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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)
2.
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)
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
____
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Exp:
12
35
23
81
46
173
92
357
184
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________
368
difference
357
⇒ 368
725
Q. No. 6 – 10 Carry One Mark Each
6.
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
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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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Q. No. 1 – 25 Carry One Mark Each
1.
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
P 2
=
= = 0.667
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)
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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)
11.
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)
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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15.
(
) (
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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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Exp:
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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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20.
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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.
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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
−∞
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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
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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

1 2 
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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34.
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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
intrinsic carrier concentration of silicon n i = 1.5 × 1010 cm −3 and
= 26mV. The electron
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
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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
(D) − I1 ( R 1 + R 2 )
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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
= XYZ + XYZ
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41.
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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)
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Exp:
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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
<z<
9
3
 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 =
1 j 210π n −1 − i 210π n
e
e
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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45.
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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)
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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  ⇒ Q c = 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]
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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
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Function of bit energy and noise PSD
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φ2 ( t )
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
H {L} = log 2
+ lg o 2
+ log 2
+ ......... = 0 + 1. + 2. + 3. + .........
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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53.
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In spherical coordinates, let aˆ θ .aˆ φ denote until vectors along the θ, φ directions.
100
sin θ cos ( ωt − βr ) aˆ θ V / m
and
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
54.
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)
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
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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 
 Z L + J50 50 − JZ L 
= 50  L
×
 = 50 

 50 + JZ L 
 50 + JZL 50 − JZ L 
Zin = λ
8
 50ZL + 50 ZL + J ( 502 − Z2L ) 

Zin = 50 
502 + Z2L


Given , Zin → Re al
So, I mg ( Zin ) = 0
502 − Z2L = 0
Z2L = 502
R 2 + X 2 = 502
R 2 = 502 − X 2 = 502 − 302
R = 40Ω
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