Sec.2.2 The Physical Layer:Channels and Modems 47 since the pulse shape can be viewed as arising from filtering sample impulses.If the combination of the pulse-shaping filter at the input with the channel filter and equalizer filter has the (sinz)/z shape of Eq.(2.11)(or,equivalently,a combined frequency response equal to the ideal low-pass filter response),the output samples will re-create the input samples. The ideal low-pass filter and the (sin )/a pulse shape suggested by the sampling theorem are nice theoretically but not very practical.What is required in practice is a more realistic filter for which the sample values at the output replicate those at the input (i.e.,no intersymbol interference)while the high-frequency noise is filtered out. An elegant solution to this problem was given in a classic paper by Nyquist [Nyq28].He showed that intersymbol interference is avoided by using a filter H'(f) with odd symmetry at the band edge;that is,H'(f+W)=1-H'(f-W)for If<W, and H'(f)=0 for f>2W (see Fig.2.7).The filter H'(f)here is the composite of the pulse-shaping filter at the transmitter,the channel filter,and the equalizer filter.In prac- tice,such a filter usually cuts off rather sharply around f=W to avoid high-frequency noise,but ideal filtering is not required,thereby allowing considerable design flexibility. It is important to recognize that the sampling theorem specifies the number of samples per second that can be utilized on a low-pass channel,but it does not specify how many bits can be mapped into one sample.For example,two bits per sample could be achieved by the mapping 11-3,10-1,00--1,and 01--3.As discussed later,it is the noise that limits the number of bits per sample. 2.2.4 Bandpass Channels So far,we have considered only low-pass channels for which H(f)is large only for a frequency band around f =0.Most physical channels do not fall into this category, and instead have the property that (H(f)is significantly nonzero only within some frequency band f<If<f2,where f>0.These channels are called bandpass channels and many of them have the property that H(0)=0.A channel or waveform with H(0)=0 is said to have no dc component.From Eq.(2.3),it can be seen that this implies thathdt.The impulse response for these channels fluctuates H'{f) 1-H'{f'+W) H'(f'-W) -W 0 W Figure 2.7 Frequency response H(f)that satisfies the Nyquist criterion for no inter- symbol interference.Note that the response has odd symmetry around the point f=W
Sec. 2.2 The Physical Layer: Channels and Modems 47 since the pulse shape can be viewed as arising from filtering sample impulses. If the combination of the pulse-shaping filter at the input with the channel filter and equalizer filter has the (sinx)/x shape of Eq. (2.11) (or, equivalently, a combined frequency response equal to the ideal low-pass filter response), the output samples will re-create the input samples. The ideal low-pass filter and the (sinx)/x pulse shape suggested by the sampling theorem are nice theoretically but not very practical. What is required in practice is a more realistic filter for which the sample values at the output replicate those at the input (i.e., no intersymbol interference) while the high-frequency noise is filtered out. An elegant solution to this problem was given in a classic paper by Nyquist [Nyq28]. He showed that intersymbol interference is avoided by using a filter H'(f) with odd symmetry at the band edge; that is, H'(j +W) = 1 - H'(j - W) for IfI W, and H'(j) = 0 for IfI > 2W (see Fig. 2.7). The filter H'(j) here is the composite of the pulse-shaping filter at the transmitter, the channel filter, and the equalizer filter. In practice, such a filter usually cuts off rather sharply around f = W to avoid high-frequency noise, but ideal filtering is not required, thereby allowing considerable design flexibility. It is important to recognize that the sampling theorem specifies the number of samples per second that can be utilized on a low-pass channel, but it does not specify how many bits can be mapped into one sample. For example, two bits per sample could be achieved by the mapping 11 -* 3, 10 -* 1,00 -* -1, and 01 -* -3. As discussed later, it is the noise that limits the number of bits per sample. 2.2.4 Bandpass Channels So far, we have considered only low-pass channels for which IH(j)1 is large only for a frequency band around f = O. Most physical channels do not fall into this category, and instead have the property that I(H(j)1 is significantly nonzero only within some frequency band f1 IfI 12, where h > O. These channels are called bandpass channels and many of them have the property that H(O) = O. A channel or waveform with H(O) = 0 is said to have no dc component. From Eq. (2.3), it can be seen that this implies that J~= h(t)dt = O. The impulse response for these channels fluctuates -W o f ---+- 1 - H'(f' + W) W Figure 2.7 Frequency response H'(j) that satisfies the Nyquist criterion for no intersymbol interference. Note that the response has odd symmetry around the point f = ~V
48 Point-to-Point Protocols and Links Chap.2 h(t) 222222222 Figure 2.8 Impulse response h(t)for which H(f)=0 for f =0.Note that the area over which h(t)is positive is equal to that over which it is negative. around 0,as illustrated in Fig.2.8;this phenomenon is often called ringing.This type of impulse response suggests that the NRZ code is not very promising for a bandpass channel. To avoid the foregoing problems,most modems for bandpass channels either di- rectly encode digital data into signals with no dc component or else use modulation techniques.The best known direct encoding of this type is Manchester coding (see Fig.2.9).As seen from the figure,the signals used to represent 0 and 1 each have no dc component and also have a transition (either I to-1 or-I to 1)in the middle of each signaling interval.These transitions simplify the problem of timing recovery at the receiver (note that with the NRZ code,timing recovery could be difficult for a long run of 0's or 1's even in the absence of filtering and noise).The price of eliminating dc components in this way is the use of a much broader band of frequencies than required. Manchester coding is widely used in practice,particularly in the Ethernet system and the corresponding IEEE 802.3 standard described in Section 4.5.There are many other direct encodings,all of a somewhat ad hoc nature,with varying frequency characteristics and no de component. 2.2.5 Modulation One of the simplest modulation techniques is amplitude modulation (AM).Here a signal waveform s(t)(called a baseband signal)is generated from the digital data as before, say by the NRZ code.This is then multiplied by a sinusoidal carrier,say cos(2 fot),to generate a modulated signal s(t)cos(2 fot).It is shown in Problem 2.6 that the frequency representation of this modulated signal is [S(f-fo)+S(f+fo)1/2 (see Fig.2.10).At the receiver,the modulated signal is again multiplied by cos(2 fot),yielding a received signal r(t)=s(t)cos(2πfot) s()s()cos(4r fot) (2.12) 2 2 n Figure 2.9 Manchester coding.A binary I is mapped into a positive pulse followed by a negative pulse.and a binary 0 is mapped into a negative pulse followed by a positive pulse.Note the transition in the middle of each signal interval
48 Point-to-Point Protocols and Links Chap. 2 r Figure 2.8 Impulse response h(t) for which H(f) = 0 for f = O. Note that the area over which h(t) is positive is equal to that over which it is negative. around 0, as illustrated in Fig. 2.8; this phenomenon is often called ringing. This type of impulse response suggests that the NRZ code is not very promising for a bandpass channel. To avoid the foregoing problems, most modems for bandpass channels either directly encode digital data into signals with no dc component or else use modulation techniques. The best known direct encoding of this type is Manchester coding (see Fig. 2.9). As seen from the figure, the signals used to represent 0 and I each have no dc component and also have a transition (either I to -lor -I to I) in the middle of each signaling interval. These transitions simplify the problem of timing recovery at the receiver (note that with the NRZ code, timing recovery could be difficult for a long run of O's or l's even in the absence of filtering and noise). The price of eliminating dc components in this way is the use of a much broader band of frequencies than required. Manchester coding is widely used in practice, particularly in the Ethernet system and the corresponding IEEE 802.3 standard described in Section 4.5. There are many other direct encodings, all of a somewhat ad hoc nature, with varying frequency characteristics and no dc component. 2.2.5 Modulation One of the simplest modulation techniques is amplitude modulation (AM). Here a signal waveform s(t) (called a baseband signal) is generated from the digital data as before, say by the NRZ code. This is then multiplied by a sinusoidal carrier, say cos(27rJot), to generate a modulated signal s(t) cos(27rfat). It is shown in Problem 2.6 that the frequency representation of this modulated signal is [S(f - fa) + S(f + fo)]/2 (see Fig. 2.10). At the receiver, the modulated signal is again multiplied by cos(27rfat), yielding a received signal r(t) = s(t) cos2 (27rfat) s(t) s(t) cos(47rfat) =2+ 2 (2.12) o o o t-. Figure 2.9 Manchester coding. A binary I is mapped into a positive pulse followed by a negative pulse. and a binary 0 is mapped into a negative pulse followed by a positive pulse. Note the transition in the middle of each signal interval
Sec.2.2 The Physical Layer:Channels and Modems 49 s(f) SIf+fo) s(f-fp) -f0 0 Figure 2.10 Amplitude modulation.The frequency characteristic of the waveform s(t) is shifted up and down by fo in frequency. The high-frequency component,at 2fo,is then filtered out at the receiver,leaving the demodulated signal s(t)/2,which is then converted back to digital data.In practice, the incoming bits are mapped into shorter pulses than those of NRZ and then filtered somewhat to remove high-frequency components.It is interesting to note that Manchester coding can be viewed as AM in which the NRZ code is multiplied by a square wave. Although a square wave is not a sinusoid,it can be represented as a sinusoid plus odd harmonics;the harmonics serve no useful function and are often partly filtered out by the channel. AM is rather sensitive to the receiver knowing the correct phase of the carrier. For example,if the modulated signal s(t)cos(2 fot)were multiplied by sin(2 fot), the low-frequency demodulated waveform would disappear.This suggests,however, the possibility of transmitting twice as many bits by a technique known as quadrature amplitude modulation (QAM).Here,the incoming bits are mapped into two baseband signals,si(t)and s2(t).Then si(t)is multiplied by cos(2 fot)and s2(t)by sin(2 fot); the sum of these products forms the transmitted QAM signal (see Fig.2.11).The received waveform is separately multiplied by cos(2fot)and sin(2fot)The first multiplication (after filtering out the high-frequency component)yields s(t)/2,and the second yields s2(t)/2. QAM is widely used in high-speed modems for voice-grade telephone channels. These channels have a useful bandwidth from about 500 to 2900 Hz (hertz,i.e.,cycles per second),and the carrier frequency is typically about 1700 Hz.The signals si(t) and s2(t)are usually low-pass limited to 1200 Hz,thus allowing 2400 sample inputs per second for each baseband waveform.Adaptive equalization is usually used at the receiver (either before or after demodulation)to compensate for variations in the channel frequency response. The next question is how to map the incoming bits into the low-pass waveforms s(t)and s2(t).The pulse shape used in the mapping is not of central importance;it will be filtered anyway,and the equalizer will compensate for its shape automatically in attempting to eliminate intersymbol interference.Thus,what is important is mapping the input bits into sample amplitudes of s(t)and s2(t).One could map one bit into a sample of s1,mapping I into +1 and 0 into-1,and similarly map a second bit into a sample of s2.This is shown in Fig.2.12(a),viewed as a mapping from two bits into two amplitudes
Sec. 2.2 The Physical Layer: Channels and Modems 49 J \ o o f / ~(f) 1 \:5(f- fo} f- Figure 2.10 Amplitude modulation. The frequency characteristic of the waveform s(t) is shifted up and down by io in frequency. The high-frequency component, at 2fo, is then filtered out at the receiver, leaving the demodulated signal s(t)j2, which is then converted back to digital data. In practice, the incoming bits are mapped into shorter pulses than those of NRZ and then filtered somewhat to remove high-frequency components. It is interesting to note that Manchester coding can be viewed as AM in which the NRZ code is multiplied by a square wave. Although a square wave is not a sinusoid, it can be represented as a sinusoid plus odd hannonics; the harmonics serve no useful function and are often partly filtered out by the channel. AM is rather sensitive to the receiver knowing the correct phase of the carrier. For example, if the modulated signal set) cos(21ffat) were multiplied by sin(27rfot), the low-frequency demodulated wavefonn would disappear. This suggests, however, the possibility of transmitting twice as many bits by a technique known as quadrature amplitude modulation (QAM). Here, the incoming bits are mapped into two baseband signals, SI(t) and S2(t). Then Sj(t) is multiplied by cos(2n1ot) and S2(t) by sin(21ffot); the sum of these products fonns the transmitted QAM signal (see Fig. 2.11). The received wavefonn is separately multiplied by cos(2n fot) and sin(2n fot) The first multiplication (after filtering out the high-frequency component) yields 051 (t)j2, and the second yields s2(t)j2. QAM is widely used in high-speed modems for voice-grade telephone channels. These channels have a useful bandwidth from about 500 to 2900 Hz (hertz, i.e., cycles per second), and the carrier frequency is typically about 1700 Hz. The signals Sj(t) and 82(t) are usually low-pass limited to 1200 Hz, thus allowing 2400 sample inputs per second for each baseband wavefonn. Adaptive equalization is usually used at the receiver (either before or after demodulation) to compensate for variations in the channel frequency response. The next question is how to map the incoming bits into the low-pass wavefonns 81 (t) and 82(t). The pulse shape used in the mapping is not of central importance; it will be filtered anyway, and the equalizer will compensate for its shape automatically in attempting to eliminate intersymbol interference. Thus, what is important is mapping the input bits into sample amplitudes of 8) (t) and S2(t). One could map one bit into a sample of 81, mapping I into +I and 0 into -1, and similarly map a second bit into a sample of 052. This is shown in Fig. 2.12(a), viewed as a mapping from two bits into two amplitudes
50 Point-to-Point Protocols and Links Chap.2 cos(2xfot) Puise shape Modulated kbits Bits to waveform samples Pulse shape sin(2mfot) (a)Modulator cos(2mfot) Carrier rec. Timing rec. Adaptive Samples equalizer to bits sin(2mt) (b)Demodulator Figure 2.11 Quadrature amplitude modulation.(a)Each sample period,bits enter the modulator,are converted into quadrature amplitudes,are then modulated by sine and cosine functions,respectively,and are then added.(b)The reverse operations take place at the demodulator. Similarly,for any given integer k,one can map k bits into two amplitudes.Each of the 2%combinations of k bits map into a different amplitude pair.This set of 2k amplitude pairs in a mapping is called a signal constellation (see Fig.2.12 for a number of examples).The first two constellations in Fig.2.12 are also referred to as phase-shift keying(PSK),since they can be viewed as simply changing the phase of the carrier rather than changing its amplitude.For voice-grade modems,where W=2400 Hz,k=2 [as in Fig.2.12(a)]yields 4800 bits per second (bps),whereas =4 [as in Fig.2.12(d)] yields 9600 bps. To reduce the likelihood of noise causing one constellation point to be mistaken for another,it is desirable to place the signal points as far from each other as possible subject to a constraint on signal power (i.e.,on the mean-squared distance from origin to constellation point,taken over all constellation points).It can be shown that the constellation in Fig.2.12(c)is preferable in this sense to that in Fig.2.12(b)for =3. Modems for voice-grade circuits are also available at rates of 14,400 and 19,200 bps.These use correspondingly larger signal constellations and typically also use error- correction coding.It is reasonable to ask at this point what kind of limit is imposed on data rate by channels subject to bandwidth constraints and noise.Shannon [Sha48]has
50 Point-to-Point Protocols and Links cos(2rrfot) Chap. 2 I k bits Bits to samples Modulated waveform sin(2rrfot) (a) Modu lator (b) Demodu lator Adaptive equalizer Samples to bits Figure 2.11 Quadrature amplitude modulation. (a) Each sample period, k bits enter the modulator, are converted into quadrature amplitudes, are then modulated by sine and cosine functions, respectively, and are then added. (b) The reverse operations take place at the demodulator. Similarly, for any given integer k, one can map k bits into two amplitudes. Each of the 2k combinations of k bits map into a different amplitude pair. This set of 2k amplitude pairs in a mapping is called a signal constellation (see Fig. 2.12 for a number of examples). The first two constellations in Fig. 2.12 are also referred to as phase-shift keying (PSK), since they can be viewed as simply changing the phase of the carrier rather than changing its amplitude. For voice-grade modems, where W = 2400 Hz, k = 2 [as in Fig. 2.12(a)] yields 4800 bits per second (bps), whereas k = 4 [as in Fig. 2.12(d)] yields 9600 bps. To reduce the likelihood of noise causing one constellation point to be mistaken for another, it is desirable to place the signal points as far from each other as possible subject to a constraint on signal power (i.e., on the mean-squared distance from origin to constellation point, taken over all constellation points). It can be shown that the constellation in Fig. 2.12(c) is preferable in this sense to that in Fig. 2.12(b) for k = 3. Modems for voice-grade circuits are also available at rates of 14,400 and 19,200 bps. These use correspondingly larger signal constellations and typically also use errorcorrection coding. It is reasonable to ask at this point what kind of limit is imposed on data rate by channels subject to bandwidth constraints and noise. Shannon [Sha48] has
Sec.2.2 The Physical Layer:Channels and Modems 51 2 a 6 Figure 2.12 Signal constellations for ● QAM.Part (a)maps two binary digits into a quadrature amplitude sample.Parts (b) and (c)each map three binary digits,and part (d)maps four binary digits.Parts (a) and (b)also can be regarded as phase-shift (c) (d) keying. shown that the capacity (i.e.,the maximum reliable data rate in bps)of such a channel is given by 1+Now (2.13) where W is the available bandwidth of the channel (essentially 2400 Hz for voice-grade telephone),S is the allowable signal power as seen by the receiver,and NoW is the noise power within the bandwidth W seen by the receiver (i.e.,No is the noise power per unit bandwidth,assumed uniformly distributed over W). This is a deep and subtle result,and the machinery has not been developed here even to state it precisely,let alone to justify its validity.The description of QAM, however,at least provides some intuitive justification for the form of the result.The available sampling rate 1/T is equal to W by the sampling theorem [recall that W is twice the low-pass bandwidth of s(t)and s2(t)].Since k bits are transmitted per sample, Wk bps can be sent.Next note that some minimum area must be provided around each point in the signal constellation to allow for noise,and this area is proportional to NoW. Also,the total area occupied by the entire constellation is proportional to the signal power S.Thus,2,the number of points in the constellation,should be proportional to S/(NoW),making k proportional to log2IS/(NoW)]. Communication engineers usually express the signal-to-noise ratio [i.e.,S/(NoW)] in terms of decibels (dB),where the number of dB is defined as 10logolS/(NoW)]. Thus,k,the number of bits per quadrature sample,is proportional to the signal-to-noise ratio in dB.The additional 1 in Eq.(2.13)is harder to explain:note,however,that it is required to keep the capacity positive when the signal-to-noise ratio is very small.The intuitive argument above would make it appear that the error probability depends on the spacing of points in the constellation and thus on k.The beauty of Shannon's theorem
Sec. 2.2 The Physical Layer: Channels and Modems 51 52 01 11 • • • • 51 • • • • 00 10 (a) (b) • • • • • • • • • • Figure 2.12 Signal constellations for • • • • • • QAM. Part (a) maps two binary digits into a quadrature amplitude sample. Parts (b) • and (e) each map three binary digits. and • • • part (d) maps four binary digits. Parts (a) and (b) also can be regarded as phase-shift lei (dl keying. c = W logz (1 + ~) (2.13) NoW shown that the capacity (i.e., the maximum reliable data rate in bps) of such a channel is given by where W is the available bandwidth of the channel (essentially 2400 Hz for voice-grade telephone), S is the allowable signal power as seen by the receiver, and NoW is the noise power within the bandwidth W seen by the receiver (i.e., No is the noise power per unit bandwidth, assumed unifonnly distributed over W). This is a deep and subtle result, and the machinery has not been developed here even to state it precisely, let alone to justify its validity. The description of QAM, however, at least provides some intuitive justification for the fonn of the result. The available sampling rate 1IT is equal to W by the sampling theorem [recall that W is twice the low-pass bandwidth of 51 (t) and 5z(t)]. Since k bits are transmitted per sample, W k bps can be sent. Next note that some minimum area must be provided around each point in the signal constellation to allow for noise, and this area is proportional to NoW. Also, the total area occupied by the entire constellation is proportional to the signal power S. Thus, 2k , the number of points in the constellation, should be proportional to SI(NoW), making k proportional to logz[SI(NoW)]. Communication engineers usually express the signal-to-noise ratio [i.e., S I(NoW)] in tenns of decibels (dB), where the number of dB is defined as 1OIoglO[SI(NoW)]. Thus, k, the number of bits per quadrature sample, is proportional to the signal-to-noise ratio in dB. The additional 1 in Eq. (2.13) is harder to explain; note, however, that it is required to keep the capacity positive when the signal-to-noise ratio is very small. The intuitive argument above would make it appear that the error probability depends on the spacing of points in the constellation and thus on k. The beauty of Shannon's theorem