Advantages , DIFFERENTIAL PULSE CODE MODULATION (DPCM) :-

**ADAPTIVE DELTA MODULATION **

**(i) Reason to use Adaptive Delta Modulation**

** **To overcome the quantization errors due to slope overlod and granular noise, the step size (D) is made adaptive to variations in the input signal x(t). Particularly in the steep segment of the signal x(t), the step size is increased. Also, if the input is varying slowly, the step size is reduced. Then, this method is known as *Adaptive Delta Modulation (ADM).*

The adaptive delta modulators can take continuous changes in step size or discrete changes in step size.

**(ii) Transmitter Part**

** **Figure 4.30 (a) shows the transmitter and 4.30 (b) shows receiver of adaptive delta modulator. The logic for step size control is added in the diagram. The step size increases or decreases according to a specified rule depending on one bit quantizer output. As an example, if one bit quantizer output is high (i.e. 1), then step size may be doubled for next sample. If one bit quantizer Output is low, then step size may be reduced by one step. Figure 4.30 shows the staircase waveforms of adaptive delta modultor and the sequence of bits to be transmitted.

**DIAGRAM**

**FIGURE 4.30** Adaptive Delta Modulator (a) Transmitter (b) Receiver

**(iii) Receiver Part**

In the receiver of adaptive delta modulator shown in figure 4.32(b), there are two portions. The first portion produces the step size from each incoming bit. Exactly the same process is followed as that in transmitter. The previous input and present input decides the step size. It is then applied to an accumulator which builds up staircase waveform. The low-pass filter then smoothens out the staircase waveform to reconstrcut the original signal.

**DIAGRAM**

**FIGURE 4.31** Waveforms for adaptive delta modulation

**4.30.1. Advantages of Adaptive Delta Modulation : Salient Features**

Adaptive delta modulation has certain advantages over delta modulation as under:

(i) the signal to noise ratio becomes better than ordinary delta modulation because of the reduction in slope overload distortion and idle noise.

(ii) because of the variable step size, the dynamic range of ADM is wider than simple DM.

(iii) utilization of bandwidth is better than delta modulation.

**NOTE:** Other advantages of delta modulation are, only one bit per sample is required and simplicity of implementation of transmitter and receiver.

**4.31 DIFFERENTIAL PULSE CODE MODULATION (DPCM)**

* (U.P. Tech., Sem. Exam., 2005-2006) *

**(i) Reason to use DPCM**

** **It may be observed that the samples of a signal are highly correlated with each other. This is due to the fact that any signal does not change fast. This means that its value from present sample to next sample does not differ by large amount. The adjacent samples of the signal carry the same information with a little difference. When these samples are encoded by a standard PCM system, the resulting encoded signal contains some redundant information. Figure 4.25 illustrates this redundant information.

**(ii) Redundant Information in PCM**

Figure 4.32 shows a continuous time signal x(t) by dotted line. This signal is sampled by flat top sampling at intervals T_{s}, 2T_{s}, 3T_{s} … nT_{s}. The sampling frequency is selected to be higher than nyquist rate. The samples are encoded by using 3 bit (7 levels) PCM. The sample is quantized to the nearest digital level as shown by small circles in the figure 4.32. The encoded binary value of each sample is written on the top of the samples. We can observe from figure 4.32 that the samples taken at 4T_{s}, 5T_{s} and 6T_{s} are encoded to same value of (110). This information can be carried only by one sample. But three samples are carrying the same information means that it is redundant. Consider another example of samples taken at 9T_{s} and 10 T_{s}. The difference between these samples s only due to last bit and first two bits are redundant, since they do not change.

If this redundancy is reduced, then overall bit rate will decrease and number of bits required to transmit one sample will also be reduced. This type of digital pulse modulation scheme is known as **Differential Pulse Code Modulation (DPCM).**

**diagram**

**figure 4.32** *Illustration of redundant information in PCM.*

*(iii)*** Working Principle **

In fact the differential pulse code modulation works on the principle of prediction. The value of the present sample is predicted from the past samples. The prediction may not be exact but it is Very close to the actual sample vlaue. Figure 4.33 shows the transmitter of Differential Pulse Code Modulation (DPCM) system. The sampled signal is denoted by (nT_{s}) and the predicted signal is denoted by (nT_{s}). The comparator finds out the difference between the actual sample value x(nT_{s}) and predicated sample value (nT_{s}). This is known as Prediction error and it is denoted by e(nT_{s}). It can be defined as,

e(nT_{s}) = x(nT_{s}) – (nT_{s})

Thus, error is the difference between unquantized input sample x(nT_{s}) and prediction of it (nT_{s}). The predicted value is produced by using a prediction filter. The quantizer output signal gap e_{q}(nT_{s}) and previous prediction is added and given as input to the prediction filter. This signal is called x_{q}(nT_{s}). This makes the prediction more and more close to the actual sampled signal. We can observe that the quantized error signal e_{q}(nT_{s}) is very small and can be encoded by using small number of bits. Thus number of bits per sample are reduced in DPCM.

**diagram**

**FIGURE 4.33 ***A Differential pulse code modulation transmitter. *

The quantizer output can be written as,

e_{q}(nT_{s}) = e(nT_{s}) + q(nT_{s}) …(4.51)

Here q(nT_{s}) is the quantization error. As shown in figure 4.26, the prediction filter input x_{q}(nT_{s}) is obtained by sum x(nT_{s}) and quantizer output i.e.,

x_{q}(nTs) = (nTs) +eq(nTs) …(4.52)

Substituting the value of e_{q}(nT_{s}) from equation (4.51) in the above equation, we get,

x_{q}(nT_{s}) =x(nT_{s}) + e(nT_{s}) + q(nT_{s}) …(4.53)

Equation (4.50) is written as,

e(nT_{s}) = x(nT_{s}) – (nT_{s})

e(nT_{s}) + (nT_{s}) = x(nT_{s}) …(4.54)

the value of e(nT_{s}) + x(nT_{s}) from above equation into equation (4.53), we get,

e_{q}(nT_{s}) = x(nT_{s}) + q(nT_{s}) …(4.55)

**NOTE: **Hence, the quantized version of the signal x_{q}(nT_{s}) is the sum of original sample value and quantization error q(nT_{s}). The quantization error can be positive or negative. Thus equation (4.55) does not depend on the prediction filter characteristics.

**(iv) Reception of DPCM Signal : Reconstruction of DPCM Signal**

Figure 4.34 shows the block diagram of DPCM receiver.

The decoder first reconstructs the quantized error signal from incoming binary signal. Thy prediction filter output and quantized error sigals are summed up to give the quantized version of the orginal signal. Thus the signal at the receiver differs from actual signal by quantization error q(nT_{s}), which is filter introduced permanently in the reconstructed signal.

**DIAGRAM**

**FIGURE 4.34*** DPCM receiver*

**4.32 EVALUATION OF OUTPUT SIGNAL TO NOISE (S/N) RATIO**

** **The output signal to quantization noise ratio for a DPCM system may be defined in a similar manner as that for a PCM system.

Therefore, (SNR) =

But, mean square value is equal to the variance.

Therefore, (SNR) = …(4.56)

where, is the variance of orignal input signal x(nT_{s}) and is the variance of the quantization error q(nT_{s}).

We can rearrange the above expression as under :

** equation** …(4.57)

where is the variance of the prediction error e(nT_{s})

Therefore, SNR = G_{p}(SNR)_{p} …(4.58)

where G_{p} = () and called as prediction gain and (SNR)_{p} = ( /) is the prediction error-to-quantization noise ratio.

**4.32.1. Importance of Prediction Gain**

** **The prediction gain G_{p} is defined as under :

Gp = …(4.59)

The prediction gain must be as high as possible. For a given baseband signal, the variance is fixed. Hence to maximize G_{p}, we have to minimize the variance of the prediction error e(nT_{s}). The predictor must be designed accordingly.

**4.32.2. Types of Predictors**

** **The predictors used for DPCM are as under :

(i) One-tap predictors

(ii) N-tap predictors

**4.32.3. Advantage of DPCM : Salient Features**

(i) As the difference between x(nT_{s}) and (nT_{s}) is being encoded and transmitted by the DPCM technique, a small difference voltage is to be quantized and encoded.

(ii) This will require less number of quantization levels and hence less number of bits to represent them.

(iii) Thus signaling rate and bandwidth of a DPCM system will be less than that of PCM.

**4.33 COMPARISON OF DIGITAL PULSE MODULATION METHODS **

After discussing all the digital pulse modulation methods in details, let us now compare all these methods from different aspects. Table 4.1 (next page) shows the comparision of PCM, Differential PCM, Delta Modulation and Adaptive Delta Modulation. This comparison is carried out on the basis of various parameters like transmission bandwidth, quantization error, number of transmitter hits per sample etc.

**MISCELLANEOUS SOLVED EXAMPLES **

**EXAMPLE 4.17. A binary channel with bit rate r= 36000 bits per second (b/s) is available for PCM voice transmission. Evaluate the appropriate values of the sampling rate f _{s}, the quantizing level q, and the number of binary digits v. Assume f_{m}= 3.2 kHz. **

**Solution:**Here, we require that

f

_{s}≥ f

_{m}= 6400

and vf

_{s}≤ r = 36000

Therefore, we have, v ≤ = 5.6

Hence, we have, v 5,

and also, q = 2

*= 2*

^{v}^{5}= 32,

and f

_{s}= = 7200 Hz = 7.2 kHz

**Ans.**

**EXAMPLE 4.18. An analog signal is sampled at the Nyquist rate f**

_{s}and quantized into q levels. Find the time duration**of 1 bit of the binary-encoded signal.**

**Solution:**Let

*v*be number of bits per sample. Then we have

*v*= log

_{2}q

where log

_{2 }q indicates the next higher integer to be taken if log, q is not integer value i.e,

*v*f

_{s}binary pulses must be transmitted per second.

Thus, we have =

where T

_{s}is the Nyquist interval.

**Ans.**

**EXAMPLE 4.19. The output signal-to-quantizing-noise ratio (SNR), in a PCM system is defined as the ratio of average signal power to average quantizing noise power. For a full-scale sinusoidal modulating signal with amplitude A, prove that**

*(U.P. Tech-Semester Exam. 2002-2003)***Table 4.1. Comparison between PCM, Delta Modulation, Adaptive Delta Modulation and Differential Pulse Code Modulation**

S.No. |
Parameter of comparison |
Pulse Code Modulation (PCM) |
Delta modulation (DM) |
Adaptive Delta Modulation (ADM) |
Defferential Pulse Code Modulation (DPCM) |

1 | Number of bits. | It can use 4, 8 or 16 bits per sample. | It uses only one bit for one sample. | Only one bit is used to encode one sample. | Bits can he more than one but are less than PCM. |

2 | Levels and step size | The number of levels de- pend on number of bits. Levle size is kept fixed. | Step size is kept fixed and cannot be varied. | According to the signal variation, step size varies (i.e. Adapted). | Here, Fixed number of levels are used. |

3 | Quantization error and distortion | Quantization error de-pends on number of levels used. | Slope overload distor-tion and granular noise are present. | Quantization noise is present but other errors are absent. | Slope overload distor-tion and quantization noise is present. |

4 | Transmission bandwidth | Highest bandwidth is re- quired since number of bits are high | Lowest bandwidth is re- quired. | Lowest bandwidth is re- quired. | Bandwidth required is lower than PCM. |

5 | Feedback | There is no feedback in transmitter or receiver. | Feedback exists in trans- miner. | Feedback exists. | Here, Feedback exists. |

6 | Complexity of im- plementation | System complex. | Simple. | Simple. | Simple. |

**EQUATION**

or **EQUATION**

** Where q is the number of quantizing levels.**

**Solution:** Since, here peak-to-peak excursion of the quatizer input is 2A. Therefore, the quantizer step size will be.

D =

Then, the average quantizing noise power is

Then, **EQUATION**

The output signal-to-quantizing-noise ratio of a PCM system for a full scale test tone is, therefore,

**EQUATION**

Expressing this in decibels, we have

* *log * q* **Hence Proved.**

**EXAMPLE 4.20 In a binary PCM system, the output signal-to-quantizing-noise ration is to be held to a minimum value of 40 dB. Determine the number of reuqired levels, and find the corresponding output signal-to-quantizing-noise ration. (GATE Examination-1997)**

**Solution:**In a binary PCM system, q = 2

*, where v is the number of binary digits. Then, we have*

^{v}Now, since

Therefore,

Thus, we have, q =

and the number of binary digits v is

v = [log2 82] = [6.36] 7

Then, the number of levles required is q =2

^{7}= 128, and corresponding output signal-to-quantizting noise ratio will be

+ 6.02 x 7 = 43.9 dB

**Ans.**

**NOTE:**Equation (i) indicates that each bit in the code word of a binary PCM system contributes 6 dB to the output signal-to-quantizing noise ratio. In fact, this is called the 6 dB rule.