Assignment 4 – EKT358
1. Consider a compact disc that uses pulse-code modulation to record audio signals
whose bandwidth 15 kHz. Specifications of the modulator include the following:
Quantization : uniform with 512 levels
Encoding: Binary
Determine:
(a) The Nyquist rate
(b) Minimum permissible bit rate
Nyquist rate, fs= 2B = 2(15 kHz) = 30 kHz
R = nfs
n = log 512/ log 2 =9
R = nfs= 9 (30kHz) = 270 kbits/s
2. A television signal (video and audio) has a bandwidth of 4.2 MHz. This signal is
sampled, quantized and binary coded to obtain a PCM signal.
(a) Determine the sampling rate if the signal is to be sampled at a rate 20% above the
Nyquist rate.
(b) If the samples are quantized into 1024 levels, determine the number of binary pulses
required to encode each sample.
(c) Determine the binary pulses rate(bits per second) of the binary-coded signal, and the
minimum bandwidth required to transmit this signal
Nyquist rate, fs= 2B = 2(4.2 MHz) = 84 MHz
20% above Nyquist rate, fs = 120/100 x 84 MHz = 10.08 MHz.
2n=1024
n = log21024 = 10
R = nfs= 10(10.08MHz) = 100.8Mbits/s
Bmin = R/2 = 100.8 MHz/ 2 = 50.4 MHz
3. Determine the minimum number of bits required for PCM codes with the following
dynamic ranges and determine the coding efficiencies:
a) 24 dB
b) 48 dB
c) 72 dB
DR(dB)  20 log( 2 n  1)
24 / 20  log( 2 n  1)
2 n  15.85  1
log 16.85
 4.07
log 2
n  4.07, n  5
4.07
efficiency 
x100%  81.5%
5
n
DR(dB)  20 log( 2 n  1)
48 / 20  log( 2 n  1)
2 n  251.19  1
log 252.19
 7.98
log 2
n  7.98, n  8
7.98
efficiency 
x100%  99.73%
8
n
DR(dB )  20 log( 2 n  1)
72 / 20  log( 2 n  1)
2 n  3981.07  1
log 3982.07
 11.96
log 2
n  11.96, n  12
11.96
efficiency 
x100%  99.66%
12
n
4. For a PCM system with the following parameters:
Maximum analog input frequency = 4 kHz
Maximum decoded voltage at the receiver= ±2.55 V
Minimum dynamic range=46 dB
Determine
(a)
(b)
(c)
(d)
(e)
Minimum sample rate
Minimum number of bits used in the PCM code
Resolution
Quantization error
Coding efficiency
(a) Nyquist rate, fs= 2B = 2(4 kHz) = 8 kHz
(b)
DR(dB)  20 log( 2 n  1)
46 / 20  log( 2 n  1)
2 n  199.53  1
log 200.53
n
 7.64
log 2
n  7.64, n  8
8 bits must be used for the magnitude but one additional sign bit is required to indicate +ve
and –ve voltage, thus the minimum number of bits=9
(c) resolution 
(d) Qe 
Vmax
V
2.55
 nmax  8
 0.01V
DR 2  1 2  1
resolution 0.01

 0.005V
2
2
(e) Coding efficiency = minimum number of bits(including sign bit) x 100%
Actual number of bits(including sign bit)
= 8.64/9 x 100% =96%
5. Briefly explain the process of digital companding.
With digital companding, the analog signal is first sampled and converted to a linear PCM
code and then the linear code is digitally compressed. In the receiver, the compressed PCM
code is expanded and then decoded (i.e., converted back to analog).
6. Determine the voltage of the input signals if the SQR = 40 dB and q =0.2 V.
⁄
[
]
⁄
7. A companding system with µ = 100 used to compand from 0V to 10 V sinusoid signal.
Draw the characteristic of the typical system.
Vout 
Vin
Vout
0
0
1
5.2
2
6.57
Vmax ln(1   Vin Vmax )
ln(1   )
3
7.44
4
8.05
5
8.52
6
8.91
7
9.24
8
9.52
12
10
8
Vin
6
Vout
4
2
0
1
2
3
4
5
6
7
8
9
10
11
9
9.77
10
10
8. Suppose analog signal has the bandwidth of 15 kHz (uniform). Find the minimum first
null bandwidth (square wave pulses) at which it would be possible to transmit a PCM
signal while maintaining average SNR of at least 58 dB. Assume Vpeak = 5V.