42 Point-to-Point Protocols and Links Chap.2 output due to the sum of two inputs is simply the sum of the corresponding outputs. As an example of these properties,the output for Fig.2.3(b)can be calculated from the output in Fig.2.3(a).In particular,the response to a negative-amplitude rectangular pulse is the negative of that from a positive-amplitude pulse (property 2),the output from a delayed pulse is delayed from the original output (property 1),and the output from the sum of the pulses is the sum of the outputs for the original pulses (prop- erty 3). Figure 2.3 also illustrates a simple way to map incoming bits into an analog waveform s(t).The virtual channel accepts a new bit from the DLC module each T seconds;the bit value 1 is mapped into a rectangular pulse of amplitude +1,and the value 0 into amplitude-1.Thus,Fig.2.3(b)represents s(t)for the data se- quence 110100.This mapping scheme is often referred to as a nonreturn to zero (NRZ) code.The notation NRZ arises because of an alternative scheme in which the pulses in s(t)have a shorter duration than the bit time T,resulting in s(t)returning to 0 level for some interval between each pulse.The merits of these schemes will become clearer later. Next,consider changing the rate at which bits enter the virtual channel.Figure 2.4 shows the effect of increasing the rate by a factor of 4(i.e.,the signaling interval is reduced from T to T/4).It can be seen that output r(t)is more distorted than before. The problem is that the response to a single pulse lasts much longer than a pulse time,so that the output at a given t depends significantly on the polarity of several input pulses; this phenomenon is called intersymbol interference. From a more general viewpoint,suppose that h(t)is the channel output correspond- ing to an infinitesimally narrow pulse of unit area at time 0.We call h(t)the impulse response of the channel.Think of an arbitrary input waveform s(t)as a superposition of very narrow pulses as shown in Fig.2.5.The pulse from s()to s(T+6)can be viewed as a small impulse of area os()at time T:this gives rise to the output 8s()h(t-T)at 0T/4 0T4 (a) r(t) 3T/2 3T/2 (b) Figure 2.4 Relation of input and output waveforms for the same channel as in Fig.2.3. Here the binary digits enter at 4 times the rate of Fig.2.3,and the rectangular pulses last one-fourth as long.Note that the output r(t)is more distorted and more attenuated than that in Fig.2.3
42 Point-ta-Point Protocols and Links Chap. 2 output due to the sum of two inputs is simply the sum of the corresponding outputs. As an example of these properties, the output for Fig. 2.3(b) can be calculated from the output in Fig. 2.3(a). In particular, the response to a negative-amplitude rectangular pulse is the negative of that from a positive-amplitude pulse (property 2), the output from a delayed pulse is delayed from the original output (property 1), and the output from the sum of the pulses is the sum of the outputs for the original pulses (property 3). Figure 2.3 also illustrates a simple way to map incoming bits into an analog waveform s(t). The virtual channel accepts a new bit from the DLe module each T seconds; the bit value I is mapped into a rectangular pulse of amplitude + I, and the value 0 into amplitude -1. Thus, Fig. 2.3(b) represents s(t) for the data sequence 110100. This mapping scheme is often referred to as a nonreturn to zero (NRZ) code. The notation NRZ arises because of an alternative scheme in which the pulses in s(t) have a shorter duration than the bit time T, resulting in s(t) returning to 0 level for some interval between each pulse. The merits of these schemes will become clearer later. Next, consider changing the rate at which bits enter the virtual channel. Figure 2.4 shows the effect of increasing the rate by a factor of 4 (i.e., the signaling interval is reduced from T to T /4). It can be seen that output r(t) is more distorted than before. The problem is that the response to a single pulse lasts much longer than a pulse time, so that the output at a given t depends significantly on the polarity of several input pulses; this phenomenon is called intersymbol interference. From a more general viewpoint, suppose that h(t) is the channel output corresponding to an infinitesimally narrow pulse of unit area at time O. We call h(t) the impulse response of the channel. Think of an arbitrary input waveform s(t) as a superposition of very narrow pulses as shown in Fig. 2.5. The pulse from S(T) to S(T +6) can be viewed as a small impulse of area 6S(T) at time T; this gives rise to the output IiS(T)h(t - T) at o T/4 r(t) o T/4 (a) 3T/2 o T/2 T~ (b) Figure 2.4 Relation of input and output waveforms for the same channel as in Fig. 2.3. Here the binary digits enter at 4 times the rate of Fig. 2.3. and the rectangular pulses last one-fourth as long. Note that the output r(t) is more distorted and more attenuated than that in Fig. 2.3
Sec.2.2 The Physical Layer:Channels and Modems 43 Area≈5s(T) Area≈islr+6) 5t) T+6 8(r)h(t-r) 6sr+6}ht-T-6) Figure 2.5 Graphical interpretation of the convolution equation.Input s(t)is viewed as the superposition of narrow pulses of width 6.Each such pulse yields an output 6s()h(t-7).The overall output is the sum of these pulse responses. time t.Adding the responses at time t from each of these input pulses and going to the limit6→0 shows that r-Jx s(T)h(t -T)dT (2.1) This formula is called the convolution integral,and r(t)is referred to as the con- volution of s(t)and h(t).Note that this formula asserts that the filtering aspects of a channel are completely characterized by the impulse response h(t);given h(t),the out- put r(t)for any input s(t)can be determined.For the example in Fig.2.3,it has been assumed that h(t)is 0 for t<0 and ae-for t0.where =2/T.Given this,it is easy to calculate the responses in the figure. From Eq.(2.1),note that the output at a given time t depends significantly on the input s(7)over the interval where h(t-T)is significantly nonzero.As a result,if this interval is much larger than the signaling interval T between successive input pulses, significant amounts of intersymbol interference will occur. Physically,a channel cannot respond to an input before the input occurs,and therefore h(t)should be 0 for t <0;thus,the upper limit of integration in Eq.(2.1) could be taken as t.It is often useful,however,to employ a different time reference at the receiver than at the transmitter,thus eliminating the effect of propagation delay.In this case,h(t)could be nonzero for t<0,which is the reason for the infinite upper limit of integration. 2.2.2 Frequency Response To gain more insight into the effects of filtering,it is necessary to look at the frequency domain as well as the time domain;that is,one wants to find the effect of filtering
.... Sec. 2.2 The Physical Layer: Channels and Modems lis(r)h(t - r) ~IiS(r+Ii)h(t -r-li) r r+1i t-- t-- 43 (2.1 ) Figure 2.5 Graphical interpretation of the convolution equation. Input s(t) is viewed as the superposition of narrow pulses of width 6. Each such pulse yields an output 6S(T)h(t - T). The overall output is the sum of these pulse responses. time t. Adding the responses at time t from each of these input pulses and going to the limit {) ---+ 0 shows that T(t) = l:x s(T)h(t - T)dT This fonnula is called the convolution integral, and T(t) is referred to as the convolution of set) and h(t). Note that this fonnula asserts that the filtering aspects of a channel are completely characterized by the impulse response h(t); given h(t), the output ret) for any input set) can be detennined. For the example in Fig. 2.3, it has been assumed that h(t) is 0 for t < 0 and ae- n / for t 2: 0, where a = 21T. Given this, it is easy to calculate the responses in the figure. From Eq. (2.1), note that the output at a given time t depends significantly on the input SeT) over the interval where h(t - T) is significantly nonzero. As a result, if this interval is much larger than the signaling interval T between successive input pulses, significant amounts of intersymbol interference will occur. Physically, a channel cannot respond to an input before the input occurs, and therefore h(t) should be 0 for t < 0; thus, the upper limit of integration in Eq. (2.1) could be taken as t. It is often useful, however, to employ a different time reference at the receiver than at the transmitter, thus eliminating the effect of propagation delay. In this case, h(t) could be nonzero for t < 0, which is the reason for the infinite upper limit of integration. 2.2.2 Frequency Response To gain more insight into the effects of filtering, it is necessary to look at the frequency domain as well as the time domain; that is, one wants to find the effect of filtering
44 Point-to-Point Protocols and Links Chap.2 on sinusoids of different frequencies.It is convenient analytically to take a broader viewpoint here and allow the channel input s(t)to be a complex function of t;that is, s(t)=Rels(t)]+j Im[s(t)],where j=v-1.The actual input to a channel is always real,of course.However,if s(t)is allowed to be complex in Eq.(2.1),r(t)will also be complex,but the output corresponding to Re[s(t)]is simply Relr(t)][assuming that h(t) is reall,and the output corresponding to Im[s(t)]is Imfr(t)].For a given frequency f,let s()in Eq.(2.1)be the complex sinusoid =cos(2f)+j sin(f).Integrating Eq.(2.1)(see Problem 2.3)yields r(t)=H(f)e (2.2) where H0= h(r)e-i2xidr (2.3) Thus,the response to a complex sinusoid of frequency f is a complex sinusoid of the same frequency,scaled by the factor H(f).H(f)is a complex function of the frequency f and is called the frequency response of the channel.It is defined for both positive and negative f.Let H(f)be the magnitude and H(f)the phase of H(f) li.e..H(f)=H The response r(t)to the real sinusoid cos(2ft)is given by the real part of Eq.(2.2),or r1(t)=|H(f)川cos2πft+∠H(f] (2.4) Thus,a real sinusoidal input at frequency f gives rise to a real sinusoidal output at the same freqency:H(f)gives the amplitude and H(f)the phase of that output relative to the phase of the input.As an example of the frequency response,if h(t)=ae-a for t≥0，then integrating Eq.(2.3)yields Hf)= (2.5) a+j2πf Equation(2.3)maps an arbitrary time function h(t)into a frequency function H(f); mathematically,H(f)is the Fourier transform of h(t).It can be shown that the time function h(t)can be recovered from H(f)by the inerse Fourier transform H(edf (2.6) Equation (2.6)has an interesting interpretation;it says that an (essentially)arbi- trary function of time h(t)can be represented as a superposition of an infinite set of infinitesimal complex sinusoids,where the amount of each sinusoid per unit frequency is H(f).as given by Eq.(2.3).Thus.the channel input s(t)(at least over any finite interval)can also be represented as a frequency function by s(= s(t)e-j2r'dt (2.7) stt)-stpezrdr (2.8)
44 Point-to-Point Protocols and Links Chap. 2 on sinusoids of different frequencies. It is convenient analytically to take a broader viewpoint here and allow the channel input set) to be a complex function of t; that is, set) = Re[s(t)] + j Im[s(t)], where j = ;=T. The actual input to a channel is always real, of course. However, if set) is allowed to be complex in Eq. (2.1), ret) will also be complex, but the output corresponding to Rels(t)] is simply Relr(t)] [assuming that h(t) is real], and the output corresponding to Im[s(t)] is Im[r(t)]. For a given frequency f, let s( T) in Eq. (2.1) be the complex sinusoid ej27r j T = cos(2rrf T) +j sin(2rrf T). Integrating Eq. (2.1) (see Problem 2.3) yields r(t) = H(f)ej27rjf (2.2) where (2.3) Thus, the response to a complex sinusoid of frequency f is a complex sinusoid of the same frequency, scaled by the factor H(j). H(j) is a complex function of the frequency f and is called the frequency response of the channel. It is defined for both positive and negative f. Let [H(j)[ be the magnitude and LH(j) the phase of H(f) [i.e., H(j) = [H(f)\ejLH1fl]. The response rl(t) to the real sinusoid cos(2rrft) is given by the real part of Eq. (2.2), or rl (t) = [H(j)[ cos[27rft + UI(j)] (2.4) (2.6) (2.8) (2.7) Thus, a real sinusoidal input at frequency f gives rise to a real sinusoidal output at the same freqency; IH(j)1 gives the amplitude and LH(j) the phase of that output relative to the phase of the input. As an example of the frequency response, if h(t) = ae-of for t 0, then integrating Eq. (2.3) yields n H(j) = +'2 f (2.5) a J rr. Equation (2.3) maps an arbitrary time function h(t) into a frequency function H(j); mathematically, H(j) is the Fourier transform of h(t). It can be shown that the time function h(t) can be recovered from H(f) by the inverse Fourier transform h(t) = .l: H(j)ej27rjf df Equation (2.6) has an interesting interpretation; it says that an (essentially) arbitrary function of time hU) can be represented as a superposition of an infinite set of infinitesimal complex sinusoids, where the amount of each sinusoid per unit frequency is H(f), as given by Eq. (2.3). Thus, the channel input set) (at least over any finite interval) can also be represented as a frequency function by S(f) = .l: S(tje-j 2TC j l dt set) = .l: S(j)ej27rjfdf
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 compressed
Sec. 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
46 Point-to-Point Protocols and Links Chap.2 in time,and this correspondingly expands S(f)in frequency (see Problem 2.5).The equalization at the receiver would then be required either to equalize H(f)over a broader band of frequencies,thus increasing the noise,or to allow more intersymbol interference. 2.2.3 The Sampling Theorem A more precise way of looking at the question of signaling rates comes from the sampling theorem.This theorem states that if a waveform s(t)is low-pass limited to frequencies at most W [i.e..S(f)=0,for f>Wl,then [assuming that S(f)does not contain an impulse at f =W],s(t)is completely determined by its values each 1/(2W)seconds; in particular, s(t)= 立() sin[2W(t-i/(2W))] (2.11) 2πW[t-i/(2W)] Also,for any choice of sample values at intervals 1/(2W),there is a low-pass waveform, given by Eq.(2.11),with those sample values.Figure 2.6 illustrates this result. The impact of this result,for our purposes,is that incoming digital data can be mapped into sample values at a spacing of 1/(2W)and used to create a waveform of the given sample values that is limited to f<W.If this waveform is then passed through an ideal low-pass filter with H(f)=1 for f<W and H(f)=0 elsewhere, the received waveform will be identical to the transmitted waveform (in the absence of noise);thus,its samples can be used to recreate the original digital data. The NRZ code can be viewed as mapping incoming bits into sample values of s(t), but the samples are sent as rectangular pulses rather than the ideal(sin )/r pulse shape of Eq.(2.11).In a sense,the pulse shape used at the transmitter is not critically important s(0)sin 2mWt 2xWt s(t) 3 2w 2W 2W 2W sin 2xW (t 1 2W 2W 2xW t- 2W Figure 2.6 Sampling theorem,showing a function s(t)that is low-pass limited to frequencies at most W.The function is represented as a superposition of(sin)/z functions.For each sample,there is one such function,centered at the sample and with a scale factor equal to the sample value
46 Point-to-Point Protocols and Links Chap. 2 r in time, and this correspondingly expands S(n in frequency (see Problem 2.5). The equalization at the receiver would then be required either to equalize H(f) over a broader band of frequencies, thus increasing the noise, or to allow more intersymbol interference. 2.2.3 The Sampling Theorem A more precise way of looking at the question of signaling rates comes from the sampling theorem. This theorem states that if a waveform set) is low-pass limited to frequencies at most W [i.e., S(n = 0, for IfI > W], then [assuming that S(n does not contain an impulse at f = W], set) is completely determined by its values each 1/(2W) seconds; in particular, set) = s (_7_') sin[21rW(t - i/(2W))] . 2W 21rW[t - i/(2W)] l=-CXJ (2.11 ) Also, for any choice of sample values at intervals 1/(2W), there is a low-pass waveform, given by Eq. (2.11), with those sample values. Figure 2.6 illustrates this result. The impact of this result, for our purposes, is that incoming digital data can be mapped into sample values at a spacing of 1/(2W) and used to create a waveform of the given sample values that is limited to IfI :::; w. If this waveform is then passed through an ideal low-pass filter with H(f) = 1 for IfI :::; Wand H(f) = ° elsewhere, the received waveform will be identical to the transmitted waveform (in the absence of noise); thus, its samples can be used to recreate the original digital data. The NRZ code can be viewed as mapping incoming bits into sample values of set), but the samples are sent as rectangular pulses rather than the ideal (sinx)/x pulse shape of Eq. (2.11). In a sense, the pulse shape used at the transmitter is not critically important 5(0) sin 21TWt 21TWt 2W sIt) 7 / / / -1'- / -........... ---.r'!!!'::=--::O::"""'/~-::::"---------::'>.~/'<:--'r--~:-----~"'::'--_::':::"-"-'....-==-===-="", t 1----~ '2' o 2W sin 21TW (t - 2W) 21TW (t - 2~ ) Figure 2.6 Sampling theorem, showing a function s(t) that is low-pass limited to frequencies at most W. The function is represented as a superposition of (sin x)/ x functions. For each sample, there is one such function, centered at the sample and with a scale factor equal to the sample value