EXAMPLE 4.5. Twenty four voice signals are sampled uniformly and then have to be time division multiplexed. The highest frequency component for each voice signal is equal to 3.4 kHz. Now
(i) If the signals are pulse amplitude modulated using Nyquist rate sampling, what would be the minimum channel bandwidth required.
(ii) If the signals are pulse code modulated with an 8 bit encoder, what would be the sampling rate? The bit rate of system is given as 1.5 x 106 bits/sec.
Solution: (i) As a matter of fact, if N channels are time division multiplexed, then minimum transmission bandwidth is expressed as,
BW = Nfm
Here, fm is the maximum frequency in the signals.
Given, fm = 3.4 kHz
Therefore BW = 24 x 3.4 kHz = 81.6 kHz Ans.
(ii) The signaling rate of the system in given as,
r = 1.5 x 106 bits/sec
Since there are 24 channels, the bit rate Of an individual channel is,
r (one channel) = = 62500 bits/sec
Further, since each samples is encoded using 8 bits, the samples per second will be,
Sample/sec =
Note that the samples per seconds is nothing but sampling frequency.
Thus, we have
Solving, we get, fs = 7812.5 Hz or samples per second Ans.
EXAMPLE 4.6. A PCM system uses a uniform quantizer followed by a 7-bit binary encoder. The bit rate of the system is equal to 50 x 106 bits/sec.
(i) What is the maximum message signal bandwidth for which the system operates satisfactorily?
(ii) Calculate the output signal to quantization noise ratio when a full load sinusoidal modulating wave of frequency 1 MHz is applied to the input. (U.P. Tech-Semester Exam. 2005-2006)
Solution: (i) Let us assume that the message bandwidth be fm, Hz. Therefore sampling frequency should be,
fs ³ 2fm
The number of bits given as v = 7 bits
We know that the signaling rate is given as,
r ³ v.fs
or r ³ 7 x 2fm
Substituting value for r, we get
50 x 106 ³ 14 fm
or fm 3.57 MHz Ans.
Thus, the maximum message bandwidth is 3.57 MHz.
(ii) The modulating wave is sinusoidal. For such signal, the signal to quantization noise ratio is expressed as,
Substituting the value of v, we get
= 1.8 + 6 x 7 = 43.8 dB Ans.
EXAMPLE 4.7. The information in an analog waveform with maximum frequency fm = 3 kHz is to be transmitted over an M-level PCM system where the number of
quantization levels is M = 16. The quantization distortion is specified not to exeed 1% of peak to peak analog signal.
(i) What would be the maximum number of bits per sample that should be used in this PCM system?
(ii) What is the minimum sampling rate and what is the resulting bit transmission rate?
Solution: (i) Since the number of quantization levels given here are M = 16,
q = M= 16
We know that the bits and levels in binary PCM are related as,
q = 2v
Here, v = number of bits in a codeword
Thus, 16 = 2v
or v = 4 bits. Ans.
(ii) Again since fm = 3 kHz
By sampling theorem, we know that
fs ≥ 2fm
Thus, fs ≥ 2 x 3 kHz ≥ 6 kHz Ans.
Hence, the minimum sampling rate is 6 kHz
Also bit transmission rate or signaling rate is given as,
r ≥ vfs ≥ 4 x 6 x 103
or r ≥ 24 x 103 bits per second Ans.
EXAMPLE 4.8. A signal having bandwidth equal to 3.5 kHz is sampled, quantized and coded by a PCM system. The coded signal is then transmitted over a transmission channel of supporting a transmission rate of 50 k bits/sec. Determine the maximum signal to noise ratio that can be obtained by this system.
The input signal has peak to peak value of 4 volts and rms value of 0.2 V.
(Pune University-1998)
Solution: The maximum frequency of the signal is given as 3.5 kHz,
i.e., fm = 3.5 kHz
Therefore sampling frequency will be
fs ≥ 2fm ≥ 2 x 3.5 kHz ≥ kHz
We know that the signaling rate is given by
r ≥ vfs
Substituting values of r = 50 x 103 bits/sec and fs ≥ 7 x 103 Hz in above equation, we get
50 x 103 ≥ v. 7 x 103
Simplifying, we get
v ≤ 7.142 bits 8 bits
The rms value of the signal is 0.2 V. Therefore the normalized signal power will be,
Normalized signal power P =
i.e., P = 0.04 W
Further, the maximum signal to noise ratio is given by,
Substituting the values of P = 0.04, v = 8 and xmax = 2 in above equation, we have
Ans.
EXAMPLE 4.9. A signal x(t) is uniformly distributed in the range ±xmax. Evaluate maximum signal to noise ratio for this signal.
Solution: Given that the signal is uniformly distributed in the range ±xmax, therefore we can write its PDF (using the Standard Uniform Distribution) as under:,
* R = 1 for normalized power.
EQUATION
Figure 4.13 shows this PDF,
The mean square valve of a random variable X is given as,
DIAGRAM
FIGURE 4.16 PDF of a uniformly distributed random variable.
EQUATION
Therefore, mean square value of x(t) will be,
EQUATION
EQUATION …(i)
The signal power
Normalized signal power [since R = 1]
Substituting the value of from (i), we get
We know that the relation between step size, maximum amplitude of signal and number of levles is given as
Step size
Therefore, normalized signal power, =
We also know that
Normalized noise power
Therefore, signal to noise power ratio
Since q = 2v, above equation will be,
or 6 v
This is required expression for maximum value of signal to noise ratio.
EXAMPLE 4.10. Given an audio signal consisting of the sinusoidal term given as
x(t) = 3 cos (500 )
(i) Determine the signal to quantizatibn noise ratio when this is quantized using 10-bit PCM.
(ii) How many bits of quantization are needed to achieve a signal to quantization noise ratio of atleast 40 dB?
Solution: Here, given that x(t) = 3 cos (500 t)
This is sinusoidal signal applied to the quantizer.
(i) Let us assume that peak value of cosine wave defined by x(t) covers the complete range of quantizer.
i.e., Am = 3V covers complete range
In example 4.1, we have derived signal to noise ratio for a sinusoidal signal. It is expressed as
Since here 10 bit PCM is used i.e.,
V= 10
Thus, = 1.8 + 6 X 10 = 61.8 dB Ans.
(ii) For sinusoidal signal, again, let us use the same relation
i.e., 1.8 + 6v dB
To get signal to noise ratio of at least 40 dB we can write above equation as,
1.8 + 6v ≥ 40 dB
Solving this, we get v ≥ 6.36 bits = 7 bits
Hence, at least 7 bits are required to get signal to noise ratio of 40 dB. Ans.
EXAMPLE 4.11. A 7 bit PCM system employing uniform quantization has an overall signaling to of 56 k bits per second. Calculate the signal to quantization noise that would result when its input is a sine wave with peak amplitude equal to 5 Volt. Find the dynamic range the sine wave inputs in order that the signal to quantization noise ratio may be less 1n 30 dBs. What is the theoretical maximum frequency that this system can handle?
(Madras University-1999)
Solution: The number of bits in the PCM system are
u = 7 bits
Assume hat 5 V peak to peak voltage utilizes complete range of quantizer. Then, we can find signal to quantization noise ratio as,
= 1.8 + 6v dB = 1.8 + 6 x 7 = 43.8 dB
We know that the signaling rate is given as,
r = v fs
Substituting r= 56 x 103 bits/second and v = 7 bits in above equation, we obtain 56 x 103 = 7.fs
Simplifying, we get
Sampling frequency, fs = 8 x 103 Hz
Further, using sampling theorem we have,
fs ≥ 2fm
Thus, maximum frequency that can be handled is given as,
Ans.