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Sec. 6–8 Matched Filters 469<br />

occur at t = t 0 and that the input signal waveshape has been distorted by the filter in order to peak<br />

up the output signal at t = t 0 .<br />

In applications to digital signaling with a rectangular bit shape, this matched filter is<br />

equivalent to an integrate-and-dump filter, as we now illustrate. Assume that we signal with one<br />

rectangular pulse and are interested in sampling the filter output when the signal level is maximum.<br />

Then the filter output at t = t 0 is<br />

q<br />

r 0 (t 0 ) = r(t 0 ) * h(t 0 ) = r(l)h(t 0 - l) dl<br />

L<br />

When we substitute for the matched-filter impulse response shown in Fig. 6–16c, this equation<br />

becomes<br />

-q<br />

r 0 (t 0 ) =<br />

L<br />

t 0<br />

t 0 -T<br />

r(l) dl<br />

(6–166)<br />

Thus, we need to integrate the digital input signal plus noise over one symbol period T (which<br />

is the bit period for binary signaling) and “dump” the integrator output at the end of the symbol<br />

period. This is illustrated in Fig. 6–17 for binary signaling. Note that for proper operation of<br />

this optimum filter, an external clocking signal called bit sync is required. (See Chapter 3 for a<br />

discussion of bit synchronizers.) In addition, the output signal is not binary, since the output<br />

sample values are still corrupted by noise (although the noise has been minimized by the<br />

matched filter). The output could be converted into a binary signal by feeding it into a comparator,<br />

which is exactly what is done in digital receivers, as described in Chapter 7.<br />

Correlation Processing<br />

THEOREM. For the case of white noise, the matched filter may be realized by correlating<br />

the input with s(t); that is,<br />

r 0 (t 0 ) =<br />

L<br />

t 0<br />

t 0 -T<br />

r(t)s(t) dt<br />

(6–167)<br />

where s(t) is the known signal waveshape and r(t) is the processor input, as illustrated<br />

in Fig. 6–18.<br />

Proof.<br />

The output of the matched filter at time t 0 is<br />

t 0<br />

r 0 (t 0 ) = r(t 0 ) * h(t 0 ) = r(l)h(t 0 - l) dl<br />

L<br />

-q<br />

But from Eq. (6–160),<br />

h(t) = e s(t 0 - t), 0 … t … T<br />

0, elsewhere

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