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Indian Institute of Technology Kharagpur
Analog Communication
Assignment 10
Question 1:
A PM signal can be generated by
a. passing the message signal to a FM modulator.
b. passing the message signal to a differentiator circuit followed by a FM modulator.
c. passing the message signal to a integrator circuit followed by a FM modulator.
d. None of the above.
Correct Answer : b
Detailed Solution:
A PM signal can be generated by passing the message signal to a differentiator circuit
followed by a FM modulator.
Question 2:
A FM signal can be generated by
a. passing the message signal to a PM modulator.
b. passing the message signal to a differentiator circuit followed by a PM modulator.
c. passing the message signal to an integrator circuit followed by a PM modulator.
d. None of the above.
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Indian Institute of Technology Kharagpur
Correct Answer : c
Detailed Solution:
A FM signal can be generated by passing the message signal to a integrator circuit
followed by a PM modulator.
Question 3:
A narrow band FM signal is passed through a band limiter followed by a band
pass filter before launching, to
a. convert the narrow band FM signal to a wide band FM signal
b. avoid amplitude variation of the narrow band FM signal
c. avoid phase variation of the narrow band FM signal
d. None
Correct Answer : b
Detailed Solution:
A narrow band FM signal is passed through a band limiter followed by a band pass
filter before launching, to avoid amplitude variation of the narrow band FM signal.
Question 4:
Let x(t) = A cos(2000πt) + Am sin(3000πt) + Am sin(1000πt). Then, x(t) will be a
narrow band FM if
A
a. >> 1
Am
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Indian Institute of Technology Kharagpur
A
b. << 1
Am
c. Am << 1
d. Can never be a narrow band FM
Correct Answer : a
Detailed Solution:
2Am
x(t) = A cos(2000πt)+Am sin(3000πt)+Am sin(1000πt) = A[cos(2000πt)+ sin(2000πt) cos(1000πt)].
A
2Am A
x(t) will be a NBFM if max | − cos(1000πt)| << 1 => >> 1.
A Am
Question 5:
For the same as Q4, find the bandwidth of the message signal.
a. 1000
b. 2000
c. 3000
d. None
Correct Answer : d
Detailed Solution:
Rt 2Am 2Am Rt
As x(t) is NBFM, kf −∞ m(α)dα = − cos(1000πt) = ×1000π −∞ sin(1000πt)dt.
A A
2Am
So, the message signal is B sin(1000πt) where kf B = × 1000π.
A
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Indian Institute of Technology Kharagpur
Question 6:
For the same as Q4, find the peak frequency deviation.
2Am
a.
A
1000Am
b.
A
2000πAm
c.
A
d. None
Correct Answer : b
Detailed Solution:
1 1000Am
Peak frequency deviation= kf B = .
2π A
Question 7:
A message signal m(t), shown below, is phase modulated on a carrier 100 cos 2π106 t,
with a phase-deviation constant kp equal to 3 rad/V.
The maximum phase deviation of the modulated signal is:
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a. 30 rad
b. 30/π rad
c. 15 rad
d. None
Correct Answer : a
Detailed Solution:
Instantaneous Phase of the PM signal θi (t) = 2π106 t + 3 ∗ m(t)
Maximum Phase Deviation = 3 ∗ max{|m(t)|} = 3 ∗ 10 = 30 rad
Question 8:
Maximum instantaneous frequency for Q7 is:
3
a. 106 + Hz
2π
3
b. 106 + Hz
π
c. 106 + 3 Hz
d. None
Correct Answer : b
Detailed Solution:
1 d
Maximum Instataneous Frequency = max{ 2π θ (t)}
dt i
|m0 (t)| has max value of 2, as shown below:
6
So, Maximum Instataneous Frequency=106 + .
2π
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Indian Institute of Technology Kharagpur
Question 9:
A block diagram of an Armstrong FM transmitter is shown in the following figure.
The parameters are as follows: f1 = 200kHz, f2 = 45.6M Hz, ∆f1 = 50Hz, ∆f2 =
16.15KHz where f2 and ∆f2 are the central frequency and frequency deviation at
the output.
Find the possible value of n1 and n2 .
a. n1 = 57, n2 = 4
b. n1 = 4, n2 = 57
c. n1 = 114, n2 = 2
d. n1 = 17, n2 = 19
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Correct Answer : d
Detailed Solution:
∆f2
n1 n2 = = 323 = 17 × 19
∆f1
Question 10:
For the same as the previous question find the possible value of fLO .
a. 0.25 MHz
b. 0.5 MHz
c. 0.75 MHz
d. None
Correct Answer : d
Detailed Solution:
f2 f2
fLO = n1 f1 − or − n1 f1 which gives fLO = 1 MHz.
n2 n2
