Kreatryx GATE 2018 ECE Solutions.pdf

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GATE 2018 Detailed Solutions BY

KREATRYX

ECE th

10 Feb 2018 Morning Session

Note:Don't forget to ll out your marks in the Rank Predictor Form at the end

Klassroom 2019 Program for GATE-EE GATE-EE and GATE-EC GATE-EC in Kalu Sarai, Sara i, New Delhi

Technical Question: 1

Answer: (C) Solution:  For differential equation degree of each term should be 1 & then A & B are linear

Hence A & B are linear

 For linear system, y  t   mu  t  But in (c), y  t   au t   b & hence system is non-linear non -linear

 Option (D) is linear as integration is linear process

Klassroom 2019 Program for GATE-EE and GATE-EC in Kalu Sarai, New Delhi. Call us – 8130183640 www.kreatryx.com

Question: 2

Answer: (0.48) Solution: min 1 156    0.48 min 2 2 325


Question: 3

Answer: (D) Solution: The input impedance of trans impedance amplifier is low & low output impedance because it

converts current to voltage


Question: 4

Answer: (0)

The coefficient of x2 in Taylor series is x





f   0  2!

2

t 2

f  x   e dt 0

f x  e



x2 2

f   x   xe



x2 2

f   0   0

Hence, coefficient of x2 is zero


Question: 5

Answer: (C) Solution: f  A,B, C  ABC  ABC  ABC 

m0, 2, 4 

In maximum term f  A,B,C  M1,3,5,6,7    A  B  C  A  B  C  A  B  C   A  B  C   A  B  C 


Question: 6

Answer: (0.125) Solution:

 1  1  P  1  P     2  1  1  1  0.5   1  0.25  0.5   1  0.25  2  2 

Hence error Pe   P 1  P 

Pe  0.125


Question: 7

Answer: (B) Solution: For a discrete time all pass system Z

1 p*

Also, complex poles exists in conjugate pair p1  0.250 p2  230 z1  z2 

1 *

0.250  1

 230 

*

 40

 0.530

p3  2  30 z3  0.5  30

Since pole lie outside the unit circle, so system is stable if ROC includes unit circle which means ROC is annular region between two concentric circles. Hence, impulse response must be two sided.


Question: 8

Answer: (C) Solution: To make the true period T  40 the signal

t y t  x   4 Scaling operation dose not affected the Fourier series coefficients & it only changes the fundamental frequency Hence bk  ak 

b

k 

k

 16


Question: 9

Answer: (C) Solution: MOS cascade is used to provide current buffering for the output of a common source

amplifying transistor. Earlier we saw that Common Gate configuration is used as current buffer. As the transistor Q 2 is in CG configuration with biasing voltage VG2 it acts as current buffer In Common Gate configuration we know that output resistance increases by a factor and the input resistance decrease by a factor. Let say factor is K R 0  r01  K  Kr02


To find R0 set Vi  0 Small signal model for the given circuit for output resistance

Vgs2  ix  r01 Vx   ix  gm2 Vgs2  r02 Vx  ix r02  1 gm2 r01  r02 ix Vx  r01  r02  r01r02gm2  ix

R 0  r01  r02  r01r02 gm2  R 0  r0 1r0 2 gm2 R 0   gm2r02  r01

gm2r02  A0 = Intrinsic gain of Common Gate Amplifier

R 0  A02r01 gm 

i0 Vi

As the Q 2 is current buffer only gm2Vgs2  gm1Vi

The overall trans-conductance would be gm 

i0  gm1 Vi


Question: 10

Answer: (D) Solution: 2 2 f  2ax  xy  y ax  by  ax2 y  by 3   x x2 y 2 x2 y 2

At x  1, y  2

f 2a  8b  4 x 2 2 f  2by  xy  ax  by  by2 x  ax3   y x2 y2 x2 y 2

At x  1, y  2

f 4b  a  y 4 Since,

f f  x y 4b  a



4

2a  8b 4

12b  3a

a  4b


Question: 11

Answer: (2) Solution:

Since infinite solution are possible for AX  0 , it means that matrix A is singular A  0 k

2k 0 k  k k2 2

k 3  2k k 2  k   0 2k 2  k 3  0

k2 2  k   0

k 0 & k 2


Question: 12

Answer: (B) Solution: Initially capacitors are unchanged. So zener diode is OFF & both capacitors are in sense.

Due to negative feedback, we consider virtual ground V0 Isup 

1  1mA 1k

This current flows through capacitors and charges them by equal amount and as soon as voltage becomes 2.5V across each zener diode turns ON In this case

The voltage becomes 2.5V when C

dv 1 dt

1f 

2.5  1mA t1


t1  2.5msec

Beyond this only C1 charges and VC2  2.5V and as soon as VC1  7.5 V , V0  10V dv 1 dt 5 1F   1mA t2 C

t2  5msec

Total time to charge upto 7.5V  t1  t 2  7.5msec


Question: 13

Answer: (0.001) Solution: For distortion-less line,

  RG and Z0  R G



G

R Z 20

0.05 R   0.001np meter Z0 50


Question: 14

Answer: (D) Solution: Close loop frequency respose can be obtained by intersection of nyquist plot with constant M and constant N circles Hence option (d) is false


Question: 15

Answer: (4.8) Solution: B

V I2

V 2 0

Short circuit terminal 2 I1 

V1 2  5112

I2  I1 



V1 2

10 7

5 7

5 7  5V1 I 2  10 24 2 7 V1 

V1

I 2



24  4.8 5


Question: 16

Answer: (0.416) Solution: As we know W

2   V0  VF   1 1     q  NA ND 

So, W  V0  VF Hence When VF  0V Then W1  V0 and W1  1m When VF  ? W2  0.6m



W1 W2

1  0.6



V0 V0  VF 0.65 0.65  VF

VF  0.416V


Question: 17

Answer: (D)


Question: 18

Answer: (B) Solution: Since P  s  has no real roots it means P  s  never becomes zero for any real values of ‘s’ thus

implies P  s  is either positive or negative for all real values of s Hence,  real s P  s  is either increasing or decreasing So it can have only one real root


Question: 19

Answer: (5) Solution: Total time period = 70 + 5 + 75 = 150 sec Clock period = 5 sec Number of clock cycles = 30 So, we need to divide clock frequency by 30 & hence there must be 30 states. 2n  30 Hence, n = 5 So, 5 flip flops are required.


Question: 20


P  x 4  x3  

 x 4  x2   x 4  x1 

All are independent

 P  x 4  x 3  P  x 4  x 2  P  x 4  x1   P  x 4  x3  P  x 4  x 2  P  x 4  x 1 

 P  x 4  x3  0  P  x 4  x 2  0  P  x 4  x 1  0  All are Gaussian Random Variable with mean 0 & variable 2 1 2

1 1 2 2

   

1  0.125 8


Question: 21

Answer: (C)


Question: 22

Answer: (C) Solution: If a matrix has n’ distinct Eigen vectors then it have n’ linearly independent Eigen vectors

So, statement 2 implies statement 1


Question: 23

Answer: (0.25) Solution: 4cos 4cos  2400t  cos  2000  cos  2800 t

Comparing with standard eg. Ac cos c t 



ma 

Pm Pc

2

cos  c  m  t 

Acma 2

cos  c  m  t

2  ma  0.5 Ac

Now ma  Now

Acma



Am Ac



1  Am  2 2

2  0.25 8


Question: 24

Answer: (16) Solution: No. of distinct code words for n-variable is given by  2n

No of distinct code words for –variable with at least ‘K’ hamming distance is 2n  2nk 1 K 1 2

 here n  5 k2

So 252 1  16


Question: 25

Answer: (D) Solution: For CMOS logic the overall logic is complement of logic of NMOS gate

In series, the logic is AND & in parallel logic is OR F  AB  AB  AB  AB

So, it behaves as XOR gate


Question: 26

Answer: (0.75) Solution: f = 1.25 fc TE01  2 TE1.0 V V  2 2b 2a

a = 2b

b 

4

  fc  1    f 

2

2  2b 1.25

 1  1    1.25  b = 0.75

2


Question: 27

Answer: (A) Solution: dy x2  y 2 y   dx 2y x

 1 1  x2 dy  y    dx  x 2  2y y

 1 1  x2 dy  y2     dx x 2 2

y

 1 1  x2 dy  y2     dx  x 2  2y

Assume y 2  t 2y y

dy dt  dx dx

dy 1 dt  dx 2 dx

1 dy 1  1 1  x2     2 dx 2  x 2  2 dt  2   t  t    x2 dx  x 

Integrating factor  e t.

 2   1  dx  x



e 

 x lnx2 



e x x2

x e x 2 e x dx  c   e x  c  2 2 x x




y 2   x2  cx 2ex

At x  1 , y  0 0   1  ce

c  e 1 y 2   x2  x2e

e

x 1 

1

x 1

y2 x2



 x  1   ln 1  

y2 

 x2 


Question: 28

Answer: (4.5) Solution: y z r  2x   x x x

Since x & y are independent

y 0 x

r z 1 x x Since, z3  xy  yz  y3  1 Differentiate both sides with respect to x z z yy 0 3z2 x x Here, Y has been treated as constant with respect to x y z  2 x 3z  y

1 z 1 2, 1,1     31 x 2 1 r  4   4.50 2 x


Question: 29

Answer: (A) Solution: 1

T  s   C sI  A  B

s  4 1.5  s   s 1.5  1    2  4 s  4  s  4s  6

 sI  A    4   sI  A 

1

2s 1  2  8  s  4s  6 1 3s  8  0.625 3s  5  2 C  sI  A  B  2 s  4s  6 s  4s  6

sI  A 

1

B  


Question: 30


t Arect    ATsinc  fT  T  f t ATsinc  tT   Arect     Arect    T T By comparing ATsinc tT 4sinc 2t  T  2, A  2

 f  4sinc 2t   2rect   2 By parseval’s theorem 

 y t

2





2 1 dt  y   d 2 



By Hilbert Transform, magnitude of Fourier transform does not change Y    X    

 y t 



2



f 1 dt  4rect2   d  2   2





  f   4rect  2  df  4  1df  8  2


Question: 31

Answer: (C) Solution: Output of 1st multiplexer  UV  0   UV 1  UV 1  UV  0

 UV  UV Output of 2nd multiplexer  UV  UV  WX  UV  UV WX  0WX  0WX

 UV  UV  W X  X   W UV  UV 


Question: 32

Answer: (-0.5) Solution:  0.5   0.5  Pe  P  P  0.5   P    P 0.5   0.5   0.5 



1 1 1 3 0.5          0.5   2 4 2 4

1 1 1 4 2 2 1  3 3     16 8 8 16 1  Pe   4 4 0.5    0.5

    0.5        0.5 

3 4

Pe will be minimum when   0.5


Question: 33

Answer: (C) Solution: Input frequency   5 rad sec Xc 

1 1   0.2 M c 5

 200 k

By voltage division

c 

5    j200  200  j200



5   j1  j  2

 2.5 1  j  2.5 2 

 4

c  t  2.5 2 s in 5t  0.25 


Question: 34

Answer: (0.8415) Solution: Since probability of data input transition is 0.3 So, it means D changes its value thrice in every 10 cycles Assume initially D = 0 & Q = 0 So, X = 0

Average value of x = 3.3 

3  TCK  T  10 TCK



9.9  0.85  0.8415 V 10


Question: 35

Answer: (0.42) Solution: IM1  Im2

Assume both is sat. 2 2 w  w  L   VGS1  VT    L   VGS2  VT   1  2 2

 VGS1  VT   2  VGS2  VT 

2

w w  L   2 L   2  1 VG1  VG2  2 2

 2  1  2  2  Vx  1 1  2 1  Vx 

2

2

1  2 1  Vx  Case (i)

2Vx  2  1

Vx  2  1

Vx  0.29

Case (ii)

2Vx  2  1

Vx  1.70

In both case Tr is not is saturation Tr  2 is sat & Tr 1 in linear region.

IDS1  IDS2


2

w    w V V  1 nCox     VGS1  VT  VDS1  VDS   nCox    GS2 T  2 1 2  L 1   L 2  

2

2 1 Vx  Vx 2  1 Vx  2 2

2Vx  Vx 2  1  Vx2  2Vx 

2Vx  Vx2  2  2Vx2  4Vx 3Vx2  6Vx  2  0 Vx  1.57 Vx  0.42 Vx = 1.57 is not valid

Vx  0.42


Question: 36

Answer: (B) Solution:



1



 10

 E1   E   2

20 dB  20 log 

E1  10 E2 E2  0.1E1

E1 ex  0.1E1 e10x  0.1

10x = 2.3 X = 0.23


Question: 37


For y=1 only of the inputs to the or gate must be 1 For second input to be 1. x3  1 & other three variables can have any value. For first input to be 1, the output of both inverter & AND gate should be 1 For inverter output to be 1, x0  0 & other three variables can have any value For AND gate output to be 1. x0  1 & other input should also be 1 & hence output of AND gate can never be 1. So, there can only be 8 combinations for which y=1


Question: 38

Answer: (0.316) Solution: Req  2, L  1H 1 2

Time constant,   sec   V 1 i  t   1  e    1  2e2t  R  2 t

At t  0.5sec i t 

1 1  e1   0.316A  2


Question: 39

Answer: (14.925) Solution: Mr 

1

2

2 1  2

42 1  2   4 4  4 2 

 4  2 

1 4

1 0 4

1 0 16

2  0.933, 0.067   0.966, 0.2588 For resonance peak,   0.707 , hence   0.2588 s2  2s  k  0

 K

2 2 k 1

2



1 k

 0.2588

 14.925


Question: 40

Answer: (B) Solution: 2 2 e t  e   f

 t   x t 





2 2

x t / 2  2 e2 f 2

e t

/2

 2e  22f 2

0.5e t

2

2



 

H t  

2 2

e2 f

2



2

H f  

2

2 2

e2  f

2

e  42f 2

Sy  f   S   f  H  f  

Sy  f  

1  4 2f 2  4 2f 2  e  e 2 2 4

Power 

2

 4





0

2







4



2

e

 2 f 

e  42f 2 df 

 8

 0.22


Question: 41


Small signal resistance of diode  Reactance of capacitance 1

26mV  1k 26A

1  1k wc 2  10  0.5  109



6

Small signal model

Zer  100  (1000)11( j1000)  100  (500  j500)  600  j500

781.02 / 39.8 Amplitude of small signal diode current



5mV  6.4A 781.02


Question: 42

Answer: (B) Solution: As we know

R  G  B Hence ER  EG  EB

  1.24  E eV    cm  

Now Built in potential  V0  is given by

 NAND   n2  V  i 

V0  VT ln 

Now EG

mi  e

Hence V0  EG So correct order is VBR  VBG  VBB


Question: 43

Answer: (-0.207) Solution:

2 y 2

dt



dy 5y  dt 4

Taking Laplace transform both sides s2 Y  s   sy  0  y  0  sY  s  y  0 

5 Y  s 4

 2 5  s  s  4  Y  s    s  1    1 1 s   2 s 1 s1 2 Y s     2  2 2 5  1  1  1 s2  s  4  s  2   1 s  2   1 s  2   1       Taking inverse Laplace transform

 t  1 t y  t    e 2 cost  e 2 sint  2   At t   y t  e



 2

cos  

1  2 e sin  2





  e 2 = -0.2078


Question: 44

Answer: (0.5) Solution: Since the opamp has negative feedback so there may be the case of virtual ground let’s

assume virtual ground of terminal  VA  then using KCL 1  0 0  V0  1k 31k

V0   31V

But this is not possible because opamp goes into saturation region and V0 can’t be greater than 15V and less than -15V Hence, V0  15V i.e., our assumption that opamp is in virtual ground condition is wrong Now 1  VA 1k



VA  V0 31k

As we have V0  15V So, VA  0.5V


Question: 45

Answer: (D) Solution: c  t  =Accos2fct

m  t   cos 2fmt 

Fc  5f m Vi  c  t   m t  V0  av i  bv i2  t   a  c(t)  m(t)  b  c(t)  m(t)

2

V0  a  c(t)  m(t)  bc2 (t)  bm2 (t)  2bm(t)c(t) V0 is passed through 1 deal band pass filter centre as f c m  t   fm ; fm c  t   fc , fc c2 t   2fc , 2fc

m2  t  2fm , 2fm m  t  c  t   fc  fm to fc  fm

fc  5m After band pass filter

a cos  fc t  2b cos2fmt cos 2 fc t

a  2b cos2fmt  cos 2fct


Pc 

a2 2

4b2 P side band   b2 4

P side band b 2 1  2 2 pc 2 a

a  2b  a  2 b


Question: 46

Answer: (13) Solution

The bandwidth of original signal is 5kHz so the spectrum will lie between -5kHz and 5kHz and when we sample the signal, the spectrum will contain the copies of original spectrum at nfs For reconstruction the filter must only filter the original spectrum and no copy of spectrum should be included The left edge of first copy will lie at f s-5 and it should lie ahead of 8kHz so that it is filtered out by filter Hence f s- 5> 8 and hence minimum value of f s is 13 kHz


Question: 47

Answer: (3) Solution: Inverse DFT, x n 

 j kn 1 N1 X k  e N N k 0



2

2 j kn 1 7 4 x n   8 k 0 K  1 e



3 4 3 7 j n j n j n j n j n j n  1 jn 4 2 2 2 4   1  2e  3ee  4e  5e  6e  7e  8e 4  8 













3

 x 2n  x 0  x 2  x 4   x 6  n 0

x 0   

36 8

x 2 

1 1 1  2j  3  4 j  5  6j  7j  8 j   4  j4  8 8

x  4  

1 4 1  2  3  4  5  6  7  8  8 8

x 6  

1 1 1  2 j  3  4 j  5  6j  7  8j   4  4 j 8 8

3

 x 2n  18 36  4  4  4   3 n 0


Question: 48

Answer: (0.1) Solution: At   PC (Phase cross-over frequency)

Gain = 20 dB For marginally stable system Gain = 0dB at   PC Hence, the gain K must be -20 dB to make overall gain 0 dB & make the system marginally stable 20 logK   20 logK  1 K  101  0.1

Hence, if K<0.1, system becomes unstable


Question: 49

Answer: (2) Solution: This is a case of oblique incidence No reflection occurs at Breusster’s angle

Hence tan B 

r2 r1

Now from diagram and equilibrium

 tan B  2 So,

r2  2 and r1  1 r1

So, r2  2


Question: 50

Answer: (2) Solution: 1

 j

.

z 0

f  z 

dz 2

1



1 z 1 2

1

j

 f z  dz 0

has 2 poles z  1 & z  1

Based on contour shown, both lie inside contour but both are in opposite direction one clockwise and one anti clockwise.

 1 1  1   2    z  1  z  1  2

f(z)  


Question: 51

Answer: (0.119) Solution: For pp junction

 NA1    NA2 

VB  VT ln 



16 VB  25.8ln 10

mV  25.8ln10 mV 0

14

10

VB  0.119V


Question: 52

Answer: (A) Solution: As w0 goes high. The first diode turns ON which pulls up the bit line & makes B0=1

But the line B1 which is connected to w, remains open & hence B1=0 Similarity when w1 is asserted, B1 gets high & B0 = 0

1 hence  0

0



1


Question: 53

Answer: (B) Solution: Range of input voltage  6v  5%  5.7v to 6.3v

The output voltage due to zener diode is 5v I0  5

 5mA 1k The minimum zener current is 2mA Hence, minimum current through source = 7mA R1 

6.3  5  186 7mA

5.7  5  100 7mA The value of R  100 should be considered so that current always remains higher 2mA & R2 

zener operates in breakdown region. For Vin  6.3v Is 

6.3  5  13mA 100

Iload  5mA Iz(max)  13.5  8mA

Pz  5  8mA  40mW


Question: 54


 2     1

V0c  Voc  VT ln  2

1

 0.2   0.65  0.026ln    0.608  1 


Question: 55

Answer: (8) Solution: The network is symmetric about line joining the terminates So, network can be folded

R eq  R 1  R1  R 1 11 R 3  0.5  0.5  0.5 11 1.5  1 2

I

2

2

2

1.5  0.5  1 0.375  1.375  8A 2

11  8A 1.375


Aptitude Question: 1

Answer: (B) Solution:

tan   tan  

1.5  0.5 3 h 33

 0.5

h  3m


Question: 2

Answer: (B) Solution: If the sum of digits is divisible by 3, then the either no is divisible by 3

Let missing numbers = x Sum  4  1  5  x  4  2  3  22  x The minimum whole numbers for

22  x

to be divisible by 3 is x  2


Question: 3

Answer: (D) Solution: First term, a = 1

Common ratio, r 

1 4

Sum of infinite series, S 

a 1r



1 1

1 4



4 3


Question: 4

Answer: (B) Solution: By giving him the last piece of cake, you will ensure lasting Peace in our house today


Question: 5

Answer: (A) Solution: Even though there is a vast scope for its improvement tourism has removed a neglected area.


Question: 6

Answer: (C) Solution: John was considered as leader from early days as finding a good captain is challenging

Over past three series, he has been scoring big score so his performance is improving.


Question: 7

Answer: (C) Solution:

P (car = blue | identify = blue) =

P  car  blue

identifly  blue

P(identify  blue)

P (identify = blue) = P (car = blue) P (correct) + P (car = green) P (incorrect)= 0.15 × 0.8 + 0.85 × 0.2 = 0.29 P (car = blue / identify = blue) = P (identify = blue/car = blue) P  car  blue  0.29



0.8  0.15  0.41  41% 0.29


Question: 8

Answer: (B) Solution: Assume mass of each alloy = x

Total gold 

2x 3 x x 7x   x  2 3 5 37 5 10 10

Total copper 

3x 7x 3x 7x 13x     3  2 7  3 5 10 10

Overall ratio  7 :13


Question: 9

Answer: (B) Solution: If tourism improves more people will visit the location & hence city will become crowded pointed


Question: 10

Answer: (B) Solution: n

 r  Amount  P  1    100  P is amount inverted R is rate of interest N is no. of years

5

 10  10,00,000  P  1    100  P  620921.3  6,21,000


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