0195136047.pdf

718 COMMUNICATION SYSTEMS

PCM + noise

Sampler

Timing

≥ 0 choose 1

< 0 choose 0

VT =

A^2 Tb

2

(a)

Σ

+

−

A∫ (.) dt D

Tb

Sampler

PCM + noise

Square-

wave clock

Timing

≥ 0 choose 1

< 0 choose 0

(c)

Σ

+

−

∫ (.) dt D

PCM + noise

Sampler

Timing

≥ 0 choose 1

< 0 choose 0

(b)

2 A∫ (.) dt D

Inverter

0

Tb

0

Tb

0

Figure 15.3.8PCM reconstruction circuits.(a)For unipolar waveform.(b)For polar waveform.(c)For

Manchester waveform.

Manchester PCM waveform becomes a polar PCM signal; the product with the clock

inverted is the negative of a polar signal.

In Figure 15.3.8(c), after differencing in the summing junction, the response is then a double-

amplitude polar PCM signal. The rest of the circuit is similar to that of a polar PCM, as in Figure

15.3.8(b).

The very presence of noise suggests that the PCM reconstruction circuits may occasion-

ally make a mistake in deciding what input pulse was received in a given bit interval. How

often an error is made is determined by thebit-error probability. Bit errors in a unipolar

system occur much more frequently than in a polar system having the same signal-to-noise

ratio.

With negligible receiver noise, only quantization error is present when Equation (15.3.3)

applies. With not so negligible receiver noise, the recovered signalfqR(t)can be expressed as

fqR(t)=fq(t)+εn(t)=f(t)+εq(t)+εn(t) (15.3.12)

in which an errorεn(t)due to noise is introduced in the reconstructed messagefqR(t). The ratio

of desired output signal power to total output noise power is given by

(

S 0

N 0

)

PCM

=

S 0 /Nq

1 +

[

ε^2 n(t)/ε^2 q(t)

] (15.3.13)