642 C H A P T E R 11: Introduction to the Design of Discrete Filters
the receiver be ideally equal within a time delay and a constant attenuation factor. To achieve this,
the transfer function of an ideal communication channel should equal that of an all-pass filter with
a linear phase.Indeed, if the output of the transmitter is a discretized baseband signalx[n] and the recovered signal
at the receiver isαx[n−N 0 ], for an attenuation factorαand a time delayN 0 , ideally the channel is
represented by a transfer functionH(z)=Z(αx[n−N 0 ])
Z(x[n])=αz−N^0 (11.5)The constant gain of the all-pass filter permits all frequency components of the input to appear in the
output. The linear phase simply delays the signal, which is a very tolerable distortion.
To appreciate the effect of linear phase, consider the filtering of a signalx[n]= 1 +cos(ω 0 n)+cos(ω 1 n) ω 1 = 2 ω 0 n≥ 0using an all-pass filter with transfer functionH(z)=αz−N^0. The magnitude response of this filter is
α, and its phase is linear, as shown in Figure 11.3(a). The steady-state output of the all-pass filter isyss[n]= 1 H(ej^0 )+|H(ejω^0 )|cos(ω 0 n+∠H(ejω^0 ))+|H(ejω^1 )|cos(ω 1 n+∠H(ejω^1 ))
=α[ 1 +cos(ωo(n−N 0 ))+cos(ω 1 (n−N 0 ))]=αx[n−N 0 ]which is the input signal attenuated byαand delayedN 0 samples.Suppose then that the all-pass filter has a phase function that is nonlinear, for instance, the one in
Figure 11.3(b). The steady-state output would then beyss[n]= 1 H(ej^0 )+|H(ejω^0 )|cos(ω 0 n+∠H(ejω^0 )+|H(ejω^1 )|cos(ω 1 n+∠H(ejω^1 ))
=α[1+cos(ω 0 (n−N 0 ))+cos(ω 1 (n−0.5N 0 ))]6=αx[n−N 0 ]In the case of the linear phase each of the frequency components is delayedN 0 samples, and thus
the output is just a delayed version of the input. On the other hand, in the case of a nonlinear phase
the frequency component of frequencyω 1 is delayed less than the other two frequency components,
creating distortion in the signal so that the output is not a delayed version of the input.FIGURE 11.3
(a) Linear and (b) nonlinear phase.θ(ω)ω 0 ω 1 π ω
−π −N 0 ω 0(b)ω 0θ(ω)−N 0 ω 0
−N 0 ω 1ω 1 π ω
−π(a)