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Sec. 3–6 Intersymbol Interference 193
PROOF. We need to show that the impulse response of this filter is 0 at t = nT s , for
n Z 0, where T s = 1f s = 1(2f 0 ). Taking the inverse Fourier transform of Eq. (3–75), we have
or
-f 0
f 0
2f 0
h e (t) = Y(f)e jvt df + [1 + Y(f)]e jvt df + Y(f)e jvt df
L -2f 0 L -f 0 L
f 0
h e (t) =
L
f 0
-f 0
e jvt df +
L
2f 0
-2f 0
Y(f)e jvt df
= 2f 0 a sin v 0
2f 0
0t
b + Y(f)e jvt df + Y(f)e jvt df
v 0 t L -2f 0 L 0
Letting f 1 = f + f 0 in the first integral and f 1 = f - f 0 in the second integral, we obtain
h e (t) = 2f 0 a sin v 0t
b + e -jv 0t
Y(f 1 - f 0 )e jv1t df 1
v 0 t
L-f 0
f 0
+ e jv 0t
Y(f 1 + f 0 )e jv1t df 1
L-f 0
From Eqs. (3–76a) and (3–76b), we know that Y(f 1 - f 0 ) =-Y(f 1 + f 0 ); thus, we have
f 0
h e (t) = 2f 0 a sin v f 0
0t
b + j2 sin v
v 0 t
0 t Y(f 1 + f 0 )e jv1t df 1
L -f 0
This impulse response is real because H e (-f) = H
*
e (f), and it satisfies Nyquist’s first
criterion because h e (t) is zero at t = n(2f 0 ), n Z 0, and ∆t = 0. Thus, if we sample at t = n(2f 0 ),
there will be no ISI.
However, the filter is noncausal. Of course, we could use a filter with a linear phase
characteristic H e (f)e -jvT d
, and there would be no ISI if we delayed the clocking by T d sec,
since the e -jvT d
factor is the transfer function of an ideal delay line. This would move the peak
of the impulse response to the right (along the time axis), and then the filter would
be approximately causal.
At the digital receiver, in addition to minimizing the ISI, we would like to minimize the
effect of channel noise by proper filtering. As will be shown in Chapter 6, the filter that minimizes
the effect of channel noise is the matched filter. Unfortunately, if a matched filter is
used for H R (f) at the receiver, the overall filter characteristic, H e (f), will usually not satisfy the
Nyquist characteristic for minimum ISI. However, it can be shown that for the case of