Drawback of Conventional Delta Modulation , use of Integrator :-

**DELTA-SIGMA MODULATION (U.P. Tech. Sera. Exam, 2004-05) (10 Marks)**

**4.29.1 Drawback of Conventional Delta Modulation**

**The quantizer input in the conventional delta modulator can be considered as an approximation to the derivative of the input message signal. Hence, the noise results in an accumulated error in the demodulated signal. This is the drawback of the conventional delta modulator. This drawback can be overcome by using the delta sigma modulator. In the delta sigma modulator, the input sign is passed through an integrator before applying it to the delta modulator circuit.**

**4.29.2 Advantages of use of Integrator**

The use of integrator has the following advantages :

(i) The low frequency components in the input signal are boosted (pre-emphasized).

(ii) The correlation between the adjacent samples of delta modulator is increased. This improves the overall system performance, due to reduction in the variance of the error signal at the quantizer input.

(iii) It simplifies the receiver design.

**4.29.3 Block Diagram of Delta-Sigma Modulator**

**Figure 4.28 (a) shows the block diagram of a delta-sigma modulator. Here, x(t) is the message signal which is a continuous time signal. It is first passed through an integrator and a comparator. The comparator output is applied to the pulse modulator block. It consists of a hard limiter, and a product modulator.**

**DIAGRAM**

**FIGURE 4.28**

The input-output characteristics of the hard limiter is shown or all the positive inputs, it p in figure 4.28 (b). Hence, if the input to this block is negative it will produce a – 1 output and for all the positive inputs, it produces +1 output. The hard limiter output (± 1) is applied to a multiplier (i.e., product modulator). The other input to the amplifier is the clock pulses produced by the external pulse generator. The frequency of clock pulses should be higher than the Nyquist rate. At the output of the product modulator, we get the sampled version of limiter output. It is transmitted over the communication channel. Thus, transmit a one bit encoded signal. The same output signal is applied to the second integrator. The output of integrator 2 is compared with the output of integrator 1 with the help of the comparator.

**Receiver**

**Figure 4.29 shows that the receiver of sigma delta system is simply a low pass filter i.e., an integrator.**

**4.29.4. Simplified Sigma-Delta System in a Simplified Manner**

**Since integration is a linear operation, we can combine the integrators 1 and 2 into one integrator which is placed after the comparator. This will simplify the transmitter design to some extent. The simplified sigma-delta system has been shown in figure 4.29.**

**DIAGRAM**

**FIGURE 4.29**Simplified sigma-delta system.

**EVALUATION OF MAXIMUM OUTPUT SIGNAL TO NOISE RATIO**

*(U.P. Tech, Sem. Exam. 2004-05) (10 Marks)*Students are advised to go through the following example (4.13) to prove that the maximum signal to noise ratio of delta modulation system is given by,

where f

_{M}= Cutoff frequency of the low pass filter in the delta modulation receiver.

**EXAMPLE 4.13. A sinusoidal modulating signal is represented by m(t) = A**cost, where .

**Show that the maximum output signal to quantization noise ratio in a delta modulation system with no slope overload distortion is given by,**

where f

_{m}= sampling frequency and

f

_{M}= cutoff frequency of a low pass filter (LPF) at the output of a receiver.

**Solution :**We know that the condition to avoid the slope overload distortion is expressed as

*…(i)*

Therefore, the maximum value of the output signal power is expressed as

(since

*P*is proportional to square of rms value).

Hence, we have

**EQUATION**

Now, we require to obtain the expression for quantization noise power. The quantization error in delta modulation is equally likely to lie anywhere in the interval (-D, D). This means that the maximum quantization error,

_{max}= ± D.

This error can be assumed to be uniformly distributed as shown in figure 4.26.

Thus, the PDF is an uniform distribution function which is defined as under:

**DIAGRAM**

**FIGURE 4.26**

*PDF of quantization error for delta modulation.*

The mean square value or the variance of the quantization noise is given by,

Hence, we have =

Therefore, normalized quantization noise power (EQUATION) …(iii)

The delta modulated signal is passed through a reconstruction low pass filter (LPF) at the output of a DM receiver. The bandwidth of this low pass filter (LPF) is f

_{m}such that,

f

_{M}≥ f

_{m}and f

_{M}<< f

_{s}

The arrangement of filter has been shown in figure 4.27. Now, assuming that the quantization noise power N

_{q}is distributed uniformly over the frequency band upto f

_{s}, the output quantization noise power within the bandwidth f

_{M}is given by,

**DIAGRAM**

**FIGURE 4.27**

*Low pass filter (LPF) at delta modulation receiver.*

Normalized noise power at the filter output,

N’

_{q}=

DO YOU KNOW? |

The signal-to-noise ratio for either PCM or delta modulation signals can often beimproved by using companding. |

Here, substituting the values from equations (ii) and (iv), we obtain the expression for output signal to quantization noise ratio as under:

**EQUATION**

** EQUATION** …(vi)

But f_{s} =

Therefore, …(vii)

This is the desired expression for the output signal to quantization noise ratio.

**EXAMPLE 4.14. A sinusoidal voice signal x(t) = cos (6000 ****t****) is to be transmitted using either PCM or DM. The sampling rate for PCM system is 8 kHz and for the transmission with DM, the step size ****D**** is decided to be of 31.25 mV. The slope overload distortion is to be avoided. Assume that the number of quantization levels for a PCM system is 64. Determine the signaling rates of both these systems and also comment on the result. **

**Solution :** We know that the signaling rate of a PCM system is given by

* r = v f _{s}*

But

*q = 2*

^{u}v = log

_{2}q = log

_{2}64 = 6

Therefore, signaling rate of PCM = r = 6 x 8 kHz = 48 kHz.

**Ans.**

**Now, let us calculate the signaling rate of DM system as under :**

The signaling rate of a delta modulation (DM) system is equal to its sampling rate f

_{s}, because in DM only one bit is transmitted per sample. We know that the condition to avoid the slope overload distortion is given by,

or

Now, let us calculate f

_{s}.

We know that

Substituting all the values, we obtain

**EQUATION**

or fs ≥ 603.18 kHz.

Therefore, signaling rate of DM ≥ 603.18 kHz.

**Ans.**

**COMMENTS:**To transmit the same voice signal, the DM needs a very large signaling rate as compared to PCM. This is the biggest drawback of DM, which makes it an impractical system.

**EXAMPLE 4.15. Determine the output signal to noise ratio of a linear delta modulation system for a 2 kHz sinusoidal input signal sampled at 64 kHz. Slope overload distortion is not present and the post reconstruction filter has a bandwidth of 4 kHz.**

Solution : We know that

(SNR)

_{0}=

Here, f

_{s}= 64 kHz, f

_{m}= 2 kHz and f

_{M}= 4kHz

Therefore, (SNR)

_{0}=

(SNR)

_{0}= 622.51 = 27.94 dB

**Ans.**

**EXAMPLE 4.16. For the same sinusoidal input of example 4.15, calculate the signal to quantization noise ratio of a PCM system which has the same data rate of 64 kbits/s. The sampling frequency is 8 kHz and the number of bits per sample is**

*N*= 8. Comment on the result.**Solution :**The signal to noise ratio of a PCM system is given by,

(SNR)q = (1.8 + 6 N) dB = 1.8 + (6 x 8) = 49.8 dB

**Ans.**

**COMMENTS:**The SNR of a DM system is 27.94 dB which is too poor as compared to 49.8 dB of an 8 bit PCM system. Thus, for all the simplicity of DM, it cannot perform as well as an 8 bit PCM.