ece gate paper 25 answers
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1
Answer Keys:
General Ability
1 B 2 C 3 D 4 14 5 B 6 B 7 C
8 2.5 9 B 10 A
Electronics and Communication Engineering
1 2 2 A 3 C 4 70 5 2786 6 8.92 7 5
8 1.59 9 16 10 B 11 C 12 B 13 D 14 A
15 A 16 C 17 D 18 A 19 1.524 20 55 21 D
22 D 23 0.2 24 0.52 25 D 26 D 27 0.5 28 16.67
29 0.139 30 0.047 31 0 32 C 33 D 34 B 35 A
36 C 37 C 38 7.465 39 C 40 C 41 B 42 B
43 B 44 A 45 28.125 46 73.6 47 2 48 2 49 0.5
50 10 51 B 52 821.467 53 2.5 54 B 55 B
Explanations:-
General Ability
1. A renegade is a disloyal person who betrays or deserts his cause or religion or political party
or friend etc.2. The given analogy is name of study: field of study. Gerontology is the branch of science that
deals with the problems of aging. Likewise, otology is the science of ear and its diseases.
4. All numbers are made up of prime factors. All perfect squares are made up of pairs of primes.
So, in order for 3150y to be a perfect square, it must consist only of pairs of prime factors.
3150 breaks up into 315 * 10 (no primes)
315 breaks up into 5 * 63 (5 is prime, put that aside)
63 breaks up into 7*9 (7 is prime, put that aside)
9 breaks up into 3*3 (both prime, put them aside).
10 breaks up into 2*5 (both prime, put them aside).So, we've put aside: 5*7*3*3*2*5
writing in order:
2*3*3*5*5*7
Our 3s and 5s are paired off, so we don't need any more of those to create a perfect square.
We have a single 2, so we need a 2.
We have a single 7, so we need a 7.
So, the minimum possible value for y is 2*7 = 14
5. Assume c=a3
c2- b2=127
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2
(c+b)(c-b) =127
since 127 is a prime number the only possible combination is 1, 127.
since b and c are positive, c+b =127 and c-b =1 solving which we get c=64, b=63
Then a = 4 (since c = a3)
6. Consider each choice. (A) is possible; for instance, 6 + 7 + 8 = 21. (B) is impossible, so it is
correct. An integer with only even prime factors cannot have 7 as a prime factor; if it doesn't
have 7 as a prime factor, it can't yield an integer when divided by 7.
(C) is possible: If either of the odd integers is 7 or divisible by 7, the result is also divisible by
7. so (E) is possible as well. (B) is the correct choice.
7. The average of the three variables is (a + b + c)/3. However, we need to solve in terms of a,
which means we must convert b and c into something in terms of a.
We're told that a = (1/4)b, which is equivalent to b = 4a. We can plug that in and simplify the
average to: (a + 4a + c)/3
We also know that c = 7a, which we can plug directly into the average expression:
(a + 4a + 7a)/3 = 12a/3 = 4a, choice (C).
8. The formula for compound interest is
(final balance) = principal * (1 + (interest rate) / N)^(time * N)
Where N is the number of times the interest is compounded annually
After one year, Ricardo's balance is
final balance = $1,000 * (1 + (.10 / 2))^(1 * 2)
final balance = $1,000 * 1.05^2
final balance = $1,000 * 1.1025
final balance = $1,102.50
After one year, Poonam's balance is
final balance = $1,000 * (1 + (.10 / 1))^(1 * 1)
final balance = $1,000 * (1.10)^1
final balance = $1,000 * 1.10
final balance = $1,100
The difference between the accounts
difference = $1,102.50 - $1,100
difference = $2.50
Electronics and Communication Engineering
1.
j5 t j3 t
j5 t j3 t
j5 t j3 t
x t e u t e u t
x t e u t e u t
x t e u t e u t
j5 t j3 t
cs
o 0
x t x t 1 1x t e e
2 2 2
2; T 2
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3
2. The characteristic equation is
1+G(S)H(S) = 0
2
2
k s 2
1 0s s 1
s3-s2+ks2+4ks+4k = 0
R-H criteria
3
2
1
0
s 1 4k
s k 1 4k
4k k 2s 0
k 1
s 4k 0
So, for stability k>0, k>1 and k>2
i.e., k>2
3. Even no. of complementer will produce the output as input. So,output N
5. Reflection coefficient at antenna end
L 0
L 0
Z ZZ Z
Radiation resistance of quarter wave dipole 36.5
So36.5 25
0.18736.5 25
2
ref incident
2
trans incident
P P
P P 1
tra
2 2
ref 2n 2
1 1p 1So 00P 2756W
6.
i n p
n
p
i
1Since,conductivity
resistivity
for,puresilicon, en
where, electronmobility
Holemobility
n carrierconcentration
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4
19 6
p p
p 19
8 2
p
11.6 10 1.5 10 3
3500
1
1.6 10 1.5 10 3500 4
2.98 10 m V sec
n p
8
n
8 2
n
since =3
=3 2.98 10
8.92 10 m V sec
7.
Z ABC
Z ABC A B C A B . C
Minimum 5 2 input NOR gates are required to Realize Z ABC.
8.
2c c1 c2 1 B 2 B
c 1 B 2 B 1 B
I I I I I
I I I I
c 1 2 1 2 B
B
I I
4I 1.59mA
30 80 30 80
9. f 2 44 t
x t 2 sint 0 otherwise
By parsavals energy theorem,
A
B
CZ
Z
C
B
A
BI
cI
C2I
EI
c1I
E1 B2I I
X
44
2
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5
2
4
4
1energy of x t X d
2
14 d
2
4 8 162
10. T0= 4; 00
2
T 2
Fundamental Fourier term1 0 1 0a cos t b sin t
the given signal is even symmetrical b1= 0
1 1 12
1 0
1 1 1
2 1 1 1 cos ta x t cos t dt cos t dt dt4 2 2 2 2
1
1 1a FundamentalFourierterm cos t.
2 2 2
12. The characteristic equation of A is
A I 0
8 6 2
6 7 4 02 4 3
3 2 18 45 0
0,3,15
The diagonal matrix corresponding to the given matrix consists the eigen values as its
principal diagonal elements
0 0 0
D 0 3 0
0 0 15
13. An invertible system must satisfy a one-one mapping between input & output
For system 1, if x t 2 & x t 3 the output y(t) = 0.
For system 2, every input gives a unique output
15. Number of samples in linear convolution = 4+5-1 = 8
If the result of linear convolution is same as circular convolution x[n] and h[n] must have 8
samples
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6
Number of zeros required for x[n] = 4
Number of zeros required for h[n] = 3
16. cot x
cot x
lim (x+1) (1 )
x 0let y = (x+1) log y = cot x log(x+1)
2
lim log y = lim cotx .log (x+1) ( 0)x 0 x 0
log (x+1) 0 = lim
tan x 0x 0
1
x+1 = lim 1 (applying L'Hospital Rule)
sec xx 0
elog y= 1 y = e
limit value of f(x) at x = 0 is "e"
given that f(x) is continuous at x = 0
limit value at x = 0 function value at x = 0
e K K= e
17. The parameter value at T = 400k are found as3 2
19
V
19 3
V
400N 1.04 10 300
N 1.601 10 cm
400kT 0.0259 0.03453eV
300
The hole concentration is given by
V F
F V
E E kT
V
E E kT19
19 0.25 0.03453
16 3
p N e
p 1.601 10 e
p 1.601 10 e
p 1.148 10 cm
18. The large value of capacitor is used in a full wave rectifier to increase peak current rating of
the diode and for low conduction period for diode rectifier.
19. Along ab, d 0 b
a
A.dl 0, A.dl 0
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7
Along bc, d 0
2A.dl d
2c
2 2
b 0
A.dl 2 4.5
Along CD
d 0, 2 2A.dl .d
2d 2 3
2
c 3 3
A.dl d 6.333
Along da
d 0 2A.dl d
0 22
2
A.dl d 22
A.dl 2.5 6.33 1.524 20. Output = Y = ABC + BCD + ACD + ABD
=4 units of 3 input gates + 1 unit of 4 input gates
4 10 1 15 55 Minimum cost of implementation is 55/-
22. Taking Laplace transform of the given equation
1
12sL(y) y(0) 2L(y)
s
given y(0) 10
10 12L(y)
s 2 s(s 2)
Ta king inverse Laplace transform onbothsides
10 12y Ls 2 s(s 2)
1 1
t
2t 2t
0
2t
1
1 s 210L 12Ls 2 s
10e 12 e dt,Since by inverseLaplaceof division by's'
4e 6
a b
c
d
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8
23. Use superposition principle.
1stshort the voltage source.
ThenX1
3I 2 1.2A
5
Now, open circuiting the
Current source
We get X25
I 1A3 2
X X1 X2I = I I 0.2A
24. Pe 32
3 2 , 6
pe
k kG 1 at large stability G 1
s
3.14k 0.52
6 6
25. The ac equivalent of the circuit is
Writing KCL
m gsi g ______(1)
gs 2iR ______(2)
Substitute (2) in (1)
m 2i g iR
2
m
1R i
g
26.b
p
a
We have
1dx converges for P < 1
(x-a)
diverges for P 1
Hence P = 2 The given integral diverges
i
R
2 D
m gsg gs
S
2A 5V
XII X2I
1
2
3
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9
27.1 2
z12
2
1 1 2
2 1 2
2 0 2
V 1I 2I
1V 2I I ..(1)
2
Fromoutput sideV z I ..(2)
Sub(2)in (1)
0 2 1 2
0 2 1
2 2 1
1
0 1 1
0 0
1z I 2I I
2
1z I 2I
2
IGiven 2 I 2I
I
1z 2I 2I
2
1 1z 1 z
2 2
28.
m
m
T2T 4
2 2 Pt
m m m0 0
4V1 4m t m t dt V dt
T T T
2
PV
3
2 22 2 22C C P
actual
L L
100 0.5A A VP m t 16.67 W
R R 3 50 3
29. Given, n
n p
J90% 0.90
J J
2 nn i
a n0
p2p i
d po
nn
a a an pn p
d d d
D1J en
N
D1J enN
DJ 25 5so, 0.90 0.90 0.90
N N NJ JD D 25 16 5 4
N N N
a
d
a
d
N5 4.5 3.6
N
N0.139
N
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10
30. Given that
2
1.67
z
2
1.67
p(X 1.67 ) 0.04746
p(Z 1.67) 0.04746
p(Z 1.67) 0.04746
i.e., f (Z)dz 0.05
1e .dz=0.0474 0.05
2
31. The two networks on both sides of the inductor are not connected to each other completely as
they have no common point between them. So, no current flows through the inductor.
32.
0 0 03 3 3
2 2 22 2 1 3 3 2 3 1 2 3 1 2
3 1 3 2
Q t 1 S RQ Q Q Q Q
Q t 1 JQ KQ Q Q Q Q Q Q Q Q Q Q Q Q
Q Q Q Q
1 2
10 0 1 0
Q t 1 D Q
Q t 1 T Q Q Q Q Q
After eight clock pulses counter goes to 0000 state.
3 2 1 0 3 2 1 0Q Q Q Q Q Q Q Q
0 0 0 0 1 0 0 1
0 0 0 1 0 0 0 0
0 0 1 0 1 0 0 0
0 0 1 1 0 0 0 1
0 1 0 0 1 0 1 1
0 1 0 1 0 0 1 0
0 1 1 0 1 0 1 0
0 1 1 1 0 0 1 1
1 0 0 0 1 0 0 1
1 0 0 1 0 0 0 0
1 0 1 0 1 1 0 0
1 0 1 1 0 1 0 1
1 1 0 0 1 1 1 1
1 1 0 1 0 1 1 0
1 1 1 0 1 1 1 0
1 1 1 1 0 1 1 1
Present state Next state
1010 1100
1111
0111
0011
0001
0000
1001
z 1.67 z 0
0.04746
z 1.67
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11
33. For input voltage, SV 0
Diode D1is of. i.e., Reverse biased
So, current through D2 is
10 0.7i20 20 k
9.3i mA
40
i 0.232 mA
and output voltage Vo will be
o
3 3
o
O
20Ki
20 10 0.232 10
4.65V
O S Sfor 0 V 4.65V
For negative values of input voltage, output is juet the negative of positive part.
34.2
C(S) k =
R(S) s + (2+ ak)s+k
Compare with characteristic equation
n n = k and 2 = 2 + ak
k 16
2 0.7 4 2 16a a = 0.225
35. 2
2
2
2
2
dy y yGiven cos
dx x x
ycosxdy - ydx dxx
x
y xdy-ydx -dxsec
x x x
y y -dxsec d
x x x
10V
20k
2D1D
SV
3D 4D20k
O
20k
10V
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12
2 y y dxIntegrating sec d cx x x
y tan ln x c
x
put x = 1 , y = tan ln(1) + c
4 4
1 = c
y tan ln x 1
x
y1-tan
x
y 1 tan ln x
x
x= e
36. L
Zi min
L
R R 1.2K 330 20V V
R 1200
i minV 25.5V
L ZL
L L
V V 20VI 16.67mA
R R 1.2K
max zm LI I I 60mA 16.67 mA 76.67 mA
max R max ZV I R V
376.67 10 330 20 25.3 20
i maxV 45.3V
i25.5 V 45.3V
37. Consider
j t
j t
1x t X e d
2
2 x t X e d
Replace with ' '
LV
20V
25.5V 45.3V I
V
i max25.5 V 45.3V
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13
j t2 x t X e d
y t 2 x t ; x t is oddsignal
y t 2 x t
proportionalityconstant 2
38. The oscillator circuit given in the figure is BJT
Colpits oscillator circuit
So, the frequency will be
r
T
1f
2 LC
1 2T
1 2
C Cwhere C
C C
129
T 6
0.1 0.01 10C 9.09 10 F
0.11 10
r3 9
1F
2 50 10 9.09 10
rF 7.465kHz
39. For given decision rule, probability of correct reception
c
P P y p, x 0 P y q, x 0 P y r, x 1
P y p / x 0 .P x 0} P y q / x 0 . P x 0} P y r / x 1 P x 1}
0.5 0.5 0.4 0.5 0.5 0.5 0.25 0.2 0.25 0.7
Thus probability of error = c1 P 1 0.7 0.3
40. The output voltage V0will be
1O Z
2
RV V 1R
Where, 1R Resistance connected to emitter.
i.e.,12k
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14
0
0
D CE C
CE i O
12V 6 1 8V
36
V 8V
P V I
V V V 15 8 7V
C E
8 8I I 0.966mA
10 48
DSo, P 7 0.966 6.762W
Now unregulated I/P voltage is increased by 30%
i
30So, V 15 15 19.5V
100
0V will remain same
CEV 19.5 8 11.5V
DP 11.5 0.966 11.11 mW
11.1 6.76% increase 64.29%
6.76
41. Odd Parity generator is an even function
1
1
1
1
1
1
1
1
f 1 0 x 1 0 1 y
f 1 x 1 0 1 y
f x 1 0 1 y
f x 0 1 y
f x 1 y
f x y
f x y
f x y
1 1f x, y,z zf z f z
f x, y,z x y z
1 1
z f
0 f zf
1 1 z
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15
42. Periodic signal ranges from to . Thus the system must be non-causal.
Stability of LTI system
t
h d
If 0h t sin t which is a periodic signal then t ,
0sin t dt
Thus the system is unstable.
43.
A B C D output q
0 0 0 0 0 1
0 0 0 1 1 0
0 0 1 0 2 0
0 0 1 1 3 0
0 1 0 0 4 0
0 1 0 1 5 0
0 1 1 0 6 0
0 1 1 1 7 0
1 0 0 0 8 1
1 0 0 1 9 0
44. x t Ax t
Atx t e x 0
h t
0
0
0
30
2 t
O
1 0 0 0
0 0 0 0
X X X X
1 0 X X
00 01 11 10CD
AB00
01
11
10
q BCD
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16
t x 0
1 1
1cos2t sin 2t cos2t sin 2tx t 2 2
12sin 2t cos2t 2sin 2t cos2t
y t cx t
1
cos2t sin 2t0 1 2
2sin 2t cos 2t
y t 2sin 2t cos 2t
45. Since, potential function V is known, so work done Q
W may be determined as
ABEd V l
Q
PQ
P
Q P
6 0 0 0 0 6
W Q E.d QV
Q V V
10 10 10 510 10 Sin90 Cos60 Sin30 Cos120 10 10 28.125 J
16 1 32 2
l
46. The voltage division is the same as it would be for full right circular cylinders. The segment
shown with angle
, will have a capacitance/ 2
times that of the complete coaxialcapacitor.
10 r
1
10
10
2
1 1 2 2 1 2
21
1 2
2 LC
2 ln(2.25/ 2.0)
L(1.5 10 )(F)
C L(4.2 10 )(F)
Q C V C V and V V V
CV V
C C
4.2(100)
1.5 4.2
74V
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17
47.
j j j j
jj
j
j
j
j j j
j j
1 ae Y e b e X e
b eH e
1 ae
1
H e 1; b a b 2
1e
12Y e X e ; X e1 1
1 e 1 e2 2
j
j
j j
1e
2Y e1 1
1 e 1 e2 2
j0n
3
2y n Y e 2
3 1
2 2
48. For
t
22
1x t e u t X
1 j
2 1 j 1Y
1 j2 2 2 j j
1
2
f 2t
2
2Y
2 j
1 d 1 jFor e u t
X j d 2 j 2 j
By using frequency differentiation property,
1
f2t
2
f 2t
2
jjte u t
2 j
2For 2te u t
2 j
k 2
49. Noise pdf is as shown below
3n
3
1s 0s
4
1
4
Nf n
n4
-
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18
Since channel noise is additive, at receiver we receive signal with pdf as shown below:
Using MAP criteria
1r 1P y x 3 P x 3 16y x n 3
1P y x 3 P x 31 r
16
For finding optimal threshold value
1r 1 3 3r r 0.5
2
50.PQ
QC
V
PQ
2 0 0
1 2 3
V E.d
2 x dx y dy z dz 10 Volts
l
51.
C V
C v
E E kT2
i C V
2E E kT2 10 1.25 0.025
C V i
n N N e
N N n e 1.6 10 e
42 6
C VN N 1.327 10 cm ___(1)
3 2*
3 2C nC V*
V p
N m4 , N 8N
N m
From equation (2)
2 42
V
20 3 21 3
V C
8N 1.327 10
N 4.07 10 cm N 3.26 10 cm
52. p, diffusion Pdp
J qD .dx
1when s is
transmitted
0when s is
transmitted
7 13 1 73 r
1
4
1
4
-
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ECTest ID: 161275 TarGATE16 www.gateforum.com
ICPIntensive Classroom ProgrameGATE-Live Internet Based ClassesDLPTarGATE-All India Test SeriesLeaders in GATE Preparation 65+ Centers across India
All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.
19
15 41 8
P' x 10 exp cmL 12
15 4
4
1 810 exp cm
12 10 12
17 44.278 10 cm
19 17P,diffusionJ 1.6 10 12 4.278 10 2821.467 mA / cm
53. We need to find the thevenin equivalent resistance (impedance) across RLby removing it.
54. O 1 1 1G 90 tan tan 2 tan 3
2 2 2
KG
1 4 1 4 1
0
0
0, G , G 90
, G 0, G 360
55. We know that,
Newton-raphson interation formula to find the reciprocal of
kk nn 1 n
xN is x (k 1) Nx
K
Putting k=3; N=M; then3n
n 1 n
xx 4 Mx
3
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