Topic: Communications (Test 6)



Topic: Communications
Q.1
A transmission line of characteristic impedance 50 W 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 λ/4 from the load. The value of ZL is
A.
B. 250Ω
C. (19.23 + j46.15)Ω
D. (19.23 - j46.15)Ω
Answer : Option A
Explaination / Solution:

Since voltage maxima is observed at a distance of λ/4 from the load and we know that the separation between one maxima and minima equals to λ/4 so voltage minima will be observed at the load, Therefore load can not be complex it must be pure resistive.

 also RL = R0/s (since voltage maxima is formed at the load)
 R= (50/5)Ω


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Q.2
x(t) is a stationary random process with auto-correlation function.   This process is passed through the system shown below. The power spectral density of the output process y(t) is

A. (4π2f2 + 1) exp (-πf2)
B. (4π2f2 - 1) exp (-πf2)
C. (4π2f2 + 1) exp (-πf)
D. (4π2f2 - 1) exp (-πf)
Answer : Option A
Explaination / Solution:


The given circuit can be simplified as

Power spectral density of output is


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Q.3
Suppose that the modulating signal is  and the carrier signal is , which one of the following is a conventional AM signal without over-modulation
A. x(t) = AC m(t) cos (2𝜋fC t)
B. x(t) = AC [1 + m(t)] cos (2𝜋fC t)
C. x(t) = AC cos (2𝜋fC t) +AC /4 m(t) cos (2𝜋fC t)
D. x(t) Acos (2𝜋fC tcos (2𝜋fm t)+Asin (2𝜋fm tsin (2𝜋fC t)
Answer : Option C
Explaination / Solution:
No Explaination.


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Q.4
Consider an angle modulated signal  The average power of x(t) is
A. 10 W
B. 18 W
C. 20 W
D. 28 W
Answer : Option B
Explaination / Solution:
No Explaination.


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Q.5
If the scattering matrix [S ] of a two port network is  then the network is
A. lossless and reciprocal
B. lossless but not reciprocal
C. not lossless but reciprocal
D. neither lossless nor reciprocal
Answer : Option C
Explaination / Solution:




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Q.6
A transmission line has a characteristic impedance of 50 Ω and a resistance of 0.1Ω/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 : Option D
Explaination / Solution:

For distortion less transmission line characteristics impedance


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Q.7
A message signal m(t) = cos 200πt + 4cosπt  modulates the carrier c(t) = coe 2πfct where fc = 1MHz to 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 < 1 ms
B. 1 μs << RC < 0.5 ms
C. RC << μs
D. RC >> 0.5 ms
Answer : Option B
Explaination / Solution:

Highest frequency component in m(t) is 
Carrier frequency fc = 1 MHz
For Envelope detector condition


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Q.8
In the circuit shown below, the network N is described by the following Y matrix:   the voltage gain (V2/V1)is

A. 1/90
B. -1/90
C. -1/99
D. -1/11
Answer : Option D
Explaination / Solution:

From given admittance matrix we get


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Q.9
A four-phase and an eight-phase signal constellation are shown in the figure below.

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. r1 = 0.707d, r2 = 2.782d
B. r1 = 0.707d, r2 = 1.932d
C. r1 = 0.707d, r2 = 1.545d
D. r1 = 0.707d, r2 = 1.307d
Answer : Option D
Explaination / Solution:

Four phase signal constellation is shown below 

d2 = r12 + r12
d2 = 2r12

r1 = d/√2 = 0.707d




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Q.10
A four-phase and an eight-phase signal constellation are shown in the figure below.

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 : Option D
Explaination / Solution:

Here Pe for 4 PSK and 8 PSK is same because Pe depends on d . Since Pe is same, d is same for 4 PSK and 8 PSK.
Additional Power SNR



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