CALIFORNIA STATE UNIVERSITY, BAKERSFIELD (CSUB)DEPARTMENT OF ELECTRICAL & COMPUTER ENGINEERING & COMPUTER SCIENCE

ECE 423: DIGITAL COMMUNICATIONS

Homework 5Solution

QUESTION 1:(40 POINTS)Answer:For FM: The instantaneous frequency is ωi = 108 +105m(t ), (ωi )min = 108 −105 = 9.99×107

rad/s, (ωi )max = 108 + 105 = 1.001× 108 rad/s. The following figure shows that the cycle issplit into four equal parts of length 10−3/4 = 2.5×10−4 second. The instantaneous frequencyincreases linearly from (ωi )min to (ωi )max, stays there then decreases linearly back to (ωi )min

and stays there for the last quarter-cycle.

For PM: The instantaneous frequency isωi = 108+25m(t ), with (m(t ))min =−8000, (m(t ))max =8000, (ωi )min = 108−2×105 = 9.98×107 rad/s, (ωi )max = 108+2×105 = 1.002×108 rad/s. Thefollowing figure shows that the waveform in this case stays at (ωi )max for one quarter-cycle,shifts to (ωi )min and then shifts back to ωc for the last quarter cycle.

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QUESTION 2:(60 POINTS)A periodic signal m(t ) is transmitted as an angle-modulated frequency withωc = 4000π rad/s.The bandwidth of m(t ) is approximated by 200 Hz.

a) Sketch an FM signal φF M in time domain with k f = 500π.

b) Sketch a PM signal φP M in time domain with kp = 0.25π.

c) Determine the approximate bandwidth of the modulated FM signal.

d) Determine the approximate bandwidth of the modulated PM signal.

Answer:

a) For FM: The instantaneous frequency is fi = 2× 103 + 250m(t ), ( fi )min = 2000− 1000 =1kHz, ( fi )max = 2000+1000 = 3kHz . The following figure shows that the cycle is splitinto four equal segments of length 0.01 second. In the first segment, the instantaneousfrequency increases linearly from ( fi )min to ( fi )max. Next, it stays there for the secondsegment. Over the third segment, the instantaneous frequency decreases linearly backto ( fi )min and the cycle repeats.

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b) For PM: In this case, (m(t ))min = −8/0.02 = −400, (m(t ))max = 8/0.01 = 800, ( fi )min =2000−50 = 1.95kHz, ( fi )max = 2000+80 = 2.08kHz. The following figure shows that thecycle is split into four equal segments of length 0.01 second. In the first segment, theinstantaneous frequency is ( fi )max and next it switches to fc for the second segment.Over the third and forth segments, the instantaneous frequency stwiches to ( fi )min andthe cycle repeats.

c) k f = 500π, mp = 4, ∆ f = k f mp /(2π) = 1000Hz. Therefore, BF M = 2(∆ f +B) = 2(1000+200) = 2400Hz.

d) kp = 0.25π, (m(t ))max = 8/0.01 = 800, and (m(t ))mi n =−8/0.02 =−400Hz.

∆ f = kp

2π

(m(t ))max − (m(t ))mi n

2= 600

8= 75 Hz.

Therefore, BP M = 2(∆ f +B) = 2(75+200) = 550Hz.

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