Gate 2014 Ece

EC-GATE-2014 PAPER-01|

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Q. No. 1 – 5 Carry One Mark Each 1.

Choose the most appropriate phrase from the options given below to complete the following sentence. The aircraft_______ take off as soon as its flight plan was filed. (A) is allowed to (B) will be allowed to (C) was allowed to (D) has been allowed to Answer: (C) Read the statements: All women are entrepreneurs. Some women are doctors Which of the following conclusions can be logically inferred from the above statements? (A) All women are doctors (B) All doctors are entrepreneurs (C) All entrepreneurs are women (D) Some entrepreneurs are doctors Answer: (D)

2.

3.

Choose the most appropriate word from the options given below to complete the following sentence. Many ancient cultures attributed disease to supernatural causes. However, modern science has largely helped _________ such notions. (A) impel (B) dispel (C) propel (D) repel Answer: (B) 4.

The statistics of runs scored in a series by four batsmen are provided in the following table, Who is the most consistent batsman of these four? Batsman

Average

Standard deviation

K

31.2

5.21

L

46.0

6.35

M

54.4

6.22

N

17.9

5.90

(A) K (B) L (C) M (D) N Answer: (A) Exp: If the standard deviation is less, there will be less deviation or batsman is more consistent 5.

What is the next number in the series? 12 35 81 Answer: 725

173

357

____

India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 1

EC-GATE-2014 PAPER-01| Exp:

12

35

23

81

46

173

92

357

184

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________

368

difference

357 ⇒ 368 725


Find the odd one from the following group: W,E,K,O

I,Q,W,A

(A) W,E,K,O

F,N,T,X

(B) I,Q,W,A

N,V,B,D

(B) F,N,T,X

(D) N,V,B,D

Answer: (D) Exp:

W

E

8

K

6

O

4

1

Q

8

W

6

A

4

F

N 8

T 6

X 4

N

V 8

B 6

D 2

Difference of position: D 7.

For submitting tax returns, all resident males with annual income below Rs 10 lakh should fill up Form P and all resident females with income below Rs 8 lakh should fill up Form All people with incomes above Rs 10 lakh should fill up Form R, except non residents with income above Rs 15 lakhs, who should fill up Form S. All others should fill Form T. An example of a person who should fill Form T is (A) a resident male with annual income Rs 9 lakh (B) a resident female with annual income Rs 9 lakh (C) a non-resident male with annual income Rs 16 lakh (D) a non-resident female with annual income Rs 16 lakh

Answer: (B) Exp:

Resident female in between 8 to 10 lakhs haven’t been mentioned.

8.

A train that is 280 metres long, travelling at a uniform speed, crosses a platform in 60 seconds and passes a man standing on the platform in 20 seconds. What is the length of the platform in metres?

Answer: 560 Exp:

For a train to cross a person, it takes 20 seconds for its 280m. So, for second 60 seconds. Total distance travelled should be 840. Including 280 train length so length of plates =840-280=560


EC-GATE-2014 PAPER-01| 9.

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The exports and imports (in crores of Rs.) of a country from 2000 to 2007 are given in the following bar chart. If the trade deficit is defined as excess of imports over exports, in which year is the trade deficit 1/5th of the exports? 120 110 100 90 80 70 60 50 40 30 20 10 0

(A) 2005

Exports

2000

2001

2002

(B) 2004

Im ports

2003

2004

(C) 2007

2005

2006

2007

(D) 2006

Answer: (D) Exp:

2004,

imports − exp orts 10 1 = = exp orts 70 7

26 2 = 76 7 20 1 2 0 0 6, = 100 5 10 1 2007, = 100 11

2 0 0 5,

10.

You are given three coins: one has heads on both faces, the second has tails on both faces, and the third has a head on one face and a tail on the other. You choose a coin at random and toss it, and it comes up heads. The probability that the other face is tails is (A) 1/4

(B) 1/3

(C) 1/2

(D) 2/3

Answer: (B)



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For matrices of same dimension M, N and scalar c, which one of these properties DOES NOT ALWAYS hold? (A) (MT)T = M (B) (cMT)T = c(M)T (D) MN = NM (C) (M + N)T = M T + NT Answer: (D) Exp: Matrix multiplication is not commutative in general. 2.

In a housing society, half of the families have a single child per family, while the remaining half have two children per family. The probability that a child picked at random, has a sibling is _____ Answer: 0.667 Exp:

Let E1 = one children family

E 2 = two children family and A = picking a child then by Baye’s theorem, required probability is 1 .x 2 E 2 = = = 0.667 P 2 A 1 x 1 . + .x 3 2 2 2 (Here ‘x’ is number of families)

( )

3.

 z2 − z + 4 j  C is a closed path in the z-plane given by z = 3. The value of the integral → ∫C  z + 2 j  dz is (A) −4π (1 + j2 )

(B) 4 π ( 3 − j2 )

(C) −4π ( 3 + j2 )

(D) 4 π (1 − j2 )

Answer: (C) Exp:

Z = −2 j is a singularity lies inside C : Z = 3 ∴ By Cauchy’s integral formula,

Z2 − Z + 4 j 2 ∫ C Z + 2 j dz = 2πj.  Z − Z + 4 j Z=−2 j = 2πj[ −4 + 2 j + 4 j] = −4π [3 + j2] A real (4 × 4) matrix A satisfies the equation A2 = I, where I is the (4 × 4) identity matrix. The positive eigen value of A is __________. Answer: 1 4.

Exp:

1 is also its eigen value. Since, we λ require positive eigen value. ∴λ = 1 is the only possibility as no other positive number is self inversed

A 2 = I ⇒ A = A −1 ⇒ if λ is on eigen value of A then



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Let X1, X2, and X3 be independent and identically distributed random variables with the uniform distribution on [0, 1]. The probability P{X1 is the largest} is ________

Answer: 0.32-0.34 6.

For maximum power transfer between two cascaded sections of an electrical network, the relationship between the output impedance Z1 of the first section to the input impedance Z2 of the second section is (A) Z2 = Zl

(C) Z2 = Z1∗

(B) Z2 = − Zl

(D) Z2 = − Z1∗

Answer: (C) Exp: Two cascaded sections

Section 1

Section Z1 Z L Z 2

2

Z1 = Output impedance of first section

Z2 = Input impedance of second section For maximum power transfer, upto 1st section is

ZL = Z1* ZL = Z2 ⇒ Z1* 7.

Consider the configuration shown in the figure which is a portion of a larger electrical network i5 i2

R R

i4 i1

R i3 i6

For R = 1Ω and currents i1 = 2A, i4 = -1A, i5 = -4A, which one of the following is TRUE? (A) i6 = 5 A (B) i 3 = −4A (C) Data is sufficient to conclude that the supposed currents are impossible (D) Data is insufficient to identify the current i 2 ,i3 , and i 6 Answer: (A) India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 5


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Exp: Given i1 = 2A

i5

i 4 = −1A

B

i2

i5 = − 4A

1Ω

1Ω

KCL at node A, i1 + i 4 = i 2 ⇒ i 2 = 2 − 1 =1A

i3

1. KCL at node B, i 2 + i5 = i3

i4

⇒ i3 =1 − 4 = − 3A

A

KCL at node C, i3 + i6 = i1

i1

1Ω

C i6

⇒ i6 = 2 − ( −3) = 5A

8.

When the optical power incident on a photodiode is 10µW and the responsivity is 0.8 A / W, the photocurrent generated ( in µA ) is ________.

Answer: 8 Exp: Responsivity ( R ) =

0.8 =

Ip P0

Ip

10 × 10−6 ⇒ I8 = 8µA

9.

In the figure, assume that the forward voltage drops of the PN diode D1 and Schottky diode D2 are 0.7 V and 0.3 V, respectively. If ON denotes conducting state of the diode and OFF denotes non-conducting state of the diode, then in the circuit, 1kΩ

10 Ω

(A) both D1 and D2 are ON (C) both D1 and D2 are OFF Answer: (D) Exp: Assume both the diode ON. Then circuit will be as per figure (2) 10 − 0.7 ∴I = = 9.3mA 1k 0.7 − 0.3 I D2 = = 20mA 20 Now, I D1 = I − ID2

20 Ω

D2

D1

(B) D1 is ON and D2 is OFF (D) D1 is OFF and D2 is ON

D1

10V

D2

Figure (1) 1K

= −10.7 mA ( Not possible ) ∴ D1is OFF and hense D 2 − ON

20Ω

1K

20Ω

I ID1

10V

0.7V

ID2 0.3V



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10.

If fixed positive charges are present in the gate oxide of an n-channel enhancement type MOSFET, it will lead to (A) a decrease in the threshold voltage (B) channel length modulation (C) an increase in substrate leakage current (D) an increase in accumulation capacitance Answer: (A) A good current buffer has (A) low input impedance and low output impedance (B) low input impedance and high output impedance (C) high input impedance and low output impedance (D) high input impedance and high output impedance Answer: (B) 11.

Exp:

Ideal current Buffer has Zi = 0

Z0 = ∞ 12.

In the ac equivalent circuit shown in the figure, if i in is the input current and RF is very large, the type of feedback is

RD

RD

υout

M2 M1 small signal input i in

RF

(A) voltage-voltage feedback (B) voltage-current feedback (C) current-voltage feedback (D) current-current feedback Answer: (B) Exp: Output sample is voltage and is added at the input or current ∴ It is voltage – shunt negative feedback i.e, voltage-current negative feedback 13.

In the low-pass filter shown in the figure, for a cut-off frequency of 5kHz, the value of R2 ( in kΩ ) is ____________. R2

C Vi

1kΩ R1

−

+

10nF Vo



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Answer: 3.18 Exp:

f = 5KHz Cut off frequency ( LPF ) = ⇒ R2 =

14.

1 = 5KHz 2πR 2 C

1 = 3.18 kΩ 2π × 5 × 103 × 10 × 10−9

In the following circuit employing pass transistor logic, all NMOS transistors are identical with a threshold voltage of 1 V. Ignoring the body-effect, the output voltages at P, Q and R are, 5V

5V

5V

5V P

(A) 4 V, 3 V, 2 V (C) 4 V, 4 V, 4 V Answer: (C) Exp: Assume al NMOS are in saturation

Q

R

(B) 5 V, 5 V, 5 V (D) 5 V, 4 V, 3 V 5V

∴ VDS ≥ ( VGS − VT ) 5V

For m1

M1

(5 − Vp ) ≥ ( 5 − Vp − 1)

(5 − V ) > ( 4 − V ) ⇒ Sat p

P

p

∴ ID1 = k ( VGS − VT )

5V

M2

2

ID1 = K ( 4 − Vp ) ........ (1) 2

Q 5V

M3

For m 2 , I D1 = K ( 5 − VQ − 1)

For m 3 ,

2

I D2 = K ( 4 − VQ ) ......( 2 ) 2

∴ ID1 = ID2

(4 − V ) = (4 − V )

2

Q

⇒ Vp = VQ & Vp + VQ = 8

⇒ Vp = VQ = 4V

2

R

∴ I D 2 = I D3

(4 − V ) = (4 − V ) 2

2

p

I D3 = K ( 5 − VR − 1)

Q

2

R

⇒ VR = VQ = 4V ∴ Vp = VQ = VR = 4V



(

) (

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)

The Boolean expression ( X + Y ) X + Y + X + Y + X simplifies to

(A) X Answer: (A)

(B) Y

(C) XY

(D) X+Y

Exp: Given Boolean Expression is ( X + Y ) ( X + Y ) + XY + X As per the transposition theorem

( A + BC ) = ( A + B )( A + C ) so, ( X + Y ) ( X + Y ) = X + YY

( X + Y ) ( X + Y ) + XY + X

= X+0

( )

= X + XY .X

= X + ( X + Y ) .X = X + XX. + Y.X = X + 0 + Y.X Apply absorption theorem = X (1 + Y ) = X.1= X

16.

Five JK flip-flops are cascaded to form the circuit shown in Figure. Clock pulses at a frequency of 1 MHz are applied as shown. The frequency (in kHz) of the waveform at Q3 is __________ .

1

J4 Q4 > clk 1 K4

1 J3 Q3 > clk 1 K2

1

J2 Q2 > clk 1 K2

1 J1 Q1 > clk 1 K1

1 J0 > clk 1 K0

clock

Answer: 62.5 Exp: Given circuit is a Ripple (Asynchrnous) counter. In Ripple counter, o/p frequency of each flip-flop is half of the input frequency if their all the states are used otherwise o/p frequency input frequency of the counter is = modulus of the counter input frequency 16 6 1×10 = H z = 62.5 kHz 16

So, the frequency at Q3 =

17.

A discrete-time signal x [ n ] = sin ( π2 n ) ,n being an integer,is (A) periodic with period π .

(C) periodic with period π / 2 . Answer: (D)

(B) periodic with period π 2 . (D) not periodic



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Assume x [ n ] to be periodic, (with period N)

⇒ x [n ] = x [n + N]

⇒ sin ( π 2 n ) = sin ( π2 ( n + N ) ) Every frigonometric function repeate after 2π interval. ⇒ sin ( π2 n + 2πk ) = sin ( π 2 h + π 2 N )

 2k  ⇒ 2πk = π2 N ⇒ N =    π  Since ‘k’ is any integer, there is no possible value of ‘k’ for which ‘N’ can be an integer, thus non-periodic. 18.

Consider two real valued signals, x(t) band-limited to [ −500 Hz, 500 Hz ] and y ( t ) band-

limited to [ −1kHz, 1kHz ] . For z ( t ) = x ( t ) . y ( t ) , the Nyquist sampling frequency (in kHz) is __________ Answer: 3 Exp:

x ( t ) is band limited to [ −500Hz, 500Hz ] y ( t ) is band limited to [ −1000Hz, 1000Hz ] z ( t ) = x ( t ) .y ( t )

Multiplication in time domain results convolution in frequency domain. The range of convolution in frequency domain is [ −1500Hz, 1500 Hz ] So maximum frequency present in z(t) is 1500Hz Nyquist rate is 3000Hz or 3 kHz 19.

A continuous, linear time-invariant filter has an impulse response h(t) described by 0≤t ≤3 h ( t ) = { 30 for otherwise

When a constant input of value 5 is applied to this filter, the steady state output is _______. Answer: 45 x (t ) Exp: y (t ) h (t ) y(t) = x (t)* h (t) x(t) =

5 t

h (t) = 3

3

t

3

y ( t ) = ∫ 3.5.dτ = 45 ( steady state output ) 0



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The forward path transfer function of a unity negative feedback system is given by G ( s) =

K ( s + 2 )( s − 1)

The value of K which will place both the poles of the closed-loop system at the same location, is ______. Answer: 2.25 Exp:

Given G ( s ) =

K s + 2 ( )( s − 1)

H (s) = 1 Characteristic equation: 1 + G ( s ) H ( s ) = 0

1+ The poles are s1,2 = − 1 ± If

( s + 2 )( s − 1)

= 0

9 − 4K 4

9 − K = 0, then both poles of the closed loop system at the same location. 4

So, K =

21.

K

9 ⇒ 2.25 4

Consider the feedback system shown in the figure. The Nyquist plot of G(s) is also shown. Which one of the following conclusions is correct? +

k −

Im G ( jω )

G (s) −1

+1 Re G ( jω )

(A) G(s) is an all-pass filter (B) G(s) is a strictly proper transfer function (C) G(s) is a stable and minimum-phase transfer function (D) The closed-loop system is unstable for sufficiently large and positive k Answer: ( D) Exp:

For larger values of K, it will encircle the critical point (-1+j0), which makes closed-loop system unstable.



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22.

In a code-division multiple access (CDMA) system with N = 8 chips, the maximum number of users who can be assigned mutually orthogonal signature sequences is ________ Answer: 7.99 to 8.01 Exp:

Spreading factor(SF)=

chip rate symbol rate

This if a single symbol is represented by a code of 8 chips Chip rate =80×symbol rate S.F (Spreading Factor) =

8 × symbol rate =8 symbol rate

Spread factor (or) process gain and determine to a certain extent the upper limit of the total number of uses supported simultaneously by a station. 23.

The capacity of a Binary Symmetric Channel (BSC) with cross-over probability 0.5 is ________ Answer: 0 Exp: Capacity of channel is 1-H(p) H(p) is entropy function With cross over probability of 0.5 1 1 1 1 H ( p ) = log 2 + log 2 =1 2 0.5 2 0.5 ⇒ Capacity = 1 − 1 = 0

24.

S12  S A two-port network has sattering parameters given by [ S] =  11  . If the port-2 of the S21 S22  two-port is short circuited, the S11 parameter for the resultant one-port network is

( A)

s11 − s11 s22 + s12s21 1 + s22

( B)

s11 − s11 s22 − s12s21 1 + s22

( C)

s11 − s11 s22 + s12s21 1 − s22

( D)

s11 − s11 s22 + s12s21 1 − s22

Answer:(B) Exp: a1

Two port Network

b1

a2 b2

b1 = s11a1 + s12 a 2 b 2 = s 21a1 + s 22 a 2  b1   s11 s12   a1   b  = s   ;  2   21 s 22  a 2 

s1 =

b1 a1

a 2 =0

By verification Answer B satisfies. India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 12


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The force on a point charge +q kept at a distance d from the surface of an infinite grounded metal plate in a medium of permittivity ∈ is (A) 0 q2 towards the plate 16π ∈ d 2

(C)

(B)

q2 away from the plate 16π ∈ d 2

(D)

q2 towards the plate 4π ∈ d 2

Answer:(C) Exp:

+q

1 Q1Q 2 F= 4π∈ R 2 1 92 92 F= = 4π ∈ ( 2d ) 2 16π∈ d 2

d

metal plate

Since the charges are opposite polarity

d

the force between them is attractive. −q

Q.No. 26 – 55 Carry Two Marks Each 26.

The Taylor series expansion of 3 sin x + 2 cos x is

( A ) 2 + 3x − x 2 −

x3 + ....... 2

( B) 2 − 3x + x 2 −

x3 + ....... 2

( C) 2 + 3x + x 2 +

x3 + ....... 2

( D ) 2 − 3x − x 2 +

x3 + ....... 2

Answer: (A) Exp:

   x2  x3 + ...  3sin x + 2cos x = 3  x − + ...  + 2 1 − 3! 2!     = 2 + 3x − x 2 −

27.

For a Function g(t),

x3 + ... 2

it is given that

t

+∞

−∞

−∞

y ( t ) = ∫ g ( τ ) dτ, then ∫

(A)0

∫

+∞

−∞

g ( t ) e − jωt dt = ωe −2 ω

2

for any real value ω . If

y ( t ) dt is

(B)-j

(C) -

j 2

(D)

j 2

Answer: (B) Exp:

Given ∞

∫ g ( t ).e

− jwt

dt = ω.e−2w ( let G ( jω) ) 2

−∞

∞

⇒

∫ g ( t ) dt = 0

−∞


EC-GATE-2014 PAPER-01| y(t) =

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t

∫ g ( z ).dz ⇒ y ( t ) = g ( t ) * u ( t ) u ( t ) in unit step function 

−∞

⇒ Y ( jω) = G ( jω) .U ( jω)

Y ( jω) =

∞

∫ y ( t ).e

− jω t

dt

−∞

∞

⇒ Y ( j0 ) =



∫ y ( t ) dt = ω.e

−2w 2

−∞

1   jω + πδ ( ω)   ω = 0  

1 = = −j j 28.

The volume under the surface z(x, y) = x + y and above the triangle in the x-y plane defined by {0 ≤ y ≤ x and 0 ≤ x ≤ 12} is___________.

Answer: 864 Exp:

Volume = ∫∫ Z ( x, y ) dydx = R

12

x

∫ ∫ ( x + y ) dydx

x =0 y =0

x

12

12  y2  3 3  x3  = ∫  xy +  .dx = ∫ x 2 dx =   = 864 2 0 2 2  3 0 x =0  0 12

29.

Consider the matrix:

0 0 0 0 01

J6 =

0 0 0 0 10 0 0 0 1 00 0 0 10 0 0 0 1 0 0 0 0 1 0 0 0 0 0

Which is obtained by reversing the order of the columns of the identity matrix I 6 . Let P = I 6 + αJ 6 , where α is a non-negative real number. The value of α for which det(P) = 0 is ___________. Answer: 1 Exp:

1 0   0 1 1 α  Consider, ( i ) Let P = I 2 + αJ 2 =  + α  =  0 1  1 0   α 1 

⇒ P = 1 − α2 1 0 0 α  0 1 α 0   ( ii ) Let P = I4 + α J 4 =  0 α 1 0   α 0 0 1



1

α

0

0

1

α

P = (1 ) α 0

1 0

0 − (α ) 0 α 1

α 0

1 0

= (1 − α 2 ) − ( α )  α (1 − α 2 )  = (1 − α 2 )

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2

S im ilarly , if P = I 6 + α J 6 th e n w e g et

P = (1 − α 2 )

3

∴ P = 0 ⇒ α = − 1, 1 ∵ α is n o n n e g ativ e ∴α =1 30.

A Y-network has resistances of 10Ω each in two of its arms, while the third arm has a resistance of 11Ω in the equivalent ∆ − network, the lowest value ( in Ω ) among the three resistances is ______________.

Answer: 29.09Ω Exp: 10Ω

10Ω

X 10Ω

11Ω

Z

11Ω

10Ω

Y Delta Connection

Star Connection

X = 29.09Ω

y = 32Ω z = 32 Ω

X= y= z=

(10 )(10 ) + (10 )(11) + (10 )(11) 11 10 10 + 10 ( )( ) ( )(11) + (10 )(11) 10 (10 )(10 ) + (10 )(11) + (10 )(11) 10

Ω Ω Ω

i.e, lowest value among three resistances is 29.09Ω 31.

A 230 V rms source supplies power to two loads connected in parallel. The first load draws 10 kW at 0.8 leading power factor and the second one draws 10 kVA at 0.8 lagging power factor. The complex power delivered by the source is (A) (18 + j 1.5) kVA

(B) (18 - j 1.5) kVA

(C) (20 + j 1.5) kVA

(D) (20 - j 1.5) kVA



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Answer: (B) Exp: + L o a d II

L o a d I

230V

−

Load 1: P = 10 kw cos φ = 0.8

   SI = P − jQ = 10 − j7.5 KVA Q = P tan φ = 7.5 KVAR 

Load 2: S = 10 KVA

cos φ = 0.8 cos φ = 0.8 =

sin φ =

Q S

P S

P → P = 8kw 10

Q = 6KVAR

SI = P + jQ = 8 + j6

Complex power delivered by the source is SI + SII = 18 − j1.5 KVA 32.

A periodic variable x is shown in the figure as a function of time. The root-mean-square (rms) value of x is_______. x 1

0

t T/2

T/2

Answer: 0.408



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T

Exp:

x rms =

2 T  t 0≤ t ≤ 2 = T T ≤ t ≤T 0 2 

x(t)

=

2 1 x ( t ) ) dt ( ∫ T0

X

1

2 T T2  2 1 2   .t  .dt + ∫ ( 0 ) .dt  T  ∫0  T   T 2  

( 0,0)

T 2

T

t

T

=

1 4  t3  2 .   T T2  3 0

x rms =

33.

4 T3 ⇒ . 3T 3 8

1 ⇒ 0.408 6

In the circuit shown in the figure, the value of capacitor C(in mF) needed to have critically damped response i(t) is____________. 40 Ω

4H i (t)

Answer: 10mF Exp: By KVL,

di ( t )

C + − VO

1 i ( t ) dt dt C∫ Differentiate with respect to time, R .di ( t ) R di ( ti ) i ( t ) 0 = + . + = 0 dt 2 L dt LC d 2i ( t ) R di ( t ) i ( t ) + . + = 0 dt 2 L dt LC v ( t ) = Ri ( t ) + L.

+

−R 4 R ±   − L  L  LC = 2 2

D1,2

−R 1  R  ±   − 2L  2L  LC For critically damped response, 2

D1,2 =

2

1 4L  R  ⇒ C= 2 F   = LC R  2L  Given, L=4H; R= 40Ω C =

4× 4

( 40 )

2

⇒ 10mF



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A BJT is biased in forward active mode, Assume VBE = 0.7V, kT / q = 25mV and reverse saturation current IS = 10−13 A. The transconductance of the BJT (in mA/V) is ________.

Answer: 5.785 Exp:

VBE = 0.7V,

KT = 25mV, Is = 10−13 q

Transconductance, g m =

IC VT

IC = IS  eVBE /VT − 1 = 10−13 e0.7/ 25mV − 1 = 144.625mA ∴gm =

35.

IC 144.625 mA = = 5.785 A / V VT 25 mV

The doping concentrations on the p-side and n-side of a silicon diode are 1 × 1016 cm −3 and 1 × 1017 cm −3 , respectively. A forward bias of 0.3 V is applied to the diode. At T = 300K, the kT = 26mV. The electron intrinsic carrier concentration of silicon n i = 1.5 × 1010 cm −3 and q concentration at the edge of the depletion region on the p-side is (A) 2.3 × 109 cm −3

(B) 1 × 1016 cm −3

(C) 1 × 1017 cm −3

(D) 2.25 × 106 cm −3

Answer:(A) Exp:

Electron concentration, n

n i 2 Vbi /VT e NA

(1.5 × 10 ) =

10 2

e0.3/26mV 1 × 1016 = 2.3 × 109 / cm 3 36.

A depletion type N-channel MOSFET is biased in its linear region for use as a voltage controlled resistor. Assume threshold voltage −8 2 2 VTH = 0.5V, VGS = 2.0 V, VDS = 5V, W / L = 100, COX = 10 F / cm and µ n = 800cm / V − s . The value of the resistance of the voltage controlled resistor ( in Ω ) is ________.

Answer:500 Exp:

Given VT = −0.5V; VGS = 2V; VDS = 5V; W

L

= 100; Cθx = 10−8 f / cm

µ n = 800cm 2 / v − s 1 W  2 ( VGS − VT ) VDS − VDS2  I D = µ n C0 x 2 L  −1

 ∂I D   ∂ 1 W   2 ( VGS − VT ) VDS − VDS2     µ n C0x   = rds  L   ∂VDS   ∂VDS  2

−1


EC-GATE-2014 PAPER-01| W W   = µ n C0 x ( VGS − VT ) − µ n C0x VDS  L L   ⇒ rds =

=

37.

µ n C0 x

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−1

1 W ( VGS − VT − VDs ) L

1 = 500Ω 800 × 10 × 100 ( 2 + 0.5 − 5 ) −8

In the voltage regulator circuit shown in the figure, the op-amp is ideal. The BJT has VBE = 0.7 V and β = 100, and the zener voltage is 4.7V. For a regulated output of 9 V, the value of R ( in Ω ) is ______ . VI = 12 V

V0 = 9 V +

1kΩ

1kΩ

−

Vz = 4.7 V

R

V = 12V i

9V

Answer:1093 Exp: Given VBE = 0.7V, β = 100, VZ = 4.7V, V0 = 9V

1K

+ −

R R + 1k R 4.7 = 9 × (∵ VR = Vz ) R + 1k R = 1093 Ω VR = 9 ×

38.

VR R

Vz

In the circuit shown, the op-amp has finite input impedance, infinite voltage gain and zero input offset voltage. The output voltage Vout is R2

(A) − I 2 ( R 1 + R 2 ) (B) I 2 R 2 (C) I1 R 2

R1

l1

−

l2 +

Vout

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Answer: (C) Exp: Given, Zi = ∞ A 0L = ∞ Vi0 = 0

R2

V2 = ( R1 / /R 2 ) I1 =

R1

R 1R 2 I1 ...... (1) R1 + R 2

V2

KCL at inverting node V2 V2 − V0 + =0 R1 R2

I1

− +

V0

V1

(∴ Zi = ∞ )

1 V0 1  = V2  +  R2  R1 R 2  V0  R 1R 2   R 2 + R 1  =  I1   R 2  R 1 + R 2   R 1R 2  ⇒ V0 = I1R 2 39.

For the amplifier shown in the figure, the BJT parameters are VBE = 0.7 V, β = 200, and thermal voltage VT = 25mV. The voltage gain ( v 0 / v i ) of the amplifier is _______. VCC = +12V

R1

RC 5kΩ

33kΩ vi

1 µF

1 µF R2 11kΩ

vo

RS 10Ω

R E1 1k Ω

CE 1mF

Answer: -237.76 Exp:

VBE = 0.7V, β = 200, VT = 25mV

DC Analysis: 11k = 3V 11k + 33k VE = 3 − 0.7 = 2.3V VB = 12 ×

IE =

2.3 = 2.277 mA 10 + 1k



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I B = 11.34 µA IC = 2.26 mA 25mV = 10.98 Ω 2.277 mA V −βR C −200 × 5k AV = 0 = = Vi β re + (1 + β )( R s ) 200 × 10.98 + ( 201)10 re =

A V = −237.76 40.

The output F in the digital logic circuit shown in the figure is XOR X AND

Y

F

Z XNOR

( A ) F = XYZ + XYZ ( C) F = XYZ + XYZ Answer: (A) Exp:

( B) F = XYZ + XYZ ( D ) F = XYZ + XYZ

XOR

X Y K F Z XNOR

Assume dummy variable K as a output of XOR gate K = X ⊕ Y = XY + XY

F = K. ( K Z )

= ( KZ + K.Z ) = K. KZ + K.K.Z

(

= 0 + K.Z ∵ K. K = 0 and K.K = K

)

Put the value of K in above expression

F = ( XY + XY ) Z

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Consider the Boolean function, F ( w, x, y,z ) = wy + xy + wxyz + wxy + xz + xyz. which one of the following is the complete set of essential prime implicants? (A) w, y, xz, x z

(B) w, y, xz

(C) y, x y z

(D) y, xz,xz

Answer: (D) Exp: Given Boolean Function is F ( w, x, y, z ) = wy + xy + wxyz + wxy + xz + xyz

By using K-map xz wx 00

yz 00

01

1

11

10

1

1

01

1

1

1

11

1

1

1

1

1

10

1

y

xz

So, the essential prime implicants (EPI ) are y, xz, xz 42.

The digital logic shown in the figure satisfies the given state diagram when Q1 is connected to input A of the XOR gate. S=0

D1 Q1 >

CLK

Q1

00

A D2 Q2 S

>

S =1 Q2

10

S =1 S=0 S=0 S =1

01 S =1 11 S=0

Suppose the XOR gate is replaced by an XNOR gate. Which one of the following options preserves the state diagram? (A) Input A is connected to Q 2 (B) Input A is connected to Q2 (C) Input A is connected to Q1 and S is complemented (D) Input A is connected to Q1 Answer: (D) India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 22


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The input of D2 flip-flop is D 2 = Q1s + Q1 s (∵ A = Q1 )

The alternate expression for EX-NOR gate is = A ⊕ B = A ⊕ B = A ⊕ B So, if the Ex-OR gate is substituted by Ex-NOR gate then input A should be connected to Q1

(∵ A

D 2 = Q1S + Q1 S = Q1S + Q1 .S

= Q1 )

= Qi S + Q1 .S n

43.

n

 1   1 Lex x [ n ] =   u ( n ) −  −  u ( − n − 1) . The Region of Convergence (ROC) of the z −9   3 transform of x[n]

(A) is z >

1 9

(B) is z <

1 3

1 1 (C) is > z > 3 9

(D) does not exist.

Answer: (C)  −1   −1  Given x [ n ] =   u [ n ] −   u [ −n − 1]  9   3  n

Exp:

n

1  −1  for   u [ n ] R oc in z > 9  9  h

(Right sided sequence, R oc in exterior of circle of radius 1 ) 9 Thus overall R oc in 44.

1 1
 πn  Consider a discrete time periodic signal x[n] = sin   . Let a k be the complex Fourier  s 

series coefficients of x[n]. The coefficients { a k } are non-zero when k = Bm ± 1, where m is any integer. The value of B is_________. Answer: 10 Exp:

 πn  Given x [ n ] = sin   ; N = 10  5  ⇒ Fourier series co-efficients are also periodic with period N = 10

x [n] = a1 =

2π 2π n −1 − i n 1 j 10 e e 10 2j 2j

1 −1 −1 ; a −1 = ⇒ a −1 = a −1+10 = a 9 = 2j 2j 2j

a1 = a1 + 10  a1 = a1 + 20  or a −1 = a −1 + 10  a −1 = a −1 + 20 ⇒ k = 10 m + 1 or k = 10.m − 1 ⇒ B = 10



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A system is described by the following differential equation, where u(t) is the input to the system and y(t) is the output of the system.

y ( t ) + 5y ( t ) = u ( t ) .

When y(0) = 1 and u(t) is a unit step function, y(t) is (A) 0.2 + 0.8e −5t

(B) 0.2 − 0.2e −5t

(C) 0.8 + 0.2e −5t

(D) 0.8 − 0.8e −5t

Answer: (A) Exp: Given y ( t ) + 5y ( t ) = u ( t ) and y ( 0 ) =1; u ( t ) is a unit step function. Apply Laplace transform to the given differential equation. S y ( s) − y ( 0) + 5 y (s ) = y ( s ) [ s + 5] =

1 s

1   dy   + y ( 0) L   = s y ( s ) − y ( 0)  L u ( t ) = 1   s  s   dt    

1 +1 s y (s) = ( s + 5) y (s) =

( s +1) s ( s + 5)

⇒

A B + s s+ 5

A= 1 ; B = 4 5 5 1 4 y (s) = + 5s 5 ( s + 5 ) Apply inverse Laplace transform, 1 4 −5t + e 5 5 y ( t ) = 0.2 + 0.8e −5t y(t) =

46.

Consider the state space model of a system, as given below

.   x 1   −1 1 0  .     x 2  =  0 −1 0   .   0 0 −2   x 3    

 x1  x  +  2  x 3 

 x1  0  4  u; y = 1 1 1  x  [ ] 2    x 3   0 

The system is (A) controllable and observable (B) uncontrollable and observable (C) uncontrollable and unobservable (D) controllable and unobservable Answer: (B) India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 24


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Exp: From the given state model,

 −1 1 0  0    A =  0 −1 0  B =  4   0   0 0 −2 

c = [1 1 1]

Controllable: Q c = c =  B AB A 2 B  if Q c ≠ 0 → controllable  0 4 −8 Q c =  4 −4 4  ⇒ Qc = 0  0 0 0  ∴ uncontrollable  C  Observable : Q 0 =  CA   CA 2  If Q 0 ≠ 0 → observable 1 1 1 Q 0 =  −1 0 −2  ⇒ Q 0 =1  1 −1 4  ∴ Observable The system is uncontrollable and observable

47.

The phase margin in degrees of G ( s ) =

10 calculated using the ( s + 0.1)( s + 1) + ( s + 10)

asymptotic Bode plot is_______. Answer: 48 10 Exp: G (s) = s + 0.1 ( )( s +1)( s +10 ) 10 G (s) = s  s   0.1 1 +  [ 1+ s ]  1+  .10  0.1   10  10 G (s) = [1+10s][ 1+ s][ 1+ 0.1s] By Approximation, G ( s ) =

10 [10s +1]


EC-GATE-2014 PAPER-01| Phase Margin = θ =180 + GH ω=ωgc

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10

ωgc = 1 =

 10 × 0.99  = 180 − tan −1   1   Phase Margin = 95° .73

100ω2 + 1

99 1ω 99 ⇒ ωgc = 0.9949r / sc 1ω

=100ω2 = ⇒ ω2

Asymptotic approximation, Phase margin = φ − 45° 48

48.

1 . The 2% settling time of the step ( s + 1) + ( s + 2 ) response is required to be less than 2 seconds.

For the following feedback system G ( s ) =

r +

C (s)

G (s)

y

−

Which one of the following compensators C(s) achieves this?

( A)

 1  3  s + 5 

( B)

 0.03  5 + 1  s 

( C) 2 ( s + 4)

( D)

 s + 8 4  s + 3

Answer: (C) Exp: By observing the options, if we place other options, characteristic equation will have 3rd order one, where we cannot describe the settling time. If C ( s ) = 2 ( s + 4 ) is considered The characteristic equation, is

s 2 + 3s + 2 + 2s + 8 = 0 ⇒ s 2 + 5s +10 = 0 Standard character equation s 2 + 2 ξωn s + ω2n = 0

ω2n = Given, 2% settling time, 49.

10; ξωn = 2.5

4 < 2 ⇒ ξw n > 2 ξw n

Let x be a real-valued random variable with E[X] and E[X2] denoting the mean values of X and X2, respectively. The relation which always holds true is

( A ) ( E [ X ])

2

( B) E  X 2  ≥ ( E [ X ])

> E  X 2 

( C) E  X 2  = ( E [ X ])

( D ) E  X 2  > ( E [ X ])

2

2

2

Answer: (B) Exp: V ( x ) = E ( x 2 ) − {E ( x )} ≥ 0 i.e., var iance cannot be negative 2

∴ E ( x 2 ) ≥ {E ( x )}

2



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Consider a random process X ( t ) = 2 sin ( 2 πt + ϕ ) , where the random phase ϕ is uniformly distributed in the interval [ 0, 2 π] . The auto-correlation E  X ( t1 ) X ( t 2 ) 

( A ) cos ( 2π ( t1 + t 2 ) ) ( C ) sin ( 2π ( t1 + t 2 ) )

( B ) sin ( 2π ( t1 − t 2 ) ) ( D ) cos ( 2π ( t1 − t 2 ) )

Answer: (D) Exp:

Given X(t) = 2 sin ( 2πt + φ )

f φ (θ)

φ in uniformly distributed in the interval [0, 2π ] E [ x(t1 )x(t 2 ) ] = ∫

2π

0

1 2π

2 sin(2 πt1 + θ) 2 sin ( 2πt 2 + θ ) f φ (θ)dθ

2π

1 .dθ 2π 1 2π 1 2π = sin(2π(t1 + t 2 ) + 2θ)dθ + cos(2π(t1 − t 2 )dθ ∫ 2π 0 2π ∫0 = 2 ∫ sin ( 2πt1 + θ ) sin ( 2πt 2 + θ ). 0

0

θ

2π

First integral will result into zero as we are integrating from 0 to 2 π. Second integral result into cos {2π(t1 − t 2 )} ⇒ E [ X(t1 )X(t 2 )] = cos ( 2π(t1 − t 2 )

51.

Let Q

( γ)

be the BER of a BPSK system over an AWGN channel with two-sided noise

power spectral density N0/2. The parameter γ is a function of bit energy and noise power spectral density. A system with tow independent and identical AWGN channels with noise power spectral density N0/2 is shown in the figure. The BPSK demodulator receives the sum of outputs of both the channels. AWGN Channel1 0 /1

BPSK Modulator

+

0 /1 BPSK Demodulator

AWGN Channel 2

(

)

If the BER of this system is Q b γ , then the value of b is _____________. Answer: 1.414 Exp:

    2E    E  Bit error rate for BPSK = Q   . Q    NO    N O   2   

⇒Y=

2E NO



Function of bit energy and noise PSD

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NO 2

Counterllation diagram of BPSK Channel is A WGN which implies noise sample as independent

−a

a

φ1 ( t )

Let 2x + n1 + n 2 = x1 + n1 where x1 = 2x +

n = n1 + n 2 1

 2E  Now Bit error rate = Q    N O1    1

x + n1

x noise in channel 1

+

+

2x + n1 + n2

x + n2

E1 is energy in x1 noise in channel 2

N O1 is PSD of h1 E1 = 4E [as amplitudes are getting doubled]

N O1 = N O [independent and identical channel]  4E   2E  ⇒ Bit error rate = Q  = Q 2   ⇒ b = 2 or 1.414  N   N O O    

52.

A fair coin is tossed repeatedly until a ‘Head’ appears for the first time. Let L be the number of tosses to get this first ‘Head’. The entropy H(L) in bits is _________.

Answer: 2 Exp:

In this problem random variable is L L can be 1, 2,.............. P {L = 1} =

1 2

P {L = 2} =

1 4

P {L = 3} =

1 8

1 1 1 1 1 1 1 1 1 + lg o 2 + log 2 + ......... = 0 + 1. + 2. + 3. + ......... H {L} = log 2 1 1 1 2 4 8 2 4 8 2 4 8 [ Arithmatic gemometric series summation] =

1 .1 2 2 + =2 2 1− 1 1   2 1 −   2



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In spherical coordinates, let aˆ θ .aˆ φ denote until vectors along the θ, φ directions. 100 and sin θ cos ( ωt − βr ) aˆ θ V / m r 0.265 H= sin θ cos ( ωt − βr ) aˆ φ A / m r represent the electric and magnetic field components of the EM wave of large distances r from a dipole antenna, in free space. The average power (W) crossing the hemispherical shell located at r = 1km,0 ≤ θ ≤ π / 2 is _______ E=

Answer: 55.5 Exp:

100 sin θ e − Jβr r

Eθ =

0.265 sin θ e− Jβr r 1 Pavg = ∫ E θ H*Q .ds 2 s 1 100 ( 0.265 ) 2 2 = ∫ sin θ r sin θ dθ dφ 2 s r2 1 Pavt = ∫ ( 26.5 ) sin 2 dθ dφ 2 s HQ =

π

=13.25

2π

2

∫

sin 3 θdθ

θ= 0

∫

Q =0

( 3 ) ( 2π )

dφ = 13.25. 2

P = 55.5 w For a parallel plate transmission line, let v be the speed of propagation and Z be the characteristic impedance. Neglecting fringe effects, a reduction of the spacing between the plates by a factor of two results in (A) halving of v and no change in Z (B) no change in v and halving of Z (C) no change in both v and Z (D) halving of both v and Z Answer: (B)

54.

Exp:

276 d log   ∈r r

Zo =

d → distance between the two plates so, zo – changes, if the spacing between the plates changes. V=

1 LC

→ independent of spacing between the plates



55.

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λ section of a lossless transmission line of characteristic 8 impedance 50Ω is found to be real when the other end is terminated by a load

The input impedance of a

Z L ( = R + jX ) Ω. if X is 30 Ω, the value of R ( in Ω ) is _________

Answer: 40 Exp: Given, = λ s Zo = 50Ω

(

Zin

)

 Z + JZo  = Zo  L   Zo + KZL   Z + J50   ZL + J50 50 − JZL  = 50  L ×  = 50    50 + JZ L   50 + JZL 50 − JZ L 

Zin = λ

8

 50ZL + 50 ZL + J ( 502 − ZL2 )   Zin = 50  502 + Z2L   Given , Zin → Re al

So, I mg ( Zin ) = 0 502 − ZL2 = 0 Z2L = 502 R 2 + X 2 = 502

R 2 = 502 − X 2 = 502 − 302 R = 40Ω



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Choose the most appropriate word from the options given below to complete the following sentence. Communication and interpersonal skills are_____ important in their own ways. (A) each

(B) both

(C) all

(D) either

Answer: (B) 2.

Which of the options given below best completes the following sentence? She will feel much better if she ________________. (A) will get some rest

(B) gets some rest

(C) will be getting some rest

(D) is getting some rest

Answer: (B) 3.

Choose the most appropriate pair of words from the options given below to complete the following sentence. She could not _____ the thought of _________ the election to her bitter rival. (A) bear, loosing

(B) bare, loosing

(C) bear, losing

(D) bare, losing

Answer: (C) 4.

A regular die has six sides with numbers 1 to 6 marked on its sides. If a very large number of throws show the following frequencies of occurrence: 1 → 0.167; 2 → 0.167; 3 → 0.152; 4 → 0.166; 5 → 0.168; 6 → 0.180. We call this die (A) irregular

(B) biased

(C) Gaussian

(D) insufficient

Answer: (B) Exp:

For a very large number of throws, the frequency should be same for unbiased throw. As it not same, then the die is baised.

5.

Fill in the missing number in the series. 2

3

6

15

___ 157.5

630

Answer: 45 Exp: 2

3

1.5

6

2

15

2.5

45

3

157.5 630

3.5

4

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Find the odd one in the following group Q,W,Z,B (A) Q,W,Z,B

B,H,K,M

(B) B,H,K,M

W,C,G,J

M,S,V,X

(C) W,C,G,J

(D) M,S,V,X

Answer: (C) Exp:

a 17

W 23

Z 26

B 2

B

H 6

6

7.

3

K 3

2

N 2

W

C 6

G 4

J 3

M 6

S

V 3

X 2

Lights of four colors (red, blue, green, yellow) are hung on a ladder. On every step of the ladder there are two lights. If one of the lights is red, the other light on that step will always be blue. If one of the lights on a step is green, the other light on that step will always be yellow. Which of the following statements is not necessarily correct? (A) The number of red lights is equal to the number of blue lights (B) The number of green lights is equal to the number of yellow lights (C) The sum of the red and green lights is equal to the sum of the yellow and blue lights (D) The sum of the red and blue lights is equal to the sum of the green and yellow lights

Answer: (D) The sum of eight consecutive odd numbers is 656. The average of four consecutive even numbers is 87. What is the sum of the smallest odd number and second largest even number? Answer: 163 8.

Exp:

Eight consecutive odd number =656 a-6, a-1, a-2, a ,a+2 ,a+4, a+6 a+8=656 a=81 Smallest m=75

… (1)

Average consecutive even numbers

a −2+a +a +2+a +4 = 87 4 ⇒ a = 86

⇒

Second largest number =88 1+2=163 9.

The total exports and revenues from the exports of a country are given in the two charts shown below. The pie chart for exports shows the quantity of each item exported as a percentage of the total quantity of exports. The pie chart for the revenues shows the percentage of the total revenue generated through export of each item. The total quantity of exports of all the items is 500 thousand tonnes and the total revenues are 250 crore rupees. Which item among the following has generated the maximum revenue per kg?



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Exports Item 6

Revenues Item 1

Item 1

Item 6 16%

11%

12%

19%

Item2

Item5

Item 2 20%

12%

20%

Item3

Item4

20%

19%

22%

(A) Item 2 Answer: (D) Item:2 Exp:

Item 5

Item3 Item 4 6%

(B) Item 3

20 × 250 × 107 100 20 × 500 × 103 100 0.5 × 104 = 5 × 103 1 = Item 2 Item: 6 19 = 1.18 = Item 6 16

(C) Item 6

23%

(D) Item 5

Item:3 23 × 250 × 107 19 × 500 × 103 1.2 = Item 3

Item:5 20 5 = = 1.6 ⇒ 1.6 = Item 5 12 3

It takes 30 minutes to empty a half-full tank by draining it at a constant rate. It is decided to simultaneously pump water into the half-full tank while draining it. What is the rate at which water has to be pumped in so that it gets fully filled in 10 minutes? (A) 4 times the draining rate (B) 3 times the draining rate (C) 2.5 times the draining rate (D) 2 times the draining rate Answer: (A) Exp: Vhalf = 30(s) drawing rate = s

10.

Total volume =60 S tank

(s1 )(10) − (s)10 = 30s s1 (s) − s = 3s s1 = 4s s1 = 4drawing rate



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The determinant of matrix A is 5 and the determinant of matrix B is 40. The determinant of matrix AB is ________. Answer: 200 Exp:

AB = A . B = ( 5 ) . ( 40 ) = 200

2.

Let X be a random variable which is uniformly chosen from the set of positive odd numbers less than 100. The expectation E[X]is __________. Answer:50 Exp:

X = 1,3,5,....,99 ⇒ n = 50 ( number of observations )

∴E(x) =

3.

1 n 1 1 2 x i = [1 + 3 + 5 + .... + 99] = ( 50 ) = 50 ∑ n i =1 50 50

For 0 ≤ t < ∞, the maximum value of the function f ( t ) = e − t − 2e −2 t occurs at

(A) t = loge4 Answer: (A) Exp:

(B) t = loge2

(C) t = 0

(D) t = loge8

(C) e

(D) ∞

f ' ( t ) = −e − t + 4e−2t = 0 1 ⇒ e − t  4e − t − 1 ⇒ e− t = ⇒ t = log e4 4 '' 4 and f ( t ) < 0 at t = log e x

4.

 1 The value of lim  1 +  is x x →∞ 

(A) ln2 Answer: (C)

(B) 1.0 x

Exp:

 1 lim  1 +  = e ( standard limit ) x →∞  x

5.

If the characteristic equation of the differential equation d2y dy + 2α +y=0 2 dx dx has two equal roots, then the values of α are

( A) ± 1

( B) 0,0

( C) ± j

( D ) ± 1/ 2

Answer: (A) Exp:

For equal roots, Discriminant B2 − 4AC = 0 ⇒ 4α 2 − 4 = 0

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Norton’s theorem states that a complex network connected to a load can be replaced with an equivalent impedance (A) in series with a current source

(B) in parallel with a voltage source

(C) in series with a voltage source

(D) in parallel with a current source

Answer: (D) Exp: Norton’s theorem

IN

7.

Zequ

Load

In the figure shown, the ideal switch has been open for a long time. If it is closed at t=0, then the magnitude of the current (in mA) through the 4 kΩ resistor at t = 0+ is _______. 5kΩ

4 kΩ

1kΩ

i 10 V −+

10µF

t=0 5 kΩ

1mH

4 kΩ

Answer: 1.2 mA Exp:

For t = o+ 10 ⇒ 1.11mA i (o +) = 9K i ( o + ) 1.2 mA

8.

10V

• •

+ −

i (0+)

A silicon bar is doped with donor impurities ND = 2.25 x 1015 atoms / cm3. Given the intrinsic carrier concentration of silicon at T = 300 K is ni = 1.5 x 1010 cm-3. Assuming complete impurity ionization, the equilibrium electron and hole concentrations are (A) n0 = 1.5 x 1016 cm-3, p0 = 1.5 x 105 cm-3 (B) n0 = 1.5 x 1010 cm-3, p0= 1.5 x 1015 cm-3 (C) n0 = 2.25 x 1015 cm-3, p0 = 1.5 x 1010 cm-3

(D) n0 = 2.25 x 1015 cm-3, p0 = 1 x 105 cm-3 Answer: (D) Exp:

N D = 2.25 × 1015 Atom / cm3

h i = 1.5 × 1010 / cm 3 Since complete ionization taken place, h 0 = N D = 2.25 × 1015 / cm3 10 n i 2 (1.5 × 10 ) = = 1 × 105 / cm 3 P0 = n0 2.25 × 1015 2



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An increase in the base recombination of a BJT will increase (A) the common emitter dc current gain β

(B) the breakdown voltage BVCEO (C) the unity-gain cut-off frequency fT (D) the transconductance gm Answer: (B) In CMOS technology, shallow P-well or N-well regions can be formed using (A) low pressure chemical vapour deposition (B) low energy sputtering (C) low temperature dry oxidation (D) low energy ion-implantation Answer: (D) 10.

11.

The feedback topology in the amplifier circuit (the base bias circuit is not shown for VCC simplicity) in the figure is Io

Rc

(A) Voltage shunt feedback

Vo

(B) Current series feedback RS

(C) Current shunt feedback

RE VS ~

(D) Voltage series feedback Answer: (B) Exp: By opening the output feed back signed becomes zero. Hence it is current sampling. As the feedback signal vf is subtracted from the signal same vs it is series mixing. 12.

In the differential amplifier shown in the figure, the magnitudes of the common-mode and differential-mode gains are Acm and Ad, respectively. If the resistance RE is increased, then (A) Acm increases VCC (B) common-mode rejection ratio increases RC RC (C) Ad increases (D) common-mode rejection ratio decreases − + V0 Answer: (B) Exp:

A d does not depend on R E A cm decreases as R E is increased

∴ CMRR =

Ad = Increases A cm

+ Vi − RE

I0 − VEE



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A cascade connection of two voltage amplifiers A1 and A2 is shown in the figure. The openloop gain Av0, input resistance Rin, and output resistance RO for A1 and A2 are as follows: A1:A v 0 = 10, R in = 10kΩ, R 0 = 1kΩ A2 : A v0 = 5, R in = 5kΩ, R 0 = 200 Ω

The approximate overall voltage gain Vout / Vin is __________.

+

+ Vin

A1

R L Vout 1kΩ −

A2

− Answer: 34.722 Exp:

Overall voltage gain, A v =

V0 = A V1 A V2 Vi

 Zi 2   R L    Zi2 + Z01   R L + Z02

  

 5k   1k  = 10 × 5     5k + 1k  1k + 200  A V = 34.722

14.

For an n-variable Boolean function, the maximum number of prime implicants is (A) 2(n-1)

(B) n/2

(C) 2n

(D) 2(n-1)

Answer: (D) = 2 ( n −1)

Exp:

For an n-variable Boolean function, the maximum number of prime implicants

15.

The number of bytes required to represent the decimal number 1856357 in packed BCD (Binary Coded Decimal) form is __________ .

Answer: 4 Exp: In packed BCD (Binary Coded Decimal) typically encoded two decimal digits within a single byte by taking advantage of the fact that four bits are enough to represent the range 0 to 9. So, 1856357 is required 4-bytes to stored these BCD digits 16.

In a half-subtractor circuit with X and Y as inputs, the Borrow (M) and Difference (N = X - Y) are given by (A) M = X, ⊕ Y, N = XY

(B) M = XY,

N = X⊕ Y

(C) M = X Y , ⊕ N = X ⊕ Y

(D) M = XY

N = X⊕Y

Answer: (C)



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Exp: Function Table for Half-subtractor is

X

Y

Difference (N)

Borrow (M)

0 0 0 Hence, N = X ⊕ Y and m = XY 0 1 1

0 1

1

0

1

0

1

1

0

0

Hence, N = X ⊕ Y and m = XY 17.

An FIR system is described by the system function 7 3 H ( z ) = 1 + z −1 + z −2 The system is 2 2 (A) maximum phase (B) minimum phase (C) mixed phase

(D) zero phase

Answer: (C) Exp:

Minimum phase system has all zeros inside unit circle maximum phase system has all zeros outside unit circle mixed phase system has some zero outside unit circle and some zeros inside unit circle. 7 3 for H ( s ) = 1 + z −1 + z −2 2 2 One zero is inside and one zero outside unit circle hence mixed phase system

Let x[n] = x[-n]. Let X(z) be the z-transform of x[n]. If 0.5 + j 0.25 is a zero o X(z), which one of the following must also be a zero of X(z). (A) 0.5 - j0.25 (B) 1/(0.5 + j0.25) (D) 2 + j4 (C) 1/(0.5 - j0.25) Answer: (B) 18.

Exp:

Given x [ n ] = x [ −n ] ⇒ x ( z ) = x ( z −1 ) [ Time reversal property in z − transform ] ⇒ if one zero is 0.5 + j0.25

then other zero will be

19.

1 0.5 + j0.25

Consider the periodic square wave in the figure shown. x

1

0

1

2

3

4

t

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The ratio of the power in the 7th harmonic to the power in the 5th harmonic for this waveform is closest in value to _______. Answer: 0.5 1 Exp: For a periodic sequence wave, nth harmonic component is α n 1 ⇒ power in nth harmonic component is α 2 n th ⇒ Ratio of the power in 7 harmonic to power in 5th harmonic for given waveform is 1 2 7 = 25 ≈ 0.5 1 2 49 5 20.

The natural frequency of an undamped second-order system is 40 rad/s. If the system is damped with a damping ratio 0.3, the damped natural frequency in rad/s is ________. Answer: 38.15 r / sec

Exp:

Given ωn = 40 r / sec

ξ = 0.3

ωd = ωn 1 − ξ 2 ωd = 40 1 − ( 0.3)

2

ωd = 38.15 r / sec 21.

For the following sytem, x1 ( s ) +

s s +1

−

When X1 ( s ) = 0 , the transfer function

( A)

s +1 s2

( B)

1 s +1

+

+

x 2 (s)

Y (s)

1 s

y ( s) is x 2 ( s)

( C)

s+2 s ( s + 1)

s +1 s ( s + 2)

( D)

Answer: (D) Exp:

If X1 ( s ) = 0 Y (s)

X2 (s ) Y (s)

X2 (s)

; The block diagram becomes

=

1 s 1 s 1+ . s ( s + 1)

X2 ( s )

1 ( s + 1) s = ⇒ (s + 2) / s + 1 s (s + 2)

+ −

Y (s)

1 s

+S

( S + 1)



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The capacity of a band-limited additive white Gaussian noise (AWGN) channel is given by P   C = W log 2  1 + 2  bits per second (bps), where W is the channel bandwidth, P is the  σ w

average power received and σ2 is the one-sided power spectral density of the AWGN. For a P fixed 2 = 1000, , the channel capacity (in kbps) with infinite bandwidth ( W → ∞ ) is σ approximately (A) 1.44 (B) 1.08 (C) 0.72 (D) 0.36 Answer: (A)

Exp:

P   ω ln 1 + 2  P    σ ω C = lim ω log 2 1 + 2  = lim ω→∞ w →∞ ln 2  σ ω

1 lim = ln 2 ω→∞

P  P    ln 1 + 2  ln 1 + 2  P P σ ω σ ω  . lim  = 2 2 P P σ σ ln 2 ω→∞ 2 σω σ2 ω ↓ This lim it is equivalent to

lim

ln [1 + x ]

ω→∞

x

=1 =

P P = ln 2 e 2 = 1.44 KGpa σ .ln 2 σ 2

23.

Consider sinusoidal modulation in an AM system. Assuming no overmodulation, the modulation index ( µ ) when the maximum and minimum values of the envelope, respectively, are 3 V and 1 V, is ________. Answer: 0.5 A ( t )max − A ( t ) min Exp: µ= A ( t ) max + A ( t ) min

µ=

24.

3 −1 1 = = 0.5 3 +1 2

To maximize power transfer, a lossless transmission line is to be matched to a resistive load impedance via a λ / 4 transformer as shown. lossless transmission line λ / 4 transformer Z L = 50Ω

Z L = 100Ω

The characteristic impedance ( in Ω ) of the λ / 4 transformer is _________. Answer: 70.7Ω



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( 4)

Here impedance is matched by using QWT λ ∴ Z'0 =

Z L Zin

= 100 × 50 = 50 2 = Z'0 = 70.7Ω 25.

Which one of the following field patterns represents a TEM wave travelling in the positive x direction? ˆ H = −4zˆ ( A ) E = +8y, ( C) E + 2z,ˆ H = +2yˆ

( B)

ˆ H = −3zˆ E = −2y,

( D)

ˆ H = +4zˆ E = −3y,

Answer: (B) Exp:

For TEM wave Electric field (E), Magnetic field (H) and Direction of propagation (P) are orthogonal to each other. Here P = + a x By verification

E = − 2a y , H = − 3a z E × H = − a y ×− a z = + a x → P

Q. No. 26 – 55 Carry Two Marks Each 26.

The system of linear equations  2 1 3  a   5   3 0 1  b =  −4 has       1 2 5  c   14 

(A) a unique solution

(B) infinitely many solutions

(C) no solution

(D) exactly two solutions

Answer: (B) Exp:

2 1 3 5  [ A / B] = 3 0 1 −4 1 2 5 14  1 3 5  1 3 5  2 2 R 2 → 2R 2 − 3R 1   R3 +R 2   0 − 3 − 7 −23 → 0 − 3 − 7 −23  R 3 → 2R 3 − R 1 0 3 0 0 7 23  0 0  Since, rank ( A ) = rank ( A / B ) < number of unknowns ∴ Equations have infinitely many solutions.



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The real part of an analytic function f(z) where z =.x + jy is given by e-y cos(x). The imaginary part of f(z) is (A) eycos(x)

(B) e-ysin(x)

(C) -eysin(x)

(D) –e-ysin(x)

Answer: (B) Exp:

real part u = e− y cos x and V = ? ∂v ∂v dx + dy ∂x ∂y ∂u ∂u = − dx + dy ( U sin g C − R equations ) = e − y cos xdx − e − y sin xdy = d  e − y sin x  ∂y ∂x

dv =

Integrating, we get V = e − y sin x 28.

The maximum value of the determinant among all 2×2 real symmetric matrices with trace 14 is ________.

Answer: 49 Exp:

y x General 2 × 2 real symmetric matrix is   x z  ⇒ det = yz − x 2 and trace is y + z = 14 ( given ) ⇒ z = 14 − y .............. ( *) Let f = yz − x 2 ( det ) = − x 2 − y 2 + 14y ( u sin g *)

Using maxima and minima of a function of two variables, we have f is maximum at x = 0, y = 7 and therefore, maximum value of the determinant is 49

29.

If r = xaˆ x + yaˆ y + zaˆ z and r = r, then div ( r 2 ∇ ( ln r ) ) = _______.

Answer: 3 Exp:

r ∇ ( ln r ) = 2 ⇒ div ( r 2∇ ( ln r ) ) = div ( r ) = 3 r

∂   1  x  1 ∇ ( ln r ) = ∑ aˆ x ∂x ( ln r ) = ∑ aˆ x  r  r  = r 2     30.

∑ aˆ x x =

r r 2 

A series LCR circuit is operated at a frequency different from its resonant frequency. The operating frequency is such that the current leads the supply voltage. The magnitude of current is half the value at resonance. If the values of L, C and R are 1 H, 1 F and 1Ω , respectively, the operating angular frequency (in rad/s) is ________.

Answer: 0.45 r/sec Exp:

The operating frequency (wx), at which current leads the supply. i.e., ωx <ωr again magnitude of current is half the value at resonance



i,e.,. at ω = ωx ⇒ I x = at ω= ωx ⇒ Iresonance =

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V z

V R

I resonance 2 V V = = Z = 2R i.e., z 2R Ix =

Given R = 1Ω;

L=1H;

 1  Z = R2 + − ωL   ωc 

C=1F

2

= 2

2

 1  =R +  − ωL  = 4  ωc  2

By substituting R, L & C values, 2

1 1  ⇒ 1 +  − ω  = 4 ⇒ ω2 = 2 = 5 ω ω  1 Assume ω2 = x, then, x + = 5 x ⇒ x 2 − 5x + 1= 0 x1, 2 = 4.791, 0.208 if x = 4.791 ⇒ ω= 2.18 r sec if x = 0.208 ⇒ ω= 0.45r sec But ωx < ωr So, operating frequency ωx = 0.45 r sec 31.

In the h-parameter model of the 2-port network given in the figure shown, the value of h22 (in S) is _____ . 3Ω

1

1'

3 Ω 3Ω

2Ω

2

2'

2Ω 2Ω

Answer: 1.24 Exp:

If two, π − n ws are connected in parallel, The y-parameter are added



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i.e., y equ = y1 + y 2 −1   2  1 −1  3 3 2   y1 = y2 = − 1   2 −1 1  3   2   3 −5  5 3 6 y equ =   −5 5  3  6 − y12   1 y11   y11 h =   ∆y  y 21  y11   y11 where ∆y = y11 y 22 − y12 − y 21  5  5     −5  −5   The value of h 22 = ∆y =     −       3  3     6  6   ∆y = 2.0833 y11 = 5 ∴ h 22 = 1.24 3

32.

In the figure shown, the capacitor is initially uncharged. Which one of the following expressions describes the current I(t) (in mA) for t > 0? R1 1kΩ 5V +−

I R2 2 kΩ

C 1 µF

(

)

( B) I ( t ) =

5 2 1 − e − t / τ , τ = msec 2 3

(

)

( D) I ( t ) =

5 1 − e − t / τ , τ = 3 msec 2

( A) I ( t ) =

5 2 1 − e − t / τ , τ = msec 3 3

( C) I ( t ) =

5 1 − e − t / τ , τ = 3msec 2

(

)

(

)

Answer: (A) Exp:

ν c ( t ) = VR 2 ( t ) = Vfinal + [ Vinitial − Vfinal ] e

−t

τ

2 ×103 ×10−6 3 2 R equ = 2K 1K ⇒ KΩ 3 Cequ = 1µF τ = R equ .Cequ ⇒


EC-GATE-2014 PAPER-02| τ =

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2 msec 3

Vinitial = 0volts 2 10 = volts 3 3 10 10 − t τ νR 2 ( t ) = − e 3 3 −t −t ν (t) 10  5 1 − e τ  volts ⇒ i R 2 (t) = R 2 νR 2 ( t ) = =  1 − e τ  mA      3 2K 3 Vfinal = Vs.s = 5.

33.

In the magnetically coupled circuit shown in the figure, 56 % of the total flux emanating from one coil links the other coil. The value of the mutual inductance (in H) is ______ . M

10 Ω

(

4H

)

60 cos 4t + 300 V ~

5H

(1/16) F

Answer: 2.49 Henry Exp: Given 56% of the total flux emanating from one coil links to other coil. i.e, K = 56% ⇒ 0.56 We have, K =

M L1L 2

L1 = 4H; L 2 = 5H M = ( 0.56 ) 20 ⇒ m = 2.50H Assume electronic charge q = 1.6×10-19 C, kT/q = 25 mV and electron mobility µ n = 1000 cm2/V-s. If the concentration gradient of electrons injected into a P-type silicon sample is 1×1021/cm4, the magnitude of electron diffusion current density (in A/cm2) is _________. Answer: 4000 kJ Exp: Given q = 1.6 × 10−19 ; = 2.5 mV, µ n = 1000cm 2 / v − s q 34.

D n kJ = µn q 2 ⇒ D n = 25mV × 1000cm / v − S ⇒ 25cm 2 / s From Einstein relation,

Diffuion current Density J = q D n = 1.6 × 10−19 × 25 × 1 × 1021 = 4000 A / cm 2

dn dx



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Consider an abrupt PN junction (at T = 300 K) shown in the figure. The depletion region width Xn on the N-side of the junction is 0.2 µm and the permittivity of silicon ( ε si ) is 1.044×10-12 F/cm. At the junction, the approximate value of the peak electric field (in kV/cm) is _________.

P + − region N A >> N D

X11

N − region N D = 1016 / cm3

Answer: 30.66 Exp:

Given x n = 0.2 µm, ∈Si = 1.044 × 10−12 F / µ n N D = 1016 / cm3

Peak Electric field, E = =

36.

q ND xn ∈

1.6 × 10−19 × 1016 × 0.00002 = 30.66 KV / cm 1.044 × 10−12

When a silicon diode having a doping concentration of NA = 9 × 1016 cm-3 on p-side and ND = 1 × 1016 cm-3 on n-side is reverse biased, the total depletion width is found to be 3µm . Given that the permittivity of silicon is 1.04 × 10–12 F/cm, the depletion width on the p-side and the maximum electric field in the depletion region, respectively, are (A) 2.7 µm and 2.3 × 105 V/cm

(B) 0.3 µm and 4.15 × 105 V/cm

(C) 0.3 µm and 0.42 × 105 V/cm

(D) 2.1 µm and 0.42 × 105 V/cm

Answer: (B) Exp:

Given N A = 9 × 1016 / cm3 ; N D = 1 × 1016 / cm3 Total depletion width x = x n + x p = 3 µm

∈= 1.04 × 10−12 F / cm x n N A 9 × 1016 = = x p N D 1 × 1016 x n = 9x p ......... (1) Total Depletion width, x n + x p = 3µm 9x p + x p = 3 µm x p = 0.3 µm 1.6 × 10−19 × 9 × 1016 × 0.3 µm ∈ 1.04 × 10−12 = 4.15 × 105 V / cm

Max. Electric field, E =

qN A x p

=



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The diode in the circuit shown has Von = 0.7 Volts but is ideal otherwise. If Vi = 5sin ( ωt ) Volts, the minimum and maximum values of VO (in Volts) are, respectively, 1kΩ Vi

Vo

R1

R2 1kΩ

+ − 2V

(A) -5 and 2.7

(B) 2.7 and 5

(C) -5 and 3.85

(D) 1.3 and 5

Answer: (C) Exp:

When Vi makes Diode 'D' OFF, V0 = Vi

∴ V0 ( min ) = −5V When Vi makes diode 'D ' ON, V0 =

( Vi − 0.7 − 2 ) + V R1 + R 2

∴ V0 ( max ) =

on

+ 2V

( 5 − 0.7 − 2 )1k + 0.7 + 2V 1k + 1k

= 3.85V

38.

For the n-channel MOS transistor shown in the figure, the threshold voltage VTh is 0.8 V. Neglect channel length modulation effects. When the drain voltage VD = 1.6 V, the drain current ID was found to be 0.5 mA. If VD is adjusted to be 2 V by changing the values of R and VDD, the new value of ID (in mA) is VDD R D G

(A) 0.625 Answer: (C)

(B) 0.75

S

(C) 1.125

(D) 1.5



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Given VTh = 0.8V

1 w 2 When VD = 1.6V, ID = 0.5mA = µ n cos ( VDS − VTh ) 2 L [∵ Device is in sat ] 1 ω ⇒ µ n cos = 0.78125 × 10−3 A / V 2 2 L When VD = 2V 1 ω 2 I D = µ n cos ( VDS − VTh ) 2 L = 078125 × 10−3 ( 2 − 0.8 )1.125mA 39.

For the MOSFETs shown in the figure, the threshold voltage Vt = 2V and K=

1  W µC∞   = 0.1 mA / V 2 . The value of ID (in mA) is ________.  L 2 VDD = +12 V R1 10 kΩ

R2 10 kΩ

ID

VDD = −5V

Answer: 0.9 Exp:

1 W = 0.1A / V 2 Given Vt = 2V, K = µ cos 2 L 2 1 W I D1 = I D2 = µ n cos VGs1 − Vt 2 L

(

= 0.1mA / V 2 ( 5 − 2 ) = 0.9 mA

)

2

12V

10 K

10 K

ID

−5V



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In the circuit shown, choose the correct timing diagram of the output (y) from the given waveforms W1, W2, W3 and W4. X1

D

Q

FF1 >

Clk

Q

X2 D

Q

>

Q

output ( y )

ClK

X1 X2

W1 W2 W3 W4

(A) W1 (B) W2 (C) W3 (D) W4 Answer: (C) Exp: This circuit has used negative edge triggered, so output of the D-flip flop will changed only when CLK signal is going from HIGH to LOW (1 to 0) 1

CLK

0

1

0

1

0

1

0

X1

X2 Y( w 3 )



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This is a synchronous circuit, so both the flip flops will trigger at the same time and will respond on falling edge of the Clock. So, the correct output (Y) waveform is associated to w3 waveform. 41.

The outputs of the two flip-flops Q1, Q2 in the figure shown are initialized to 0,0. The sequence generated at Q1 upon application of clock signal is

J1 Q1

Q1

J2 Q2 > K2Q2

> K1 Q1

CLK

(A) 01110… Answer: (D) Exp: Clock Initial → 1st CP → 2nd CP → 3rd CP → 4th CP→

(B) 01010…

(C) 00110…

(D) 01100…

J1 ( Q 2 )

K1 ( Q 2 )

J 2 ( Q1 )

K 2 ( Q1 )

Q1

Q2

-

-

-

-

0

0

1

0

0

1

1

0

1

0

1

0

1

1

0

1

1

0

0

1

0

1

0

1

0

0

So, the output sequence generated at Q1 is 01100…. 42.

For the 8085 microprocessor, the interfacing circuit to input 8-bit digital data (DI0 – DI7) from an external device is shown in the figure. The instruction for correct data transfer is

(A) MVI A, F8H

A2 A1 A0

(B) IN F8H

(C) OUT F8H

3 − to − 8 7 Decoder 6 C 5 4 B 3 2 A 1 0 G 2A G 2B G1

10 / M RD A3 A A5 4 A6 A7

I / O Device

Digital Inputs

DI 0 − DI 7

DS1

Data Bus DO0 − D7 ( D − D ) 0 7

DS2

A8 A9 A10 A11 A12 A

13

(D) LDA F8F8H

A14

A15



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Answer: (D) Exp:

This circuit diagram indicating that it is memory mapped I/O because to enable the 3-to-8 decoder G 2A is required active low signal through ( Io m ) and G 2B is required active low through

(R ) D

it means I/o device read the status of device LDA instruction is

appropriate with device address. Again to enable the decoder o/p of AND gate must be 1 and Ds 2 signal required is 1 which is the o/p of multi-i/p AND gate to enable I/O device. So, A15 A14 A13 A12 A11 A10 A 9 A 8 A 7 A 6 A 5 A 4 A 3 A 2 A1 A 0 1 0 0 0 1 1 1 1 1 0 0 0 1 1 1 1 F

8

8

F

Device address = F8F8H The correct instruction used → LDA F8F8H 43.

Consider a discrete-time signal n for 0 ≤ n ≤ 10 x [n] =  0 otherwise

If y[n] is the convolution of x[n] with itself, the value of y[4] is _________. Answer: 10 Exp:

 n for 0 ≤ n ≤ 10  Given x [ n ] =   elsewhere  0

y [n ] = x [n ] * x [n ] n

y [ n ] = ∑ x [ k ].x [ n − k ] k =0

4

⇒ y [ 4] = ∑ x [ k ].x [ G − k ] k =0

= x ( 0 ) .x ( 4 ) + x (1) x ( 3) + x ( 2 ) x ( 2 ) + x ( 3) x (1) + x ( 4 ) .x ( 0 ) = 0 + 3 + 4 + 3 + 0 = 10 44.

The input-output relationship of a causal stable LTI system is given as y [ n ] = αy [ n − 1] + βx [ n ]

If the impulse response h[n] of this system satisfies the condition Σ ∞n =0 h [ n ] = 2, the relationship between α and β is

( A)

α = 1− β/ 2

( B)

α = 1+ β / 2

( C)

α = 2β

( D)

α = −2β



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Answer: (A) Exp:

Given system equation as

y [ n ] = α y [ n − 1] + βx [ n ] y (z)

=

β 1 − α z −1

⇒ H (z) =

β 1 − α z −1

⇒

x (z)

h [n] = β (α ) u [n]

[ causal system ]

h

∞

Also given that

∑ h [n] = 2 h =0

 1  β =2 1 − α  β 1− α = 2 α =1−

45.

β 2

The value of the integral

∫

∞ −∞

sin c 2 ( 5t ) dt is ____________.

Answer: 0.2 Exp:

We can use pasrevalis theorem sin 5πt 5πt ⇒ in frequency domain

Let x ( t ) sin ( 5t ) =

1

−2.5

−2.5 ∞

X (f )

5

∞

f 2.5

1 Now, ∫ x ( t ) dt = ∫ x ( t ) df = ∫   5 −∞ −∞ −2.5   2

=

2

2

1 1 × 5 = = 0.2 25 5



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An unforced liner time invariant (LTI) system is represented by

 x 1   −1 0   x1   =      x 2   0 −2   x 2  If the initial conditions are x1(0) = 1 and x2(0) = -1, the solution of the state equation is

( A)

x1 ( t ) = −1, x 2 ( t ) = 2

( B)

x1 ( t ) = −e − t , x 2 ( t ) = 2e − t

(C)

x1 ( t ) = e − t , x 2 ( t ) = −e−2t

( D)

x1 ( t ) = −e − t , x 2 ( t ) = −2e − t

Answer: (C) Exp:

Solution of state equation of X ( t ) = L−1 SI − A −1  .X ( 0 )  1  −1 0  X ( 0) =   A =    −1  0 − 2

[SI − A ]

−1

0  S + 1 = S + 2   0 =

−1

0 S + 2  (S + 1)(S + 2 )  0 S + 1

1

 1  −1 [SI − A ] =  S + 1  0 

 0   1  S + 2 

 −1  1  L  S + 1 −1   −1  L ( SI − A )  =     0  

e− t −1 L−1 ( SI − A )  =    0   X1 ( t )  e − t  =  X 2 ( t )   0

    −1  1  L    S + 2 

0

0   e−2t 

0   1   e −2t   −1

 X1 ( t )   −e t    =  −2t   X 2 ( t )   −e 

∴

X1 ( t ) = e− t

X 2 ( t ) = −e −2t



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The Bode asymptotic magnitude plot of a minimum phase system is shown in the figure.

26.02 G ( jω )

6.02

( dB)

0 − 6.02

10

2

1

0.1

20

ω ( rad / s ) in log scale

If the system is connected in a unity negative feedback configuration, the steady state error of the closed loop system, to a unit ramp input, is_________. Answer: 0.50 Exp:

G ( jw ) dB

( −20 dB / dec )

26.02

6.02 20 0.1

1

10

2

w (r / sec )

−6.02

( −20 db (dec ) ) → Due to initial slope , it is a type-1 system, and it has non zero velocity error coefficient (KV ) → The magnitude plot is giving 0dB at 2r/sec.

Which gives k v ∴kv = 2

The steady state error ess =

A kv

given unit ramp input; A = 1

ess =

1 2

ess = 0.50



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Consider the state space system expressed by the signal flow diagram shown in the figure. C3 S−1

1 u

S−1

S−1

x3

x1

x2

a3

c1

y

a2 a1

The corresponding system is (A) always controllable (B) always observable (C) always stable (D) always unstable Answer: (A) Exp: From the given signal flow graph, the state model is  0 1 X 0   X1  0  1         X 2  = 0 0 1   X 2  + 0 u X  a a a1   X 3  1   3  3 2

 X1  Y = [ C1 C 2 C3 ]  X 2   X 3  0 0 1 0   A = 0 0 1  ; B = 0  ;C = [ C1 C 2 C3 ] a 3 a 2 a1  1  Controllability: Q c =  B

AB

A 2 B

0  QC = 0 1 

0

  a1  a 2 + a12 

1 a1

1

QC = 1 ≠ 0 Observability C1 C      Q O =  CA  ⇒  a 3c3   2  CA  c 2 a 3 + c3 ( a1a 3 )

C2 c1 + a 2 c3 a 2 c 2 + c 3 ( a 1a 2 + a 3 )

Q 0 ⇒ depends on a 1 ,a 2 ,a 3 & c1 & c 2 & c 3

  c 2 + a1c3   2 c1 + a1c 2 + c3 ( a1 + a 2 ) 

C3

.

It is always controllable India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 25


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The input to a 1-bit quantizer is a random variable X with pdf f x ( x ) = 2e −2x for x ≥ 0 and

f x ( x ) = 0 for x < 0 , for x < 0 For outputs to be of equal probability, the quantizer threshold should be _____.

Answer: 0.35 Exp:

one bit

X

Quantizer

Q (x )

One bit quantizer will give two levels. Both levels have probability of

1 2

Pd of input X is

fx ( π )

xT

Let x T be the thsuhold x ≥ xT  x < x T 

 x1 Q(x) =  x2

Where x1 and x 2 are two levels P {Q ( r ) = x1 } =

1 2 ∞ 1 ⇒ ∫ 2.e −2x dx = 2 xT 2.

e −2x −2

∞

= xT

1 2

1 2 1 = 2

−e −2∞ + e −2x T = e −2x T

1 2 − 2x T = −0.693

− 2x T = ln

x T = 0.35



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Coherent orthogonal binary FSK modulation is used to transmit two equiprobable symbol waveforms s1 ( t ) = α cos 2 πf1t and s2 ( t ) = cos 2 πf 2 t, where α = 4mV. Assume an AWGN channel with two-sided noise power spectral density optimal receiver and the relation Q ( v ) = rate of 500 kbps is

( A) Q ( 2)

( B)

(

Q 2 2

)

1 2π

∫

∞ v

e− u

2

/2

N0 = 0.5 × 10−12 W / Hz. Using an 2

du the bit error probability for a data

( C) Q ( 4)

( D)

(

Q 4 2

)

Answer: (C) Exp: For Binary FSK

 E  Bit error probability = Q   N  O   E → Energy per bit [No. of symbols = No. of bits]

A2T 1 , A = 4 × 10−3 ,T = [inverse of data rate] 2 500 × 103 16 × 10−6 × 2 × 10−6 ⇒E= = 16 × 10−12 2 N 0 = 1 × 10−12 E=

 16 × 10−12 ⇒ Pe = Q   1 × 10−12  51.

  = Q ( 4)  

The power spectral density of a real stationary random process .X(t) is given by  1, Sx ( f ) =  0,w 

f ≤w f >w

1    The value of the expectation E  π X ( t )  t −   is ___________.  4w    Answer: 4 1  , f ≤ w  Exp: Given Sx ( f ) =  w  0 , f ≥ w   

R x ( τ) =

w

1

∫ w .e

j2 πft

df

−w

=

1 e j2 πwt − e − j2 πwt 1  sin ( 2πwt )  =   w j2πt w πt 

 1  1   1  Now, E  π × ( t ) .x  t −  = π R x   ⇒ π. . w  4w    4w  

1   sin  2πw .  4 4w  = 1 1 π. 4w



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In the figure, M(f) is the Fourier transform of the message signal .m(t) where A = 100 Hz and B = 40 Hz. Given v(t) = cos ( 2πf c t ) and w ( t ) = cos ( 2 π ( f c + A ) t ) , where f c > A The cutoff

frequencies of both the filters are f C M (f ) −1

−A m (t)

−B

B

High Pass Filter

v (t)

A f

w (t)

Low Pass Filter

s(t)

The bandwidth of the signal at the output of the modulator (in Hz) is _____. Answer: 60 Exp: m ( t ) ↔ M ( f )

M (f )

1

−A

−B

B

A

f

After multiplication with V ( t ) = cos ( 2πf c t ) Let w1 ( t ) = m ( t ) .V ( t )

⇒ W1 ( f ) ( specturm of w1 ( t ) ) is

−fc − A

− fc − B

−fc

− fc + B

fc + A

−fc − A

− fC − A

− fC − B

fC fC + B

− fc − B

fc

− fc + B

fc + A

After high pass filter

fC + A

After multiplication with cos ( 2π ( f c + A ) t ) and low pass filter of cut off f c India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 28


− (A − B)

Bandwidth = A − B = 100 − 40 = 60

53.

O

f

(A − B)

If the electric field of a plane wave is

E ( Z, t ) = x3cos ωt − kz + 30O − y4sin ωt − kz + 45O ( mV / m ) ,

(

)

(

the polarization state of the plane wave is (A) left elliptical (C) right elliptical Answer: (A) Exp:

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)

(B) left circular (D) right circular

E ( z1t ) = 3cos ( cot − kz + 3o° ) a x − 4 − sin ( ωt − kz + 45° ) a y

E x = 3cos ( ωt − kz + 30° ) E y = − 4cos ( ωt − kz + 45° )

At z = 0 E x = 3cos ( ωt + 30o )

E y = −4sin ( ωt + 45o )

E x ≠ E y → so Elliptical polarization Q = 30° −135° = − 105° ∴ left hand elliptical (LEP)

54.

In the transmission line shown, the impedance Zin (in ohms) between node A and the ground is _________. A Z0 = 50Ω, L = 0.5λ 100 Ω

Zin = ?

50 Ω

Answer: 33.33Ω Exp:

Here =

(

λ 2

Zin = λ

2

)=Z

L

= 50Ω

∴ Zin = (100 50 ) =

100 = 33.33Ω 3



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For a rectangular waveguide of internal dimensions a × b ( a > b ) , the cut-off frequency for the TE11 mode is the arithmetic mean of the cut-off frequencies for TE10 mode and TE20 mode. If a = 5 cm, the value of b (in cm) is _____.

Answer: 2 Exp:

t c10 =

C 1   2 a

2

1 t c10 = K   ; a

t c11 = K

2 t c20 = K   a

1 1 + a 2 b2 f c10 + f c20 2 K 1 2 = + 2  a a 

Given t c11 = K

1 1 + a 2 b2

1 1 3 + 2 = 2 a b 2a 1 1 9 1 9 1 + 2 = ⇒ − + = 2 5 b 4 ( 5) 5 20 b − 0.2 + 0.45 = ∴

1 b2

1 1 = 2 ⇒ b = 2cm 2 b 2



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“India is a country of rich heritage and cultural diversity.” Which one of the following facts best supports the claim made in the above sentence? (A) India is a union of 28 states and 7 union territories. (B) India has a population of over 1.1 billion. (C) India is home to 22 official languages and thousands of dialects. (D) The Indian cricket team draws players from over ten states. Answer: C Exp: Diversity is shown in terms of difference language 2.

The value of one U.S. dollar is 65 Indian Rupees today, compared to 60 last year. The Indian Rupee has ____________. (A) Depressed (B) Depreciated (C) Appreciated (D) Stabilized Answer: B 3.

'Advice' is ________________. (A) a verb (C) an adjective Answer: B

(B) a noun (D) both a verb and a noun

The next term in the series 81, 54, 36, 24 … is ________ 4. Answer: 16 Exp:

81 − 54 = 27;27 ×

2 = 18 3

2 = 12 3 2 36 − 24 = 12;12 × = 8 3 ∴ 24 − 8 = 16 54 − 36 = 18;18 ×

5.

In which of the following options will the expression P < M be definitely true? (A) M < R > P > S (B) M > S < P < F (C) Q < M < F = P (D) P = A < R < M Answer: D


Find the next term in the sequence: 7G, 11K, 13M, ___ (A) 15Q (B) 17Q (C) 15P Answer: B

(D) 17P



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The multi-level hierarchical pie chart shows the population of animals in a reserve forest. The correct conclusions from this information are:

Re d − ant

Beetle

Tiger Elephant Mammal

Honey − bee In sec t

Leopard Re ptile Snake

Moth

Bird Crocadile Drongo

Hawk

Bulbul

Butterfly

(i) Butterflies are birds (ii) There are more tigers in this forest than red ants (iii) All reptiles in this forest are either snakes or crocodiles (iv) Elephants are the largest mammals in this forest (A) (i) and (ii) only (B) (i), (ii), (iii) and (iv) (C) (i), (iii) and (iv) only (D) (i), (ii) and (iii) only Answer: D Exp: It is not mentioned that elephant is the largest animal 8.

A man can row at 8 km per hour in still water. If it takes him thrice as long to row upstream, as to row downstream, then find the stream velocity in km per hour. Answer: 4 Exp: 4 km/hr. Speed of man=8 Left distance =d Time taken=

d 8

Upstream: Speed of stream=s ⇒ speed upstream = S' = (8 − s)

 d  t' =   8−s  Downstream: Given speed downstream = t '' =

d 8+s



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⇒ 3t ' = t '' 3d d ⇒ = 8−s 8+s 3d d ⇒ = 8−s 8+s ⇒ s = 4km / hr

9.

A firm producing air purifiers sold 200 units in 2012. The following pie chart presents the share of raw material, labour, energy, plant & machinery, and transportation costs in the total manufacturing cost of the firm in 2012. The expenditure on labour in 2012 is Rs. 4,50,000. In 2013, the raw material expenses increased by 30% and all other expenses increased by 20%. If the company registered a profit of Rs. 10 lakhs in 2012, at what price (in Rs.) was each air purifier sold? Trans Labour portat 15% ion

Plant and machinery

10%

Raw Material

20%

30%

Energy 25%

Answer: 20,000 Exp:

Total expenditure= =

15 x = 4,50,000 100

x=3×106 Profit=10 lakhs So, total selling price =40,00,000 Total purifies=200 … (2) S.P of each purifier=(1)/(2)=20,000

… (1)

10.

A batch of one hundred bulbs is inspected by testing four randomly chosen bulbs. The batch is rejected if even one of the bulbs is defective. A batch typically has five defective bulbs. The probability that the current batch is accepted is _________ Answer: 0.8145 Exp:

Probability for one bulb to be non defective is

95 100 4

 95  ∴ Probabilities that none of the bulbs is defectives   = 0.8145  100 



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Q.No. 1 – 25 Carry One Mark Each 1. The maximum value of the function f(x) = ln(1 + x)- x (where .x > - 1) occurs at x=______. Answer: 0 Exp:

f1 (x) = 0 ⇒

1 −1 = 0 1+ x

−x =0⇒x =0 1+ x −1 and f 11 ( x ) = < 0 at x = 0 2 (1 + x )

⇒

2.

Which ONE of the following is a linear non-homogeneous differential equation, where x and y are the independent and dependent variables respectively?

( A)

dy + xy = e − x dx

( B)

dy + xy = 0 dx

( C)

dy + xy = e − y dx

( D)

dy − y + e = e− y = 0 dx

Answer: A Exp:

(A)

dy + xy = e − x is a first order linear equation (non-homogeneous) dx

dy + xy = 0 is a first order linear equation (homogeneous dx (C), (D) are non linear equations

(B)

3.

Match the application to appropriate numerical method.

Application

Numerical |Method

P1: Numerical integration

M1: Newton-Raphson Method

P2: Solution to a transcendental equation

M2: Runge-Kutta Method

P3: Solution to a system of linear equations

M3: Simpson’s 1/3-rule

P4: Solution to a differential equation

M4: Gauss Elimination Method

(A) P1—M3, P2—M2, P3—M4, P4—M1

(B) P1—M3, P2—M1, P3—M4, P4—M2

(C) P1—M4, P2—M1, P3—M3, P4—M2 Answer: B

(D) P1—M2, P2—M1, P3—M3, P4—M4

Exp: 4.

P1 − M3,P2 − M1, P3 − M4,P4 − M2

An unbiased coin is tossed an infinite number of times. The probability that the fourth head appears at the tenth toss is (A) 0.067 (B) 0.073 (C) 0.082 (D) 0.091



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Answer: C Exp: P[fourth head appears at the tenth toss] = P [getting 3 heads in the first 9 tosses and one head at tenth toss]

  1 9   1  21 =  9 C3 .    ×   = = 0.082   2    2  256 5.

If z = xyln(xy), then

( A)

x

∂z ∂z +y =0 ∂x ∂y

( B)

y

∂z ∂z =x ∂x ∂y

( C)

x

∂z ∂z =y ∂x ∂y

( D)

y

∂z ∂z +x =0 ∂x ∂y

Answer: C Exp:

  ∂z 1 = y  x × × y + ln xy  = y (1 + ln xy ) ∂x xy   and

6.

∂z ∂z ∂z = x (1 + ln xy ) ⇒ x =y ∂y ∂x ∂y

A series RC circuit is connected to a DC voltage source at time t = 0. The relation between the source voltage VS, the resistance R, the capacitance C, and the current i(t) is given below: 1 t i ( u ) du c ∫0 Which one of the following represents the current f(t)? Ve = Ri ( t ) +

( B)

( A)

i (t)

i (t)

0

( C)

t

0

t

( D)

i (t)

i (t)

0

t

0

t



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Answer: A Exp:

In a series RC circuit, VS and in steady state at t = ∞ , R capacitor behaves like open circuit and no current flows through the circuit → Initially at t = 0, capacitor charges with a current of

→ So the current i(t) represents an exponential decay function i (t )

0

7.

→t

In the figure shown, the value of the current I (in Amperes) is __________. 5Ω

5Ω I

5V ±

↑ 1A

10 Ω

Answer: 0.5 Exp: V

5V

+ −

5Ω

5Ω 1A

I

10Ω

V−5 V −1 + = 0 5 15 30 ⇒ V = volts 4 V 2 ⇒ current I = ⇒ ⇒ 0.50 Amperes 15 4 Apply KCL at node V,

8.

In MOSFET fabrication, the channel length is defined during the process of (A) Isolation oxide growth (B) Channel stop implantation (C) Poly-silicon gate patterning (D) Lithography step leading to the contact pads

Answer: C India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 6


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9.

A thin P-type silicon sample is uniformly illuminated with light which generates excess carriers. The recombination rate is directly proportional to (A) The minority carrier mobility (B) The minority carrier recombination lifetime (C) The majority carrier concentration (D) The excess minority carrier concentration Answer: D Exp:

(

Recombination rate, R = B n n o + n n'

)( P

no

+ Pn'

)

n n0 & Pn0 = Electron and hole concentrations respectively under thermal equilibrium

n 'n & p 'n = Excess elements and hole concentrations respectively

10.

At T = 300 K, the hole mobility of a semiconductor µ P = 500cm 2 / V − s and

kT = 26 mV. q

The hole diffusion constant D P in cm2/s is ________ Answer: 13 Exp: From Einstein relation, D P kJ = µp q ⇒ D P = 26 mV × 500cm 2 / v − s = 13cm 2 / s

11.

The desirable characteristics of a transconductance amplifier are (A) High input resistance and high output resistance (B) High input resistance and low output resistance (C) Low input resistance and high output resistance (D) Low input resistance and low output resistance Answer: A Transconductance amplifier must have z i = ∞ and z 0 = ∞ ideally Exp: 12.

In the circuit shown, the PNP transistor has VBE = 0.7 and β = 50. Assume that R B = 100kΩ For V0 to be 5 V, the value of R C ( in kΩ ) _________________

RC V0

RB

VEE = 10 V



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Answer: 1.075 Exp:

KVL in base loop gives, IB =

10 − 0.7 = 93 µA 100K

⇒ IC = β IB = 50 × 93 µA = 4.65mA from figure, V0 = IC R C ⇒ RC =

13.

V0 5V = = 1.075 Ω IC 4.65 mA

The figure shows a half-wave rectifier. The diode D is ideal. The average steady-state current (in Amperes) through the diode is approximately ____________. D 10sin ωt ~ f = 50 Hz

R 100 Ω

C 4 mF

Answer: 0.09 Exp:

Vdc = Vm −

Idc 4fc

Idc R L = Vm −

Idc 4fc

1   Idc  R L + = Vm 4fc   ⇒ Idc =

14.

10 1 100 + 4 × 50 × 4 × 10−3

= 0.09A

An analog voltage in the range 0 to 8 V is divided in 16 equal intervals for conversion to 4-bit digital output. The maximum quantization error (in V) is _________________

Answer: 0.25 Exp:

Maximum quantization error is

step − size =

step − size 2

8−0 1 = = 0.5V 16 2

Quantization error = 0.25 V India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 8


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The circuit shown in the figure is a D

Q D Latch En Q

Q D Latch En Q

Clk

(A) Toggle Flip Flop (B) JK Flip Flop (C) SR Latch (D) Master-Slave D Flip Flop Answer: D Exp: Latches are used to construct Flip-Flop. Latches are level triggered, so if you use two latches in cascaded with inverted clock, then one latch will behave as master and another latch which is having inverted clock will be used as a slave and combined it will behave as a flip-flop. So given circuit is implementing Master-Slave D flip-flop 16.

Consider the multiplexer based logic circuit shown in the figure. W

0 MUX 1

0 MUX 1

F

S1 S2

Which one of the following Boolean functions is realized by the circuit?

( A) ( C) Answer: D Exp:

( B) ( D)

F = WS1 S2 F = W + S1 + S2

W

0 MUX

F = WS1 + WS2 + S1S2 F = W ⊕ S1 ⊕ S2

Y 0 MUX

1

F

1

S1 S2

Output of first MUX =

ws1 + ws1 = w ⊕ s1

Let Y = w ⊕ s1 Output of second MUX = Ys2 + Y s 2 = Y ⊕ s2 = w ⊕ s1 + s 2 India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 9


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Let x(t)= cos (10πt ) + cos ( 30πt ) be sampled at 20 Hz and reconstructed using an ideal lowpass filter with cut-off frequency of 20 Hz. The frequency/frequencies present in the reconstructed signal is/are (A) 5 Hz and 15 Hz only

(B) 10 Hz and 15 Hz only

(C) 5 Hz, 10 Hz and 15 Hz only

(D) 5 Hz only

Answer: (A) Explanation: x ( t ) = cos (10πt ) + cos ( 30πt ) , Fs = 20Hz Spectrum of x(t)

−15 − 5

5

t

15

Spectrum of sampled version of x(t)

−35

25

− 15 − 5

5

15

25 35

After LPF, signal will contain 5 and 15Hz component only

18.

For

an

all-pass

system

H (z) =

Re ( a ) ≠ 0, Im ( a ) ≠ 0, then b equals

(A) a

( z − b) , where H e = 1, ( ) (1 − az )

(B) a*

−1

− jω

−1

(C) 1/a*

for

all

ω .If

(D) 1/a

Answer: (B) Exp:

For an all pass system, pole =

1 1 or zero = zero * pole*

pole = a 1 zero = b 1 1 ⇒ = or b = a * b a* 19.

A modulated signal is y(t) = m.(t)cos(40000 πt ), where the baseband signal m(t) has frequency components less than 5 kHz only. The minimum required rate (in kHz) at which y(t) should be sampled to recover m(t) is __________________.



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Answer: 10 KHz. Exp: Since m(t) is a base band signal with maximum frequency 5 KHz, assumed spreads as follows: M(f ) f

m(t)

−5k

+5k

7 ∵ y(t) = m(t) cos(40000π t)  → m(f )

M(f )

f (Hz)

*1 [δ(f − 20k) + δ(f + 20k)] 2

1 [ M(f − 20k) + M(f + 20 k)] 2 Thus the spectrum of the modulated signal is as follows: ∴ y(f ) =

y(f )

−20k

−25k

15k

−15k

25k

20k

f (Hz)

If y(t) is sampled with a sampling frequency ‘fs’ then the resultant signal is a periodic extension of successive replica of y(f) with a period ‘fs’. It is observed that 10 KHz and 20 KHz are the two sampling frequencies which causes a replica of M(f) which can be filtered out by a LPF. Thus the minimum sampling frequency (fs) which extracts m(t) from g(f) is 10 KHz. 20.

Consider the following block diagram in the figure. R (s)

The transfer function

( A)

G1G 2 1 + G1G 2

+

G1

+

G2

C ( s)

+ +

C ( s) is R ( s)

( B)

G1G 2 + G1 + 1

( C)

G1G 2 + G 2 + 1

( D)

G1 1 + G1G 2



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Answer: C Exp: By drawing the signal flow graph for the given block diagram G1 R (s)

G2

1

C (s)

1

1 1

Number of parallel paths are three Gains P1 = G1G 2 , P2 = G 2 , P3 = 1 By mason’s gain formula,

C(s)

R (s)

= P1 + P2 + P3

⇒ G1G 2 + G 2 + 1

21.

The input −3e2 t u ( t ) , where u(t) is the unit step function, is applied to a system with transfer

s−2 . If the initial value of the output is -2, then the value of the output at steady s+3 state is__________________. Answer: 0 Exp: 1

function .

Y (s )

X (s )

=

S− 2 S+3

⇒ SY ( s ) + 3Y ( s ) = S × ( s ) − 2X ( s ) Due to initial condition, we can write above equation as Sy ( s ) − y ( 0 ) + 3y ( s ) = sx ( s ) − x ( 0 − ) − 2x ( s )

y ( 0− ) = −2, x ( 0− ) = 0

 x ( t ) = 3e2t u ( t )   −3  ⇒ Sy ( s ) + 2 + 3y ( s ) = ( s − 2 )   s−2 −5 ( s + 3) y ( s ) = −3 − 2 ⇒ y ( s ) = 5+3 −3t ⇒ y ( t ) = −5e u ( t )

y ( ∞ )( steady sate ) = 0 Exp:

2

s−2 ;X ( t ) = −3e2 t .u ( t ) s+3 −3 −3 ∴ X (s) = ⇒ Y (s) = s−2 s+3 H (s) =

y ( t ) at t =∞ ⇒ y ( ∞ ) = lim S.y ( s ) = lim s →0

y(∞) = 0

s →0

−3s s+3



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The phase response of a passband waveform at the receiver is given by φ ( f ) = −2πα ( f − f c ) − 2πβ f c

Where fc is the centre frequency, and α and β are positive constants. The actual signal propagation delay from the transmitter to receiver is

(A)

α −β α+β

αβ α +β

( B)

( C)

α

( D)

β

Answer: C Exp:

Phase response of pass band waveform φ ( f ) = −2πα ( f − f c ) − 2πβ f c

Group delay t y =

− dφ ( f ) 2π df

=α

Thus ' α ' is actual signal propagation delay from transmitter to receiver 23.

Consider an FM signal f ( t ) = cos [ 2 πf c t + β1 sin 2 πf1t + β 2 sin 2πf 2 t.] . The maximum deviation of the instantaneous frequency from the carrier frequency fc is

( A)

( B)

β1f1 + β2f 2

β1f 2 + β 2f1

( C)

β1 + β2

( D)

f1 + f 2

Answer: A Exp:

Instantaneous phase φi ( t ) = 2π f c t + β1 sin 2πf1 + β 2 sin 2πf 2 t d 1 φi ( t ) × dt 2π = f c + β1f1 cos 2πf1t + β2 f 2 cos 2πf 2 t

Instantaneous frequency f i ( t ) =

Instantaneous frequency deviation = β1f1 cos 2 πf1 t + β2 f 2 cos 2 πf 2 t Maximum ∆f = β1f1 + β2 f 2

24.

Consider an air filled rectangular waveguide with a cross-section of 5 cm × 3 cm. For this waveguide, the cut-off frequency (in MHz) of TE21 mode is _________.

Answer: 7810MHz. 2

Exp:

C 2 1 f c ( TE 21 ) =   +  2 9 b

3 × 1010  2   1    +  2  5 3 2

=

2

2

= 1.5 × 1010 0.16 + 0.111 = 0.52 × 1.5 × 1010 = 7.81 GHz = 7810 MHz.



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In the following figure, the transmitter Tx sends a wideband modulated RF signal via a coaxial cable to the receiver Rx. The output impedance ZT of Tx, the characteristic impedance Z0 of the cable and the input impedance ZR of Rx are all real. Transmitter

Characteristic Im pedance = Z0

Re ceiver

ZT

ZR

TX

RX

Which one of the following statements is TRUE about the distortion of the received signal due to impedance mismatch? (A) The signal gets distorted if ZR ≠ Z0, irrespective of the value of ZT (B) The signal gets distorted if ZT ≠ Z0, irrespective of the value of ZR (C) Signal distortion implies impedance mismatch at both ends: ZT ≠ Z0 and ZR ≠ Z0 (D) Impedance mismatches do NOT result in signal distortion but reduce power transfer efficiency Answer: C Exp:

Signal distortion implies impedance mismatch at both ends. i.e., ZT ≠ Z 0 ZR ≠ Z0


The maximum value of f(x)=2x3 – 9x2 +12x - 3 in the interval 0 ≤ x ≤ 3 is _______.

Answer: 6 Exp:

f 1 ( x ) = 6x 2 − 18x + 12 = 0 ⇒ x = 1, 2 ∈ [ 0,3] Now f ( 0 ) = −3 ; f ( 3) = 6 and f (1) = 2 ; f ( 2 ) = 1

Hence, f(x) is maximum at x = 3 and the maximum value is 6

27.

Which one of the following statements is NOT true for a square matrix? (A) If A is upper triangular, the eigenvalues of A are the diagonal elements of it (B) If A is real symmetric, the eigenvalues of A are always real and positive (C) If A is real, the eigenvalues of A and AT are always the same (D) If all the principal minors of A are positive, all the eigenvalues of A are also positive

Answer: B



Exp:

 −1 Consider, A  1

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1 which is real symmetric matrix − 1

Characteristic equation is A − λI = 0

⇒ (1 + λ ) − 1 = 0 2

⇒ λ + 1 = ±1

∴λ = 0, − 2

( not positive )

(B) is not true (A), (C), (D) are true using properties of eigen values 28. Exp:

A fair coin is tossed repeatedly till both head and tail appear at least once. The average number of tosses required is __________________. Let the first toss be Head. Let x denotes the number of tosses( after getting first head) to get first tail. We can summarize the even as: Event

x

Probability(p(x))

T

1

1/2

HT

2

1/2*1/2=1/4

HHT

3

1/8

(After getting first H)

and so on……….. ∞ 1 1 1 E ( x ) = ∑ xp ( x ) = 1x + 2x + 3x 2 4 8 x =1 1 1 1 Let, S =1x + 2x + 3x 2 4 8 1 1 1 1 ⇒ S= + 2x + 3x 2 4 8 16 I − II gives ( )

( I) ( II )

1 1 1 1  1 1 −  S = + + + + 2 4 8 16  2 1 1 ⇒ S = 2 =1 1 2 1− 2 ⇒S= 2 ⇒ E(x) = 2 i.e. The expected number of tosses (after first head) to get first tail is 2 and same can be applicable if first toss results in tail. Hence the average number of tosses is 1+2 = 3. India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 15


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29.

Let X1, X2, and X3 be independent and identically distributed random variables with the uniform distribution on [0, 1]. The probability P{X1+ X2 ≤ X3} is ________. Answer: 0.16 Exp: Given x1 x 2 and x 3 be independent and identically distributed with uniform distribution on

[0,1]

Let z = x1 + x 2 − x 3

⇒ P {x1 + x 2 ≤ x 3 } = P {x1 + x 2 − x 3 ≤ 0}

= P {z ≤ 0} Let us find probability density function of random variable z. Since Z is summation of three random variable x1 , x 2 and − x 3 Overall pdf of z is convolution of the pdf of x1 x 2 and − x 3 pdf of {x1 + x 2 } is 1

O

1

2

1

pdf of − x 3 is −1

P {z ≤ 0} =

( z + 1)

0

∫

2

−1

30.

2

( z + 1) dz =

3 0

6

= −1

1 = 0.16 6

Consider the building block called ‘Network N’ shown in the figure. Let C = 100µF and R = 10kΩ Network N +

+

C

V1 ( s)

R

V2 ( s )

−

−

Two such blocks are connected in cascade, as shown in the figure.

+

V1 ( s) −

+

+

Network N

Network N −

V3 ( s ) −



The transfer function

( A)

s 1+ s

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v 3 ( s) of the cascaded network is v1 ( s )

( B)

s2 1 + 3s + s2

( C)

 s    1+ s

2

( D)

s 2+s

Answer: B Exp:

Two blocks are connected in cascade, Represent in s-domain, + V1 ( s )

+

1 SC

1 SC

R

V3 ( s )

R

−

−

V3 ( s ) V1 ( s )

R . R 1 1   1  R+R+ +R + R sc  SC  SC  

=

R. R 1 1 R .  2R ( SC ) + 1 + [1 + RSC] SC SC SC

=

S2 C 2 .R.R 2 2 2 1 + 2R ( SC )  + RSC + R S C

=

S2 .100 × 100 × 10−6 × 10−6 × 10 × 10 × 103 × 103 S2 × 100 × 106 × 104 × 10−12 + 3S + 100 × 10−6 × 104 + 1

V3 ( s ) V1 ( s )

31.

=

=

S2 1 + 3S + S2

In the circuit shown in the figure, the value of node voltage V2 is 10∠0O V + −

V1

V2 4Ω

4∠0OA

(A) 22 + j 2 V

− j3Ω

(B) 2 + j 22 V

6Ω

j6Ω

(C) 22 – j 2 V

(D) 2 – j 22 V



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Answer: D 4Ω

Exp:

Super node

0

V

10 0 V + −

4 0o A

V2

6Ω

− j3Ω

j6 Ω

KVL for V1 & V2 : V1 +

+ −

−

10 0

o

V1 − V2 = 10 0o

V2 + −

....(1)

V1 = V2 + 10 0o KCL at super node: V1 V2 V2 + + =0 − j3 6 j6 V1 V2 V2 + + = 4 0o − j3 V6 j6

... ( 2 )

−4 0o +

from (1) & ( 2 ) ,

V2 + 10 0o V2 V2 + + = 4 0o − j3 6 j6

 1 1 1 10 V2  + +  = 4 0o + j3  − j3 6 j6  ∴ V2 = ( 2 − j22 ) Volts 32.

In the circuit shown in the figure, the angular frequency ω (in rad/s), at which the Norton equivalent impedance as seen from terminals b-b' is purely resistive, is __________________. 1Ω

1F b

+ 10cos ωt ~ ( Volts) −

0.5 H

b'

Answer: 2 r/sec



Norton’s equivalent impedance 1 1* jω. 2 + 1 ZN = 1 jω.1 1 + jω. 2 jω 1 = + 2 + jω jω

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b

1Ω

1F 0.5H b1

Zu

( ω2 − 2 ) − jω .  ω2 + 2 jω     = N 2 4 2  2 jω − ω   ω + 4ω  Equating imaginary term to zero i.e., ω3 − 4ω = 0

( 2 − ω ) + jω ⇒ Z 2

ZN =

⇒ ω ( ω2 − 4 ) = 0 ⇒ ω = 2 r / sec 33.

For the Y-network shown in the figure, the value of R 1 ( in Ω ) in the equivalent ∆ -network is _____________________. R1 5Ω

3Ω

7.5Ω

Answer: 10Ω Exp:

R1

3Ω

5Ω

7.5Ω

R1 =

( 7.5)( 5) + ( 3)( 5 ) + ( 7.5)( 3) Ω 7.5

R 1 = 10Ω 34.

The donor and accepter impurities in an abrupt junction silicon diode are 1 x 1016 cm-3 and 5 x 1018 cm-3, respectively. Assume that the intrinsic carrier concentration in silicon ni = 1.5 x kT 1010 cm-3 at 300 K, = 26 mV and the permittivity of silicon εsi = 1.04 × 10−12 F / cm. The q built-in potential and the depletion width of the diode under thermal equilibrium conditions, respectively, are (A) 0.7 V and 1 x 10-4 cm

(B) 0.86 V and 1 x 10-4 cm

(C) 0.7 V and 3.3 x 10-5 cm

(D) 0.86 V and 3.3 x 10-5 cm



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Answer: D Exp:

  NA ND 5 × 1018 × 1 × 1016   Vbi = VT ln = 26 mv ln  (1.5 × 1010 )2  ni2   = 0.859V

2εS Vbi  N A + N D  −5   = 3.34 × 10 cm q  NA ND 

W=

35.

The slope of the ID vs VGS curve of an n-channel MOSFET in linear regime is 10−3 Ω −1 at VDS = 0.1V. . For the same device, neglecting channel length modulation, the slope of the

(

)

I D vs VGS curve in A / V under saturation regime is approximately ___________.

Answer: 0.07 Exp:

 V 2 In linear region, I D = k ( VGS − VT ) VDS − DS  2   ∂I D = 10−3 = kVDS ∂VGS ⇒K=

∵ VDS is small,

10−3 = 0.01 0.1

In saturation region, I D = ID = ∂ ID ∂VGS

36.

VDS2 is neglected 2

=

1 2 k ( VGS − VT ) 2

k ( VGS − VT ) 2 k 0.01 = = 0.07 2 2

An ideal MOS capacitor has boron doping-concentration of 1015 cm-3 in the substrate. When a gate voltage is applied, a depletion region of width 0.5 µm is formed with a surface (channel) potential of 0.2 V. Given that εo = 8.854 × 10-14 F/cm and the relative permittivities of silicon and silicon dioxide are 12 and 4, respectively, the peak electric field (in V/µm) in the oxide region is __________________.

Answer: 2.4 Exp:

Es =

2 × 0.2 = 0.8 v / µm 0.5

E ox =

Es E s = 2.4 v / µm E ox



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In the circuit shown, the silicon BJT has β = 50 . Assume VBE =0.7 V and VCE(sat) = 0.2 V. Which one of the following statements is correct? 10 V RC 50kΩ 5V

RB

(A) For RC = 1 kΩ, the BJT operates in the saturation region (B) For RC = 3 kΩ, , the BJT operates in the saturation region (C) For RC =20 kΩ, , the BJT operates in the cut-off region (D) For RC =20 kΩ, , the BJT operates in the linear region Answer: B Exp: KVL in base loop, 5 − I B ( 50 k ) − 0.7 = 0 5 − 0.7 = 80 µA 50 k ⇒ IC = βIB = 50 × 86 µA = 4.3mA IB =

∴RC =

10 − VCE ( sat ) IC

=

10 − 0.2 4.3mA

R C = 2279 Ω and the BJT is in saturation

38.

Assuming that the Op-amp in the circuit shown is ideal, VO is given by 3R V1

R 2R

( A)

5 V1 − 3V2 2

( B)

V2

5 ZV1 − V2 2

− +

VO R

( C)

3 7 − V1 + V2 2 2

( D)

− 3V1 +

11 V2 2

Answer: D Exp: Virtual ground and KCL at inverting terminal gives V2 − V1 V2 V2 − V0 + + =0 R 2R 3R V0 V2 V2 V2 V1 = + + − 3R R 3R 2R R V0 = −3V1 +

3R

• V1

R

V2 • 2R

− • V2

+

• Vo R

11 V2 2



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For the MOSFET M1 shown in the figure, assume W/L = 2, VDD = 2.0 V, µ n Cox = 100µA / V 2 and VTH = 0.5 V. The transistor M1 switches from saturation region to linear region when Vin (in Volts) is_________________.

39.

VDD

R = 10 kΩ Vout Vin

M1

Answer: 1.5 Transistor m1 switch from saturation to linear

Exp:

⇒ VDS = VGS − VT ; where VDS = V0 and VGS = Vi

∴ VDS = V0 = Vi − VT 1 w 2 Drain current ID = µ n cos ( VGS − VT ) 2 L VDD − Vo 1 2 = × 100 × 10−6 × 2 ( VGS − 0.5 ) 10K 2 2 − ( Vi − 0.5 ) 10 K

= 100 × 10−6 ( Vi − 0.5 )

2

⇒ Vi = 1.5V

If WL is the Word Line and BL the Bit Line, an SRAM cell is shown in

40.

WL

WL

VDD BL

(A)

VDD BL

BL

BL

( B)



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WL

WL

VDD

VDD BL

BL

( C)

BL

BL

( D)

Answer: B Exp: For an SRAM construction four MOSFETs are required (2-PMOS and 2-NMOS) with interchanged outputs connected to each CMOS inverter. So option (B) is correct. 41.

In the circuit shown, W and Y are MSBs of the control inputs. The output F is given by 4 :1 MUX

4 :1 MUX I0

I0

I1

I1

VCC

Q

Q

I2

I2

F

I3

I3 W

Y

X

Z

(A)

F = W X + WX + Y Z

( B)

F = W X + WX + YZ

( C)

F = W XY + WXY

( D)

F = W + X YZ

Answer: C Exp:

I0 I1

Vcc

I2

(

)

I0 4 :1 MUX

Q

I3

I1 I2

4 :1 MUX

Q

F

I3

W

Y

X

The output of the first MUX =

Z

W × Vcc + WX.Vcc WX + WX

(∵ Vcc = log ic1)

=W ⊕ X Let Q = W ⊕ X India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 23


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The output of the second MUX = Q.Y Z + Q. Y Z = Q.Y ( Z + Z ) = Q.Y.1 = Q.Y Put the value of Q in above expression = ( WX + WX ) .Y = W X.Y + WX.Y 42.

If X and Y are inputs and the Difference (D = X – Y) and the Borrow (B) are the outputs, which one of the following diagrams implements a half-subtractor? Y

I0

( A)

I1

X

2 :1 MUX

Y

( C)

X B

2 :1 MUX

B

I1

Y

I0 I1

X

I1

B

2 :1 MUX

B

I0

S

Y

S I0

2 :1 MUX

I1

( B)

S X

D

S

2 :1 MUX

I0

2 :1 MUX S

Y

S I0 I1

Y

I1

( B)

S X

I0

D

S X

2 :1 MUX

I0

D I1

2 :1 MUX

D

Answer: A Exp: X

Y

D

B

0

0

0

0

0

1

1

1

1

0

1

0

1

1

0

0

So, D = X ⊕ Y = XY + XY and B = X.Y India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 24


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I0

Y

2 :1 MUX

D = X. Y + XY =X ⊕ Y

2 :1 MUX

B = X. Y + X.0 = X.Y + 0

I1

X

I0

Y

I1

43.

= X .Y

Let H1 ( z ) = (1 − pz −1 ) , H 2 ( z ) = (1 − qz −1 ) , H ( z ) = H1 ( z ) + r H 2 ( z ) . The quantities p, q, r −1

−1

1 1 are real numbers. Consider p = ,q = , r < 1. If the zero of H(z) lies on the unit circle, then 2 4 r = ________ Answer: -0.5 Exp:

H1 ( z ) = (1 − Pz −1 )

−1

H 2 ( z ) = (1 − qz −1 )

−1

1 − qz −1 + r (1 − Pz −1 ) (1 + r ) − ( q + rp ) z −1 1 1 H (z) = +r = = 1 − Pz −1 (1 − qz −1 ) (1 − Pz −1 )(1 − Pz −1 ) (1 − Pz −1 )(1 − Pz −1 ) zero of H ( z ) =

q + rp 1+ r

Since zero is existing on unit circle ⇒

q + rp q + rp = 1 or = −1 1+ r 1+ r

1 r 1 r − + − + 4 2 = 1 or 4 2 = −1 1+ r 1+ r 1 r − + =1+ r 4 2 ⇒r=−

r=−

5 r 5 ⇒ =− 2 2 4

or or

−

1 r + = −1 − r 4 2

3 −3r = 4 2

r = − 1 ⇒ r = −0.5 2

5 is not possible 2



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Let h(t) denote the impulse response of a causal system with transfer function

1 . Consider s +1

the following three statements. S1: The system is stable. S2:

h ( t + 1) is independent of t for t 0. h ( t)

S3: A non-causal system with the same transfer function is stable. For the above system, (A) Only S1 and S2 are true

(B) only S2 and S3 are true

(C) Only S1 and S3 are true

(D) S1, S2 and S3 are true

Answer: A Exp:

h ( t ) ↔ H (s) =

1 ⇒ h ( t ) = e− t u ( t ) s +1

S1 : System is stable (TRUE)

Because h(t) absolutely integrable h ( t + 1)

S2 :

e

h(t)

− ( t +1)

e− t

is independent of time (TRUE)

⇒ e −1 (independent of time)

S3 : A non-causal system with same transfer function is stable

1 ↔ −e − t u ( − t ) (a non-causal system) but this is not absolutely integrable thus s +1 unstable. Only S1 and S2 are TRUE

45.

The z-transform of the sequence x[n] is given by X ( z ) = convergence z > 2. Then, x [ 2 ] is ______.

1

(1 − 2z )

−1 2

, , with the region of

Answer: 12 Exp(1):

X ( z) =

1

(1 − 2z )

−1 2

=

1 1 −1 (1 − 2z ) (1 − 2z−1 )

x [ n ] = 2n u [ n ] * 2n u [ n ] n

x [ n ] = ∑ 2K.2(

n −k )

k =0

2

⇒ x [ 2] = ∑ 2k.2(

2− k )

= 20.22 + 21.21 + 22.20 = 4 + 4 + 4 = 12

k =0



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Exp(2): X (z) =

1

Z2

=

(1 − 2Z ) ( Z − 2) −1 2

2

    Z  −1  Z X(n) = Z .  Z−2 Z − 2  ↓  ↓ v( z )    u ( z ) = ∑ u m .Vn − m ( u sin g conduction theorem and u n = 2n ; v n = 2n ) n

m =0 n

= ∑ 2m.2n − m = 2n ( n + 1) m =0

∴ x ( 2 ) = 12 46.

The steady state error of the system shown in the figure for a unit step input is _______. E ( s)

R (s) + r (t)

−

e (t)

1 s+2

K=4

C ( s)

C(t)

2 s+4

Answer: 0.5 Exp:

4 2 ;H ( s ) = s+2 s+4 For unit step input,

Given G ( s ) =

k p = lim G ( s ) H ( s ) s→0

 4  2  k p = lim    s→0 s + 2   s + 4  kp = 1 Steady state error ess = ess =

1 1+1

ess =

1 ⇒ 0.50 2

A 1 + kp



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The state equation of a second-order linear system is given by x ( t ) = Ax ( t ) , x ( 0) = x 0  e− t   e − t − e −2 t  0 1 For x 0 =   , x ( t ) =  − t  and for x 0 =   , x ( t ) =  − t −2 t   −1  1   −e   − e + 2e 

3 when x 0 =   , x ( t ) is 5

( A)

 −8e − t + 11e −2 t   −t −2 t   8e − 22e 

( B)

 11e − t − 8e −2t   −t −2t   −11e + 16e 

(C)

 3e − t − 5e −2t   −t −2t   −3e + 10e 

( D)

 5e − t − 3e −2 t   −t −2 t   −5e + 6e 

Answer: B Exp:

Apply linearity principle, 3   1 0 5 = a  −1 + b 1  s       a =3; b =8  e− t  e − t − e −2 t  ⇒ x ( t ) = 3  −t  +  −t −2t   −e   −e + 2e 

 11e− t − 8e−2t  ⇒ x(t) =  −t −2 t   −11e + 16e  48.

In the root locus plot shown in the figure, the pole/zero marks and the arrows have been removed. Which one of the following transfer functions has this root locus? jω

2 1 σ

( A)

s +1 ( s + 2 )( s + 4)( s + 7 )

( B)

s+4 ( s + 1)( s + 2)( s + 7)

( C)

s+7 ( s + 1)( s + 2)( s + 4)

( D)

( s + 1)( s + 2) ( s + 7)( s + 4)



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Answer: B Exp::

For transfer function

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

From pole zero plot jω

−7

49.

−4

−2

σ

−1

Let X(t) be a wide sense stationary (WSS) random process with power spectral density SX(f). If Y(t) is the process defined as Y ( t ) = X ( 2t − 1) , the power spectral density SY(f) is

(A)

1 f  SY ( f ) = SX   e− jπf 2 2

( B)

1 f  SY ( f ) = SX   e − jπf / 2 2 2

(C)

1 f  SY ( f ) = SX   2 2

( D)

1 f  SY ( f ) = SX   e − j2 πf 2 2

Answer: C Exp: Shifting in time domain does not change PSD. Since PSD is Fourier transform of autocorrelation function of WSS process, autocorrelation function depends on time difference.

X ( t ) ↔ R x ( z ) ↔ Sx ( f ) 1 f  Y ( t ) = X ( 2t − 1) ↔ R y ( 2ζ ) ↔ Sx   2 2 [time scaling property of Fourier transform] 50.

A real band-limited random process X(t) has two-sided power spectral density

Sx (f ) =

{

10 − 6 ( 3000 − f 0

) W atts / H z

for f ≤ 3 kH z otherw ise

Where f is the frequency expressed in Hz. The signal X(t)modulates a carrier cos16000 πt and the resultant signal is passed through an ideal band-pass filter of unity gain with centre frequency of 8 kHz and band-width of 2 kHz. The output power (in Watts) is _______. Answer: 2.5 Exp: Sx ( f )

Sx ( f )

3 × 10−3 watts

−3

+3

f (in KHz )



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After modulation with cos (16000πt ) 1 Sx ( f − f c ) + Sx ( f + f c )  4 This is obtain the power spectral density Random process y(t), we shift the given power spectral density random process x(t) to the right by fc shift it to be the left by fc and the two shifted power spectral and divide by 4. Sy ( f ) =

1.5 × 10 −3 / 2

−11

−8

−5

5

8

11

f (in KHz )

After band pass filter of center frequency 8 KHz and BW of 2 kHz 1.5 × 10 −3 / 2

−9 − 8 − 7

7

8

9

Total output power is area of shaded region = 2 [ Area of one side portion ] = 2 [ Area of triangle + Area of rec tan gle ]  1  2  − × 2 × 103 × 0.5 × 10−3 + 2 × 103 × 1 × 10−3  2  =  2 = [ 0.5 + 2] = 2.5 watts

51.

In a PCM system, the signal m ( t ) = { sin (100πt ) + cos (100πt )} V is sampled at the Nyquist

rate. The samples are processed by a uniform quantizer with step size 0.75 V. The minimum data rate of the PCM system in bits per second is _____. Answer: 200 Exp: Nyquist rate = 2 × 50 Hz

= 100 samples / sec

∆=

m ( t ) max − m ( t ) min L

⇒L=

(

2− − 2

)

0.75

2 2 = 3.77 = 4 0.75 No. of bits required to encode ‘4’ levels = 2 bits/level Thus data rate = 2 × 100 = 200 bits / sec L=



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A binary random variable X takes the value of 1 with probability 1/3. X is input to a cascade of 2 independent identical binary symmetric channels (BSCs) each with crossover probability 1/2. The outputs of BSCs are the random variables Y1 and Y2 as shown in the figure. X

BSC

Y1

BSC

Y2

The value of H(Y1) + H(Y2) in bits is___________. Answer: 2

1 Let P {x = 2} = , 3

Exp:

P {x = 0} =

2 3

to find H ( Y1 ) we need to know P { y1 = 0} and P { y 2 = 1} P {Y1 = 0} = P {Y1 = 0 / x1 = 0} P {x1 = 0} + P { y1 = 0 / x1 = 1} P {x1 = 1}

1 1 1 2 1 = . + × = 2 3 2 3 2 P {y1 = 1} =

1 2

1 1 ⇒ H ( y1 ) = log 2 2 + log 2 2 = 1 2 2 Similarly P {y 2 = 0} =

1 1 and P { y 2 = 1} = 2 2

⇒ H {y 2 } = 1 ⇒ H {y1} + H {y 2 } = 2 bits

53.

Given the vector A = ( cos x )( sin y ) aˆ x + ( sin x )( cos y ) aˆ y , where aˆ x , aˆ y denote unit vectors along x,y directions, respectively. The magnitude of curl of A is ________

Answer: 0 Exp (1): aˆ x

aˆ y

aˆ z

∂ ∂ ∂ Curl A = ∂x ∂y ∂z cos x sin y sin x cos y 0 =0 ∴ Curl A = 0



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Exp(2): Given A = cos x sin yaˆ x + sin x cos y aˆ y ax ∇×A =

∂ / ∂x cos x sin y

ay

az

∂ / ∂y sin x cos y

∂ / ∂z 0

= a x ( 0 ) − a y ( 0 ) + a z ( cos x cos y − cos x cos y ) = 0

∴ ∇×A = 0

54.

A region shown below contains a perfect conducting half-space and air. The surface current ˆ amperes per meter. The tangential H K s on the surface of the perfect conductor is K s = x2 field in the air just above the perfect conductor is y

KS

Air

x

Perfect conductor

(A) ( xˆ + zˆ ) 2 amperes per meter

(B) xˆ 2 amperes per meter

(C) − zˆ 2 amperes per meter

(D) zˆ 2 amperes per meter

Answer: Exp:

D

Given medium (1) is perfect conductor Medium (2) is air ∴ H1 = 0

From boundary conditions

( H1 − H 2 ) × a n = K S H1 = 0 an = ay

K S = 2aˆ x

− H 2 × a y = 2aˆ x − ( H x a x + H y a y + H z a z ) × a y = 2a x − H x a z + H z a x = 2a x ∴ Hz = 2 H = 2a z India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 32


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(

)

Assume that a plane wave in air with an electric field E = 10cos ωt − 3x − 3z aˆ y V/m is

incident on a non-magnetic dielectric slab of relative permittivity 3 which covers the region. Z > 0 The angle of transmission in the dielectric slab is _________________ degrees. Answer: 30 Exp:

(

)

Given E = 10cos ωt − 3x − 3z a y

E = E 0e

(

− Jβ x cos θx + y cos θ y + z cos θz

)

So, β x = β cos θx = 3 β y = β cos θ y = 0 βz = β cos θz = 3 β x + β y 2 + βz 2 = β2 2

9 + 3 = β2 ⇒ β = 13 β cos θz = 3 ⇒ cos θz =

3 ⇒ θz = 61.28 = θi 13

sin θi E2 sin 61.28 3 0.8769 = ⇒ = ⇒ = sin θt sin θt E1 sin θt 1 3 θt = 30.4 ⇒ θt 30o



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Which of the following options is the closest in meaning to the word underlined in the sentence below? In a democracy, everybody has the freedom to disagree with the government. (A) Dissent

(B) Descent

(C) Decent

(D) Decadent

Answer: A Exp:

Dissent is to disagree

2.

After the discussion, Tom said to me, 'Please revert!’ He expects me to _________. (A) Retract

(B) Get back to him

(C) Move in reverse

(D) Retreat

Answer: B Exp:

Revert means set back

3.

While receiving the award, the scientist said, "I feel vindicated". Which of the following is closest in meaning to the word ‘vindicated’? (A) Punished

(B) Substantiated

(C) Appreciated

(D) Chastened

Answer: B Exp:

Vindicate has 2 meanings 1. Clear of blain 2. Substantiate, justify

4.

Let f ( x, y ) = x n y m = P. If x is doubled and y is halved, the new value of f is

( A)

( B)

2n−m P

2 m− n P

( C) 2 ( n − m) P

( D) 2 ( m − n ) P

Answer: A m

Exp:

5.

y (2x) ×   = 2n − m × x n y m 2 n

In a sequence of 12 consecutive odd numbers, the sum of the first 5 numbers is 425. What is the sum of the last 5 numbers in the sequence?

Answer: 495 Exp:

Let consecutive odd numbers be a-10, a-8, a-6, a-4, a-2, a, ……a+12 Sum of 1st 5 number = 5a-30=425 ⇒ a=91 Last 5 numbers=(a+4)+(a+6)+…….+(a+12) =(95+97+99+101+103)= 495



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Find the next term in the sequence: 13M, 17Q, 19S, ___ (A) 21W

(B) 21V

(C) 23W

(D) 23V

Answer: C Exp:

13

M

17(13 + 4) Q(M + 4) 19(17 + 2) S(Q + 2) 23(19 + 4) W = (s+ 4)

⇒ 23W

7.

If ‘KCLFTSB’ stands for ‘best of luck’ and ‘SHSWDG’ stands for ‘good wishes’, which of the following indicates ‘ace the exam’? (A) MCHTX

(B) MXHTC

(C) XMHCT

(D) XMHTC

Answer: B Exp:

KCLFTSB

SHSWDG

Reverse order:

Reverse order:

BCS TOF LUCK

GO OD W I S HES

Ace the exam Reverse order should be MAXE EHT ECA Looking at the options we have M X H T C

8.

Industrial consumption of power doubled from 2000-2001 to 2010-2011. Find the annual rate of increase in percent assuming it to be uniform over the years. (A) 5.6

(B) 7.2

(C) 10.0

(D) 12.2

Answer: B Exp:

r   A = P 1 +   100  A = 2P

n

10

r   2 = 1+   100  ∴ r = 7.2



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A firm producing air purifiers sold 200 units in 2012. The following pie chart presents the share of raw material, labour, energy, plant & machinery, and transportation costs in the total manufacturing cost of the firm in 2012. The expenditure on labour in 2012 is Rs. 4,50,000. In 2013, the raw material expenses increased by 30% and all other expenses increased by 20%. What is the percentage increase in total cost for the company in 2013?

Transpor Labour − tation 15% 30%

Plant and machinery 30%

Raw Material 20% Energy 25%

Answer: 22% Exp: Let total cost in 2012 is 100 Raw material increases in 2013 to 1.3 x 20=26 Other Expenses increased in 2013 to 1.2 x 80=96 Total Cost in 2013 =96+26=122 Total Cost increased by 22% Hint:Labour cost (i.e, 4,50,000) in 2012 is redundant data. A five digit number is formed using the digits 1,3,5,7 and 9 without repeating any of them. What is the sum of all such possible five digit numbers? (A) 6666660 (B) 6666600 (C) 6666666 (D) 6666606 Answer: B Exp: 1 appears in units place in 4! Ways Similarly all other positions in 4! Ways Same for other digits. Sum of all the numbers = (11111) X 4! (1+3+5+7+9) = 6666600 10.



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Q.No. 1 – 25 Carry One Mark Each ∞

1.

The series

1

∑ n! converges to n =0

(A) 2 ln2

(B)

(C) 2

2

(D) e

Answer: D ∞

Exp:

1

1

1

∑ n! = 1 + 1! + 2! + ......... n =0

= e as e x = 1 +

x x2 + + ......., ∀ x in R 1! 2!

The magnitude of the gradient for the function f ( x, y, z ) = x 2 + 3y 2 + z 3 at the point (1,1,1) is _________. Answer: 7 Exp: ( ∇f )P(1,1,1) = i ( 2x ) + j ( 6y ) + k ( 3z 2 ) 2.

(

)(

P 1,1,1)

= 2i + 6 j + 3k

( ∇f ) P 3.

= 4 + 36 + 9 = 7

Let X be a zero mean unit variance Gaussian random variable. E  X  is equal to _____

Answer: 0.8 Exp: X ~ N ( 0,1) ⇒ f ( x ) = ∴ E{ x } = ∫

∞

−∞

4.

=

1

=

2

2π

∞

x .f ( x ) dx

x2 ∫ x e

−x2

0

∫ 2π

∞

0

1 − x2 2 e , −∞ < x < ∞ 2π

e− u du =

2

dx 2 = 0.797 0.8 π

If a and b are constants, the most general solution of the differential equation d2x dx + 2 + x = 0 is 2 dt dt

( A ) ae − t

( B) ae− t + bte − t

( C) ae t + bte − t

( D ) ae −2t

Answer: B Exp:

A.E : − m 2 + 2m + 1 = 0 ⇒ m = −1, −1 ∴ general solution is x = ( a + bt ) e − t



5.

The directional derivative of f ( x, y ) = an angle of

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xy ( x + y ) at (1,1) in the direction of the unit vector at 2

π with y-axis, is given by _____ . 4

Answer: 3  2xy + y 2   x 2 + 2xy  1 x 2 y + xy 2 ) ⇒ ∇f = i  (  + j  2 2  2    3 3 at (1,1) , ∇f = i+ j 2 2

Exp: f =

eˆ = unit vector in the direction i.e., making an angle of

π with y-axis 4

π   π   =  sin  i +  cos  j 4  4   3  1  ˆ ∇f = 2  ∴ directional derivative = e. =3  2  2  6.

The circuit shown in the figure represents a

Ii

A l Ii

(A) Voltage controlled voltage source (C) Current controlled current source Answer: C Exp:

R

(B) Voltage controlled current source (D) Current controlled voltage source

A1I1

The dependent source represents a current controlled current source 7.

The magnitude of current (in mA) through the resistor R2 in the figure shown is_______. R2 1kΩ

10 mA

R1 2 kΩ

R3 4 kΩ

2 mA

R 4 3kΩ India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 5


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Answer: 2.8 Exp: By source transformation R 2 = 1kΩ

20V

+ −

2 kΩ

4 kΩ

I

− +

8V

3kΩ

By KVL, 20 − 10k.I + 8 = 0 28 ⇒I= 10 k ⇒ I = 2.8mA

8.

At T = 300 K, the band gap and the intrinsic carrier concentration of GaAs are 1.42 eV and 106 cm-3, respectively. In order to generate electron hole pairs in GaAs, which one of the wavelength ( λ C ) ranges of incident radiation, is most suitable? (Given that: Plank’s constant is 6.62 × 10-34 J-s, velocity of light is 3 × 1010 cm/s and charge of electron is 1.6 × 10-19 C) (A) 0.42 µm < λ C < 0.87 µm

(B) 0.87 µm < λ C < 1.42 µm

(C) 1.42 µm < λ C < 1.62 µm

(D) 1.62 µm < λ C < 6.62 µm

Answer: A Exp: E =

9.

hC 6.62 × 10−34 × 3 × 108 ⇒λ= = 0.87 µm λ 1.42 × 1.6 × 10−19

In the figure ln ( ρi ) is plotted as a function of 1/T, where ρi the intrinsic resistivity of silicon, T is is the temperature, and the plot is almost linear. ln ( ρi )

1/ T

The slope of the line can be used to estimate (A) Band gap energy of silicon (Eg)

(

(B) Sum of electron and hole mobility in silicon µ n + µ p

)

(

(C) Reciprocal of the sum of electron and hole mobility in silicon µ n + µ p

)

−1

(D) Intrinsic carrier concentration of silicon ( n i ) India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 6


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Answer: A 3

n i α T 2 e − Eg /kT

Exp:

ρι α

and

1 ηi

∴ From the graph, Energy graph of Si can be estimated

The cut-off wavelength (in µm) of light that can be used for intrinsic excitation of a semiconductor material of bandgap Eg= 1.1 eV is ________ Answer: 1.125 hC Exp: E= λ 6.6 × 10−34 × 3 × 108 ⇒λ= = 1.125 µm 1.1 × 1.6 × 10−19

10.

If the emitter resistance in a common-emitter voltage amplifier is not bypassed, it will (A) Reduce both the voltage gain and the input impedance (B) Reduce the voltage gain and increase the input impedance (C) Increase the voltage gain and reduce the input impedance (D) Increase both the voltage gain and the input impedance Answer: B Exp: When a CE amplifier’s emitter resistance is not by passed, due to the negative feedback the voltage gain decreases and input impedance increases

11.

12.

Two silicon diodes, with a forward voltage drop of 0.7 V, are used in the circuit shown in the figure. The range of input voltage Vi for which the output voltage V0 = Vi , is +

Vi

R +

D1 −1V ±

−

( A ) − 0.3V < Vi < 1.3V ( C) − 1.0 V < Vi < 2.0V

D2

VO

± 2V −

( B) − 0.3V < Vi < 2V ( D ) − 1.7 V < Vi < 2.7V

Answer: D Exp: When Vi < −1.7V ; D1 − ON and D 2 − OFF ∴ V0 = −1.7V

When Vi > 2.7V; D1 − OFF & D 2 − ON ∴ V0 = 2.7V When − 1.7 < Vi < 2.7V, Both D1 & D 2 OFF

∴ V0 = Vi India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 7


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The circuit shown represents:

vi

C2

−

+ 12 V v0

+

R2

− 12 V

R1 −2 V

C1

(A) A bandpass filter (C) An amplitude modulator Answer: D

(B) A voltage controlled oscillator (D) A monostable multivibrator

For a given sample-and-hold circuit, if the value of the hold capacitor is increased, then (A) Droop rate decreases and acquisition time decreases (B) Droop rate decreases and acquisition time increases (C) Droop rate increases and acquisition time decreases (D) Droop rate increases and acquisition time increases Answer: B 14.

Exp:

Capacitor drop rate =

dv dt

dv 1 ∝ dt c ∴ Drop rate decreases as capacitor value is increased For a capacitor, Q = cv = i × t ⇒ t ∝ c

For a capacitor,

∴ Acquisition time increases as capacitor value increased

15.

In the circuit shown in the figure, if C=0, the expression for Y is C

A B Y A B

(A) Y = A B + A B

( B) Y = A + B

( C) Y = A + B

( D) Y = A B


EC-GATE-2014 PAPER-04| Answer: A Exp: C=0

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1 A⋅B

A B

A⋅B

Y

A ⋅ B + A ⋅ B = AB

A B

A⋅B

Y = 1.A B =A B = A ⊕ B = A B + AB + AB 16.

The output (Y) of the circuit shown in the figure is

( A)

VD0

A+ B+C A

( B)

B

A + B . C + A .C A

( C)

C Output ( Y )

B

A+ B+ C

C

( D)

A.B .C

Answer: A Exp:

VDD

A

B

A

C

output ( Y )

B

C



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This circuit is CMOS implementation If the NMOS is connected in series, then the output expression is product of each input with complement to the final product. So, Y = A.B .C

= A + B +C

17.

A Fourier transform pair is given by − j6 πf

n

FT Ae 2   u [ n + 3] ⇔ 3 2   1 −   e − j2 πf 3 where u[n] denotes the unit step sequence. The value of A is _________. Answer: 3.375 n

Exp:

2 x [ n ] =   u [ n + 3] 3 −3

X ( e jΩ ) =

∞

2

∑  3 

n

.e − jΩn

n =−3

2 j3Ω   .e 3  = 2 1 − e − jΩ 3

3

 3  27 ⇒A=  = = 3.375 8 2

18.

A real-valued signal x(t) limited to the frequency band f ≤ invariant system whose frequency response is w  − j4 πf , f ≤ e 2 H (f ) =  w  0, f >  2 The output of the system is ( A ) x ( t + 4) ( B) x ( t − 4 ) ( C) x ( t + 2)

w is passed through a linear time 2

( D) x ( t − 2)

Answer: D Exp: Let x ( t ) Fourier transform be x ( t ) x (t )

h (t )

y (t )

y ( t ) = x ( t ) * h ( t ) [ convolution ]

⇒ Y ( f ) = X ( f ) .H ( f ) ⇒ Y ( f ) = e − j4 πf .X ( f ) ⇒ y ( t ) = x ( t − 2) India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 10


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The sequence x[n] = 0.5n u[n], where u[n] is the unit step sequence, is convolved with itself to ∞

∑ y ( n ) _________________.

obtain y[n]. Then

n =−∞

Answer: 4 Exp:

y [n ] = x [n ] * x [n ]

Let Y ( e jΩ ) is F.T. pair with y [ n ]

⇒ Y ( e jΩ ) = X ( e jΩ ) .X ( e jΩ ) Y ( e jΩ ) =

1 1 . − jΩ 1 − 0.5e 1 − 0.5e − jΩ

also Y ( e jΩ ) =

∞

∑ y [ n ].e

∞

⇒

− jΩn

h =−∞

∑ y [ n ] = Y ( e ) = 0.5 . 0.5 = 4 1

j0

1

n =−∞

20.

In a Bode magnitude plot, which one of the following slopes would be exhibited at high frequencies by a 4th order all-pole system? (A) – 80 dB/decade (B) – 40 dB/decade (C) +40 dB/decade (D) +80 dB/decade Answer: A Exp: → In a BODE diagram, in plotting the magnitude with respect to frequency, a pole introduce a line 4 slope −20dB / dc → If 4th order all-pole system means gives a slope of ( −20 ) * 4 dB / dec i.e. − 80dB / dec

21.

For the second order closed-loop system shown in the figure, the natural frequency (in rad/s) is U (s) + −

(A) 16 Answer: C Exp:

Transfer function

(B) 4 Y (s)

U (s)

=

4 S (S + 4 )

Y (s)

(C) 2

(D) 1

4 S + 4s + 4 2

If we compare with standard 2nd order system transfer function i.e.,

wn2 s 2 + 2ξw n s + w n 2

w n 2 = 4 ⇒ w n = 2 rad / sec India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 11


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22.

If calls arrive at a telephone exchange such that the time of arrival of any call is independent of the time of arrival of earlier or future calls, the probability distribution function of the total number of calls in a fixed time interval will be (A) Poisson (B) Gaussian (C) Exponential (D) Gamma Answer: A Exp: Poisson distribution: It is the property of Poisson distribution. In a double side-band (DSB) full carrier AM transmission system, if the modulation index is doubled, then the ratio of total sideband power to the carrier power increases by a factor of _________________. Answer: 4 23.

Exp:

Ratio of total side band power α µ2 Carrier power

If it in doubled, this ratio will be come 4 times 24. For an antenna radiating in free space, the electric field at a distance of 1 km is found to be 12 mV/m. Given that intrinsic impedance of the free space is 120 π Ω, the magnitude of average power density due to this antenna at a distance of 2 km from the antenna (in nW/m2 ) is________________. Answer: 47.7 Exp: Electric field of an antenna is     ηI0 dl 1 J   Jβ Eθ = sin θ  + 2 − βr 3  4π r r  ↓  ↓ ↓ Radiation inductive Electrostatic  field  field field   1 ∴E α r E1 r2 = ⇒ E 2 = 6 mv / m E 2 r1

P=

25.

1 E 2 1 36 × 10−8 = = 47.7 nw / m 2 2 η 2 120π

Match column A with column B.

Column A

Column B

(1) Point electromagnetic source

(P) Highly directional

(2) Dish antenna

(Q) End free

(3) Yagi-Uda antenna

(R) Isotropic

1→ P

1→ R

1→ Q

1→ R

(A) 2 → Q 3→ R

(B) 2 → P 3→ Q

(C) 2 → P 3→ R

(D) 2 → Q 3→ P



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Answer: B Exp: 1. Point electromagnetic source, can radiate fields in all directions equally, so isotropic. 2. Dish antenna → highly directional 3. Yagi – uda antenna → End fire

R

F

Figure: Yagi-uda antenna


With initial values d2y dy + 4 + 4y = 0 2 dx dx at x = 1 is ________ Answer: 0.54 Exp:

y(0)

=

y’(0)=1

the

solution

of

the

differential

equation

A.E : m 2 + 4m + 4 = 0 ⇒ m = −2, −2 ∴ solutions is y = ( a + bx ) e −2x

y ' = ( a + bx ) ( −2e −2 x ) + e −2 x ( b )

.......(1) ........ ( 2 )

u sin g y ( 0 ) = 1; y ( 0 ) = 1, (1) and ( 2 ) gives a = 1 and b = 3 ∴ y = (1 + 3x ) e−2x '

at x = 1, y = 4e −2 = 0.541 0.54 27.

Parcels from sender S to receiver R pass sequentially through two post-offices. Each post1 office has a probability of losing an incoming parcel, independently of all other parcels. 5 Given that a parcel is lost, the probability that it was lost by the second post-office is_________. Answer: 0.44 Exp: Parcel will be lost if a. it is lost by the first post office b. it is passed by first post office but lost by the second post office 1 4 1 9 + x = 5 5 5 25 P (Parcel lost by second post if it passes first post office)= P (Parcel passed by first post office) x P(Parcel lost by second post office)

Prob(parcel is lost) =



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4 1 4 = × = 5 5 25

Prob(parcel lost by 2nd post office | parcel lost)=

28.

The unilateral Laplace transform of f ( t ) is

4 / 25 4 = = 0.44 9 / 25 9

1 . Which one of the following is the s + s +1 2

unilateral Laplace transform of g(t) = t. f(t)?

( A)

(s

−s 2

)

+ s +1

2

( B)

− ( 2s + 1)

(s

2

)

+ s +1

2

( C)

(s

S 2

)

+ s +1

2

( D)

(s

2S + 1 2

)

+ s +1

2

Answer: D Exp: (1) If f ( t ) ↔ F ( s )

−d F(s) ds 1 −d   =  2  ds  s + s + 1  − ( 2s + 1) 2s + 1 =− = 2 2 ( s2 + s + 1) ( s2 + s + 1)

Then tf ( t ) ↔

Exp: (2) F (s ) =

1 s + s +1 2

L  g ( t ) = t.f ( t )  = −

=

29.

(s

d  F ( s )  ( u sin g multiplication by t ) ds 

2s + 1 2

+ s + 1)

2

For a right angled triangle, if the sum of the lengths of the hypotenuse and a side is kept constant, in order to have maximum area of the triangle, the angle between the hypotenuse and the side is (A) 12O

(B) 36O

(C) 60O

(D) 45O

Answer: (C) ( As per IIT Website) Exp:

Let x (opposite side), y (adjacent side) and z (hypotenuse) of a right angled triangle. Given Z + y = K ( cons tan t ) ......(1) and angle between them say ' θ ' then Area,

1 1 z2 xy = ( z sin θ )( z cos θ ) = sin 2θ 2 2 4 k Now (1) ⇒ z + z sin θ = k ⇒ z = 1 + sin θ

A=



∴A =

k2 4

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 sin 2θ    2  (1 + sin θ ) 

In order to have max imum area,

dA =0 dθ

2 k 2  (1 + sin θ ) ( 2cos 2θ ) − sin 2θ ( cos θ ) .2 (1 + sin θ )  ⇒  =0 4 4  (1 + sin θ ) 

⇒θ=

30.

π = 30o , Answer obtained is different than official key 6

The steady state output of the circuit shown in the figure is given by y ( t ) = A ( ω) sin ( ωt + φ ( ω) ) . If the amplitude A ( ω ) = 0.25, then the frequency ω is C y (t)

R

± sin ( ωt )

C C

( A)

1 3RC

2 3RC

( B)

Answer: B Exp:

( C)

1 RC

( D)

2 RC

V

+ −

C

R sin ωt

y (t )

C C

By nodal method,

V − 1 0o V V + + =0 R 1 2 j ωc j ωc

(

) (

)

j ωc  1 0 1 V  + j ωc + = 2  R R 2 V= 2 + 3jωRC V 1 Y= ⇒ 2 2 + jω3RC 1 1 given A ( ω) = ⇒ 4 4 + 9R 2 c2 .ω2 o

⇒ ω=

2 3 RC



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In the circuit shown in the figure, the value of V0(t) (in Volts) for t → ∞ is ______. ix 2H + 2i x −

10 u ( t ) A

+ Vo ( t ) −

5Ω

5Ω

Answer: 31.25 Exp:

ix

B

10x ( t )

2H

+ −

+ 5Ω V0 ( t ) −

2ix

5Ω

A

For t → ∞ , i.e., at steady state, inductor will behave as a shot circuit and hence VB = 5.i X

By KCL at node B, − 10 + VB − 2i x + i x = 0 ⇒ i x = V0 ( t ) = 5i x ( t ) ⇒ V0 ( t ) =

32.

50 8

250 = 31.25 volts 8

The equivalent resistance in the infinite ladder network shown in the figure is Re. 2R Re

R

R

R R

R

R

R

The value of Re/R is ________ Answer: 2.618 Exp: R

R R

R

R R

R

R

R e qu

→ For an infinite ladder network, if all resistance are having same value of R 1+ 5  Then equivalent resistance is   .R  2 

→ For the given network, we can split in to R is in series with Requivalent India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 16


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R Re q =

(1 + 5 ) R 2

R e qu

⇒ R equ = R + 1.618R ⇒

33.

R equ R

= 2.618

For the two-port network shown in the figure, the impedance (Z) matrix (in Ω ) is 30 Ω 1

+

+

10 Ω 1'

( A)

60 Ω

−

 6 24   42 9   

−

( B)

9 8  8 24   

2

( C)

2'

9 6  6 24   

( D)

 42 6   6 60  

Answer: C Exp: For the two-part network 1 1  30 + 10 Y matrix =   −1  30

1  30   1 1 + 60 30  −

Zmatrix = [ Y ]

−1

 0.1333 Z=  −0.0333 9 6  Z=   6 24 

− 0.0333 0.05 

−1

Consider a silicon sample doped with ND = 1×1015/cm3 donor atoms. Assume that the intrinsic carrier concentration ni = 1.5×1010/cm3. If the sample is additionally doped with NA = 1×1018/cm3 acceptor atoms, the approximate number of electrons/cm3 in the sample, at T=300 K, will be _________________. Answer: 225.2

34.

Exp: P = N A − N D = 1 × 1018 − 1 × 1015 = 9.99 × 1017 10 η 2 (1.5 × 10 ) η= i = = 225.2 / cm3 P 9.99 × 1017 2



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Consider two BJTs biased at the same collector current with area A1 = 0.2µm × 0.2 µm and A 2 = 300 µm × 300 µm . Assuming that all other device parameters are identical, kT/q = 26 mV, the intrinsic carrier concentration is 1 × 1010 cm-3, and q = 1.6 × 10-19 C, the difference between the base-emitter voltages (in mV) of the two BJTs (i.e., VBE1 – VBE2) is _________________. Answer: 381

35.

Exp: IC1 = IC2 ( Given ) VBE1

IS1 e VT = IS2 e BE2 ( VBE1 − VBE2 ) IS VT e = 2 IS1 V

/VT

VBE1 − VBE 2 = VT ln

(V

BE1

36.

)

IS2 IS1

 300 × 300  = 26 × 10−3 ln    0.2 × 0.2 

∵ I S αA

− VBE2 = 381mV

An N-type semiconductor having uniform doping is biased as shown in the figure. V N − type semiconductor

If EC is the lowest energy level of the conduction band, EV is the highest energy level of the valance band and EF is the Fermi level, which one of the following represents the energy band diagram for the biased N-type semiconductor?

( A)

EC EF

EC

( B)

EF EV

EV

( C)

EC EF

( D) EC EF

EV EV

Answer: D 37.

Consider the common-collector amplifier in the figure (bias circuitry ensures that the transistor operates in forward active region, but has been omitted for simplicity). Let IC be the collector current, VBE be the base-emitter voltage and VT be the thermal voltage. Also, g m and r0 are the small-signal transconductance and output resistance of the transistor, respectively. Which one of the following conditions ensures a nearly constant small signal voltage gain for a wide range of values of RE?



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VCC

Vin

Vout RE

( A)

g m R E << 1

( B)

( C)

I C R E >> VT

g m r0 >> 1

( D)

VBE >> Vr

Answer: B Exp:

AV =

RE RE IE R E = = V re + R E T + R E VT + I E R E IE

∴ AV

IC R E (∵ IC IE ) VT + IC R E

∴ IC R E >> U T ⇒ A V in almost cons tan t 38.

A BJT in a common-base configuration is used to amplify a signal received by a 50Ω antenna. Assume kT/q = 25 mV. The value of the collector bias current (in mA) required to match the input impedance of the amplifier to the impedance of the antenna is________. Answer: 0.5 Exp:

39 .

Input impedance of CB amplifier, z i = re =

VI IE

⇒ 50 =

25 mV (∵ signal is received from 50Ω antenna and VT = 25 mV ) IE

⇒ IE =

25mV = 0.5 mA 50Ω

For the common collector amplifier shown in the figure, the BJT has high β , negligible VCE(sat), and VBE = 0.7 V. The maximum undistorted peak-to-peak output voltage vo (in Volts) is______. VCC = +12V

1 µF

R1 5kΩ

vi

1 µF

R2 10 kΩ

vo RE 1kΩ



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Answer: 9.4 Exp:

∵ β = high, I B is neglected

10 k = 8V 10 k + 5k VE = VB − 0.7 = 7.3V

∴ VB = 12 ×

∴ VCE = 12 − 7.3 = 4.7V

∴ Maximum undistorted V0 ( p − p ) = 2 × 4.7V = 9.4V 40.

An 8-to-1 multiplexer is used to implement a logical function Y as shown in the figure. The output Y is given by 0 D 0 D 0 0 1 0

I0 I1 I2 I3 I4 I5 I6 I7

Y

S2 S1 S0 A

( A) ( C)

B

C

( B) ( D)

Y= A BC + ACD Y= A BC + ACD

Y= A BC + A BD Y= A BD + A BC

Answer: C Exp: Y = ABCD + ABCD + ABC

0 D 0 D 0 0 1 0

Remaining combinations of the select lines will produce output 0. So, Y = ACD ( B + B ) + ABC

= ACD + ABC

Io I1 I2 I3 I4 I5 I6 I7

= ABC + ACD 41.

8 :1 MUX

Y

S 2 S1

S0

B

C

A

A 16-bit ripple carry adder is realized using 16 identical full adders (FA) as shown in the figure. The carry-propagation delay of each FA is 12 ns and the sum-propagation delay of each FA is 15 ns. The worst case delay (in ns) of this 16-bit adder will be __________. A0

B0

FA 0

S0

A1 C0

B1

FA1

S1

A14 C1

FA14

S14

A15

B14 C14

B15

FA15

C15

S15



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Answer: 195 Exp:

A0

A1

B0

FA 0

C0

A14

B1

C1 FA1

........ .

S1

S0

A15

B14

B15

C14

C15

FA14

FA15

S14

S15

This is 16-bit ripple carry adder circuit, in their operation carry signal is propagating from 1st stage FA0 to last state FA15, so their propagation delay is added together but sum result is not propagating. We can say that next stage sum result depends upon previous carry. So, last stage carry (C15) will be produced after 16 ×12 ns = 192 ns Second last stage carry (C14) will be produced after 180 ns. For last stage sum result (S15) total delay = 180ns + 15ns = 195ns So, worst case delay = 195 ns 42.

An 8085 microprocessor executes “STA 1234H” with starting address location 1FFEH (STA copies the contents of the Accumulator to the 16-bit address location). While the instruction is fetched and executed, the sequence of values written at the address pins A15-A8 is (A) 1FH, 1FH, 20H, 12H

(B) 1FH, FEH, 1FH, FFH, 12H

(C) 1FH, 1FH, 12H, 12H

(D) 1FH, 1FH, 12H, 20H, 12H

Answer: A Exp: Let the opcode of STA is XXH and content of accumulator is YYH. Instruction:

STA 1234 H

Starting address given = 1FFEH So, the sequence of data and addresses is given below: Address (in hex) : Data (in hex) A 15 − A 8 A 7 − A 0

1F FE H → XXH 1F FF H → 34H 20 00 H → 12 H 12 34 H → YYH

43.

1 . To make s2 + s − 6 this system causal it needs to be cascaded with another LTI system having a transfer function H1(s). A correct choice for H1(s) among the following options is

A stable linear time invariant (LTI) system has a transfer function H ( s) =

(A) s + 3

(B) s - 2

(C) s - 6

(D) s + 1



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Answer: B Exp:

Given, H ( s ) =

1 1 = s + s − 6 ( s + 3)( s − 2 ) 2

It is given that system is stable thus its ROC includes jω axis . This implies it cannot be causal, because for causal system ROC is right side of the rightmost pole. ⇒ Poles at s = 2 must be removes so that it can be become causal and stable simultaneously.

⇒

1

( s + 3)( s − 2)

( s − 2) =

1 s+3

Thus H1 ( s ) = s − 2 44.

A causal LTI system has zero initial conditions and impulses response h(t). Its input x(t) and output y(t) are related through the linear constant-coefficient differential equation d2 y ( t ) dy ( t ) +a + a2y ( t) = x ( t) 2 dt dt Let another signal g(t) be defined as g ( t ) = a 2 ∫ h ( τ ) dτ + t

0

dh ( t ) + ah ( t ) dt

If G(s) is the Laplace transform of g(t), then the number of poles of G(s) is _______. Answer: 1 Exp: Given differential equation

s 2 y ( s ) + α sy ( s ) + α 2 y ( s ) = x ( s ) ⇒

y (s )

x (s )

1 = H (s ) s + αs + α 2

=

2

t

g ( t ) = α 2 ∫ h ( z ) dz + 0

= α2 = α2

=

H (s) s

d h ( t ) + αh ( t ) dt

+ SH ( s ) + α H ( s )

1 1 α +s 2 + 2 2 2 s s + α2 + α s ( s + αs + α ) ( s + 2s + α ) 2

α 2 + αs + s 2 1 = 2 2 s ( s + αs + α ) s

No. of poles = 1



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The N-point DFT X of a sequence x [ n ] ,0 ≤ n ≤ N − 1 is given by 1 N

X[k] =

N −1

∑ x [ n]e

−j

2π nk, N

0≤ k ≤N − L

n=0

Denote this relation as X = DFT(x). For N = 4, which one of the following sequences satisfies DFT(DFT (x))=x ?

( A) ( C)

x = [1 2 3 4 ]

x = [1 2 3 2 ]

( B) ( D)

x = [1 3 2 2 ]

x = [1 2 2 3]

Answer: B Exp: This can be solve by directly using option and satisfying the condition given in question

X = DFT ( x ) D FT ( D FT ( x ) ) = DFT ( X ) =

N −1

1 N

∑ X [n ] e

−j

2π nk N

n =0

DFT y [1 2 3 4] 1  1  1 1 1  10  1 − j − 1 j   2    1     = 1  2 + 2 j X= 2  4 1 − 1 1 − 1  3  4      1 + j − 1 − j  4   −2 − j2  DFT of ( x ) will not result in [1 2 3 4] Try with DFT of y [1 2 3 2] 1  1   8  4  1 1 1     1 − j − 1 j   2  1     = 1  −2  =  −1 X= 4  0  0  4 1 − 1 1 − 1  3          −2   −1 1 + j − 1 − j  2   4 1 −   1 1   DFT of =  0 4 1     −1 1 Same as x Then ‘B’ is right option

46.

1   4  2  1     − j − 1 j   −1 1  4   2  = = − 1 1 − 1   0  2  6  3        + j − 1 − j  −1 4 2 1

1

.  x1  0 1   x1  The state transition matrix φ ( t ) of a system  .  =    0 0  x 2  x 2   t 1  1 0

( A) 

1 0   t 1

( B) 

0 1  1 t 

( C) 

1 t   0 1

( D) 



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Answer: D Exp:

Given state model,  x 1 ( t )   0  =  x 2 ( t )   0

1   x1 ( t )    0   x 2 ( t ) 

0 1  A=  0 0 φ ( t ) ⇒ state transistion matrix −1 φ ( t ) = L−1 ( SI − A )    −1

s − 1 1 ⇒ 2 [SI − A ] =   s 0 s  1 2 1 s s  φ ( t ) = L−1  0 1   s  1 t  φ( t ) =   0 1  −1

47.

s 0 

1 s 

ps2 + 3ps − 2 with p a positive real parameter. s2 + ( 3 + p ) s + ( 2 − p ) The maximum value of p until which GP remains stable is ________. Consider a transfer function G p ( s ) =

Answer: 2 Exp:

Given G p ( s ) =

ps 2 + 3ps − 2 s2 + ( 3 + p ) s + ( 2 − p )

By R − H criteria The characteristic equation is s 2 + ( 3 + p ) s + ( 2 − p ) = 0

i.e. s 2 + ( 3 + p ) s + ( 2 − p ) = 0 By forming R-H array, s2 1 s1 ( 3 + φ ) s0 ( 2 − p )

(2 − p) 0

For stability, first column elements must be positive and non-zero i.e. (1)( 3 + p ) > 0 ⇒ p > −3 and ( 2 )( 2 − p ) > 0 ⇒ p < 2

i.e. −3 < p < 2 The maximum value of p unit which G p remains stable is 2 India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 24


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The characteristic equation of a unity negative feedback system 1 + KG(s) = 0. The open loop transfer function G(s) has one pole at 0 and two poles at -1. The root locus of the system for varying K is shown in the figure. jω

ζ = 0.5 A x −1

1 − 3

( 0,0) x O

σ

The constant damping ratio line, for ζ = 0.5 , intersects the root locus at point A. The distance from the origin to point A is given as 0.5. The value of K at point A is ________ . Answer: 0.375 Exp:

We know that the co-ordinate of point A of the given root locus i.e., magnitude condition G (s) H (s) = 1

Here, the damping factor ξ = 0.5 and the length of 0A = 5 ξ = 0.5 A

*

Then in the right angle triangle cos θ =

OX OX 1 ⇒ cos 60 = ⇒ OX = OA 0.5 4 AX AX 3 ⇒ sin 60 = ⇒ AX = OA 0.5 4

⇒ sin θ =

−1

*

−2

3

* −1

θ X

*O

3

So, the co-ordinate of point A is −1 + j 3 4 4 Substituting the above value of A in the transfer function and equating to 1

i.e. by magnitude condition, k s ( s + 1)

=1

2 s = −1 + j 3 4 4

1 3  9 3  k= + . +  16 16  16 16 

2

k = 0.375 India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 25


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Consider a communication scheme where the binary valued signal X satisfies P{X = + 1}=0.75 and P{X = -1}= 0.25. The received signal Y = X + Z, where Z is a Gaussian random variable with zero mean and variance σ 2 . The received signal Y is fed to the threshold ˆ is: detector. The output of the threshold detector X

ˆ = { +1. X −1.

Y>τ Y ≤ τ.

{

}

ˆ ≠ X , the threshold τ should be To achieve a minimum probability of error P X (A) Strictly positive (B) Zero (C) Strictly negative (D) Strictly positive, zero, or strictly negative depending on the nonzero value of σ 2 Answer: C Exp: C H1 : x = +1; H 0 : x = −1

P ( H1 ) = 0.75; P ( H 0 ) = 0.25 Received signal γ =X+Z Where Z ∼ N ( 0, −2 ) ; f Z ( z ) =

1

e− Z

2

2 σ2

σ 2π if X = 1  1+ Z Received signal γ =   −1 + Z if X = −1 f γ ( y H1 ) = f γ ( y H0 ) =

−

1 σ 2π

e

1 2 σ2

−

1 σ 2π

e

1 2 σ2

( γ−1)2

( γ+1)2

At optimum threshold yopt: for minimum probability of error

f γ ( y H1 )

f γ ( y H0 ) −

e

= y = yopt

P ( H0 ) P ( H1 )

1  ( γ−1)2 − ( γ+1)2  2 σ 2 

= yopt

e

+ 2 yopt σ2

y opt =

=

P ( H0 ) P ( H1 )

P ( H0 ) P ( H1 )

σ 2  P ( H 0 )  −1.1σ 2 ln  = −0.55σ 2  = 2  P ( H1 )  2

y opt = Optimum threshold y opt < 0 ∴ Threshold is negative. India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 26


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Consider the Z-channel given in the figure. The input is 0 or 1 with equal probability. 1.0

0

0

INPUT

0.25 OUTPUT

1

0.75

1

If the output is 0, the probability that the input is also 0 equals ______________ Answer: 0.8 Exp:

Given channel X=0

1.0

input

0.25

Y =0 output

X =1

Y =1 0.75

We have to det er min e, P {x = 0 / y = 0} P {x = 0 / y = 0} =

P { y = 0 / x = 0} P {x = 0} P { y = 0}

=

1. 1

2

1 1. 1 + 0.25 × 2 2

=

4 = 0.8 5

51.

An M-level PSK modulation scheme is used to transmit independent binary digits over a band-pass channel with bandwidth 100 kHz. The bit rate is 200 kbps and the system characteristic is a raised-cosine spectrum with 100% excess bandwidth. The minimum value of M is ________. Answer: 16 Exp:

1 (1 + α ) T [Where T is symbol duration. α is roll of factor]

Bandwidth requirement for m-level PSK =

1 (1 + α ) = 100 × 103 T α = 1 [100% excess bandwidth ]

⇒

1 ( 2 ) = 100 × 103 T Bit duration 2 ⇒T= 1 100 × 103 = = 0.5 × 10−5 = 5 × 10−6 sec 3 200 × 10 = 20 µ sec ⇒

Bit duration =

Symbol duration 20 × 10−6 sec ⇒ log 2 m = = 4 ⇒ M = 16 log 2 m 5 × 10−6



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Consider a discrete-time channel Y =- X + Z, where the additive noise z is signal-dependent. In particular, given the transmitted symbol X ∈{ − a, +a} . at any instant, the noise sample Z is chosen independently from a Gaussian distribution with mean β X and unit variance. Assume a threshold detector with zero threshold at the receiver. When β = 0, the BER was found to be Q(a) = 1 × 10-8   Q ( v ) =

1 2π

∫

∞ v

e − u / 2du, and for v > 1, use Q ( v ) ≈ e − v 2

2

/2

 

When β = −0.3, the BER is closest to (A) 10-7 (B) 10-6 Answer: C X∈[-a,a] and P(x = -a) = P(x = a) = ½ Exp:

(C) 10-4

(D) 10-2

γ =X+Z → Received signal

Z ∼ N ( βX,1) fZ ( z ) =

1

e

1 2 − ( Z −βX ) 2

2π  −a + z if x = −a γ=  a + z if x = + a

Q ( a ) = 1 × 10−8 Q ( a ) ≈ e−ϑ

2

2

H1 : x = + a H 0 : x = −a and Threshold = 0 f γ ( y H1 ) =

1

fγ ( y H0 ) =

1

2π 2π

e

−

e

1 ( y − a (1+β ) )2 2

−

1 ( y + a (1+β ) )2 2

BER : Pe = P ( H1 ) P ( e H1 ) + P ( H 0 ) P ( e H 0 ) ∞

1 1 − 12 ( y − a (1+β) )2 1 1 − 12 ( y + a (1+β ))2 e dy e dy = Q ( a (1 + β ) ) + ∫ 2π 2 −∞ 2 −∫0 2π β=0 0

=

Pe = Q ( a ) = 1 × 10−8 = e− a

2

2

⇒ a = 6.07

β = −0.3 Pe = Q ( 6.07 (1 − 0.3) ) = Q ( 4.249 ) Pe = e

− ( 4.249 ) 2 2

= 1.2 × 10−4

Pe 10−4.



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The electric field (assumed to be one-dimensional) between two points A and B is shown. Let ψ A and ψ B be the electrostatic potentials at A and B, respectively. The value of ψ B − ψ A in Volts is ________.

40 kV / cm

20 kV / cm

0 kV / cm

A 5µ m

B

Answer: -15 Exp:

A

B

(5 × 10

( 0 kV / cm, 20kV / cm ) E − 20 =

−4

kV / cm, 40 kV / cm )

40 − 20 x − 0 ) ⇒ E = 4 × 104 x + 20 −4 ( 5 × 10 B

5×10−4 / cm

A

0

VAB = − ∫ E.dl = − ∫

( 4 × 10 x + 20 ) dx

5×10−4

  x2 = −  4 × 104 + 20x  2  0

4

= − ( 2 × 104 × 25 × 10−8 + 20 × 5 × 10−4 )

= − ( 50 × 10−4 + 100 × 10−4 ) = −150 × 10−4 kV ⇒ VAB = −15V

54.

Given F = zaˆ x + xaˆ y + yaˆ z . If S represents the portion of the sphere x 2 + y 2 + z 2 = 1 for z ≥ 0 , then ∫ ∇ × F . ds is ___________. S

Answer: 3.14 Exp:

∫ ∇ × F.ds = ∫ S

C

F.dr(u sin g stoke 's theorem and C is closed curve i.e.,

x 2 + y 2 = 1, z = 0 ⇒ x = cos θ, y = sin θ and θ : 0 to 2π = ∫ zdx + xdy + ydz C

2π

= ∫ xdy = ∫ cos θ ( cos θ dθ ) C

0

2π

1 sin 2θ  = θ +  = π 3.14 2 2 0



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If E = − ( 2y3 − 3yz 2 ) xˆ − ( 6xy 2 − 3xz 2 ) yˆ + ( 6xyz ) zˆ is the electric field in a source free region, a valid expression for the electrostatic potential is

( A)

xy 3 − yz 2

( B)

2xy 3 − xyz 2

(C)

y3 + xyz 2

( D)

2xy 3 − 3xyz 2

Answer: D Exp:

Given E = − ( 2y3 − 3yz 2 ) a x − ( 6xy 2 − 3xz 2 ) a y + 6xyz.a z By verification option (D) satisfy E = −∇V


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