( e - 1) 2

(D) None of these

1

z

x+ z

-1

0

x -z

ò ò ò

( x + y + z) dy dx dz is equal to

(A) 4

(B) -4

(C) 0

(D) None of these

is given by (A) 1

(B)

1 2

3 2

(D)

5 4

(C)

*************

45. The area of the cardioid r = a (1 + cos q) is given by (A) 2 ò

p

ò

a (1 + cos q)

ò

a (1 + cos q)

q= 0 r = 0

(C) 2 ò

p2

0

Page 548

r=0

rdrdq

(B) 2 ò

rdrdq

(D) 2 ò

p

0 p4

0

ò

a (1 + cos q)

r=a

ò

a (1 + cos q)

r=0

3 ( e - 1) 2

(A) ( e - 1) 3

51.

101 (D) 6

1 1 1

0 0 0

rdrdq rdrdq www.gatehelp.com


Integral calculus

SOLUTIONS 1. (A)

ò

Þ

F ( a) =

ax +K log a

aa +K log a

1 aa 1 - aa K = = log a log a log a F ( x) =

3. (C)

a 1-a 1 + = [ a x - a a + 1] log a log a log a x

a

dx ò 1 + sin x

dx x x x x ö 2 + cos 2 ÷ + 2 sin cos ç sin 2 2ø 2 2 è x sec 2 dx 2 =ò =ò dx 2 2 x xö xö æ æ ç 1 + tan ÷ ç cos + sin ÷ 2 2ø 2ø è è x Put 1 + tan = t 2 x 2 dt 2 Þ sec2 dx = 2 dt Þ ò 2 dt = - + K 2 t t x -2 cos -2 2 +K = +K = x x x cos + sin 1 + tan 2 2 2 x x x -2 cos cos - sin 2 2 2 = ´ +K x x x x cos + sin cos - sin 2 2 2 2 x x 2 x -2 cos + 2 sin cos 2 2 2 +K = 2 x 2 x - sin cos 2 2 -(1 + cos x) + sin x = + k = tan x - sec x - 1 + K cos x

òæ

= tan x - sec x + c

3x + 1 dx 2x -2x + 3 2

=

p=

Þ

5 dx 3 log (2 x 2 - 2 x + 3) + ò 2 2 4 æ 4 æ 5ö 1ö ÷ ç x - ÷ + çç ÷ 2ø è è 2 ø

3 5 1 = log (2 x 2 - 2 x + 3) + tan -1 4 4æ 5ö ç ÷ ç 2 ÷ è ø 5. (C) Let I = ò =ò

1 2 5 2

x-

dx 1 + 3 sin 2 x

cosec2 x dx cosec2 x dx = cosec 2 x + 3 ò (1 + cot2 x) + 3

Put cot x = t Þ I =ò =

3 5 , q = 4 2

3 4x -2 5 dx dx + ò 2 4 ò 2 x2 - 2 x + 3 2 2x -2x + 3

I=

1 1 log t = log ( x 2 + 1) 2 2

2. (A) F ( x) = ò a x dx + K =

=

4. (A) Let I = ò

Let 3 x + 1 = p( 4 x - 2) + q

x dx 2 x +1

Put x 2 + 1 = t Þ 2 xdx = dt 1 1 x ò x 2 + 1 dx = ò 2 × t dt =

Chap 9.3

- cosec2 x dx = dt

1 -dt t 1 æ cot x ö = cot-1 = cot-1 ç ÷ 4 + t2 2 2 2 è 2 ø

1 tan -1 (2 tan x) 2

6. (C) Let I = ò

2 sin x + 3 cos x dx 3 sin x + 4 cos x

Let (2 sin x + 3 cos x) = p( 3 cos x - 4 sin x) + q( 3 sin x + 4 cos x) p=

1 18 , q= 25 25

I=

1 25

=

3 cos x - 4 sin x

3 sin x + 4 cos x

18

ò 3 sin x + 4 cos x dx + 25 ò 3 sin x + 4 cos x dx

1 18 log ( 3 sin x + 4 cos x) + x 25 25 2

ò

1 = 3 2

ì 4 öü æ 2 2 2 ç x - ÷ï ïæ 4 ö æ5 ö æ 4ö æ5 ö -1 3 í ç x - ÷ ç ÷ - ç x - ÷ + ç ÷ sin ç 5 ÷ý 3 3 3 3 è ø è ø è ø è ø çç ÷÷ï ï è 3 øþ î

=

3 + 8 x - 3 x 2 dx = 3

ò

2

4ö æ5 ö æ ç ÷ - ç x - ÷ dx 3 3ø è ø è

7. (B)

3x - 4 25 3 3x - 4 3 + 8 x - 3 x2 + sin -1 6 18 5

8. (B)

www.gatehelp.com

ò

dx 2 x + 3x + 4 2

=

1 2

ò

dx æ 23 ö 3ö æ ÷ ç x + ÷ + çç ÷ 4 è ø è 4 ø 2

2

Page 549


UNIT 9

1

=

sinh

2

æ ç ç è

2x + 3

ò

9. (B)

3 4 = 1 sinh -1 4 x + 3 2 23 23 ö ÷ ÷ 4 ø

x+

-1

=ò

x2 + x + 1

=ò

x2 + x + 1

=

dx +

2x + 1

ò

dx + 2 ò

= 2 x 2 + x + 1 + 2 sinh -1

dx

ò

x 1- x

13. (D)

x2 + x + 1 æ 3ö 1ö æ ÷ ç x + ÷ + çç ÷ 2ø è è 2 ø 2

2

1 x+ 2 3 2

2x + 1

I =ò

( x + 1) 1 - 2 x - x 2

Þ

dx = -

1 dt t2

ò

dt æ 1 t 2 - çç è 2

ö ÷ ÷ ø

2

=-

2

2

ò

1 sin( a - b)


=

1 sin ( a - b)

sin [( x - b) - ( x - a)]

dx

sin( x - b) cos( x - a) - cos( x - b) sin( x - a) dx sin( x - a) sin( x - b)

=

1 [cot( x - a) - cot( x - b)]dx sin( a - b) ò

=

1 [log sin ( x - a) - log sin ( x - b)] dx sin ( a - b)

=

ì sin( x - a) ü 1 log í ý sin ( a - b) î sin( x - b) þ

14. (D) Let I =

òe

dx e - x dx =ò 1 - e- x -1

x

dx 15. (B) Let I = ò =ò dt 2 t2 - 1

dx 1 + x + x2 + x3

dx (1 + x) (1 + x 2 ) A Bx + C 1 = + 2 (1 + x)(1 + x ) 1 + x 1 + x 2

Let

1 = A(1 + x 2 ) + ( Bx + C)(1 + x) 1 2

æ 2 ö ÷ cosh -1 çç ÷ 2 è x + 1ø

1

=-ò

sin( a - b) dx

Put 1 - e - x = t Þ e - x dx = dt dt I =ò = log t = log (1 - e - x ) t

1 dt t2

1

12. (C)

Page 550

1

1 ö æ1 ö æ1 1 - 2ç - 1 ÷ - ç - 1 ÷ t t t è ø è ø 2

=-

=

1 t -

1

=-

x+c

dx


=

=I

(2 x - 1) + c

Put x + 1 =

2

pö

æ

ò cosecçè x + 4 ÷ødx


3

I = ò 2 dq = 2 q + c = 2 sin -1

11. (D) Let I = ò

1

1 sin( a - b)

´ò

Put x = sin q Þ dx = 2 sin q cos q dq 2 sin q cos q 2 sin q cos q I =ò dq = ò dq 2 sin q cos q sin q 1 - sin q

I = sin

pö æ sinç x + ÷ 4ø è

=

=

dx

2

-1

dx

1 é 1æ p öù 1 æ x pö log tan ç + ÷ - log cot ç x + ÷ú = ê 2è 4 øû 2 ë 2 è2 8ø

=

2 dx

( x 2 + x + 1)1 2 + 2 sinh -1 1 2

10. (B)

2

ò

dx

x2 + x + 1

2x + 1

1

=


dx

ò sin x + cos x dx p p sin x cos + cos x sin 4 4

t

cosh -1 1

Comparing the coefficients of x 2 , x and constant terms, 2

A + B = 0, B + C = 0, C + A = 1 Solving these equations, we get 1 1 1 A = , B=- , C= 2 2 2 1 1 1 x -1 I = dx - ò 2 dx 2 ò1+ x 2 x +1 =

1 1 1 log (1 + x) - log ( x 2 + 1) + tan -1 x 2 2 2

=

1 4

é ( x + 1) 2 log + 2 tan -1 ê x2 + 1 ë

www.gatehelp.com

ù xú û


Integral calculus

16. (B) Let I = ò

sin x dx 1 - sin x

ò

=

=ò

1 - (1 - sin x) dx 1 - sin x

=ò

=ò

1 1 + sin x dx - ò dx = ò dx - x 1 - sin x 1 - sin 2 x

=ò

=ò

1 + sin x dx - x = ò (sec2 x + sec x tan x) dx - x cos 2 x

= tan x + sec x - x

Chap 9.3

sin x + cos x (sin x + cos 2 x) + 2 sin x cos x 2

sin x + cos x (cos x + cos x) 2

dx

sin x + cos x dx = ò dx = x sin x + cos x 35

35

ò 5 x - 3 dx = - ò 5 x - 3 dx +

22. (D)

0

0

= ò e x f ( x) dx + ò e x f ¢( x) dx = { f ( x) e x - ò f ¢( x) ex dx} + ò ex f ¢( x) dx = f ( x) × ex

9ö æ 9 =ç+ ÷+ è 10 5 ø =

æ 1 + sin x ö 18. (A) Let I = ò e çç ÷÷ dx è 1 + cos x ø x

23. (B)

24. (D)

c

dt = [tan -1 t ]1e +1

2

p 4

c

ò x(1 - x) dx = ò ( x - x ) dx 2

0

1 ö 1 æ1 = ç x 2 - x 3 ÷ = c 2 ( 3 - 2 c) 2 3 è ø0 6

x3 x × x2 dx = ò 2 dx x +1 x +1

c

x( x + 1 - 1) x dx = ò xdx - ò 2 dx 2 x +1 x +1 2

20. (A) Let I = ò sin = sin -1 x × x - ò

-1

1 1 - x2 x 1 - x2

x dx = ò sin

-1

x × 1 × dx

× x dx dx

xdx = - tdt = x sin -1 x + ò dt = x sin -1 x + t = x sin -1 x + 1 - x 2 + c sin x + cos x 1 + sin 2 x

dx

ò

x(1 - x) dx = 0

1 2 c ( 3 - 2 c) = 0 6

Þ

0

Þ

c=

3 2

25. (D) Put x 2 + x = t

In second part put 1 - x 2 = t 2

ò

e

òt

e x dx = dt =

Þ

c

1 2 1 x - log ( x 2 + 1) + c 2 2

21.

e x dx 2x +1

0

0

2

ò

1

òe

= tan -1 e - tan -1 1 = tan -1 e -

x = e tan + c 2

x-

dx = + e- x

1

x

= x sin

x

Put e x = t

1ì x x x x ü x x í e × 2 tan - ò e × 2 tan dx ý + ò e tan dx 2î 2 2 2 þ

-1

òe 0

1 x x = ò e x sec 2 dx + ò e x tan dx 2 2 2

=

éæ 5 ö æ 9 9 öù êç 2 - 3 ÷ - ç 10 - 5 ÷ú è ø è øû ë

9 æ 1 9 ö 13 + ç- + ÷= 10 è 2 10 ø 10 1

x xö æ ç 1 + 2 sin cos ÷ x 2 2 ÷ dx =òe ç x çç ÷÷ 2 cos 2 ø è 2

=ò

ò 5 x - 3 dx

35

1

35

19. (C) I = ò

1

ö æ 5 x2 æ 5 ö = ç - x 2 + 3 x ÷ + çç - 3 x ÷÷ è 2 ø0 ø3 5 è 2

17. (B) Let I = ò e { f ( x) + f ¢( x)} dx x

=

dx

1

ò

2x + 1

0

x+ x

26. (A)

2

2

dx = ò 0

dt t

Þ

(2 x + 1) dx = dt

= 2( t1 2 ) 20 = 2 2

p

òx

4

sin 5 xdx

p

Since, f ( - x) = ( - x) 4 sin 5 ( - x) = -x 4 sin 5 x f ( x) is odd function thus p

òx

4

sin 5 x dx = 0

p

27. (A) www.gatehelp.com

p2

p2

0

0

2 ò cos x dx =

ò

1 (cos 2 x + 1) dx 2 Page 551


UNIT 9


1 æ1 ö ç sin 2 x + x ÷ 2 è2 ø0

pö æp = ò 2 sinç (1 - t) - ÷dt = 2 4 è ø 0

=

1 é1 æp öù (sin p - sin 0) + ç - 0 ÷ú 2 êë2 è2 øû

pö æp = - ò 2 sin ç t - ÷dt = - 1 2 4 è ø 0

=

1 é1 pù p (0 - 0) - 0 + ú = ê 2 ë2 2û 4

2I = 0

1

p2

=

p ö

ò 2 sinçè 4 - 2 t ÷ødt 0

1

æ 3ö æ 1ö Gç ÷ Gç ÷ 1 p p2 p 2 2 Aliter 1. ò cos 2 x dx = è ø è ø = 2 = 2 4 æ4ö 0 2 Gç ÷ è2 ø p2

Aliter 2. Use Walli’s Rule

æp

1

ò cos

2

x =

0

Þ

I =0

31. (C) Let I =

2a

f ( x)

ò f ( x) + f (2 a - x) dx ....(1) 0

I=

f (2 a - x)

2a

ò f (2 a - x) + f ( x) dx....(2) 0

1 p p × = 2 2 4

Adding (1) and (2), we get 2I =

2a

f ( x) + f (2 a - x)

ò f ( x) + f (2 a - x) dx

=

0

a

28. (B) Let I = ò a 2 - x 2 dx

Þ

Put x = a sin q Þ dx = a cos q dq when x = 0, q = 0, p when x = a, q = 2

32. (C) Let I =

2a

ò 1 × dx = [ x ]

2a 0

= 2a

0

I = a

0

Put

0

Þ

p2

= a 2 ò cos 2 q dq = a 2 × 0

1 p (By Walli’s Formula) × 2 2

1 - x2

1 - x2

xdx

1 - x2 = t 1 2 1 - x2

( -2 x) dx = dt

when x = 0, t = 1, when x = 1, t = 0 0

pa 2 4

I = ò -e t dt = -[ e t ]10 = -[ e 0 - e1 ] = e - 1 1

a

Aliter:

ò

e

0

p2

I = ò a 2 - a 2 sin 2 q a cos q dq

=

1

ò

a 2 - x 2 dx

1

0

a

é 1 pa 2 ù pa 2 xù é1 = = ê x a 2 - x 2 + a 2 sin -1 ú = ê0 + 4 2 a û0 ë 4 úû ë2 p2

dx 1 x + x2 0

33. (B) Let I = ò

1

1

=ò 0

29. (D) Let I = ò log (tan x) dx ....(1) 0

dx æ 3ö 1ö æ ÷ ç x - ÷ + çç ÷ 2ø è è 2 ø 2

p2

æp ö I = ò log tan ç - x ÷ dx 2 è ø 0

=

0

=

æ 2 é 1 1 öù 2 æ p p ö -1 ÷ú= - tan -1 çç ç + ÷ êtan 3 è6 6ø 3 ë 3 3 ÷øû è

=

p2

I = ò log (cot x)....(2)

2

é 1ù x- ú 1 ê -1 2ú êtan 3 ê 3 ú 2 êë 2 úû 0

2p 3 3

=

2p 3 9

Adding (1) and (2), we get p2

2 I = ò [log (tan x) + log (cot x)]dx

1

34. (B) Let I =

-1

0

0

p2

= ò log (tan x × cot x) dx 0

p2

= ò log 1 dx = 0

Þ

I =0

=

1

Page 552

x

-x ò-1 x dx + 0

dx =

1

ò -1dx + ò 1 × dx = -[ x ]

0 -1

-1

+ [ x ]10

0

= - [0 - ( -1)] + [1 - 0 ] = 0

0

æ pt p ö 30. (D) Let I = ò 2 sin ç - ÷dt ....(i) è 2 4ø 0

ò

x

35. (C)

100 p

p

0

0

ò |sin x|dx = 100 ò |sin x|dx

[ . .. sin x is periodic with period p] www.gatehelp.com

1

x

ò x dx 0


Integral calculus

p

1

= 100 ò sin x dx = 100( - cos x) 0p

40. (D)

0

1 + x2

ò ò 0

= 100( - cos p + cos 0) = 100(1 + 1) = 200. p

p

0

0

= ( - cos x) m (sin x) n

1 [ x 1 + x 2 + log( x + 1 + x 2 )]10 2 1 = [ 2 + log (1 + 2 )] 2 =

41. (A) Let I = òò ydxdy,

= -cos m x sin n x, if m is odd I = ò cos

dx

0

f ( p - x) = cos m ( p - x) sin n ( p - x)

m

x2

0

1

Where f ( x) = cos m x sin n x

p

0

1

dydx = ò [ y ]01 +

= ò 1 + x 2 dx

36. (C) Let I = ò cos m x sin nx dx = ò f ( x) dx

A

x sin x dx = 0, if m is odd

Solving the given equations y 2 = 4 x and x 2 = 4 y , we get

n

x = 0, x = 4 . The region of integration A is given by

0

2 x

4 é y2 ù = ydydx ò ò ê 2 ú dx û x2 4 0 x2 4 0 ë 4 2 x

p

A =ò

37. (A) Let I = ò xF (sin x) dx ....(1) 0

p

4

é 48 x5 ù 1æ x4 ö ÷÷dx = ê x 2 çç 4 x ú = 5 160 2 10 ø ë û0 è 0 4

= ò ( x - p) F [sin ( p - x)]dx

=ò

0

I =

Chap 9.3

p

ò ( p - x) F(sin x) dx ....(2)

42. (A) The curves are

0

Adding (1) and (2), we get p

2 I = ò pF (sin x) dx

x 2 + y 2 = a 2 ...

....(i)

x + y = a...

....(ii)

0

Þ

I=

The curves (i) and (ii) intersect at A (a, 0) and B (0,a)

p

1 pF (sin x) dx 2 ò0 p2

38. (B) Let I = ò 0

The required area A =

ex æ 2 x xö + 2 tan ÷dx ç sec 2 è 2 2ø

p2

x 1 = ò e xsec 2 dx + 2 2 0

4 2 x

A =ò

x 1 Here, I1 = ò e x sec 2 dx 2 2 0

òe 0

= e - I 2 , I1 + I 2 = e p2

I = I1 + I 2 = e

x ò0 e × 2 tan 2 dx x

tan

ò dydx

y = - x....(ii)

2 x -x

1

4

= ò [2 x + x ]dx 1

æ 32 ö æ 4 1 ö 101 =ç + 8÷ -ç + ÷ = 6 è 3 ø è3 2ø

p2

x

4

ò dydx = ò [ y ]

1 -x

p2

y= a - x

ò

If a figure is drawn then from fig. the required area is

p2

p æ ö = ç e p 2 tan - 0 ÷ 4 è ø

x =0

y = 2 x i.e., y 2 = 4 x....(i)

x ò0 e tan 2 dx = I1 + I 2 x

p2

a2 - x 2

43. (D) The given equations of the curves are

p2

1 xù é1 = ê e x × 2 tan ú 2 û0 2 ë2

a

44. (B) The equations of the given curves are

x dx 2

y 2 = 9 x....(i)

x - y + 2 = 0....(ii)

The curves (i) and (ii) intersect at

p2

A(1, 3) and B(4, 6)

p2

If a figure is drawn then from fig. the required area is 1

39. (B)

ò 0

1

x

x

1 3ù é 2 2 2 òx ( x + y ) dy dx = ò0 êë x y + 3 y úû x dx

1

1 1 ù é = ò ê x 5 2 + x 3 2 - x 3 - x 3 ú dx 3 3 û 0 ë 1

2 52 1 4 ù 3 é2 = ê x7 2 + x - x ú = 15 3 û 0 35 ë7

4 3 x

A =ò

4

ò dydx = ò [ y ]

1 x+ 2

3 x x+ 2

dx

1

4

4

1 é ù = ò [ 3 x - ( x + 2)]dx = ê2 x 3 2 - x 2 - 2 x ú 2 ë û1 1 1 ö 1 æ = (16 - 8 - 8) - ç 2 - - 2 ÷ = 2 è ø 2

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Page 553


UNIT 9

Thus the equation volume is V = 4 ò zdxdy

45. (A) The equation of the cardioid is r = a (1 + cos q)

A

....(i)

If a figure is drawn then from fig. the required area is p a (1 + cos q)

ò

Required area A = 2

= =

1 2 1 4

cos 2 qdq =

0

2 ò q dq + 0

1 4

1 4

dq

1 1 1

50. (A)

p2

ò q (1 + cos 2 q)dq 2

òòòe

1 1

= ò ò [ ex +

0

2 ò q cos 2 q dq

dxdydz 1 1

] dydz = ò ò [ e1 + 0 0

0

1

= ò [( e 2 + z - e1 + z ) - ( e1 + z - e z )]dz 0

1

= ò ( e 2 + z - 2 e1 + z + e z ) dz = [ e 2 + z - 2 e1 + z + e z ]10 0

=

p3 1 æ -p ö 1 + ç -0÷ 96 4 è 4 ø 8

= ( e 3 - 2 e 2 + e) - ( e 2 - 2 e + 1)

p2

= e 3 - 3e 2 + 3e - 1 = ( e - 1) 3

p2

1

ò cos 2 q dq

51. (C)

0

z x+ z

ò ò ò ( x + y + z) dydxdz

-1 0 x - z p2

p p3 p 1æ1 ö - ç sin 2 q ÷ = 96 16 8 è 2 ø0 16

ö æ p2 çç - 1 ÷÷ 16 ø è

47. (A) The curve is r 2 = a 2 cos 2 q If a figure is drawn then from fig. the required area is p4

a

é1 2 ù òr = 0rdrdq = 4 ò0 êë2 r úû 0

p4

x+ z

=

é( x + y + z) 2 ù ò-1 ò0 êë ú dxdz 2 ûx- y

=

é(2 x + 2 z) 2 æ 2 x ö2 ù -ç ÷ ú dxdz òòê 2 è 2 ø úû -1 0 ê ë

1 z

1 z

z

1 1 3 é ù é( x + z) 3 x 3 ù = 2 ò êò (( x + z) 2 - x 2 ) dx ú dz = 2 ò ê - ú dz 3 3 û0 -1 ë 0 -1 ë û

cos 2 q

dq

1

ésin 2 q ù = 2 ò a cos 2 q dq= 2 a ê = a2 ú 2 ë û0 0

2 2 = ò [(2 z) 3 - z 3 - z ]dz = 3 -1 3

48. (C) The equations of given curves are

æ1 1ö = 4ç - ÷ = 0 è4 4ø

p4

2

y( x 2 + 2) = 3 x....(i)

1

éz4 ù z dz = 6 4 ò ê4ú ë û -1 -1 1

3

and 4 y = x 2 ....(ii)

The curve (i) and (ii) intersect at A (2, 1). If a figure is drawn then from fig. the required area is The required area A =

2

3x ( x 2 + 2 )

x =0

y= x 2 4

ò

ò dxdy

49. (B) The equation of the cylinder is x 2 + y 2 = a 2 The equation of surface CDE is z = h. If a figure is drawn then from fig. the required area is Page 554

- e y + z ]dydz

= ò [ e1 + y + z - e y + z ]10 dz

p3 1 éæ cos 2 q ö æ cos 2 q ö ù - êç -q ÷ - òç÷d q ú 96 4 ëêè 2 ø0 2 ø úû 0è

2

y+ z

1

0

p 4 a cos 2 q

x + y+ z

y+ z 1 0

0 0

p2

=

ò

1 p × = pa 2 h. 2 2

0 0 0

p2

q= 0

dx = a cos q dq,

0

p2 ù p3 1 é = + ê- ò q sin 2 q dq ú 96 4 ë 0 û

A =4

0

0

p2 p2 p2 ù 1 é1 ù 1 éæ sin 2 q ö sin 2 q = ê q 3 ú + êç q 2 dq ú ÷ - ò 2q 4 ë3 û0 4 êëè 2 ø0 2 úû 0

=

a

dx = 4 hò a 2 - x 2 dx

q cos q

p2

p2

p2

Þ

= 4 ha 2 ò cos 2 q dq = 4 ha 2 ×

é1 2 ù òr = 0rdrdq = ò0 êë2 r úû o

2

0

0

p2

p 2 q cos q

òq

a2 - x 2

0

Volume V = 4 h ò a 2 - a 2 sin 2 q × a cos q dq

The required area

ò

a

p2

r = q cos q....(i)

q= 0

a

ò ò hdydx = 4 hò [ y ]

Let x = a sin q,

r=0

46. (C) The equation of the given curve is

A=

=4

a2 - x 2

0

ò rdrdq

q= 0


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********


Complex Variables

Chap 9.5

10. The integration of f ( z) = x 2 + ixy from A(1, 1) to B(2,

17. The value of f ( 3) is

4) along the straight line AB joining the two points is

(A) 6

(B) 4i

-29 (A) + i11 3

(C) -4i

(D) 0

(C)

29 (B) - i11 3

23 + i6 5

(D)

18. The value of f ¢(1 - i) is

23 - i6 5

11.

e2 z òc ( z + 1) 4 dz = ? where c is the circle of z = 3

(A)

4 pi -3 e 9

(B)

4 pi 3 e 9

(C)

4 pi -1 e 3

(D)

8 pi -2 e 3

12.

1 - 2z òc z( z - 1)( z - 2) dz = ? where c is the circle z = 15.

(A) 2 + i 6 p

(B) 4 + i 3p

(C) 1 + ip

(D) i3p

(A) 7 ( p + i2)

(B) 6 (2 + ip)

(C) 2 p (5 + i13)

(D) 0

Statement for 19–21: Expand the given function in Taylor’s series. 19. f ( z) =

(A) 1 + 2( z + z 2 + z 3......)

(B) -1 - 2( z - z 2 + z 3......)

(C) -1 + 2( z - z 2 + z 3......)


20. f ( z) =

13. ò ( z - z 2 ) dz = ? where c is the upper half of the circle

(B)

-3 (D) 2

3 (C) 2 14.

2 3

-1 é 1 1 ù 1 - ( z - 1) + 2 ( z - 1) 2 .......ú ê 2 ë 2 2 û

(B)

1é 1 1 ù 1 - ( z - 1) + 2 ( z - 1) 2 .......ú 2 êë 2 2 û

(C)

1é 1 1 ù 1 + ( z - 1) + 2 ( z - 1) 2 .......ú 2 êë 2 2 û

cos pz dz = ? where c is the circle z = 3 c z -1 (B) - i2 p

(C) i6 p2

(D) - i6 p2

15.


ò

(A) i2 p

21. f ( z) = sin z about z =

sin pz 2 òc ( z - 2)( z - 1) dz = ? where c is the circle z = 3

(A) i6p

(B) i2p

(C) i4p

(D) 0

16. The value of

1 cos pz dz around a rectangle with 2 pi òc z 2 - 1

vertices at 2 ± i , -2 ± i is (A) 6

(B) i2 e

(C) 8

(D) 0

2 ù 1 é pö 1 æ pö æ + 1 z z ç ÷ ç ÷ - .......ú ê 4 ø 2 !è 4ø 2 êë è úû

(B)

2 ù 1 é pö 1 æ pö æ ê1 + ç z - ÷ + ç z - ÷ + .......ú 4 ø 2 !è 4ø 2 êë è úû

(C)

2 ù 1 é æ pö 1 æ pö ê1 - ç z - ÷ - ç z - ÷ - .......ú 4 ø 2 !è 4ø 2 êë è úû

(D) None of the above 22. If z + 1 < 1, then z -2 is equal to (A) 1 +

x + y = 4.

¥

å ( n + 1)( z + 1)

n -1

n =1

(B) 1 +

¥

å ( n + 1)( z + 1)

n +1

n =1

3z 2 + 7 z + 1 f ( z0 ) = ò dz , where c is the circle ( z - z0 ) c 2

p 4

(A)

Statement for Q. 17–18:

2

1 about z = 1 z +1

(A)

c

z =1 -2 (A) 3

z -1 about the points z = 0 z +1

(C) 1 +

¥

å n( z + 1)

n

n =1

(D) 1 +

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¥

å ( n + 1)( z + 1)

n

n =1

Page 565


UNIT 9

Statement for Q. 23–25.

28. The Laurent’s series of f ( z) = 1 in Laurent’s ( z - 1)( z - 2)

Expand the function

series for the condition given in question. 23. 1 < z < 2 1 2 3 (A) + 2 + 3 + ....... z z z (B) K - z -3 - z -2 - z -1 (C)


1 1 1 1 3 - z - z2 z -K 2 4 8 18

where z < 1 1 5 3 21 5 (A) z z + z .......... 4 16 64 (B)

1 1 2 5 4 21 6 + z + z + z .......... 2 4 16 64

(C)

1 3 15 5 z - z3 + z .......... 2 4 8

(D)

1 1 2 3 4 15 6 + z + z + z .......... 2 2 4 8

1 3 7 + 2 + 4 ........... 2 z z z

1 - e Zz at its pole is z4 -4 (B) 3


29. The residue of the function

24. z > 2

(A)

4 3

(C)

-2 3

(A)

6 13 20 + + 3 + ........ z z2 z

(B)

1 8 13 + + + ......... z z2 z3

(C)

1 3 7 + 3 + 4 + ......... 2 z z z

(D)

2 3 4 - 3 + 4 - ........ 2 z z z

25. z < 1 (A) 1 + 3z

+7 2 15 2 z + z ..... 2 4

(B)

1 3 7 15 3 + z + z2 + z ... 2 4 8 16

(C)

1 3 z2 z3 + + + ....... 4 4 8 16

26. If z - 1 < 1 , the Laurent’s series for

30. The residue of z cos

1 is z( z - 1)( z - 2)

( z - 1) 3 ( z - 1) 5 - ......... 2! 5!

(D) - ( z - 1)

-1

1 at z = 0 is z -1 (B) 2

1 2

(C)

1 3

31.

ò z(1 - z)( z - 2) dz = ? where c is

(D) 1 - 2z

-1 3 z = 15 .

(A) - i3p

(B) i3 p

(C) 2

(D) -2

32.

z cos z dz = ? where c is z - 1 = 1 pö c z ç ÷ 2ø è

òæ

(A) 6 p

(B) - 6p

(C) i2p


(C) - ( z - 1) - ( z - 1) - ( z - 1) - .......... 3

2 3

(A)

( z - 1) 3 ( z - 1) 5 - ........... 2! 5!

(B) - ( z - 1) -1 -

(D)

c


(A) - ( z - 1) -

z is, ( z + 1)( z 2 + 4) 2

5

1

- ( z - 1) - ( z - 1) - ( z - 1) - ......... 3

5

33.

2 ò z e z dz = ? where c is z = 1 c

27. The Laurent’s series of

1 for z < 2 is z( e z - 1)

1 1 1 1 2 (A) 2 + + + 6z + z + .......... z 2 z 12 720 1 1 1 1 2 (B) 2 + z + .......... z 2 z 12 720 (C)

1 1 1 1 2 + + z2 + z + .......... z 12 634 720

(D) None of the above Page 566

(B) - i3p

(A) i3p C)

34.

ip 3


2p

dq

ò 2 + cos q = ? 0

(A)

-2 p 2

(C) 2 p 2

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(B)

2p 3

(D) -2 p 3


Complex Variables

¥

35.

x2 ò ( 2 + a 2 )( x 2 + b2 ) dx = ? -¥ x

(A)

p ab a+b

(B)

(C)

p a+b

(D) p ( a + b)

36.

¥

dx

ò1+ x

6

p ( a + b) ab

=?

0

(A) (C)

p 6

(B)

2p 3

(D)

p 2

***************

p 3

Chap 9.5

SOLUTIONS 1. (C) Since, f ( z) = u + iv = Þ

u=

x3 - y3 ; x2 + y2

v=

x 3(1 + i) - y 3(1 - i) ; z ¹0 x2 + y2

x3 + y3 x2 + y2

Cauchy Riemann equations are ¶u ¶v ¶u ¶v and = =¶x ¶y ¶y ¶x By differentiation the value of we get

¶u ¶y ¶v ¶v at(0, 0) , , , ¶x ¶y ¶x ¶y

0 , so we apply first principle method. 0

At the origin, ¶u u(0 + h, 0) - u(0, 0) h3 h2 = lim = lim = 1 h ®0 h ®0 ¶x h h ¶u u(0, 0 + k) - u(0, 0) - k3 k2 = lim = lim = -1 k ®0 ¶v h ®0 k k ¶v v(0 + h, 0) - v(0, 0) h3 h2 = lim = lim =1 h ®0 ¶x h ®0 h h ¶v v(0, 0 + k), v(0, 0) k3 k2 = lim = lim =1 k ®0 ¶y k ®0 k k Thus, we see that

¶u ¶v ¶u ¶v and = =¶x ¶y ¶y ¶x

Hence, Cauchy-Riemann equations are satisfied at z = 0. Again, f ¢(0) = lim z ®0

f ( z) - f (0) z

é( x 3 - y 3) + i( x 3 + y 3) 1 ù = lim ê 2 2 z ®0 (x + y ) ( x + iy) úû ë Now let z ® 0 along y = x, then é( x 3 - y 3) + i( x 3 + y 3) 1 ù 2i 1+ i f ¢ (0) = lim ê = = 2 2 ú z ®0 (x + y ) ( x + iy) û 2(1 + i) 2 ë Again let z ® 0 along y = 0, then é x 3 + i( x 3) 1 ù f ¢ (0) = lim ê =1 + i 2 x ®0 x úû ë (x ) So we see that f ¢(0) is not unique. Hence f ¢(0) does not exist. Df df = lim dz Dz ®0 Dz Du + iDv or f ¢( z) = lim Dz ®0 Dx + iDy

2. (A) Since, f ¢( z) =

....(1)

Now, the derivative f ¢( z) exits of the limit in equation (1) is unique i.e. it does not depends on the path along which Dz ® 0. www.gatehelp.com

Page 567


UNIT 9

Let Dz ® 0 along a path parallel to real axis Þ

Dy = 0 \ Dz ® 0

Þ

Now let v be the conjugate of u then ¶v ¶v ¶u ¶u dv = dx + dy = dx + dy ¶x ¶y ¶y ¶x

Dx ® 0

Now equation (1) Du + iDv Du Dv f ¢( z) = lim = lim + i lim Dx ®0 Dx ®0 Dx Dx ®0 Dx Dx ¶u ¶v f ¢( z) = +i ¶x ¶x


(by Cauchy-Riemann equation) Þ ....(2)

dv = 2 x dx + 2(1 - y) dy

On integrating v = x 2 - y 2 + 2 y + C

Again, let Dz ® 0 along a path parallel to imaginary

5. (C) Given f ( z) = u + i v

....(1)

axis, then Dx ® 0 and Dz ® 0 ® Dy ® 0

Þ

....(2)

Thus from equation (1) Dz + iDv Du Dv ¶u ¶v = lim + i lim = + f¢( z) = lim Dy ®0 Dy ®0 iDy Dy ®0 iDz iD y i ¶y ¶y

add equation (1) and (2)

f ¢( z) =

-i ¶ u ¶ v + ¶y ¶y

....(3)

Now, for existence of f ¢( z) R.H.S. of equation (2) and (3) must be same i.e., ¶u ¶v ¶v ¶u +i = -i ¶x ¶x ¶y ¶y

if ( z) = -v + iu

Þ

(1 + i) f ( z) = ( u - v) + i( u + v)

Þ

F ( z) = U + iV

where, F ( z) = (1 + i) f ( z); U = ( u - v); V = u + v Let F ( z) be an analytic function. Now, U = u - v = e x (cos y - sin y) ¶U ¶U and = e x (cos y - sin y) = e x ( - sin y - cos y) ¶x ¶y Now, dV =

¶u ¶v ¶v -¶u and = = ¶x ¶y ¶x ¶y

-¶U ¶U dx + dy....(3) ¶y ¶x

= e x (sin y + cos y) dx + e x (cos y - sin y) dy

¶u ¶u ¶v ¶v f ¢( z) = -i = +i ¶x ¶y ¶y ¶x

= d[ e x (sin y + cos y)] on integrating V = e x (sin y + cos y) + c1

3. (A) Given f ( z) = x 2 + iy 2 since, f ( z) = u + iv

F ( z) = U + iV = e x (cos y - sin y) + ie x (sin y + cos y) + ic1

Here u = x 2 and v = y 2 ¶u ¶u Now, u = x 2 Þ = 2 x and =0 ¶x ¶y

= e x (cos y + i sin y) + ie x (cos y + i sin y) + ic1

and v = y 2

and f ¢( z) =

(1 + i) f ( z) = (1 + i) e z + ic1 ( i + 1) i i(1 - i) Þ f ( z) = e z + c1 = e z + c1 = ez + c1 2 1+ i (1 + i)(1 - i)

¶v ¶v = 0 and =2y ¶x ¶y

Þ

we know that

F ( z) = (1 + i) e x + iy + ic1 = (1 + i) ez + ic1

f ¢( z) =

¶u ¶u -i ¶x ¶y

....(1)

Þ

f ( z) = e z + (1 + i) c

6. (C) u = sinh x cos y

¶v ¶v + i ....(2) ¶y ¶x

Now, equation (1) gives f ¢( z) = 2 x

....(3)

and equation (2) gives f ¢( z) = 2 y

....(4)

Now, for existence of f ¢( z) at any point is necessary that the value of f ¢( z) most be unique at that point, whatever be the path of reaching at that point

¶u = cosh x cos y = f( x, y) ¶x ¶u and = - sinh x sin y = y( x, y) ¶y by Milne’s Method f ¢( z) = f( z, 0) - iy( z, 0) = cosh z - i × 0 = cosh z On integrating f ( z) = sinh z + constant

From equation (3) and (4) 2 x = 2 y Hence, f ¢( z) exists for all points lie on the line x = y.

Þ

f ( z) = w = sinh z + ic

¶u ¶2u 4. (B) = 2(1 - y) ; =0 ¶x ¶x 2

....(1)

the function x and hence i.e. in w).

¶u ¶2u = -2 x ; =0 ¶y ¶y 2

....(2)

7. (A)

(As u does not contain any constant, the constant c is in

Þ Page 568

¶2u ¶2u + = 0, Thus u is harmonic. ¶x 2 ¶y 2

¶v ¶v = 2 y = h( x, y), = 2 x = g( x, y) ¶x ¶y

by Milne’s Method f ¢( z) = g( z, 0) + ih( z, 0) = 2 z + i 0 = 2 z On integrating f ( z) = z 2 + c www.gatehelp.com


Complex Variables

8. (D)

¶v -( x 2 + y 2 ) - ( x - y)2 y = ¶y ( x2 + y2 )2

f ¢¢( z o) =

y - x - 2 xy = g( x, y) ( x2 + y2 )2 2

=

2

Chap 9.5

n! 2 pi

f ( z) dz

ò (z - z ) o

c

Taking n = 3,

or

n +1

c

f ( z) dz

ò (z - z )

4

=

o

c

f ( z) dz

ò (z - z ) pi f ¢¢( z o) 3

n +1

=

o

2pi n f ( z o) n!

....(1)

e 2 z dz e 2 z dz = ( z + 1) 4 òc [ z - ( -1)]4

¶v ( x 2 + y 2 ) - ( x - y)2 x y 2 - x 2 + 2 xy = = = h( x, y) ¶x ( x2 + y2 )2 ( x2 + y2 )2

Given fc

By Milne’s Method

Taking f ( z) = e 2 z , and z o = -1 in (1), we have

f ¢( z) = g( z, 0) + ih( z, 0) = -

e 2 z dz

1 1 æ 1 ö + iç - 2 ÷ = - (1 + i) 2 z2 z z è ø

ò ( z + 1)

=

c

pi f ¢¢¢( -1)....(2) 3

Þ

f ¢¢¢( -1) = 8 e

Þ

e 2 z dz

ò ( z + 1)

4

Since, f ( z) is analytic within and on z = 3 8 pi - z e 2 z dz = e ò 4 3 |z |= 3 ( z + 1)

By Milne’s Method

12. (D) Since,

f ¢( z) = f( z, 0) - iy( z, 0) 2 cos 2 z - 2 -2 = - i(0) = = - cosec2 z (1 - cos 2 z) 2 1 - cos 2 z

1 - 2z

c

Formula 1 I1 = ò dz = 2 pi c z

On A, z = 1 + i and On B, z = 2 + 4 i Let z = 1 + i corresponds to t = 0

é 1 ê f ( z o) = 2pi ë

x = b, y = d

b = 1, d = 1

ò f ( z) dz = ò ( x c

2

x = a + b, y = c + d a = 1, c = 3

+ ixy)( dx + idy)

ò [( t + 1)

everywhere in c i.e. z = 15 . , hence by Cauchy’s integral

+ i( t + 1)( 3t + 1)][ dt + 3i dt ]

t= 0

theorem 1 I3 = ò dz = 0....(4) -2 z c

1

= ò [( t 2 + 2 t + 1) + i( 3t 2 + 4 t + 1)](1 + 3i) dt 0

1

é t3 ù 29 = (1 + 3i) ê + t 2 + t + i( t 3 + 2 t 2 + t) ú = + 1 1i 3 3 ë û0

using equations (2), (3), (4) in (1), we get 1 - 2z 1 3 òc z( z - 1)( z - 2) dz = 2 (2 pi) + 2 pi - 2 (0) = 3pi

11. (D) We know by the derivative of an analytic function that

ò

the circle z = 15 . , so the function f ( z) is analytic

1

=

....(2)

inside z = 15 . , therefore I 2 = 2 pi....(3) 1 For I 3 = ò dz, the singular point z = 2 lies outside -2 z c

dx = dt ; dy = 3 dt

c

2

and it

f ( z) dz ù ú [Here f ( z) = 1 = f ( z o) and z o = 0] c z - zo û 1 Similarly, for I 2 = ò dz, the singular point z = 1 lies -1 z c

and z = 2 + 4 i corresponding to t = 1

2 = a + 1, 4 = c + 1 Þ

1

ò z dz

z = 15 . , therefore by Cauchy’s integral

lies inside

AB is , y = 3t + 1 Þ

1 3 I1 + I 2 - I 3....(1) 2 2

c

10. x = at + b, y = ct + d

Þ

=

Since, z = 0 is the only singularity for I1 =

f ( z) = - ò cosec2 z dz + ic = cot z + ic

Þ

1 - 2z 1 1 3 = + z( z - 1)( z - 2) 2 z z - 1 2( z - 2)

ò z( z - 1)( z - 2) dz

On integrating

and t = 1

....(3)

If is the circle z = 3

¶u 2 sin 2 x sinh 2 y = = y( x, y) ¶y (cosh 2 y - cos 2 x) 2

Þ

8 pi -2 e 3

=

c

2 cos 2 x cosh 2 y - 2 = f( x, y) (cosh 2 y - cos 2 y) 2

Þ

-2

equation (2) have

¶u 2 cos 2 x (cosh 2 y - cos 2 x) - 2 sin 2 2 x 9. (A) = ¶x (cosh 2 y - cos 2 x) 2

then, t = 0

f ¢¢¢( z) = 8 e 2 z

Þ

Now, f ( z) = e 2 z

On integrating 1 1 f ( z) = (1 + i) ò 2 dz + c = (1 + i) + c z z

=

4

13. (B) Given contour c is the circle z = 1

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Page 569


UNIT 9

Þ

z = e iq

3z 2 + 7 z + 1 òc z - zo dz = 2 pif( zo)

dz = ieiqdq

Þ

Now, for upper half of the circle, 0 £ q £ p 2 ò ( z - z ) dz = c

p

ò (e

iq

Þ

- e 2 iq) ie iqdq

Þ

f ¢( z o) = 2pif¢( z o)

f¢ ( z) = 6 z + 7 and f¢¢( z) = 6

f ¢(1 - i) = 2 pi[ 6(1 - i) + 7 ] = 2 p (5 + 13i)

1 é1 1 ù 2 × ( e 2 pi - 1) - ( e 3px - 1) ú = i êë2 3 û 3

19. (C) f ( z) =

14. (B) Let f ( z) = cos pz then f ( z) is analytic within and on z = 3, now by Cauchy’s integral formula 1 f ( z) dz f ( z) f ( z o) = dz Þ ò = 2pif ( z o) 2pi òc z - z o c z - zo

Þ Þ

take f ( z) = cos pz, z o = 1, we have cos pz òz = 3 z - 1 dz = 2 pif (1) = 2pi cos p = -2pi

z -1 2 =1 z +1 z +1

f (0) = -1, f (1) = 0 2 Þ f ¢( z) = ( z + 1) 2

f ¢¢( z) =

-4 ( z - 1) 3

f ¢¢¢( z) =

12 ( z + 1) 4

Þ Þ

f ¢(0) = 2;

f ¢¢(0) = -4; f ¢¢¢(0) = 12; and so on.

Now, Taylor series is given by

sin pz 2 15. (D) ò dz c ( z - 1)( z - 2)

f ( z) = f ( z 0 ) + ( z - z 0 ) f ¢( z 0 ) +

( z - z0 ) 2 f ¢¢( z 0 ) + 2!

sin pz 2 sin pz 2 dz - ò dz z -2 z -1 c c

( z - z0 ) 3 f ¢¢¢( z 0 ) + ..... 3!

ò

= 2 pif (2) - 2 pif (1) since, f ( z) = sin pz 2

about z = 0

Þ

f ( z) = -1 + z(2) +

f (2) = sin 4 p = 0 and f (1) = sin p = 0

16. (D) Let, I = =

Þ

since, f( z) = 3z 2 + 7 z + 1

p é e 2 iq e 3iq ù = i ò ( e 2 iq - e 3iq)dq = i ê 3i úû 0 ë 2i 0

=

f ( z o) = 2pif( z o)

and f ¢¢( z o) = 2pi f¢¢( z o)

q= 0

p

=i×


1 2 pi

òz c

2

1 cos pz dz -1

z2 z3 ( -4) + (12) + .... 2! 3!

= -1 + 2 z - 2 z 2 + 2 z 3.... f ( z) = -1 + 2( z - z 2 + z 3 ....)

1 1 ö æ 1 ç ÷ cos pz dz ò 2 × 2 pi c è z - 1 z + 1 ø

20. (B) f ( z) =

1 z +1

f ¢( z) =

-1 ( z + 1) 2

Þ

f ¢(1) =

-1 4

3z 2 + 7 z + 1 òc z - 3 dz , since zo = 3 is the only

f ¢¢( z) =

2 ( z + 1) 3

Þ

f ¢¢(1) =

1 4

3z 2 + 7 z + 1 singular point of and it lies outside the z -3

f ¢¢¢( z) =

-6 ( z + 1) 4

Þ

f ¢¢¢(1) = -

Or I =

1 æ cos nz cos nz ö ç ÷dz z +1 ø 4 pi òc è z - 1

17. (D) f ( 3) =

circle x 2 + y 2 = 4 i.e., z = 2, therefore

3z 2 + 7 z + 1 is z -3

Þ

f (1) =

f ( z) = f ( z 0 ) + ( z - z 0 ) f ¢( z 0 ) +

( z - z0 ) 2 f ¢¢( z 0 ) 2! +

Hence by Cauchy’s theorem—

c

3z 2 + 7 z + 1 dz = 0 z -3

( z - z0 ) 3 f ¢¢¢( z 0 ) + K 3!

about z = 1 f ( z) =

18. (C) The point (1 - i) lies within circle z = 2 ( . .. the distance of 1 - i i.e., (1, 1) from the origin is 2 which is less than 2, the radius of the circle). Let f( z) = 3z 2 + 7 z + 1 then by Cauchy’s integral formula Page 570

3 and so on. 8

Taylor series is

analytic everywhere within c. f ( 3) = ò

1 2

=

2 1 æ -1 ö ( z - 1) + ( z - 1)ç ÷+ 2 2! è 4 ø

3 æ 1 ö ( z - 1) æ 3 ö ç - ÷+K ç ÷+ 3! è 8 ø è4ø

1 1 1 1 ( z - 1) + 3 ( z - 1) 2 - 4 ( z - 1) 3 +.... 2 22 2 2

or f ( z) =

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1 1 1é 1 ù 1 - ( z - 1) + 2 ( z - 1) 2 - 3 ( z - 1) 3 + ....ú 2 êë 2 2 2 û


Complex Variables

21. (A) f ( z) = sin z f ¢( z) = cos z

p 1 æ pö f ç ÷ = sin = 4 2 è4ø

Þ

ö 1æ 1æ 1 1 1 z z2 z3 ö çç 1 + + + + .. ÷÷ - ç 1 + + 2 + 3 + K÷ 2è 2 4 9 z z z z è ø ø 1 1 1 1 3 or f ( z) = K-z -4 - z -2 - z -1 - - z - z 2 z -K 2 4 8 18 f ( z) = -

1 æ pö f ¢ç ÷ = 2 è4ø

Þ

f ¢¢( z) = - sin z

Þ

1 æ pö f ¢¢ç ÷ = 4 2 è ø

f ¢¢¢( z) = - cos z

Þ

1 æ pö and so on. f ¢¢¢ç ÷ = 4 2 è ø

24. (C)

( z - z0 ) f ¢¢( z 0 ) 2!

about z =

( z - z0 ) 3 f ¢¢¢( z 0 ) + .... 3!

p 4 2

æ 1 ö ÷ çç 2 ÷ø è

pö æ çz - ÷ 4ø +è 3!

22. (D) Let f ( z) = z -2 =

1 1 = 2 z [1 - (1 + z)]2

=

-1

=

1æ 2 4 8 ö ç 1 + + 2 + 3 + .... ÷ zè z z z ø

f ( z) =

1 3 7 + + +K z2 z3 z4

25. (B) z < 1,

zö 1 1 1æ = - ç1 - ÷ z -2 z -1 2è 2ø

1 1 1 1 = + z( z - 1)( z - 2) 2 z z - 1 2( z - 2)

Since, 1 + z < 1, so by expanding R.H.S. by binomial

For z - 1 < 1 Let z - 1 = u

theorem, we get

Þ

+ ( n + 1)(1 + z) n + K or f ( z) = z -2 = 1 +

å ( n + 1)( z + 1)

z = u + 1 and u < 1 1 1 1 1 = + z( z - 1)( z - 2) 2 z z - 1 2( z - 2)

=

n

n =1

1 1 1 23. (B) Here f ( z) = ....(1) = ( z - 1)( z - 2) z - 2 z - 1 Since, z > 1

Þ

1 < 1 and z < 2 z

1 1 1æ 1ö = = ç1 - ÷ 1ö zè z -1 zø æ zç 1 - ÷ zø è =

Þ

z 2

<1

and

-1 æ zö 1 1 = ç1 - ÷ = z -2 2 è 2ø 2

equation (1) gives—

1 1 1 1 1 - + = (1 + u) -1 - u-1 - (1 - u) -1 2 2( u + 1) u 2( u - 1) 2

1 1 [1 - u + u2 - u3 + ... ] - u-1 - (1 + u + u2 + u3 + ...) 2 2 1 3 -1 = ( -2 u - 2 u - ...) - u = -u - u3 - u5 - K - u-1 2 =

Required Laurent’s series is f ( z) = -( z - 1) -1 - ( z - 1) - ( z - 1) 3 - ( z - 1) 5 - K

-1

27. (B) Let f ( z) =

1æ 1 1 1 ö ç 1 + + 2 + 3 + K÷ zè z z z ø -1

+ (1 - z) -1

1 2

26. (D) Since,

¥

-1

é ù z z2 z3 1 + + + + Kú + (1 + z + z 2 + z 3 + ...) ê 2 4 8 ë û 1 3 7 15 3 f ( z) = + z + z 2 + z +K 2 4 8 16

=-

f ( z) = [1 - (1 + z)]-2

f ( z) = 1 + 2(1 + z) + 3(1 + z) 2 + 4(1 + z) 3 + K

1 <1 z

1æ 1 1 1 ö ç 1 + + 2 + 3 + K÷ z z z 2è ø

1 1æ 2ö = ç1 - ÷ z -2 z è zø

3

2 3 ù 1 é 1æ pö 1 æ pö pö æ ê1 + ç z - ÷ - ç z - ÷ - ç z - ÷ -...ú 4 ø 2 !è 4ø 3! è 4ø 2 êë è úû

-1

Þ

1æ1 3 7 ö ç + 2 + 3 + K÷ zèz z z ø

Þ

æ 1 ö ÷ +K çç 2 ÷ø è

1 1 < <1 z 2

Þ

Laurent’s series is given by 1æ 2 4 98 1 1 1 ö ö 1æ f ( z) = ç 1 + + 2 + 3 + .. ÷ - ç 1 + + 2 + 3 + .. ÷ zè z z z z z z z ø ø è =

pö æ çz - ÷ 1 pö 1 4ø æ f ( z) = + çz - ÷ +è 2! 4ø 2 2 è

f ( z) =

and

2

+

2 <1 z

1 1æ 1ö = ç1 - ÷ z -1 z è zø

Taylor series is given by f ( z) = f ( z 0 ) + ( z - z 0 ) f ¢( z 0 ) +

Chap 9.5

= é ù z z z ê1 + 2 + 4 + 9 + Kú ë û 2

3

1 z( e z - 1)

1 é ù z z3 z4 + + + K - 1ú z ê1 + z + 2 ! 3! 4 ! ë û

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2

Page 571


Complex Variables

1

p

1

ò f ( z) dz = 2 pi ´ 6 = 3 pi

r

-idz dq = ; z £ q £ 2p z

Þ

0

çç e 2 iq è

ò b( z) dz = 0 r

¥

c: z =1

ò (x

dz z + 4z + 1

x p dz = 2 2 + a )( x + b ) a+b 2

2

-¥

2

dz = ò f ( z) dz 1 + z6 c c

36. (C) Let I = ò

2

Let f ( z) =

òæ

e 3iq dq R a 2 öæ b2 ö + 2 ÷÷çç e 2 iq + 2 ÷÷ R øè R ø

Now when R ® ¥,

-idz dq ò0 2 + cos q = òc 1 æz 1 ö; 2 + çz + ÷ 2è zø

2p

c

p

=

1æ 1ö and cos q = ç z + ÷ 2è zø

= - 2iò

ie 2 iqiRe iqdq 2 2 iq + a 2 )( R 2 e2 iq + b2 ) 0 (R e

ò f ( z) dz = ò

Now

c

34. (B) Let z = eiq

Chap 9.5

1 z2 + 4z + 1

c is the contour containing semi circle r of radius R and

f ( z) has poles at z = - 2 + 3, -2 - 3 out of these only

segment from -R to R. For poles of f ( z), 1 + z6 = 0

z = -2 + 3 lies inside the circle c : z = 1

Þ

ò f ( z) dz = 2pi(Residue at z = -2 +

where n = 0, 1, 2, 3, 4, 5, 6

3)

z = ( -1) p 6 = e i ( 2 n + 1 ) p 6

c

=

lim

z ®-2 +

3

( z + 2 - 3) f ( z) = lim z ®-2 +

ò f ( z) dz = 2 pi ´ 2

1

c

2p

dq

ò 2 + cos q = -2 i ´ 0

=

3 pi 3

1 3

( z + 2 + 3)

=

1 2 3

pi 3 =

- 3+i , i, 2

Only poles z =

Now, residue at z = -2 + 3

2p

+ 3+i 2 1 = ( z1 - z 2 )( z1 - z 3)( z1 - z 4 )( z1 - z 5)( z1 - z6 ) Residue at z =

1

=

3

3i(1 + 3 i)

=

where c is be semi circle r with segment on real axis from -R to R. The poles are z = ± ia, z = ± ib. Here only z = ia and z = ib lie within the contour c

ò f ( z) dz = 2pi c

Residue at z =

1 + 3i is 12 i

ò f ( z) dz +

Now

Residue at z = ib z2 -b = lim ( z - ib) = z ®ib ( z - ia)( z + ia)( z + ib)( z - ib) 2 i( a 2 - b2 ) R

ò f ( z) dz = ò f ( z) dz + ò f ( z) dz c

=

r

ò f ( z) dz

ò f ( z) dz =

-R

2p ....(1) 3

iq

iRe dq

ò f ( z) dz = ò 1 + R e

6 6 iq

=

0

where R ® ¥,

1 + 3i 12 i

-R

R

c

=

R

2 pi 2p (1 - 3i + 1 + 3i + 2 i) = 12 i 3

r

p

z a = ( z - ia)( z - ia)( z 2 + b2 ) 2 i( a 2 - b2 )

1 3i(1 - 3i)

=

c

Residue at z = ia, z ®ia

=

ò

r

2

1 6i

f ( z) dz = ò f ( z) dz +

or

(sum of residues at z = ia and z = ib)

= lim ( z - ia)

1 - 3i 12 i

Residue at z = i is

z2 35. (C) I = ò 2 dz = ò f ( z) dz 2 2 2 c ( z + a )( z + b ) c

3+i lie in the contour 2

p

ò 0

ie iqdq R5 1 + e6 iq R6

ò f ( z) dz ® 0 r

(1) ®

¥

ò 0

ax 2p = 1 + x6 3

-R

p 2 pi ( a - b) = a+b 2 i ( a 2 - b2 )

********

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Page 573


CHAPTER

9.6 PROBABILITY AND STATISTICS

1. In a frequency distribution, the mid value of a class is

6. A distribution consists of three components with

15 and the class interval is 4. The lower limit of the

frequencies 45, 40 and 15 having their means 2, 2.5 and

class is

2 respectively. The mean of the combined distribution is

(A) 14 (C) 12

(A) 2.1 (C) 2.3

(B) 13 (D) 10

2. The mid value of a class interval is 42. If the class

7. Consider the table given below

size is 10, then the upper and lower limits of the class are (A) 47 and 37 (C) 37.5 and 47.5

(B) 37 and 47 (D) 47.5 and 37.5

3. The following marks were obtained by the students in a test: 81, 72, 90, 90, 86, 85, 92, 70, 71, 83, 89, 95, 85,79, 62. The range of the marks is

(A) 9

(B) 17

(C) 27

(D) 33

(B) 2.2 (D) 2.4

Marks

Number of Students

0 – 10

12

10 – 20

18

20 – 30

27

30 – 40

20

40 – 50

17

50 – 60

6

The arithmetic mean of the marks given above, is

4. The width of each of nine classes in a frequency

(A) 18

(B) 28

distribution is 2.5 and the lower class boundary of the

(C) 27

(D) 6

lowest class is 10.6. The upper class boundary of the highest class is (A) 35.6

(B) 33.1

(C) 30.6

(D) 28.1

5.

In

a

monthly

test,

the

marks

8. The following is the data of wages per day: 5, 4, 7, 5, 8, 8, 8, 5, 7, 9, 5, 7, 9, 10, 8 The mode of the data is (A) 5 (C) 8

obtained

in

(B) 7 (D) 10

9. The mode of the given distribution is

mathematics by 16 students of a class are as follows: 0, 0, 2, 2, 3, 3, 3, 4, 5, 5, 5, 5, 6, 6, 7, 8

Weight (in kg)

40

43

46

49

52

55

Number of Children

5

8

16

9

7

3

The arithmetic mean of the marks obtained is (A) 3

(B) 4

(C) 5

(D) 6

Page 574

(A) 55

(B) 46

(C) 40

(D) None

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Probability and Statistics

10. If the geometric mean of x, 16, 50, be 20, then the

16. The mean deviation of the following distribution is

value of x is (A) 4 (C) 20

(B) 10 (D) 40

11. If the arithmetic mean of two numbers is 10 and

Chap 9.6

x

10

11

12

13

14

f

3

12

18

12

3

(A) 12 (C) 1.25

(B) 0.75 (D) 26

their geometric mean is 8, the numbers are (A) 12, 18 (C) 15, 5

17. The standard deviation for the data 7, 9, 11, 13,

(B) 16, 4 (D) 20, 5

15 is

12. The median of 0, 2, 2, 2, -3, 5, -1, 5, 5, -3, 6, 6, 5, 6 is (B) -1.5 (D) 3.5

(A) 0 (C) 2

(A) 2.4

(B) 2.5

(C) 2.7

(D) 2.8

18. The standard deviation of 6, 8, 10, 12, 14 is

13. Consider the following table

(A) 1

(B) 0

(C) 2.83

(D) 2.73

Diameter of heart (in mm)

Number of persons

19. The probability that an event A occurs in one trial of

120

5

an experiment is 0.4. Three independent trials of

121

9

122

14

123

8

124

5

125

9

experiment are performed. The probability that A occurs at least once is (A) 0.936

(B) 0.784

(C) 0.964

(D) None

20.

Eight

coins

are

tossed

simultaneously.

The

probability of getting at least 6 heads is The median of the above frequency distribution is (A) 122 mm

(B) 123 mm

(C) 122.5 mm

(D) 122.75 mm

(A)

7 64

(B)

37 256

(C)

57 64

(D)

249 256

21. A can solve 90% of the problems given in a book and 14. The mode of the following frequency distribution, is Class interval

Frequency

3–6

2

6–9

5

9–12

21

12–15

23

15–18

10

18–21

12

21–24

3

(A) 11.5

(B) 11.8

(C) 12

(D) 12.4

B can solve 70%. What is the probability that at least one of them will solve a problem, selected at random from the book? (A) 0.16

(B) 0.63

(C) 0.97

(D) 0.20

22. A speaks truth in 75% and B in 80% of the cases. In what percentage of cases are they likely to contradict each other narrating the same incident ? (A) 5%

(B) 45%

(C) 35%

(D) 15%

23. The odds against a husband who is 45 years old, living till he is 70 are 7:5 and the odds against his wife

15. The mean-deviation of the data 3, 5, 6, 7, 8, 10,

who is 36, living till she is 61 are 5:3. The probability

11, 14 is

that at least one of them will be alive 25 years hence, is

(A) 4

(B) 3.25

(A)

61 96

(B)

(C) 2.75

(D) 2.4

(C)

13 64

(D) None

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5 32

Page 575


UNIT 9


24. The probability that a man who is x years old will

30. If 3 is the mean and (3/2) is the standard deviation

die in a year is

of a binomial distribution, then the distribution is

A1 , A2 , K, An each x

p. Then amongst n

persons

years old now, the probability

that A1 will die in one year is (A)

1 n2

(B) 1 - (1 - p) n

(C)

1 [1 - (1 - p) n ] n2

(D)

1 [1 - (1 - p) n ] n

drawn from each bag, the probability that both are white is

5 24

60

æ1 4ö (D) ç + ÷ è5 5 ø

the distribution is 12

æ1 3ö (B) ç + ÷ è4 4ø

24

æ1 1ö (D) ç + ÷ è2 2 ø

32

1 (B) 4

æ1 5ö (C) ç + ÷ è6 6ø

(D) None

32. A die is thrown 100 times. Getting an even number is considered a success. The variance of the number of

contains 4 white and 2 red balls. If one ball is drawn

(A) 50 (C) 10

from each bag, the probability that one is white and one is red, is

8 27

16

æ1 1ö (A) ç + ÷ è7 8 ø

successes is

(C)

5

binomial distribution are 24 and 18 respectively. Then,

26. A bag contains 5 white and 4 red balls. Another bag

13 (A) 27

12

31. The sum and product of the mean and variance of a

bag contains 3 white and 5 black balls. If one ball is

(C)

æ 1 3ö (B) ç + ÷ è2 2 ø

æ4 1ö (C) ç + ÷ è5 5ø

25. A bag contains 4 white and 2 black balls. Another

1 (A) 24

12

æ3 1ö (A) ç + ÷ è4 4ø

(B) 25 (D) None

33. A die is thrown thrice. Getting 1 or 6 is taken as a 5 (B) 27

success. The mean of the number of successes is 3 2 (A) (B) 2 3

(D) None

(C) 1

(D) None

27. An anti-aircraft gun can take a maximum of 4 shots

34. If the sum of mean and variance of a binomial

at an enemy plane moving away from it. The

distribution is 4.8 for five trials, the distribution is

probabilities of hitting the plane at the first, second,

æ1 4ö (A) ç + ÷ è5 5 ø

third and fourth shot are 0.4, 0.3, 0.2 and 0.1 respectively. The probability that the gun hits the plane is (A) 0.76

(B) 0.4096

(C) 0.6976

(D) None of these

æ2 3ö (C) ç + ÷ è5 5 ø

5

æ1 2ö (B) ç + ÷ è 3 3ø

5

5

(D) None of these

35. A variable has Poission distribution with mean m. The probability that the variable takes any of the

28. If the probabilities that A and B will die within a

values 0 or 2 is

year are p and q respectively, then the probability that

æ m2 ö (A) e - m çç 1 + m + ÷ 2 ! ÷ø è

(B) e m (1 + m) -3 2

(C) e 3 2 (1 + m 2 ) -1 2

æ m2 ö (D) e - m çç 1 + ÷ 2 ! ÷ø è

only one of them will be alive at the end of the year is (A) pq

(B) p(1 - q)

(C) q(1 - p)

(D) p + 1 - 2 pq

29. In a binomial distribution, the mean is 4 and

36.

variance is 3. Then, its mode is

P (2) = 9 P ( 4) + 90 P ( 6), then the mean of X is

(A) 5

(B) 6

(A) ± 1

(B) ± 2

(C) 4

(D) None

(C) ± 3

(D) None

Page 576

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If

X

is

a

Poission

variate

such

that



Chap 9.6

37. When the correlation coefficient r = ± 1, then the

43. If Sxi = 30,

two regression lines

å y = 318 and n = 6, then the regression coefficient bxy

(A) (B) (C) (D)

is

å yi = 42,

å xi yi = 199, å xi2 = 184,

2 i

are perpendicular to each other coincide are parallel to each other do not exist

(A) -0.36 (C) 0.26

(B) -0.46 (D) None

44. Let r be the correlation coefficient between x and y

38. If r = 0, then (A) there is a perfect correlation between x and y (B) x and y are not correlated.

and byx , bxy be the regression coefficients of y on x and x on y respectively then (A) r = bxy + byx

(B) r = bxy ´ byx

(D) there is a negative correlation between x and y

(C) r = bxy ´ byx

(D) r =

39. If Sxi = 15,

45. Which one of the following is a true statement.

(C) there is a positive correlation between x and y

Syi = 36,

Sxi yi = 110 and n = 5, then

cov ( x, y) is equal to (A) 0.6 (C) 0.4

(B) 0.5 (D) 0.225

40. If cov ( x, y) = -16.5, var ( x) = 2.89 and var ( y) = 100, then the coefficient of correlation r is equal to

(A)

1 2

( bxy + byx ) = r

(B)

(C)

1 2

( bxy + byx ) > r

(D) None of these

1 2

( bxy + byx ) < r

46. If byx = 1.6 and bxy = 0.4 and q is the angle between two regression lines, then tan q is equal to (A) 0.18 (C) 0.16

(B) -0.64 (D) -0.97

(A) 0.36 (C) 0.97

1 ( bxy + byx ) 2

(B) 0.24 (D) 0.3

41. The ranks obtained by 10 students in Mathematics

47. The equations of the two lines of regression are :

and Physics in a class test are as follows

4 x + 3y + 7 = 0

and 3 x + 4 y = 8 = 0. The correlation

coefficient between x and y is

Rank in Maths

Rank in Chem.

1

3

2

10

3

5

48. If cov( X , Y ) = 10, var ( X ) = 6.25 and var( Y ) = 31.36,

4

1

then r( X , Y ) is

5

2

6

9

7

4

49. If å x = å y = 15,

8

8

n = 5, then bxy = ?

9

7

(A) - 13

(B) -

10

6

(C) - 14

(D) - 12

The coefficient of correlation between their ranks is (A) 0.15 (C) 0.625 42. If Sxi = 24,

(B) 0.224 (D) None å yi = 44,

(A) 1.25

(B) 0.25

(C) -0.75

(D) 0.92

(A)

5 7

(B)

(C)

3 4

(D) 0.256

50. If å x = 125,

4 5

å x 2 = å y 2 = 49, å xy = 44 and 2 3

å y = 100, å x 2 = 1650, å y 2 = 1500,

å xy = 50 and n = 25, then the line of regression of x on y is

Sxi yi = 306, å xi2 = 164,

å yi2 = 574 and n = 4, then the regression coefficient byx

(A) 22 x + 9 y = 146

(B) 22 x - 9 y = 74

(C) 22 x - 9 y = 146

(D) 22 x + 9 y = 74

is equal to (A) 2.1 (C) 1.225

(B) 1.6 (D) 1.75

*********

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Page 577


UNIT 9

SOLUTION


A.M. = A +

S( fd) æ 300 ö = ç 25 + ÷ = 28. Sf 100 ø è

1. (B) Let the lower limit be x. Then, upper limit is x + ( x + 4) x + 4. = 15 Þ x = 13. 2

8. (C) Since 8 occurs most often, mode =8.

2. (A) Let the lower limit be x. Then, upper limit x + 10.

10. (B) ( x ´ 16 ´ 50)1 3 = 20

x + ( x + 10) = 42 2

Þ

Þ

x = 37.

9. (B) Clearly, 46 occurs most often. So, mode =46.

Lower limit = 37 and upper limit =47. 3. (D) Range = Difference between the largest value = (95 - 62) = 33. 4. (B) Upper class boundary = 10.6 + (2.5 ´ 9) = 331 ..

x ´ 16 ´ 50 = (20) 3

æ 20 ´ 20 ´ 20 ö x = çç ÷÷ = 10. è 16 ´ 50 ø

11. (B) Let the numbers be a and b Then, a+b = 10 2 ab = 8

Þ Þ

( a + b) = 20

and

ab = 64

a - b = ( a + b) 2 - 4 ab = 44 - 256 = 144 = 12.

5. (B)

Solving a + b = 20 and a - b = 12 we get a = 16 and

Marks

Frequency f

f ´1

0

2

0

2

2

4

3

3

9

4

1

4

5

4

20

6

2

12

7

1

7

8

1

8

Diameter of heart (in mm)

Number of persons

Cumulative frequency

å f = 16

å( f ´ x) = 64

120

5

5

121

9

14

122

14

28

123

8

36

124

5

41

125

9

50

A.M. =

b = 4. 12. (D) Observations in ascending order are -3, -3, -1, 0, 2, 2, 2, 5, 5, 5, 5 6, 6, 6 Number of observations is 14, which is even. Median =

1 1 [7 the term +8 the term] = (2 + 5) = 35 . . 2 2

13. (A) The given Table may be presented as

å( f ´ x) 64 = = 4. åf 16

6. (B) Mean =

45 ´ 2 + 40 ´ 2.5 + 15 ´ 2 220 = = 2.2. 100 100

7. (B)

n n = 25 and + 1 = 26. 2 2

Class

Mid value x

Frequenc yf

Deviation d = x- A

f ´d

Here n = 50. So,

0–10

5

12

-20

-240

Medium =

10–20

15

18

-10

-180

20–30

25 = A

27

0

0

30–40

35

20

10

200

14. (B) Maximum frequency is 23. So, modal class is

40–50

45

17

20

320

12–15.

50–60

55

6

30

180

L1 = 12, L2 = 15, f = 23, f1 = 21 and f2 = 10.

Sf = 100

Page 578

Þ

S ( f ´ d) = 390

1 122 + 122 (25th term +26 th term) = = 122. 2 2

[ . .. Both lie in that column whose c.f. is 28]

Thus Mode = L1 +

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f - f1 ( L2 - L1 ) 2 f - f1 - f2



= 12 +

(23 - 21) (15 - 12) = 12.4. ( 46 - 21 - 10)

Chap 9.6

6

æ 3 + 5 + 6 + 7 + 8 + 10 + 11 + 14 ö 15. (C) Mean = ç ÷ = 8. 8 è ø

7

2

æ1ö æ1ö æ1ö 1 = 8 C6 × ç ÷ × ç ÷ + 8 C7 × ç ÷ × + 8 C8 2 2 è ø è ø è2 ø 2 8´7 1 1 1 37 = ´ + 8´ + = 2 ´ 1 256 256 256 256

8

æ1ö ×ç ÷ è2 ø

Sd = 3 - 8 + 5 - 8 + 8 - 8 + 10 - 8 + 11 - 8 + 14 - 8

21. (C) Let E = the event that A solves the problem. and

= 22

F = the event that B solves the problem.

Sd 22 Thus Mean deviation = = = 2.75. n 8

Clearly E and F are independent events. 90 70 P ( E) = = 0.9, P ( F ) = = 0.7, 100 100

16. (B)

P ( E º F ) = P ( E) × P ( F ) = 0.9 ´ 0.7 = 0.63

x

f

f ´x

d = x-M

f ´d

10

3

30

2

6

11

12

132

1

12

12

18

216

0

0

13

12

156

1

12

14

3

42

2

6

Sf = 48

Sfx = 576

Thus M =

= P ( E) + P ( F ) - P ( E º F ) = (0.9 +0.7 - 0.63) =0.97.

Sfd = 36

576 = 12. 48

B speaks the truth)] = P (E and F ) + P (E and F)

2

2

= P ( E) × P ( F ) + P ( E ) × P ( F )

2

2

Sd2 = 7 - 11 + 9 - 11 + 11 - 11 + 13 - 11 + 15 - 11 = 40 s=

Sd 2 40 = = 8 = 2 2 = 2 ´ 1.41 = 2.8. n 5

2

2

2

=

3 1 1 4 3 1 7 æ 7 ö =ç ´ 100 ÷% = 35%. ´ + ´ = + = 4 5 4 5 20 5 20 è 20 ø

23. (A) Let E = event that the husband will be alive 25 years hence and F =event that the wife will be alive 25

6 + 8 + 10 + 12 + 14 50 18. (C) M = = = 10. 5 5

years hence. 2

Sd2 = 6 - 10 2 + 8 - 10 + 10 - 10 + 12 - 10 + 14 - 10 = 40

6=

F =event that B speaks the truth. 75 3 80 4 Then, P ( E) = = , P( F) = = 100 4 100 5 3ö 1 4ö 1 æ æ P ( E) = ç 1 - ÷ = , P( F ) = ç 1 - ÷ = 4ø 4 5ø 5 è è = P[(A speaks truth and B tells a lie) or (A tells a lie and

7 + 9 + 11 + 13 + 15 55 = = 11. 5 5

2

22. (C) Let E =event that A speaks the truth.

P (A and B contradict each other).

Sfd 36 So, Mean deviation = = = 0 .75 n 48 17. (D) m =

Required probability = P ( E È F )

Then,

P ( E) =

5 12

and P ( F ) =

3 8

5 ö 7 3ö 5 æ æ and P ( F ) = ç 1 - ÷ = . Thus P ( E) = ç 1 ÷= 12 12 8 è ø è ø 8

Sd2 40 = n 5

Clearly, E and F are independent events.

= 8 = 2 2 = 2 ´ 1.414 = 2.83 (app.)

So, E and F are independent events. 19. (B) Here p = 0.4, q = 0.6 and n = 3.

P(at least one of them will be alive 25 years hence)

Required probability = P(A occurring at least once)

= 1 - P(none will be alive 24 years hence) 7 5 ö 61 æ = 1 - P ( E º F ) = 1 - P ( E) × P ( F ) = ç 1 ´ ÷= 12 8 ø 96 è

= 3C1 × (0.4) ´ (0.6) 2 + 3C2 × (0.4) 2 ´ (0.6) + 3C3 × (0.4) 3 4 36 16 6 64 ö 784 æ =ç3´ ´ + 3´ ´ + = 0.784. ÷= 10 100 100 10 1000 ø 1000 è

20. (B) p =

1 , 2

q=

1 , 2

24. (D) P(none dies) = (1 - p) (1 - p)....n times = (1 - p) n

n = 8. Required probability

= P (6 heads or 7 heads or 8 heads)

P(at least one dies) = 1 - (1 - p) n . 1 P(A1 dies) = {1 - (1 - p) n }. n

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Page 579



Sxi 15 å yi 36 = = 3, y = = = 7.2 n 5 n 5 æ Sx y ö æ 110 ö cov( x, y) = ç i i - x y ÷ = ç - 3 ´ 7.2 ÷ = 0.4 n 5 è ø è ø

47. (C) Given lines are : y = -2 -

39. (C) x =

40. (D) r =

cov ( x, y) var ( x) × var ( y)

=

-16.5 2.89 ´ 100

Chap 9.6

æ 7 3 ö and x = ç - - y ÷ è 4 4 ø -3 -3 and . byx = bxy = 4 4 3 æ -3 -3 ö 9 or r = - = -0.75. So, r 2 = ç ´ ÷= 4 ø 16 4 è 4

= -0.97.

41. (B) Di = -2, - 8, - 2, 3, 3, - 3, 3, 0, 2, 4.

[. .. byx and bxy are both negative ® r is negative]

SDi2 = ( 4 + 64 + 4 + 9 + 9 + 9 + 9 + 0 + 4 + 16) = 128. é 6( SDi2 ) ù æ 6 ´ 128 ö 37 R = ê1 2 ú = çç 1 - 10 ´ 99 ÷÷ = 165 = 0.224. ( ) n n 1 ø ë û è

48. (A) r( X , Y ) =

( Sxi )( Syi ) n 42. (A) byx = é 2 ( Sxi ) 2 ù êSxi - n ú ë û

49. (C) byx =

Sxi yi -

50. (B) bxy = =

( Sxi )( Syi ) ù æ 199 - 30 ´ 42 ö é ç ÷ êëSxi yi úû è 6 ø n 43. (B) byx = = é 2 ( Syi ) 2 ù 42 ´ 42 ù é êSyi - n ú êë318 úû 6 ë û (199 - 210) -11 = = = -0.46. ( 318 - 294) 24

r 2 = bxy ´ byx 45. (C)

cov( X , Y ) var( X ) var( Y )

=

10 6.25 ´ 31.36

=

5 7

nSxy - ( Sx)( Sy) nSx 2 - ( Sx) 2

æ 5 ´ 44 - 15 ´ 15 ö 1 = çç ÷÷ = 5 ´ 49 15 ´ 15 4 è ø

24 ´ 44 ö æ ç 306 ÷ 4 ø = ( 306 - 264) = 42 = 2.1 =è 2 (164 - 144) 20 é (24) ù ê164 - 4 ú ë û

44. (C) byx = r ×

3 x 4

nSxy - ( Sx)( Sy) nSy 2 - ( Sy) 2

25 ´ 50 - 125 ´ 100 9 = 25 ´ 1500 - 100 ´ 100 22

Also,

x=

125 = 5, 25

y=

100 = 4. 25

Required line is x = x + bxy ( y - y) 9 Þ x =5 + ( y - 4) Þ 22 x - 9 y = 74. 22

sy sx and bxy = r × sx sy

Þ

r = bxy ´ byx .

1 1 é sy sx ù ( bxy + byx ) > r is true if êr × +r× >r sy úû 2 2 ë sx

i.e. if s2y + sx2 > 2 sx s y i.e. if ( s y - sx ) 2 > 0, which is true. 46. (A) r = 1.6 ´ 0.4 = .64 = 0.8 byx = r × m1 =

sy sx

Þ

sy sx

=

byx r

1 sy 1 5 , × = ´2= r sx 0.8 2

æ m - m2 tan q = çç 1 è 1 + m1 m2

=

1.6 =2 0.8 m2 = r ×

sy sx

= 0.8 ´ 2 = 1.6.

ö æ 2.5 - 1.6 ö 0.9 ÷÷ = çç = 0.18. ÷÷ = ø è 1 + 2.5 ´ 1.6 ø 5

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Page 581


(B)

x2 x5 x8 x11 + + + 2 20 160 4400

(C)

x2 x5 x8 x11 + + + 2 20 160 2400 2

(D)

5

8

Chap 9.7

Statement for Q. 18–19: dy For = 1 + y 2 given that dx x:

0

0.2

0.4

0.6

y:

0

0.2027

0.4228

0.6841

11

x x x x + + + 2 40 480 2400

12. For dy dx = xy given that y = 1 at x = 0. Using Euler method taking the step size 0.1, the y at x = 0.4 is

Using Milne’s method determine the value of y for x given in question.

(A) 1.0611

(B) 2.4680

18. y (0.8) = ?

(C) 1.6321

(D) 2.4189

(A) 1.0293

(B) 0.4228

(C) 0.6065

(D) 1.4396

Statement for Q. 13–15. For dy dx = x 2 + y 2 given that y = 1 at x = 0.

19. y (10 . ) =?

Determine the value of y at given x in question using

(A) 1.9428

(B) 1.3428

modified method of Euler. Take the step size 0.02.

(C) 1.5555

(D) 2.168

13. y at x = 0.02 is


(A) 1.0468

(B) 1.0204

(C) 1.0346

(D) 1.0348

Apply Runge Kutta fourth order method to obtain y (0.2), y (0.4) and y (0.6) from dy dx = 1 + y 2 , with y = 0 at x = 0. Take step size h = 0.2.

14. y at x = 0.04 is (A) 1.0316

(B) 1.0301

(C) 1.403

(D) 1.0416

20. y (0.2) = ?

15. y at x = 0.06 is (A) 1.0348

(B) 1.0539

(C) 1.0638

(D) 1.0796

(A) 0.2027

(B) 0.4396

(C) 0.3846

(D) 0.9341

21. y (0.4) = ? (A) 0.1649

(B) 0.8397

(C) 0.4227

(D) 0.1934

16. For dy dx = x + y given that y = 1 at x = 0. Using

22. y (0.6) = ?

modified Euler’s method taking step size 0.2, the value

(A) 0.9348

(B) 0.2935

(C) 0.6841

(D) 0.563

of y at x = 1 is (A) 3.401638

(B) 3.405417

(C) 9.164396

(D) 9.168238

23. For dy dx = x + y 2 , given that y = 1 at x = 0. Using Runge Kutta fourth order method the value of y

17. For the differential equation dy dx = x - y 2 given

x = 0.2 is (h = 0.2)

that

(A) 1.2735

(B) 2.1635

(C) 1.9356

(D) 2.9468

x:

0

0.2

0.4

0.6

y:

0

0.02

0.0795

0.1762

24. For dy dx = x + y given that y = 1 at x = 0. Using Runge Kutta fourth order method the value of y

Using Milne predictor–correction method, the y at next value of x is (A) 0.2498

(B) 0.3046

(C) 0.4648

(D) 0.5114

at

x = 0.2 is

at

(h = 0.2)

(A) 1.1384

(B) 1.9438

(C) 1.2428

(D) 1.6389 *********

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Page 583


UNIT 9

SOLUTIONS

x2 = x0 = 35 . -

1. (B) Let f ( x) = x 3 - 4 x - 9


x1 - x0 f ( x0 ) f ( x1 ) - f ( x0 )

0.5 ( - 0.5441) = 37888 . 0.3979 + 0.5441

Since f (2) is negative and f ( 3) is positive, a root lies

Since f ( 37888 . ) = - 0.0009 and f ( 4) = 0.3979, therefore

between 2 and 3.

the root lies between 3.7888 and 4.

First approximation to the root is 1 x1 = (2 + 3) = 2.5. 2

Taking x0 = 37888 . , x1 = 4, we obtain 0.2112 x3 = 37888 . ( - .009) = 37893 . 0.3988

Then f ( x1 ) = 2.5 3 - 4(2.5) - 9 = - 3.375

Hence the required root correct to three places of

i.e. negative\The root lies between x1 and 3. Thus the

decimal is 3.789.

second approximation 1 x2 = ( x1 + 3) = 2.75. 2

4. (D) Let f ( x) = xe x - 2, Then f (0) = - 2, and

to

the

root

is

Then f ( x2 ) = (2.75) 3 - 4(2.75) - 9 = 0.7969 i.e. positive. The root lies between x1 and x2 . Thus the third 1 approximation to the root is x3 = ( x1 + x2 ) = 2.625. 2 Then

f ( x3) = (2.625) 3 - 4(2.625) - 9 = - 1.4121

i.e.

negative. The root lies between x2 and x3 . Thus the fourth 1 approximation to the root is x4 = ( x2 + x3) = 2.6875. 2 Hence the root is 2.6875 approximately. 2. (B) Let f ( x) = x 3 - 2 x - 5

f (1) = e - 2 = 0.7183 So a root of (i ) lies between 0 and 1. It is nearer to 1. Let us take x0 = 1. Also f ¢( x) = xe x + e x and f ¢(1) = e + e = 5.4366 By Newton’s rule, the first approximation x1 is f ( x0 ) 0.7183 x1 = x0 =1 = 0.8679 f ¢( x0 ) 5.4366 f ( x1 ) = 0.0672, f ¢( x1 ) = 4.4491. Thus the second approximation x2 is f ( x1 ) 0.0672 x2 = x1 = 0.8679 = 0.8528 f ( x1 ) 4.4491 Hence the required root is 0.853 correct to 3 decimal

So that f (2) = - 1 and f ( 3) = 16

places.

i.e. a root lies between 2 and 3. Taking

x0 = 2, x1 = 3, f ( x0 ) = - 1, f ( x1 ) = 16,

in

the

method of false position, we get x1 - x0 1 x2 = x0 f ( x0 ) = 2 + = 2.0588 f ( x1 ) - f ( x0 ) 17

5. (B) Let y = x + log10 x - 3.375 To obtain a rough estimate of its root, we draw the graph of (i ) with the help of the following table :

Now, f ( x2 ) = f (2.0588) = - 0.3908 i.e., that root lies between 2.0588 and 3. Taking x0 = 2.0588, x1 = 3, f ( x0 )

x

1

2

3

4

y

-2.375

-1.074

0.102

1.227

= - 0.3908, f ( x1 ) = 16 in (i), we get 0.9412 x3 = 2.0588 ( - 0.3908) = 2.0813 16.3908

Taking 1 unit along either axis = 0.1, The curve crosses

Repeating this process, the successive approximations

approximation to the root.

are

Now let us apply Newton–Raphson method to

x4 = 2.0862, x5 = 2.0915, x6 = 2.0934, x7 = 2.0941,

f ( x) = x + log10 x - 3.375 1 f ¢( x) = 1 + log10 e x

the x–axis at x0 = 2.9, which we take as the initial

x8 = 2.0943 etc. Hence the root is 2.094 correct to 3 decimal places.

Taking x0 = 35 . , x1 = 4, in the method of false position,

f (2.9) = 2.9 + log10 2.9 - 3.375 = - 0.0126 1 f ¢(2.9) = 1 + log10 e = 11497 . 2.9

we get

The first approximation x1 to the root is given by

3. (C) Let f ( x)2 x - log10 x - 7

Page 584

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x1 = x0 -

f ( x0 ) 0.0126 = 2.9 + = 2.9109 ¢ f ( x0 ) 11497 .

at x = 0,

Thus the second approximation x2 is given by

d2 y =1 + 2 = 3 dx 2

at x = 0, y = 1,

f ( x1 ) 0.0001 = 2.9109 + = 2.91099 f ¢( x1 ) 11492 .

2

Hence the desired root, correct to four significant figures, is 2.911

d 3y d2 y æ dy ö y = 2 2 ç ÷ dx 3 dx 2 è dx ø

6. (B) Let x = 28 so that x 2 - 28 = 0 Taking f ( x) = x 2 - 28, Newton’s iterative method gives 28 ö f ( xn ) x 2 - 28 1 æ ÷ = xn - n = çç xn + ¢ 2 xn 2è f ( xn ) xn ÷ø

é dy d 2 y d 3y ù y 3 + ê dx dx 2 dx 3 úû ë

d4 y = -2 dx 4

at x = 0, y = 1

Now since f (5) = - 3, f ( 6) = 8, a root lies between 5 and 6.

d4 y = 34 dx 4

The Taylor series expression gives y( x + h) = y( x) + h

Taking x0 = 5.5, x1 =

1æ 28 ö 1 æ 28 ö ÷÷ = ç 5.5 + çç x0 + ÷ = 5.29545 2è 5.5 ø x0 ø 2 è

x2 =

1æ 28 ö 1 æ 28 ö ÷÷ = çç 5.29545 + çç x1 + ÷ = 5.2915 2è 5.29545 ÷ø x1 ø 2 è

d 3y =-8 dx 3

x = 0, y = 1,

at

xn + 1 = xn -

dy = -1 dx

y = 1,

d2 y dy =1 -2y 2 dx dx

f ( x1 ) = - 0.0001, f ¢( x1 ) = 11492 .

x2 = x1 -

Chap 9.7

dy h2 d 2 y h3 d 3 y h4 d 4 y +K + + + 3 ! dx 3 4 ! dx 4 dx 2 ! dx 2

y(0.1) = 1 + 0.1( -1) +

(0.1) 2 (0.1) 3 (0.1) 4 3+ ( -8) + 34 + ...... 2! 3! 4!

= 1 - 0.1 + 0.015 - 0.001333 + 0.0001417 = 0.9138 9. (C) Here f ( x, y) = x 2 + y 2 , x0 = 0

1æ 28 ö 1 æ 28 ö ÷ = ç 5.2915 + x3 = çç x2 + ÷ = 5.2915 2è 5.2915 ÷ø x2 ÷ø 2 çè

y0 = 0

We have, by Picard’s method

Since x2 = x3 upto 4 decimal places, so we take

y = y0 +

28 = 5.2915.

x

ò f ( x, y) dx

....(1)

x0

The first approximation to y is given by 7. (B) Let h = 0.1, dy = 1 + xy Þ dx 3

given x0 = 0, d2 y dy =x +y dx 2 dx

2

d y d y dy , =x +2 dx 3 dx 2 dx given that Þ

x1 = x0 + h = 0.1

x = 0,

4

3

2

d y d y d y =x +3 dx 4 dx 3 dx 2 y =1

The Taylor series expression gives : dy h2 d 2 y h3 d 3 y + + + 3 ! dx 3 dx 2 ! dx 2

Þ

y (0.1) = 1 + 0.1 ´ 1 +

Þ

y(0.1) = 1 + 0.1 +

(0.1) 2 (0.1) 3 ×1 + 2 +K 2! 3!

0.01 0.001 + +K 2 3

= 1 + 0.1 + 0.005 + 0.000033 ......... 8. (B) Let h = 0.1,

given

dy = x - y2 x1 = x0 + h = 0.1, dx

x

ò f ( x, y ) dx 0

x0

dy d2 y d 3y d4 y , = 1 ; 2 = 1, = 2 = 3 and so on dx dx dx 3 dx 4

y( x + h) = y( x) + h

y (1 ) = y0 +

x0 = 0,

Where y0 = 0 +

x

x

ò f ( x, 0) dx = ò x dx. 2

0

...(2)

0

The second approximation to y is given by x

ò f ( x, y

y ( 2 ) = y0 +

(1 )

) dx = 0 +

0

x0

æ x = 0 + ò çç x 2 + 9 0è x

Now,

6

æ

x3 ö ÷dx 3 ÷ø

ö x x ÷÷dx = + 3 63 ø 3

y (0.4) =

10. (C) Here

x

ò f ççè x,

7

(0.4) 3 (0.4) 7 + = 0.02135 3 63

f ( x, y) = y - x ; x0 = 0, y0 = 2

We have by Picard’s method

= 11053 . y0 = 1

y = y0 +

ò

x

x0

f ( x, y) dx

The first approximation to y is given by y (1 ) = y0 +

x

ò f ( x, y0 ) dx = 2 +

x0

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x

ò f ( x, 2) dx 0

Page 585


UNIT 9

x

= 2 + ò (2 - x) dx

=2 + 2x -

0

x2 2

....(1)

x

ò f ( x, y

(1 )

=2 +

Here

æ

ò f ççè x, 2 + 2 x -

x0 2

= 2 + ò (2 + 2 x 0

x 2

....(2)

æ

ò f ççè x, 2 + 2 x +

=2 +

x0

y4 = y3 + hf ( x3 , y3) = 10302 . + 0.1 f (0.3 , 10302 . )

x x ö ÷dx 2 6 ÷ø 2

= 10302 . + 0.03090

3

. y4 = y( 0 .4 ) = 10611 Hence y( 0 .4 ) = 10611 .

ö æ x x = 2 + ò çç 2 + 2 x + - ÷÷dx 2 6 ø 0è x

2

3

13. (B) The Euler’s modified method gives

x2 x3 x4 + 2 6 24

=2 + 2x +

11. (B) Here

f ( x, y) = x + y 2 , x0 = 0

y1* = y0 + hf ( x0 , y0 ), h y1 = y0 + [ f ( x0 , y0 ) + f ( x1 , y1*)] 2

y0 = 0

Now, here h = 0.02, y0 = 1, x0 = 0

We have, by Picard’s method y = y0 +

. y1* = 1 + 0.02 f (0, 1), y1* = 1 + 0.02 = 102 h Next y1 = y0 + [ f ( x0 , y0 ) + f ( x , y1*)] 2 0.02 =1 + [ f (0, 1) + f (0.02, 102 . )] 2

x

ò f ( x, y0 ) dx

x0

The first approximation to y is given by y (1 ) = y0 +

x

ò f ( x, y ) dx 0

=0 +

x0 x

x

ò f ( x, 0) dx

= 1 + 0.01 [1 + 10204 . ] = 10202 .

0

2

x = 0 + ò xdx = 2 0

So,

x

ò f ( x, y

(1 )

) dx = 0 +

x0

x

æ

ò f ççè x, 0

x2 2

ö ÷÷dx ø

= 10202 . + 0.02 [ f (0.02, 10202 . )] = 10202 . + 0.0204

= 10406 . h Next y2 = y1 + [ f ( x, y) + f ( x2 , y2* )] 2 0.02 y2 = 10202 . + [ f (0.02, 10202 . ) + f (0.04, 10406 . )] 2

æ x x x ö ÷÷dx = + = ò çç x + 2 50 4 ø 0è x

4

2

5

The third approximation is given by y ( 3) = y0 +

x

ò f ( x, y

(2 )

= 10202 . + 0.01 [10206 . + 10422 . ] = 10408 .

) dx

x0

. y2 = y( 0 .04 ) = 10408

æ x2 x5 ö ÷dx = 0 + ò f çç x, + 2 20 ÷ø 0 è x

15. (C) y3* = y2 + hf ( x2 , y2 )

x æ 2 x7 ö x2 x5 x8 x11 x4 x10 ÷÷dx = + + + = ò çç x + + + 2 20 160 4400 4 400 40 ø 0è

12. (A) x: 0 Page 586

0.1

0.2

. y1 = y (0.02) = 10202

14. (D) y2* = y1 + h f ( x1 , y1 )

The second approximation to y is given by y ( 2 ) = y0 +

y3 = y2 + hf ( x2 , y2 ) = 101 . + 0.1 f (0.2 , 101 . ) n = 3 in (1) gives

) dx

x0 x

n = 2 in (1) gives . + 0.0202 = 10302 . y3 = y( 0 .3) = 101

x

(2 )

h = 0.1

Thus y2 = y( 0 .2 ) = 101 .

3

ò f ( x, y

y0 = 1,

= 1 + 0.1 f (0.1 , 1) = 1 + 0.1 (0.1) = 1 + 0.01

The third approximation to y is given by y ( 3) = y0 +

x0 = 0,

n = 0 in (1) gives y2 = y1 + h f ( x1 , y1 )

x x 2 6

=2 + 2x +

....(1)

y1 = 1 + 0.1 f (0, 1) = 1 + 0 = 1

ö ÷÷dx ø

2

x2 - x) dx 2

2

yn + 1 = yn + h( xn , yn ) y1 = y0 + hf ( x0 , y0 )

) dx

x0 x

Euler’s method gives n = 0 in (1) gives

The second approximation to y is given by y ( 2 ) = y0 +


0.3

0.4

= 10416 . + 0.02 f (0.04, 10416 . ) = 10416 . + 0.0217 = 10633 . h Next y3 = y2 + [ f ( x2 , y2 ) + f ( x3 , y3*)] 2

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UNIT 9


1 1 ö æ k2 = hf ç x0 + h, y0 + k1 ÷ = (0.2) f (0.1, 0.1) = 0.202 2 2 ø è

h k ö æ k2 = hf ç x0 + , y0 + 1 ÷ 2 2ø è

1 1 ö æ k3 = hf ç x0 + h, y0 + k2 ÷ = (0.2) f (0.1, 0.101) = 0.2020 2 2 è ø

= (0.2) f (0.1, 11 . ) = 0.2(1.31) = 0.262 h k ö æ k3 = hf ç x0 + , y0 + 2 ÷ 2 2 ø è

k4 = hf ( x0 + h, y0 + k3) = 0.2 f (0.2, 0.2020) = 0.20816 1 k = [ k1 + 2 k2 + 2 k3 + k4 ] 6 1 = [0.2 + 2 (.202) + 2 (.20204) + 0.20816 ], 6 k = 0.2027 such that

y1 = y(0.2) = y0 + k = 0 + 0.2027 = 0.2027

21. (C) We now to find y2 = y(0.4), k1 = hf ( x1 , y1 ) = (0.2) f (0.2, 0.2027) = 0.2 (10410 . )

= .2082

= 0.2 f (0.1, 1131 . ) = 0.2758 k4 = hf ( x0 + h, y0 + k3) = (0.2) f (0.2, 12758 . ) = 0.3655 1 k = [ k1 + 2 k2 + 2 k3 + 2 k4 ] 6 1 = [0.2 + 2 (0.262) + 2 (0.2758) + 0.3655 ] = 0.2735 6 Here

y1 = y( 0 .2 ) = y0 + k

= 1 + 0.2735

Þ

12735 .

1 1 ö æ k2 = hf ç x1 + h , y1 + k1 ÷ 2 2 ø è

24. (C) Here f ( x, y) = x + y h = 0.2

= (0.2) f (0.3, 0.3068)

k1 = hf ( x0 , y0 ) = 0.2 f (0, 1) = 0.2 h k ö æ . ) = 0.24 k2 = hf ç x0 + , y0 + 1 ÷ = (0.2) f (0.1, 11 2 2ø è

To find y1 = y( 0 .2 ) ,

= 0.2188

1 1 ö æ k3 = hf ç x1 + h , y1 + k2 ÷ 2 2 è ø

h k ö æ . ) = 0.244 k3 = hf ç x0 + , y0 + 2 ÷ = (0.2) f (0.1, 112 2 2 ø è

= 0.2 f (0.3, 0.3121) = .2194 k4 = hf ( x1 + h, y1 + k3) = 0.2 f (0.4, .4221) = 0.2356 1 k = [ k1 + 2 k2 + 2 k3 + k4 ] 6 1 = [0.2082 + 2(.2188) + 2(.2194) + 0.356 ] = 0.2200 6

k4 = hf ( x0 + h, y0 + k3) = (0.2) f (0.2, 1244 . ) = 0.2888 1 k = [ k1 + 2 k2 + 2 k3 + k4 ] 6 1 = [0.2 + 2(0.24) + 2(0.244) + 0.2888 ] = 0.2428 6

y2 = y( 0 .4 ) = y1 + k = 0.2200 + .2027 = 0.4227

y1 = y( 0 .2 ) = y0 + k = 1 + 0.2428

= 12428 .

22. (C) We now to find y3 = y( 0 .6 ) , k1 = hf ( x2 , y2 ) = (0.2) f (0.4, 0.4228)

= 0.2357

1 1 ö æ k2 = hf ç x2 + h, y2 + k1 ÷ 2 2 è ø

***********

= (0.2) f (0.5, 0.5406) = 0.2584 1 1 ö æ k3 = hf ç x2 + h, y2 + k2 ÷ 2 2 ø è = 0.2 f (0.5, .5520) = 0.2609 1 k4 = [ k1 + 2 k2 + 2 k3 + k4 ] 6 1 = [0.2357 + 2(.2584) + 2(0.2609) + 0.2935 ] 6 1 = [0.2357 + 0.5168 + 0.5218 + 0.2935 ] = 0.2613 6 y3 = y( 0 .6 ) = y2 + k = .4228 + 0.2613 = 0.6841 23. (A) Here given f ( x, y) = x + y

x0 = 0

y0 = 1,

h = 0.2

2

To find y1 = y( 0 .2 ) , k1 = hf ( x0 , y0 ) Page 588

= (0.2) f (0, 1) = (0.2) ´ 1 = 0.2 www.gatehelp.com


CHAPTER

10.5 EC-07

1. If E Denotes expectation, the variance of a random

6. For the function e- x , the linear approximation around

variable X is given by

x = 2 is

(A) E[ X 2 ] - E 2 [ X ]

(B) E[ X 2 ] + E 2 [ X ]

2

2

(C) E[ X ]

(D) E [ X ]

2. The following plot shows a function y which varies 2

linearly with X. The value of the integral I = ò y dx is y

1

(B) 1 - x

(C) [ 3 + 2 2 - 1(1 + 2 x ]e -2

(D) e -2

7. An independent voltage source in series with an impedance Z s = Rs + jX s delivers a maximum average power to a load impedance Z L when

3 2 1 -1

(A) ( 3 - x) e -2

1

2

x

3

(A) Z L = RS + jX S

(B) Z L = RS

(C) Z L = jX S

(D) Z L = RS - jX S

8. The RC circuit shown in the figure is R

(A) 1.0

(B) 2.5

(C) 4.0

(D) 5.0

C +

+

Vi

R

C

Vo

3. For x << 1, coth( x) can be approximated as

(C)

1 x

4. lim

(D) sin( q/2)

q®0

q

1 x2

is

(C) 2

(D) not defined

(C) a band-pass filter

(D) a band-reject filter

impurities are introduced with a concentration of N A per cm 3(where N A >> ni ) the electron concentration per

5. Which of the following functions is strictly bounded ?

(C) x 2

(B) a high-pass filter

semiconductor are ni per cm 3 at 300 K. Now, if acceptor

(B) 1

1 x2

(A) a low-pas filter

9. The electron and hole concentrations in an intrinsic

(A) 0.5

(A)

-

-

(B) x 2

(A) x

(B) e x (D) e - x

2

cm 3 at 300 K will be (A) ni

(B) ni + N A

(C) N A - ni

(D)

www.gatehelp.com

ni2 NA Page 639


UNIT 10

10. In a

p+ n junction diode under reverse biased the

Previous year Papers

15. If closed-loop transfer function of a control system is

magnitude of electric field is maximum at

given as T( s) =

(A) the edge of the depletion region on the p -side

(A) an unstable system

(B) the edge of the depletion region on the n -side

(B) an uncontrollable system

(C) the p+ n junction

(C) a minimum phase system

(D) the center of the depletion region on the n-side

(D) a non-minimum phase system

11. The correct full wave rectifier circuit is

16. If the Laplace transform of a signal y( t) is

Output

(B)

Input

Output

(A)

Input

Y ( s) =

1 s ( s -1 )

s-5 ( s + 2 )( s + 3)

then It is

, then its final value is

(A) -1

(B) 0

(C) 1

(D) unbounded

17. If R( t) is the auto correlation function of a real, wide-sense stationary random process, then which of (A) R( t) = R( -t)

Output

(D)

Input

Output

C

Input

the following is NOT true (B) R( t) £ R(0) (C) R( t) = - R( -t) (D) The mean square value of the process is R(0) 12. In a trans-conductance amplifier, it is desirable to

18. If S( f )is the power spectral density of a real,

have

wide-sense stationary random process, then which of the following is ALWAYS true?

(A) a large input resistance and a large output resistance (B) a large input resistance and a small output resistance

(A) S(0) £ S( f )

(B) S( f ) ³ 0

(C) S( - f ) = -S( f )

(D)

¥

ò S( f )df

=0

-¥

(C) a small input resistance and a large output resistance

19. A plane wave of wavelength l is traveling in a direction making an angle 30 o with positive x -axis. The

(D) a small input resistance and a small output resistance

®

E field of the plane wave can be represented as (E0 is constant)

13. X = 01110 and Y = 11001 are two 5-bit binary numbers represented in two's complement format. The sum of X and Y represented in two's complement format using 6 bits is (A) 100111

(B) 0010000

(C) 000111

(D) 101001

®

æ 3p p ÷ö j çç wt x z l l ÷ø è

®

æ 3p p ö÷ j çç wt + x z l l ÷ø è

(A) E = yE0 e (C) E = yE0 e

®

æ p 3p ÷ö j çç wt - x z l l ø÷ è

®

æ p 3p ö÷ j çç wt - x + z l l ÷ø è

(B) E = yE0 e (D) E = yE0 e

20. If C is close curve enclosing a surface S, then the ®

®

the electric flux density D are related by 14. The Boolean function Y = AB + CD is to be realized using only 2-input NAND gates. The minimum number of gates required is (A) 2

(B) 3

(C) 4

(D) 5

Page 640

®

magnetic field intensity H , the current density j and

® æ® ö ¶D÷ ® ç (A) ò ò H . d s = ò j + .d l ç ¶t ÷ s c è ø ®

®

® æ® ö ¶D ÷ ® ç (B) ò H . d l = òò j + d .d s ç ¶t ÷ s s è ø ®

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®


UNIT 10


29. For the circuit shown in the figure, the Thevenin

33. Group I lists four types of p - n junction diodes.

voltage and resistance looking into X - Y are

match each device in Group I with one of the option in

1W X

Group II to indicate the bias condition of the device in its normal mode of operation. Group-I (P) Zener Diode (Q) Solar cell (R) LASER diode (S) Avalanche Photodiode

2W

2A

1W

2i

Y

(A) (C)

4 V, 2W 3

2 (B) 4 V, W 3

4 2 V, W 3 3

(D) 4 V, 2W

Group-II (1) Forward bias (2) Reverse bias

(A) P - 1 Q - 2 R - 1 S - 2 (B) P - 2 Q - 1 R - 1 S - 2 (C) P - 2 Q - 2 R - 1 S - 2

30. In the circuit shown, Vc is 0 volts at t = 0 sec. for t > 0, the capacitor current ic ( t), where t is in seconds, is given by

(D) P - 2 Q - 1 R - 2 S - 2 34. The DC current gain (b) of a BJT is 50. Assuming that the emitter injection efficiency is 0.995, the base

iC

20 kW

4 mF

20 kW

10 V

transport factor is + VC -

(A) 0.980

(B) 0.985

(C) 0.990

(D) 0.995

35. group I lists four different semiconductor devices. (A) 0.50 exp( -25 t) mA

match each device in Group I with its characteristic

(B) 0.25 exp( -25 t) mA

property in Group II.

(C) 0.50 exp( -25 t) mA (D) 0.25 exp( -25 t) mA 31. In the AC network shown in the figure, the phasor

(B) P - 1 Q - 4 R - 3 S - 2

A

~

Group-II (1) Population inversion (2)Pinch-off voltage (3) Early effect (4) Fat-band voltage

(A) P - 3 Q - 1 R - 4 S - 2

voltage V AB (in volts) is

A

Group-I (P)BJT (Q)MOS capacitor (R) LASER diode (S) JFET

5W

(C) P - 3 Q - 4 R - 1 S - 2

5W

(D) P - 3 Q - 2 R - 1 S - 4 -j3

j3

36. For the Op-Amp circuit shown in the figure, Vo is

B

(A) 0 (C) 12.5 Ð30 o

2 kW

(B) 5 Ð30 o (D) 17 Ð30 o

1 kW vo

1V

32. A p+ n junction has a built-in potential of 0.8 V. The

1 kW 1 kW

depletion layer width at reverse bias of 1.2V is 2 mm. For a reverse bias of 7.2 V, the depletion layer width will be (A) 4 mm

(B) 4.9 mm

(C) 8 mm

(D) 12 mm

(A) -2 V

(B) -1 V

(C) -0.5 V

(D) 0.5 V

37. For the BJT circuit shown, assume that the b of the transistor is very large and VBE = 0.7 V . The mode of operation of the BJT is

Page 642

www.gatehelp.com


EC-07

10 kW

2V

10 V

Chap 10.5

(A) 7.00 to 7.29 V

(B) 7.14 to 7.29 V

(C) 7.14 to 7.43 V

(D) 7.29 to 7.43 V

41.

1 kW

The

Boolean

expression

Y = AB CD + ABCD+ ABCD+ ABCD (A) cut-off

(B) saturation

minimized to

(C) normal active

(D) reverse active

(A) Y = A B CD + A B C + A C D

38. In the Op-Amp circuit shown, assume that the diode current follows the equation I = I s exp(V/VT ). For Vi = 2 V , V0 = V01 ,

and

Vi = 4 V , V0 = V02 .

for

The

relationship between V01 and V02 is

can

be

(B) Y = A B CD + B C D + A B C D (C) Y = AB C D + B C D + AB CD (D) Y = AB C D + B C D + A B C D 42. The circuit diagram of a standard TTL NOT gate is

D

shown in the figure. Vi = 2.5 V , the modes of operation of vi

2 kW

the transistors will be vo

VCC =5 V

(A) V02 = 2 V01

(B) V02 = e V01

(C) V02 = V01 ln 2

(D) V01 - V02 = VT ln 2

100 W

1.4 kW

4 kW 2

Q4 +

39. In the CMOS inverter circuit shown, if the

Q1

Q2

D +

transconductance parameters of the NMOS and PMOS transistors are kn = kp = m n Cox

Wn Ln

= m p Cox

Wp Lp

Q3

= 40 mA/V 2

1 kW

and their threshold voltages are VTHn = VTHp = 1V , the

-

-

current I is 5V

(A) Q1 : revere active;Q2 : normal active; Q3: saturation; Q4 :cut-off

PMOS 2.5 V

I

(B) Q1 : revere active;Q2 : saturation; Q3: saturation; Q4 :cut-off

NMOS

(A) 0 A

(B) 25 mA

(C) 45 mA

(D) 90 mA

(C) Q1 : normal active;Q2 : cut-off; Q3: cut-off; Q4 :saturation (D) Q1 : saturation;Q2 : saturation; Q3: saturation; Q4 :normal active

40. For the Zener diode shown in the figure, the Zener

43. In the following circuit, X is given by

voltage at knee is 7V, the knee current is negligible and 0

the Zener dynamic resistance is 10W. if the input

1 1 0

voltage ( Vi ) range is from 10 to 16V, the output voltage

(V0 ) ranges

from

S1 S0

200 W +

vi

I0 4-to-1 MUX I1 Y I2 I3

A

0 1 1 0

I0 4-to-1 I1 MUX I2 Y I3 S1

B

vo

(A) X = A B C + A B C + A BC + ABC

_

(B) X = ABC + A B C + ABC + ABC www.gatehelp.com

X

S0

C

Page 643


UNIT 10


(C) X = AB + BC + AC

47. (A) The 3-dB bandwidth of the low-pas signal e- t u( t),

(D) X = A B + B C + AC 44. The following binary values were applied to the X

where u( t) is the unit step function, is given by 1 1 (A) Hz (B) 2 - 1 Hz 2p 2p

and Y inputs of NAND latch shown in the figure in the

(C) ¥

sequence indicated below

(D) 1 Hz

48. A Hilbert transformer is a

X = 0, Y = 1; X = 0, Y = 0; X = 1, Y = 1.

(A) non-linear system (C) time-varying system

X P

(B) non-causal system (D) low-pass system

49. The frequency response of a linear, time-invariant system is given by H ( f ) = 1 + j510 pf . The step response of

Q Y

the system is

The corresponding stable P , Q outputs will be (A) P = 1, Q = 0; P = 1, Q = 0; P = 1, Q = 0 or P = 0, Q = 1 (B) P = 1, Q = 0; P = 0, Q = 1; or P = 0 Q = 1; P = 0, Q = 1

(C)

(C) P = 1, Q = 0; P = 1, Q = 1; P = 1, Q = 0 or P = 0, Q = 1 (D)P = 1, Q = 0; P = 1, Q = 1; P = 1, Q = 1

t - ö æ (B) 5çç 1 - e 5 ÷÷u( t) è ø

(A) 5( 1 - e -5t ) u( t) 1 (1 - e-5t )u(t) 2

50.

A

5-point

(C)

sequence

t - ö 1æ ç 1 - e 5 ÷u( t) ÷ 5 çè ø

x[ n]

is

given

as

45. For the circuit shown, the counter state (Q1Q0 )

x[ -3] = 1, x[ -2 ] = 1, x[ -1] = 0, x[0 ] = 5, x[1] = 1. Let X ( e jw )

follows the sequence

denote the discrete-time Fourier transform of x[ n]. The p value of

ò X (e

jw

)dw is

-p

(A) 5

(B) 10p (D) 5 + j10 p

(C) 16p Q0

D0

D1

Q1

51. The z-transform x[ z ] of a sequence x[ n] is given by X ( z) = 1 -02.z5-1 . It is given that the region of convergence of x[ n] includes the unit circle. The value of x[0] is

(A) 00, 01, 10, 11, 00

(B) 00, 01, 10, 00, 01

(C) 00, 01, 11, 00, 01

(D) 00, 10, 11, 00, 10

(A) -0.5

(B) 0

(C) 0.25

(D) 0.5

52. A Control system with PD controller is shown in 46.

8085

the figure If the velocity error constant K V = 1000 and

microprocessor system as an I/O mapped I/O as show in

the damping ration z = 0.5, then the value of K P and K D

the figure. The address lines A0 and A1 of the 8085 are

are

An

8255

chip

is

interfaced

to

an

used by the 8255 chip to decode internally its thee ports

R(s)

and the Control register. The address lines A3 to A7 as

C(s)

+

well as the IO/M signal are used for address decoding. The range of addresses for which the 8255 chip would get selected is 8255

(A) K P = 100, K D = 0.09

(B) K P = 100, K D = 0.9

(C) K P = 10, K D = 0.09

(D) K P = 10, K D = 0.9

53. The transfer function of a plant is T( s) = (A) F8H - FBH

(B) F8H - FCH

(C) F8H - FFH

(D) F0H - F7H

Page 644

www.gatehelp.com

5 (s + 5)( s2 + s + 1)


EC-07

The second-order approximation of T( s) using dominant pole concept is 1 (A) (s + 5)(s + 1) (C)

5

(B)

5 s2 + s + 1

(s + 5)(s + 1) 1 s2 + s + 1

(D)

1 s 2 -1

. If the plant is operated in a unity feedback

stabilize this control system is 10( s - 1) 10( s + 4) (A) (B) s+2 s+2 10( s + 2) s + 10

(B)

1 s + 11s + 11

(C)

10 s + 10 s + 11s + 11

(D)

1 s + s + 11

2

(A) (B) (C) (D)

decreasing decreasing decreasing increasing

2

the the the the

step size granular noise sampling rate step size

digital communications. The expression for p( t) with unity roll-off facto is given by

s + 10

p( t) =

55. A unity feedback control system has an open-loop

sin 4 pWt 4 pWt(1 - 16W 2 t 2 )

The value of p( t) at t =

transfer function G( s) =

2

59. The raised cosine pulse p( t) is used for zero ISI in

2(s + 2)

(D)

10 s + 11s + 11 2

can be reduced by

configuration, then the lead compensator that an

(C)

(A)

58. In delta modulation, the slope overload distortion

54. The open-loop transfer function of a plant is given as G( s) =

Chap 10.5

K . s( s 2 + 7 s + 12)

The gain K for which s = 1 + j1 will lie on the root

1 is 4W

(A) -0.5

(B) 0

(C) 0.5

(D) ¥

60. In the following scheme, if the spectrum M ( f ) of

locus of this system is (A) 4

(B) 5.5

(C) 6.5

(D) 10

m( t) is as shown, then the spectrum Y ( f ) of y( t) will be M

m(t)

56. The asymptotic Bode plot of a transfer function is as S

shown in the figure. The transfer function G( s) Hilbert Transform

corresponding to this Bode plot is 0

G(jw)dB 60 dB

-20 dB/dec

(A)

40 dB

(B)

-40 dB/dec

20 dB

w 1

10

20 100

-60 dB/dec

(A)

1 ( s + 1)( s + 20)

(B)

1 s( s + 1)( s + 20)

(C)

100 s( s + 1)( s + 20)

(D)

100 s( s + 1)(1 + 0.05 s)

0

0

(D)

C

0

61.

During

0

transmission

over

a

certain

binary

57. The state space representation of a separately

communication channel, bit errors occurs independently

excited DC servo motor dynamics is given as

with probability p. The probability of AT MOST one bit

é ê ë

dw dt dia dt

1ù é wù é 0 ù ù é-1 ú = ê-1 -10 ú êi ú + ê10 ú u ûë a û ë û û ë

in error in a block of n bits is given by (A) pn (C) np(1 - p) n -1 + (1 - p) n www.gatehelp.com

(B) 1 - pn (D) 1 - (1 - p) n Page 645


UNIT 10


62. In a GSM system, 8 channels can co-exist in 200

67. A load of 50W is connected in shunt in a 2-wire

KHz bandwidth using TDMA. A GSM based cellular

transmission line of Z 0 = 50W as shown in the figure.

operator is allocated 5 MHz bandwidth. Assuming a

The 2-port scattering parameter matrix (s-matrix) of

1 5

frequency reuse factor of , i.e. a five-cell repeat pattern,

the shunt element is

the maximum number of simultaneous channels that

é- 1 (A) ê 12 ë 2

can exist in one cell is (A) 200 (C) 25

(B) 40 (D) 5

ù - úû

é0 1 ù (B) ê ú ë1 0 û

1 2 1 2

2 é- 1 3ù (C) ê 23 1ú ë 3 - 3û

63. In a Direct Sequence CDMA system the chip rate is 1.2288 ´ 106 chips per second. If the processing gain is desired to be at Least 100, the data rate (A) must be less than or equal to 12.288 ´ 10 3 bits/sec (B) must be greater than 12.288 ´ 10 3 bits per sec (C) must be exactly equal to 12.288 ´ 10 3 bits per sec

é 1 (D) ê 43 ë- 4

- 43 ù 1ú 4û

68. The parallel branches of a 2-wire transmission line are terminated in 100 W and 200 W resistors as shown in the figure. The characteristic impedance of the line is Z 0 = 50W and each section has a length of 4l . The voltage reflection coefficient G at the input is

(D) can take any value less than 122.88 ´10 3 bits/sec

l/4 l/4

200

64. An air-filled rectangular waveguide has inner

W

dimensions of 3 cm ´ 2 cm. The wave impedance of the TE20 mode of propagation in the waveguide at a frequency

of

30

GHz

is

(free

space

impedance

h0 = 377 W) (A) 308W

(B) 355W

(C) 400W

(D) 461W

200 W

l/4

®

65. The H field (in A/m) of a plane wave propagating in (A) - j

free space is given by ®

H=x

pö 5 3 5 æ cos( wt - bz) + y sinç wt - bz + ÷ h0 h0 2 è ø

The time average power flow density in Watts is h0 100 (B) (A) 100 h0 (C) 50h20

(D)

7 5

(C) j

5 7

69. A

l 2

(B)

-5 7

(D)

5 7

dipole is kept horizontally at a height of

l0 2

above

a perfectly conducting infinite ground plane. The ®

radiation pattern in the lane of the dipole (E plane)

50 h0

looks approximately as (A)

®

y

(B)

y

66. The E field in a rectangular waveguide of inner dimensions a ´ b is given by ®

E=

z

2

wm æ p ö æ 2 px ö ç ÷ H 0 sinç ÷ sin( wt - bz) y h2 è 2 ø è a ø

y

Where H 0 is a constant, and a and b are the dimensions

along

the

x -axis

and

the

(C) TM 20

(D) TE10

Page 646

(D)

y -axis

is (B) TM11

y

C

respectively. The mode of propagation in the waveguide (A) TE20

z

www.gatehelp.com

z

z


EC-07

70. A right circularly polarized (RCP) plane wave is

Common Data for Questions 74, 75 :

incident at an angle of 60 o to the normal, on an air-dielectric interface. If the reflected wave is linearly polarized, the relative dielectric constant xr 2 is

Two 4-ray signal constellations are shown. It is given that f1 and f2 constitute an orthonormal basis for the two constellations. Assume that the four symbols in both the constellations are equiprobable. Let N0 / 2

Linearly Polarized

RCP

Chap 10.5

denote the power spectral density of white Gaussian

air

noise.

Dielectric

(B) 3 (D) 3

(A) 2 (C) 2

0

Common Data for Questions 71, 72, 73: The

figure

shows

the

high-frequency

capacitance-voltage(C-V) characteristics of a Metal/ SiO2/silicon

(MOS)

capacitor

having

an

area

of

1 ´ 10 -4 cm 2 . Assume that the perimitivities ( e 0 e r ) of silicon and Si O2 are 1 ´ 10 -12 F/cm and 35 . ´ 10 -13 F/cm respectively. C

Constellation 1

74. The ratio of the average energy of constellation 1 to the average energy of constellation 2 is (A) 4 a 2

(B) 4

(C) 2

(D) 8

75.

7 pF

Constellation 2

If

these

constellations

are

used

for

digital

communications over an AWGN channel, then which of the following statements is true ? (A) Probability of symbol error for Constellation 1 is lower

1 pF V

0

71. The gate oxide thickness in the MOS capacitor is (A) 50 nm

(B) 143 nm

(C) 350 nm

(D) 1 mm

72. The maximum depletion layer width in silicon is

(B) Probability of symbol error for Constellation 1 is higher (C) Probability of symbol error is equal for both the constellations (D) The value of N 0 will determine which of the two constellations has a lower probability of symbol error,

(A) 0.143 mm

(B) 0.857 mm

Linked Answer Questions: Q. 76 to Q. 85 Carry

(C) 1 mm

(D) 1.143 mm

Two marks Each.

73. Consider the following statements about the C-V characteristics plot:

Statement for Linked Answer Questions 76 & 77: Consider the Op-Amp circuit shown in the figure.

S1: The MOS capacitor has an n-type substrate.

R1

S2: If positive charges are introduced in the oxide, the

R1

C-V plot will shift to the left. vi

Then which of the following is true? (A) Both S1 and S2 are true

vo R C

(B) S1 is true and Se is false (C) S1 is false and S2 is true (D) Both S1 and S2 are false www.gatehelp.com

Page 647


UNIT 10

76. The transfer function V0 ( s)/Vi ( s) is

80. The eigenvalue and eigenvector pairs ( li Vi ) for the

(A)

1 - sRC 1 + sRC

(B)

1 + sRC 1 - sRC

(C)

1 1 - sRC

(D)

1 1 + sRC

system are æ æ é 1ù ö é 1ù ö (A) çç -1, ê ú ÷÷ and çç -2, ê ú ÷÷ ë-1û ø è è ë-2 û ø

77. If Vi = V1 sin( wt) and V0 = V2 sin( wt - f), then the minimum and maximum values of f (in radians) are respectively (A) - 2p and

p 2

(C) -p and 0

(B) 0 and


p 2

æ æ é 1ù ö é 1ù ö (B) çç -1, ê ú ÷÷ and çç -2, ê ú ÷÷ ë-1û ø è è ë-2 û ø æ é 1ù ö æ é 1ù ö (C) çç -1ê ú ÷÷ and çç -2, ê ú ÷÷ 1 è ë-2 û ø è ë ûø æ æ é 1ù ö é 1ù ö (D) çç -2, ê ú ÷÷ and çç 1, ê ú ÷÷ 1 ë ûø è è ë-2 û ø

(D) - 2p and 0

Statement for Linked Answer Questions 78 & 79. An 8085 assembly language program is given below.

81. The system matrix A is é 0 1ù (A) ê ú ë-1 1û

1ù é 1 (B) ê ú ë-1 -2 û

1ù é 2 (C) ê ú ë-1 -1û

1ù é 0 (D) ê ú ë-2 -3û

Line 1:

MVI A, B5H

2:

MVI B, OEH

3:

XRI 69H

4:

ADD B

5:

ANI 9BH

6:

CPI 9FH

7:

STA 3010H

density function f ( x) as shown in the figure. Decision

8:

HLT

boundaries of the quantizer are chosen so as t maximize

Statement fo Linked Answer Questions 82 & 83: An input to a 6-level quantizer has the probability

78. The contents of the accumulator just after execution of the ADD instruction in line 4 will be (A) C3H

(B) EAH

(C) DCH

(D) 69H

the entropy of the quantizer output. It is given that 3 consecutive decision boundaries are ' -1' , '0 ' and '1'. f(x) a

b

79. After execution of line 7 of the program, the status -5

of the CY and Z flags will be (A) CY = 0, Z = 0

(B) CY = 0, Z = 1

(C) CY = A, Z = 0

(D) CY = 1, Z = 1

-1

0

1

5

82. The values of a and b are (A) a =

1 1 and b = 6 12

(B) a =

1 3 and b = 5 40

(C) a =

1 1 and b = 4 16

(D) a =

1 1 and b = 3 24

Statement for Linked Answer Questions 80 & 81. Consider a linear system whose state space representation is x( t) = Ax( t). If the initial state vector of é 1ù the system is x(0) = ê ú, then the system response is ë-2 û é e -2 x ù . If the itial state vector of the system x( t) = ê -2 t ú ë-2 e û

83. Assuming that the reconstruction levels of the quantizer are the mid-points of the decision boundaries, the ratio of signal power to quantization noise power is

é 1ù changes to x(0) = ê ú, then the system response ë-2 û

(A)

152 9

(B)

é e- t ù becomes x( t) = ê - t ú ë -e û

(C)

76 3

(D) 28

Page 648

x

www.gatehelp.com

64 3


EC-07

Statement for Linked Answer Questions 84 & 85.

ANSWER

In the digital-to Analog converter circuit shown in the figure below, VR = 10 V and R = 10 kW. R

2R

R

2R

2R

i

R

R

Chap 10.5

1. A

2. B

3. C

4. A

5. D

6. A

7. D

8. C

9. D

10 .C

11. C

12. A

13. C

14. B

15. D

16. A

17. C

18. B

19. A

20. D

21. C

22. A

23. C

24. D

25. B

26. B

27. A

28. D

29. D

30. A

31. D

32. A

33. B

34. B

35. C

2R R vo +

84. The current is (A) 3125 . mA

(B) 62.5mA

36. C

37. B

38. D

39. D

40. C

(C) 125mA

(D) 250mA

41. D

42. B

43. A

44. C

45. B

46. C

47. A

48. A

49. B

50. B

51. D

52. B

53. C

54. A

55. D

56. D

57. A

58. D

59. C

60. A

61. C

62. B

63. A

64. C

65. D

66. A

67. B

68. D

69. C

70. D

71. A

72. B

73. C

74. B

75. B

76. A

77. C

78. B

79. C

80. A

81. D

82. A

83.

84. B

85. C

85. The voltage V0 is (A) -0.781 V

(B) -1.562 V

(C) -3.125 V

(D) -6.250 V

************

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Page 649


CHAPTER

10.1 EC-03

Duration : Three Hours

Maximum Marks : 150

4. The Laplace transform of i( t) is given by I ( s) =

Q.1—30 carry one mark each

As t ® ¥, The value of i( t) tends to

1. The minimum number of equations required to analyze the circuit shown in Fig. Q. 1 is C

C

2 s(1 + s)

(A) 0

(B) 1

(C) 2

(D) ¥

5. The differential equation for the current i( t) in the R

R

~

C

R

circuit of Fig. Q.5 is R

i1(t)

Fig. Q1

(A) 3

(B) 4

(C) 6

(D) 7

sin t

2W

2H

~

1F

Fig. Q5

impedance consisting of 1W resistance in series with 1

d 2i di (A) 2 2 + 2 + i( t) = sin t dt dt

H inductance. The load that will obtain the maximum

(B) 2

d 2i di +2 + 2 i( t) = cos t dt 2 dt

(C) 2

d 2i di +2 + i( t) = cos t 2 dt dt

(D) 2

d 2i di +2 + 2 i( t) = sin t dt 2 dt

2.. A source of angular frequency 1 rad/sec has a source

power transfer is (A) 1 W resistance (B) 1 W resistance in parallel with 1 H inductance (C) 1 W resistance in series with 1 F capacitor (D) 1 W resistance in parallel with 1 F capacitor

6. n-type silicon is obtained by doping silicon with 3. A series RLC circuit has a resonance frequency of

(A) Germanium

(B) Aluminium

1 kHz and a quality factor Q = 100. If each of R, L and C

(C) Boron

(D) Phosphorus

is doubled from its original value, the new Q of the circuit is

7. The bandgap of silicon at 300 K is

(A) 25

(B) 50

(C) 100

(D) 200

(A) 1.36 eV

(B) 1.10 eV

(C) 0.80 eV

(D) 0.67 eV

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Page 591


EC-03

Chap 10.1

20. A 0 to 6 counter consists of 3 flip flops and a

25. A PD controller is used to compensate a system.

combination circuit of 2 input gate(s). The combination

Compared

circuit consists of

compensated system has

(A) one AND gate

(A) a higher type number

(B) one OR gate

(B) reduced damping

(C) one AND gate and one OR gate

(C) higher noise amplification

(D) two AND gates

(D) larger transient overshoot

21. The Fourier series expansion of a real periodic

26. The input to a coherent detector is DSB-SC signal

signal with fundamental frequency f0 is given by

plus noise. The noise at the detector output is

g p ( t) =

åc

n

n = -¥

e j 2 pf0t . It is given that c3 = 3 + j5. Then c-3 is

to

the

uncompensated

system,

the

(A) the in-phase component

(A) 5 + j 3

(B) -3 - j5

(B) the quadrature component

(C) -5 + j 3

(D) 3 - j5

(C) zero (D) the envelope

22. Let x( t) be the input to a linear, time-invariant system. The required output is 4 x( t - 2). The transfer

27. The noise at the input to an ideal frequency detector

function of the system should be

is white. The detector is operating above threshold. The

(A) 4 e

j 4 pf

(C) 4 e

- j 4 pf

23.

A

(B) 2 e

- j8 pf

power spectral density of the noise at the output is

j8 pf

(A) raised-cosine

(B) flat

(C) parabolic

(D) Gaussian

(D) 2 e sequence

x( n)

with

the

z-transform

X ( z) = z 4 + z 2 - 2 z + 2 - 3z -4 is applied as an input to a

28. At a given probability of error, binary coherent FSK

linear, time-invariant system with the impulse response

is inferior to binary coherent PSK by

h( n) = 2 d( n - 3) where

(A) 6 dB

(B) 3 dB

(C) 2 dB

(D) 0 dB

ì 1, n = 0 d( n) = í î 0, otherwise

29. The unit of Ñ ´ H is

The output at n = 4 is

(A) Ampere

(A) -6

(B) zero

(C) 2

(D) -4

(C) Ampere/meter

(B) Ampere/meter 2

(D) Ampere-meter

30. The depth of penetration of electromagnetic wave in 24. Fig. Q.24 shows the Nyquist plot of the open-loop

a medium having conductivity s at a frequency of 1

transfer function G( s) H ( s) of a system. If G( s) H ( s) has

MHz is 25 cm. The depth of penetration at a frequency

one right-hand pole, the closed-loop system is

of 4 MHz will be

Im GH - plane

w=0

(-1, 0)

Re

(A) 6.25 cm

(B) 12.50 cm

(C) 50.00 cm

(D) 100.00 cm

Q.31—90 carry two marks each.

w is positive

31. Twelve 1 W resistance are used as edges to form a

Fig. Q24

cube. The resistance between two diagonally opposite

(A) always stable (B) unstable with one closed-loop right hand pole (C) unstable with two closed-loop right hand poles (D) unstable with three closed-loop right hand poles

corners of the cube is 5 (A) W 6 (C)

6 5

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(B) 1 W (D)

3 W 2 Page 593


UNIT 10

32. The current flowing through the resistance R in the circuit in Fig. Q.32 has the form P cos 4 t, where P is 1 F 10.24

Previous Year Papers

éV ù ù - Ls ú é I s ù ê ú 1 = s 1 ú êëI 2 s úû ê ú R + Ls + ê0 ú ú Cs û ë û

1 é ê R + Ls + Cs (D) ê - Ls ê ë

M=0.75 H

35. An input voltage 3W

R = 3.92 W V=2cos 4t

v( t) = 10 2 cos ( t + 10) + 10 3 cos (2 + 10 ° ) V

~

is applied to a series combination of resistance R = 1W and an inductance L = 1 H. The resulting steady

Fig. Q32

state current i( t) in ampere is

(A) (0.18 + j 0.72)

(B) (0.46 + j 190 . )

(C) -(0.18 + j 190 . )

(D) -(0.192 + j 0.144)

The circuit for Q.33–34 are given in Fig. Q.33–34. For both the questions, assume that the switch S is in position 1 for a long time and thrown to position 2 at t = 0. 1

(B) 1 - cos ( t + 55 ° ) + 10

3 cos (2 t + 55 ° ) 2

(C) 10 cos ( t - 55 ° ) + 10 cos (2 t + 10 ° - tan -1 2) (D) 1 - cos ( t - 55 ° ) + 10

3 cos (2 t - 35 ° ) 2

C

S

36. The driving-point impedance Z ( s) of a network has

2

the pole-zero locations as shown in Fig. Q.36. If Z(0) = 3, R

V

(A) 10 cos ( t + 55 ° ) + 10 cos (2 t + 10 ° + tan -1 2)

R

i1

L

then Z ( s) is

i2

Im

C 1

Fig. Q33-34

-3

Re

-1

33. At t = 0 + , the current i1 is (A)

-V 2R

-V (C) 4R

(B)

s - plane

-1

-V R

Fig. Q36

(D) zero

34. I1 ( s) and I 2 ( s) are the Laplace transforms of i1 ( t) and

(A)

3( s + 3) s + 2s + 3

(B)

2( s + 3) s + 2s + 2

(C)

3( s - 3) s2 - 2 s - 2

(D)

2( s - 3) s2 - 2 s - 3

2

2

i2 ( t) respectively. The equations for the loop currents I1 ( s) and I 2 ( s) for the circuit shown in Fig. Q.33–34,

37. The impedance parameters Z11 and Z12

after the switch is brought from position 1 to position 2

two-port network in Fig. Q.37 are

at t = 0, are 1 é ê R + Ls + Cs (A)ê - Ls ê ë

2W

1 é ê R + Ls + Cs (B) ê - Ls ê ë

ù é Vù - Ls ú é I s ù ê- ú 1 s = ú 1 ú êëI 2 s úû ê R+ ê 0 ú ú Cs û û ë

1 é ê R + Ls + Cs (C) ê - Ls ê ë

ù é Vù - Ls ú é I s ù ê- ú 1 s = ú 1 ú êëI 2 s úû ê R + Ls + ê 0 ú ú ë û Cs û

Page 594

2W

3W

2

1

éV ù ù - Ls ú é I s ù ê ú 1 = s 1 ú êëI 2 s úû ê ú R+ ê0 ú ú Cs û ë û

1W

1W

2’

1’

Fig. Q37

(A) Z11 = 2.75W and Z12 = 0.25 W (B) Z11 = 3W

and Z12 = 0.5 W

(C) Z11 = 3W

and Z12 = 0.25 W

(D) Z11 = 2.25 W www.gatehelp.com

of the

and Z12 = 0.5 W


EC-03

Chap 10.1

38. An n-type silicon bar 0.1 cm long and 100 mm 2 in

(A) 2.26 eV

(B) 1.98 eV

cross-sectional

(C) 1.17 eV

(D) 0.74 eV

area

has

a

majority

carrier

concentration of 5 ´ 10 20 / m 3 and the carrier mobility is 0.13 m 2 /V-s at 300 K. If the charge of an electron is 1.5 ´

43..When the gate-to-source voltage ( VGS ) of a MOSFET

10 -19 coulomb, then the resistance of the bar is

with threshold voltage of 400 mV, working in saturation

(A) 106 Ohm

is 900 mV, the drain current is observed to be 1 mA.

(C) 10

-1

(B) 10 4 Ohm

Ohm

(D) 10

-4

Ohm

Neglecting the channel width modulation effect and assuming that the MOSFET is operating at saturation,

39. The electron concentration in a sample of uniformly doped n-type silicon at 300 K varies linearly from 1017 cm 3 at x = 0 to 6 ´ 1016 cm 3 at x = 2 mm. Assume a situation that electrons are supplied to keep this concentration gradient constant with time. If electronic charge is 1.6 ´ 10 -19 coulomb and the diffusion constant Dn = 35 cm 2 s, the current density in the silicon, if no electric field is present, is

the drain current for an applied VGS of 1400 mV is (A) 0.5 mA

(B) 2.0 mA

(C) 3.5 mA

(D) 4.0 mA

44. If P is Passivation, Q is n-well implant, R is metallization and S is source/drain diffusion, then the order in which they are carried out in a standard n-well CMOS fabrication process, is

(A) zero

(B) -112 A cm 2

(C) +1120 A cm 2

(D) -1120 A cm 2

(A) P–Q–R–S

(B) Q–S–R–P

(C) R–P–S–Q

(D) S–R–Q–P

40. Match items in Group 1 with items in Group 2, most suitably.

45. An amplifier without feedback has a voltage gain of

Group 1

Group 2

50, input resistance of 1 kW and output resistance of 2.5

P. LED

1. Heavy doping

kW. The input resistance of the current-shunt negative

Q. Avalanche photo diode

2. Coherent radiation

feedback amplifier using the above amplifier with a

R.Tunnel diode

3.Spontaneous

feedback factor of 0.2, is

S. LASER

4. Current gain

emission

(A) 1/11 kW

(B) 1/5 kW

(C) 5 kW

(D) 11 kW

(A)

(B)

(C)

(D)

P-1

P-2

P-3

P-2

Q-2

Q-3

Q-4

Q-1

46. In the amplifier circuit shown in Fig. Q.46, the

R-4

R-1

R-1

R-4

values of R1 and R2 are such that the transistor is

S-3

S-4

S-2

S-3

operating at VCE = 3 V and I C = 15 . mA when its b is

41. At 300 K, for a diode current of 1 mA, a certain germanium diode requires a forward bias of 0.1435 V,

150. For a transistor with b of 200, the operating point ( VCE , I C ) is VCC = 6 V

whereas a certain silicon diode requires a forward bias R1

of 0.718 V. Under the conditions stated above, the

R2

closest approximation of the ratio of reverse saturation current in germanium diode to that in silicon diode is (A) 1 (C) 4 ´ 10

(B) 5 (D) 8 ´ 10 3

3

Fig. Q46

42. A particular green LED emits light of wavelength 5490 A°. The energy bandgap of the semiconductor material

used

there

is

(Plank’s

(A) (2 V, 2 mA)

(B) (3 V, 2 mA)

(C) (4 V, 2 mA)

(D) (4 V, 1 mA)

constant

= 6.626 ´ 10 -34 J – s) www.gatehelp.com

Page 595


UNIT 10


47. The oscillator circuit shown in Fig. Q.47 has an ideal

51. Three identical amplifiers with each one having a

inverting amplifier. its frequency of oscillation (in Hz) is

voltage gain of 50, input resistance of 1 kW and output resistance of 250 W, are cascaded. The open circuit voltage gain of the combined amplifier is

C

C

C

R

R

R

(A) 49 dB

(B) 51 dB

(C) 98 dB

(D) 102 dB

52. An ideal sawtooth voltage waveform of frequency 500 Hz and amplitude 3 V is generated by charging a capacitor of 2 mF in every cycle. The charging requires

Fig. Q47

1

(A) (2 p 6 RC) 1 (C) ( 6 RC)

1 (B) (2 pRC)

(A) constant voltage source of 3 V for 1 ms

6 (D) (2 pRC)

(C) constant current source of 3 mA for 1 ms

(B) constant voltage source of 3 V for 2 ms

(D) constant current source of 3 mA for 2 ms 48. The output voltage of the regulated power supply 53. The circuit shown in Fig. Q.53 has 4 boxes each

shown in Fig. Q.48 is

described by inputs, P, Q, R and outputs Y, Z with

+

Y = P Å Q Å R, Z = RQ + PR + QP . The circuit acts as

1 kW

a 15 V DC Unregulated Power source

Q Vz = 3 V 40 kW

P

Regulated DC Output

20 kW _

P

Z Y R

Fig. Q48

(A) 3 V (C) 9 V

Q

P

P

Q

Z Y R

(B) 6 V (D) 12 V

Q

Z Y R

P

Q

Z Y R

Output

Fig. Q53

49. The action of a JFET in its equivalent circuit can best be represented as a (A) (B) (C) (D)

(B) 4 bit substracter giving P - Q

Current Controlled Current Source Current Controlled Voltage Source Voltage Controlled Voltage Source Voltage Controlled Current Source

(C) 4 bit substracter giving Q - R (D) 4 bit adder giving P + Q + R

50. If the op-amp in Fig. Q.50 is ideal, the output voltage Vout will be equal to

54. If the functions W , X , Y and Z are as follows W = R + PQ + RS X = PQRS + P Q R S + PQ R S

5 kW

Y = RS + PR + PQ + P Q

1 kW 2V Vout

1 kW 3V 8 kW

Fig. Q50

(A) 1 V

(B) 6 V

(C) 14 V

(D) 17 V

Page 596

(A) 4 bit adder giving P + Q

Z = R + S + PQ + P × Q × R + PQ × S Then (A) W = Z , X = Z

(B) W = Z , X = Y

(C) W = Y

(D) W = Y = Z

55. A 4 bit ripple counter and a 4 bit synchronous counter are made using flip flops having a propagation delay of 10 ns each. If the worst case delay in the ripple

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EC-03

counter and the synchronous counter be R and S respectively, then (A) R =10 n, S =40 ns

(B) R =40 ns,

S =10 ns

(C) R =10 ns, S =30 ns

(D) R =30 ns, S =10 ns

56. The DTL, TTL, ECL and CMOS families of digital ICs are compared in the following 4 columns P

Q

R

S

Fanout is minimum

DTL

DTL

TTL

CMOS

Power consumption is minimum

TTL

CMOS

ECL

DTL

Propagation delay is minimum

CMOS

ECL

TTL

TTL

(A) (B) (C) (D)

Chap 10.1

BCD to binary code Binary to excess -3 code Excess -3 to Gray code Gray to Binary code

59. In the circuit shown in Fig. Q.59, A is a parallel-in, parallel-out 4 bit register, which loads at the rising edge of the clock C. The input lines are connected to a 4 bit bus, W. Its output acts as the input to a 16 ´ 4 ROM whose output is floating when the enable input E is 0. A partial table of the contents of the ROM is as follows MSB

CLK

The correct column is (A) P

(B) Q

(C) R

(D) S

A

1

57. The circuit shown in Fig. Q.57 is a 4 bit DAC. The input bits 0 and 1 are represented by 0 and 5 V respectively. The OP AMP is ideal, but all the resistance and the 5 V inputs have a tolerance of ±10%. The specification (rounded to the nearest multiple of 5%) for the tolerance of the DAC is (A) ±35%

(B) ±20%

(C) ±10%

(D) ±5%

E

ROM

R R

CLK

t

2R t1 4R

Fig. Q59

Vout 8R R

Fig. Q57

58. The circuit shown in Fig. Q.58 converts MSB

+

+

t2

+

Address

Data

0

0011

2

1111

4

0100

6

1010

8

1011

10

1000

12

0010

14

1000

The clock to the register is shown, and the data on the W bus at time t1 is 0110. The data on the bus at time t2 is

MSB

Fig. Q58

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Page 597


UNIT 10


(A) 1111

(B) 1011

Data for Q.65–66 are given below. Solve the

(C) 1000

(D) 0010

problems and choose the correct answers. X ( t) is a random process with a constant mean

60. In an 8085 microprocessor, the instruction CMP B has been executed while the content of the accumulator

value of 2 and the autocorrelation function R X ( t) = 4[ e

is less than that of register B. As a result

-0 .2 t

+ 1].

(A) Carry flag will be set but Zero flag will be reset (B) Carry flag will be reset but Zero flag will be set (C) Both Carry flag and Zero flag will be reset

65. Let X be the Gaussian random variable obtained by sampling the process at t = ti and let ¥

Q ( a) = ò

(D) Both Carry flag and Zero flag will be set

a

61. Let X and Y be two statistically independent random variables uniformly distributed in the ranges ( -1, 1) and (( -2, 1) respectively. Let Z = X + Y . Then the probability that ( Z £ - 2) is (A) zero (C)

1 3

1 2p

e

-

y2 2

dy

The probability that [ x £ 1] is (A) 1 - Q(0.5)

(B) Q(0.5)

(C) Q( 2 1 2 )

(D) 1 - Q( 2 1 2 )

(B)

1 6

66. Let Y and Z be the random variables obtained by

(D)

1 12

sampling X ( t)

at t = 2

and t = 4

respectively. Let

W = Y - Z . The variance of W is 62. Let P be linearity, Q be time-invariance, R be causality and S be stability. A discrete time system has

(A) 13.36

(B) 9.36

(C) 2.64

(D) 8.00

the input-output relationship, 67. Let x( t) = 2 cos ( 800 pt) + cos (1400 pt). x( t) is sampled

n ³1 ì x( n) ï y( n) = í 0, n =0 ï x( n + 1) n £ -1 î

with the rectangular pulse train shown in Fig. Q.67. The only spectral components (in kHz) present in the

where x( n) is the input and y( n) is the output. The

sampled signal in the frequency range 2.5 kHz to 3.5 kHz are

above system has the properties

-3

p(t)

(A) P, S but not Q, R

(B) P, Q, S but not R

(C) P, Q, R, S

(D) Q, R, S but not P

T0 = 10 sec

3

Data for Q.63–64 are given below. Solve the

-T0

t

-T0/6 0 T0/6

problems and choose the correct answers.

T0

Fig. Q67

(A) 2.7, 3.4

(B) 3.3, 3.6

filter (RC-LPF) with R = 1 kW and C = 10 . mF.

(C) 2.6, 2.7, 3.3, 3.4, 3.6

(D) 2.7, 3.3

63. Let H ( f ) denote the frequency response of the

68. The signal flow graph of a system is shown in Fig.

RC-LPF. Let f1 be the highest frequency such that H ( f1 ) 0.95. Then f1 (in Hz) is 0 £ | f | £ f1 H (0)

Q.68. The transfer function

The system under consideration is an RC low-pass

(A) 327.8

(B) 163.9

(C) 52.2

(D) 104.4

R(s)

C(s ) R(s )

1 s

1

1

of the system is 1 s

6 -4

-2

-3

C(s)

Fig. Q68

64. Let t g ( f ) be the group delay function of the given RC-LPF and f2 = 100 Hz. Then t g ( f2 ) in ms, is (A) 0.717

(B) 7.17

(C) 71.7

(D) 4.505

Page 598

6 (A) 2 s + 29 s + 6

(B)

6s s + 29 s + 6

s( s + 2) s + 29 s + 6

(D)

s( s + 27) s + 29 s + 6

(C)

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2

2

2


EC-03

69 The root locus of the system G( s) H ( s) =

Chap 10.1

72. The gain margin and the phase margin of a feedback system with

K s( s + 2)( s + 3)

G( s) H ( s) =

has the break-away point located at

s are ( s + 100) 3

(A) (-0.5, 0)

(B) (-2.548, 0)

(A) - dB, 0°

(B) ¥, ¥

(C) (-4, 0)

(D) (-0.784, 0)

(C) ¥, 0°

(D) 88.5 dB, ¥

70. The approximate Bode magnitude plot of a

73. The zero-input response of a system given by the

minimum phase system is shown in Fig. Q.70. The

state-space equation é x& 1 ù é1 0 ù é x1 ù é x1 (0) ù é1 ù ê x& ú = ê1 1 ú ê x ú and ê x (0) ú = ê0 ú is ûë 2 û ë 2û ë ë 2 û ë û

transfer function of the system is dB 160 140

20

0.1

100

10

( s + 0.1) ( s + 10) 2 ( s + 100)

ée t ù (C) ê t ú ëte û

ét ù (D) ê t ú ëte û

frequency fc = 1 MHz using a nonlinear device with the

( s + 0.1) 3 (B) 10 7 ( s + 10)( s + 100)

2

(C) 108

ée t ù (B) ê ú ët û

74. A DSB-SC signal is to be generated with a carrier

Fig. Q70

( s + 0.1) 3 (A) 108 ( s + 10) 2 ( s + 100)

éte t ù (A) ê ú ët û

( s + 0.1) ( s + 10)( s + 100) 2 3

(D) 10 9

input-output characteristic v0 = a0 v1 + a1 vi3 where a0 and a1

are constants. The output of the nonlinear device

can be filtered by an appropriate band-pass filter. Let vi = Acl cos (2pfcl t) + m( t)

where

m( t) is the message

l c

signal. Then the value of f (in MHz) is 71. A second-order system has the transfer function C (s) 4 = 2 R (s) s + 4s + 4

(A) 1.0

(B) 0.333

(C) 0.5

(D) 3.0

The data for Q.75-76 are given below. Solve the With r ( t) as the unit-step function, the response c( t) of the system is represented by (A) Step Response

Let m( t) = cos [( 4 p ´ 10 3) t ] be the message signal

(B)

and c( t) = 5 cos [(2 p ´ 106 ) t ] be the carrier.

Step Response

2

1

75. c( t) and m( t) are used to generate an AM signal. The Amplitude

Amplitude

1.5 1 0.5 0

0

5

10 15 Time (sec)

20

0

25

modulation index of the generated AM signal is 0.5. Total side band power Then the quantity is Carrier power

0.5

0

(C)

10

5 Time (sec)

(D) 1

Amplitude

1

(A)

1 2

(B)

1 4

(C)

1 3

(D)

1 8

76. c( t) and m( t) are used to generate an FM signal. If

Step Response

Step Response 1.5

Amplitude

problems and choose the correct answers.

the peak frequency deviation of the generated FM is three times the transmission bandwidth of the AM

0.5

signal,

0.5

then

the

coefficient

of

the

term

cos [2 p(1008 ´ 10 t)] in the FM signal (in terms of the 3

0

0 0

2 4 Time (sec)

6

0

2

4 Time (sec)

6

Bessel coefficients) is (A) 5 J 4 ( 3)

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(B)

5 2

J8 ( 3) Page 599


EC-03

Chap 10.1

86. A uniform plane wave traveling in air is incident on

90. Two identical antennas are placed in the q = p

the plane boundary between air and another dielectric

plane as shown in Fig. Q.90. The elements have equal

medium with e r = 4. The reflection coefficient for the

amplitude excitation with 180° polarity difference,

normal incidence, is

operating at wavelength l. The correct value of the

2

magnitude of the far-zone resultant electric field

(A) zero

(B) 0.5 Ð180 °

(C) 0.333 Ð0 °

(D) 0.333Ð180 °

strength normalized with that of a single element, both computed for f = 0, is

87. If the electric field intensity associated with a perfect

dielectric

medium

E( z, t) = 10 cos (2 p ´ 10 t - 0.1pz) 7

is

given

by

then

the

volt/m,

s

uniform plane electromagnetic wave traveling in a f s

velocity of the traveling wave is (A) 3.00 ´ 108 m/sec

(B) 2.00 ´ 108 m/sec

(C) 6.28 ´ 10 7 m/sec

(D) 2.00 ´ 10 7 m/sec

Fig. Q.90

88. A short-circuited stub is shunt connected to a transmission line as shown in Fig. Q.88. If Z 0 = 50 ohm,

æ 2 ps ö (A) 2 cos ç ÷ è l ø

æ 2 ps ö (B) 2 sin ç ÷ è l ø

æ ps ö (C) 2 cos ç ÷ è lø

æ ps ö (D) 2 sin ç ÷ è lø

the admittance Y seen at the junction of the stub and

**************

the transmission line is

l/8

Zo

Zo

Z L 100 W Zo

l/2 Y

Fig. Q.88

(A) (0.01 - j0.02) mho

(B) (0.02 - j0.01) mho

(C) (0.04 - j0.02) mho

(D) (0.02 + j0) mho

89. A rectangular metal wave guide filled with a dielectric material of relative permitivity e r = 4 has the inside dimensions 3.0 cm ´ 1.2 cm. The cut-off frequency for the dominant mode is (A) 2.5 GHz

(B) 5.0 GHz

(C) 10.0 GHz

(D) 12.5 GHz

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Page 601


UNIT 10

ANSWER SHEET

Page 602

1.

(B)

2.

(C)

3.

(B)

4.

(C)

5.

(C)

6.

(D)

7.

(B)

8.

(A)

9.

(C)

10.

(A)

11.

(B)

12.

(D)

13.

(B)

14.

(C)

15.

(A)

16.

(D)

17.

(C)

18.

(B)

19.

(B)

20.

(D)

21.

(D)

22.

(C)

23.

(B)

24.

(A)

25.

(C)

26.

(A)

27.

(A)

28.

(D)

29.

(B)

30.

(B)

31.

(A)

32.

(*)

33.

(D)

34.

(D)

35.

(C)

36.

(B)

37.

(A)

38.

(C)

39.

(C)

40.

(C)

41.

(C)

42.

(A)

43.

(D)

44.

(B)

45.

(A)

46.

(A)

47.

(A)

48.

(C)

49.

(D)

50.

(B)

51.

(D)

52.

(D)

53.

(B)

54.

(A)

55.

(B)

56.

(C)

57.

(A)

58.

(D)

59.

(C)

60.

(A)

61.

(A)

62.

(A)

63.

(C)

64.

(B)

65.

(A)

66.

(C)

67.

(A)

68.

(A)

69

(D)

70

(A)

71.

(B)

72.

(D)

73.

(C)

74

(A)

75

(D)

76.

(D)

77.

(B)

78.

(A)

79

(C)

80

(D)

81.

(B)

82.

(D)

83.

(B)

84.

(C)

85.

(C)

86.

(D)

87.

(B)

88.

(A)

89

(B)

90.

(D)

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UNIT 10


5. For the R-L circuit shown in Fig. Q.5, the input

(C) current controlled current source

voltage vi ( t) = u( t). The current i( t) is

(D) current controlled voltage source

1H

i(t)

10. Voltage series feedback (also called series-shunt feedback) results in vi(t)

2W

(A) increase in both input and output impedances (B) decrease in both input and output impedances (C) increase in input impedance and decrease in output impedance

Fig Q.5

(A)

(B)

i(t)

(D) decrease in input impedance and increase in output impedance

i(t)

0.5 0.31

1 0.63 t(sec)

2

t(sec)

½

(C)

11. The circuit in Fig. Q.11 is a

(D)

i(t)

i(t)

0.5 0.31

vi

1 0.63

R

vo

R

C

C ½

t(sec)

2

t(sec)

6. The impurity commonly used for realizing the base

Fig Q.11

region of a silicon n-p-n transistor is (A) Gallium

(B) Indium

(C) Boron

(D) Phosphorus

7. If for a silicon n-p-n transistor, the base-to-emitter voltage ( VBE ) is 0.7 V and the collector-to-base voltage

(A) low-pass filter

(B) high-pass filter

(C) band-pass filter

(D) band-reject filter

12. Assuming VCEsat = 0.2 V and b = 50, the minimum base current ( I B ) required to drive the transistor in Fig. Q.12 to saturation is

( VCB ) is 0.2 V, then the transistor is operating in the (A) normal active mode

(B) saturation mode

(C) inverse active mode

(D) cutoff mode

3 V IC 1 kW IB

8. Consider the following statements S1 and S2. S1 : The b of a bipolar transistor reduces if the base Fig Q12.

width is increased. S2 : The b of a bipolar transistor increases if the doping concentration in the base is increased. Which one of the following is correct ?

(A) 56 mA

(B) 140 mA

(C) 60 mA

(D) 3 mA

(A) S1 is FALSE and S2 is TRUE

13. A master-slave flip-flop has the characteristic that

(B) Both S1 and S2 are TRUE

(A) change in the input is immediately reflected in the output

(C) Both S1 and S2 are FALSE (D) S1 is TRUE and S2 is FALSE

(B) change in the output occurs when the state of the master is affected

9. An ideal op-amp is an ideal

(C) change in the output occurs when the state of the slave is affected

(A) voltage controlled current source (B) voltage controlled voltage source Page 604

(D) both the master and the slave states are affected at the same time www.gatehelp.com


EC-04

Chap 10.2

14. The range of signed decimal numbers that can be

19. The impulse response h[ n] of a linear time-invariant

represented by 6-bit 1’s complement numbers is

system is given by

(A) -31 to +31

(B) -63 to +64

(C) -64 to +63

(D) -32 to +31

15.

A digital

system is

required

h[ n] = u[ n + 3] + u[ n - 2 ] - 2 u[ n - 7 ] where u[ n] is the unit step sequence. The above

to amplify

a

system is

binary-encoded audio signal. The user should be able to

Vout

control the gain of the amplifier from a minimum to a maximum in 100 increments. The minimum number of bits required to encode, in straight binary, is (A) 8

(B) 6

(C) 5

(D) 7

Vin 0

Fig Q.18

16. Choose the correct one from among the alternatives A, B, C, D after matching an item from Group 1 with the most appropriate item in Group 2.

(A) stable but not causal (B) stable and causal (C) causal but unstable

Group 1

Group 2

P: Shift register

1: Frequency division

Q: Counter

2: Addressing in memory chips

R: Decoder

3: Serial to parallel data conversion

(D) unstable and not causal 20. The distribution function FX ( x) of a random variable X is shown in Fig. Q.20. The probability that X = 1 is Fx(x) 1.0

(A)

(B)

(C)

(D)

P–3

P–3

P–2

P–1

Q–2

Q–1

Q–1

Q–3

R–1

R–2

R–3

R–2

0.55 0.25

-2

AND-OR-Invert (AOI) gate. For the inputs shown in Fig. Q.17, the output Y is

(A) zero

(B) 0.25

(C) 0.55

(D) 0.30

H ( z) = Fig Q.17

(A) 0

(B) 1

(C) AB

(D) AB

x

21. The z-transform of a system is

y

Input are Floating

2

1

Fig Q.20

17. Fig. Q.17 shows the internal schematic of a TTL

A B

0

z z - 0.2

If the ROC is |z |< 0.2, then the impulse response of the system is

18. Fig. Q.18 is the voltage transfer characteristic of (A) an NMOS inverter with enhancement mode transistor as load (B) an NMOS inverter with depletion mode transistor as load

(A) (0.2) n u[ n]

(B) (0.2) n u[ -n - 1]

(C) -(0.2) n u[ n]

(D) -(0.2) n u[ -n - 1]

22. The Fourier transform of a conjugate symmetric function is always

(C) a CMOS inverter

(A) imaginary

(B) conjugate anti-symmetric

(D) a BJT inverter

(C) real

(D) conjugate symmetric

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Page 605


EC-04

Chap 10.2

33. Consider the Bode magnitude plot shown in Fig.

37. Consider the following statements S1 and S2.

Q.33. The transfer function H ( s) is

S1 : At the resonant frequency the impedance of a series R–L–C circuit is zero.

20 log H(jw) 0

S2 : In a parallel G–L–C circuit, increasing the conductance G results in increase in its Q factor.

20 dB/dec -20 dB/dec

-20

Which one of the following is correct ?

40 dB/dec

(A) S1 is FALSE and S2 is TRUE 1

w

100

10

(B) Both S1 and S2 are TRUE

Fig Q.33

(C) S1 is TRUE and S2 is FALSE

(A)

( s + 10) ( s + 1)( s + 100)

(B)

10( s + 1) ( s + 10)( s + 100)

(D) Both S1 and S2 are FALSE

(C)

10 2 ( s + 1) ( s + 10)( s + 100)

(D)

10 3( s + 100) ( s + 1)( s + 10)

38..

In

an

N A = 9 ´ 10

16

V ( s) of an R–L–C 34. The transfer function H ( s) = o Vi ( s) circuit is given by 6

H ( s) =

10 s + 20 s + 106 2

The Quality factor (Q-factor) of this circuit is (A) 25

(B) 50

(C) 100

(D) 5000

abrupt

concentrations cm

on 3

p–n

the

junction,

p-side

and

the

reverse biased and the total depletion width is 3 mm. The depletion width on the p-side is (A) 2.7 mm

(B) 0.3 mm

(C) 2.25 mm

(D) 0.75 mm

39. The resistivity of a uniformly doped n-type silicon sample is 0.5 W-cm. If the electron mobility (m n ) is 1250 cm 2 /V-sec and the charge of an electron is 1.6 ´ 10 -19 Coulomb, the donor impurity concentration ( N D) in the sample is

conditions

(A) 2 ´ 1016 cm 3

(B) 1 ´ 1016 cm 3

(C) 2.5 ´ 1015 cm 3

(D) 5 ´ 1015 cm 3

zero.

Its

transfer

function

H ( s) = VC ( s) Vi ( s) is 1 (A) 2 s + 10 3 s + 106

10 mH

10 kW

are

respectively. The p-n junction is

35. For the circuit shown in Fig. Q.35, the initial are

doping

n-side

106 (B) 2 s + 10 3 s + 106

40. Consider an abrupt p-n junction. Let Vbi be the built-in potential of this junction and VR be the applied

vi(t)

reverse bias. If the junction capacitance ( C j ) is 1 pF for

vo(t)

100 mF

Vbi + VR = 1 V, then for Vbi + VR = 4 V, C j will be

Fig Q35.

(C)

10 3 s 2 + 10 3 s + 106

(D)

(A) 4 pF

(B) 2 pF

(C) 0.25 pF

(D) 0.5 pF

41. Consider the following statements S1 and S2.

106 s 2 + 106 s + 106

S1 : The threshold voltage ( VT ) of a MOS capacitor decreases with increase in gate oxide thickness.

36. A system described by the following differential equation d2 y dy +3 + 2 y = x( t) dt 2 dt

S2 : The threshold voltage ( VT ) of a MOS capacitor decreases with increase in substrate doping concentration. Which one of the following is correct ?

is initially at rest. For input x( t) = 2 u( t), the output

(A) S1 is FALSE and S2 is TRUE (B) Both S1 and S2 are TRUE

y( t) is -t

-2 t

(A) (1 - 2 e + e ) u( t) -t

-2 t

(C) (0.5 + e + 15 . e ) u( t)

-t

-2 t

(B) (1 + 2 e - e ) u( t) -t

-2 t

(D) (0.5 + 2 e + 2 e ) u( t)

(C) Both S1 and S2 are FALSE (D) S1 is TRUE and S2 is FALSE

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Page 607


UNIT 10


42. The drain of an n-channel MOSFET is shorted to

46. A bipolar transistor is operating in the active region

the gate so that VGS = VDS . The threshold voltage ( VT ) of

with a collector current of 1 mA. Assuming that the b of

the MOSFET is 1 V. If the drain current ( I D) is 1 mA for

the transistor is 100 and the thermal voltage ( VT ) is 25

VGS = 2 V, then for VGS = 3 V, I D is

mV, the transconductance ( g m ) and the input resistance

(A) 2 mA

(B) 3 mA

( rp)

(C) 9 mA

(D) 4 mA

of

the

transistor

in

the

common

emitter

configuration, are (A) g m = 25 mA/V and rp = 15.625 kW

43. The longest wavelength that can be absorbed by

(B) g m = 40 mA/V and rp = 4.0 kW

silicon, which has the bandgap of 1.12 eV, is 1.1 mm. If

(C) g m = 25 mA/V and rp = 2.5 kW

the longest wavelength that can be absorbed by another material is 0.87 mm, then the bandgap of this material is

(D) g m = 40 mA/V and rp = 2.5 kW 47. The value of C required for sinusoidal oscillations of

(A) 1.416 eV

(B) 0.886 eV

(C) 0.854 eV

(D) 0.706 eV

frequency 1 kHz in the circuit of Fig. Q.47 is 2.1 kW

1 kW

44. The neutral base width of a bipolar transistor, biased in the active region, is 0.5 mm. The maximum

C

electron concentration and the diffusion constant in the

1 kW

base are 1014 cm 3 and Dn = 25 cm 2 sec respectively. Assuming negligible recombination in the base, the

C

1 kW

collector current density is (the electron charge is 1.6 ´ 10 -19 Coulomb) (A) 800 A/cm2

(B) 9 A/cm2

(C) 200 A/cm2

(D) 2 A/cm2

Fig Q.47

(A)

45. Assume that the b of the transistor is extremely

1 mF 2p

(C)

large and VBE = 0.7 V, I C and VCE in the circuit shown in Fig. Q.45 are

1 2p 6

(B) 2p mF mF

(D) 2 p 6 mF

48. In the op-amp circuit given in Fig. Q.48, the load

5 V

current iL is IC

R1

2.2 kW

4 kW

vi

+ VEC

R1 vo

1 kW

300 kW R1

R2 iL

RL

Fig Q.45 Fig Q.48

(A) I C = 1 mA, VCE = 4.7 V (B) I C = 0.5 mA, VCE = 375 . V

(A) -

vs R2

(B)

vs R2

(C) -

vs RL

(D)

vs R1

(C) I C = 1 mA, VCE = 2.5 V (D) I C = 0.5 mA, VCE = 39 . V

Page 608

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UNIT 10


57. Consider the sequence of 8085 instructions given

60. A 1 kHz sinusoidal signal is ideally sampled at 1500

below

samples /sec and the sampled signal is passed through

LXI H, 9258

an ideal low-pass filter with cut-off frequency 800 Hz.

MOV A, M

The output signal has the frequency

CMA MOV M , A Which one of the following is performed by this

(A) zero Hz

(B) 0.75 kHz

(C) 0.5 kHz

(D) 0.25 kHz

61. A rectangular pulse train s( t) as shown in Fig. Q.61

sequence?

is convolved with the signal cos 2 ( 4 p ´ 10 3 t). The

(A) Contents of location 9258 are moved to the accumulator

convolved signal will be a s(t)

(B) Contents of location 9258 are compared with the contents of the accumulator

1

(C) Contents of location 8529 are complemented and stored in location 8529 (D) Contents of location 5892 are complemented and stored in location 5892 58. A Boolean function f of two variables x and y is

1 ms

Fig Q.61

(A) DC

(B) 12 kHz sinusoid

(C) 8 kHz sinusoid

(D) 14 kHz sinusoid

62. Consider the sequence

defined as follows :

x[ n] = [ -4 - j5 1 + j2 5 ]

f (0, 0) = f (0, 1) = f (1, 1) = 1; f (1, 0) = 0 Assuming complements of x

and y

are not

available, a minimum cost solution for realizing f using only 2-input NOR gates and 2-input OR gates (each having unit cost) would have a total cost of (A) 1 unit

(B) 4 units

(C) 3 units

(D) 2 units

The conjugate anti-symmetric part of the sequence is (A) [-4 - j2.5

and store the result in the accumulator. The numbers are available in registers B and C respectively. A part of

4 - j2.5]

j2

(B) [- j2.5

1

(C) [- j2.5

j2

(D) [-4

59. It is desired to multiply the numbers 0AH by 0BH

1

j2.5] 0] 4]

63. A causal LTI system is described by the difference equation 2 y[ n] = ay[ n - 2 ] - 2 x[ n] + bx[ n - 1]

the 8085 program for this purpose is given below:

The system is stable only if

MVI A, 00H LOOP:

t

0

––––––––––

(A) a = 2, b < 2

–––––––––––

(B) a > 2, b > 2

–––––––––––

(C) a < 2, any value of b

HLT

(D) b < 2, any value of a

END

64. A causal system having the transfer function

The sequence of instructions to complete the

H ( s) =

program would be (A) JNZ LOOP, ADD B, DCR C

1 s+2

is excited with 10u( t). The time at which the

(B) ADD B, JNZ LOOP, DCR C

output reaches 99% of its steady state value is

(C) DCR C, JNZ LOOP, ADD B

(A) 2.7 sec

(B) 2.5 sec

(C) 2.3 sec

(D) 2.1 sec

(D) ADD B, DCR C, JNZ LOOP Page 610

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EC-04

65. The impulse response h[ n] of a linear time invariant system is given as ì -2 2 ï h[ n] = í 4 2 ï 0 î

n = 2, - 2

sin(2 t) ù ésin( -2 t) (B) ê sin( t ) sin( -3t) úû ë

otherwise

e jpn 4 , then the output is (C) 4 e jpn

(B) 4 2 e - jpn

4

(D) -4 e jpn

4

2ù é-2 69. If A = ê ú, then sin At is ë 1 -3û é sin( -4 t) + 2 sin( -t) - sin( -4 t) + 2 sin( -t) ù (A) ê 2 sin( -4 t) + sin( -t) úû ë - sin( -4 t) + sin( -t)

n = 1, - 1

If the input to the above system is the sequence (A) 4 2 e jpn

Chap 10.2

sin( 4 t) + 2 sin( t) 2 sin( -4 t) - 2 sin( -t) ù - sin( -4 t) + sin( t) 2 sin( 4t) + sin( t) úû

é (C) ê ë

cos( -t) + 2 cos( t) 2 cos( -4 t) - 2 sin( -t) ù é (D) ê -2 cos( 4 t) + cos( t) úû ë - cos( -4 t) + sin( -t)

4

4

66. Let x( t) and y( t) with Fourier transforms F ( f ) and

70. The open-loop transfer function of a unity feedback

Y ( f ) respectively be related as shown in Fig. Q.66. Then

system is

Y ( f ) is

G( s) = x(t)

y(t)

1

-2

0

-2

t

2

0

The range of K for which the system is stable is 21 (A) (B) 13 > K > 0 > K >0 4

t

(C)

-1

Fig Q.66

1 (A) - X ( f 2) e - j 2 pf 2

1 (B) - X ( f 2) e j 2 pf 2

(C) - X ( f 2) e j 2 pf

(D) - X ( f 2) e - j 2 pf

the number of roots which lie in the right half of

phase of the system response at 20 Hz is (B) 0°

(C) 90°

(D) –180°

x3

c

x4

d

(A) 4

(B) 2

(C) 3

(D) 1

x& 1 = -3 x1 - x2 = u, x& 2 = 2 x1 , y = x1 + u

68. Consider the signal flow graph shown in Fig. Q.68. x The gain 5 is x1 b

the s-plane is

72. The state variable equations of a system are :

(A) –90°

x2

(D) -6 < K < ¥

P ( s) = s 5 + s 4 + 2 s 3 + 2 s 2 + 3s + 15

zeros at 5 Hz, 100 Hz and 200 Hz. The approximate

a

21
71. For the polynomial

67. A system has poles at 0.01 Hz, 1 Hz and 80 Hz;

x1

K s( s + s + 2)( s + 3) 2

x5

1

The system is (A) controllable but not observable (B) observable but not controllable (C) neither controllable nor observable

e

f

g

(D) controllable and observable

Fig Q.68

(A) (B) (C)

1 - ( be + cf + dg) abcd bedg 1 - ( be + cf + dg) abcd 1 - ( be + cf + dg) + bedg

1 - ( be + cf + dg) + bedg (D) abcd

é1 0 ù At 73. Given A = ê ú, the state transition matrix e is ë0 1 û given by é 0 e- t ù (A) ê - t ú 0û ëe

é 0 et ù (B) ê t ú ëe 0 û

ée - t 0ù (C) ê -t ú ë 0 e û

ée t 0 ù (D) ê tú ë0 e û

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Page 611


UNIT 10

74. Consider the signal x( t) shown in Fig. Q.74. Let h( t) denote the impulse response of the filter matched to x( t), with h( t) being non-zero only in the interval 0 to 4 sec. The slope of h( t) in the interval 3 < t < 4 sec is x(t) 1

0

3

2

1

4

t(sec)

Fig. Q.74

(A) ½ sec -1

(B) –1 sec -1

(C) –1/2 sec -1

(D) 1 sec -1


78. Consider a binary digital communication system with equally likely 0’s and 1’s. When binary 0 is transmitted the voltage at the detector input can lie between the levels -0.25 V and +0.25 V with equal probability; when binary 1 is transmitted, the voltage at the detector can have any value between 0 and 1 V with equal probability. If the detector has a threshold of 0.2V (i.e. if the received signal is greater than 0.2V, the bit is taken as 1), the average bit error probability is (A) 0.15

(B) 0.2

(C) 0.05

(D) 0.5

79. A random variable X with uniform density in the interval 0 to 1 is quantized as follows:

75. A 1 mW video signal having a bandwidth of 100

if 0 £ X £ 0.3,

xq = 0

MHz is transmitted to a receiver through a cable that

if 0.3 £ X £ 1,

xq = 0.7

has 40 dB loss. If the effective one-sided noise spectral density at the receiver is 10 -20 Watt/Hz, then the signal-to-noise ratio at the receiver is (A) 50 dB

(B) 30 dB

(C) 40 dB

(D) 60 dB

76. A 100 MHz carrier of 1V amplitude and a 1 MHz

where xq

is the quantized value of X. The

root-mean square value of the quantization noise is (A) 0.573

(B) 0.198

(C) 2.205

(D) 0.266

80. Choose the correct one from among the alternatives A, B, C, D after matching an item from Group 1 with

modulating signal of 1V amplitude are fed to a balanced

the most appropriate item in Group 2.

modulator. The output of the modulator is passed

Group 1

Group 2

1 : FM

P : Slope overload

2 : DM

Q : m-law

3 : PSK

R : Envelope detector

4 : PCM

S : Capture effect

through an ideal high-pass filter with cut-off frequency of 100 MHz. The output of the filter is added with 100 MHz signal of 1V amplitude and 90° phase shift as shown in Fig. Q.76. The envelope of the resultant signal is 1 Mhz, 1 V

HPF 100 Mhz

Balanced Modulator

y(t)

T : Hilbert transfer U : Matched filter

100 Mhz, 1 V 90

100 Mhz, 1 V

o

Fig Q.76

(A) constant (C)

5 4 - sin(2 p ´ 106 t)

(B)

1 + sin(2 p ´ 106 t)

(D)

5 4 + cos(2 p ´ 106 t)

77. Two sinusoidal signals of same amplitude and frequencies 10 kHz and 10.1 kHz are added together. The combined signal is given to an ideal frequency detector. The output of the detector is (A) 0.1 kHz sinusoid

(B) 20.1 kHz sinusoid

(C) a linear function of time (D) a constant Page 612

(A)

(B)

(C)

(D)

1–T

1–S

1–S

1–U

2–P

2–U

2–P

2–R

3–U

3–P

3–U

3–S

4–S

4–T

4–Q

4–Q

81. Three analog signals, having bandwidth 1200 Hz, 600 Hz and 600 Hz, are sampled at their respective Nyquist rates, encoded with 12 bit words, and time division multiplexed. The bit rate for the multiplexed signal is (A) 1, 15.2 kbps

(B) 28.8 kbps

(C) 27.6 kbps

(D) 38.4 kbps

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EC-04

Chap 10.2

82. Consider a system shown in Fig. Q.82. Let X ( f ) and

(C) B1 = 20 kHz, B2 = 10 kHz

Y ( f ) denote the Fourier transforms of x( t) and y( t)

(D) B1 = 10 kHz, B2 = 10 kHz

respectively. The ideal HPF has the cutoff frequency 10 85. Consider a 300 W, quarter-wave long (at 1 GHz)

kHz.

transmission line as shown in Fig. Q.85. It is connected

X(f )

to a 10 V, 50W source at one end and is left open circuited at the other end. The magnitude of the voltage at the open circuit end of the line is -3

-1

HPF 10 kHz

Balanced Modulator

x(t)

3

1

f (kHz)

Balanced Modulator

~

~

10 kHz

13 kHz

y(t)

10 V, 50 W source

Zo = 300 W

l/4

Fig Q.85

Fig Q.82

The positive frequencies where Y ( f ) has spectral

(A) 10 V

(B) 5 V

(C) 60 V

(D) 60/7 V

peaks are (A) 1 kHz and 24 kHz

(B) 2 kHz and 24 kHz

86. In a microwave test bench, why is the microwave

(C) 1 kHz and 14 kHz

(D) 2 kHz and 14 kHz

signal amplitude modulated at 1 kHz ? (A) To increase the sensitivity of measurement

83. A parallel plate air-filled capacitor has plate area of 10 -4 m 2 and plate separation of 10 -3 m. It is connected to a 0.5 V, 3.6 GHz source. The magnitude of the (B) 100 mA

(C) 10 A

(D) 1.59 mA

(C) To study amplitude modulation (D) Because crystal detector fails at microwave frequencies

displacement current is ( e o = 1 36 p ´ 10 -9 F m) (A) 10 mA

(B) To transmit the signal to a far-off place

84. A source produces binary data at the rate of 10

r 87. If and E = (a$ x + ja$ y) e jkz - jwt r jkz - jwt ,the time-averaged Poynting H = ( k wm )(a$ y + ja$ x ) e vector is

kbps. The binary symbols are represented as shown in

(A) null vector

Fig.Q.84

(B) ( k wm )a$ z

(C) (2k wm )a$ z

(D) ( k 2wm )a$ z

Binary 1

Binary 1

88. Consider an impedance Z = R + jX marked with

1V

point P in an impedance Smith chart as shown in Fig. 0

0.1

t(ms)

0

0.1

t(ms)

Q.88. The movement from point P along a constant resistance circle in the clockwise direction by an angle

-1 V

45° is equivalent to

Fig Q.84 .5

r=0

The source output is transmitted using two modulation schemes, namely Binary PSK (BPSK) and Quadrature PSK (QPSK). Let B1 and B2 be the bandwidth

requirements

of

BPSK

x=0

respectively.

Assuming that the bandwidth of the above rectangular

x = -0.5 P

pulses is 10 kHz, B1 and B2 are

x = -1

(A) B1 = 20 kHz, B2 = kHz Fig. Q.88

(B) B1 = 10 kHz, B2 = 10 kHz www.gatehelp.com

Page 613


CHAPTER

10.3 EC-05

Duration : Three Hours 150

Maximum Marks :

5. The function x( t) is shown in the figure. Even and odd parts of a unit step function u( t) are respectively, x(t)

Question 1- 30 Carry one Mark each.

1

-1

3

d2 y æ dy ö 2 3 2 + 4ç ÷ + y +2= x dt è dt ø

Fig. Q5

1 1 (A) , x( t) 2 2

(A) degree = 2, order = 1 (B) degree = 3, order = 2

(C)

(C) degree = 4, order = 3 (D) degree = 2, order =3

1 1 (B) - , x( t) 2 2

1 1 , - x( t) 2 2

1 1 (D) - , - x( t) 2 2

6. The region of convergence of z - transform of the n

(B) 4 cos(20 t + 3) + 2 sin(710 t) (D) 1

(C)

2. Choose the function f ( t); - ¥ < t < ¥ for which a

(C) e

-t

sin(25 t)

n

æ5ö æ6ö sequence ç ÷ u( n) - ç ÷ u( -n - 1) must be è6ø è5ø 5 5 (B) z > (A) z < 6 6

Fourier series cannot be defined. (A) 3 sin(25 t)

t

0

1. The following differential equation has

5 6
(D)

6 < z <¥ 5

3. A fair dice is rolled twice. The probability that an odd 7. The condition on R, L and C such that the step

number will follow on even number is 1 1 (B) (A) 2 6 (C)

1 3

(D)

response y( t) in the figure has no oscillations, is L

1 4

R

u(t)

C

y(t)

4. A solution of the following differential equation is given by Fig. Q7

d2 y dy -5 + 6y =0 dt 2 dt (A) y = e 2 x + e -3x

(B) y = e 2 x + e 3x

(C) y = e -2 x + e -3x

(D) y = e -2 x + e -3x

1 (A) R ³ 2

L C L (C) R ³ 2 C

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L C 1 (D) R = LC (B) R ³

Page 615


UNIT 10

8. The ABCD parameters of an ideal n:1 transformer é n 0ù shown in the figure are ê ú. The value of x will be ë 0 Xû i i1


(A) abundance of Silicon on the surface of the Earth. (B) larger bandgap of Silicon in comparison to Germanium.

2

(C) favorable properties of Silicon - dioxide (SiO2)

+

+ n:1

(D) lower melting point.

v1

v2

-

-

14. The effect of current shunt feedback in an amplifier is to

Fig. Q8

(A) n

1 (B) n

(C) n2

(D)

(A) increase the input resistance and decrease the output resistance. (B) increase both input and output resistance

1 n2

(C) decrease both input and output resistance.

9. In a series RLC circuit, R = 2 kW, L = 1 H and 1 m. The resonant frequency is 400 1 (A) 2 ´ 10 4 Hz (B) ´ 10 4 Hz p

(D) decrease the input resistance and increase the output resistance. 15. The input resistance of the amplifier shown in the

C=

figure is 30 kW 10 kW

(D) 2 p ´ 10 Hz

4

4

(C) 10 Hz

vo

~

10. The maximum power that can be transferred to the load resistor RL from the voltage source in the figure is

Ri

100 W

Fig. Q15

RL

10 V

(B) 10 W

(C) 0.25 W

(D) 0.5 W

(B) 10 kW

(C) 40 kW

(D) infinite

16. The first and the last critical frequency of an RC -

Fig. Q10

(A) 1 W

30 (A) kW 4

driving point impedance function must respectively be (A) a zero and a pole (B) a zero and a zero

11. The bandgap of Silicon at room temperature is

(C) a pole and a pole

(A) 1.3 eV

(B) 0.7 eV

(D) a pole and a zero

(C) 1.1 eV

(D) 1.4 eV

17. The cascode amplifier is a multistage configuration of

12. A Silicon PN junction at a temperature of 20° C has a reverse saturation current of 10 pico - Amperes (pA).

(A) CC - CB

(B) CE - CB

(C) CB - CC

(D) CE - CC

The reserve saturation current at 40° C for the same bias is approximately

18. Decimal 43 in Hexadecimal and BCD number

(A) 30 pA

(B) 40 pA

(C) 50 pA

(D) 60 pA

system is respectively (A) B2, 0100 011 (B) 2B, 0100 0011

13. The primary reason for the widespread use of

(C) 2B, 0011 0100

Silicon in semiconductor device technology is

(D) B2, 0100 0100

Page 616

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EC-05

Chap 10.3

19. The Boolean function f implemented in the figure

23.

using two input multiplexes is

pö æ s( t) = 8 cosç 20 p - ÷ + 4 sin(15 pt) is 2ø è

0 MUX f

A

C

0 MUX

C

1

1

The

power

in

(A) 40

(B) 41

(C) 42

(D) 82

the

signal

24. Which of the following analog modulation scheme requires

the

minimum

transmitted

power

and

minimum channel bandwidth? B

Fig. Q19

(A) ABC + ABC

(B) ABC + ABC

(C) ABC + A B C

(D) A BC + ABC

(A) VSB

(B) DSB - SC

(C) SSB

(D) AM

25. A linear system is equivalently represented by two sets of state equations: X& = AX + BU And W& = CW + DU

20. Which of the following can be impulse response of a The eigenvalues of the representations are also

causal system?

computed as [ l] and [m ]. Which one of the following

(A)

(B) h(t)

statements is true?

h(t)

t

t

(C)

(D) h(t)

(A) [ l] = [m ] and X = W

(B) [ l] = [m ] and X ¹ W

(C) [ l] ¹ [m ] and X = W

(D) [ l] = [m ] and X ¹ W

26. Which one of the following polar diagrams

h(t)

corresponds to a lag network? (A)

t

t

(B) Im

Im w= 0

w=¥

Re

n

æ1ö 21. Let x( n) = ç ÷ u( n), y( n) = x 2 ( n) and Y ( e jw ) be the è2 ø

w= 0

w=¥

Re

j0

Fourier transform of y( n) then Y ( e ) is 1 (A) 4

(B) 2

(C) 4

4 (D) 3

(C)

(D) Im

Im

w=¥

w= 0 w=¥

w= 0

Re

Re

22. Find the correct match between group 1 and group 2 Group 1

27. Despite the presence of negative feedback, control

Group II

systems still have problems of instability because the

P. {1 + km( t)} A sin( wc t)

W. Phase Modulation

(A) Components used have non-linearities

Q. km( t) A sin( wc t)

X.Frequency Modulation

R. A sin( wc t + km( t))

Y. Amplitude Modulation

(B) Dynamic equations of the subsystem are not known exactly. (C) Mathematical analysis involves approximations.

(A) P-Z, Q-Y, R-X, S-W

(B) P-W, Q-X, R-Y, S-Z

(C) P-X, Q-W, R-Z, S-Y

(D) P-Y, Q-Z, R-W, S-X

(D) System has large negative phase angle at high frequencies.

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Page 617


EC-05

37. Given an orthogonal matrix é1 1 1 1ù ê1 1 -1 -1ú ú, [ AA T ]-1 is A =ê 1 1 0 0 ê ú ê0 0 1 -1úû ë

(A) 5 V and 2 W

(B) 7.5 V and 2.5 W

(C) 4 V and 2 W

(D) 3 V and 2.5 W

41. If R1 = R2 = R3 = R and R3 = 11 . R in the bridge circuit shown in the figure, then the reading in the ideal

é 14 0 0 0 ù ê0 1 0 0 ú 4 ú (A) ê 1 0 0 0ú ê 2 ê0 0 0 1 ú ë 2û

é 12 0 0 0 ù ê0 1 0 0 ú 2 ú (B) ê 1 0 0 0ú ê 2 ê0 0 0 1 ú ë 2û

é1 ê0 (C) ê ê0 ê0 ë

é 0 0 0ù ê0 1 0 0 ú 4 ú (D) ê ê0 0 14 0 ú ê0 0 0 1 ú ë 4û

0 0 0ù 1 0 0ú ú 0 1 0ú 0 0 1 úû

Chap 10.3

voltmeter connected between a and b is R1

R4

a

10 V

1 4

b

V R3

R2

Fig. Q41

38. For the circuit show in the figure, the instantaneous

(A) 0.238 V

(B) 0.138 V

current ii ( t) is

(C) -0.238 V

(D) 1 V.

-j2

j2

42. The h parameters of the circuit shown in the figure

i1

~

~

O

5 0 A

3W

are

10 60O A

I1

I2

10 W

+ 20 W

V1

Fig. Q38

10 3 (A) Ð90 ° Amps. 2

10 3 (B) Ð - 90 ° Amps. 2

(C) 5 Ð60 ° Amps

(D) 5 Ð - 60 ° Amps

-

V2 -

Fig. Q42

39. Impedance Z as shown in the given figure is j2 W

j5 W

+

j10 W

é 0.1 0.1ù (A) ê ú ë-0.1 0.3û

-1ù é10 (B) ê ú 1 0 . 05 ë û

é30 20 ù (C) ê ú ë20 20 û

1ù é10 (D) ê ú 1 0 . 05 ë û

j2 W

j10 W

43. A square pulse of 3 volts amplitude is applied to C-R circuit shown in the figure. The capacitor is initially Fig. Q39

uncharged. The output voltage VO at time t = 2 sec is

(A) j29 W

(B) j9 W

(C) j19 W

(D) j39 W

0.1 mF

I1 Vi

-j2

+

I2 +

3V 1 kW

V1

40. For the circuit shown in figure, Thevenin's voltage and Thevenin's equivalent resistance at terminals a - b

2 sec

t

-

V2 -

is Fig. Q43

5W

0.5I1

5W

a

10 V

(A) 3 V

(B) -3 V

(C) 4 V

(D) -4 V

b

44. A Silicon sample A is doped with 1018 atoms/cm 3 of Fig. Q40

boron. Another sample b of identical dimension is doped with 1018 atoms/cm 3 phosphorus. The ratio of electron to

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Page 619


UNIT 10


hole mobility is 3. The ratio of conductivity of the

48. The OP-amp circuit shown in the figure is filter. The

sample A to B is

type of filter and its cut. Off frequency are respectively. 1 3

(A) 3

(B)

2 (C) 3

3 (D) 2

10 kW 10 kW vo vi 1 mF

45. A Silicon PN junction diode under reverse bias has depletion region of width 10 mm. The

relative

permitivity of Silicon, e r = 117 . and the preemptively of free

space

e o = 8.85 ´ 10

-12

F/m.

The

depletion

capacitance of the diode per square meter is (A) 100 mF

(B) 10 mF

(C) 1 mF

(D) 20 mF

1 kW

Fig. Q48

(A) high pass, 1000 rad/sec. (B) Low pass, 1000 rad/sec. (C) high pass, 1000 rad/sec. (D) low pass, 10000 rad/sec.

46. For an npn transistor connected as shown in figure VBE = 0.7 volts. Given that reverse saturation current of the junction at room temperature 300 K is 10 -13 A, the

49. In an ideal differential amplifier shown in the figure, a large value of ( RE ) VCC

emitter current is

RC

RC

IC _

V1

VBE +

V2

RE -VEE

Fig. Q46

(A) 30 mA

(B) 39 mA

(C) 49 mA

(D) 20 mA

Fig. Q49

(A) increase both the differential and common - mode gains

47. The voltage eo is indicated in the figure has been

(B) increases the common mode gain only

measured by an ideal voltmeter. Which of the following

(C) decreases the differential mode gain only

can be calculated ?

(D) decreases the common mode gain only. 1 MW

50. For an n-channel MOSFET and its transfer curve shown in the figure, the threshold voltage is eo

VD=5 V ID

D

1 MW Transfer Characteristics

VG=1 V

G S

Fig. Q47 1V

(A) Bias current of the inverting input only

VGS

VS=1 V

Fig. Q50

(B) Bias current of the inverting and non-inverting inputs only

(A) 1 V and the device is in active region

(C) Input offset current only

(B) -1 V and the device is in saturation region

(D) Both the bias currents and the input offset current.

(C) 1 V and the device is in saturation region

Page 620

(D) -1 V and the device is in active region. www.gatehelp.com


UNIT 10


57. The given figure shows a ripple counter using

60. Match

positive edge triggered flip-flops. If the present state of

combination.

the following and choose the correct

Group 1

the counter is Q2Q1Q0 = 001 then its next state Q2Q1Q

E. Continuos and periodic signal

will be 1

1 T0

Q0

F. Continuous and periodic signal

1 T1

Q1

T2

G. Discrete and aperiodic signal

Q2

H. Discrete and periodic signal CLK

Q0

Q1

Group 2

Q2

1. Fourier representation is continuous and aperiodic

Fig. Q57

(A) 010

(D) 100

(B) 111

(D) 101

2. Fourier representation is discrete and aperiodic 3. Fourier representation is continuous 4. Fourier representation is discrete and periodic

58. What memory address range is NOT represents by chip # 1 and chip # 2 in the figure A0 to A15 in this

(A) E-3, F-2, G-4, H-1

(A) E-1, F-3, G-2, H-4

(C) E-1, F-2, G-3, H-4

(D) E-2, F-1, G-4, H-3

figure are the address lines and CS means chip select. 61. A signal x( n) = sin( w0 n + f) is the input to a linear A0 - A7

256 bytes

time- invariant system having a frequency response

Chip #1

H ( e jw ). If the output of the system Ax( n - n0 ) then the most general form of will be (A) -n0 w0 + b for any arbitrary real

A8

(B) -n0 w0 + 2pk for any arbitrary integer k

A9

(C) n0 w0 + 2 pk for any arbitrary integer k

A9

(D) -n0 w0 f

A8

62. For a signal the Fourier transform is X ( f )). Then A0 - A7

the inverse Fourier transform of X ( 3 f + 2) is given by

256 bytes

(A)

Chip #2

1 æ t ö j 3pt xç ÷e 2 è2 ø

(B)

(C) 3 x( 3t) e - j 4 pt A10 - A16

not used

(B) 1500 - 16FF

(C) F900-FAFF

(D) F800 - F9FF

j 4 pt 3

(D) x( 3t + 2)

63. The polar diagram of a conditionally stable system

Fig. Q58

(A) 0100 - 02FF.

1 ætö xç ÷e 3 è 3ø

for open loop gain K = 1 is shown in the figure. The open loop transfer function of the system is known to be stable. The closed loop system is stable for Im

59. The output y( t) of a linear time invariant system is related to its input x( t) by the following equation

-8

-2 -0.2

y( t) = 0.5 x( t - td + T) + x( t - td ) + 0.5 x( t - td + T) The filter transfer function H( w) of such a system is given by (A) (1 + cos wT) e - jwt d

(B) (1 + 0.5 cos wT) e - jwt d

(C) (1 - cos wT) e - jwt d

(D) (1 - 0.5 cos wT) e - jwt d

Page 622

www.gatehelp.com

Fig. Q63

Re


EC-05

1 1
(A) K < 5 and

(B) K <

1 (C) K < and 5 < K 8

1 1 and < K < 5 2 8

1 (D) K > and 5 > K 8

66. A ramp input applied to an unity feedback system results in 5% steady state error. The type number and zero frequency gain of the system are respectively (A) 1 and 20

64. In the derivation of expression for peak percent

(C) 0 and

overshoot æ -px M p = expç ç 1 - x2 è

Chap 10.3

ö ÷ ´ 100% ÷ ø

(B)0 and 20

1 20

(D) 1 and

1 20

67. A double integrator plant G( s) = K s 2 , H ( s) = 1 is to be compensated to achieve the damping ratio and and undamped natural frequency, w = 5 rad/s which one of

Which one of the following conditions is NOT required? (A) System is linear and time invariant (B) The system transfer function has a pair of complex conjugate poles and no zeroes.

the following compensator Ge ( s) will be suitable ? s+3 s + 9.9 (A) (B) s + 9.9 s+3 (C)

s-6 s + 8.33

(D)

s-6 s

(C) There is no transportation delay in the system. 68. An unity feedback system is given as

(D) The system has zero initial conditions.

G( s) = 65. Given the ideal operational amplifier circuit shown in the figure indicate the correct transfer characteristics assuming ideal diodes with zero cut-in voltage.

Indicate the correct root locus diagram. (A)

10 V

K (1 - s) . s( s + 3)

jw

jw

(B)

vi vo

s

s

-10 V 2 kW

(C)

jw

jw

(D)

0.5 kW 2 kW

s

s

Fig. Q65

(A)

Vi Vo

(B)

69. A MOS capacitor made using P type substrate is in the accumulation mode. The dominant charge in the

Vo +10 V

channel is due to the presence of

+10 V

(A) holes -8 V

+5 V

Vi

-5 V

+8 V

Vi

(B) electrons (C) positively charged ions

(C)

(D) negatively charged ions

-10 V

-10 V

(D)

Vo

70. A device with input x( t) and output y( t) is

Vo +10 V

characterized y( t) = x 2 ( t). An FM signal with frequency

+5 V -5 V

deviation of 90 kHz and modulating signal bandwidth +5 V

Vi

+5 V

-5 V -5 V

-10 V

Vi

of 5 kHz is applied to this device. The bandwidth of the output signal is (A) 370 kHz

(B) 190 kHz

(C) 380 kHz

(C) 95 kHz

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Page 623


EC-05

Chap 10.3

COMMON DATA QUESTION 78, 79, 80:

81b. If in addition following code exists from 019H

Given, rd = 20 kW, I DSS = 10 mA, Vp = -8 V

onwards, ORI 40 H

20 V

ADD M 2 kW

What will be the result in the accumulator after the last instruction is executed?

+ +

2 MW vi

vo 2V

–

–

(B) 20 H

(C) 60 H

(D) 42 H

Statement for Linked Answer Question 82a

Zo

Zi

(A) 40 H

and

82b: The dopen loop transfer function of a unity

Fig. Q78

feedback system is given by

78. Z i and Z O of the circuit are respectively (A) 2 MW and 2 kW

(B) 2 MW and

(C) ¥ and 2 kW

(D) ¥ and

20 kW 11

82a. The gain and phase crossover frequencies in rad/sec are, respectively

20 kW 11

79. I D and VDS under DC conditions are respectively (A) 5.625 mA and 8.75 V

(B) 7.500 mA and 5.00 V

(C) 4.500 mA and 11.00 V

(D) 6.250 mA and 7.50 V

80. Transconductance in milli-Siemens (mS)

and

(A) 0.632 and 1.26

(B) 0.632 and 0.485

(C) 0.485 and 0.632

(D) 1.26 and 0.632

82b. Based on the above results, the gain and phase margins of the system will be (A) -7.09 dB and 87.5 °

(B) 7.09 dBand 87.5 °

(C) 7.09 dB and -87.5 °

(D) -7.09 and -87.5 °

voltage gain of the amplifier are respectively (A) 1.875 mS and 3.41

(B) 1.875 mS and -3.41

(C) 3.3 mS and -6

(D) 3.3 mS and 6

Statement for linked answer question 83a and 83b Asymmetric three - level midtread quantizer is to be designed assuming equiprobable occurrence of all

Linked Answer Questions : Q.81a to 85b Carry

quantization levels.

Two Marks Each 83a. If the probability density function is divided into Statement For Linked Answer Questions 81a and

three regions as shown in the figure, the value of a in

81b:

the figure is p(x)

Consider an 8085 microprocessor system. 81a. The following program starts at location 0100H.

Region 1

LXI SP, OOFF

-3

LXI H, 0701

Region 2

-1 -a

Region 3

a1

3

x

Fig. Q83

MVI A, 20H

(A)

1 3

(B)

2 3

SUB M

(C)

1 2

(D)

1 4

The content of accumulator when the program counter reaches 0109 H is (A) 20 H

(B) 02 H

(C) 00 H

(D) FF H

83b. The quantization noise power for the quantization region between - a and + a in the figure is (A)

4 81

(B)

1 9

(C)

5 81

(D)

2 81

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Page 625


UNIT 10

Statement of Linked Answer Questions 84a


y[n]

and

84b

2

(C)

1

Voltage standing wave pattern in a lossless

½

transmission line with characteristic impedance 50 and

-6

-5

-4

-3

-2

-2 -1

a resistive load is shown in the figure.

0 1 y[n]

2

3

4

5

6

2

3

4

5

6

2

V(z)

1

(D)

4

½ -6

-5

-4

-3

-2

-1

0

1

1 l/2

l

85b. The Fourier transform of y[2 n] will be

Fig. Q84a and Q84b

(A) e-2 jw [cos 4 w + 2 cos 2 w + 2 ]

84a. The value of the load resistance is (A) 50 W

(B) 200 W

(C) 12.5 W

(D) 0

(B) [cos 2 w + 2 cos w + 2 ] (C) e - jw [cos 2 w + 2 cos w + 2 ] (D) e -2 jw [cos 2 w + 2 cos w + 2 ]

84b. The reflection coefficient is given by (A) -0.6

(B) -1

(C) 0.6

(D) 0

************

Statement of Linked Answer Question 85a and 85b: A sequence x (n) has non-zero values as shown in the figure.(A) x[n] 2 1 ½ -6

-5

-4

-3

-2

-1

0

1

2

3

4

5

6

n

Fig. Q85

85a. The sequence ì æn ö ï xç - 1 ÷ y[ n] = í è 2 ø ï 0 î

for n even

will be

for n odd y[n] 2 1

(A)

½ -6

-5

-4

-3

-2

-1

0

1

2

3

4

5

6

1

2

3

4

5

6

n

y[n] 2 1

(B)

½ -6

Page 626

-5

-4

-3

-2

-1

0

n

n

www.gatehelp.com

n


CHAPTER

10.4 EC-06

Duration : Three Hours

Maximum Marks : 150

Q.1 to carry Q.20 one marks each and Q.21 to Q.85 carry two marks each.

(A) 0

(B) 1

(C) 2

(D) 3

1ù 0 ú is ú 1úû

(D) e t u( t)

H ( jw) = A( w) e jf( w )

does

not

produce

any

phase

distortions if (A) A( w) = Cw2 , f( w) = kw3

(B) A( w) = Cw2 , f( w) = kw

(C) A( w) = Cw, f( w) = kw2

(D) A( w) = C, f( w) = kw-1

corresponding to peak and valley currents are Vp , VD respectively. The range of tunnel-diode voltage for

(B) Ñ P + Ñ (Ñ × P)

VDwhich the slope of its I - VD characteristics is

(D) Ñ (Ñ × P) - Ñ 2P

negative would be

2

(C) Ñ 2P + Ñ ´ P

(C) e - t u( t)

7. The values of voltage ( VD) across a tunnel-diode

2. Ñ ´ Ñ ´ P Where P is a vector, is equal to (A) P ´ Ñ ´ P - Ñ P

(B) e 2 t u( t)

6. A low-pass filter having a frequency response

é 1 1 1. The rank of the matrix ê 1 -1 ê êë 1 1

2

(A) e-2 t u( t)

(A) VD < 0

(B) 0 £ VD < Vp

òò (Ñ ´ P) × ds where P is a vector, is equal to (A) ò P × dl (B) ò Ñ ´ Ñ ´ P × dl

(C) Vp £ VD < Vv

(D) VD ³ Vv

(C) ò Ñ ´ P × dl

semiconductor under equilibrium is

3.

(D) ò Ñ × Pdv

8. The concentration of minority carriers in an extrinsic (A) Directly proportional to the doping concentration

4. A probability density function is of the form p( x) = Ke

-a x

, x Î ( -¥, ¥)

(B) Inversely proportional to the doping concentration (C) Directly proportional to the intrinsic concentration

(A) 0.5

(B) 1

(D) Inversely proportional to the intrinsic concentration

(C) 0.5

(D) a

9. Under low level injection assumption, the injected

The value of K is

5. A solution for the differential equation x( t) + 2 x( t) = d( t) With initial condition x(0 -) = 0 Page 628

minority carrier current for an extrinsic semiconductor is essentially the (A) Diffusion current

(B) Drift current

(C) Recombination current

(D) Induced current

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EC-06

10. The phenomenon known as “Early Effect” in a bipolar transistor refers to a reduction of the effective base-width caused by

(B) The reverse biasing of the base – collector junction (C) The forward biasing of emitter-base junction (D) The early removal of stored base charge during saturation-to-cut off switching 11. The input impedance ( Z i ) and the output impedance ( Z 0 ) of an ideal trans-conductance (voltage controlled current source) amplifier are (B) Z i = 0, Z 0 = ¥

(C) Z i = ¥, Z 0 = 0

(D) Z i = ¥, Z 0 = ¥

15. The Dirac delta function is defined as ì1 t =0 (A) d( t) = í î 0 otherwise ì1 t =0 (B) d( t) = í î 0 otherwise

(A) Electron – Hole recombination at the base

(A) Z i = 0, Z 0 = 0

Chap 10.4

t =0

ì¥ (D) d( t) = í î0

t =0 and otherwise

otherwise

points on its I D - VGS curve : (i) VGS = 0 at I D = 12 mA and

(C)

3 < z <3 2

ò d( t) dt = 1

-¥

(B)

2 < z <3 3

(D)

1 2
control system is given by G( s) =

highest trans – conductance gain for small signals? (D) VGS = 3 Volts

¥

16. If the region of convergence of x1 [ n] + x2 [ n] is 1 2 < z < then the region of convergence of x1 [ n] - x2 [ n] 3 3

Which of the following Q – point will give the

(C) VGS = 0 Volts

ò d( t) dt = 1

-¥

17. The open-loop function of a unity-gain feedback

(ii) VGS = -6 Volts at I D = 0 mA

(B) VGS = -3 Volts

and

includes 1 (A) < z < 3 3

12. An n-channel depletion MOSFET has following two

(A) VGS = -6 Volts

¥

ì1 (C) d( t) = í î0

K ( s + 1)( s + 2)

The gain margin of the system in dB is given by (A) 0

(B) 1

13. The number of product terms in the minimized

(C) 20

(D)

sum-of-product

18. In the system shown below, x( t) = (sin t) u( t) In

expression

obtained

through

the

following K – map is (where, “d” denotes don’t care states)

steady-state, the response y( t) will be x(t)

1

0

0

1

0

d

0

0

0

0

d

1

1

0

0

1

(A) 2

(B) 3

(C) 4

(D) 5

y(t)

1 s+1

Fig Q. 18

(A) (C)

pö æ sinç t - ÷ 4ø 2 è

1 1

2

e - t sin t

(B)

pö æ sinç t + ÷ 4ø 2 è

1

(D) sin t - cos t

19. The electric field of an electromagnetic wave propagation in the positive direction is given by

14. Let x( t) « X ( jw) be Fourier Transform pair. The

E = a$ x sin( wt - bz) + a$ y sin( wt - bz + p 2)

Fourier Transform of the signal x(5 t - 3) in terms of

The wave is

X ( jw) is given as j 3w

j 3w

(A)

1 e 5

æ jw ö Xç ÷ è 5 ø

(B)

1 e 5

(C)

1 - j 3w æ jw ö e Xç ÷ 5 è 5 ø

(D)

1 j 3w æ jw ö e Xç ÷ 5 è 5 ø

5

5

æ jw ö Xç ÷ è 5 ø

(A) Linearly polarized in the z–direction (B) Elliptically polarized (C) Left-hand circularly polarized (D) Right-hand circularly polarized

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Page 629


EC-06

Chap 10.4

30. A two-port network is represented by ABCD,

34. In the figures shown below, assume that all the

parameters given by

capacitors are initially uncharged. If vi ( t) = 10 u( t) Volts,

éV1 ù é A ê I ú = êC ë 1û ë

vo( t)is given by

B ù é V2 ù D úû êë-I 2 úû

1k

If port – 2 is terminated by the input impedance

DRL + A (C) C + BRL

4 mF

Vi(t)

1 mF

4k

VO(t) -

-

ARL + B (D) D + CRL

Fig. Q.34

31. In the two port network shown in the figure below z12 and z 21 are respectively

(A) 8 e -0 .004 t Volts

(B) 8(1 - e -0 .004 t ) Volts

(C) 8u( t) Volts

(D) 8 Volts

35. Consider two transfer functions

I1

I2

G1 ( s) = bI1

re

+

+

seen at port – 1 is given by A + BRL ARL + C (A) (B) C + DRL BRL + D

ro

1 s And G2 ( s) = 2 s 2 + as + b s + as + b

The 3–dB bandwidths of their frequency responses are, respectively

Fig Q.31

(A) re and bro

(B) 0 and -bro

(C) 0 and bro

(D) re and -bro

32.

The

first

and

the

last

critical

(A)

a 2 - 4 b,

a2 + 4 b

(B)

a 2 + 4 b,

a2 - 4 b

(C)

a 2 - 4 b,

a2 - 4 b

(D)

a 2 + 4 b,

a2 + 4 b

36. A negative resistance Rneg is connected to a passive

frequencies

(singularities) of a driving point impedance function of a

network N having driving point impedance Z1 ( s) as shown below. For Z 2 ( s) to be positive real, Rneg

passive network having two kinds of elements, are a pole and a zero respectively. The above property will be

N

satisfied by (A) RL network only (B) RC network only Z1(s)

Z2(s)

(C) LC network only

Fig Q.36

(D) RC as well as RL networks 33. A 2 mH inductor with some initial current can be represented as shown below, where s is the Lap lace Transform variable. The value of initial current is Is

(A) Rneg £ Re Z1 ( jw), "w

(B) Rneg £ Z1 ( jw) , "w

(C) Rneg £ Im Z1 ( jw), "w

(D) Rneg £ ÐZ1 ( jw), "w

37. In the circuit shown below, the switch was connected to position 1 at t < 0 and at t = 0, it is changed to position 2. Assume that the diode has zero voltage

0.002s

drop and a storage time ts . For 0 < t £ ts , vR is given by (all in Volts)

1

1 mV

5V

(A) 0.5A

(B) 2.0A

(C) 1.0 A

(D) 0.0 A

+

2

Fig Q.33 5V

1 kW

vR _

Fig Q.37

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Page 631


UNIT 10

(A) vR = -5

(B) vR = +5

(C) 0 £ vR < 5

(D) -5 £ vR < 0


The voltage VC across the capacitor at t = 1 is

38. The majority carriers in an n–type semiconductor have an average drift velocity v in a direction perpendicular to a uniform magnetic field B. The electric field E induced due to Hall effect acts in the direction.

(A) 0 Volt

(B) 6.3 Volts

(C) 9.45 Volts

(D) 10 Volts

42. For the circuit shown below, assume that the zener diode is ideal with a breakdown voltage of 6 volts. The waveform observed across R is 6V

(A) v ´ B

(B) B ´ v

(C) along v

(D) opposite to v

+ 12sin wt

39. Find the correct match between Group 1 and Group

~

R

VR

2.

-

Group 1

Group 2

Fig Q.42 6V

E-Varactor diode

1-Voltage reference

F-PIN diode

2-High frequency switch

G-Zener diode

3-Tuned circuits

H-Schottky diode

4-Current controlled attenuator

(A)

(A) E-4, F-2, G-1, H-3

(B) E-2, F-4, G-1, H-3

(C) E-3, F-4, G-1, H-2

(D) E-1, F-3, G-2, H-4

6V

(B)

40. A heavily doped n- type semiconductor has the following data: -12 V

Hole-electron ratio

:0.4

Doping concentration

:4.2 ´ 108 atoms/m 3

Intrinsic concentration

:15 . ´ 10 4 atoms/m 3

The

ratio

of

conductance

of

the

12 V

n-type

(D)

semiconductor to that of the intrinsic semiconductor of same material and ate same temperature is given by (A) 0.00005

(B) 2,000

(C) 10,000

(D) 20,000

-6 V

(C) 41. For the circuit shown in the following figure, the

-6 V

capacitor C is initially uncharged. At t = 0 the switch S is closed. In the figures shown the OP AMP is supplied

Q. 43 A new Binary Coded Pentary (BCP) number

with and the ground has been shown by the symbol

system is proposed in which every digit of a base–5

S

number is represented by its corresponding 3–bit

C=1 mF

binary code. For example, the base–5 number 24 will be 1 kW

– VC +

represented by its BCP code 010100. In this numbering vo

system, the BCP code 10001001101 corresponds of the following number is base–5 system

10 V

Fig Q.41

Page 632

(A) 423

(B) 1324

(C) 2201

(D) 4231

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EC-06

44. An I / O peripheral device shown in Fig.(b) below is

Chap 10.4

(A)

(B)

(C)

(D)

to be interfaced to an 8085 microprocessor. To select the I/O device in the I/O address range D4 H – D7 H, its chip–select ( CS) should be connected to the output of the decoder shown in as below: LSB

A2

3-to-8 Decoder

A3 A4

MSB

D0 D1 D2 D3 D4 D5 D6 D7

DATA IORD IOWR

I/O Peripheral

A1

47. Two D – flip – flops, as shown below, are to be

A0

connected as a synchronous counter that goes through

A7 A6 A5

the following sequence 00 ® 01 ® 11 ® 10 ® 00 ® K

CS

Fig Q.44

(A) output 7

(B) output 5

(C) output 2

Q0

D0 Clock

D1

Q1

Q0

45. For the circuit shown in figures below, two 4 – bit parallel – in serial – out shift registers loaded with the

MSB

CK

CK

(D) output 0

Q1

Fig Q.47

The inputs D0 and D1 respectively should be

data shown are used to feed the data to a full adder.

connected as,

Initially, all the flip – flops are in clear state. After

(A) Q1 and Q0

(B) Q0 and Q1

(C) Q1Q0 and Q1Q0

(D) Q1 Q0 and Q1Q0

applying two clock pulses, the outputs of the full-adder should be 1

0

1

1

D

Q

S

A

48. Following is the segment of a 8085 assembly

Full Adder

CLK

language program

B

0

0

1

1

D

Q

CLK

Ci

CO

Q

D CLK

3000 H CLK

Fig Q.45

(A) S = 0 C0 = 0

(B) S = 0 C0 = 1

(C) S = 1 C0 = 0

(D) S = 1 C0 = 1

46. A 4 – bit D / A converter is connected to a free – running 3 – big UP counter, as shown in the following figure. Which of the following waveforms will be observed at VO ? Q2

D3

Q1

D1

Q0

D3

3 - Bit Counter

On completion of RET execution, the contents of SP is (A) 3CF0 H

(B) 3CF8 H

(C) EFFD H

(D) EFFF H

1 kW

49. The point P in the following figure is stuck at 1.

D2 Clock

LXI SP, EFFF H CALL 3000 H : : : LXI H, 3CF4 PUSH PSW SPHL POP PSW RET

D/A Converter

vo

The output f will be A B

P

f

1 kW C

Fig Q.46

Fig Q.49

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Page 633


EC-06

(A)

p 2

(B)

p 3

(C)

p 4

(D)

p 6

(C) RC <

Chap 10.4

1 wc

(D) RC >

1 wc

64. In the following figure the minimum value of the constant “C” , which is to be added to y1 ( t) and y2 ( t)

59. A linear system is described by the following state

such that y1 ( t) and y2 ( t)

equation

Q is quantizer with L levels, stepwise D allowable signal dynamic range [-V, V]

é 0 1ù X& ( t) = AX ( t) + BU ( t), A = ê ú ë-1 0 û

x(t) with range -V , V 2 2

The state transition matrix of the system is sin t ù cos t úû

sin t ù é- cos t (B) ê sin t cos t úû ë

é- cos t - sin t ù (C) ê cos t úû ë- sin t

é- cos t - sin t ù (D) ê sin t úû ë cos t

é cos t (A) ê ë- sin t

(A) D

60. The minimum step- size required for a Delta – Modulator operating at 32 K , samples/sec to track the

D2 12

(B)

D 2

(D)

D L

(A) 4 ´ 10 4 Hz

(B) 2 ´ 106 Hz

(C) 2 ´ 10 9 Hz

(D) 2 ´ 1010 Hz

61. A zero mean white Gaussian noise is passed through an ideal lowpass filter of bandwidth 10 kHz. The output is then uniformly sampled with sampling period ts = 0.03 msec. The samples so obtained would be (A) correlated

(B) statistically independent

(C) uncorrelated

(E) orthogonal

62. A source generates three symbols with probabilities 0.25, 0.25, 0.50 at a rate of 3000 symbols per second. Assuming independent generation of symbols, the most efficient source encoder would have average bit rate as (A) 6000 bits/sec

(B) 4500 bits/sec

(C) 3000 bits/sec

(D) 1500 bits/sec

63. The diagonal clipping in Amplitude Demodulation

66. A medium of relative permitivity e r 2 = 2 forms an interface with free – space. A point source of electromagnetic energy is located in the medium at a depth of 1 meter from the interface. Due to the total internal reflection, the transmitted beam has a circular cross-section over the interface. The area of the beam cross-section at the interface is given by (A) 2 p m 2 (C)

p 2

(B) p2 m 2 (D) p m 2

m2

67. A medium is divide into regions I and II about x = 0 plane, as shown in the figure below. An electromagnetic wave with electric field E1 = 4 a$ x + 3a$ y + 5 a$ z is incident normally on the interface from region I. The electric file E2 in region II at the interface is Region I m1=mo er1=4 s1=0

(using envelope detector) can be avoided if RC time – constant of the envelope detector satisfies the following condition, (here W is message bandwidth and wc is (B) RC >

Region II m2=mo er2=4 s2=0 E2

E1

carrier frequency both in rad /sec) 1 W

C

(D) 2 -4

So that slope overload is avoided, would be

(A) RC <

y2(t)

(B) 2 -8

x( t) = 125 t( u( t) - u( t - 1)(250 - 125 t)( u( t - 1) - u( t - 2))

(C) 2 -6

Q

65. A message signal with 10 kHz bandwidth is lower side Band SSB modulated with carrier fc1 = 106 Hz frequency the resulting signal is then passed through a Narow Band Frequency Modulator with carrier frequency fc 2 = 10 9 Hz. The bandwidth of the output would be

signal (here u( t) is the unit function)

(A) 2

y1(t)

Q

Fig Q.64

(C)

-10

and are different, is

1 W

x<0

www.gatehelp.com

x=0

Fig Q.67

x>0

Page 635


UNIT 10


(A) E2 = E1

(B) 4 a$ x + 0.75 a$ y - 125 . a$ z

72. If bDC is increase by 10%, the collector – to- emitter

(C) 3a$ x + 3a$ y + 5 a$ z

(D) -3a$ x + 3a$ y + 5 a$ z

voltage drop (A) increases by less than or equal to 10%

68. When a planes wave traveling in free-space is

(B) decreases by less than or equal to 10%

incident normally on a medium having the fraction of

(C) increases by more than 10%

power transmitted into the medium is given by

(D) decreases by more than 10%

(A) 8/9

(B) 1/2

(C) 1/3

(D) 5/6

73. The small signal gain of the amplifier vc vs is

69. A rectangular wave guide having TE10 mode as

(A) 10

(B) –5.3

(C) 5.3

(D) 10

dominant mode is having a cut off frequency 18-GHz for the mode TE30 . The inner broad – wall dimension of

Common Data for Question 74, 75 :

the rectangular wave guide is

Let g( t) = p( t)* p( t) where * denotes convolution

(A) 5/3cms

(B) 5 cms

and p( t) = u( t) - u( t - 1) with u( t) being the unit step

(C) 5/2 cms

(D) 10 cms

function.

70. A mast antenna consisting of a 50 meter long

74. The impulse response of filter matched to the signal

vertical conductor operates over a perfectly conducting

s( t) = g( t) - d( t - 2)* g( t) is given as:

ground plane. It is base-fed at a frequency of 600 kHz.

(A) s(1 - t)

(B) -s(1 - t)

The radiation resistance of the antenna in Ohms is

(C) -s( t)

(D) s( t)

2 p2 (A) 5 (C)

p2 (B) 5

4 p2 5

75. An Amplitude Modulated signal is given as x AM = 100( p( t) + 0.5 g( t)) cos wc t

(D) 20 p2

In the interval. One set of possible values of the modulating signal and modulation index would be

Common Data for Question 71,72,73: In the transistor amplifier circuit show in the figure

below,

the

transistor

has

the

(B) t, 10 .

(C) t, 2.0

(D) t 2 , 0.5

following Linked Answer Question : Q.75 to Q.85 carry two

parameters:

marks each.

bDC = 60, VBE = 0.7 V, hie ® ¥, hfe ® ¥ The capacitance can be assumed to be infinite. 12 V

1 kW 5.3 kW

~

+ vC –

CC

Statement of Linked Answer Question 76 & 77: A regulated power supply, shown in figure below, has an unregulated input (UR) of 15 volts and generates a regulated output Use the component values shown in the figure.

53 kW

vS

(A) t, 0.5

Q1 15 V (UR) + 10 kW

1 kW

Fig Q.70

12 kW

71. Under the DC conditions, the collector – to- emitter voltage drop is (A) 4.8 Volts

(B) 5.3 Volts

(C) 6.0 Volts

(D) 6.6 Volts

6V

24 kW

Fig Q.76

Page 636

www.gatehelp.com

vo –


UNIT 10

ANSWER

take 400 ms for an electromagnetic wave to travel from source end to load end and vice – versa. At t = 400 ms, the voltage at the load end is found to be 40 volts. 84. The load resistance is (A) 25 Ohms

(B) 50 Ohms

(C) 75 Ohms

(D) 100 Ohms


1. C

2. D

3. A

4. C

5. A

6. B

7. C

8. B

9. A

10 .B

11. D

12. D

13. A

14. A

15. D

16. D

17.D

18. A

19. C

20. A

21. A

22. B

23. D

24. C

25. D

26. C

27. A

28. C

29. A

30. D

31. B

32. B

33. A

34. B

35. B

36. B

37. D

38. A

39. C

40. D

41. D

42. B

43. D

44. B

45. D

46. B

47. A

48. B

49. D

50. B

51. C

52. B

53. C

54. B

55. D

56. B

57. C

58. D

59.A

60. B

61. B

62. B

63. D

64. C

65. B

66. D

67. C

68. A

69. C

70. A

71. C

72. B

73. A

74. D

75. A

76. C

77. B

78. A

79.

80. C

81. B

82. C

83.C

84.

85.

85. The steady state current through the load resistance is (A) 1.2 Amps

(B) 0.3 Amps

(C) 0.6 Amps

(D) 0.4 Amps

***********

Page 638

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CHAPTER

10.7 EC-09

Q.1 - Q. 20 carry one mark each. 12 V

1. The order of the differential equation 3

d 2 y æ dy ö 4 -1 +ç ÷ + y = e is dr 2 è dt ø (A) 1

(B) 2

(C) 3

(D) 4

10 V

2. The Fourier series of a real periodic function has only P. Q. R. S.

10 min

0

Cosine terms if it is even sine terms if it is even cosine terms if it is odd sine terms if it is odd

(A) 220 J

(B) 12 kJ

(C) 13.2 kJ

(D) 14.4 kJ

5. In an n-type silicon crystal at room temperature, which of the following can have a concentration of 4 ´ 1019 cm-3?

Which of the above statements are correct?

(A) Silicon atoms

(B) Holes

(C) Dopant atoms

(D) Valence electrons

(A) P and S

(B) P and R

6. The full forms of the abbreviations TTL and CMOS in

(C) Q and S

(D) Q and R

reference to logic families are

3. A function is given by f ( t) = sin t + cos 2 t. Which of 2

the following is true ? (A) f has frequency components at 0 and 1 / 2 p Hz (B) f has frequency components at 0 and 1 / p Hz (C) f has frequency components at 1 / 2 p and 1 / p Hz

(A) Triple Transistor Logic and Chip Metal Oxide Semiconductor (B) Tristate Transistor Logic and Chip Metal Oxide Semiconductor (C) Transistor Transistor Logic and Complementary Metal Oxide Semiconductor

(D) f has frequency components at 0.1 / 2 p and 1 / p Hz

(D) Tristate Transistor Logic and Complementary Metal Oxide Silicon

4. A fully charged mobile phone with a 12 V battery is

7. The ROC of Z-transform of the discrete time sequence

good for a 10 minute talk-time. Assume that, during the

æ1ö æ1ö x( n) = ç ÷ u( n) - ç ÷ u( -n - 1) is è 3ø è2 ø 1 1 1 (B) z > (A) < z < 3 2 2

talk-time, the battery delivers a constant current of 2 A and its voltage drops linearly from 12 V to 10 V as shown in the figure. How much energy does the battery deliver during this talk-time ? Page 662

www.gatehelp.com

p

n


EC-09

(C) z <

1 3

Chap 10.7

11. A fair coin is tossed 10 times. What is the

(D) 2 < z < 3

probability that ONLY the first two tosses will yield

8. The magnitude plot of a rational transfer function G( s) with real coefficients is shown below. Which of the following compensators has such a magnitude plot ?

heads ? æ1ö (A) ç ÷ è2 ø

2

(B)

10

(D)

10

2

10

10

æ1ö (C) ç ÷ è2 ø

20 dB

æ1ö C2 ç ÷ è2 ø

æ1ö C2 ç ÷ è2 ø

12. If the power spectral density of stationary random process is a sinc-squared function of frequency, the shape of its autocorrelation is

- 40 dB

(A)

(A) Lead compensator

(B)

(B) Lag compensator (C) PID compensator| (D) Lead-lag compensator 9. A white noise process X ( t) with two-sided power

(C)

(D)

spectral density 1 ´ 10 -10 W/Hz is input to a filter whose magnitude squared response is shown below. 1

-10 kHz

10 kHz

13. If f ( z) = c0 + c1 z -1 , then

f

The power of the output process Y ( t) is given by (A) 5 ´ 10 -7 W

(B) 1 ´ 10 -6 W

(C) 2 ´ 10 -6 W

(D) 1 ´ 10 -5 W

1 + f ( z) dz is given by z unit circle

ò

(A) 2 pc1

(B) 2 p(1 + c0 )

(C) 2 pjc1

(D) 2 pj(1 + c0 )

14. In the interconnection of ideal sources shown in the figure, it is known that the 60 V source is absorbing

10. Which of the following statements is true regarding

power.

the fundamental mode of the metallic waveguides

Which of the following can be the value of the current source I ?

shown ?

20 V .P: Coaxial

Q: Cylindrical I

60 V

12 A

R: Rectangular

(A) Only P has no cutoff-frequency (B) Only Q has no cutoff-frequency

(A) 10 A

(B) 13 A

(C) Only R has no cutoff-frequency

(C) 15 A

(D) 18 A

(D) All three have cut-off frequencies

15. The ratio of the mobility to the diffusion coefficient in a semiconductor has the units www.gatehelp.com

Page 663


UNIT 10

(A) V -1 (C) V.cm

-1


(B) cm. V -1

20. Two infinitely long wires carrying current are as

(D) V.s

shown in the figure below. One wire is in the y - z plane and parallel to the y -axis. The other wire is in the x - y

16. In a microprocessor, the service routine for a certain

plane and parallel to the x -axis. Which components of

interrupt starts from a fixed location of memory which

the resulting magnetic field are non-zero at the origin ?

cannot be externally set, but the interrupt can be

z

delayed or rejected Such an interrupt is

1A

(A) non-maskable and non-vectored (B) maskable and non-vectored (C) non-maskable and vectored

y

(D) maskable and vectored 17. If the transfer function of the following network is V0 ( s) 1 the value of the load resistance RL is = Vi ( s) 2 + sCR +

+

1A x

(A) x, y, z components

(B) x, y components

(C) y, z components

(D) x, z components

Q. 21 to Q. 60 carry two marks each. –

–

(A) R/4

(B) R/2

(C) R

(D) 2 R

18. Consider the system

21. Consider two independent random variables X and Y with identical distributions. The variables X and Y 1 1 1 take values 0, 1 and 2 with probabilities , and 2 4 4 respectively. What is the conditional probability

é1 0 ù dx = Ax + Bu with A = ê ú dt ë0 1 û

é pù and B = ê ú where p and q are arbitrary real numbers. ëq û Which

of

the

following

statements

about

P ( X + Y = 2 X - Y = 0) ? (A) 0 (C)

the

(B)

1 6

1 16

(D) 1

controllability of the system is true ?

22. The Taylor series expansion of

(A) The system is completely state controllable for any nonzero values of p and q

by

sin x at x = p is given x-p

(A) 1 +

( x - p) 2 +... 3!

(B) -1 -

( x - p) 2 +... 3!

(C) The system is uncontrollable for all values of p and q

(C) 1 -

( x - p) 2 +... 3!

(D) -1 +

( x - p) 2 +... 3!

(D) We cannot conclude about controllability from the given data

23. If a vector field V is related to another vector field A

(B) Only p = 0 and q = 0 result in controllability

®

®

®

®

through V = Ñ ´ A, which of the following is true? Note : 19. For a message signal m( t) = cos(2pfm t) and carrier of

C and SC refer to any closed contour and any surface

frequency fc , which of the following represents a single

whose boundary is C.

side-band (SSB) signal ?

(A)

(A) cos(2 pfm t) cos(2 pfc t) (B) cos(2pfc t) (C) cos[2p( fc + fm ) t ]

®

®

C

®

(B)

SC

®

®

C

www.gatehelp.com

SC

®

®

®

®

®

ò A. dl = òò V . dS C

®

(C) ò Ñ ´ V . dl = òò Ñ ´ A. dS

(D) [1 + cos(2 pfm t) cos(2 pfc t)] Page 664

®

ò V . dl = òò A. dS

SC

®

®

®

®

(D) ò Ñ ´ A. dl = òò V . dS C

SC


EC-09

Chap 10.7

24. Given that F ( s) is the one-sided Laplace transform of f ( t), the Laplace transform of (A) sF ( s) - f (0) (C)

ò

s

0

(B)

ò

t

0

f ( t) dt is

1 F ( s) s

100 V

1 (D) [ F ( s) - f (0)] s

F ( t) dt

25. Match each differential equation in Group I to its

(A) 0.2 e -125t u( t) mA

(B) 20 e -1250 t u( t) mA

family of solution curves from Group II.

(C) 0.2 e -1250 t u( t) mA

(D) 20 e -1000 t u( t) mA

P. Q. R. S.

Group 1 dy y = dx x dy y =dx x dy x = dx y

Group II 29. In the circuit shown, what value of RL maximizes

1. Circles

the power delivered to RL ?

2. Straight lines 3. Hyperbolas

dy x =dx y

100 V

(A) P - 2, Q - 3, R - 3, S - 1 (B) P - 1, Q - 3, R - 2, S - 1 (C) P - 2, Q - 1, R - 3, S - 3 (D) P - 3, Q - 2, R - 1, S - 2 26. The eigen values of the following matrix are

8 W 3

(A) 2.4 W

(B)

(C) 4 W

(D) 6 W

30. The time domain behavior of an RL circuit is

é-1 3 5 ù ê-3 -1 6 ú ê ú êë 0 0 3 úû

represented by di + Ri= V0 (1 + Be - Rt / L sin t) u( t). dt V For an initial current of i(0) = 0 , the steady state R value of the current is given by V 2 V0 (B) i( t) ® (A) i( t) ® 0 R R L

(A) 3, 3 + 5 j, 6 - j

(B) -6 + 5 j, 3 + j, 3 - j

(C) 3 + j, 3 - j, 5 + j

(D) 3, - 1 + 3 j, - 1 - 3 j

27. An AC source of RMS voltage 20 V with internal impedance Z s = (1 + 2 j)W feeds a load of impedance Z L = (7 + 4 j)W in the figure below. The reactive power

(C) i( t) ®

consumed by the load is

V0 (1 + B) R

(D) i( t) ®

2 V0 (1 + B) R

31. In the circuit below, the diode is ideal. The voltage V is given by

1A

(A) 8 VAR

(B) 16 VAR|

(C) 28 VAR

(D) 32 VAR

28. The switch in the circuit shown was on position a for a long time, and is move to position b at time t = 0.

(A) min( Vi ,1)

(B) max( Vi 1)

(C) min( -Vi ,1)

(C) max( -Vi ,1)

The current i( t) for t > 0 is given by www.gatehelp.com

Page 665


EC-09

(A) NAND: first (0,1) then (0,1) NOR: first (1,0) then (0,0)

Chap 10.7

(

(B) NAND: first (1,0) then (1,0) NOR: first (1,0) then (1,0)

0

(C) NAND: first (1,0) then (1,0) NOR: first (1,0) then (0,0) (D) NAND: first (1,0) then (1,1) NOR: first (0,1) then (0,1)

A) Causal, LP

(B) BIBO, LTI

(C) BIBO, Causal, LTI

(D) LP, LTI

39. What are the counting states (Q1 , Q2 ) for the counter

42. The 4-point Discrete Fourier Transform (DFT) of a

shown in the figure below?

discrete time sequence {1,0,2,3} is Q

Q

Q JK Flip Flop

Clock

(A) [0,-2+2j,2,-2-2j]

(B) [2,2+2j,6,2-2j]

(C) [6,1-3j,2,1+3j]

(D) [6-1+3j,0,-1,-3j]

JK Flip Flop

43. The feedback configuration and the pole-zero

1

K

locations of G( s) = (A) 11, 10, 00, 11, 10,...

(B) 01, 10, 11, 00, 01,...

(C) 00, 11, 01, 10, 00,...

(D) 01, 10, 00, 01, 10,...

s2 - 2 s + 2 are shown below. The root s2 + 2 s + 2

locus for negative values of k, i.e. for -¥ < k < 0, has breakaway/break-in points and angle of departure at pole P (with respect to the positive real axis) equal to Im (s)

40. A system with transfer function H ( z) has impulse

+

response h(.) defined as h(2) = 1, h( 3) = -1 and h( k) = 0

+ -

otherwise. Consider the following statements.

k

G(s)

O

X

Q

P

Re (s)

S1: H ( z) is a low-pass filter. S2: H ( z) is an FIR filter. Which of the following is correct? (A) Only S2 is true (B) Both S1 and S2 are false

(A) ± 2 and 0°

(B) ± 2 and 45°

(C) ± 3 and 0°

(D) ± 3 and 45°

44. An LTI system having transfer function

s2 + 1 s2 + 2 s + 1

(C) Both S1 and S2 are true, and S2 is a reason for S1

and input x( t) = sin( t + 1) is in steady state. The output

(D) Both S1 and S2 are true, but S2 is not a reason for S1

{ x( k)}. Which of the following is true?

41. Consider a system whose input x and output y are related by the equation

where h( t) is shown in the graph. the

following

(B) y(.) is nonzero for all sampling frequencies ws (D) y(.) is zero for ws > 2, but nonzero for w2 < 2

-¥

of

(A) y(.) is zero for all sampling frequencies ws (C) y(.) is nonzero for ws > 2, but zero for ws < 2

¥

y( t)= ò x( t - t) h(2 t) dt

Which

is sampled at a rate ws rad/s to obtain the final output

four

45. The unit step response of an under-damped second

properties

are

possessed by the system ? BIBO: Bounded input gives a bounded output. Causal: The system is causal,

order system has steady state value of -2. Which one of the following transfer functions has these properties ? -2.24 -3.82 (A) 2 (B) 2 s + 2.59 s + 112 . s + 191 . s + 191 . (C)

LP: The system is low pass.

-2.24 s - 2.59 s + 112 . 2

(D)

-382 s - 191 . s + 191 . 2

LTI: The system is linear and time-invariant.

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Page 667


UNIT 10

46. A discrete random variable X takes values from 1 to 5 with probabilities as shown in the table. A student

®

(D) j =


ö 1 B0 z$ æ ç ÷, r ¹ 0 m 0 çè x 2 + y 2 ÷ø

calculates the mean of X as 3.5 and her teacher calculates the variance of X as 1.5. Which of the

50. A transmission line terminates in two branches,

following statements is true ?

each of length l / 4, as shown. The branches are terminated by 50W loads. The lines are lossless and

k

1

2

3

4

5

P ( X = k)

0.1

0.2

0.4

0.2

0.1

have the characteristic impedances shown. Determine the impedance Z i as seen by the source.

(A) Both the student and the teacher are right (B) Both the student and the teacher are wrong (C) The student is wrong but the teacher is right (D) The student is right but the teacher is wrong Q. 47 A message signal given by æ1ö æ1ö 1n( t)= ç ÷ cos w1 t - ç ÷ sin w2 t è2 ø è2 ø

(A) 200W

(B) 100W

(C) 50W

(D) 25W

Common Date Questions

is amplitude-modulated with a carrier of frequency

Common Date for Questions 51 and 52:

wc to generate

Consider s( t)= [1 + m( t)]cos wc t

modulation scheme ? (B) 11.11%

(C) 20%

(D) 25%

the resulting capacity C2 is given by (D) C2 = C1 + 0.3B

®

2 B0 z$ æ çç 2 m 0 è x + y2

ö ÷÷, r ¹ 0 ø

Permittivity of free space = 8.85 ´ 10 -14 F.cm -1 Dielectric constant of silicon = 12 51 The built-in potential of the junction

(D) cannot be estimated from the data given 52. The peak electric field in the device is (A) 0.15 MV . cm -1 , directed from p-region to n-region

[ Hint : The algebra is trivial in cylindrical coordinates.]

(B) j = -

Depletion width on the n-side = 0.1 mm

(C) is 0.82

What current distribution leads to this field ? ö 1 B0 z$ æ ç ÷, r ¹ 0 m 0 çè x 2 + y 2 ÷ø

Doping on the n-side = 1 ´ 1017 cm -3

(B) is 0.76 V

® æ ö x y B= B0 çç 2 y$ - 2 x$ ÷÷ 2 2 x +y ø èx +y

®

(B) 0.15 MV . cm -1 , directed from n-region to p-region (C) 1.80 MV . cm -1 , directed from p-region to n-region (D) 1.80 MV . cm -1 , directed from n-region to p-region Common Data for Questions 53 and 54:

®

(C) j = 0, r ¹ 0 Page 668

room

(A) is 0.70 V

49. A magnetic field in air is measured to be

(A) j =

at

Thermal voltage = 26 mV

capacity C1 . If the SNR is doubled keeping B constant,

(C) C2 = C1 + 2 B

junction

Intrinsic carrier concentration = 1.4 ´ 1010 cm -3

a signal to noise ratio SNR >> 1 and bandwidth B has

(B) C2 = C1 + B

p-n

Depletion width on the p-side = 10 . mm

48. A communication channel with AWGN operating at

(A) C2 = 2 C1

silicon

temperature having the following parameters:

What is the power efficiency achieved by this (A) 8.33%

a

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EC-09

Chap 10.7

The Nyquist plot of a stable transfer function G( s)

Consider the CMOS circuit shown, where the gate

is shown in the figure. We are interested in the stability

voltage VG of the n-MOSFET is increased from zero,

of the closed loop system in the feedback configuration

while the gate voltage of the p-MOSFET is kept

shown.

constant at 3 V. Assume that, for both transistors, the

Im

magnitude of the threshold voltage is 1 V and the product of the trans-conductance parameter and the +

-1-0.5

Re

+

(W/L) ratio, i.e. the quantity mCax (W / L), is 1 mA . V -2 .

G(s) -

-j

53. Which of the following statements is true? (A) G( s) is an all-pass filter (B) G( s) has a zero in the right-half plane (C) G( s) is the impedance of a passive network 57. For small increase in VG beyond 1 V, which of the

(D) G( s) is marginally stable

following gives the correct description of the region of 54. The gain and phase margins of G( s) for closed loop

operation of each MOSFET ?

stability are

(A) Both the MOSFETs are in saturation region

(A) 6 dB and 180°

(B) 3 dB and 180°

(B) Both the MOSFETs are in triode region

(C) 6 dB and 90°

(D) 3 dB and 90°

(C) n-MOSFET is in triode and p-MOSFET is in saturation region

Common Data for Questions 55 and 56: The amplitude of a random signal is uniformly

(D) n-MOSFET is in saturation and p-MOSFET is in triode region.

distributed between -5 V and 5 V. 55. If the signal to quantization noise ratio required in

58. Estimate the output voltage V0 for VG = 15 . V.

uniformly quantizing the signal is 43.5 dB, the step size

[Hint : Use the appropriate current-voltage equation for

of the quantization is approximately

each MOSFET, based on the anser to Q. 57.]

(A) 0.0333 V

(B) 0.05 V

(C) 0.0667 V

(D) 0.10 V

Statement for Linked Answer Questions 59 and 60:

56. If the positive values of the signal are uniformly quantized with a step size of 0.05 V, and the negative

Two products are sold from a vending machine,

values are uniformly quantized with a step size of 0.1 V,

which has two push buttons P1 and P2 . When a button is

the resulting signal to quantization noise ratio is

pressed, the price of the corresponding product is

approximately

displayed in a 7-segment display.

(A) 46 dB

(B) 43.8 dB

(C) 42 dB

(D) 40 dB

If no buttons are pressed, '0' is displayed, signifying 'Rs. 0'. If only P1 is pressed, '2' is displayed, signifying 'Rs.

Linked Answer Questions Statement for Linked Answer Questions 57 and 58

2'. If only P2 is pressed, '5' is displayed, signifying 'Rs.

: 5'.

If both P1 and P2 are pressed, 'E' is displayed, signifying 'Error'.

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Page 669

MULTIPLE CHOICE QUESTION Electronics & Communication Engineering Fifth Edition

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