ece gate paper 19 answers
TRANSCRIPT
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Answer Key
General Aptitude
1 A 2 C 3 C 4 544 5 C 6 D 7 B
8 0.2 9 A 10 C
Electronics and Communication Engineering
1 A 2 B 3 B 4 0.06 5 0 6 C 7 D
8 B 9 1 10 3.766 11 2 12 5 13 64 14 31
15 D 16 C 17 A 18 A 19 C 20 C 21 B
22 0.029 23 B 24 D 25 A 26 -2.5 27 A 28 A
29 C 30 A 31 B 32 B 33 A 34 B 35 16.4336 3.5 37 D 38 B 39 C 40 5 41 1923 42 A
43 A 44 C 45 1 46 B 47 D 48 1.73 49 B
50 A 51 9 52 A 53 58.1 54 188.49 55 3.324
Explanations:-
General Aptitude
4. Required value 2 2 327 11 4 729 121 64 544
5. Anshu takes x days to do the work
4 8xAtul takes 2 x = days
5 5
1But given (Atul+ Anshu) =1daywork =
16
1 1 1 1 5 1+ = + =
8xx 16 x 8x 16
5
1 5 1 13x 1
+ = =x 8x 16 8 16
13x =16 = 26
8
8. We know that without any restriction she has to go 3 blocks south and 3 blocks east to get to
work. Hence, we need to find the number of arrangements of the anagram SSSEEE
Total arrangements : 6!/3!*3! = 20
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2
we know that she has to take 2 blocks south 1st. So, 1st 2 are fixed SS. The remaining 4 steps
we will have to find the number of arrangements of SEEE
= 4!/3!*1! = 4
Hence Probability = No of fav/total = 4/20 = 1/5
9. We know that 2x/y is a prime greater than 2. Therefore, x/y must be greater than 1, soeliminate (I).
We can quickly eliminate II and III by picking numbers. If we set 2x/y = 3 (a prime number
greater than 2), we get:
x/y = 3/2
So, any two numbers in a ratio of 3:2 will suffice. We can let x=6 and y=4 to see that neither
statement II nor III (nor I, for that matter) mustbe true.
As an aside, although it wasn't necessary for this question, we should remember to be careful
about the assumptions we make. Nowhere in the question stem does it say that x and y have
to be integers - we certainly could have picked non-integer values to eliminate all 3
statements as well.
The answer is A (none).
10. P(only 1 letter in the correct envelope) = P(1st correct and 2,3,4 in wrong) OR P(2nd correct
and 1,2,3 in wrong) OR P(3rd correct and 1,2,4 in wrong) OR P(4th correct and 1,2,3 in
wrong)
P(1st correct) = 1/4 (1 correct out of 4 envelopes)
P(2nd wrong) = 2/3
P(3rd wrong) = 1/2
P(4th wrong) = 1/1
Hence, P(1st correct and 2,3,4 in wrong) = 1/4*2/3*1/2 = 1/12
Same will be the probability for the other 3 casesHence required Probability = 4*1/12 = 1/3.
Electronics and Communication Engineering
1. Since both lines of regression pass through the point x, y , we have
8 x 10 y 66 0
40x 18y 214
Solving these we get x 13, y 17
2. Given 3f (x) x 2x 5
A=2; b=2;
By regula falsi method; we have
1stapproximation of the root is
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1
af (b) bf (a) 2f (3) 3f(2) 2(16) 3( 1)x
f (b) f (a) f (3) f(2) 16 ( 1)
32 32.0588
17
3. The directional derivative of f in the direction ofp
aa is f .
a
Where P is given point,
2 2
p(1, 1,1)
f f ff i j k i (y z) j (2xyz) k(xy )
x y z
f i 2 j k
Directional derivative ( i j 2k) 3
i 2 j k .1 1 4 6
4.2
1L[sin t]
s 1
222
3t
s 30
d 1 2sL[tsin t]
ds s 1 s 1
e (tsin t)dt L[tsint]
By definition of Laplace transform
30.06
50
5. t 0 sX(0) limX(t) lims.X(s) 2ss(808)
lim 0s(s 2s 101)
8. When the switch is at 1, the steady state becomes
i = 100/10+20 = 3.33 A
As soon as the switch is moved to position 2 at t = 0, the LR circuit starts decaying and the
decay current is given by
t
i Ie
Where I is the initial steady state current and is time constant L/Rt
40t0.025i 3.33e 3.33 e A
9. Diode D1will turn on first
0
5 1I 2mA
2k
V 1V
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10. Transistor Q2will be off as BEV 1.3V
E E2 0.7 3k(I ) 10 0 I 3.766mA
11. TH 1V (T ) 4V
TH 2
TH
V (T ) 6V
R 2.4k
B
5
B1
E1
E2 C1
B2 CE1 E1
E1 CE1
CE1
4 0.7 (2.4k 100 1k)I 0
3.3I 3.22265 10
102400
I 3.2226mA
I I 3.1903mA
6 2.4kI 0.7 V 1kI 0
6 0.0765 0.7 1k(I ) V
V 2V
CE2 CE210 3.3(1) V 2 3.3(1) 0 V 1.4V
12. For Half adder
Sum S A Bandcarry,C A.B
13. Conversion time N clockperiod
clockperiod 8 s
conversion time 8 8 s 64 s
14. n2 1 32 1 31
S A B
C AB
B
A AB
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15.
16.GE
2 kTi Cn N N e
17. y t h x t d
10
6
y 3 h x 3 d=4
5 10
4 5
y 1 2+ 1=7
18. M=2 M+N-1=5 N=4
19. An open loop system can be changed into stable/unstable using closed loop technique
21. s
3
term is not present. So, the given system is Unstable.
22. 2 2H 0.4log 1 0.4 0.6log 1 0.6
0.971bits symbol
C 1 H 0.029
0.6
0.4 0.4
0.6
drift
310
510 E
a a b b c c d d
a b c d
a b c d
a b c d
-1 1
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24. c
1 c
2
1
c c
c
120
2 2
120 120
25.
x y
o
x y
E 5cos t z a 5sin t z a
5cos t z a 5cos t z 90 a
We can see that x & y component of electric field have same magnitude and y componentleads x-component by 90
o. So it is left circular polarized.
26. 1 1 3
A B2.5 b 7.5
2 2 1R R 2.5 R ; then
1 1 3
A B ~0 b 2.5 0
Per the system to have infinitely many solutions;
b 2.5 0 b 2.5
27. Given P=
1 1 1 3
1 2 3 4
2 3 4 9
Reduce the matrix A to Echelon form.
Apply 2 2 1 3 3 1R R R ; R R 2R ;
P ~
1 1 1 3
0 1 2 3
0 1 2 3
Apply 3 3 2R R R ;
P ~
1 1 1 3
0 1 2 1
0 0 0 2
Number of non-zero rows = 3 = no. of given vectors
Rank of P 3
So, the given vectors are linearly independent
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28. 2 2 2f z x y 2ixy z is analytic Everywhere
1 i
0f z dz
is independent of the path joining
Z = 0 and z = I + i
29. 3 3 2 2x D 3x D xD 1 .y 0 cauchy Euler equation
3
3
1.z
z 21 2 3
1 /2
1 2
d d( 1)( 2) 3. ( 1) 1 .y 0where , D
dz dx
1 y 0 L.D.Ewith constan t coefficients
1 3A.E : m 1 0 m 1, i
2 2
solution is y
3 3c e e c cos z c sin .z
2 2
3y c x x c cos n x
2
l 3
3c sin n x
2
l
30. 2f x 1 2cosx 2sin x in 0,
f ' x 2sinx 4sin x cosx
f ' x 0 2sin x 1 2cos x 0
1sin x 0; cosx2
x 0 , ,3
f 0 1 2 1 ; f 1 2 1 3
and 3f 3 1 2 1 2 2 3 4 2
f x has absolute maximum at x
Now f" x 2cosx 4cos2x
1 1at x , f " x 2 43 2 2
f x has local minimum at x 0,3
31. Taking the laplace transform of the whole circuit
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Applying KVL,
15 I S S
S S
25
I S
S 1
i t 5sin t
32. 3O
5
1 12 20 10 L 6.33 H
LC L 10
3
5
1 1BW 10 10 R 10
RC R 10
33.
A B
A
B
[T] [T ][T ] (cascade connection)
J5 J51 J5 J51
T 11 1
1 1
J51.5 J51 J5
J2T J
1 11 2
J2
1.5 J51 J5 J5
[T] J 11 1
2
1.5 J7.5 2.5 J10 25
3 J
2 2 J5 1
4 J7.5 J10 25
1.5 0.5J J5 1
5s
I S
S
1S
J5
J2
J5
1
AT
BT
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35.
H-12
1 2 1 2
1 1f = = 16.43kHz
2 R R C C 2 30 25 0.5 0.25 10
36. 0 11L 1 2
0 1
1 1 1L
V - V7 - VI = I +I
2K 2K
V 2V
7 - V 2V - V 7I 3.5mA
2K 2K 2K
38. J X
k Y
n 1
n
n
J K Q
0 0 Q
0 1 0
1 0 1
1 1 Q
39. BC+D2
1011 11001101 0010
1 1000 1110
Z = 0, CY = 1, MSB = 1 so S = 1
Number of 1s = 4 so P = 1.
40.
15 3
d B
11
2142B
n 19 15
d
N 5 10 cm & V 100V
2 V 2(11.7)(8.85 10 )(100)x
eN 1.6 10 5 10
n(min)x 5.09 m.
41.2 1
R k k 12516 8
I
I
I
Loa
7
2k
2k
2k
2k
V
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10
2
0
15
0
12
RA
2 2125 10 ohmcm
2
2.5ohmcm
1
en
n 1.923 10
1923 10
43.
-1
-1
-1
-1
b[n]=x[n] b[n-1]
b[n] 1
x[n] 1 z
1y[n]=b[n] b[n-1]
2
y[n] 11 zb[n] 2
11 z
y[n] 2
x[n] 1 z
44. 2
2
S y(S) Sy(S)-2y(S) x(S)
y(S) 1
x(S) S -S-2
poles are 2,-1
It contains j axis, So ROC should be between two poles.
So assertion is true.
Reason: It is not always true that for stable system
ROC should lie between two poles for causal stable system. ROC should be right side of right
most poles
x n y n+ +
1/ 2
-1z
b(n)
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11
45. 1x 2t 1 x 2 t 2
46. 1 2 1 3 4 2 2 3 3 1 2 1 3 4 2dr 1 g g h g g h g g h g g h g g h
k k 1 2 3 4P g g g g
1 2 3 4g g g g
R dr
47. Char equation 1+ G = 0
2 2
2
s 2 s 4 k s 3 s 5 0
s 6s 8 k s 8s 15 0
s 1 k 3 8k 6 8 18k 0
for all coefficient ve
1 k 0
k 1
k 6 8
8 15k 0
k 18 5
k 6 8
8 15k 01 k 0 8k 6 0
k 1 k 6 8 0.75k 1
2
1
1 22
1.50.5 2.5
x 2t
x 2 t 1 2 1
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So k< - 1 and k>0.75
Char equation: all coefficient must have same sign either +ve orve
48. t
c t k 1 e Given ; k 1 2.5
dat t t ; c t 0.5
dt 1.732sec
49. O O 1 OGH 180 90 3tan 1802
1 O pc3tan 2 270
pcM| 0 GM
2which will not be satisfied for any frequency
4
From 0 to
gc does not exist PM
51. 2
ONNoise power1 H f 2B2
0j4 ftON
9e 2B2
O9N B
Noise power 2 = ON
2B2
ON B
Therefore Ratio of 0
0
Noise power 1
Noise
9
power
N
B
B
N2 = 9
52. BPSK Rx
c2
sin t6T
mB
mA
detection
c
2sin t
T 6
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When received signal is
1 c
2ES sin t
T c
2sin t
T 6
S
O
m C c C
T
2E 2 EA sin t 30 sin t dt cos30 cos 2 t 30 dt
T T T
S
O
M C c C
T T
2E 2 EB sin t 30 sin t dt cos 2 t 30 cos30 dt
T T T
When2S is received
m
m
A E cos50
B E cos30
1
e
0 0
d E cos30P G Q
N 2 N / 2
0 0
3 2 EQ E Q
2 N 2N
53. Wave gets attenuated to 37% of its original value when it travels distance equal to depth ofpenetration
2
Here
6 6
0
258.11 m
3 2 5 10 5 10
54. For 0
Wave impedance is given by
r
r
120 12016
60 188.49
55. Reflection coefficient
L 0
L 0
Z Z 50 75j 100s 0.53748 97.12
Z Z 50 75j 100
1 s 1 0.53748VSWR 3.324
1 s 1 0.53748
m
m
A E cos30
B E cos30
d '
x
x