Natural Sampling

Natural Sampling

Natural Sampling –

It was convenient, for the purpose of introducing some basic ideas, to begin our discussion of time multiplexing by assuming instantaneous commutation and decommutation.

Such instantaneous sampling, however, is hardly feasible.

Even if it were possible to construct switches which could operate in an arbitrarily short time, we would be disinclined to use them.

The reason is that instantaneous samples at the transmitting end of the channel have infinitesimal energy, and when transmitted through a bandlimited channel give rise to signals having a peak value which is infinitesimally small.

We recall that in Fig.  I1 = m1 (0) dr.

Such infinitesimal signals will inevitably be lost in background noise.

A much more reasonable manner of sampling, referred to as natural sampling, is shown in Fig.

Here the sampling waveform S(t) consists of a train of pulses having duration τ and separated by the sampling time Ts.

The baseband signal is m(t), and the sampled signal S(t)m(t) is shown in Fig. .

Observe that the sampled signal consists of a sequence of pulses of varying amplitude whose tops are not flat but follow the waveform of the signal m(t).

With natural sampling, as with instantaneous sampling, a signal sampled at

the Nyquist rate may be reconstructed exactly by passing the samples through an ideal low-pass filter with cutoff at

the frequency f where is the highest-frequency spectral component of the signal.

To prove this, we note that the sampling waveform S(t) shown in Fig. is given by  A = 1 and T0 = Ts

S(t) = τ / T+ 2τ / Ts   ( C1 cos 2π t/Ts + C2 cos 2 × 2πt/Ts + ……). ….(1)

with the constant Cn given by

Cn  = sin ( nπτ/Ts) / nπτ/Ts.  ……(2)

This sampling waveform differs from the sampling waveform of Eq. for instantaneous sampling only in that dt is replaced by τ and by the fact that the amplitudes of the various harmonics are not the same.

The sampled baseband signal S(t)m(t) is, for Ts- 1/2Fm

S(t)m(t) = τ / Ts m(t) + 2τ / T[m(t)C1 cos 2π(2fm)t + m(t)C2 cos 2π(4fm)t+. …. ] …(3)

Therefore, as in instantaneous sampling, a low-pass filter with cutoff at fm will deliver an output signal s0(t) given by

s0(t) = τ / Ts m(t) ….(4)

which is the same as is given by the first term of Eq.  except with dt replaced by τ.

With samples of finite duration, it is not possible to completely eliminate the crosstalk generated in a channel, sharply bandlimited to a bandwidth fc .

If N signals are to be multiplexed, then the maximum sample duration is τ = Ts/N.

It is advantageous, for the purpose of increasing the level of the output signal, to make τ as large as possible.

For, as is seen in Eq.  ,  s0(t) increases with τ.

However, to help suppress crosstalk, it is ordinarily required that the samples be limited to a dura tion much less than Ts/N.

The result is a large guard time between the end of one sample and the beginning of the next.

What is the Op Amp ?

What is Flip flop ?

IC Operational Amplifier –

Bragg law

Ripple counter

ANALOG TO DIGITAL : THE NEED

Sampling Theorem and Low Pass Signal

Sampling of Bandpass Signal

PULSE AMPLITUDE MODULATION

Channel Bandwidth for PAM

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