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Q. No. 1 – 25 Carry One Mark Each
1. Consider the following statements regarding the  complex Poynting vector  P

for
the power radiated by a point source in an infinite homogeneous and lossless
medium.  Re P( )

denotes the real part of  P

S denotes a spherical surface whose .
centre is at the point source, and
ɵ
n denotes the unit surface normal on S. Which
of the following statements is TRUE?
(A) Re P( )

remains constant at any radial distance from the source
(B) Re P( )

increases with increasing radial distance from the source
(C)  ( )

s
Re P .n dS
∫∫

 remains constant at any radial distance from the source
(D)  ( )

s
Re P .n dS
∫∫

 decreases with increasing radial distance form the source
Answer: – (D)
Exp: –  ( )
S
Re P .nds ˆ
∫∫

 gives average power and it decreases with increasing radial
distance from the source
2. A transmission line of characteristic impedance 50Ω is terminated by a 50Ω load.
When excited by a sinusoidal voltage source at 10GHz, the phase difference
between two points spaced 2mm apart on the line is found to be
4
π
radians. The
phase velocity of the wave along the line is
(A)
8
× 0.8 10 m / s (B)
8
× 1.2 10 m / s (C)
8
× 1.6 10 m / s (D)
8
× 3 10 m / s
Answer: – (C)
Exp: –
0 Z 50 = Ω ;
L Z 50 = Ω
For
4
π
radians the distance is 2mm
The phase velocity
10
7 8
P
3
2 10
v 16 10 1.6 10 m / s
2
16 10
ω × π ×
× = × = = =
β π
×
3. An analog signal is band-limited to 4kHz, sampled at the Nyquist rate and the
samples are quantized into 4 levels. The quantized  levels are assumed to be
independent and equally probable. If we transmit two quantized samples per
second, the information rate is ________ bits / second.
(A) 1 (B) 2 (C) 3 (D) 4
Answer: – (D)
Exp: – Since two samples are transmitted and each sample has 2 bits of information,
then the information rate is 4 bits/sec. EC-Paper Code-B      GATE 2011                   www.gateforum.com
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4. The root locus plot for a system is given below. The open loop transfer function
corresponding to this plot is given by
(A)  ( ) ( )
( )
( ) ( )
s s 1
G s H s k
s 2 s 3
+
=
+ +
(B)  ( ) ( )
( )
( ) ( )
2
s 1
G s H s k
s s 2 s 3
+
=
+ +
(C)  ( ) ( )
( ) ( ) ( )
1
G s H s k
s s 1 s 2 s 3
=
+ + −
(D)  ( ) ( )
( )
( ) ( )
s 1
G s H s k
s s 2 s 3
+
=
+ +
Answer: – (B)
Exp: –  ‘ x ‘ → indicates pole
‘O’ → indicates zero
The point on the root locus when the number of poles and zeroes on the real axis
to the right side of that point must be odd
5. A system is defined by its impulse response  ( ) ( )
n
h n 2 u n 2 = − . The system is
(A) stable and causal  (B) causal but not stable
(C) stable but not causal (D) unstable and non-causal
Answer: – (B)
Exp: –  ( ) ( )
n
h n 2 u n 2 = −
h n( ) is existing for n>2 ; so that  ( ) h n 0;n 0 = < ⇒ causal
( ) ( )
n n
n n n 2
h n 2 u n 2 2
∞ ∞ ∞
= ∞= ∞−=
∑ ∑ ∑ = − = = ∞ ⇒ System is unstable
6. If the unit step response of a network is  ( )
t
1 e
−α
− , then its unit impulse response
is
(A)
t
e
−α
α (B)
1 t
e
− −α
α (C) ( )
1 t
1 e
− −α
− α (D) ( )
t
1 e
−α
− α
Answer: – (A)
Exp: – S t( ) → step response
Impulse response  ( ) ( ) ( ) ( )
d d t t
h t S t 1 e e
dt dt
α α
= = − = α
7. The output Y in the circuit below is always ‘1’  when
× × o ×
σ
−3 −2 −1 0
P
Q
R
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(A) two or more of the inputs P,Q,R are ‘0’
(B) two or more of the inputs P,Q,R are ‘1’
(C) any odd number of the inputs P,Q,R is ‘0’
(D) any odd number of the inputs P,Q,R is ‘1’
Answer: – (B)
Exp: – The output Y expression in the Ckt
= + + Y PQ PR RQ
So that two or more inputs are ‘1’, Y is always ‘1’.
8. In the circuit shown below, capacitors C1 and C2 are very large and are shorts at
the input frequency.  vi
is a small signal input. The gain magnitude
o
i
v
v
at 10M
rad/s is
(A) maximum (B) minimum (C) unity (D) zero
Answer: – (A)
Exp: – In the parallel RLC Ckt
L 10 H = µ and = C 1nF
7
g
6 9
1 1
10 rad / s 10Mrad / s
LC 10 10 10
− −
ω = = = =
× ×
So that for a tuned amplifier, gain is maximum at resonant frequency
9. Drift current in the semiconductors depends upon
(A) only the electric field
(B) only the carrier concentration gradient
(C) both the electric field and the carrier concentration
(D) both the electric field and the carrier concentration gradient
Answer: – (C)
Exp: – Drift current, J E = σ
( = µ + µ J n p qE n P )
So that it depends on carrier concentration and electric field.
~
+
+
2kΩ
+
C2
Q1
o
v
C1
2kΩ
i
v
2.7V
2kΩ
2kΩ
10 Hµ
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10. A Zener diode, when used in voltage stabilization circuits, is biased in
(A) reverse bias region below the breakdown voltage
(B) reverse breakdown region
(C) forward bias region
(D) forward bias constant current mode
Answer: – (B)
Exp: –
For Zener diode
Voltage remains constant in break down region and current carrying capacity in
high.
11. The circuit shown below is driven by a sinusoidal input  =
i p v V cos t / RC ( ) . The
steady state output vo is
(A) ( p ) V / 3 cos t / RC ( )   (B)  ( p ) V / 3 sin t / RC ( )
(C) ( p ) V / 2 cos t / RC ( )   (D)  ( p ) V / 2 sin t / RC ( )
Answer: – (A)
Exp: –
0 2
i 1 2
v z
v z z
=
+
where
2
1
z R ||
j c
=
ω
and
1
z R
j c
1
+ =
ω
( )
2
R
z
R jcw 1
=
+
Given
i p
1 t
w v v cos
RC RC
   
= =
   
   
∵ 2
R
z
1 j
= ⇒
+
1
1
z R
j c
+ =
ω
( )
1
R R 1 j
jR
− ⇒ + =
( )
0
i
R
v 1 j 1 1
R v 1 2 3
R 1 j
1 j
+
= = =
+
− +
+
v
i ~
R C
C
R
+
o
v
+
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12. Consider a closed surface S surrounding volume V. If  r

is the position vector of a
point inside S, with
ɵ
n the unit normal on S, the value of the integral
s
5r.ndS
∫∫

ɵ
 is
(A) 3V (B) 5V (C) 10V (D) 15V
Answer: – (D)
Exp: – Apply the divergence theorem
S v
5r.n.dx 5 .rdV = ∇
∫∫∫ ∫∫
  

( )
v
= 5 3 dv
∫∫∫ = 15 V ( ∇ = .r 3 and r  is the position vector)
 
13. The modes in a rectangular waveguide are denoted by
mn
mn
TE
TM
where m and n are
the eigen numbers along the larger and smaller dimensions of the waveguide
respectively. Which one of the following statements is TRUE?
(A) The TM10 mode of the wave does not exist
(B) The TE10 mode of the wave does not exist
(C) The TM10  and the TE10  modes both exist and have the same cut-off
frequencies
(D) The TM10 and TM01  modes both exist and have the same cut-off frequencies
Answer: – (A)
Exp: –  TM10
mode doesn’t exist in rectangular waveguide.
14. The solution of the differential equation  ( )
dy
ky, y 0 c
dx
= = is
(A)
ky
x ce
= (B)
cy
= x ke  (C)
kx
= y ce  (D)
kx
y ce
=
Answer: – (C)
Exp: – Given  ( ) y 0 C = and
dy
ky,
dx
=
dy
kdx
y
= ⇒
kx c
= + ln y kx c y e e ⇒ =
When ( ) y 0 C = ,
0
y k e1
( ) ∴ =
kx
y c e k C 1
∵ = =
15. The  Column-I lists the attributes and the  Column-II lists the modulation
systems. Match the attribute to the modulation system that best meets it
Column-I  Column-II
P Power efficient transmission of signals 1 Conventional AM
Q
Most bandwidth efficient transmission of
voice signals
2 FM
R Simplest receiver structure 3 VSB
S
Bandwidth efficient transmission of
signals with Significant dc component
4 SSB-SC
(A) P-4;Q-2;R-1;S-3  (B) P-2;Q-4;R-1;S-3
(C) P-3;Q-2;R-1;S-4  (D) P-2;Q-4;R-3;S-1 EC-Paper Code-B      GATE 2011                   www.gateforum.com
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Answer: – (B)
Exp: – Power efficient transmission → FM
Most bandwidth efficient → SSB-SC
Transmission of voice signal
Simplest receives structure → conventional AM
Bandwidth efficient transmission of → VSB
Signals with significant DC component
16. The differential equation  ( )
2
2
d y dy
100 20 y x t
dt dt
− + = describes a system with an
input x(t) and an output y(t). The system, which is initially relaxed, is excited by
a unit step input. The output y(t) can be represented by the waveform
(A)    (B)
(C)    (D)
Answer: – (A)
Exp: –  ( )
2
2
100d y 20dy
y x t
dt dt
= + −
Apply L.T both sides
( ) ( )
2 1
100s 20s 1 Y s
s
( ) ( ) ( ) = + −
1
x t u t x s
3
 
= =
 
 
( )
( )
2
1
Y s
s 100s 20s 1
=
+ −
So we have poles with positive real part ⇒ system is unstable.
y t( )
t
y t( )
t
y t( )
t
y t( )
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17. For the transfer function  ( ) G j 5 j ω = + ω , the corresponding Nyquist plot for
positive frequency has the form
(A)    (B)
(C)    (D)
Answer: – (A)
Exp: – As we increases real part ‘5’ is fixed only imaginary part increases.
18. The trigonometric Fourier series of an even function does not have the
(A) dc term   (B) cosine terms
(C) sine terms   (D) odd harmonic terms
Answer: – (C)
Exp: –  f t( ) is even function, hence
k b 0 =
Where
k
‘b ‘ is the coefficient of sine terms
19. When the output Y in the circuit below is ‘1’, it implies that data has
(A) changed from 0 to 1 (B) changed from 1 to 0
(C) changed in either direction (D) not changed
Answer: – (A)
Exp: – When data is ‘0’, Q is ‘0’
And Q’ is ‘1’ first flip flop
Data is changed to 1
Q is 1 → first ‘D’
Q’ is connected to 2
nd
flip flop
So that
2 Q 1 =
So that the inputs of AND gate is ‘1’ ⇒ = y ‘1’
σ
5
σ
j5
σ
1 / 5
σ
1 / 5
D Q
Q
D Q
Q
Clock
Data
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20. The logic function implemented by the circuit below is (ground implies logic 0)
(A)  = F AND P, Q ( ) (B)  = F OR P, Q ( ) (C)  = F XNOR P, Q ( ) (D)  = F XOR P, Q ( )
Answer: – (D)
Exp: – From the CKT
O is connected to
0 3
‘I ‘ & ‘ I ‘
And ‘1’ is connected to
1 2
I & I ∴ = + F PQ PQ = XOR P, Q ( )
21. The circuit below implements a filter between the input current i
i and the output
voltage vo. Assume that the opamp is ideal. The filter implemented is a
(A) low pass filter   (B) band pass filter
(C) band stop filter   (D)  high pass filter
Answer: – (D)
Exp: – When W=0; inductor acts as a  ⇒ S.C V 0 0
=
And when ω = ∞ , inductor acts as a  ⇒ O.C V i R 0 1 1
=
So it acts as a high pass filter.
22. A silicon PN junction is forward biased with a  constant current at room
temperature. When the temperature is increased by 10ºC, the forward bias
voltage across the PN junction
(A) increases by 60mV  (B) decreases by 60mV
(C) increases by 25mV  (D) decreases by 25mV
Answer: – (D)
Exp: – For Si forward bias voltage change by -2.5mv
0
/ C
For
0
10 C increases, change will be  − × = − 2.5 10 25mV
× 4 1MUX
F
0
I
1
I
2
I
3
I
0
S
1
S
P Q
Y
1
L
i
i
1
R
+
+
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23. In the circuit shown below, the Norton equivalent current in amperes with respect
to the terminals P and Q is
(A)  − 6.4 j4.8 (B)  − 6.56 j7.87 (C) 10 j0 + (D) 16 j0 +
Answer: – (A)
Exp: – When terminals P & Q are S.C
Then the CKT becomes
From current Division rules
( ) ( ) ( )
N
16 25 16 25
I
25 15 j30 40 j30
= =
+ + +
( ) ( )
( )
16 25
6.4 j4.8
10 4 j3
− = =
+
24. In the circuit shown below, the value of RL such that the power transferred to RL
is maximum is
(A) 5Ω (B) 10 Ω (C) 15 Ω (D) 20 Ω
Answer: – (C)
Exp: – For maximum power transmission
*
R R L TH
=
For the calculation of RTH
+
+
10Ω 10Ω
10Ω
RL
1A
5V 2V
j30Ω
25Ω
15Ω
j50 − Ω
P
Q
O
∠ 16 0 A
j 30Ω
25Ω
P
15Ω
Q
SC
IN I =
0
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TH ( R 10 || 10 10 15 ) = + = Ω
25. The value of the integral
( )
2
c
3z 4
dz
z 4z 5
+ −
+ +
∫
where c is the circle  z 1 = is given by
(A) 0 (B) 1/10 (C) 4/5 (D) 1
Answer: – (A)
Exp: –
2
C
3z 4
dz 0
z 4z 5
+ −
=
+ +
( ( ) ) ∫
2
2 ∵ z 4z 5 z 2 1 0 + + = + + =
z 2 j = − ± will be outside the unit circle
So that integration value is ‘zero’.
Q. No. 26 – 51 Carry Two Marks Each
26. A current sheet
ɵ
= J 10u A/m y

lies on the dielectric interface x=0 between two
dielectric media with
r1 r1
ε = µ = 5, 1 in Region -1 (x<0) and
r2 r2
ε = µ = 5, 2 in
Region -2 (x>0). If the magnetic field in Region-1 at x=0
is
ɵ ɵ
1 = + H 3u 30u A / m x y

the magnetic field in Region-2 at x=0
+
is
(A)
ɵ ɵ ɵ
2 = + − H 1.5u 30u 10u A / m x y z

(B)
ɵ ɵ ɵ
2 = + − H 3u 30u 10u A / m x y z

(C)
ɵ ɵ
2 = + H 1.5u 40u A / m x y

(D)
ɵ ɵ ɵ
2 = + + H 3u 30u 10u A / m x y z

Answer: – (A)
Exp: –
2 1 2 1
− = × ⇒ H H J a H H 10u 30u 10u t t n t t z y z = − = − ˆ ˆ 
And Bn Bn 1 2
=
µ = µ H H 1 1 2 2
1
2 2
2
H H
µ
= ⇒
µ
Normal component in x direction
2 x ( )
1
H 3 uˆ
2
= 1.5uˆx
= ;  = + − ˆ ˆ 2 x y z H 1.5u 30u 10u A / m
J

x 0 =
x
y
( x 0 Re gion 2 : , 2 ) r2 r2
> − ε µ =
( x 0 Re gion 1 : , 1 ) r1 r1
< − ε µ =
10Ω
Q
RTH
10Ω
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27. A transmission line of characteristic impedance 50W is terminated in a load
impedance ZL.  The VSWR of the line is measured as 5 and the first of the voltage
maxima in the line is observed at a distance of
4
λ
from the load. The value of ZL
is
(A) 10Ω   (B) 250 Ω
(C) (19.23 j46.15 + Ω)  (D) (19.23 j46.15 − Ω)
Answer: – (A)
Exp: – Voltage maximum in the line is observed exactly at
4
λ
Therefore
L
‘ z ‘ should be real
0
L
z
VSWR
z
=
L
50
z 10
5
⇒ = = Ω (∵ Voltage minimum at load)
28. X(t) is a stationary random process with autocorrelation function
( ) ( )
2
x R exp r τ = π . This process is passed through the system shown below. The
power spectral density of the output process Y(t) is
(A) ( ) ( )
2 2 2
4 f 1 exp f π + −π
(B) ( ) ( )
2 2 2
4 f 1 exp f π − −π
(C) ( ) ( )
2 2
4 f 1 exp f π + −π
(D) ( ) ( )
2 2
4 f 1 exp f π − −π
Answer: – (A)
Exp: – The total transfer function  ( H(f) j2 f 1 = π − )
( ) ( ) ( )
2
= S f H f S f
X x ( ) ( )
F
R S f
x x
τ ←→
( )
2
2 2 f
4 f 1 e
−π
= π + ( )
2 2
t F f
e e ∵
−π − π ←→
29. The output of a 3-stage Johnson (twisted ring)  counter is fed to a digital-toanalog (D/A) converter as shown in the figure below. Assume all the states of the
counter to be unset initially. The waveform which represents the D/A converter
output Vo is
X t( ) ∑
( ) H f j2 f = π
Y t( )
+
Vref D / A
Converter
D2 D1 D0
Q2 Q1 Q0
Johnson
Counter
Vo
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(A)    (B)
(C)    (D)
Answer: – (A)
Exp: – For the Johnson counter sequence
D D D V 2 1 0 0
0 0 0 1
1 0 0 4
1 1 0 6
1 1 1 7
0 1 1 3
0 0 1 1
0 0 0 0
30. Two D flip-flops are connected as a synchronous counter that goes through the
following QBQA sequence 0011011000…
The combination to the inputs DA and DB are
(A) D Q ; D Q A B B A
= =  (B) D Q ; D Q A A B B
= =
(C)  ( D Q Q Q Q ; D Q = + = A A B A B B A ) (D)  ( D Q Q Q Q ; D Q = + = A A B A B B B )
Answer: – (D)
Exp: – Q (present) Q(next)
QB QA
1
QB
1
QA DB DA
0  0 1  1 1  1
1  1 0  1 0  1
0  1 1  0 1  0
1  0 0  0 0  0
o
V Vo
o
V
o
V
1 0
1 0
1
0
1
QA
QB
DB
0
1
1
1
0
1
QA
B
Q DA
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31. In the circuit shown below, for the MOS transistors,
2
n ox
µ = µ C 100 A / V and the
threshold voltage
T
= V 1V. The voltage Vx at the source of the upper transistor is
(A) 1V (B) 2V (C) 3V (D) 3.67V
Answer: – (C)
Exp: – The transistor which has
W
4
L
=
V 6 V DS x
= − and  V 5 V GS x
− =
GS T x V V 5 V 1 − = − − 4 Vx
− =
V V V DS GS T
− <
So that transistor in saturation region.
The transistor which has
W
1
L
=
Drain is connected to gate
So that transistor in saturation
V V V DS GS T
> > (∵ V V =
DS GS )
The current flow in both the transistor is same
( ) ( )
2 2
1 GS T 2 GS T
n 0x n 0x
1 2
w w V V V V
c c .
L 2 L 2
    − −
µ = µ    
   
( ) ( )
2 2
5 V 1 V 4 x x
4 1
2 2
− − −
= (∵ GS x V V 0 = − )
( )
2 2
4 V 8V 16 V 2V 1 x x x x − + = − +
2 ⇒ x x 3V 30V 63 0 − + = ⇒ x
= V 3V
32. An input  ( ) ( x t exp 2t u t t 6 ) = − + δ − ( ) ( ) is applied to an LTI system with impulse
response  ( ) h t u t = ( ) . The output is is
(A)   ( ) − − + + 1 exp 2t u t u t 6 ( ) ( )
 
(B)   ( ) − − + − 1 exp 2t u t u t 6 ( ) ( )
 
(C)    ( 0.5 1 exp 2t u t u t 6 ) − − + + ( ) ( )
 
(D)    ( 0.5 1 exp 2t u t u t 6 ) − − + − ( ) ( )
 
5V
6V
W / L 4 =
x
V
W / L 1 =EC-Paper Code-B      GATE 2011                   www.gateforum.com
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Answer: – (D)
Exp: –  ( )
1 6s
x s e
s 2
+ =
+
and  ( )
1
H s
s
=
( ) Y s H s s = × ( ) ( )
( )
6s
1 e
s s 2 s
+ =
( ) +
6s
1 1 1 e
2 s 2 2 s s
+ − =
+
( ) ( ) ( ) ( )
2t
y t 0.5 1 e u t u t 6 ⇒ −
− + − =
33. For a BJT the common base current gain  α = 0.98 and the collector base junction
reverse bias saturation current
CO
I 0.6 A. = µ This BJT is connected in the common
emitter mode and operated in the active region with a base drive current
IB=20XA. The collector current IC for this mode of operation is
(A) 0.98mA (B) 0.99mA (C) 1.0mA (D) 1.01mA
Answer: – (D)
Exp: – I I 1 I = β + + β ( C B CB0 )
0.98
49
1 1 0.98
α
= β = = =
− α −
B CB0
I 20 A, I 0.6 A = µ = µ
C ∴ = I 1.01mA
34. If  ( ) ( )
( )
2
2 s 1
F s L f t
s 4s 7
+
= =  
 
+ +
then the initial and final values of f(t) are
respectively
(A) 0,2 (B) 2,0 (C) 0,2/7 (D) 2/7,0
Answer: – (B)
Exp: – ( )
( )
2
t 0 s
s 2s 1
Lt f t Lt 2
→ →∞ s 4s 7
+
= =
+ +
( )
( )
2
t s 0
s 2s 1
Lt f t Lt 0
→∞ → s 4s 7
+
= =
+ +
35. In the circuit shown below, the current I is equal to
(A) 14 0ºA (B) 2.0 0ºA (C) 2.8 0ºA (D) 3.2 0ºA
Answer: – (B)
Exp: – Apply the delta – to – star conversion
The circuit becomes
~
+
14 0ºV
j4Ω
j4 − Ω
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The net Impedance  ( ) = + − + ( 2 j4 || 2 j4 2 )
4 16
2 7
4
+
= + = Ω
0
14 0 0
I 2 0 A
7
∠ = =
36. A numerical solution of the equation  ( ) f x x x 3 0 = + − = can be obtained using
Newton- Raphson method. If the starting value is  x  = 2 for the iteration, the
value of x that is to be used in the next step is
(A) 0.306 (B) 0.739 (C) 1.694 (D) 2.306
Answer: – (C)
Exp: –
( )
( )
n
n 1 n 1
n
f x
x x
f x
+
− =
( ) ( f 2 2 2 3 2 1 = + − = − ) and  ( )
1 1 2 2 1
f 2 1
2 2 2 2
+
= + =
( )
n 1
2 1
x 2 1.694
x 2 1
2 2
+
= − = ⇒
+
37. The electric and magnetic fields for a TEM wave of frequency 14 GHz in a
homogeneous medium of relative permittivity
r
ε and relative permeability  µ =
r 1
are given by
j t 280 y ( ) j t 280 y ( )
p z E E e u V / m H 3e u A / m ˆ ˆx
ω − π ω − π
= =
 
Assuming the speed of light in free space to be 3  x 10
8
m/s, the intrinsic
impedance of free space to be  120π , the relative permittivity
r
ε of the medium
and the electric field amplitude Ep are
(A)
r p
ε = = π 3, E 120  (B)
r p
ε = = π 3, E 360
(C)
r p
ε = = π 9, E 360  (D)
r p
ε = = π 9, E 120
Answer: – (D)
Exp: –
r
r
E
120
H
µ µ
= η = = π
∋ ∋
P r
r
E
120
3
µ
= η = π
Only option ‘D’ satisfies
~
+
14 0ºV
j4Ω
j4 − Ω
I
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38. A message signal  ( ) m t cos 200 t 4 cos t = π + π modulates the carrier
( ) c c
= π = c t cos 2 f t where f 1 MHZto produce an AM signal. For demodulating the
generated AM signal using an envelope detector, the time constant RC of the
detector circuit should satisfy
(A) 0.5 ms < RC < 1ms (B)  µ << < 1 s RC 0.5 ms
(C) RC s << µ   (D)  >> RC 0.5 ms
Answer: – (B)
Exp: – Time constant should be length than
m
1
f
And time constant should be far greater than
c
1
f
m
4000a
f 2000
2a
= =
C
1 1
Rc
f 2000
> >>
µ << << 1 s RC 0.5ms
39. The block diagram of a system with one input it and two outputs y1 and y2 is given
below.
A  state  space model of the above system in terms of the state vector   x   and
the output vector
T
y y y 1 2
=  
 
is
(A) x 2 x 1 u; y 1 2 x
= + =      
     
(B)
1
x 2 x 1 u; y x
2
 
    = − + =
     
 
(C)
2 0 1
x x u; y 1 2 x
0 2 1
• −   
= + =  
     
−    
(D)
2 0 1 1
x x u; y x
0 2 1 2
     
= + =
     
     
Answer: – (B)
Exp: – Draw the signal flow graph
1
+ s 2
2
+ s 2
y1
y2
1
+ s 2
2
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From the graph
ɺx 2x 4 = − + &  y x 1 1
= ;  y 2x 2 1
=
1
2
y 1
x 2 x 1 u; x
y 2
   
   
   
= − + =    
   
ɺ
40. Two systems H1 (z)  and H2 (z) are connected in cascade as shown below. The
overall output y(n) is the same as the input x(n) with a one unit delay. The
transfer function of the second system H2 (z) is
(A)
( )
( )
1
1 1
1 0.6z
z 1 0.4z
− −
(B)
( )
( )
1 1
1
z 1 0.6z
1 0.4z
− −
(C)
( )
( )
1 1
1
z 1 0.4z
1 0.6z
− −
(D)
( )
( )
1
1 1
1 0.4z
z 1 0.6z
− −
Answer: – (B)
Exp: – The overall transfer function
1
z
= (∵ unit day T.F
1
z
( =
( ) ( )
1
1 2 H z H z z
( )  ; =
( )
( )
( )
1
1
1
2 1
1
1 0.6z
z
H z z
H z 1 0.4z
= =
41. An 8085 assembly language program is given below. Assume that the carry flag
is initially unset. The content of the accumulator  after the execution of the
program is
MVI  A,07H
RLC
MOV B,A
RLC
RLC
ADD B
RRC
(A) 8CH (B) 64H (C) 23H (D) 15H
4 1
y
2
y
−1
2 / S
x 1 / S
−2
x
X n( ) ( )
( )
( )
1
1 1
1 0.4z
H z
1 0.6z
=
H z
2 ( ) y n( )EC-Paper Code-B      GATE 2011                   www.gateforum.com
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Answer: – (C)
Exp: – MVI A, 07 H  ⇒ 0000 0111  ← The content of ‘A’
RLC  ⇒ 0000 1110  ← The content of ‘A’
MOV B, A  ⇒ 0000 1110  ← The content of ‘B’
RLC  ⇒ 0001 1100  ← The content of ‘B’
RLC  ⇒ 0011 1000  ← The content of ‘B’
ADD B
A 0000 1110
B 0011 1000
0100 0110
+
0010 0011
RRC 23H
2 3
42. The first six points of the 8-point DFT of a real valued sequence
are5, 1 j3, 0,3 j4, 0 and 3 j4. − − + . The last two points of the DFT are respectively
(A) 0, 1-j3 (B) 0, 1+j3 (C) 1+j3, 5 (D) 1 – j3, 5
Answer: – (C)
Exp: – DFT points are complex conjugates of each other and they one symmetric to the
middle point.
( ) ( )
( ) ( )
( ) ( )
( ) ( )
*
*
*
*
x 0 x 7
x 1 x 6
x 2 x 5
x 3 x 4
=
=
=
=
⇒ Last two points will be  ( )
*
x 0 and  ( )
*
x 1 = + 1 j3 and 5
43. For the BJT QL in the circuit shown below,
BEon 0.7V, VCEsat
, V 0.7V. β = ∞ = =
The switch
is initially closed. At time  t = 0,  the switch is opened. The time  t  at which Q1
leaves the active region is
(A) 10 ms (B) 25 ms (C) 50 ms (D) 100 ms
5V
0.5mA
−5V
Q1
= t 0
µ5 F
4.3kΩ
−10VEC-Paper Code-B      GATE 2011                   www.gateforum.com
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Answer: – (C)
Exp: – Apply KVL at the BE junction
E
5 0.7 10 4.3
I 1mA
4.3k 4.3k
+ − −
= = =
Ω Ω
Always
E
= I 1mA ; At collector junction
Cap ( + = I 0.5mA 1mA ) (∵ β = ∞ = ;I I
E C )
Cap
= − = I 1 0.5 0.5mA always constant
V V V CE C E
= − ⇒ V V V C CE E
+ =
( )
3
0.7 4.3 3 1 10
= + × × = + 0.7 4.3 (∵ = V I R E E E )
V 5V V C cap
= =
cap Cap
t
V I
c
=   Or
( ) ( )
6
Cap
3
Cap
V C 5 5 10
t
I 0.5 10
× ×
= =
×
= 50ms
44. In the circuit shown below, the network N is described by the following Y matrix:
0.1S 0.01S
Y .
0.01S 0.1S
  −
=
 
 
the voltage gain
2
1
V
is
V
(A) 1/90 (B) –1/90 (C) –1/99 (D) –1/11
Answer: – (D)
Exp: – N 100V 25I
1 1
= + ; V 100I
2 2
− =
2 3 1 4 2
= + I Y V Y V ⇒ 0.01V 0.01V 0.1V 2 1 2
− = + ⇒ 2
1
V 1
V 11
=
45. In the circuit shown below, the initial charge on the capacitor is 2.5 mC, with the
voltage polarity as indicated. The switch is closed at time t=0. The current i(t) at
a time t after the switch is closed is
(A)  ( ) ( )
3
= − × i t 15 exp 2 10 t A
(B)  ( ) ( )
3
= − × i t 5 exp 2 10 t A
(C)  ( ) ( )
3
= − × i t 10 exp 2 10 t A
(D)  ( ) ( )
3
= − − × i t 5 exp 2 10 t A
5V
0.5mA
−5V
= t 0
5mF
4.3kΩ
−10V
V I R E E E
=
VC
+
+
100V −
25Ω
V1 N
+
V2 100Ω
2
I
1 I
100V
+
+
i t( )
10Ω
50 FµEC-Paper Code-B      GATE 2011                   www.gateforum.com
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Answer: – (A)
Exp: – = Q 2.5mC
3
initial 6
2.5 10 C
V 50V
50 10 F
×
= =
×
⇒ Thus net voltage  = + = 100 50 150V
( )
150
i t
10
= exp  ( )
t
− × 2 10 t A =15 exp  ( )
t
− × 2 10 t A
46. The system of equations
x y z 6
x 4y 6z 20
x 4y z
= + +
= + +
+ + λ = µ
has NO solution for values of λ and µ given by
(A) λ = µ = 6, 20 (B) λ = µ ≠ 6, 20 (C) λ ≠ µ = 6, 20 (D) λ ≠ µ ≠ 6, 20
Answer: – (B)
Exp: – Given equations are  + + = x y z 6 , + + = x 4y 6z 20 and  x 4y z + + λ = µ
If  λ = 6 and  µ = 20 , then  + + = x 4y 6z 20
+ + = x 4y 6z 20 infinite solution
If  λ = 6 and  µ ≠ 20 , the
x 4y 6z 20
x 4y 6z
= + +
+ + = µ
(µ ≠ 20) no solution
If  λ ≠ 6 and  µ = 20
x 4y 6z 20
x 4y z 20
= + +
+ + λ =
will have solution
λ ≠ 6 and  µ ≠ 20 will also give solution
47. A fair dice is tossed two times. The probability that the second toss results in a
value that is higher than the first toss is
(A) 2/36 (B) 2/6 (C) 5/12 (D) ½
Answer: – (C)
Exp: – Total number of cause = 36
Total number of favorable causes = 5+ 4 + 3 + 2 + 1 = 15
Then probability
15 5
36 12
= =
(1,1) (2,1) (3,1) (4,1) (5,1)
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(1, 2) (2, 2) (3, 2) (4, 2) (5, 2)
(6, 2)
(1, 3) (2, 3) (3, 3) (4, 3) (5, 3)
(6, 3)
(1, 4) (2, 4) (3, 4) (4, 4) (5, 4)
(6, 4)
(1, 5) (2, 5) (3,5) (4, 5) (5, 5)
(6,5)
(1, 6) (2, 6) (3, 6) (4, 6) (5, 6)
(6, 6)
Common Data Questions: 48 & 49
The channel resistance of an N-channel JFET shown in the figure below is 600 Ω
when the full channel thickness (tch) of  10 mµ is available for conduction. The
built-in voltage of the gate P
+
N junction (Vbi) is -1 V. When the gate to source
voltage (VGS) is 0 V, the channel is depleted by  µ1 m on each side due to the builtin voltage and hence the thickness available for conduction is only  µ8 m
48. The channel resistance when VGS = -3 V is
(A) 360Ω (B) 917Ω (C) 1000Ω (D) 3000Ω
Answer: – (C)
Exp: – Width of the depletion largeW V V α +
bi GS
2
1
W 1 3
W 1
− −
⇒ =
w 2w 2 1
= ( = µ = µ ) 2 1 m 2 m
So that channel thickness = 10 – 4 =  µ6 m
µ − 8 m 750
µ − 6 m ?
d
8
r 750
6
= × 1000 = Ω
49. The channel resistance when VGS = 0 V is
(A) 480Ω (B) 600Ω (C) 750Ω (D) 1000Ω
+
VGS
Source
Gate
P
+
P
+
N
ch
t
DrainEC-Paper Code-B      GATE 2011                   www.gateforum.com
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Answer: – (C)
Exp: –
don
oh
1
r
t
α
At VGS= 0,
ch d ( t 10 m; Given r 600 = µ = Ω)
d
10
r 600 at 8 m
8
= × ← µ = Ω 750
Common Data Questions: 50 & 51
The input-output transfer function of a plant   ( )
( )
2
100
H s .
s s 10
=
+
The plant is
placed in a unity negative feedback configuration as shown in the figure below.
50. The gain margin of the system under closed loop unity negative feedback is
(A) 0dB (B) 20dB (C) 26 dB (D) 46 dB
Answer: – (C)
Exp: – ( )
( )
2
100
H s
s s 10
=
+
Phase cross over frequency=
1 0
90 2 tan 180
10
−   ω
− = − −
 
 
1 0
2 tan 90
10
−   ω
− = − ⇒
  ⇒
 
1 0
tan 45
10
−   ω
=
 
 
⇒ ω = 10 rad / sec
( ) ( )
( )
2
100
H jw
j10 j10 10
=
+
1 1
10.2 20
= =
1
GM 20 log 20 log20 26dB
1 / 20
= = =
51. The signal flow graph that DOES NOT model the plant transfer function H(s) is
(A)    (B)
(C)    (D)
+
r u
( )
( )
2
100
H s
s s 10
=
+
y
plant
Σ
1 1 / s 1 / s 1 / s 100
u
−10 −10
y
1 / s 1 / s 1 / s 100
u
−20
y
−100
1 / s 1 / s 1 / s 100
u
−20
y
−100
1 / s 1 / s 1 / s 100
u y
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Answer: – (D)
Exp: -(D) Option (D) does not fix for the given transfer function.
Linked Answer Questions: Q.52 to Q.55 Carry Two Marks Each
Statement for Linked Answer Questions: 52 & 53
In the circuit shown below, assume that the voltage drop across a forward biased
diode is 0.7 V. The thermal voltage
t
= = V kT / q 25mV. The small signal input
( i p p ) = ω = v V cos t where V 100mV .
52. The bias current IDC through the diodes is
(A) 1 mA (B) 1.28 mA (C) 1.5 mA (D) 2 mA
Answer: – (A)
Exp: –
( )
DC
12.7 0.7 0.7 0.7 0.7
I 1mA
9900
+ + + −
= =
53. The ac output voltage vac is
(A) 0.25 cos t mV (ω )  (B)  ( 1cos t mV ω )
(C)  ( 2 cos t mV ω )   (D) 22 cos t mV (ω )
Answer: – (C)
Exp: – AC dynamic resistance,
T
d
D
V 2 25mV
r 50
I 1mA
η ×
= = = Ω
η = 2 for Si (∵ forward drop = 0.7V)
The ac dynamic resistance offered by each diode = 50Ω
ac i ( )
4 50
V V ac
9900 50
  × Ω
= ∴
 
  +
3 100
200 10 cos wt
10000
−  
× =
 
 
ac
= V 2 cos wt mV ( )
+
+
DC ac
I i +
V v DC ac
+
12.7V
vi
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Statement for Linked Answer Questions: 54 & 55
A four-phase and an eight-phase signal constellation are shown in the figure
below.
54. For the constraint that the minimum distance between pairs of signal points be d
for both constellations, the radii r1, and r2 of the circles are
(A)
1 2
= = r 0.707d, r 2.782d (B)
1 2
= = r 0.707d, r 1.932d
(C)
1 2
= = r 0.707d, r 1.545d (D)
1 2
= = r 0.707d, r 1.307d
Answer: – (D)
Exp:-  For 1
st
constellation
2 2 2
1 1
r r d + =
2 2
1 ⇒ = r d / 2 1 ⇒ = r 0.707d
For 2
nd
constellation
2
d
r cos 67.5
2
=
2
= r 1.307d
55. Assuming high SNR and that all signals are equally probable, the additional
average transmitted signal energy required by the 8-PSK signal to achieve the
same error probability as the 4-PSK signal is
(A) 11.90 dB (B) 8.73 dB (C) 6.79 dB (D) 5.33 dB
Answer: – (D)
Exp: – Energy
2 2
1 2
= r and r
( )
( )
2
2
1
2 2
2
r 0.707d
r 1.307d
= ⇒
Energy  ( )
( )
( )
2
2
1.307
in dD 10log 5.33dB
0.707
= =
Q. No. 56 – 60 Carry One Mark Each
56. There are two candidates P and Q in an election. During the campaign, 40% of
the voters promised to vote for P, and rest for Q. However, on the day of election
15% of the voters went back on their promise to vote for P and instead voted for
Q. 25% of the voters went back on their promise to vote for Q and instead voted
for P. Suppose, P lost by 2 votes, then what was the total number of voters?
(A) 100 (B) 110 (C) 90 (D) 95
Answer: – (A)
Q
d r
1 I
Q
d
2
r
I
2
r
0
45
2
r
d / 2 d d / 2
0
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Exp: –  P Q
40% 60%
6% 6%
15% 15%
49% 51%
2% 2
100% 100
+ −
− +
= ∴
=
57. Choose the most appropriate word from the options given below to complete the
following sentence:
It was her view that the country’s problems had been_________ by
foreign technocrats, so that to invite them to come back would be
counter-productive.
(A) Identified   (B) ascertained   (C) Texacerbated  (D) Analysed
Answer: – (C)
Exp: -The clues in the question are —foreign technocrats did something negatively to
the problems – so it is counter-productive to invite them. All other options are
non-negative. The best choice is exacerbated which means aggravated or
worsened.
58. Choose the word from the options given below that is most nearly opposite in
meaning to the given word:
Frequency
(A) periodicity   (B) rarity
(C) gradualness   (D) persistency
Answer: – (B)
Exp: – The best antonym here is rarity which means shortage or scarcity.
59. Choose the most appropriate word from the options given below to complete the
following sentence:  Under ethical guidelines recently adopted by the
Indian Medical Association, human genes are to be manipulated only to
correct diseases for which______________ treatments are
unsatisfactory.
(A) Similar (B) Most (C) Uncommon (D) Available
Answer: – (D)
Exp: – The context seeks to take a deviation only when the existing/present/current/
alternative treatments are unsatisfactory. So the word for the blank should be a
close synonym of existing/present/current/alternative. Available is the closest of
all.
60. The question below consists of a pair of related words followed by four pairs of
words. Select the pair that best expresses the relation in the original pair:
Gladiator : Arena
(A) dancer : stage   (B) commuter: train
(C) teacher : classroom  (D) lawyer : courtroom
Answer: – (D) EC-Paper Code-B      GATE 2011                   www.gateforum.com
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Exp: – The given relationship is worker: workplace. A gladiator is (i) a person, usually a
professional combatant trained to entertain the public by engaging in mortal
combat with another person or a wild.(ii) A person engaged in a controversy or
debate, especially in public.
Q. No. 61 – 65 Carry Two Marks Each
61    The fuel consumed by a motorcycle during a journey while traveling at various
speeds is indicated in the graph below.
The distances covered during four laps of the journey are listed in the table below
Lap Distance (kilometers) Average speed
(kilometers per hour)
P 15 15
Q 75 45
R 40 75
S 10 10
From the given data, we can conclude that the fuel consumed per kilometre was
least during the lap
(A) P (B) Q (C) R (D) S
Answer: – (A)
Exp: –         Fuel consumption Actual
15 1
P 60km / l l
60 4
75 5
Q 90km / l l
90 6
40 8
R 75km / l l
75 15
10 1
S 30 km / l l
30 3
=
=
=
=
120
90
60
30
0
0 15 30 45 60 75 90
Speed
(kilometers per hour)
consumption  Fuel
(kilometers per litre) EC-Paper Code-B      GATE 2011                   www.gateforum.com
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62. Three friends, R, S and T shared toffee from a bowl. R took 1/3
rd
of the toffees,
but returned four to the bowl. S took 1/4
th
of what was left but returned three
toffees to the bowl. T took half of the remainder but returned two back into the
bowl. If the bowl had 17 toffees left, how many toffees-were originally there in
the bowl?
(A) 38 (B) 31 (C) 48 (D) 41
Answer: – (C)
Exp: – Let the total number of toffees is bowl e x
R took
1
3
of toffees and returned 4 to the bowl
∴ Number of toffees with
1
R x 4
3
− =
Remaining of toffees in bowl =
2
x 4
3
+
Number of toffees with S =
1 2
x 4 3
4 3
 
− +
 
 
Remaining toffees in bowl =
3 2
x 4 4
4 3
 
+ +
 
 
Number of toffees with
1 3 2
T x 4 4 2
2 4 3
 
= + + +  
 
Remaining toffees in bowl =
1 3 2
x 4 4 2
2 4 3
   
+ + +
   
   
Given,
1 3 2
x 4 4 2 17
2 4 3
   
= + + +
   
   
3 2
x 4 27
4 3
 
= + ⇒
 
 
⇒ = x 48
63. Given that f(y) = | y | / y, and q is any non-zero real number, the value of
| f(q) – f(-q) | is
(A) 0 (B) -1 (C) 1 (D) 2
Answer: – (D)
Exp: – Given,
y
f(y)
y
( ) ( ) =
q q q
f q ; f q
q q q
− −
= = − = ⇒
( ) ( )
q q 2 q
f q f q
q q q
− = + = = 2
64. The sum of n terms of the series 4+44+444+…. is
(A) ( )
n 1
4 / 81 10 9n 1
+
  − −
 
(B) ( )
n 1
4 / 81 10 9n 1
  − −
 
(C) ( )
n 1
4 / 81 10 9n 10
+
  − −
 
(D) ( )
n
4 / 81 10 9n 10   − −
 EC-Paper Code-B      GATE 2011                   www.gateforum.com
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Answer: – (C)
Exp: – Let S=4 (1 + 11 + 111 + ……..) ( )
4
9 99 999 …….
9
+ + + =
{ ( ) ( ) ( )}
{ } ( )
( )
{ }
2 3
n
2 n n 1
4
10 1 10 1 10 1 ………
9
4 4 4 10 1
10 10 ……10 n 10 n 10 9n 10
9 9 9 81
+
+ − + − + − =
  −
 
= + + − = − = − −  
 
 
65. The horse has played a little known but very important role in the field of
medicine. Horses were injected with toxins of diseases until their blood built up
immunities. Then a serum was made from their blood. Serums to fight with
diphtheria and tetanus were developed this way.
It can be inferred from the passage that horses were
(A) given immunity to diseases (B) generally quite immune to diseases
(C) given medicines to fight toxins (D) given diphtheria and tetanus serums
Answer: – (B)
Exp: – From the passage it cannot be inferred that horses are given immunity as in (A),
since the aim is to develop medicine and in turn immunize humans. (B) is correct
since it is given that horses develop immunity after some time. Refer “until their
blood built up immunities”. Even (C) is invalid since medicine is not built till
immunity is developed in the horses. (D) is incorrect since specific examples are
cited to illustrate and this cannot capture the essence.

EC-Paper Code-B      GATE 2011                   www.gateforum.com© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the written permission. Discuss GATE 2011 question paper at www.gatementor.com.                                                                 1 Q. No. 1 – 25 Carry One Mark Each 1. Consider the following statements regarding the  complex Poynting vector  P for the power radiated by a point source in an infinite homogeneous and lossless medium.  Re P( ) denotes the real part of  P S denotes a spherical surface whose .centre is at the point source, and ɵn denotes the unit surface normal on S. Which of the following statements is TRUE?  (A) Re P( ) remains constant at any radial distance from the source  (B) Re P( ) increases with increasing radial distance from the source  (C)  ( )sRe P .n dS∫∫ remains constant at any radial distance from the source  (D)  ( )sRe P .n dS∫∫ decreases with increasing radial distance form the source Answer: – (D) Exp: –  ( )SRe P .nds ˆ∫∫ gives average power and it decreases with increasing radial distance from the source 2. A transmission line of characteristic impedance 50Ω is terminated by a 50Ω load. When excited by a sinusoidal voltage source at 10GHz, the phase difference between two points spaced 2mm apart on the line is found to be 4π radians. The phase velocity of the wave along the line is  (A) 8× 0.8 10 m / s (B) 8× 1.2 10 m / s (C) 8× 1.6 10 m / s (D) 8× 3 10 m / sAnswer: – (C) Exp: – 0 Z 50 = Ω ; L Z 50 = Ω For 4π radians the distance is 2mm  The phase velocity 107 8P32 10v 16 10 1.6 10 m / s216 10−−ω × π ×× = × = = =β π×3. An analog signal is band-limited to 4kHz, sampled at the Nyquist rate and the samples are quantized into 4 levels. The quantized  levels are assumed to be independent and equally probable. If we transmit two quantized samples per second, the information rate is ________ bits / second.  (A) 1 (B) 2 (C) 3 (D) 4 Answer: – (D) Exp: – Since two samples are transmitted and each sample has 2 bits of information,  then the information rate is 4 bits/sec. EC-Paper Code-B      GATE 2011                   www.gateforum.com© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the written permission. Discuss GATE 2011 question paper at www.gatementor.com.                                                                 2 4. The root locus plot for a system is given below. The open loop transfer function corresponding to this plot is given by  (A)  ( ) ( )( )( ) ( )s s 1G s H s ks 2 s 3+=+ + (B)  ( ) ( )( )( ) ( )2s 1G s H s ks s 2 s 3+=+ +   (C)  ( ) ( )( ) ( ) ( )1G s H s ks s 1 s 2 s 3=+ + −   (D)  ( ) ( )( )( ) ( )s 1G s H s ks s 2 s 3+=+ +Answer: – (B) Exp: –  ‘ x ‘ → indicates pole ‘O’ → indicates zero The point on the root locus when the number of poles and zeroes on the real axis to the right side of that point must be odd 5. A system is defined by its impulse response  ( ) ( )nh n 2 u n 2 = − . The system is   (A) stable and causal  (B) causal but not stable   (C) stable but not causal (D) unstable and non-causal Answer: – (B) Exp: –  ( ) ( )nh n 2 u n 2 = −h n( ) is existing for n>2 ; so that  ( ) h n 0;n 0 = < ⇒ causal  ( ) ( )n nn n n 2h n 2 u n 2 2∞ ∞ ∞= ∞= ∞−=∑ ∑ ∑ = − = = ∞ ⇒ System is unstable  6. If the unit step response of a network is  ( )t1 e−α− , then its unit impulse response is  (A) te−αα (B) 1 te− −αα (C) ( )1 t1 e− −α− α (D) ( )t1 e−α− αAnswer: – (A) Exp: – S t( ) → step response  Impulse response  ( ) ( ) ( ) ( )d d t th t S t 1 e edt dtα α= = − = α7. The output Y in the circuit below is always ‘1’  when × × o ×jωσ−3 −2 −1 0PQRYEC-Paper Code-B      GATE 2011                   www.gateforum.com© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the written permission. Discuss GATE 2011 question paper at www.gatementor.com.                                                                 3  (A) two or more of the inputs P,Q,R are ‘0’  (B) two or more of the inputs P,Q,R are ‘1’  (C) any odd number of the inputs P,Q,R is ‘0’  (D) any odd number of the inputs P,Q,R is ‘1’ Answer: – (B) Exp: – The output Y expression in the Ckt = + + Y PQ PR RQ So that two or more inputs are ‘1’, Y is always ‘1’. 8. In the circuit shown below, capacitors C1 and C2 are very large and are shorts at the input frequency.  vi is a small signal input. The gain magnitude oivv at 10M rad/s is  (A) maximum (B) minimum (C) unity (D) zero Answer: – (A) Exp: – In the parallel RLC Ckt L 10 H = µ and = C 1nF7g6 91 110 rad / s 10Mrad / sLC 10 10 10− −ω = = = =× × So that for a tuned amplifier, gain is maximum at resonant frequency 9. Drift current in the semiconductors depends upon   (A) only the electric field    (B) only the carrier concentration gradient   (C) both the electric field and the carrier concentration  (D) both the electric field and the carrier concentration gradient Answer: – (C) Exp: – Drift current, J E = σ( = µ + µ J n p qE n P ) So that it depends on carrier concentration and electric field. ~+−+−2kΩ+C2Q1ovC12kΩiv2.7V2kΩ2kΩ10 Hµ5VEC-Paper Code-B      GATE 2011                   www.gateforum.com© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the written permission. Discuss GATE 2011 question paper at www.gatementor.com.                                                                 4 10. A Zener diode, when used in voltage stabilization circuits, is biased in  (A) reverse bias region below the breakdown voltage  (B) reverse breakdown region  (C) forward bias region  (D) forward bias constant current mode Answer: – (B) Exp: –     For Zener diode  Voltage remains constant in break down region and current carrying capacity in high. 11. The circuit shown below is driven by a sinusoidal input  =i p v V cos t / RC ( ) . The steady state output vo is  (A) ( p ) V / 3 cos t / RC ( )   (B)  ( p ) V / 3 sin t / RC ( )    (C) ( p ) V / 2 cos t / RC ( )   (D)  ( p ) V / 2 sin t / RC ( )  Answer: – (A) Exp: – 0 2i 1 2v zv z z=+ where 21z R ||j c=ω and 1z Rj c1+ =ω( )2RzR jcw 1=+ Given i p1 tw v v cosRC RC   = =      ∵ 2Rz1 j= ⇒+11z Rj c+ =ω( )1R R 1 jjR− ⇒ + =( )0iRv 1 j 1 1R v 1 2 3R 1 j1 j+= = =+− ++vi ~R CCR+−ov+−V2EC-Paper Code-B      GATE 2011                   www.gateforum.com© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the written permission. Discuss GATE 2011 question paper at www.gatementor.com.                                                                 5 12. Consider a closed surface S surrounding volume V. If  r is the position vector of a point inside S, with ɵn the unit normal on S, the value of the integral s5r.ndS∫∫ɵ is  (A) 3V (B) 5V (C) 10V (D) 15V Answer: – (D) Exp: – Apply the divergence theorem  S v5r.n.dx 5 .rdV = ∇∫∫∫ ∫∫  ( )v= 5 3 dv∫∫∫ = 15 V ( ∇ = .r 3 and r  is the position vector) ∵13. The modes in a rectangular waveguide are denoted by mnmnTETM where m and n are the eigen numbers along the larger and smaller dimensions of the waveguide respectively. Which one of the following statements is TRUE?  (A) The TM10 mode of the wave does not exist  (B) The TE10 mode of the wave does not exist  (C) The TM10  and the TE10  modes both exist and have the same cut-off frequencies  (D) The TM10 and TM01  modes both exist and have the same cut-off frequencies Answer: – (A) Exp: –  TM10 mode doesn’t exist in rectangular waveguide.  14. The solution of the differential equation  ( )dyky, y 0 cdx= = is  (A) kyx ce−= (B) cy= x ke  (C) kx= y ce  (D) kxy ce−=Answer: – (C) Exp: – Given  ( ) y 0 C = and dyky,dx=dykdxy= ⇒kx c= + ln y kx c y e e ⇒ = When ( ) y 0 C = , 0y k e1( ) ∴ =kxy c e k C 1∵ = =15. The  Column-I lists the attributes and the  Column-II lists the modulation systems. Match the attribute to the modulation system that best meets it Column-I  Column-II P Power efficient transmission of signals 1 Conventional AM Q Most bandwidth efficient transmission of voice signals 2 FM R Simplest receiver structure 3 VSB S Bandwidth efficient transmission of signals with Significant dc component 4 SSB-SC  (A) P-4;Q-2;R-1;S-3  (B) P-2;Q-4;R-1;S-3  (C) P-3;Q-2;R-1;S-4  (D) P-2;Q-4;R-3;S-1 EC-Paper Code-B      GATE 2011                   www.gateforum.com© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the written permission. Discuss GATE 2011 question paper at www.gatementor.com.                                                                 6 Answer: – (B) Exp: – Power efficient transmission → FM  Most bandwidth efficient → SSB-SC  Transmission of voice signal   Simplest receives structure → conventional AM  Bandwidth efficient transmission of → VSB  Signals with significant DC component 16. The differential equation  ( )22d y dy100 20 y x tdt dt− + = describes a system with an input x(t) and an output y(t). The system, which is initially relaxed, is excited by a unit step input. The output y(t) can be represented by the waveform  (A)    (B)    (C)    (D)  Answer: – (A) Exp: –  ( )22100d y 20dyy x tdt dt= + − Apply L.T both sides ( ) ( )2 1100s 20s 1 Y ss( ) ( ) ( ) = + −1x t u t x s3 = =  ∵( )( )21Y ss 100s 20s 1=+ − So we have poles with positive real part ⇒ system is unstable. y t( )ty t( )ty t( )ty t( )tEC-Paper Code-B      GATE 2011                   www.gateforum.com© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the written permission. Discuss GATE 2011 question paper at www.gatementor.com.                                                                 7 17. For the transfer function  ( ) G j 5 j ω = + ω , the corresponding Nyquist plot for positive frequency has the form  (A)    (B)    (C)    (D)  Answer: – (A) Exp: – As we increases real part ‘5’ is fixed only imaginary part increases. 18. The trigonometric Fourier series of an even function does not have the  (A) dc term   (B) cosine terms  (C) sine terms   (D) odd harmonic terms Answer: – (C) Exp: –  f t( ) is even function, hence k b 0 =   Where k’b ‘ is the coefficient of sine terms 19. When the output Y in the circuit below is ‘1’, it implies that data has  (A) changed from 0 to 1 (B) changed from 1 to 0  (C) changed in either direction (D) not changed Answer: – (A) Exp: – When data is ‘0’, Q is ‘0’  And Q’ is ‘1’ first flip flop   Data is changed to 1  Q is 1 → first ‘D’   Q’ is connected to 2nd flip flop   So that 2 Q 1 = So that the inputs of AND gate is ‘1’ ⇒ = y ‘1’σjω5σjωj5σjω1 / 5σjω1 / 5D QQD QQClockDataYEC-Paper Code-B      GATE 2011                   www.gateforum.com© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the written permission. Discuss GATE 2011 question paper at www.gatementor.com.                                                                 8 20. The logic function implemented by the circuit below is (ground implies logic 0)  (A)  = F AND P, Q ( ) (B)  = F OR P, Q ( ) (C)  = F XNOR P, Q ( ) (D)  = F XOR P, Q ( )Answer: – (D) Exp: – From the CKT  O is connected to 0 3’I ‘ & ‘ I ‘ And ‘1’ is connected to 1 2I & I ∴ = + F PQ PQ = XOR P, Q ( )21. The circuit below implements a filter between the input current ii and the output voltage vo. Assume that the opamp is ideal. The filter implemented is a             (A) low pass filter   (B) band pass filter   (C) band stop filter   (D)  high pass filter Answer: – (D) Exp: – When W=0; inductor acts as a  ⇒ S.C V 0 0= And when ω = ∞ , inductor acts as a  ⇒ O.C V i R 0 1 1= So it acts as a high pass filter. 22. A silicon PN junction is forward biased with a  constant current at room temperature. When the temperature is increased by 10ºC, the forward bias voltage across the PN junction  (A) increases by 60mV  (B) decreases by 60mV  (C) increases by 25mV  (D) decreases by 25mV Answer: – (D) Exp: – For Si forward bias voltage change by -2.5mv0/ C For 010 C increases, change will be  − × = − 2.5 10 25mV× 4 1MUXF0I1I2I3I0S1SP QY1Lii1R−++voEC-Paper Code-B      GATE 2011                   www.gateforum.com© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the written permission. Discuss GATE 2011 question paper at www.gatementor.com.                                                                 9 23. In the circuit shown below, the Norton equivalent current in amperes with respect to the terminals P and Q is    (A)  − 6.4 j4.8 (B)  − 6.56 j7.87 (C) 10 j0 + (D) 16 j0 +Answer: – (A) Exp: – When terminals P & Q are S.C  Then the CKT becomes  From current Division rules ( ) ( ) ( )N16 25 16 25I25 15 j30 40 j30= =+ + +( ) ( )( )16 256.4 j4.810 4 j3− = =+24. In the circuit shown below, the value of RL such that the power transferred to RLis maximum is  (A) 5Ω (B) 10 Ω (C) 15 Ω (D) 20 ΩAnswer: – (C) Exp: – For maximum power transmission *R R L TH= For the calculation of RTH+−+−10Ω 10Ω10ΩRL1A5V 2Vj30Ω25Ω15Ωj50 − ΩPQO∠ 16 0 Aj 30Ω25ΩP15ΩQSCIN I =0∠ 16 0 AEC-Paper Code-B      GATE 2011                   www.gateforum.com© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the written permission. Discuss GATE 2011 question paper at www.gatementor.com.                                                                 10 TH ( R 10 || 10 10 15 ) = + = Ω25. The value of the integral ( )2c3z 4dzz 4z 5+ −+ +∫ where c is the circle  z 1 = is given by  (A) 0 (B) 1/10 (C) 4/5 (D) 1 Answer: – (A) Exp: – 2C3z 4dz 0z 4z 5+ −=+ +( ( ) ) ∫22 ∵ z 4z 5 z 2 1 0 + + = + + =z 2 j = − ± will be outside the unit circle   So that integration value is ‘zero’. Q. No. 26 – 51 Carry Two Marks Each 26. A current sheet ɵ= J 10u A/m y lies on the dielectric interface x=0 between two dielectric media with r1 r1ε = µ = 5, 1 in Region -1 (x<0) and r2 r2ε = µ = 5, 2 in Region -2 (x>0). If the magnetic field in Region-1 at x=0- is ɵ ɵ1 = + H 3u 30u A / m x ythe magnetic field in Region-2 at x=0+ is    (A) ɵ ɵ ɵ2 = + − H 1.5u 30u 10u A / m x y z (B) ɵ ɵ ɵ2 = + − H 3u 30u 10u A / m x y z   (C) ɵ ɵ2 = + H 1.5u 40u A / m x y (D) ɵ ɵ ɵ2 = + + H 3u 30u 10u A / m x y zAnswer: – (A) Exp: – 2 1 2 1− = × ⇒ H H J a H H 10u 30u 10u t t n t t z y z = − = − ˆ ˆ  And Bn Bn 1 2=µ = µ H H 1 1 2 212 22H Hµ= ⇒µ Normal component in x direction 2 x ( )1H 3 uˆ2= 1.5uˆx= ;  = + − ˆ ˆ 2 x y z H 1.5u 30u 10u A / mJx 0 =xy( x 0 Re gion 2 : , 2 ) r2 r2> − ε µ =( x 0 Re gion 1 : , 1 ) r1 r1< − ε µ =10Ω−QRTH10Ω10ΩEC-Paper Code-B      GATE 2011                   www.gateforum.com© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the written permission. Discuss GATE 2011 question paper at www.gatementor.com.                                                                 11 27. A transmission line of characteristic impedance 50W is terminated in a load impedance ZL.  The VSWR of the line is measured as 5 and the first of the voltage maxima in the line is observed at a distance of 4λ from the load. The value of ZLis  (A) 10Ω   (B) 250 Ω   (C) (19.23 j46.15 + Ω)  (D) (19.23 j46.15 − Ω)Answer: – (A) Exp: – Voltage maximum in the line is observed exactly at 4λ   Therefore L’ z ‘ should be real 0LzVSWRz=L50z 105⇒ = = Ω (∵ Voltage minimum at load) 28. X(t) is a stationary random process with autocorrelation function ( ) ( )2x R exp r τ = π . This process is passed through the system shown below. The power spectral density of the output process Y(t) is  (A) ( ) ( )2 2 24 f 1 exp f π + −π (B) ( ) ( )2 2 24 f 1 exp f π − −π   (C) ( ) ( )2 24 f 1 exp f π + −π   (D) ( ) ( )2 24 f 1 exp f π − −πAnswer: – (A) Exp: – The total transfer function  ( H(f) j2 f 1 = π − )( ) ( ) ( )2= S f H f S fX x ( ) ( )FR S fx xτ ←→( )22 2 f4 f 1 e−π= π + ( )2 2t F fe e ∵−π − π ←→29. The output of a 3-stage Johnson (twisted ring)  counter is fed to a digital-toanalog (D/A) converter as shown in the figure below. Assume all the states of the counter to be unset initially. The waveform which represents the D/A converter output Vo is X t( ) ∑( ) H f j2 f = πY t( )+−Vref D / AConverterD2 D1 D0Q2 Q1 Q0JohnsonCounterVoClockEC-Paper Code-B      GATE 2011                   www.gateforum.com© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the written permission. Discuss GATE 2011 question paper at www.gatementor.com.                                                                 12  (A)    (B)    (C)    (D)  Answer: – (A) Exp: – For the Johnson counter sequence D D D V 2 1 0 00 0 0 11 0 0 41 1 0 61 1 1 70 1 1 30 0 1 10 0 0 0−−−−−−−30. Two D flip-flops are connected as a synchronous counter that goes through the following QBQA sequence 0011011000…   The combination to the inputs DA and DB are  (A) D Q ; D Q A B B A= =  (B) D Q ; D Q A A B B  = = (C)  ( D Q Q Q Q ; D Q = + = A A B A B B A ) (D)  ( D Q Q Q Q ; D Q = + = A A B A B B B )Answer: – (D) Exp: – Q (present) Q(next) QB QA1QB1QA DB DA 0  0 1  1 1  1   1  1 0  1 0  1  0  1 1  0 1  0   1  0 0  0 0  0   oV VooVoV1 01 0101QAQBDB0  11101QABQ DA=EC-Paper Code-B      GATE 2011                   www.gateforum.com© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the written permission. Discuss GATE 2011 question paper at www.gatementor.com.                                                                 13 31. In the circuit shown below, for the MOS transistors, 2n oxµ = µ C 100 A / V and the threshold voltage T= V 1V. The voltage Vx at the source of the upper transistor is  (A) 1V (B) 2V (C) 3V (D) 3.67V Answer: – (C) Exp: – The transistor which has W4L=V 6 V DS x= − and  V 5 V GS x− =GS T x V V 5 V 1 − = − − 4 Vx− =V V V DS GS T− < So that transistor in saturation region.  The transistor which has W1L= Drain is connected to gate  So that transistor in saturation   V V V DS GS T> > (∵ V V =DS GS ) The current flow in both the transistor is same ( ) ( )2 21 GS T 2 GS Tn 0x n 0x1 2w w V V V Vc c .L 2 L 2    − −µ = µ       ( ) ( )2 25 V 1 V 4 x x4 12 2− − −= (∵ GS x V V 0 = − )( )2 24 V 8V 16 V 2V 1 x x x x − + = − +2 ⇒ x x 3V 30V 63 0 − + = ⇒ x= V 3V32. An input  ( ) ( x t exp 2t u t t 6 ) = − + δ − ( ) ( ) is applied to an LTI system with impulse response  ( ) h t u t = ( ) . The output is is  (A)   ( ) − − + + 1 exp 2t u t u t 6 ( ) ( )  (B)   ( ) − − + − 1 exp 2t u t u t 6 ( ) ( )  (C)    ( 0.5 1 exp 2t u t u t 6 ) − − + + ( ) ( )  (D)    ( 0.5 1 exp 2t u t u t 6 ) − − + − ( ) ( ) 5V6VW / L 4 =xVW / L 1 =EC-Paper Code-B      GATE 2011                   www.gateforum.com© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the written permission. Discuss GATE 2011 question paper at www.gatementor.com.                                                                 14 Answer: – (D) Exp: –  ( )1 6sx s es 2−+ =+ and  ( )1H ss=( ) Y s H s s = × ( ) ( )( )6s1 es s 2 s−+ =( ) +6s1 1 1 e2 s 2 2 s s−+ − =+( ) ( ) ( ) ( )2ty t 0.5 1 e u t u t 6 ⇒ −− + − =33. For a BJT the common base current gain  α = 0.98 and the collector base junction reverse bias saturation current COI 0.6 A. = µ This BJT is connected in the common emitter mode and operated in the active region with a base drive current IB=20XA. The collector current IC for this mode of operation is  (A) 0.98mA (B) 0.99mA (C) 1.0mA (D) 1.01mA Answer: – (D) Exp: – I I 1 I = β + + β ( C B CB0 )0.98491 1 0.98α= β = = =− α −B CB0I 20 A, I 0.6 A = µ = µ  C ∴ = I 1.01mA34. If  ( ) ( )( )22 s 1F s L f ts 4s 7+= =   + + then the initial and final values of f(t) are respectively  (A) 0,2 (B) 2,0 (C) 0,2/7 (D) 2/7,0 Answer: – (B) Exp: – ( )( )2t 0 ss 2s 1Lt f t Lt 2→ →∞ s 4s 7+= =+ +( )( )2t s 0s 2s 1Lt f t Lt 0→∞ → s 4s 7+= =+ +35. In the circuit shown below, the current I is equal to  (A) 14 0ºA (B) 2.0 0ºA (C) 2.8 0ºA (D) 3.2 0ºAAnswer: – (B) Exp: – Apply the delta – to – star conversion  The circuit becomes ~+−14 0ºVj4Ω6Ω6Ω6Ωj4 − ΩIEC-Paper Code-B      GATE 2011                   www.gateforum.com© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the written permission. Discuss GATE 2011 question paper at www.gatementor.com.                                                                 15  The net Impedance  ( ) = + − + ( 2 j4 || 2 j4 2 )4 162 74+= + = Ω014 0 0I 2 0 A7∠∠ = =36. A numerical solution of the equation  ( ) f x x x 3 0 = + − = can be obtained using Newton- Raphson method. If the starting value is  x  = 2 for the iteration, the value of x that is to be used in the next step is  (A) 0.306 (B) 0.739 (C) 1.694 (D) 2.306 Answer: – (C) Exp: – ( )( )nn 1 n 1nf xx xf x+− =( ) ( f 2 2 2 3 2 1 = + − = − ) and  ( )1 1 2 2 1f 2 12 2 2 2+  = + =( )n 12 1x 2 1.694x 2 12 2+−= − = ⇒+37. The electric and magnetic fields for a TEM wave of frequency 14 GHz in a homogeneous medium of relative permittivity rε and relative permeability  µ =r 1are given by j t 280 y ( ) j t 280 y ( )p z E E e u V / m H 3e u A / m ˆ ˆxω − π ω − π= =  Assuming the speed of light in free space to be 3  x 108 m/s, the intrinsic impedance of free space to be  120π , the relative permittivity rε of the medium and the electric field amplitude Ep are  (A) r pε = = π 3, E 120  (B) r pε = = π 3, E 360 (C) r pε = = π 9, E 360  (D) r pε = = π 9, E 120Answer: – (D) Exp: – rrE120Hµ µ= η = = π∋ ∋P rrE1203µ= η = π∋  Only option ‘D’ satisfies ~+−14 0ºVj4Ω2Ωj4 − ΩI2Ω2ΩEC-Paper Code-B      GATE 2011                   www.gateforum.com© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the written permission. Discuss GATE 2011 question paper at www.gatementor.com.                                                                 16 38. A message signal  ( ) m t cos 200 t 4 cos t = π + π modulates the carrier ( ) c c= π = c t cos 2 f t where f 1 MHZto produce an AM signal. For demodulating the generated AM signal using an envelope detector, the time constant RC of the detector circuit should satisfy  (A) 0.5 ms < RC < 1ms (B)  µ << < 1 s RC 0.5 ms (C) RC s << µ   (D)  >> RC 0.5 msAnswer: – (B) Exp: – Time constant should be length than m1f And time constant should be far greater than c1fm4000af 20002a= =C1 1Rcf 2000> >>µ << << 1 s RC 0.5ms39. The block diagram of a system with one input it and two outputs y1 and y2 is given below.  A  state  space model of the above system in terms of the state vector   x   and the output vector Ty y y 1 2=   is  (A) x 2 x 1 u; y 1 2 x•= + =            (B) 1x 2 x 1 u; y x2•     = − + =       (C) 2 0 1x x u; y 1 2 x0 2 1• −   = + =       −     (D) 2 0 1 1x x u; y x0 2 1 2•     = + =          Answer: – (B) Exp: – Draw the signal flow graph 1+ s 22+ s 2y1y21+ s 22+ s 2EC-Paper Code-B      GATE 2011                   www.gateforum.com© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the written permission. Discuss GATE 2011 question paper at www.gatementor.com.                                                                 17  From the graph ɺx 2x 4 = − + &  y x 1 1= ;  y 2x 2 1=12y 1x 2 x 1 u; xy 2         = − + =       ɺ40. Two systems H1 (z)  and H2 (z) are connected in cascade as shown below. The overall output y(n) is the same as the input x(n) with a one unit delay. The transfer function of the second system H2 (z) is (A) ( )( )11 11 0.6zz 1 0.4z−− −−−   (B) ( )( )1 11z 1 0.6z1 0.4z− −−−−(C) ( )( )1 11z 1 0.4z1 0.6z− −−−−   (D) ( )( )11 11 0.4zz 1 0.6z−− −−−Answer: – (B) Exp: – The overall transfer function 1z−= (∵ unit day T.F 1z− ( =( ) ( )11 2 H z H z z−( )  ; =( )( )( )1112 111 0.6zzH z zH z 1 0.4z−−−−−= =−41. An 8085 assembly language program is given below. Assume that the carry flag is initially unset. The content of the accumulator  after the execution of the program is MVI  A,07H RLCMOV B,ARLCRLCADD BRRC  (A) 8CH (B) 64H (C) 23H (D) 15H 4 1y2y−12 / Sx 1 / S−2xX n( ) ( )( )( )11 11 0.4zH z1 0.6z−−−=−H z2 ( ) y n( )EC-Paper Code-B      GATE 2011                   www.gateforum.com© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the written permission. Discuss GATE 2011 question paper at www.gatementor.com.                                                                 18 Answer: – (C) Exp: – MVI A, 07 H  ⇒ 0000 0111  ← The content of ‘A’  RLC  ⇒ 0000 1110  ← The content of ‘A’  MOV B, A  ⇒ 0000 1110  ← The content of ‘B’  RLC  ⇒ 0001 1100  ← The content of ‘B’  RLC  ⇒ 0011 1000  ← The content of ‘B’  ADD B        A 0000 1110B 0011 10000100 0110+0010 0011RRC 23H2 3→42. The first six points of the 8-point DFT of a real valued sequence are5, 1 j3, 0,3 j4, 0 and 3 j4. − − + . The last two points of the DFT are respectively (A) 0, 1-j3 (B) 0, 1+j3 (C) 1+j3, 5 (D) 1 – j3, 5 Answer: – (C) Exp: – DFT points are complex conjugates of each other and they one symmetric to the middle point. ( ) ( )( ) ( )( ) ( )( ) ( )****x 0 x 7x 1 x 6x 2 x 5x 3 x 4====⇒ Last two points will be  ( )*x 0 and  ( )*x 1 = + 1 j3 and 5 43. For the BJT QL in the circuit shown below, BEon 0.7V, VCEsat, V 0.7V. β = ∞ = = The switch is initially closed. At time  t = 0,  the switch is opened. The time  t  at which Q1leaves the active region is  (A) 10 ms (B) 25 ms (C) 50 ms (D) 100 ms 5V0.5mA−5VQ1= t 0µ5 F4.3kΩ−10VEC-Paper Code-B      GATE 2011                   www.gateforum.com© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the written permission. Discuss GATE 2011 question paper at www.gatementor.com.                                                                 19 Answer: – (C) Exp: – Apply KVL at the BE junction E5 0.7 10 4.3I 1mA4.3k 4.3k+ − −= = =Ω Ω Always E= I 1mA ; At collector junction Cap ( + = I 0.5mA 1mA ) (∵ β = ∞ = ;I IE C )Cap= − = I 1 0.5 0.5mA always constant   V V V CE C E= − ⇒ V V V C CE E+ =( )30.7 4.3 3 1 10−= + × × = + 0.7 4.3 (∵ = V I R E E E )V 5V V C cap= =cap CaptV Ic=   Or ( ) ( )6Cap3CapV C 5 5 10tI 0.5 10−−× ×= =×= 50ms44. In the circuit shown below, the network N is described by the following Y matrix: 0.1S 0.01SY .0.01S 0.1S  −=   the voltage gain 21VisV (A) 1/90 (B) –1/90 (C) –1/99 (D) –1/11 Answer: – (D) Exp: – N 100V 25I1 1= + ; V 100I2 2− =2 3 1 4 2= + I Y V Y V ⇒ 0.01V 0.01V 0.1V 2 1 2− = + ⇒ 21V 1V 11−=45. In the circuit shown below, the initial charge on the capacitor is 2.5 mC, with the voltage polarity as indicated. The switch is closed at time t=0. The current i(t) at a time t after the switch is closed is  (A)  ( ) ( )3= − × i t 15 exp 2 10 t A (B)  ( ) ( )3= − × i t 5 exp 2 10 t A (C)  ( ) ( )3= − × i t 10 exp 2 10 t A (D)  ( ) ( )3= − − × i t 5 exp 2 10 t A5V0.5mA−5V= t 05mF4.3kΩ−10VV I R E E E=VC+−+100V −25ΩV1 N−+V2 100Ω2I1 I100V+−−+i t( )10Ω50 FµEC-Paper Code-B      GATE 2011                   www.gateforum.com© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the written permission. Discuss GATE 2011 question paper at www.gatementor.com.                                                                 20 Answer: – (A) Exp: – = Q 2.5mC3initial 62.5 10 CV 50V50 10 F−−×= =×⇒ Thus net voltage  = + = 100 50 150V( )150i t10= exp  ( )t− × 2 10 t A =15 exp  ( )t− × 2 10 t A46. The system of equations x y z 6x 4y 6z 20x 4y z= + += + ++ + λ = µ has NO solution for values of λ and µ given by  (A) λ = µ = 6, 20 (B) λ = µ ≠ 6, 20 (C) λ ≠ µ = 6, 20 (D) λ ≠ µ ≠ 6, 20Answer: – (B) Exp: – Given equations are  + + = x y z 6 , + + = x 4y 6z 20 and  x 4y z + + λ = µ If  λ = 6 and  µ = 20 , then  + + = x 4y 6z 20+ + = x 4y 6z 20 infinite solution  If  λ = 6 and  µ ≠ 20 , the  x 4y 6z 20x 4y 6z= + ++ + = µ(µ ≠ 20) no solution  If  λ ≠ 6 and  µ = 20x 4y 6z 20x 4y z 20= + ++ + λ = will have solution λ ≠ 6 and  µ ≠ 20 will also give solution 47. A fair dice is tossed two times. The probability that the second toss results in a value that is higher than the first toss is  (A) 2/36 (B) 2/6 (C) 5/12 (D) ½ Answer: – (C) Exp: – Total number of cause = 36  Total number of favorable causes = 5+ 4 + 3 + 2 + 1 = 15  Then probability 15 536 12= =(1,1) (2,1) (3,1) (4,1) (5,1)(6,1)EC-Paper Code-B      GATE 2011                   www.gateforum.com© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the written permission. Discuss GATE 2011 question paper at www.gatementor.com.                                                                 21 (1, 2) (2, 2) (3, 2) (4, 2) (5, 2)(6, 2)(1, 3) (2, 3) (3, 3) (4, 3) (5, 3)(6, 3)(1, 4) (2, 4) (3, 4) (4, 4) (5, 4)(6, 4)(1, 5) (2, 5) (3,5) (4, 5) (5, 5)(6,5)(1, 6) (2, 6) (3, 6) (4, 6) (5, 6)(6, 6)Common Data Questions: 48 & 49  The channel resistance of an N-channel JFET shown in the figure below is 600 Ωwhen the full channel thickness (tch) of  10 mµ is available for conduction. The built-in voltage of the gate P+ N junction (Vbi) is -1 V. When the gate to source voltage (VGS) is 0 V, the channel is depleted by  µ1 m on each side due to the builtin voltage and hence the thickness available for conduction is only  µ8 m48. The channel resistance when VGS = -3 V is  (A) 360Ω (B) 917Ω (C) 1000Ω (D) 3000ΩAnswer: – (C) Exp: – Width of the depletion largeW V V α +bi GS21W 1 3W 1− −⇒ =−w 2w 2 1= ( = µ = µ ) 2 1 m 2 m So that channel thickness = 10 – 4 =  µ6 mµ − 8 m 750            µ − 6 m ?d8r 7506= × 1000 = Ω49. The channel resistance when VGS = 0 V is  (A) 480Ω (B) 600Ω (C) 750Ω (D) 1000Ω+−VGSSourceGateP+P+NchtDrainEC-Paper Code-B      GATE 2011                   www.gateforum.com© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the written permission. Discuss GATE 2011 question paper at www.gatementor.com.                                                                 22 Answer: – (C)  Exp: -donoh1rtα At VGS= 0,ch d ( t 10 m; Given r 600 = µ = Ω)          d10r 600 at 8 m8= × ← µ = Ω 750  Common Data Questions: 50 & 51  The input-output transfer function of a plant   ( )( )2100H s .s s 10=+ The plant is placed in a unity negative feedback configuration as shown in the figure below. 50. The gain margin of the system under closed loop unity negative feedback is  (A) 0dB (B) 20dB (C) 26 dB (D) 46 dB Answer: – (C) Exp: – ( )( )2100H ss s 10=+ Phase cross over frequency=1 090 2 tan 18010−   ω− = − −  1 02 tan 9010−   ω− = − ⇒  ⇒ 1 0tan 4510−   ω=  ⇒ ω = 10 rad / sec( ) ( )( )2100H jwj10 j10 10=+1 110.2 20= =1GM 20 log 20 log20 26dB1 / 20= = =51. The signal flow graph that DOES NOT model the plant transfer function H(s) is  (A)    (B)   (C)    (D)  +−r u( )( )2100H ss s 10=+yplantΣ1 1 / s 1 / s 1 / s 100u−10 −10y1 / s 1 / s 1 / s 100u−20y−1001 / s 1 / s 1 / s 100u−20y−1001 / s 1 / s 1 / s 100u y−100EC-Paper Code-B      GATE 2011                   www.gateforum.com© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the written permission. Discuss GATE 2011 question paper at www.gatementor.com.                                                                 23 Answer: – (D) Exp: -(D) Option (D) does not fix for the given transfer function. Linked Answer Questions: Q.52 to Q.55 Carry Two Marks Each Statement for Linked Answer Questions: 52 & 53  In the circuit shown below, assume that the voltage drop across a forward biased diode is 0.7 V. The thermal voltage t= = V kT / q 25mV. The small signal input ( i p p ) = ω = v V cos t where V 100mV . 52. The bias current IDC through the diodes is  (A) 1 mA (B) 1.28 mA (C) 1.5 mA (D) 2 mA Answer: – (A) Exp: -( )DC12.7 0.7 0.7 0.7 0.7I 1mA9900+ + + −= =53. The ac output voltage vac is  (A) 0.25 cos t mV (ω )  (B)  ( 1cos t mV ω ) (C)  ( 2 cos t mV ω )   (D) 22 cos t mV (ω )Answer: – (C) Exp: – AC dynamic resistance, TdDV 2 25mVr 50I 1mAη ×= = = Ωη = 2 for Si (∵ forward drop = 0.7V)  The ac dynamic resistance offered by each diode = 50Ωac i ( )4 50V V ac9900 50  × Ω= ∴   +3 100200 10 cos wt10000−  × =  ac= V 2 cos wt mV ( )+−+−DC acI i +V v DC ac+12.7Vvi9900ΩEC-Paper Code-B      GATE 2011                   www.gateforum.com© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the written permission. Discuss GATE 2011 question paper at www.gatementor.com.                                                                 24 Statement for Linked Answer Questions: 54 & 55  A four-phase and an eight-phase signal constellation are shown in the figure below. 54. For the constraint that the minimum distance between pairs of signal points be d for both constellations, the radii r1, and r2 of the circles are  (A) 1 2= = r 0.707d, r 2.782d (B) 1 2= = r 0.707d, r 1.932d (C) 1 2= = r 0.707d, r 1.545d (D) 1 2= = r 0.707d, r 1.307dAnswer: – (D) Exp:-  For 1st constellation  2 2 21 1r r d + =2 21 ⇒ = r d / 2 1 ⇒ = r 0.707d For 2nd constellation  2dr cos 67.52=2= r 1.307d55. Assuming high SNR and that all signals are equally probable, the additional average transmitted signal energy required by the 8-PSK signal to achieve the same error probability as the 4-PSK signal is  (A) 11.90 dB (B) 8.73 dB (C) 6.79 dB (D) 5.33 dB Answer: – (D) Exp: – Energy 2 21 2= r and r( )( )2212 22r 0.707dr 1.307d= ⇒ Energy  ( )( )( )221.307in dD 10log 5.33dB0.707= =  Q. No. 56 – 60 Carry One Mark Each 56. There are two candidates P and Q in an election. During the campaign, 40% of the voters promised to vote for P, and rest for Q. However, on the day of election 15% of the voters went back on their promise to vote for P and instead voted for Q. 25% of the voters went back on their promise to vote for Q and instead voted for P. Suppose, P lost by 2 votes, then what was the total number of voters?  (A) 100 (B) 110 (C) 90 (D) 95 Answer: – (A) Qd r1 IQd2rI2r0452rd / 2 d d / 2067.5EC-Paper Code-B      GATE 2011                   www.gateforum.com© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the written permission. Discuss GATE 2011 question paper at www.gatementor.com.                                                                 25 Exp: –  P Q40% 60%6% 6%15% 15%49% 51%2% 2100% 100+ −− += ∴=57. Choose the most appropriate word from the options given below to complete the following sentence:  It was her view that the country’s problems had been_________ by foreign technocrats, so that to invite them to come back would be counter-productive.  (A) Identified   (B) ascertained   (C) Texacerbated  (D) Analysed  Answer: – (C) Exp: -The clues in the question are —foreign technocrats did something negatively to the problems – so it is counter-productive to invite them. All other options are non-negative. The best choice is exacerbated which means aggravated or worsened. 58. Choose the word from the options given below that is most nearly opposite in meaning to the given word:  Frequency (A) periodicity   (B) rarity  (C) gradualness   (D) persistency Answer: – (B) Exp: – The best antonym here is rarity which means shortage or scarcity.59. Choose the most appropriate word from the options given below to complete the following sentence:  Under ethical guidelines recently adopted by the Indian Medical Association, human genes are to be manipulated only to correct diseases for which______________ treatments are unsatisfactory.  (A) Similar (B) Most (C) Uncommon (D) Available Answer: – (D) Exp: – The context seeks to take a deviation only when the existing/present/current/ alternative treatments are unsatisfactory. So the word for the blank should be a close synonym of existing/present/current/alternative. Available is the closest of all. 60. The question below consists of a pair of related words followed by four pairs of words. Select the pair that best expresses the relation in the original pair: Gladiator : Arena  (A) dancer : stage   (B) commuter: train  (C) teacher : classroom  (D) lawyer : courtroom Answer: – (D) EC-Paper Code-B      GATE 2011                   www.gateforum.com© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the written permission. Discuss GATE 2011 question paper at www.gatementor.com.                                                                 26 Exp: – The given relationship is worker: workplace. A gladiator is (i) a person, usually a professional combatant trained to entertain the public by engaging in mortal combat with another person or a wild.(ii) A person engaged in a controversy or debate, especially in public.Q. No. 61 – 65 Carry Two Marks Each 61    The fuel consumed by a motorcycle during a journey while traveling at various speeds is indicated in the graph below.    The distances covered during four laps of the journey are listed in the table below Lap Distance (kilometers) Average speed (kilometers per hour) P 15 15 Q 75 45 R 40 75 S 10 10  From the given data, we can conclude that the fuel consumed per kilometre was least during the lap  (A) P (B) Q (C) R (D) S Answer: – (A) Exp: –         Fuel consumption Actual  15 1P 60km / l l60 475 5Q 90km / l l90 640 8R 75km / l l75 1510 1S 30 km / l l30 3====12090603000 15 30 45 60 75 90Speed (kilometers per hour)consumption  Fuel(kilometers per litre) EC-Paper Code-B      GATE 2011                   www.gateforum.com© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the written permission. Discuss GATE 2011 question paper at www.gatementor.com.                                                                 27 62. Three friends, R, S and T shared toffee from a bowl. R took 1/3rd of the toffees, but returned four to the bowl. S took 1/4th of what was left but returned three toffees to the bowl. T took half of the remainder but returned two back into the bowl. If the bowl had 17 toffees left, how many toffees-were originally there in the bowl?  (A) 38 (B) 31 (C) 48 (D) 41 Answer: – (C) Exp: – Let the total number of toffees is bowl e x  R took 13 of toffees and returned 4 to the bowl ∴ Number of toffees with 1R x 43− = Remaining of toffees in bowl = 2x 43+ Number of toffees with S = 1 2x 4 34 3 − +   Remaining toffees in bowl = 3 2x 4 44 3 + +  Number of toffees with 1 3 2T x 4 4 22 4 3 = + + +    Remaining toffees in bowl = 1 3 2x 4 4 22 4 3   + + +       Given, 1 3 2x 4 4 2 172 4 3   = + + +      3 2x 4 274 3 = + ⇒  ⇒ = x 4863. Given that f(y) = | y | / y, and q is any non-zero real number, the value of   | f(q) – f(-q) | is  (A) 0 (B) -1 (C) 1 (D) 2 Answer: – (D) Exp: – Given, yf(y)y( ) ( ) =q q qf q ; f qq q q− −= = − = ⇒−( ) ( )q q 2 qf q f qq q q− = + = = 2 64. The sum of n terms of the series 4+44+444+…. is  (A) ( )n 14 / 81 10 9n 1+  − −  (B) ( )n 14 / 81 10 9n 1−  − −  (C) ( )n 14 / 81 10 9n 10+  − −  (D) ( )n4 / 81 10 9n 10   − − EC-Paper Code-B      GATE 2011                   www.gateforum.com© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the written permission. Discuss GATE 2011 question paper at www.gatementor.com.                                                                 28 Answer: – (C) Exp: – Let S=4 (1 + 11 + 111 + ……..) ( )49 99 999 …….9+ + + ={ ( ) ( ) ( )}{ } ( )( ){ }2 3n2 n n 1410 1 10 1 10 1 ………94 4 4 10 110 10 ……10 n 10 n 10 9n 109 9 9 81++ − + − + − =  − = + + − = − = − −    65. The horse has played a little known but very important role in the field of medicine. Horses were injected with toxins of diseases until their blood built up immunities. Then a serum was made from their blood. Serums to fight with diphtheria and tetanus were developed this way.  It can be inferred from the passage that horses were  (A) given immunity to diseases (B) generally quite immune to diseases  (C) given medicines to fight toxins (D) given diphtheria and tetanus serums Answer: – (B) Exp: – From the passage it cannot be inferred that horses are given immunity as in (A), since the aim is to develop medicine and in turn immunize humans. (B) is correct since it is given that horses develop immunity after some time. Refer “until their blood built up immunities”. Even (C) is invalid since medicine is not built till immunity is developed in the horses. (D) is incorrect since specific examples are cited to illustrate and this cannot capture the essence.

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