Sec.2.2 The Physical Layer:Channels and Modems 45 The channel output r(t)and its frequency function ROf)are related in the same way. Finally,since Eq.(2.2)expresses the output for a unit complex sinusoid at frequency f, and since s(t)is a superposition of complex sinusoids as given by Eq.(2.8),it follows from the linearity of the channel that the response to s(t)is r(t)=H(f)S(f)ej2rf'df (2.9) Since r(t)is also the inverse Fourier transform of R(f),the input-output relation in terms of frequency is simply R(f)=H(f)s(f) (2.10) Thus,the convolution of h(t)and s(t)in the time domain corresponds to the multiplication of H(f)and S(f)in the frequency domain.One sees from Eq.(2.10)why the frequency domain provides insight into filtering. If H(f)=I over some range of frequencies and S(f)is nonzero only over those frequencies,then R(f)=S(f),so that r(t)=s(t).One is not usually so fortunate as to have H(f)=I over a desired range of frequencies,but one could filter the received signal by an additional filter of frequency response (f)over the desired range: this additional filtering would yield a final filtered output satisfying r(t)=s(t).Such filters can be closely approximated in practice subject to additional delay [i.e.,satisfying r(t)=s(t-r).for some delay ]Such filters can even be made to adapt automatically to the actual channel response H(f);filters of this type are usually implemented digitally, operating on a sampled version of the channel output,and are called adaptive equalizers. These filters can and often are used to overcome the effects of intersymbol interference. The question now arises as to what frequency range should be used by the signal s(t):it appears at this point that any frequency range could be used as long as H(f) is nonzero.The difficulty with this argument is that the additive noise that is always present on a channel has been ignored.Noise is added to the signal at various points along the propagation path,including at the receiver.Thus,the noise is not filtered in the channel in the same way as the signal.If H(f)is very small in some interval of frequencies.the signal is greatly attenuated at those frequencies.but typically the noise is not attenuated.Thus.if the received signal is filtered at the receiver by H(f),the signal is restored to its proper level,but the noise is greatly amplified. The conclusion from this argument(assuming that the noise is uniformly distributed over the frequency band)is that the digital data should be mapped into s(t)in such a way that S(f)is large over those frequencies where H(f)is large and S(f)is small (ideally zero)elsewhere.The cutoff point between large and small depends on the noise level and signal level and is not of great relevance here.particularly since the argument above is qualitative and oversimplified.Coming back to the example where h(t)=ae for t>0,the frequency response was given in Eq.(2.5)as H(f)=a/(a+j2f).It is seen that H(f)is approximately I for small f and decreases as 1/f for large f.We shall not attempt to calculate S(f)for the NRZ code of Fig.2.3,partly because s(t)and S(f)depend on the particular data sequence being encoded,but the effect of changing the signaling interval T can easily be found.If T is decreased,then s(t)is compressedSec. 2.2 The Physical Layer: Channels and Modems 45 The channel output ret) and its frequency function R(j) are related in the same way. Finally, since Eq. (2.2) expresses the output for a unit complex sinusoid at frequency j, and since s(t) is a superposition of complex sinusoids as given by Eq. (2.8), it follows from the linearity of the channel that the response to s(t) is ret) = I: H(j)S(j)eJ27Cftdj (2.9) Since ret) is also the inverse Fourier transform of R(j), the input-output relation in terms of frequency is simply R(j) = H(j)S(j) (2.10) Thus, the convolution of h(t) and 8(t) in the time domain corresponds to the multiplication of H(j) and S(j) in the frequency domain. One sees from Eq. (2.10) why the frequency domain provides insight into filtering. If H(j) = lover some range of frequencies and S(j) is nonzero only over those frequencies, then R(j) = S(j), so that ret) = s(t). One is not usually so fortunate as to have H(j) = lover a desired range of frequencies, but one could filter the received signal by an additional filter of frequency response H- 1 (j) over the desired range; this additional filtering would yield a final filtered output satisfying r(t) = .s(t). Such filters can be closely approximated in practice subject to additional delay [i.e., satisfying r(t) = 8(t - T), for some delay T]. Such filters can even be made to adapt automatically to the actual channel response H(!); filters of this type are usually implemented digitally, operating on a sampled version of the channel output, and are called adaptive equalizers. These filters can and often are used to overcome the effects of intersymbol interference. The question now arises as to what frequency range should be used by the signal .s(t); it appears at this point that any frequency range could be used as long as H(j) is nonzero. The difficulty with this argument is that the additive noise that is always present on a channel has been ignored. Noise is added to the signal at various points along the propagation path, including at the receiver. Thus, the noise is not filtered in the channel in the same way as the signal. If IH(j)1 is very small in some interval of frequencies, the signal is greatly attenuated at those frequencies, but typically the noise is not attenuated. Thus, if the received signal is filtered at the receiver by H- 1 (j), the signal is restored to its proper level, but the noise is greatly amplified. The conclusion from this argument (assuming that the noise is uniformly distributed over the frequency band) is that the digital data should be mapped into 8(t) in such a way that IS(j)1 is large over those frequencies where IH(j)1 is large and IS(j)1 is small (ideally zero) elsewhere. The cutoff point between large and small depends on the noise level and signal level and is not of great relevance here, particularly since the argument above is qualitative and oversimplified. Coming back to the example where 1I(t) = ae- of for t 2' 0, the frequency response was given in Eq. (2.5) as H(j) = 0:/(0 +j2K!). It is seen that IH(j)1 is approximately I for small/and decreases as l/! for large f. We shall not attempt to calculate S(j) for the NRZ code of Fig. 2.3, partly because .s(t) and S(j) depend on the particular data sequence being encoded, but the effect of changing the signaling interval T can easily be found. If T is decreased, then s(t) is compressed