Assignment6 Solutions

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4400 341: Introduction to Communication Systems 
Spring 2017 
Solution to Homework Assignment #6: 
 
1) Figure 1 shows the Fourier spectra of signals and . Determine the Nyquist intervals and 
the sampling rate for the signals , , 
 , 
 , and . (25 points) 
Hint: Use the frequency convolution and the width property of convolution 
 
Figure 1 
The bandwidth of and are 5 and 12 kHz respectively. Therefore the Nyquist sampling 
rate is 10 kHz for and is 24 kHz for . 
Also 
 
 , 
and from the width property of convolution the bandwidth of 
 is twice of and that of 
 
 is times the bandwidth of . Similarly the bandwidth of is the sum of 
the bandwith of and . Therefore the Nyquist rate is 20 kHz for 
 , kHz for 
 
 , and 34 kHz for . 
 
 
2) Signals 
 and are applied at the inputs of the ideal low-pass filters 
 and , as shown in Figure 2. The outputs and 
 of these filters are multiplied to obtain the signals , Find the Nyquist rate of 
 , , and . Use the convolution property and the width property of convolution to 
determine the bandwidth of . (30 points) H1(f)
g1(t) y1(t)
X
H2(f)
g2(t) y2(t)
y(t)=y1(t)y2(t)
Figure 2
 
 
 
 
 , 
 
From figure above, it’s obvious that the bandwidth of and are 10 and 5 kHz and from 
the width property of convolution the bandwidth of is the summation of 
two signal’s bandwidth: kHz. Hence, the Nyquist rates for 
three signals are 20, 10 and 30 kHz respectively. 
3) A compact disc (DC) records audio signals digitally by using PCM. Assume that the audio signal 
bandwidth equals 15 KHz. 
3-a) If the Nyquist samples are uniformly quantized into L= 65,536 levels then binary-coded, 
determine the number of binary digits required to encode a sample. (5 points) 
65,536=2
16
, so that 16 binary digits are needed to encode each sample. 
3-b) If the audio signal has average power of 0.1W and peak voltage of 1V, find the resulting 
signal to quantization noise ratio (SQNR) of the uniform quantizer output in part (a). (15 
points) 
Given and , . Thus 
 
 
 
 , 
 
 
 
 . 
3-c) Determine the number of binary digits per second (bits/s) required to encode the audio signal. 
(5 points) 
The bandwidth is 15 kHz. Hence, the Nyquist rate is 30 kHz. 
30×10
3
 [kS/s] × 16 [bit/S]= 480 kbit/s. 
3-d) For practical reasons discussed in class, signals are sampled at a rate above the Nyquist rate. 
Practical CDs use 44,100 samples per second. If L= 65,536, determine the number of bits per 
second required to encode the signal. (5 points) 
65,536=2
16
, so that 16 binary digits are needed to encode each sample. 
44.1×10
3
 [kS/s] × 16 [bit/S]= 705.6 kbit/s. 
 
 
4) A television signal (video and audio) has a bandwidth of 4.5MHz. The signal is sampled, quantized 
and binary-coded to obtain a PCM signal. 
4-a) Determine the sampling rate if the signal is to be sampled at a rate 20% above the Nyquist 
rate. (5 points) 
The bandwidth is 4.5 MHz. Hence, the Nyquist rate is 9 MHz. The actual 
sampling rate is 1.2×9=10.8 MHz. 
4-b) If the samples are quantized into 1024 levels, determine the number of binary pulses required 
to encode each sample. (5 points) 
1024=2
10
, 10 bit or binary pulses are needed to encode each sample. 
4-c) Determine the binary pulse rate (bits per second) of the binary-coded signal. (5 points) 
10.8×10
6
 [MS/s] × 10 [bit/S]= 108 Mbit/s.