Channel Bandwidth for PAM

Channel Bandwidth for PAM

Channel Bandwidth for PAM

Suppose that we have independent baseband signals m1(t), m2(t), etc.,each of which is bandlimited to f_M.

What must be the bandwidth of the communications channel which will allow all N signals to be transmitted simultaneously using PAM time-division multiplexing?

We shall now show that, in principle at least, the channel need not have a bandwidth larger than Nf_M.

The baseband signal, say m1(t), must be sampled at intervals not longer than Ts = 1/2f_M .

Between successive samples of m1(t) will appear samples of the other N-1 signals.

Therefore, the interval of separation between successive samples of different baseband signals is 1/2f_MN.

The composite signal, then, which is presented to the transmitting end of the communications channel, consists of a sequence of samples, that is, a sequence of impulses.

If the bandwidth of the channel were arbitrarily great, the waveform at the receiving end would be the same as at the sending end and demultiplexing could be achieved in a straightforward manner.

If, however, the bandwidth of the channel is restricted, the channel response to an instantaneous sample will be a waveform which may well persist with significant amplitude long after the time of selection of the sample.

In such a case, the signal at the receiving end at any particular sampling time may well have significant contributions resulting from previous samples of other signals.

Consequently the signal which appears at any of the output terminals in Fig. will not be a single baseband signal but will be instead a combination of many or even all the baseband signals.

Such combining of baseband signals at a communication system output is called crosstalk and is to be avoided as far as possible.

Let us assume that our channel has the characteristics of an ideal low-pass filter with angular cutoff frequency ω_c =2πf_c unity gain, and no delay.

Let a sample be taken, say, of m1(t), at t=0.

Then at t=0 there is presented at the transmitting end of the channel an impulse of strength I₁ m1(0) dt.

The response at the receiving end is S_R1 (t) given by

S_R1 = I₁ ω_c  sin ω_c t  / π ω_c t ……(1)

The normalized response πS_R (t) /ω_c is shown in Fig. by the solid plot.

At t=0 the response attains a peak value proportional to the strength of the impulse I₁ = m (0) dt, which is in turn proportional to the value of the sample m(0).

This response persists indefinitely.

Observe, however, that the response passes through zero at intervals which are multiples of πω_c = 1/2f_c .

Suppose, then, that a sample of m2(t) is taken and transmitted at t = 1/2 f_c. 

If I_{2  =}m₂(t=1/2f_c) dt,

S_R2 (t)= I₂ ω_c  sin ω_{c (}t  -1/2f_c) / π ω_c (t  -1/2f_c) ……(2)

This response is shown by the dashed plot.

Suppose, finally, that the demultiplexing is done also by instantaneous sampling at the receiving end of the channel, for m1 (t) at t=0 and for m2(t) at t =1/2fc.

Then, in spite of the persistence of the channel response,

there will be no crosstalk, and the signals m1 (t) and m2(t) may be completely separated and individually recovered.

Similarly, additional signals may be sampled and multiplexed, provided that each new sample is taken synchronously, every 1/2f_cs.

The sequence must, of course, be continually repeated every 1/2fM  s, so that each signal is properly sampled.

We have then the result that with a channel of bandwidth fc we need to separate samples by intervals 1/2f_c.

The sampling theorem requires that the samples of an individual baseband signal be separated by intervals not longer than 1/2Fm.

Hence the total number of signals which may be a multiplexed is N=fc/fM or fc = NfM as indicated earlier.

In principle then,

multiplexing a number of signals by PAM time division requires no more bandwidth than would be required to multiplex these signals by frequency-division multiplexing using single-sideband transmission.

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