**Encoding in PCM**

** **We know that encoding is the process that follows the sampling and quantization. Encoding process converts the quantized samples into codewords. In a binary code, each symbol may have either a 0 value or a 1 value. There are various formats (i.e., waveforms) to represent the binary sequence. They are known as line codes. Figure 4.13 shows two of such formats called unipolar NRZ and polar NRZ, where the NRZ stands for non-return to zero.

**DIAGRAM**

**FIGURE 4.13*** Two binary formats.*

**4.15.5. Multiplexing in PCM Systems**

** **It is possible to multiplex the PCM signals using the time division multiplexing (TDM) principle. With increase in number of independent message sources, the time interval allotted to each source has to be reduced to accommodate all the sources. This reduces the duration of each binary hit in PCM codeword. This increases the bandwidth requirement of the system. If the pulses become, very short then the impairments in the transmission medium start to interfere with the proper operation of the system. Therefore, in practice, it is essential to restrict the number of message sources.

**4.15.6. Synchronization in PCM**

** **For a PCM system with TDM, it is essential to synchronize the transmitter and receiver for proper operation of the system. For synchronization, it is essential to synchronize the clocks at the transmitter and receiver.

**4.16 PERFORMANCE EVALUATION OF PCM : PROBABILITY OF ERROR FOR PCM**

To evaluate the performance of the PCM system, we have to consider two major sources of noise as under:

(i) channel noise

(ii) quantization noise.

**(i) Channel Noise** gets introduced anywhere along the transmission path. It is also known as decoding noise.

**(ii) Quantization Noise** as discussed earlier is introduced at the transmitter and is carried along to the receiver output.

Both of them are present simultaneously but we shall consider them one by one to find their effect on the PCM system.

**Channel Noise and its Effect**

** **The major effect of channel noise is introduction of transmission errors at the receiver when the PCM signal is being reconstructed. Due to such errors, the receiver will make mistake in making the decision about whether a 0 was received or a 1 was received. A 0 may be mistaken as 1 and al may be mistaken as 0. Such errors must be minimized so as to improve the fidelity of PCM system.

**Probability of Error (Pe)**

** **The fidelity of a PCM system in presence of channel noise is measured in terms of error or probability of error. The probability of error or error rate is the probability that the symbol at the receiver output differs from that transmitted.

**Expression for Probability for Error (Pe)**

** **To obtain the expression for probability for error (Pe), let us use the matched filter shown in figure 4.14 and assume that the type of noise is AWGN, i.e., additive while gaussian noise*.

**DIAGRAM**

**FIGURE 4.14 ***Matched filter receiver for PCM*

The deviation for the probability of error for a PCM system (P_{e}) has been given in chapter 5. Here, let us only state the expression for the error probability (P_{e}) of PCM system as under:

**EQUATION **

Where E_{max} = Peak signal energy

N_{0} = Noise spectral density

We can substitue E_{max} = P_{max} Tb where Pmax is the maximum or peak signal power and T_{b} is the bit duration. Hence, the error probatility in terms of power is as under:

**EQUATION** …(4.34)

The ratio N_{0}/T_{b} can be seen as the average noise power contained in a transmission bandwidth equal to the bit rate (1/T_{b}). Hence, E_{max}/N_{0} may be viewed as the peak signal to noise power ration.

**Observation**

(i) The everage probability of error in PCM receiver depends only on the ratio of peak signal energy E_{max} to the noise power spectral density N_{0} measured at the receiver input.

(ii) The complementary error function *erfc* is a monotonically decreasing function. Hence, erfc decreases with increase in the ratio (E_{max}/N_{0}) as shown in figure 4.15

DIAGRAM

**FIGURE 4.15** Probability of error in a PCM system.

**Error Threshold**

** **From figure 4.15, it may be observed that the probability of error decreases very tepidly as the value of this ratio E_{max}/N_{0} is increased. A very small increase in transmitted signal energy or power will make the reception of binary pulses almost error free. The effect of increase in the ratio E_{max}/N_{0} has been illustrated in Table 4.1

**Table 4.1 Effect of E _{max}/N_{0} on probability of error **

S.No. | E_{max}/N_{0} |
Probability of Error, P _{e} |
For a bit rate of 105 bits per second. This is about 1 error every |

1. 2. 3. 4. 5. 6. |
10.3 dB 14.4 dB 16.6 dB 18 dB 19 dB 20 dB |
10^{-2}10 ^{-4}10 ^{-6}10 ^{-8}10 ^{-10}10 ^{-12} |
1 msec 0.1 sec 10 sec 20 min 1 day 3 months |

Table 4.1 shows that there is an error threshold at about 17 dB.

Above this value of E_{max}/N_{0}, the probability of error is very small. Whereas below this threshold, the probability of error is high and hence the effect of noise is quite significant. The effect of channel noise may be reduced by using the regenerative repeaters. Another important characteristics of PCM system is its raggedness to interference. As discussed earlier, for on off signaling, there is no effect of noise unless the peak amplitude is greater than half the pulse height.

Thus, if adequate margin over the error threshold is provided, then the system can successfully withstand large amount of noise and interference. Hence we can say that PCM is a noise resistant or rugged system.

**4.17 COMPARISON OF PCM AND ANALOG MODULATION**

** **The threshold effect in PCM is similar to a property of analog modulation methods such as FM or PPM. The property is that, these systems tend to reduce the wideband noise above the threshold levels. The PCM also provides the wideband noise reduction if it is operated above its threshold which is given by,

where q = 2* ^{v}* for binary PCM and q = M

*for M-ary PCM.*

^{v}We assume that the sampling frequency is close to the Nyquist rate and BW = Nf

_{m}Hz. Then, q = M

*= M*

^{v}*where b = BW/f*

^{b}_{m}is known as the bandwidth ratio.

Therefore, we have

Here, Signal to noise ratio at the destination

= Signal power at the destination

Here, it may be noted that the signal to noise ratio (S/N)

_{D}, is proportional to M

^{2b}which is much higher than the (S/N)

_{D}of the wideband FM which is proportional to only b or b

^{2}. Hence, PCM performs better than FM. Figure 4.16 shows the performance of various modulation types as a function of . All the curves in figure 4.16 have been plotted for S

_{x}= 1/2. The dots indicate the threshold points. The PCM curves have been drawn for M = 2 and v = b.

**DIAGRAM**

**FIGURE 4.16**Comparison of PCM and analog communication.

**Conclusions**

Some of the important observations from figure 4.16 may be listed as under :

(i) For PCM if b is constant, then increase in beyond the threshold value th (corresponding to the threshold point) does not increase (S/N)

_{D}at all. Let us observe the flat PCM curves in figure 4.16. Hence, PCM must be operated just above the threshold.

(ii) Near threshold, the PCM does offer some advantages over FM and PPM, with the same values of b and (S/N)

_{D}.

(iii) However, this advantage is gained at the expense of more complicated and expensive circuitry.

(iv) The (S/N)

_{D}for FM and PPM increases linearly with increase in the value of and becomes better than that of PCM for the higher values of .

**4.17.1. Benefits of PCM**

**Figure 4.16 reveals the following benefits of using the PCM :**

(i) PCM allows the use of regenerative repeaters. This improves its noise performance.

(ii) PCM allows the transmission of analog signals in the form of digital signals.

**4.17.2. PCM is not used for Radio Broadcasting**

**In ratio broadcasting, a relatively large signal to noise ratio (typically of the order of 60 dB) is required. To get this level of (S/N)**

_{D}, the PCM with b > 8 is needed. However, we can obtain the same performance with an FM system with b = 6 and with much simpler transmitter and receiver circuits. Therefore, higher bandwidth requirement and complicated circuitry are the drawbacks of PCM which does not allow its use for the radio, TV broadcasting applications.

**EXAMPLE 4.1. Derive an expressin for signal to quantization noise ratio for a PCM system which emplys linear (i.e., uniform) quantization technique. Given that input to the PCM system is a sinusoidal signal.**

*(U.P.S.C.I.E.S. Examination-1999)*or

**A PCM system uses a uniform quantizer followed by a**

*v*bit encoder. Show that rms signal to quantization noise ratio is approximately given as (1.8 + 6*v*) dB.*(U.P. Tech-Semester Exam. 2002-2003)***Solution:**Let us assume that the modulating signal is a sinusoidal voltage, having a peak amplitude equal to A

_{m}. Also, let this signal cover the complete excursion of representation levels.

Then, the power of this signal will be,

P =

Here, V = rms value i.e.,

Therefore, we have …(i)

In case when R = 1, the power P is normalized, i.e.,

Normalized power with R = 1 in equation (i).

We know that, signal to quantization noise ratio is given as,

x

Here

and x

_{max}= A

_{m}

Substituting, these values in the equation (ii), we get

**EQUATION**

Expressing signal to noise power ratio in dB, we get

x

or

or x 10 x 0.3

Therefore, we have

(for sinusoidal signal)

**Ans.**

EXAMPLE 4.2. A Television signal having a bandwidth of 4.2 MHz is transmitted using binary PCM system. Given that the number of quantization levels is 512. Determine:

(i) Code word length

(ii) Transmission bandwidth

(iii) Final bit rate

(iv) Output signal to quantization noise ratio.

**Solution:**Givn that the bandwidth is 4.2 MHz. This means that highest frequency component will have frequency of 4.2 MHz i.e.,

f

_{m}= 4.2 MHz

Also, given that Quantization levels, q = 512

(i) We know that the number of bits and quantization levels are related in binary PCM as under:

q = 2

^{v}i.e., 512 = 2

^{v}or log

_{10}512 =

*v*log

_{10}2

or

Simplifying, we get, v = 9 bits

Hence, the code word length is 9 bits.

**Ans.**

(ii) We know that the transmission channel bandwidth is given as,

BW

*v*f

_{m}9 x 4.2 x 10

^{6}Hz 37.8 MHz

**Ans.**

(iii) The final bit rate is equal to signaling rate.

We know that the signaling rate is given as,

*r = vf*

_{s }….(i)Here, sampling frequency is given as f

_{s}³ 2f

_{m}

Thus, f

_{s}³ 2 x 4.2 MHz since f

_{m}= 4.2 MHz

Or f

_{s}³ 8.4 MHz

Substituting this value of ‘f

_{s}‘ in equation (1) for signaling rate, we get

Or r = 9 x 8.4 X 10

^{6}bits/sec = 75.6 x 10

^{6}bits/sec

**Ans.**

The transmission bandwidth may also be obtained as,

x 75.6 x 10

^{6}bits/sec

or BW ³ 37.8 MHz which is same as the value obtained earlier.

(iv) The output signal to noise ratio is expressed as

But v = 9

Therefore, x 9

or

**Ans.**

**EXAMPLE 4.3. The bandwidth of an input signal to the PCM is restricted to 4 kHz. The input signal varies in amplitude from – 3.8 V to + 3.8 V and has the average power of 30mW. The required signal to noise ratio is given as 20 dB. The PCM modulator produces binary output. Assuming uniform quantization,**

**(i) Find the number of bits required per sample.**

**(ii) Outputs of 30 such PCM coders are time multiplexed. What would be the minimum required transmission bandwidth for this multiplexed signal?**

*(Punjab Technical University-1999)***Solution:**The given value of signal to noise ratio is 20 dB.

This means that.

Hence,

(i) We know that the signal to quantization noise ratio is given as.

Here, we are given

x

_{max}= 3.8V

P = 30 mW

and

Therefore,

**EQUATION**

Solving, we get v = 6.98 bits = 7 bits

**Ans.**

(ii) The maximum frequency is given as

f

_{m}= 4 kHz

We know that the transmission bandwidth is expressed as,

*BW*

*³*

*vf*

_{m}Since there are 30 PCM coders which are time multiplexed, the transmission bandwidth must be,

BW ³ 30 x

*v*x f

_{m}³ 30 x 7 x 4 kHz ³ 840 kHz

**Ans.**

We also know that the signaling rate is two times the transmission bandwidth, i.e.

Signaling rate, r = 840 x 2 bits/sec = 1680 bits/sec.

**Ans.**

**EXAMPLE 4.4.**The information in an analog signal voltage waveform is to be transmitted over a PCM system with an accuracy of ± 0.1% (full scale). The analog voltage waveform has a bandwidth of 100 Hz and an amplitude range of – 10 to + 10 volts.

(i) Find the minimum sampling rate required.

(ii) Find the number of bits in each PCM word.

(iii) Find minimum bit rate required in the PCM signal.

(iv) Find the minimum absolute channel bandwidth required for the transmission of the PCM signal.

*(West Bengal-1999)***Solution:**Here an accuracy is given as ± 0.1%. This means that the quantization error must be ± 0.1% or the maximum quantization error must be ± 0.1%.

Thus,

_{max}= ± 0.1% = ± 0.001

We know that the maximum quantization error for an uniform quantizer is expressed as,

_{max}=

or

Therefore,

Step size = 2 x 0.001 = 0.002

We know that the step size, number of quantization levels and maximum value of the signal are related as

Here, given 10 volts

Substituting, values of D and x

_{max}in equation (i), we get

or

Hence, the number of levels are 10,000.

(i) The maximum frequency in the signal is given as 100 Hz, i.e.,

f

_{m}= 100Hz

By sampling theorem minimum sampling frequency should be,

f

_{m}³ 2f

_{m}³ 2 x 100 ³ 200 Hz

**Ans.**

(ii) We know that minimum 10,000 levels should be used to quantize the signal. If binary PCM is used, then number of bits for each samples may be calculated as under, i.e.

*q = 2*

^{v}Here, q = number of levels

*v*= bits in PCM.

Thus, 10,000 = 2

^{v}log

_{10}10,000 =

*v*log

_{10}2

or v =

**Ans.**

(iii) The bit rate or signaling rate is expressed as,

r ³

*vf*³ 14 x 200 ³ 2800bits per second.

_{s}**Ans.**

(iv) The transmission bandwidth for PCM is expressed as,

x 2800 ³ 1400 Hz

**Ans.**