exam questions-section 1

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1. In the following non planar graph number of independent loop equations are A. 8 B. 12 C. 7 D. 5 Answer & Explanation Answer:  Option D Explanation:  Number of independent loop equations are given by, L = B - N  + 1 where L : No. of loop equations B : No. of branches = 12 N  : No. of nodes = 8 L = 12 - 8 + 1 = 5. View Answer  Workspace Report Discuss in Forum 2. A transmission line has a characteristic impedance of 50 Ω and a resistance of 0.1 Ω/m, if the line is distortion less, the attenuation constant (in Np/m) is A. 500 B. 5 C. 0.014 D. 0.002 Answer & Explanation 

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8/11/2019 Exam Questions-Section 1

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1. In the following non planar graph number of independent loop equations are

A. 8 B. 12

C. 7 D. 5

Answer & Explanation 

Answer: Option D 

Explanation: 

Number of independent loop equations are given by,

L = B - N  + 1

where L : No. of loop equations

B : No. of branches = 12

N  : No. of nodes = 8

L = 12 - 8 + 1 = 5.

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2. A transmission line has a characteristic impedance of 50 Ω and a resistance of 0.1 Ω/m, if the

line is distortion less, the attenuation constant (in Np/m) is

A. 500 B. 5

C. 0.014 D. 0.002

Answer & Explanation 

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Answer: Option D 

Explanation: 

Distortion less

 = RG and Z0 = RG → 

.

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3. A certain 8 bit uniform quantization PCM system can accommodate a signal ranging from - 1 V to

+ V. The rms value of the signal is V . The signal to quantization noise ratio is:

A. 30 dB

B. 46.91 dB

C. 40 dB

D. 50.79 dB

Answer & Explanation 

Answer: Option D 

Explanation: 

(SNR)dB = 1.76 + 6n 

n = 8

Thus (SNR)dB = 1.76 + 6 x 8 = 50.76 dB.

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4. A rectangular waveguide, in dominant TE mode, has dimensions 10 cm x 15 cm. The cut off

frequency is

A. 10 GH z  

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B. 1 GH z  

C. 15 GH z  

D. 25 GH z  

Answer & Explanation 

Answer: Option B 

Explanation: 

For TE10 mode

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5. A radar receiver has a noise figure of 10 dB at 300 K having a bandwidth of 2.5 MHz. The

minimum power it can receive is

A. 3.45 x 10-15 W

B. 1.38 x 10-15 W

C. 7.5 x 10-15 W

D. 93.15 x 10-15 W

Answer & Explanation 

Answer: Option D 

Explanation: 

Pm = k T0Δf (F - 1) = 93.15 x 10-15 W.

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6. In the circuit shown, in switch S is open for a long time and is closed at t  = 0. The current i (t )

for t  ≥ 0+ is

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A. i (t ) = 0.5 - 0.125e-1000t  A

B. i (t ) = 1.5 - 0.125e-1000t  A

C. i (t ) = 0.5 - 0.5e-1000t  A

D. i (t ) = 0.375e-1000t  A

Answer & Explanation 

Answer: Option A 

Explanation: 

i (f ) = 0.5, i (i ) = 0.75

i (t ) = Vr  + (i i  - i  j )e-1/ = 0.5 - 0.125e-1000t  .

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7. In a semiconductor material. The hole concentration is found to be 2 x 2.5 x 1015 cm-3. If mobility

of carriers is 0.13 m2 / v-s. Then find the current density if electric field intensity is 3.62 x 10-19 

A. 7.6237 x 10-4 A/cm2 

B. 7.6237 x 10-5 A/cm2 

C. 7.6237 x 10-3 A/cm2 

D. none of these

Answer & Explanation 

Answer: Option A 

Explanation: 

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Current density J = σE 

Where σ = conductivity 

Given -> μ = 0.13 m2 /v-s = 0.13 x 104 cm2 /V sec  

P = 2.25 x 1015 /cm3 

We have, ni  = 1.5 x 1010 

Also n.p. =

n = / p 

= (1.6 x 10-19 x 0.13 x 104 x 2.25 x 1015) x

= (0.468) (4.5 x 1015)

σ = 2.106 x 1015 μ/cm 

J = σE 

Current density = 2.106 x 1015 x 3.620 x 10-19 

= 7.6237 x 10-4 A/m2.

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8.

Consider the system with and where p and q are

arbitrary real numbers. Which of the following statements about the controllability of the system

is true?

A. The system is completely state controllable for any non zero values p and q 

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B. Only p = 0 and q = 0 result in controllability

C. The system is uncontrollable for all values of p and q 

D. We cannot conclude about controllability from the given data

Answer & Explanation 

Answer: Option C 

Explanation: 

Use the condition of controllability.

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9. A MOSFET has a threshold voltage of 1 V and oxide thickness of 500 x 10-8 [εr  = 3.9; ε0 = 8.85

x 10-14 F/cm, q = 1.6 x 10-19 c ]. The region under the gate is ion implanted for threshold voltage

tailoring. The base and type of impant required to shift threshold voltage to - 1 V are __________

.

A. 8.6 x 1011 /cm2, p-type

B. 8.6 x 1011 /cm2, n-type

C. 0.86 x 109 /cm2, p-type

D. 1.02 x 1012 /cm2, n-type

Answer & Explanation 

Answer: Option A 

Explanation: 

VT(new ) = VT(odd ) +

= 6.903 x 10-8 

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f B = - 8.6 x 1011 

The threshold voltage is always negative for p-channel and hence implant is of p-type.

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10. What is the propagation constant for air filled wave guide with dimensions a = 1.59" andb =

0.795" at 4.95 GHz?

A. 0.6698 B. 0.7698

C. 0.503 D. 0.6598

Answer & Explanation 

Answer: Option A 

Explanation: 

Here,

a = 1.59'' = 40.386 mm = 4.04 cm

.

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11. Radiation resistance of an antenna is 54 Ω and loss resistance is 6 Ω. If antenna has power

gain of 10, then directivity is:

A. 9

B. 11.11

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C. data not sufficient

D. 10

Answer & Explanation 

Answer: Option B 

Explanation: 

Efficiency of antenna

Power gain = 10

Thus directivity .

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12. If the power spectral density of stationary random process is a sine-squared function of

frequency, the shape of its autocorrelation is

A.

 

B.

 

C.

 

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D.

 

Answer & Explanation 

Answer: Option B 

Explanation: 

Since autocorrelation function and power spectral density bears a Fourier transform relation, then

since required in frequency domain will five rectangular convolutions in time domain thus it is

triangular function.

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13. The inverse of given Laplace transform is

A. sin t  

B. cos t  

C. et  

D. e2t  

Answer & Explanation 

Answer: Option B 

Explanation: 

s = x 2 - e x  

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(t ) = cos t .

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14. Consider the amplitude modulated (AM) signal Ac  cos ωc t  + 2 cos ωmt  cos ωc t  For demodulating

the signal using envelope detector, the minimum value of Ac should be

A. 2 B. 1

C. 0.5 D. 0

Answer & Explanation 

Answer: Option A 

Explanation: 

AC cos ωc t  + 2 cos ωmt  cos ωc t  

AC cosωc t   

for envelope detection μ < 1 ⇒  < 1 ⇒ Ac  should be at least-2.

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15.In a 4 bit weighted resistor D/A converter, the resistor value corresponding to LSB is 32 kΩ. Theresistor value corresponding to MSB will be

A. 32 KΩ 

B. 16 KΩ 

C. 8 KΩ 

D. 4 KΩ 

Answer & Explanation 

Answer: Option D 

Explanation: 

2n - 1 R = 32 K&Omeg;

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n = 4

⇒ 8R = 32

R = 4 KΩ. 

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16. Find 'X'  in the circuit below :

f 1(A, B, C, D) = Σ(6, 7, 13, 14); 

f 2(A, B, C, D) = Σ(3, 6, 7); f 3(A, B, C, D) = Σ(5, 6, 7, 14, 15) 

A. (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 15)

B. 0

C. Σ(14) 

D. 1

Answer & Explanation 

Answer: Option C 

Explanation: 

f 1(A, B, C, D) = ∑(6, 7, 13, 14) 

f 2(A, B, C, D) = ∑(3, 6, 7) 

f 1 ⊕ f 2 = ∑(3, 13, 14) 

f 3 x (f 1 ⊕ f 2) = ∑(14) = y  

X = f 1 x y  ⇒ ∑14.

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17. Suppose that the modulating signal is m(t ) = 2cos (2f mt ) and the carrier signal is x C(t ) =

AC cos(2f c t ), which one of the following is a conventional AM signal without over modulation?

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A.  x (t ) = Ac m(t )cos(2f c t )

B.  x (t ) = Ac  [1 + m(t )]cos (2f c t )

C. 

D.  x (t ) = Ac cos(2f mt )cos(2f c t ) + Ac sin(2f mt )sin(2f c t )

Answer & Explanation 

Answer: Option C 

Explanation: 

(C) is without over modulation.

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18. What are the values of Emax and Emin displayed on the oscilloscope, when a 1 kV P -P carries is

modulated to 50%?

A. 2 kV, 0.5 kV

B. 1 kV, 0.5 kV

C. 0.75 kV, 0.25 kV

D. 0.5 kV, 1.5 kV

Answer & Explanation 

Answer: Option C 

Explanation: 

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Emax - Emin = 0.5 x 1 kV = 0.5 kV

2Emax = 1.5 kVi, Emax = 0.75 kV

Emin = 0.25 kV.

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19. Consider the common emitter amplifier shown below with the following circuit parameters: β =

100, gm = 0.3861 A/V, r 0 = ∞ r  p = 259 Ω, Rs = 1 kΩ, RB = 93 kΩ, RC= 250 Ω, RL = 1 kW, C1 = ∞

and C2 = 4.7mF.

The resistance seen by the source Vs is

A. 258 Ω 

B. 1258 Ω 

C. 93 kΩ 

D. ∞ 

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Answer & Explanation 

Answer: Option B 

Explanation: 

Zs = Rs + (RB || Br s)

r c  = 2.475 = 1.258 kV

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20. As the temperature increases the mobility of electrons __________ .

A. decreases because the number of carries increases with increased collisions 

B. increases because the number of carries decreases with decreased collisions

C. remains same

D. none of the above

Answer & Explanation 

Answer: Option A 

Explanation: 

Mobility of electrons

m = 2.5 for electrons and 2.7 for holes in Si.

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21. Consider the following

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1.  Maximum energy of electrons liberated photoelectrically is independent of light intensity2.  Maximum energy of electrons liberated photoelectrically varies nonlinearly with frequency

of incident light.

A. (1) is true and (2) is false

B. (1) and (2) both are true

C. (1) is false and (2) is true

D. (1) and (2) both are false

Answer & Explanation 

Answer: Option A 

Explanation: 

Statement (ii) is false because maximum energy of photons varies linearly with frequency of

incident light. It can be shown as follows :

F1 > F2 > F3 

= light intensity = constant

Photocurrent Vs Anode voltage with frequency and incident light as a parameter.

The light intensity is constant.

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22. When phase-lag compensation is used in a system, then gain cross over frequency, bandwidth

and undamped frequency are respectively

A. Increased, Increased, Increased

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B. Increased, Increased, Decreased

C. Increased, Decreased, Decreased

D. Decreased, Decreased, Decreased

Answer & Explanation 

Answer: Option D 

Explanation: 

Phase lag is delay between a correcting signal of control system and response to it.

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23. The equivalent form of the logical expression ABC + A BC + ABC + ABC + AB C is

A. C + (A ⊕ B)

B. (A + B + C)(A + B + C)

C. (A + B + C)(A + B + C)(A + B + C)

D. (A + B + C)(A + B + C)(A + B + C)

Answer & Explanation 

Answer: Option D 

Explanation: 

This can be solve by using Boolean identity but using k map it will be more easy

(A + B + C) (A + B + C) (A + B + C).

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24. The gradient of any scalar field always yields __________ .

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A. a solenoidal field

B. a conservative field

C. an irrotational field

D. none of these

Answer & Explanation 

Answer: Option B 

Explanation: 

Dot product is conservative. Gradient is nothing but dot product.

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25. In a flash type ADC, employing 15 comparators resolution with 10 volts reference; is expected to

be :

A. 0.625 V

B. 0.666 V

C. 0.525 V

D. insufficient data

Answer & Explanation 

Answer: Option A 

Explanation: 

Since number of comparators = 2n - 1 = 15;

Then number of bits are '4'

Resolution

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26.

Consider the signal flow graph shown in figure below. The gain is :

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A. 

B.

 

C. 

D. 

Answer & Explanation 

Answer: Option C 

Explanation: 

P1 = abcd  

L1 = be, L2 = cf , L3 = dg 

Δ1 = 1

.

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27. During transmission over a communicate channel, bit errors occur independently with

probability . If a block of 3 bits are transmitted, the probability of at most one bit erroris equal

to

A. 

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B. 

C. 

D. none

Answer & Explanation 

Answer: Option D 

Explanation: 

Probability of no-error + probability one error

.

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28.If x 1[n]  x 1( z ) with ROC = R1 and x 2[n]  x 2( z ) with ROC = R2 then ROC for x 1[n]

+ x 2[n] will be __________ (where Ri  → Region of convergence and x 1[n] and x 2[n] are causal).

A. R1 ∩ R2 

B. R1 ∪ R2 

C. (R1 ∪ R2) ∩ (R1)

D. (R1 ∩ R2) ∪ (R1)

Answer & Explanation 

Answer: Option A 

Explanation: 

R1 ∩ R2, Suppose R1 for x 1(n) be

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Then ROC for x 1(n) + x 2(n)

R1 ∩ R2 (R1 < R2)

Then ROC for x 1(n) + x 2(n)

R1 ∩ R2(R1 < R2)

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29.

Consider a unity feedback control system with open-loop transfer function The

steady state error of the system due to a unity step input is :

A. zero B. K

C. 1/K D. infinite

Answer & Explanation 

Answer: Option A 

Explanation: 

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.

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30. The analog signal given below is sampled at 1200 samples per second. m(t) = 2 sin 120 t  +

3 sin 240 t  + 4 sin 360 t  + 5 sin 480 t  + 6 sin 600 t  + 7 sin 720 t . The folding (maximum)

frequency is __________ .

A. 300 H z  

B. 360 H z  

C. 600 H z  

D. 480 H z  

Answer & Explanation 

Answer: Option C 

Explanation: 

Folding frequency (max.) = = 600 H z .

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31.

If the system T.F. is The differential equation representing the system

is

A. 

B. 

C. 

D. none of these

Answer & Explanation 

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Answer: Option C 

Explanation: 

(s3 + 2s2 + 5s + 6)y(s) = (s2 + 4) x (s)

s3y(s) + 2s2y(s) + 5sy(s) + 6y(s) = s2 x (s) + 4 x (s)

Replacing s by and y(s) by y(t) and x(s) by x(t) we get

y(t) + 2   y(t) + 5   y(t) + 6y(t) 

=  x(t) + 4 x(t) 

+ 2 + 5 + 6 = + 4 x  

This is required differential equation.

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32. In the circuit below, delay of the EXOR and AND

Logic gates are 20 μs and 10 μs respectively, then the wave y ' and y  (with initial condition y  = 1)

will be :

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A.

 

B.

 

C.

 

D.

 

Answer & Explanation 

Answer: Option B 

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Explanation: 

Since y  is at logic '1' initially, y ' must be at logic '1'.

Also, one input of the EXOR gate is tied to zero therefore 'f ' will be directly transferred to the

output.

The logic gates will introduce gate delays which will lead to the answer .

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33. The antenna current of an AM transmitter is 8 A, when only the carrier is sent, but it increases to

8.93 A when the carrier is modulated by a single sine wave. Determine the antenna current when

the percent of modulation changes to 0.8

A. 9.19 A

B. 9.91 A

C. 10.56 A

D. 8.61 A

Answer & Explanation 

Answer: Option A 

Explanation: 

Ic  = 8 A

m = 0.8

= 8 x 1.149 = 9.19 A.

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34. A noiseless 3 KH z  channel can transmit binary at the rate __________ .

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A. 12000 bps 

B. 10000 bps 

C. 6000 bps 

D. 3000 bps 

Answer & Explanation 

Answer: Option C 

Explanation: 

Maximum data rate for noiseless channel is given by Nyquist Theorem

= 2H log2 V bits/sec  

V - discrete levels (Here Binary i.e. 2)

H - Bandwidth

Maximum data rate = 2H log2 V

= 2(3 KH z ) log2 2 = 2 x 3 x 103 x 1

= 6 x 103 = 6000 bps.

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35. Find V x  Vy  V z  

A. V x  = -6 Vy  = 3 V z  = -3

B. V x  = -6 Vy  = -3 V z  = 1

C. V x  = 6 Vy  = 3 V z  = 3

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D. V x  = 6 Vy  = 1 V z  = 3

Answer & Explanation 

Answer: Option D 

Explanation: 

Apply Mesh analysis

Apply KVL to mesh 2

-2 + Vy  + 1 = 0

Vy  = 1

Apply KVL to mesh 1

-8 + V x  + 2 = 0

V x  = 6 V

Apply KVL to mesh 3

-1 + V z  -2 = 0

V z  = 3 V.

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36.

If x[n]  then the sequence x[n] is

A. non-periodic

B. periodic

C. depends on n 

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D. none of these

Answer & Explanation 

Answer: Option A 

Explanation: 

Consider

Number of samples per period = 10 

This is not a rational number. Hence signal is non periodic.

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37. The impulse response and the excitation function of a linear time invariant causal system are

shown in figure (a) and (b) respectively. The output of the system at t  = 2 sec is equal to

A. 0

B. 2

C. 

D. 

Answer & Explanation 

Answer: Option D 

Explanation: 

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The area of the shaded region

.

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38. Consider a stable and causal system with impulse response h(t) and system function H(S).

Suppose H(S) is rational, contains a pole at S = - 2, and does not have a zero at the origin. Thelocation of all other poles and zero is unknown for each of the following statements. Let us

determine whether statement is true or false.

1.  f[h(t) e-3t  ] converges

2. 3.  h(t) has finite duration4.  H(s) = H(- s)

Choose correct option.

A. 1 - True, 2 - False, 3 - True, 4 - False

B. 1 - False, 2 - False, 3 - False, 4 - True

C. 1 - False, 2 - False, 3 - False, 4 - False

D. 1 - True, 2 - can't say, 3 - True, 4 - can't say

Answer & Explanation 

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Answer: Option C 

Explanation: 

Statement 1 is false, since f{h(t)e3t  } corresponds to the value of the Laplace transform

of h(t) at s = 3.

If this converges, it implies that s = - 3 is in the ROC.

A casual and stable system must always have its ROC to the right of all its poles. However, s = - 3

is not to the right of the pole at s = - 2.

Statement 2 is false, because it is equivalent to stating that H(0) = 0. This contradicts the fact

that H(s) does not have a zero at the origin.

Statement 3 is false. If h(t) is of finite duration, then if its Laplace transform has any points in its

ROC, ROC must be the entire s-plane.

However, this is not consistent with H(s) having a pole at s = - 2.

Statement 4 is false. If it were true, then H(s) has a pole at s = - 2, it must also have a pole

at s = 2.

This is inconsistent with the fact that all the poles of a causal and stable system must be in the

left half of the s-plane.

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39. The specific gravity of tungsten is 13.8 and its atomic weight is 184.0 Assume that there are

two free elements per atom. Then the Fermi level or characteristic energy for the crystal in eV will

be

A. 8.95 eV

B. -8.95 eV

C. 7.326 eV

D. -7.326 eV

Answer & Explanation 

Answer: Option C 

Explanation: 

A quantity of any substance equal to its molecular weight in grams is a mole of that substance

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contains the same number of molecules as one mole of any other substance. This Avogadro's

number and equals 6.02 x 1023 molecules per mole. Thus

= 9.03 x 1028 electrons/m3 

Since for tungsten the atomic and the molecular weights are the same. Therefore for tungsten

EF = 3.64 x 10-19(η)2/3 

= 3.64 x 10- 19(9.03 x 1028)2/3 

EF = 7.326 eV.

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40. An air-filled rectangular waveguide has dimensions 8cm x 10cm . The phase velocity of guided

wave at frequency of 3 GH z  for TE10 mode is

A. 1.5 x 108 m/s 

B. 5.5 x 108 m/s 

C. 3.46 x 108 m/s 

D. 0.18 x 107 m/s 

Answer & Explanation 

Answer: Option C 

Explanation: 

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VP = 3.46 x 108 m/s.

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41. For  = - 1, find the value of V2 

A. ∞ 

B. 0

C. V1 

D. none of the above

Answer & Explanation 

Answer: Option B 

Explanation: 

V1 = i 1R1 + i 2R2 

i 1 + i 1 = i 2 

V2 = i 2R2 

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Putting  = - 1

= 0 V.

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42.

A unity-feedback control system has the open loop transfer function  . If the

input system is a unit ramp, the steady state error will be

A. 0 B. 0.5

C. 2 D. infinity

Answer & Explanation 

Answer: Option C 

Explanation: 

.

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43. For a given voltage, four heating coils will produce maximum heat when connected

A. all in parallel

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B. all in series

C. with two parallel pairs in series

D. one pair in parallel with the other two in series

Answer & Explanation 

Answer: Option A 

Explanation: 

.

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44. Denominator polynomial of a transfer function of certain network is:

s3 + s2 + 2s + 24

Then the network is:

A. stable

B. oscillatory

C. unstable

D. depends on numerator polynomial

Answer & Explanation 

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Answer: Option C 

Explanation: 

Routh array

There is negative number present in first column. Thus network is unstable.

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45. A TV picture is to be transmitted over a channel of 6 MH z  bandwidth at a 35 dB S/N ratio. The

capacity of the channel is

A. 50 Mbps 

B. 100 Mbps 

C. 36 Mbps 

D. 72 Mbps 

Answer & Explanation 

Answer: Option D 

Explanation: 

W = B = Bandwidth

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C = 6 x 106 x 12 = 72 Mbps.

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46. The ratio of the diffusion coefficient in a semiconductor has the units

A. V-1  B. em.V-1 

C. V.cm-1  D. V.s 

Answer & Explanation 

Answer: Option A 

Explanation: 

.

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47. Consider the probability density f(x) = ae-b| x | where x  is a random variable whose allowable

values range from x  = - ∞ to  x  = + ∞. P(1 ≤  x  ≤ 2) 

A. 

B. 

C. 

D. (e-b - e-2b)

Answer & Explanation 

Answer: Option C 

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Explanation: 

P( x ) = ae-b|x|d  x  

to get apply aebx dx  + ae-bx d  x ,

we get = .

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48. The amplitude of a pair of composite sinusoidal signal y (n) = x 1(n) + x 2(n) with x 1(n) = sin

(5n) x 2(n) = 3 sin (5n) is __________ .

A. 2 B. 3

C. 4 D. 1

Answer & Explanation 

Answer: Option A 

Explanation: 

Amplitude = .

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49. The amplifier circuit shown below uses a silicon transistor. The capacitors Cc and CEcan be

assumed to be short at signal frequency and the effect of output resistance r 0can be ignored. If

CE is disconnected from the circuit, which one of the following statements is TRUE?

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A. The input resistance Ri  increases and the magnitude of voltage gain Av decreases

B. The input resistance Ri  decreases and the magnitude of voltage gain Av decreases

C. Both input resistance Ri  and the magnitude of voltage gain Av decrease

D. Both input resistance Ri  and the magnitude of voltage gain Av increase

Answer & Explanation 

Answer: Option A 

Explanation: 

Given circuit after removing CE will behave as current-series feedback.

Overall voltage gain will decrease as feedback signal comes into picture and since it is current-

series feedback, input impedance increases.

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50. A n-channel JFET has IDSS = 2 mA, Gate to source voltage VGS = - 2 V and trans-conductance is

0.5 mW then pinch-off voltage is __________ .

A. - 2 V

B. 2 V

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C. - 4 V

D. 4 V

Answer & Explanation 

Answer: Option C 

Explanation: 

= - 8VP - 16

+ 8VP + 16 = 0

(VP + 4)2 = 0

VP = -4 V.

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