APPLICATIONS OF PCM , ADVANTAGES of pulse code modulation ppt , DRAWBACKS OF PCM

By   November 26, 2019

ADVANTAGES of pulse code modulation ppt , DRAWBACKS OF PCM :-
APPLICATIONS OF PCM
Some of the applications of PCM may be listed as under:
(i)         With the advent of fibre optic cables, PCM is used in telephony.
(ii)        In space communication, space craft transmits signals to earth. Here, the transmitted power is quite small (i.e., 10 or 15 W) and the distances are very large (i.be., a few million km). However, due to the high noise immunity, only PCM systems can be used in such applications.
4.25 ADVANTAGES OF PCM : SALIENT FEATURES OF PCM
Following are the advantages of a PCM system:
(i)         PCM provides high noise immunity.
(ii)        Due to digital nature of the signal, we can place repeaters between the transmitter and ct, the receivers. Infact, the repeaters regenerate the received PCM signal. This can not be possible in analog systems. Repeaters further reduce the effect of noise.
(iii)       We can store the PCM signal due to its digital nature.
(iv)       We can use various coding techniques so that only the desired person can decode the received signal.
4.26 DRAWBACKS OF PCM
            A PCM system has few drawbacks as under:
(i)         The encoding, decoding and quantizing circuitry of PCM is complex.
(ii)        PCM requires a large bandwidth as compared to the other systems.
4.27 DELTA MODULATION                    (U.P. Tech-Semester Exam. 2002-2003)
(i)         Reason to use Delta Modulation
We have observed in PCM that it transmits all the bits which are used to code a sample. Hence, signaling rate and transmission channel bandwidth are quite large in PCM. To overcome problem, Delta Modulation is used.
(ii)        Working Principle
            Delta modulation transmits only one bit per sample. Here, the present sample value is compared with the previous sample value and this result whether the amplitude is increased or decreased is transmitted. Input signal x(t) is approximated to step signal by the delta modulator. This step size is kept fixed. The difference between the input signal x(t) and staircase approximated signal is confined to two levels, i.e., +D and -D. Now, if the difference is positive, then approximated signal is increased by one step, i.e., ‘D’. If the difference is negative, then approximated signal is reduced by ‘D’.
When the step is reduced, ‘0’ is transmitted and if the step is increased, ‘1’ is transmitted. Hence, for each sample, only one binary bit is transmitted. Figure 4.20 shows the amalog signal x(t) and its staircase approximated signal by the delta modulator.
 
Thus, The principle of delta modulation can be explained with the help of few equations as under:
The error between the sampled value of x(t) and last approximated sample is given as,
d(nTs) = x(nTs) – (nTs)                                                           …(4.45)
where              e(nTs) = error at present sample
x(nTs) = sampled signal of x(t)
(nTs) = last sample approximation of the staircase waveform.
DIAGRAM
FIGURE 4.23 Delta modulation waveform.

DO YOU KNOW?
Delta modulation transmits only one bit per sample, indicating whether the signal level is increasing or decreasing, but it needs a higher sampling rate than PCM for equivalent results.

If we assume u(nTs) as the present sample approximation of staricase output,
then,    u[(n-1) Ts] = (nTs)      …(4.46)
= last sample approximation of staircase waveform
Let us define a quantity b(nTs) in such a way that,
b(nTs) = D sgn [e(nTs)]                                                …(4.47)
This means that depending on the sign of error e(nTs), the sign of step size D is decided. In other words, we can write
EQUATION
Also, If                                                b(nTs) = + D then a binary ‘1’ is transmitted
and if                                       b(nTs) = – D then a binary ‘0’ is transmitted.
Here,                                       Ts = Sampling interval.
(iv)       Transmitter Part
Figure 4.21 (a) shows the transmitter (i.e., generation of Delta Modulated signal).
The summer in the accumulator adds quantizer output (± D) with the previous sample approximation. This gives present sample approximation. i.e.,
u(nTs) = u(nTs – Ts) + [±D]
or                                             u(nTs) = u(n – 1) Ts] + b(nTs)                           …(4.49)
The previous sample approximation u[(n-1)Ts] is restored by delaying one sample period Ts. The sampled input signal x(nTs) and staircease approximated signal (nTs) are subtracted to get error signal e(nTs).
Thus, depending on the sign of e(nTs), one bit quantizer generates an output of +D or – D. If the step size is + D, then binary ‘1’ is transmitted and if it is -D, then binary ‘0’ is transmitted.
(v)        Receiver Part
            At the receiver end, shown in figure 4.21(b), the accumulator and low-pass filter (1,I)14) an, used. The accumulator generates the staircase approximated signal output and is delayed by one sampling period Ts. It is then added to the input signal. If input is binary ‘1’ then it adds +D step to the previous output (which is delayed). If input is binary ‘0’ then one step ‘D’ is subtracted from the delayed signal. Also, the low-pass filter has the cutoff frequency equal to highest frequency in x(t). This low-pass filter smoothens the staircase signal to reconstruct orignal message signal x(t).
FIGURE 4.24 (a) A Delta modulation transmitter (b) A Delta modulation receiver
4.27.1. Advantages of Delta Modulation : Salient Features of Delta Modulation
The delta modulation has certain advantages over PCM as under:
(i)         Since, the delta modulation transmits only one bit for one sample, therefore the signaling rate and transmission channel bandwidth is quite small for delta modulation compared to PCM.
(ii)        The transmitter and receiver implementation is very much simple for delta modulation. There is no analog to digital converter required in delta modulation.
4.27.2. Drawbacks of Delta Modulation
(U.P. Tech., Sem. Examination, 2003-04, 2005-06)
The delta modulation has two major drawbacks as under:
(i)         Slope overload distortion,
(ii)        Granular or idle noise
Now, let us discuss these two drawbacks in detail.
(i) Slope Overload Distortion
This distortion arises because of large dynamic range of the input signal.
As can be observed from figure 4.25, the rate of rise of input signal x(t) is so high that the staircase signal cannot approximate it, the steep size ‘D’ becomes too small for staircase signal and the original input signal x(t).
This error or noise is known as slope overload distrotion. To reduce this error, the step size must be increased when slope of signal x(t) is high.
Since the step size of delta modulator remains fixed, its maximum or minimum slopes occur along straight lines. Therefore, this modulator is also know as Linear Delta Modulator (LDM).
DIAGRAM
FIGURE 4.25 Quantization errors in delta modulation.
(ii)        Granular or Idle Noise
            Granular or Idle noise occurs when the step size is too large compared to small variations in the input signal. This means that for very small variations in the input signal, the staircase signal is changed by large amount (D) because of large step size. Figure 4.22 shows that when the input signal is almost flat, the staircase signal u(t) keeps on oscillating by ±D around the signal. The error between the input and approximated signal is called granular noise. The solution to this problem is to make step size small.
NOTE: Therefore, a large step size is required to accommodate wide dynamic range of the input signal (to reduce slope overload distortion) and small steps are required to reduce granular noise. Infact, Adaptive delta modulation is the modification to overcome these errors.
4.27.3. Bit Rate (i.e., Signaling Rate) of Delta Modulation
            Delta modulation bit rate (r) = Number of bits transmitted/second
= Number of samples/sec x Number of bits/sample = fs x 1 =fs
                Therefore, the delta modulation bit rate is (1/N) times the bit rate of a PCM system, where N is the number of bits per transmitted PCM codeword. Hence, we can say that the channel bandwidth for the Delta modulation system is reduced to a great extent as compared to that for the PCM system.
EXAMPLE 4.12 Given a sine wave of frequency fm and amplitude Am applied to a delta modulator having step size D. Show that the slope overlod distortion will occur if
Am >
here Ts is the sampling period.
(GATE Examination-1999, U.P. Tech. Sem., Exam. 2004-2005)
Solution: Let us consider that the sine wave is represented as,
x(t) = Am sig(2fmt)
It may be noted that the slope of x(t) will be maximum when derivation of x(t) with respect to ‘t’ will be maximum. The maximum slope of delta modulator may be given as,
Maximum slope =                                     …(i)
We know that, slope overload distortion will take place if slope of sine wave is greater thin, slope of delta modulator i.e.,
max
or                                                         max
or                                                         max
or
or                                                                                    Hence Proved.
EXAMPLE 4.13. A delta modulator system is designed to operate at five times the Nyquist rate for a signal having a bandwidth equal to 3 kHz bandwidth. calculate the maximum amplitude of a 2 kHz input sinusoid for which the delta modulator does not have slope over load. Given that the quantizing step size is 250 mV. Also, derive the formula that you use.
(U.P.S.C. I.E.S. Engg. Examination-1999)
Solution: In last example, we have derived the relation for slope overload distortion which will occur if,
 
Thus, slope overload will not occur if,
 
The maximum frequency in the signal is,
fm = 3 kHz
Hence,             Nyquist rate = 2fm = 2 x 3 kHz = 6 kHz
Sampling frequency    = 5 times Nyquist rate
or                                                         f, = 5 x 6 kHz = 30 kHz Hence,
Hence, Sampling interval,
Given that step size D = 250 mV = 250 x 10-3 V = 0.25 V
Again, Given that fm = 2 kHz = 2 x 103 Hz
Substituting all these values in equation (ii), we get
Simplifying, we obtain
Am < 0.6 volts                         Ans.

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