gate mock test

2015 Wiley India Pvt. Ltd.


1

100 0 100

300 100 200I A

j j j

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, 150j

(b) 158.11108.43, 150

(c) 128.11108.43, 150j

(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

Now,

ab1 1( 100) 50 0V I j

Considering the 10090 source and short circuiting the 1000 source, the equivalent circuit is shown in the following figure.


2

100 90

300 100I

j j

The current I2 is given by

Now,

ab2 2( 300) 150VV I j j

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,

TH 100 300 150Z j j j

Ans. (a)

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

be ideal.)


(a) 311 1 12 21 22 42

1 1, 0, 1 ,

Rz R z z z R

sC R sC

(b) 311 12 1 21 22 42

1 10, , 1 ,

Rz z R z z R

sC R sC

(c) 311 1 12 21 4 222

1 1, 0, , 1

Rz R z z R z

sC R sC

(d) None of these

Solution.

Applying Kirchhoffs voltage law, we get

11 1 1 2 4 2 3 3 2 3 and

IV R I V R I R I R I

sC

From the concept of virtual earth for opamps, we get

12 3

IR I

sC

Therefore,

2 3 12 4 2

2

R R IV R I

sCR

Comparing with the equations for z-parameters, we get

11 1

12

321

2

22 4

1

0

11

z RsC

z

Rz

R sC

z R

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


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

3

2

3

1

10 101 1 11

1 10

R

R

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

OT T f SB f2 2V V V

Where A S OX A S

11

OX

2 2

3.45 10

qN t qN

C

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


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

iLC

i s i

1

2 ( )f

R R C

where 3 3 3

i 1 2 ie 40 10 10 10 1.5 10 1.26 kR R R h

Therefore,

LCi 3 3 6

1 14.08 Hz2 1.26 10 1 10 5 10( )

f

The cut-off frequency due to capacitor CE is

ELC

e E

1

2f

R C

where


e E s 1 2 ie fe

3 3 3 3 3

[( )/ ]

1 10 [(1 10 40 10 10 10 1.5 10 )/100]

23.33

R R R R R h h

Therefore,

LCE 6

1682 Hz

2 (23.33) 10 10f

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

oLC 3 3 6

C L o

1 126 53 Hz

2 ( ) 2 (4 10 2 10 ) 1 10f .

R R C

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?

(a) 2

rad

(b) 2 rad (c) 3 rad

(d) 3

rad

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 10.0416 4.16% 4.2%

2424

Ans. (b)


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

(b) READY


(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 2

2 2( )

2

n

n n

H ss s

is given by

(a) 21 2n

(b) 2 4 2(1 ) 1n

(c) 2 4 2(1 2 ) 4 4 2n

(d) 2 2 2(1 ) 4 4 2n

Solution.

The Given transfer function is 2

2 2( )

2

n

n n

H ss s

Substituting s j in the above equation, we get

2

2 2

2

2 2

2 2

( )( ) 2 ( )

2

1

1 ( / ) (2 / )

n

n n

n

n n

n n

H jj j

j

j

Let us consider that

n

D

Therefore,

2

1( )

(1 ) 2H j

D j D

and

2 2 2

1( )

(1 ) (2 )H j

D D

We know that at 3 dB frequency c,

1( )

2cH j


Therefore,

2 2 2

1 1

2 (1 ) (2 )D D

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

Hence, 4 2 2 2 4 2 21 2 4 2 or (4 2) 1 0D D D D D

Therefore,

2 2 2

2 2 4 2(4 2) (4 2) 4(1)( 1) (1 2 ) 4 4 22

D

Therefore,

2 4 2(1 2 ) 4 4 2D

The value of D at the 3 dB frequency c is given by

c

n

D

2 4 2

2 4 2

(1 2 ) 4 4 2

(1 2 ) 4 4 2

c

n

c n

Since c cannot be negative, we get

2 4 2(1 2 ) 4 4 2c n

Ans. (c)

13. The following figure shows a cascaded LTI system. The impulse response of the system is

(a) h1

(b) h2

(c) 1 1

1 2 2 1( ) ( )h h h h

(d) 1 1

2 1 1 2( ) ( )h h h h

Solution.

The impulse response of the cascaded system is


1 1 1 11 2 2 2 2 1 1 1 2 2 1 1 2( ) ( ) ( ) ( ) ( )h h h h h h h h h h h h h Ans. (d)

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

equation.

( 1)( 3)( )

( 2)( 4)

s sP s

s s s

(a) 3 2 2

3 2 26 8 4 3

d y d y dy d x dxx

dt dt dt dt dt

(b) 3 2 2

3 2 28 6 4 3

d y d y dy d x dxx

dt dt dt dt dt

(c) 3 2 2

3 2 26 8 3 4

d y d y dy d x dxx

dt dt dt dt dt

(d) 3 2 2

3 2 2

d y d y dy d x dxx

dt dt dt dt dt

Solution. 2 2

2 2 2

( 1)( 3) 4 8 4 3( )

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

s s s s s sP s

s s s s s s s s s

or 2

3 2

( ) 4 3

( ) 6 8

Y s s s

X s s s s

or 2

3 2

4 3( ) ( )

6 8

D Dy t x t

D D D

where d

Ddt

or 3 2 26 8 4 3D y D Dy D x Dx x

or 3 2 2

3 2 26 8 4 3

d y d y dy d x dxx

dt dt dt dt dt

Ans. (a)

15. For a feedback control system, G(s) = 20/s2 and H(s) = (s + 3). The steady-state output for a unit step

input will be

(a) 1 (b) 0.5

(c) 0 (d) 0.33

Solution.

The transfer function is


2

2 2

2

( ) 20 / 20

( ) 1 (20 / )( 3) 20 60

20( ) ( s transfer function of unit step 1/ )

( 20 60)

C s s

R s s s s s

C s a ss s s

The steady-state error is

0

20

lim ( ) lim ( )

20lim

( 20 60)

10.33

3

t s

s

C t sC s

s

s s s

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

21 3 3 1( ) =

( 1) +3[ ] ( 1 )

s sGH s

s ss s

Ans. (a)

17. In a DSB AM system, the carrier is c( ) cos(2 )c t A f t and the message signal is given by

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


(a) c c c c[ ( ) ( ) ( ) ( )]2

Af f f f f f f f

(b) c c c c[ ( ) ( ) ( ) ( )]4

Af f f f f f f f

(c) c c c c[ ( ) ( ) ( ) ( )]A f f f f f f f f

(d) None of the above

Solution.

The modulated signal is

2 c( ) ( ) ( ) sin ( ) sin ( ) cos 2u t m t c t c t c t A f t

Taking Fourier transform on both sides, we get

c

c c

c c c

( ) ( ) ( ) ( ) ( )2

( ) ( ) ( ) ( )2

AU f f f f f f f

Af f f f f f f f

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

108 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 )

(d) 8L 50cos 2 102

V


Solution.

We have

8

6

2 10100MHz, 2m

100 10f

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

in L

8

300

50cos2 10

Z Z

t

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

8L 50cos 2 10V t

Ans. (b)

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