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# 2015 Wiley India Pvt. Ltd.

SAMPLE MCQS
1. For the circuit shown in the following figure, the Thevenins equivalent voltage and the Thevenins

## equivalent resistance, respectively, are

(a) 158.11108.43, j150
(b) 158.11108.43, 150
(c) 128.11108.43, j150
(d) 128.11108.43, 150

Solution.
The Thevenins equivalent voltage can be found using the superposition theorem. It is the voltage Vab
at the terminal a-b. Considering the 1000 source and short circuiting the 10090 source, the
equivalent circuit is shown in the following figure.

## The current I1 is given by

I1

1000
100

A
j300 j100 j 200

Now,
Vab1 I1 ( j100) 500
Considering the 10090 source and short circuiting the 1000 source, the equivalent circuit is
shown in the following figure.

## The current I2 is given by

I2

10090
j300 j100

Now,
Vab2 I 2 ( j300) j150V

Also,
VTH = Vab1 + Vab2 = 50 + j150 = 158.11108.43
To find Thevenins equivalent voltage all the active sources are deactivated. The equivalent
circuit is shown in the following figure.

Therefore,

## ZTH j100 j300 j150

Ans. (a)
2. For the opamp circuit shown in the following figure, the z-parameters are (Assume that the opamp to

be ideal.)

(a) z11 R1

R3 1
1
, z12 0, z21 1
, z22 R4
sC
R2 sC

## (b) z11 0, z12 R1

(c) z11 R1

R3 1
1
, z21 1
, z22 R4
sC
R
sC

R3 1
1
, z12 0, z21 R4 , z22 1

sC
R2 sC

## (d) None of these

Solution.
Applying Kirchhoffs voltage law, we get
I1
and V2 R4 I 2 R3 I 3 R2 I 3
sC
From the concept of virtual earth for opamps, we get
I
R2 I 3 1
sC
Therefore,
R R3 I1 R I
V2 2
4 2
sCR2
Comparing with the equations for z-parameters, we get
1

z11 R1
sC

z12 0
V1 R1I1

R 1
z21 1 3

R2 sC
z22 R4
Ans. (a)
3. For the circuit shown in the following figure, find the amplitude of the output voltage pulse when the

light pulse having wavelength of 1000 nm, pulse width of 1 s and energy of 10 mJ is incident on the
active area of the photodiode. The responsivity of the photodiode is 0.5 A/W at 1000 nm.
(a) 2.5 V
(b) 2.25 V
(c) 2.75 V
(d) 2.0 V

## 2015 Wiley India Pvt. Ltd.

Solution.
The incident light pulse has an energy of 10 mJ and a pulse width of 1 s.
Therefore, the input peak power = 10 103/1 = 10 mW
Output current from the photodiode = 0.5 10 103 = 5 mA
Voltage across the resistance R = 50 5 103 = 250 mV
Gain of the amplifier

R2
10 103
1
1
11
R1
1103

## Amplitude of the output pulse = 250 103 11 = 2.75 V.

Ans. (c)
4. The transit time of the current carriers through the channel of a JFET decides its_____characteristics.

(a) source
(b) drain
(c) gate
(d) source and drain
Solution:
For the N-channel MOS, threshold voltage is given by
VT VTO 2f VSB 2f

Where

2qN AS
COX

tOX 2qN AS
3.45 1011

Therefore, the threshold voltage of an N-channel MOSFET can be increased by reducing the channel
dopant concentration.
Ans. (b)
5. Which of the following cannot be fabricated on an IC?

(a) Transistors
(b) Diodes
(c) Resistors
(d) Large inductors and transformers

## 2015 Wiley India Pvt. Ltd.

Solution.
large components cannot be fabricated on an IC therefore Lagre inductors and transformers cannot be
fabricated on an IC
Ans. (d)
6. What is the lower cut-off frequency of the BJT amplifier shown in the following figure. Given

that the h-parameters of the transistor are hie = 1.5 k and hfe = 100.

(a) 14.1 Hz
(b) 26.5 Hz
(c) 682 Hz
(d) None of these
Solution.
The cut-off frequency due to the capacitor Ci is
f LCi

1
2 (Ri Rs )Ci

where
Ri R1 R2 hie 40 103 10 103 1.5 103 1.26 k

Therefore,
f LCi

1
14.08 Hz
2 (1.26 10 1103 ) 5 10 6
3

f LCE
where

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1
2 ReCE

Re RE [( Rs R1 R2 hie )/hfe ]
1 103 [(1 103 40 103 10 103 1.5 103 )/100]
23.33
Therefore,
f LCE

1
682 Hz
2 (23.33) 10 106

## The cut-off frequency due to capacitor Co is given by

1
1
f LCo

26.53 Hz
2 (RC RL )Co 2 (4 103 2 103 ) 1 106
As we can see fLCE is significantly higher than fLCo and fLCi, hence, fLCE is the predominant factor in
determining the low-frequency response for the complete system. Hence, the cut-off frequency for the
overall system is approximately equal to 682 Hz.
Ans. (c)
7. If in an oscillator, the amplifier portion is a two-stage common-emitter configuration, what should be

the phase-shift to be introduced by the feedback network at the oscillations frequency for sustained
oscillations?

2

(d)
3
(a)

Solution.
A two-stage common-emitter amplifier provides a phase-shift of 2 rad. Therefore, the feedback
network must not introduce any more phase-shift or introduce phase-shift equal to multiples of 2 rad
in order to satisfy Barkhausen criterion for sustained oscillations.
Ans. (b)
8. A regulated power supply operates from 220 20 VAC. It produces a no load regulated output

voltage of 24 0.5 VDC. Also, the regulated output voltage falls from 24 VDC to 23.8 VDC as the
load changes from no load to full load condition for the nominal value of input voltage. What is the
value of line regulation?
(a) 1%
(b) 4.2%
(c) 5.7%
(d) 1.5%
Solution.
The line regulation is
24.5 23.5 1

24
24

Ans. (b)

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9. Of the various commonly used logic families, the one with highest speed and the one with least power

## dissipation, respectively, are

(a) TTL and CMOS
(b) CMOS and TTL
(c) CMOS and ECL
(d) ECL and CMOS
Solution.
ECL being a non-saturating bipolar logic family is the fastest and CMOS inherently dissipates least
power due to use of MOS devices.
Ans. (d)
10. Identify the flipflop whose function table is given in the following figure.

(a) Positive edge-triggered J-K flipflop with active-HIGH J and K inputs and active-LOW
PRESET and CLEAR inputs
(b) Positive edge-triggered J-K flipflop with active-HIGH J and K inputs and active-HIGH
PRESET and CLEAR inputs
(c) Positive edge-triggered J-K flipflop with active-LOW J and K inputs and active-HIGH
PRESET and CLEAR inputs
(d) Positive edge-triggered J-K flipflop with active-LOW J and K inputs and active-LOW PRESET
and CLEAR inputs
Solution.
The first three entries of the function table indicate that the J-K flipflop has active HIGH PRESET
and CLEAR inputs. Referring to fourth and fifth entries of the function table, it has active LOW J
and K inputs. The seventh row of the function table confirms this. The output responds to positive
LOW-to-HIGH edges of the clock input. Thus, the flipflop represented by the given function table
is presentable, clearable, positive edge-triggered flipflop with active HIGH PRESET, CLEAR and
active-LOW J and K inputs.
Ans. (c)
11. The following signal is used when a peripheral device requests the microprocessor to have a DMA

operation.
(a) INTR and INTA

## (c) HOLD and HLDA

(d) (d) RD and WR
Solution.
These are READ and WRITE signals. A LOW on the READ signal indicates that the selected
memory or I/O device is ready to be read and the data bus is available for data transfer. A LOW on
the WRITE signal indicates that data on the data bus are to be written into a memory or I/O location.
Data is set up at the trailing edge of the WRITE signal.
Ans. (d)
12. The 3 dB bandwidth of a typical second-order system with the transfer function

H ( s)

n2
s 2 2n s n2

H ( s)

n2
s 2 2n s n2

is given by
(a) n 1 2 2
2
4
2
(b) n (1 ) 1
2
4
2
(c) n (1 2 ) 4 4 2
2
2
2
(d) n (1 ) 4 4 2

Solution.
The Given transfer function is

## Substituting s j in the above equation, we get

H ( j )

n2
( j ) 2 2n ( j ) n2
n2

n2 2 2 jn
1

2
2
1 ( / n ) j (2 / n )
Let us consider that

Therefore,

H ( j )

1
1
and H ( j )
(1 D ) j 2 D
(1 D 2 )2 (2 D)2
2

H ( jc )

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1
2

Therefore,

1
1

2 2
2
(1 D ) (2 D)2
Therefore,

(1 D2 )2 (2 D)2 2
Hence,

1 D4 2D2 4 2 D2 2 or D4 D2 (4 2 2) 1 0
Therefore,

(4 2 2) (4 2 2) 2 4(1)(1)
D
(1 2 2 ) 4 4 4 2 2
2
2

Therefore,

D (1 2 2 ) 4 4 4 2 2
The value of D at the 3 dB frequency c is given by

D c
n

c (1 2 2 ) 4 4 4 2 2
n
c n (1 2 2 ) 4 4 4 2 2
Since c cannot be negative, we get

c n (1 2 2 ) 4 4 4 2 2
Ans. (c)
13. The following figure shows a cascaded LTI system. The impulse response of the system is

(a) h1
(b) h2
(c) (h11 h2 1 ) (h2 h1 )
(d) (h2 h1 ) (h11 h2 1 )
Solution.
The impulse response of the cascaded system is

## 2015 Wiley India Pvt. Ltd.

(h h ) (h h ) (h h ) h h h
1

1
1

(h2 h1 ) (h11 h2 1 )
Ans. (d)

14. A control system is represented by the following transfer function. Determine the system differential

equation.
P( s )

( s 1)( s 3)
s( s 2)( s 4)

(a)

d 2 y dy d 2 x
d3y
dx

6
2 8 2 4 3x
3
dt
dt
dt dt dt

(b)

d 2 y dy d 2 x
d3y
dx

8
2 6 2 4 3x
3
dt
dt
dt dt dt

(c)

d 2 y dy d 2 x
d3y
dx

6
2 8 2 3 4 x
3
dt
dt
dt dt dt

(d)

d 3 y d 2 y dy d 2 x dx

x

dt 3 dt 2 dt dt 2 dt

Solution.

P( s )

( s 1)( s 3)
s 2 4s 8
s 2 4s 3

s( s 2)(s 4) s(s 2 6s 8) s 2 6s 2 8s

or

Y (s)
s 2 4s 3
3
X ( s ) s 6 s 2 8s

or

D2 4D 3
y (t ) 3
x(t )
2
D 6 D 8D

d
dt
3
D y 6D2 8Dy D2 x 4Dx 3x
D

where
or

d3y
d2y
dy d 2 x
dx

8
2 4 3x
3
2
dt
dt
dt dt
dt

or

Ans. (a)
15. For a feedback control system, G(s) = 20/s and H(s) = (s + 3). The steady-state output for a unit step
2

input will be
(a) 1
(c) 0

(b) 0.5
(d) 0.33

Solution.
The transfer function is

## 2015 Wiley India Pvt. Ltd.

C ( s)
20 / s 2
20

R( s ) 1 (20 / s 2 )( s 3) s 2 20s 60
20
C ( s)
s ( s 2 20s 60)
lim C (t ) lim sC ( s )
t

## (as transfer function of unit step 1 / s)

s 0

lim
s 0

20s
s ( s 20s 60)
2

1
0.33
3

Ans. (d)
16. The following figures (a) and (b) show the Nyquist stability plots of a control system represented by

GH(s) = 1/s(s + 1). What would be the transfer function of the control system whose Nyquist stability
plot is shown in Fig. (b)?

## (a) (3s2 + 3s + 1)/[s(s + 1)]

(b) 3/s(s + 1)
(c) 1/3s(s + 1)
(d) None of these
Solution.
The Nyquist stability plot shown in the given figure (b) is the same as the Nyquist stability plot
shown in figure (a) with the origin of coordinate system shifted to (3, 0). Therefore, the transfer
function of this control system would be given by
1
3s 2 3s 1
GH ( s)
=
[ s( s 1)]+3
s( s 1)
Ans. (a)
17. In a DSB AM system, the carrier is

## c(t ) A cos(2fct ) and the message signal is given by

m(t ) sin c(t ) sin c 2 (t ) . The frequency domain representation of the modulated signal is

## 2015 Wiley India Pvt. Ltd.

A
[( f f c ) ( f f c ) ( f f c ) ( f f c )]
2
A
(b) [( f f c ) ( f f c ) ( f f c ) ( f f c )]
4
(c) A[( f fc ) ( f fc ) ( f fc ) ( f fc )]
(a)

## (d) None of the above

Solution.
The modulated signal is
u(t ) m(t )c(t ) sin c(t ) sin c 2 (t ) A cos 2f ct
Taking Fourier transform on both sides, we get

U( f )

A
( f ) ( f ) ( f fc ) ( f fc )
2

A
( f f c ) ( f f c ) ( f f c ) ( f f c )
2
Ans. (a)

18. The following figure shows a PCM waveform in which the amplitude levels of +1 V and 1 V are

used to represent binary symbols 1 and 0, respectively. The code word used comprises of three bits.
The sampled version of the analog signal from which this PCM is derived is

## (a) rising staircase

(b) falling staircase
(c) straight line
(d) sine wave
Solution.
The sampled version of the analog signal is shown in the following figure.

## Therefore, it is a rising staircase waveform.

Ans. (a)
19. For an electromagnetic wave propagating in a conducting medium having = 60 mS/m and r = 1,

## attenuation coefficient at 1 GHz would be equal to

(a) 48.6 104/m
(b) 38.7 106/m
(c) 58.7 104/m
(d) 78.5 104/m
Solution.
We have

= 1/ = (f)
where = 0r
Substituting the values of different parameters, we get
= 48.6 104/m
Ans. (a)

20. Refer to the transmission line section shown in the following figure. If the propagation velocity is 2
8

## 10 m/s, write the expression for VL.

(a) VL = 50 cos 2 108t
(b) VL = 50 cos (2 108t )
(c) VL = 50 sin (2 108t )

2

Solution.
We have

2 108
f 100 MHz,
2m
100 106

## Therefore, the given line is /2 long, which implies that

Zin Z L 300
50cos 2108 t

As the line is /2 long, the signal undergoes a phase delay of l = (2/) /2 = radians as it travels
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