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# Pulse Shaping and Receive Filtering

Module by: William Sethares, Andy Klein. E-mail the authors

See first that the design is wise and just: that ascertained, pursue it resolutely; do not for one repulse forego the purpose that you resolved to effect.

— William Shakespeare

When the message is digital, it must be converted into an analog signal in order to be transmitted. This conversion is done by the “transmit” or “pulse-shaping” filter, which changes each symbol in the digital message into a suitable analog pulse. After transmission, the “receive” filter assists in recapturing the digital values from the received pulses. This chapter focuses on the design and specification of these filters.

The symbols in the digital input sequence w(kT)w(kT) are chosen from a finite set of values. For instance, they might be binary ±1±1, or they may take values from a larger set such as the four-level alphabet ±1±1, ±3±3. As suggested in Figure 1, the sequence w(kT)w(kT) is indexed by the integer kk, and the data rate is one symbol every TT seconds. Similarly, the output m(kT)m(kT) assumes values from the same alphabet as w(kT)w(kT) and at the same rate. Thus the message is fully specified at times kTkT for all integers kk. But what happens between these times, between kTkT and (k+1)T(k+1)T? The analog modulation of Chapter (Reference) operates continuously, and some values must be used to fill in the digital input between the samples. This is the job of the pulse shaping filter: to turn a discrete-time sequence into an analog signal.

Each symbol w(kT)w(kT) of the message initiates an analog pulse that is scaled by the value of the signal. The pulse progresses through the communications system, and if all goes well, the output (after the decision) should be the same as the input, although perhaps with some delay. If the analog pulse is wider than the time between adjacent symbols, the outputs from adjacent symbols may overlap, a problem called intersymbol interference, which is abbreviated ISI. A series of examples in "Intersymbol Interference" shows how this happens, and the eye diagram is used in "Eye Diagrams" to help visualize the impact of ISI.

What kinds of pulses minimize the ISI? One possibility is to choose a shape that is one at time kTkT and zero at mTmT for all mkmk. Then the analog waveform at time kTkT contains only the value from the desired input symbol, and no interference from other nearby input symbols. These are called Nyquist pulses in "Nyquist Pulses". Yes, this is the same fellow who brought us the Nyquist sampling theorem and the Nyquist frequency.

Besides choosing the pulse shape, it is also necessary to choose a receive filter that helps decode the pulses. The received signal can be thought of as containing two parts: one part is due to the transmitted signal and the other part is due to the noise. The ratio of the powers of these two parts is a kind of signal-to-noise ratio that can be maximized by choice of the pulse shape. This is discussed in "Matched Filtering". The chapter concludes in "Matched Transmit and Receive Filters" by considering pulse shaping and receive filters that do both: provide a Nyquist pulse and maximize the signal-to-noise ratio.

The transmit and receive filter designs rely on the assumption that all other parts of the system are working well. For instance, the modulation and demodulation blocks have been removed from Figure 1, and the assumption is that they are perfect: the receiver knows the correct frequency and phase of the carrier. Similarly, the downsampling block has been removed, and the assumption is that this is implemented so that the decision device is a fully synchronized sampler and quantizer. Chapter (Reference) examines methods of satisfying these synchronization needs, but for now, they are assumed to be met. In addition, the channel is assumed benign.

## Spectrum of the Pulse: Spectrum of the Signal

Probably the major reason that the design of the pulse shape is important is because the shape of the spectrum of the pulse shape dictates the spectrum of the whole transmission. To see this, suppose that the discrete-time message sequence w(kT)w(kT) is turned into the analog pulse train

w a ( t ) = k w ( k T ) δ ( t - k T ) = w ( k T ) t = k T 0 t k T w a ( t ) = k w ( k T ) δ ( t - k T ) = w ( k T ) t = k T 0 t k T
(1)

as it enters the pulse shaping filter. The response of the filter, with impulse response p(t)p(t), is the convolution

x ( t ) = w a ( t ) * p ( t ) , x ( t ) = w a ( t ) * p ( t ) ,
(2)

as suggested by Figure 1. Since the Fourier transform of a convolution is the product of the Fourier transforms (from (Reference)), it follows that

X ( f ) = W a ( f ) P ( f ) . X ( f ) = W a ( f ) P ( f ) .
(3)

Though Wa(f)Wa(f) is unknown, this shows that X(f)X(f) can have no energy at frequencies where P(f)P(f) vanishes. Whatever the spectrum of the message, the transmission is directly scaled by P(f)P(f). In particular, the support of the spectrum X(f)X(f) is no larger than the support of the spectrum P(f)P(f).

As a concrete example, consider the pulse shape used in Chapter (Reference), which is the “blip” function shown in the top plot of Figure 2. The spectrum of this pulse shape can readily be calculated using freqz, and this is shown in the bottom plot of Figure 2. It is a kind of mild lowpass filter. The following code generates a sequence of N 4-PAM symbols, and then carries out the pulse shaping using the filter command.

N=1000; m=pam(N,4,5);                    % 4-level signal of length N
M=10; mup=zeros(1,N*M); mup(1:M:N*M)=m;  % oversample by M
ps=hamming(M);                           % blip pulse shape of width M
x=filter(ps,1,mup);                      % convolve pulse shape with data

The program pulsespec.m represents the “continuous-time” or analog signal by oversampling both the data sequence and the pulse shape by a factor of M. This technique was discussed in (Reference), where an “analog” sine wave sine100hzsamp.m was represented digitally at two sampling intervals, a slow symbol interval T=MTsT=MTs and a faster rate (shorter interval) TsTs representing the underlying analog signal. The pulse shape ps is a blip created by the hamming function, and this is also oversampled at the same rate. The convolution of the oversampled pulse shape and the oversampled data sequence is accomplished by the filter command. Typical output is shown in the top plot of Figure 3, which shows the “analog” signal over a time interval of about 25 symbols. Observe that the individual pulse shapes are clearly visible, one scaled blip for each symbol.

The spectrum of the output x is plotted in the bottom of Figure 3. As expected from the previous discussion, the spectrum X(f)X(f) has the same contour as the spectrum of the individual pulse shape in Figure 2.

## Intersymbol Interference

There are two situations when adjacent symbols may interfere with each other: when the pulse shape is wider than a single symbol interval TT, and when there is a nonunity channel that “smears” nearby pulses, causing them to overlap. Both of these situations are called intersymbol interference (ISI). Only the first kind of ISI will be considered in this chapter; the second kind is postponed until Chapter (Reference). Before tackling the general setup, this section provides an instructive example.

[ISI Caused by an Overly Wide Pulse Shape]

Suppose that the pulse shape in pulsespec.m is stretched so that its width is 3T3T. This triple-wide Hamming pulse shape is shown in Figure 4, along with its spectrum. Observe that the spectrum has (roughly) one-third the null-to-null bandwidth of the single-symbol wide Hamming pulse. Since the width of the spectrum of the transmitted signal is dictated by the width of the spectrum of the pulse, this pulse shape is three times as parsimonious in its use of bandwidth. More FDM users can be active at the same time.

As might be expected, this boon has a price. Figure 5 shows the output of the pulse shaping filter over a time of about 25 symbols. There is no longer a clear separation of the pulse corresponding to one data point from the pulses of its neighbors. The transmission is correspondingly harder to properly decode. If the ISI caused by the overly wide pulse shape is too severe, symbol errors may occur.

Thus, there is a tradeoff. Wider pulse shapes can occupy less bandwidth, which is always a good thing. On the other hand, a pulse shape like the Hamming blip does not need to be very many times wider before it becomes impossible to decipher the data because the ISI has become too severe. How much wider can it be without causing symbol errors? The next section provides a way of picturing ISI that answers this question. Subsequent sections discuss the practical issue of how such ISI can be prevented by a better choice of pulse shape. Yes, there are good pulse shapes that are wider than TT.

### Exercise 1

Modify pulsespec.m to reproduce Figure 4 and Figure 5 for the double-wide pulse shape.

### Exercise 2

Modify pulsespec.m to examine what happens when Hamming pulse shapes of width 4T4T, 6T6T, and 10T10T are used. What is the bandwidth of the resulting transmitted signals? Do you think it is possible to recover the message from the received signals? Explain.

## Eye Diagrams

While the differences between the pulse shaped sequences in Figure 3 and Figure 5 are apparent, it is difficult to see directly whether the distortions are serious; that is, whether they cause errors in the reconstructed data (i.e., the hard decisions) at the receiver. After all, if the reconstructed message is the same as the real message, then no harm has been done, even if the values of the received analog waveform are not identical. This section uses a visualization tool called eye diagrams that show how much smearing there is in the system, and whether symbol errors will occur. Eye diagrams were encountered briefly in Chapter (Reference) (refer back to (Reference)) when visualizing how the performance of the idealized system degraded when various impairments were added.

Imagine an oscilloscope that traces out the received signal, with the special feature that it is set to retrigger or restart the trace every nTnT seconds without erasing the screen. Thus the horizontal axis of an eye diagram is the time over which nn symbols arrive, and the vertical axis is the value of the received waveform. In the ideal case, the trace begins with nn pulses, each of which is a scaled copy of p(t)p(t). Then the n+1n+1st to 2n2nth pulses arrive, and overlay the first nn, though each is scaled according to its symbol value. When there is noise, channel distortion, and timing jitter, the overlays will differ.

As the number of superimposed traces increases, the eye diagram becomes denser, and gives a picture of how the pulse shape, channel, and other factors combine to determine the reliability of the recovered message. Consider the n=2n=2 symbol eye diagram shown in Figure 6. In this figure, the message is taken from the 4-PAM alphabet ±1±1±3±3, and the Hamming pulse shape is used. The center of the “eye” gives the best times to sample, since the openings (i.e., the difference between the received pulse shape when the data value is -1-1 and the received pulse shape when the data value is 1, or between the received pulse shape when the data value is 1 and the received pulse shape when the data value is 3) are the largest. The width marked “sensitivity to timing error” shows the range of time over which the samples quantize correctly. The noise margin is the smallest vertical distance between the bands, and is proportional to the amount of additive noise that can be resisted by the system without reporting erroneous values.

Thus, eye diagrams such as Figure 6 give a clear picture of how good (or how bad) a pulse shape may be. Sometimes the smearing in this figure is so great that the open segment in the center disappears. The eye is said to be closed, and this indicates that a simple quantizer (slicer) decision device will make mistakes in recovering the data stream. This is not good!

For example, reconsider the 4-PAM example of the previous section that used a triple-wide Hamming pulse shape. The eye diagram is shown in Figure 7. No noise was added when drawing this picture. In the top two plots there are clear regions about the symbol locations where the eye is open. Samples taken in these regions will be quantized correctly, though there are also regions where mistakes will occur. The other plots show the closed eye diagrams using 3T3T-wide and 5T5T-wide Hamming pulse shapes. Symbol errors will inevitably occur, even if all else in the system is ideal. All of the measures (noise margin, sensitivity to timing, and the distortion at zero crossings) become progressively worse, and ever smaller amounts of noise can cause decision errors.

The following code draws eye diagrams for the pulse shapes defined by the variable ps. As in the pulse shaping programs of the previous section, the N binary data points are oversampled by a factor of M and the convolution of the pulse shapes with the data uses the filter command. The reshape(x,a,b) command changes a vector x of size a*b into a matrix with a rows and b columns, which is used to segment x into b overlays, each a samples long. This works smoothly with the Matlabplot function.

N=1000; m=pam(N,2,1);                   % random signal of length N
M=20; mup=zeros(1,N*M); mup(1:M:N*M)=m; % oversampling by factor of M
ps=hamming(M);                          % hamming pulse of width M
x=filter(ps,1,mup);                     % convolve pulse shape with mup
neye=5; c=floor(length(x)/(neye*M));    % number of eyes to plot
xp=x(end-neye*M*c+1:end);               % dont plot transients at start
plot(reshape(xp,neye*M,c))              % overlay in groups of size neye

Typical output of eyediag.m is shown in Figure 8. The rectangular pulse shape in the top plot uses ps=ones(1,M), the Hamming pulse shape in the middle uses ps=hamming(M), and the bottom plot uses a truncated sinc pulse shape ps=srrc(L,0,M) for L=10 that is normalized so that the largest value is one. The rectangular pulse is insensitive to timing errors, since sampling almost anywhere (except right at the transition boundaries) will return the correct values. The Hamming pulse shape has a wide eye, but may suffer from a loss of SNR if the samples are taken far from the center of the eye. Of the three, the sinc pulse is the most sensitive, since it must be sampled near the correct instants or erroneous values will result.

### Exercise 3

Modify eyediag.m so that the data sequence is drawn from the alphabet ±1±1, ±3±3, ±5±5. Draw the appropriate eye diagram for the rectangular, Hamming, and sinc pulse shapes.

### Exercise 4

Modify eyediag.m to add noise to the pulse shaped signal x. Use the Matlab command v*randn for different values of v. Draw the appropriate eye diagrams. For each pulse shape, how large can v be and still have the eye remain open?

### Exercise 5

Combine the previous two Exercises. Modify eyediag.m as in Exercise 3 so that the data sequence is drawn from the alphabet ±1±1, ±3±3, ±5±5. Add noise, and answer the same question as in Exercise 4. Which alphabet is more susceptible to noise?

### Exercise 6

TRUE or FALSE: For two rectangular impulse responses with the same maximum magnitude but different time widths with T1>T2T1>T2, the half-power bandwidth of the frequency response of the pulse with width T1T1 exceeds that of the pulse with width T2T2.

### Exercise 7

TRUE or FALSE: For the PAM baseband signals created by a rectangular pulse and a triangular pulse with the same time width and the same maximum amplitude, the half-power bandwidth of the sequence using the triangular pulse exceeds that of the rectangular pulse.

### Exercise 8

Exercise (Reference) asked the following question:

TRUE or FALSE: The flatter the top of the pulse shape, the less sensitive the receiver is to small timing offsets.

In the absence of noise and without matched filtering, this is TRUE. Describe a noisy situation and a matched filtering that might cause the answer to be FALSE.

### Exercise 9

Consider the baseband communication system in Figure 9. The difference equation relating the symbols m[k]m[k] to the TT-spaced equalizer input u[k]u[k] for the chosen baud-timing factor ϵϵ is

u [ k ] = 0 . 04 m [ k - ρ ] + 1 . 00 m [ k - 1 - ρ ] + 0 . 60 m [ k - 2 - ρ ] + 0 . 38 m [ k - 3 - ρ ] u [ k ] = 0 . 04 m [ k - ρ ] + 1 . 00 m [ k - 1 - ρ ] + 0 . 60 m [ k - 2 - ρ ] + 0 . 38 m [ k - 3 - ρ ]
(4)

where ρρ is a nonnegative integer. The finite-impulse-response equalizer (filter) is described by the difference equation

y [ k ] = u [ k ] + α u [ k - 1 ] . y [ k ] = u [ k ] + α u [ k - 1 ] .
(5)
1. Suppose α=-0.4α=-0.4 and the message source is binary ±1±1. Is the system from the source symbols m[k]m[k] to the equalizer output y[k]y[k] open-eye? Justify your answer.
2. If the message source is 4-PAM (±1,±3±1,±3), can the system from m[k]m[k] to the equalizer output y[k]y[k] be made open-eye by selection of αα? If so, provide a successful value of αα. If not, explain.

It is now easy to experiment with various pulse shapes. pulseshape2.m applies a sinc shaped pulse to a random binary sequence. Since the sinc pulse extends infinitely in time (both backward and forward), it cannot be represented exactly in the computer (or in a real communication system) and the parameter L specifies the duration of the sinc, in terms of the number of symbol periods.

N=2000; m=pam(N,2,1);                   % 2-PAM signal of length N
M=10; mup=zeros(1,N*M); mup(1:M:N*M)=m;  % oversample by M
L=10; ps=srrc(L,0,M);                   % sinc pulse shape 2L symbols wide
sc=sum(ps)/M; x=filter(ps/sc,1,mup);    % convolve pulse shape with data

Figure 10 plots the output of pulseshape2.m. The top figure shows the pulse shape while the bottom plot shows the “analog” pulse-shaped signal x(t)x(t) over a duration of about 25 symbols. The function srrc.m first appeared in the discussion of interpolation in (Reference) (and again in Exercise (Reference)), and is used here to generate the sinc pulse shape. The sinc function that srrc.m produces is actually scaled, and this effect is removed by normalizing with the variable sc. Changing the second input argument from beta=0 to other small positive numbers changes the shape of the curve, each with a “sinc-like” shape called a square root raised cosine. This will be discussed in greater detail in Sections "Nyquist Pulses" and "Matched Transmit and Receive Filters". Typing help srrc in Matlab gives useful information on using the function.

Observe that, though the signal oscillates above and below the ±1±1 lines, there is no intersymbol interference. When using the Hamming pulse as in Figure 3, each binary value was clearly delineated. With the sinc pulse of Figure 10, the analog waveform is more complicated. But at the correct sampling instances, it always returns to ±1±1 (the horizontal lines at ±1±1 are drawn to help focus the eye on the crossing times). Unlike the TT-wide Hamming shape, the signal need not return near zero with each symbol.

### Exercise 10

In pulseshape2.m, examine the effect of using different oversampling rates M. Try M= 1, 5, 100.

### Exercise 11

Change pulseshape2.m so that the data sequence is drawn from the alphabet ±1,±3,±5±1,±3,±5. Can you visually identify the correct values in the pulse shaped signal?

### Exercise 12

In pulseshape2.m, examine the effect of using sinc approximations of different lengths L. Try L= 1, 5, 100, 1000.

### Exercise 13

In pulseshape2.m, examine the effect of adding noise to the received signal x. Try Matlab commands randn and rand. How large can the noise be and still allow the data to be recognizable?

### Exercise 14

The goal is to design a frequency-division multiplexed (FDM) system with a square root raised cosine as the transmitter pulse shape. The symbol period is TT = 137 msec. The design uses T/4T/4 sampling, pulse lengths of 8T8T, and a rolloff factor of 0.9, but it does not work, since only three modulated carrier signals fit into the alloted bandwidth without multiuser interference. Five are needed. What parameters in the design would you change and why?

### Exercise 15

Using the code from Exercise 11, examine the effects of adding noise in pulseshape2.m. Does the same amount of noise in the 6-level data have more or less effect than in the 2-level data?

### Exercise 16

Modify pulseshape2.m to include the effect of a nonunity channel. Try both a highpass channel and a bandpass channel. Which appears worse? What are reasonable criteria for “better” and “worse” in this context?

### Exercise 17

A Matlab question: In pulseshape2.m, examine the effect of using the filtfilt command for the convolution instead of the filter command. Can you figure out why the results are different?

### Exercise 18

Another Matlab question: In pulseshape2.m, examine the effect of using the conv command for the convolution instead of the filter command. Can you figure out how to make this work?

## Nyquist Pulses

Consider a multilevel signal drawn from a finite alphabet with values w(kT)w(kT), where TT is the sampling interval. Let p(t)p(t) be the impulse response of the linear filter representing the pulse shape. The signal just after pulse shaping is

x ( t ) = w a ( t ) * p ( t ) , x ( t ) = w a ( t ) * p ( t ) ,
(6)

where wa(t)wa(t) is the pulse train signal Equation 1.

The corresponding output of the received filter is

y ( t ) = w a ( t ) * p ( t ) * h c ( t ) * h R ( t ) , y ( t ) = w a ( t ) * p ( t ) * h c ( t ) * h R ( t ) ,
(7)

as depicted in Figure 1, where hc(t)hc(t) is the impulse response of the channel and hR(t)hR(t) is the impulse response of the receive filter. Let hequiv(t)=p(t)*hc(t)*hR(t)hequiv(t)=p(t)*hc(t)*hR(t) be the overall equivalent impulse response. Then the equivalent overall frequency response (i.e., F{hequiv(t)}F{hequiv(t)}) is

H equiv ( f ) = P ( f ) H c ( f ) H R ( f ) . H equiv ( f ) = P ( f ) H c ( f ) H R ( f ) .
(8)

One approach would be to attempt to choose HR(f)HR(f) so that Hequiv(f)Hequiv(f) attained a desired value (such as a pure delay) for all ff. This would be a specification of the impulse response hequiv(t)hequiv(t) at all tt, since the Fourier transform is invertible. But such a distortionless response is unnecessary, since it does not really matter what happens between samples, but only what happens at the sample instants. In other words, as long as the eye is open, the transmitted symbols are recoverable by sampling at the correct times. In general, if the pulse shape is zero at all integer multiples of kTkT but one, then it can have any shape in between without causing intersymbol interference.

The condition that one pulse does not interfere with other pulses at subsequent TT-spaced sample instants is formalized by saying that hNYQ(t)hNYQ(t) is a Nyquist pulse if there is a ττ such that

h N Y Q ( k T + τ ) = c k = 0 0 k 0 h N Y Q ( k T + τ ) = c k = 0 0 k 0
(9)

for all integers kk, where cc is some nonzero constant. The timing offset ττ in Equation 9 will need to be found by the receiver.

A rectangular pulse with time-width less than TT certainly satisfies Equation 9, as does any pulse shape that is less than TT wide. But the bandwidth of the rectangular pulse (and other narrow pulse shapes such as the Hamming pulse shape) may be too wide. Narrow pulse shapes do not utilize the spectrum efficiently. But if just any wide shape is used (such as the multiple-TT-wide Hamming pulses), then the eye may close. What is needed is a signal that is wide in time (and narrow in frequency) that also fulfills the Nyquist condition Equation 9.

One possibility is the sinc pulse

h s i n c ( t ) = sin ( π f 0 t ) π f 0 t , h s i n c ( t ) = sin ( π f 0 t ) π f 0 t ,
(10)

with f0=1/Tf0=1/T. This has the narrowest possible spectrum, since it forms a rectangle in frequency (i.e., the frequency response of a lowpass filter). Assuming that the clocks at the transmitter and receiver are synchronized so that τ=0τ=0, the sinc pulse is Nyquist because hsinc(0)=1hsinc(0)=1 and

h s i n c ( k T ) = sin ( π k ) π k = 0 h s i n c ( k T ) = sin ( π k ) π k = 0
(11)

for all integers k0k0. But there are several problems with the sinc pulse:

• It has infinite duration. In any real implementation, the pulse must be truncated.
• It is noncausal. In any real implementation, the truncated pulse must be delayed.
• The steep band edges of the rectangular frequency function Hsinc(f)Hsinc(f) are difficult to approximate.
• The sinc function sin (t)/t sin (t)/t decays slowly, at a rate proportional to 1/t1/t.

The slow decay (recall the plot of the sinc function in (Reference)) means that samples that are far apart in time can interact with each other when there are even modest clock synchronization errors.

Fortunately, it is not necessary to choose between a pulse shape that is constrained to lie within a single symbol period TT and the slowly decaying sinc. While the sinc has the smallest dispersion in frequency, there are other pulse shapes that are narrower in time and yet are only a little wider in frequency. Trading off time and frequency behaviors can be tricky. Desirable pulse shapes

1. (i): have appropriate zero crossings (i.e., they are Nyquist pulses),
2. (ii): have sloped band edges in the frequency domain, and
3. (iii): decay more rapidly in the time domain (compared with the sinc), while maintaining a narrow profile in the frequency domain.

One popular option is called the raised cosine-rolloff (or raised cosine) filter. It is defined by its Fourier transform

H R C ( f ) = 1 | f | < f 1 1 2 1 + cos π ( | f | - f 1 ) 2 f Δ , f 1 < | f | < B , 0 | f | > B H R C ( f ) = 1 | f | < f 1 1 2 1 + cos π ( | f | - f 1 ) 2 f Δ , f 1 < | f | < B , 0 | f | > B
(12)

where

• BB is the absolute bandwidth,
• f0f0 is the 6 dB bandwidth, equal to 12T12T, one half the symbol rate,
• fΔ=B-f0fΔ=B-f0, and
• f1=f0-fΔf1=f0-fΔ.

The corresponding time domain function is

h R C ( t ) = F - 1 { H R C ( f ) } = 2 f 0 sin ( 2 π f 0 t ) 2 π f 0 t cos ( 2 π f Δ t ) 1 - ( 4 f Δ t ) 2 . h R C ( t ) = F - 1 { H R C ( f ) } = 2 f 0 sin ( 2 π f 0 t ) 2 π f 0 t cos ( 2 π f Δ t ) 1 - ( 4 f Δ t ) 2 .
(13)

Define the rolloff factorβ=fΔ/f0β=fΔ/f0. Figure 11 shows the magnitude spectrum HRC(f)HRC(f) of the raised cosine filter in the bottom and the associated time response hRC(t)hRC(t) on the top, for a variety of rolloff factors. With T=12f0T=12f0, hRC(kT)hRC(kT) has a factor sin (πk)/πk sin (πk)/πk which is zero for all integer k0k0. Hence the raised cosine is a Nyquist pulse. In fact, as β0β0, hRC(t)hRC(t) becomes a sinc.

The raised cosine pulse hRC(t)hRC(t) with nonzero ββ has the following characteristics:

• zero crossings at desired times,
• band edges of HRC(f)HRC(f) that are less severe than with a sinc pulse,
• an envelope that falls off at approximately 1/|t|31/|t|3 for large tt (look at Equation 13). This is significantly faster than 1/|t|1/|t|. As the rolloff factor ββ increases from 0 to 1, the significant part of the impulse response gets shorter.

Thus, we have seen several examples of Nyquist pulses: rectangular, Hamming, sinc, and raised cosine with a variety of roll off factors. What is the general principle that distinguishes Nyquist pulses from all others? A necessary and sufficient condition for a signal v(t)v(t) with Fourier transform V(f)V(f) to be a Nyquist pulse is that the sum (over all nn) of V(f-nf0)V(f-nf0) be constant. To see this, use the sifting property of an impulse (Reference) to factor V(f)V(f) from the sum:

n = - V ( f - n f 0 ) = V ( f ) * n = - δ ( f - n f 0 ) . n = - V ( f - n f 0 ) = V ( f ) * n = - δ ( f - n f 0 ) .
(14)

Given that convolution in the frequency domain is multiplication in the time domain (Reference), applying the definition of the Fourier transform, and using the transform pair (from (Reference) with w(t)=1w(t)=1 and W(f)=δ(f)W(f)=δ(f))

F { k = - δ ( t - k T ) } = 1 T n = - δ ( f - n f 0 ) , F { k = - δ ( t - k T ) } = 1 T n = - δ ( f - n f 0 ) ,
(15)

where f0=1/Tf0=1/T, this becomes

n = - V ( f - n f 0 ) = t = - [ v ( t ) ( T k = - δ ( t - k T ) ) ] e - j 2 π f t d t = k = - T v ( k T ) e - j 2 π f k T . n = - V ( f - n f 0 ) = t = - [ v ( t ) ( T k = - δ ( t - k T ) ) ] e - j 2 π f t d t = k = - T v ( k T ) e - j 2 π f k T .
(16)

If v(t)v(t) is a Nyquist pulse, the only nonzero term in the sum is v(0)v(0), and

n = - V ( f - n f 0 ) = T v ( 0 ) . n = - V ( f - n f 0 ) = T v ( 0 ) .
(17)

Thus, the sum of the V(f-nf0)V(f-nf0) is a constant if v(t)v(t) is a Nyquist pulse. Conversely, if the sum of the V(f-nf0)V(f-nf0) is a constant, then only the DC term in Equation 16 can be nonzero, and so v(t)v(t) is a Nyquist pulse.

### Exercise 19

Write a Matlab routine that implements the raised cosine impulse response Equation 13 with rolloff parameter ββ. Hint: If you have trouble with “divide by zero” errors, imitate the code in srrc.m. Plot the output of your program for a variety of ββ. Hint 2: There is an easy way to use the function srrc.m.

### Exercise 20

Use your code from the previous exercise, along with pulseshape2.m to apply raised cosine pulse shaping to a random binary sequence. Can you spot the appropriate times to sample “by eye?”

### Exercise 21

Use the code from the previous exercise and eyediag.m to draw eye diagrams for the raised cosine pulse with rolloff parameters r=0r=0, 0.50.5, 0.90.9, 1.01.0, 5.05.0. Compare these to the eye diagrams for rectangular and sinc functions. Consider

1. Sensitivity to timing errors
2. Peak distortion
3. Distortion of zero crossings
4. Noise margin

### Exercise 22

TRUE or FALSE: The impulse response of a series combination of any αα-second-wide pulse shape filter and its matched filter form a Nyquist pulse shape for a TT-spaced symbol sequence for any TαTα.

### Exercise 23

Consider the 1.2 msec wide pulse shape p(t)p(t) shown in Figure 12.

1. Is p(t)p(t) a Nyquist pulse for the symbol period T=0.35T=0.35 msec? Justify your answer.
2. Is p(t)p(t) a Nyquist pulse for the symbol period T=0.70T=0.70 msec? Justify your answer.

### Exercise 24

Neither s1(t)s1(t) nor s2(t)s2(t) are Nyquist pulses.

1. Can the product s1(t)s2(t)s1(t)s2(t) be a Nyquist pulse? Explain.
2. Can the convolution s1(t)*s2(t)s1(t)*s2(t) be a Nyquist pulse? Explain.

Intersymbol interference occurs when data values at one sample instant interfere with the data values at another sampling instant. Using Nyquist shapes such as the rectangle, sinc, and raised cosine pulses removes the interference, at least at the correct sampling instants, when the channel is ideal. The next sections parlay this discussion of isolated pulse shapes into usable designs for the pulse shaping and receive filters.

## Matched Filtering

Communication systems must be robust to the presence of noises and other disturbances that arise in the channel and in the various stages of processing. Matched filtering is aimed at reducing the sensitivity to noise, which can be specified in terms of the power spectral density.

Consider the filtering problem in which a message signal is added to a noise signal and then both are passed through a linear filter. This occurs, for instance, when the signal g(t)g(t) of Figure 1 is the output of the pulse shaping filter (i.e., no interferers are present), the channel is the identity, and there is noise n(t)n(t) present. Assume that the noise is “white”; that is, assume that its power spectral density Pn(f)Pn(f) is equal to some constant ηη for all frequencies.

The output y(t)y(t) of the linear filter with impulse response hR(t)hR(t) can be described as the superposition of two components, one driven by g(t)g(t) and the other by n(t)n(t); that is,

y ( t ) = v ( t ) + w ( t ) , y ( t ) = v ( t ) + w ( t ) ,
(18)

where

v ( t ) = h R ( t ) * g ( t ) and w ( t ) = h R ( t ) * n ( t ) . v ( t ) = h R ( t ) * g ( t ) and w ( t ) = h R ( t ) * n ( t ) .
(19)

This is shown in block diagram form in Figure 13. In both components, the processing and the output signal are the same. The bottom diagram separates out the component due to the signal (v(kT)v(kT), which contains the message filtered through the pulse shape and the receive filter), and the component due to the noise (w(kT)w(kT), which is the noise filtered through the receive filter). The goal of this section is to find the receive filter that maximizes the ratio of the power in the signal v(kT)v(kT) to the power in the noise w(kT)w(kT) at the sample instants.

Consider choosing hR(t)hR(t) so as to maximize the power of the signal v(t)v(t) at time t=τt=τ compared with the power in w(t)w(t) (i.e., to maximize v2(τ)v2(τ) relative to the total power of the noise component w(t)w(t)). This choice of hR(t)hR(t) tends to emphasize the signal v(t)v(t) and suppress the noise w(t)w(t). The argument proceeds by finding the transfer function HR(f)HR(f) that corresponds to this hR(t)hR(t).

The total power in w(t)w(t) is

P w = - P w ( f ) d f . P w = - P w ( f ) d f .
(20)

From the inverse Fourier transform,

v ( τ ) = - V ( f ) e j 2 π f τ d f , v ( τ ) = - V ( f ) e j 2 π f τ d f ,
(21)

where V(f)=HR(f)G(f)V(f)=HR(f)G(f). Thus,

v 2 ( τ ) = - H R ( f ) G ( f ) e j 2 π f τ d f 2 . v 2 ( τ ) = - H R ( f ) G ( f ) e j 2 π f τ d f 2 .
(22)

Says that for Y(f)=HR(f)U(f)Y(f)=HR(f)U(f), Py(f)=|HR(f)|2Pu(f)Py(f)=|HR(f)|2Pu(f). Thus,

P w ( f ) = | H R ( f ) | 2 P n ( f ) = η | H R ( f ) | 2 . P w ( f ) = | H R ( f ) | 2 P n ( f ) = η | H R ( f ) | 2 .
(23)

The quantity to be maximized can now be described by

v 2 ( τ ) P w = | - H R ( f ) G ( f ) e j 2 π f τ d f | 2 - η | H R ( f ) | 2 d f . v 2 ( τ ) P w = | - H R ( f ) G ( f ) e j 2 π f τ d f | 2 - η | H R ( f ) | 2 d f .
(24)

Schwarz's inequality (Reference) says that

- a ( x ) b ( x ) d x 2 - | a ( x ) | 2 d x - | b ( x ) | 2 d x , - a ( x ) b ( x ) d x 2 - | a ( x ) | 2 d x - | b ( x ) | 2 d x ,
(25)

and equality occurs only when a(x)=kb*(x)a(x)=kb*(x). This converts Equation 24 to

v 2 ( τ ) P w - | H R ( f ) | 2 d f - | G ( f ) e j 2 π f τ | 2 d f η - | H R ( f ) | 2 d f , v 2 ( τ ) P w - | H R ( f ) | 2 d f - | G ( f ) e j 2 π f τ | 2 d f η - | H R ( f ) | 2 d f ,
(26)

which is maximized with equality when

H R ( f ) = k ( G ( f ) e j 2 π f τ ) * . H R ( f ) = k ( G ( f ) e j 2 π f τ ) * .
(27)

HR(f)HR(f) must now be transformed to find the corresponding impulse response hR(t)hR(t). Since Y(f)=X(-f)Y(f)=X(-f) when y(t)=x(-t)y(t)=x(-t), (use the frequency scaling property of Fourier transforms (Reference) with a scale factor of -1-1),

F - 1 { W * ( - f ) } = w * ( t ) F - 1 { W * ( f ) } = w * ( - t ) . F - 1 { W * ( - f ) } = w * ( t ) F - 1 { W * ( f ) } = w * ( - t ) .
(28)

Applying the time shift property (Reference) yields

F - 1 { W ( f ) e - j 2 π f T d } = w ( t - T d ) . F - 1 { W ( f ) e - j 2 π f T d } = w ( t - T d ) .
(29)

Combining these two transform pairs yields

F - 1 { ( W ( f ) e j 2 π f T d ) * } = w * ( - ( t - T d ) ) = w * ( T d - t ) . F - 1 { ( W ( f ) e j 2 π f T d ) * } = w * ( - ( t - T d ) ) = w * ( T d - t ) .
(30)

Thus, when g(t)g(t) is real,

F - 1 { k ( G ( f ) e j 2 π f τ ) * } = k g * ( τ - t ) = k g ( τ - t ) . F - 1 { k ( G ( f ) e j 2 π f τ ) * } = k g * ( τ - t ) = k g ( τ - t ) .
(31)

Observe that this filter has the following characteristics:

• This filter results in the maximum signal-to-noise ratio of v2(t)/Pwv2(t)/Pw at the time instant t=τt=τ for a noise signal with a flat power spectral density.
• Because the impulse response of this filter is a scaled time reversal of the pulse shape p(t)p(t), it is said to be “matched” to the pulse shape, and is called a “matched filter.”
• The shape of the magnitude spectrum of the matched filter HR(f)HR(f) is the same as the magnitude spectrum G(f)G(f).
• The shape of the magnitude spectrum of G(f)G(f) is the same as the shape of the frequency response of the pulse shape P(f)P(f) for a broadband m(kT)m(kT), as in "Spectrum of the Pulse: Spectrum of the Signal".
• The matched filter for any filter with an even symmetric (about some tt) time-limited impulse response is a delayed replica of that filter. The minimum delay is the upper limit of the time-limited range of the impulse response.

The following code allows hands-on exploration of this theoretical result. The pulse shape is defined by the variable ps (the default is the sinc function srrc(L,0,M) for L=10). The receive filter is analogously defined by recfilt. As usual, the symbol alphabet is easily specified by the pam subroutine, and the system operates at an oversampling rate M. The noise is specified in n, and the ratio of the powers is output as powv/poww. Observe that, for any pulse shape, the ratio of the powers is maximized when the receive filter is the same as the pulse shape (the fliplr command carries out the time reversal). This holds no matter what the noise, no matter what the symbol alphabet, and no matter what the pulse shape.

N=2^15; m=pam(N,2,1);                    % 2-PAM signal of length N
M=10; mup=zeros(1,N*M); mup(1:M:N*M)=m;  % oversample by M
L=10; ps=srrc(L,0,M);                    % define pulse shape
ps=ps/sqrt(sum(ps.^2));                  % and normalize
n=0.5*randn(size(mup));                  % noise
g=filter(ps,1,mup);                      % convolve ps with data
recfilt=srrc(L,0,M);                     % receive filter H sub R
recfilt=recfilt/sqrt(sum(recfilt.^2));   % normalize the pulse shape
v=filter(fliplr(recfilt),1,g);           % matched filter with data
w=filter(fliplr(recfilt),1,n);           % matched filter with noise
vdownsamp=v(1:M:N*M);                    % downsample to symbol rate
wdownsamp=w(1:M:N*M);                    % downsample to symbol rate
powv=pow(vdownsamp);                     % power in downsampled v
poww=pow(wdownsamp);                     % power in downsampled w
powv/poww                                % ratio

In general, when the noise power spectral density is flat (i.e., Pn(f)=ηPn(f)=η), the output of the matched filter may be realized by correlating the input to the matched filter with the pulse shape p(t)p(t). To see this, recall that the output is described by the convolution

x ( α ) = - s ( λ ) h ( α - λ ) d λ x ( α ) = - s ( λ ) h ( α - λ ) d λ
(32)

of the matched filter with the impulse response h(t)h(t). Given the pulse shape p(t)p(t) and the assumption that the noise has flat power spectral density, it follows that

h ( t ) = p ( α - t ) , 0 t T 0 , otherwise , h ( t ) = p ( α - t ) , 0 t T 0 , otherwise ,
(33)

where αα is the delay used in the matched filter. Because h(t)h(t) is zero when tt is negative and when t>Tt>T, h(α-λ)h(α-λ) is zero for λ>αλ>α and λ<α-Tλ<α-T. Accordingly, the limits on the integration can be converted to

x ( α ) = λ = - α - T α s ( λ ) p ( α - ( α - λ ) ) d λ = λ = - α - T α s ( λ ) p ( λ ) d λ . x ( α ) = λ = - α - T α s ( λ ) p ( α - ( α - λ ) ) d λ = λ = - α - T α s ( λ ) p ( λ ) d λ .
(34)

This is the cross-correlation of pp with ss as defined in (Reference).

When Pn(f)Pn(f) is not a constant, Equation 24 becomes

v 2 ( τ ) P w = | - H ( f ) G ( f ) e j 2 π f τ d f | 2 - P n ( f ) | H ( f ) | 2 d f . v 2 ( τ ) P w = | - H ( f ) G ( f ) e j 2 π f τ d f | 2 - P n ( f ) | H ( f ) | 2 d f .
(35)

To use the Schwarz inequality (Reference), associate aa with HPnHPn and bb with Gej2πfτ/PnGej2πfτ/Pn. Then Equation 26 can be replaced by

v 2 ( τ ) P w - | H ( f ) | 2 P n ( f ) d f - | G ( f ) e j 2 π f τ | 2 P n ( f ) d f - | H ( f ) | 2 P n ( f ) d f , v 2 ( τ ) P w - | H ( f ) | 2 P n ( f ) d f - | G ( f ) e j 2 π f τ | 2 P n ( f ) d f - | H ( f ) | 2 P n ( f ) d f ,
(36)

and equality occurs when a(·)=kb*(·)a(·)=kb*(·); that is,

H ( f ) = k G * ( f ) e - j 2 π f τ P n ( f ) . H ( f ) = k G * ( f ) e - j 2 π f τ P n ( f ) .
(37)

When the noise power spectral density Pn(f)Pn(f) is not flat, it shapes the matched filter. Recall that the power spectral density of the noise can be computed from its autocorrelation.

### Exercise 25

Let the pulse shape be a TT-wide Hamming blip. Use the code in matchfilt.m to find the ratio of the power in the downsampled vv to that in the downsampled ww when:

1. the receive filter is a SRRC with beta= 0, 0.1, 0.5,
2. the receive filter is a rectangular pulse, and
3. the receive filter is a 3T3T-wide Hamming pulse.

When is the ratio largest?

### Exercise 26

Let the pulse shape be a SRRC with beta=0.25. Use the code in matchfilt.m to find the ratio of the power in the downsampled vv to that in the downsampled ww when

1. the receive filter is a SRRC with beta= 0, 0.1, 0.25, 0.5,
2. the receive filter is a rectangular pulse, and
3. the receive filter is a TT-wide Hamming pulse.

When is the ratio largest?

### Exercise 27

Let the symbol alphabet be 4-PAM.

1. Repeat Exercise 25.
2. Repeat Exercise 26.

### Exercise 28

Create a noise sequence that is uniformly distributed (using rand) with zero mean.

1. Repeat Exercise 25.
2. Repeat Exercise 26.

### Exercise 29

Consider the baseband communication system in Figure 14. The symbol period TT is an integer multiple of the sample period TsTs, i.e. T=NTsT=NTs. The message sequence is nonzero only each NNth kk, i.e. k=NTs+δTsk=NTs+δTs where the integer δδ is within 0δ<N0δ<N and δTsδTs is the “on-sample” baud timing offset at the transmitter.

1. Suppose that f[k]={0.2,0,0,1.0,1.0,0,0,-0.2}f[k]={0.2,0,0,1.0,1.0,0,0,-0.2} for k=0,1,2,3,4,5,6,7k=0,1,2,3,4,5,6,7 and zero otherwise. Determine the causal g[k]g[k] that is the matched filter to f[k]f[k]. Arrange g[k]g[k] so that the final nonzero element is as small as possible.
2. For the g[k]g[k] specified in part (a), determine the smallest possible NN so the path from m[k]m[k] to y[k]y[k] forms an impulse response that would qualify as a Nyquist pulse for some well-chosen baud-timing offset σσ when yy is downsampled by NN.
3. For the g[k]g[k] chosen in part (a) and the NN chosen in part (b), determine the downsampler offset σσ for s[i]s[i] to be the nonzero entries of m[k]m[k] when w[k]=0w[k]=0 for all kk.

## Matched Transmit and Receive Filters

While focusing separately on the pulse shaping and the receive filtering makes sense pedagogically, the two are intimately tied together in the communication system. This section notes that it is not really the pulse shape that should be Nyquist, but rather the convolution of the pulse shape with the receive filter.

Recall the overall block diagram of the system in Figure 1, where it was assumed that the portion of the system from upconversion (to passband) to final downconversion (back to baseband) is done perfectly and that the channel is just the identity. Thus, the central portion of the system is effectively transparent (except for the intrusion of noise). This simplifies the system to the baseband model in Figure 15.

The task is to design an appropriate pair of filters: a pulse shape for the transmitter, and a receive filter that is matched to the pulse shape and the presumed noise description. It is not crucial that the transmitted signal itself have no intersymbol interference. Rather, the signal after the receive filter should have no ISI. Thus, it is not the pulse shape that should satisfy the Nyquist pulse condition, but the combination of the pulse shape and the receive filter.

1. (i): allow no intersymbol interference at the receiver, and
2. (ii): maximize the signal-to-noise ratio.

Hence, it is the convolution of the pulse shape and the receive filter that should be a Nyquist pulse, and the receive filter should be matched to the pulse shape. Considering candidate pulse shapes that are both symmetric and even about some time tt, the associated matched filter (modulo the associated delay) is the same as the candidate pulse shape. What symmetric pulse shapes, when convolved with themselves, form a Nyquist pulse? Previous sections examined several Nyquist pulse shapes, the rectangle, the sinc, and the raised cosine. When convolved with themselves, do any of these shapes remain Nyquist?

For a rectangle pulse shape and its rectangular matched filter, the convolution is a triangle that is twice as wide as the original pulse shape. With precise timing, (so that the sample occurs at the peak in the middle), this triangular pulse shape is also a Nyquist pulse. This exact situation will be considered in detail in (Reference).

The convolution of a sinc function with itself is more easily viewed in the frequency domain as the point-by-point square of the transform. Since the transform of the sinc is a rectangle, its square is a rectangle as well. The inverse transform is consequently still a sinc, and is therefore a Nyquist pulse.

The raised cosine pulse fails. Its square in the frequency domain does not retain the odd symmetry around the band edges, and the convolution of the raised cosine with itself does not retain its original zero crossings. But the raised cosine was the preferred Nyquist pulse because it conserves bandwidth effectively and because its impulse response dies away quickly. One possibility is to define a new pulse shape that is the square root of the raised cosine (the square root is taken in the frequency domain, not the time domain). This is called the square-root raised cosine filter (SRRC). By definition, the square in frequency of the SRRC (which is the raised cosine) is a Nyquist pulse.

The time domain description of the SRRC pulse is found by taking the inverse Fourier transform of the square root of the spectrum of the raised cosine pulse. The answer is a bit complicated:

v ( t ) = 1 T ( 1 - β + ( 4 β / π ) ) t = 0 β ( π + 2 ) 2 T π sin π 4 β + β ( π - 2 ) 2 T π cos π 4 β t = ± T 4 β sin ( π ( 1 - β ) t / T ) + ( 4 β t / T ) cos ( π ( 1 + β ) t / T ) T ( π t / T ) ( 1 - ( 4 β t / T ) 2 ) otherwise . v ( t ) = 1 T ( 1 - β + ( 4 β / π ) ) t = 0 β ( π + 2 ) 2 T π sin π 4 β + β ( π - 2 ) 2 T π cos π 4 β t = ± T 4 β sin ( π ( 1 - β ) t / T ) + ( 4 β t / T ) cos ( π ( 1 + β ) t / T ) T ( π t / T ) ( 1 - ( 4 β t / T ) 2 ) otherwise .
(38)

### Exercise 30

Plot the SRRC pulse in the time domain and show that it is not a Nyquist pulse (because it does not cross zero at the desired times). The Matlab routine srrc.m will make this easier.

Though the SRRC is not itself a Nyquist pulse, the convolution in time of two SRRCs is a Nyquist pulse. The square root raised cosine is the most commonly used pulse in bandwidth-constrained communication systems.

### Exercise 31

Consider the baseband communication system with a symbol-scaled impulse train input

s ( t ) = i = - s i δ ( t - i T - ϵ ) s ( t ) = i = - s i δ ( t - i T - ϵ )
(39)

where TT is the symbol period in seconds and 0<ϵ<0.250<ϵ<0.25 msec. The system contains a pulse shaping filter P(f)P(f), a channel transfer function C(f)C(f) with additive noise n(t)n(t), and a receive filter V(f)V(f), as shown in Figure 16. In addition, consider the time signal x(t)x(t) shown in the top part of Figure 16, where each of the arcs is a half-circle. Let X(f)X(f) be the Fourier transform of x(t)x(t).

1. If the inverse of P(f)C(f)V(f)P(f)C(f)V(f) is X(f)X(f), what is the highest symbol frequency with no intersymbol interference supported by this communication system when Δ=0Δ=0?
2. If the inverse of P(f)C(f)V(f)P(f)C(f)V(f) is X(f)X(f), select a sampler time offset T/2>Δ>-T/2T/2>Δ>-T/2 to achieve the highest symbol frequency with no intersymbol interference. What is this symbol rate?
3. Consider designing the receive filter under the assumption that a symbol period TT can be (and will be) chosen so that samples ykyk of the receive filter output suffer no intersymbol interference. If the inverse of P(f)C(f)P(f)C(f) is X(f)X(f), and the power spectral density of n(t)n(t) is a constant, plot the impulse response of the causal, minimum delay, matched receive filter for V(f)V(f).

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