**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 10 ^{6} bits/sec.**

**Solution:**(i) As a matter of fact, if

*N*channels are time division multiplexed, then minimum transmission bandwidth is expressed as,

*BW = Nf*

_{m}Here,

*f*is the maximum frequency in the signals.

_{m}Given, f

_{m}= 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 10

^{6}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, f

_{s}= 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 10**

^{6}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 f

_{m}, Hz. Therefore sampling frequency should be,

*f*³ 2f

_{s}_{m}

The number of bits given as v = 7 bits

We know that the signaling rate is given as,

*r*

*³*

*v.f*

_{s}or r ³ 7 x 2f

_{m}

Substituting value for r, we get

50 x 10

^{6}³ 14 f

_{m}

or f

_{m}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 f**

_{m}= 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 = 2*

^{v}Here, v = number of bits in a codeword

Thus,

*16 = 2*

^{v}or v = 4 bits.

**Ans.**

(ii) Again since f

_{m}= 3 kHz

By sampling theorem, we know that

f

_{s}≥ 2f

_{m}

Thus, f

_{s}≥ 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 ≥

*v*f

_{s}≥ 4 x 6 x 10

^{3}

or r ≥ 24 x 10

^{3}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., f

_{m}= 3.5 kHz

Therefore sampling frequency will be

f

_{s}≥ 2f

_{m}≥ 2 x 3.5 kHz ≥ kHz

We know that the signaling rate is given by

r ≥ vf

_{s}

Substituting values of r = 50 x 10

^{3}bits/sec and fs ≥ 7 x 10

^{3 }Hz in above equation, we get

50 x 10

^{3}≥ v. 7 x 10

^{3}

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 x

_{max}= 2 in above equation, we have

**Ans.**

**EXAMPLE 4.9. A signal x(t) is uniformly distributed in the range ±x**

_{max}. Evaluate maximum signal to noise ratio for this signal.**Solution:**Given that the signal is uniformly distributed in the range ±x

_{max}, 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., A

_{m}= 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 + 6

*v*dB = 1.8 + 6 x 7 = 43.8 dB

We know that the signaling rate is given as,

r = v f

_{s}

Substituting r= 56 x 10

^{3}bits/second and v = 7 bits in above equation, we obtain 56 x 10

^{3}= 7.f

_{s}

Simplifying, we get

Sampling frequency, f

_{s}= 8 x 103 Hz

Further, using sampling theorem we have,

f

_{s}≥ 2f

_{m}

Thus, maximum frequency that can be handled is given as,

**Ans.**