CS3282 Section 6 6.1 BMGC/ 25/08/06
University of Manchester School of Computer Science CS3282: Digital Communications ’05-06 Section 6: Inter-symbol interference and pulse shaping
Rectangular symbols are not suitable for transmitting data at the highest possible bit-rates over band- limited channels. A rectangular pulse or any other shaped pulse which is time-limited (i.e. is non-zero for a finite period of time, say T seconds) will require infinite bandwidth if it is not to be distorted, possibly unrecognizably. A symbol with finite bandwidth must have, in theory, infinite time duration. Although using a symbol which really does exist from t = -∞ to t = +∞ to send a single “1” or a “0” (or maybe two or three of them if we are using multi-level signaling) may seem impossible in practice, we have to keep this in mind as an ideal and produce approximations to this form of signaling. The pulses we use in practice may not actually go on, and back in time for ever. But they must definitely be non- zero for considerably more than T seconds when the signaling rate is 1/T symbols per second. Inevitably this means that one symbol will run into the previous and next symbol, and significantly affect several more besides. The result could be “inter-symbol interference” (ISI) where the data conveyed by one symbol causes the data of other symbols to be misinterpreted. Although we cannot avoid the overlap of symbols in the time-domain, we must find ways of making sure that the data carried by the symbols is not affected by this overlap. The solution to this challenge lies in “pulse shaping” which means that we must carefully choose the time-domain shape of the symbols and hence their spectral shape. A convenient way of generating symbols with the time-domain shape we require is to generate an impulse of the appropriate strength for each symbol and then to shape this impulse by passing it through a “shaping” filter. The impulse-response of the shaping filter is the symbol shape we wish to launch into the channel. An FIR digital filter followed by a digital to analogue converter will do this job nicely. The channel will inevitably affect the shape of the symbol and noise will be added. At the receiver, to optimize the detection process, filtering tasks are required as illustrated below.
Samples at A 3A -A intervals T
Shaping Matched Equal- Channel T T Filter filter iser n(t)
CS3282 Section 6 6.2 BMGC/ 25/08/06
Inter-symbol interference (ISI) can occur due to the ringing of one symbol into the next. However, ISI can be avoided if the transmitter's pulse shaping filter shapes the symbols so that zero-crossings at the output of the receiving filters occur T seconds, 2T seconds, and so on after (and before) the centre of the symbol. So when we sample at t=0, T, 2T, etc, we only see the centre of one symbol, all the other symbols being zero at those instants. This is nice in theory and possible to a fair degree in practice. If we combine the transmitter's shaping filter, the channel and the receiving filters (i.e. the matched filter and the equaliser) into a single frequency-response HN((ƒ)) say, then our goal is achieved if HN((ƒ)) is a “Nyquist frequency-response". To be a classed as a 'Nyquist frequency-response', as well as band- limiting from -1/T to 1/T Hz, HN((f)) must have a form of odd-symmetry about 1/(2T ) in that:
* H N (( f )) + H N ()1/T − f = cons tant for 0 ≤ | f | ≤ 1 T Two purely real frequency-responses satisfying this property are shown below. It may be shown
(Exercise 6.2 below) that this property guarantees that the impulse-response corresponding to HN((f)) , i.e. its inverse Fourier transform, has zero crossings at t = ±T, ±2T, ±3T, ….
|HN((ƒ))|
1
0.5
ƒ -1/T -1/2T 0 1/4T 1/2T 3/4T 1/T
1 : | f |< 1/(2T) The “brick-wall” filter: H B (( f )) = 0 : | f |≥ 1/(2T) has the property of having a 'Nyquist frequency-response' and we already know that its impulse response is a sinc function with zero crossings at t = ±T, ±2T, etc. The 'brick-wall' frequency- response achieves the required zero-crossing property with the minimum possible bandwidth, but is difficult to deal with because the side-lobes of its "sinc" impulse response do not die away very quickly. To design a pulse-shaping filter with a reasonable approximation to this 'brick-wall' frequency-response, we would need a very long impulse-response and hence a high order FIR digital filter. Also, assuming we could implement a good enough approximation, if there is the slightest error in the timing of our sampling point, significant ISI will occur.
CS3282 Section 6 6.3 BMGC/ 25/08/06
Raised cosine frequency response A commonly used family of Nyquist frequency-responses is a range of “raised cosine” frequency- responses parameterized by r (sometimes called α) and defined by the following formula:
1 : | f | ≤ (1− r) /(2T) H rc (( f )) = 0.5[1+ cos(πT{| f | −(1− r) /(2T)}/ r)] : (1− r) /(2T) ≤ | f | ≤ (1+ r) /(2T) 0 : (1+ r) /(2T) <| f | The impulse-response of a filter with such a frequency-response may be shown to be: sin c(t /T )cos(πrt /T) sin(πx) h(t) = where sinc(x) = lim T − 4r 2t 2 /T πx When r=0, this becomes the “brick-wall” filter mentioned earlier with the narrowest bandwidth of all the family: -1/(2T) to 1/(2T) Hz. When r = 0.5, Hrc(f) is as shown below, assuming T=0.001 seconds so that the symbol rate 1/T = 1 kHz.. From the general formula above, its spectrum is: 1 : | f | ≤ 250 Hz H rc (( f )) = 0.5[1+ cos(2π{| f | −250}/1000)] : 250 ≤ | f | ≤ 750 0 : 750 < | f |
1.2
1
50%RC spectrum 0.8
0.6 Modulus
0.4
0.2
0 Frequency -1000 -750 -500 -250 0 250 500 750 1000
In the graph above, drawn using "Microsoft Excel" , note the "odd-symmetry" about f=500 Hz for f >0 and about f = -500 Hz = - 1/(2T) for f <0. It may be shown that Hrc((f)) + H*rc(( 1000 - f ) ) = 1.
CS3282 Section 6 6.4 BMGC/ 25/08/06
When r=1, Hrc((ƒ)) becomes a pure raised cosine shaped frequency-response with no “flat-top” and the widest bandwidth of the family: -1/T to 1/T Hz. With 1/T=1 kHz, the bandwidth would be -1000 to 1000 Hz.
The impulse-responses hrc(t) corresponding to Hrc((ƒ)) with r=0 and r=1 are shown below taking T to be 10 seconds. Such graphs are easily drawn using a spreadsheet such as Microsoft Excel or MATLAB. Notice the reduction of side-lobe ripples when r=1, but this is at the expense of a doubling of the bandwidth. A raised cosine spectrally shaped pulse with a given value of r is called a "100r per- cent RC symbol" and has band-width from -(1+r)/T Hz to (1+r)/T. When r=0.5, we have a 50% RC spectrally shaped symbol (see spectrum above) with bandwidth from -750 Hz to 750 Hz if 1/T = 1 kHz.. This is 50% more than the absolute minimum band-width needed to avoid ISI with the techniques discussed up to now. The minimum bandwidth would be achieved with a 0% RC symbol which has a "brick-wall" spectrum from -1/(2T) to +1/(2T) Hz. Apart from the practical difficulties of generating a pulse with such a spectrum, the disadvantage of the brick-wall spectrum , as mentioned above, is that its time-domain shape is a "sinc" pulse which does not die away quickly enough for our liking. The 100% RC spectrum (r=1) produces a time-domain pulse which dies away much faster. So by increasing r from 0 to 1 we improve the rate of dying away at the expense of extra bandwidth.
0.1
0.08
0.06
h(t) for T=10, r=0 & 1 0.04
0.02
0 -40 -30 -20 -10 0 10 20 30 40
-0.02 t
A good way to implement a shaping filter is to use an FIR digital filter with multiplier coefficients equal to samples of h(t). A succession of shaped pulses generated by a the transmission of a continuous bit-stream are often viewed on an oscilloscope as “eye-diagrams”. The oscilloscope is
CS3282 Section 6 6.5 BMGC/ 25/08/06 triggered approximately at the beginning of each pulse, and an open eye is hopefully seen as each new shaped pulse is superimposed on the previous pulses. If noise is added, the eye is seen to close, making threshold comparisons for detection difficult.
Remember that HN((ƒ)) is the required overall response of transmitting filter, channel and receiving filters. Within the receiver an equalising filter is introduced to cancel out the filtering effect of the channel. We discuss this later.
If the overall symbol pulse shape as seen at the output of the receiving filters is to have a raised cosine spectrum and the receiving filters include a matched filter, it would not be correct for the transmitter to send symbols with raised cosine spectra. If we did send such a symbol its matched filter would have gain-response equal to the magnitude spectrum of the pulse, and the resulting output would have this spectrum squared. Its magnitude spectrum would no longer be raised cosine. It would be raised cosine squared. This may not be a problem as far as minimizing the effect of AWGN is concerned. But it would be a problem as far as ISI is concerned. We want the detector to see, at the threshold detector, not a 'raised cosine squared' spectrally shaped pulse but a raised cosine spectrally shaped pulse. It is therefore common to distribute the desired raised cosine frequency response equally between the transmitting and receiving filter, making each a “100r % root raised cosine” (RRC) frequency response. So the transmitter sends into the channel RRC spectrally shaped symbols. In the time-domain, such RRC symbols are not too dissimilar from RC symbols in that they exist for all time and are symmetric about t=0. But, perhaps surprisingly, they do not have zero crossings at t = ±T, ±2T, ±3T, … To produce such symbols in practice, we simply have to define:
1 : | f | ≤ (1− r) /(2T ) H rrc (( f )) = 0.5[1+ cos(πT{| f | −(1− r) /(2T )}/ r)] : (1− r) /(2T) ≤ | f | ≤(1+ r) /(2T) 0 : (1+ r) /(2T ) <| f | for our chosen value of r in the range 0 to 1, and apply an inverse FT to calculate the corresponding impulse response hrrc(t). This would normally be done in sampled data form and hrrc(t) would be sampled, windowed and delayed (this involves an approximation) to give us the impulse-response
(and hence the tap weights) of an FIR digital filter of manageable order. A graph of Hrrc(f) against f when 1/T=1000 Hz and r=0.5 is given below.
CS3282 Section 6 6.6 BMGC/ 25/08/06
1.2
1
50%RRC spectrum 0.8 Modulus
0.6
0.4
0.2
0 Frequency -1000 -750 -500 -250 0 250 500 750 1000
Note the lack of "odd symmetry" about f = ±500. It is clear that Hrrc((f))+H*rrc(( 1000 - f ) ) ≠constant.
The matched filter at the receiver performs two roles at the same time. The first role is to minimise the error probability PB as usual for a matched filter. To do this, the matched filter's impulse response must be equal to s(t) time-reversed and suitably delayed. Bearing in mind that s(t) = s(-t) for a RRC symbol, this is also exactly what we need to do in order to square the spectrum of s(t) to make it raised cosine rather than RRC. Hence as well as minimising PB, the matched filter also completes the job of generating a raised cosine type Nyquist frequency response as required to eliminate ISI. Strictly, as we have seen, the matched filter has impulse-response s(T-t) i.e. the time-reversed symbol delayed by a single signalling interval T. In practice we have to delay this by several more intervals of T to allow some of s(-t) ( = h(-t) in the diagram above to be included) without making the filter non- causal. We must also delay the decision against the threshold until several intervals beyond t=0 have been received. All this is because each symbol now extends beyond a single interval of T seconds.
Exercise 6.1: Does a symbol whose spectrum is '100r % RC squared' have zero-crossings at ±T, ±2T, ±3T, … as required for zero ISI? Answer: Nope (except when r=0). This is the problem.
CS3282 Section 6 6.7 BMGC/ 25/08/06
Exercise 6.2:
A “Nyquist frequency-response” HN((ƒ)) with band-width from -1/T Hz to 1/T Hz has symmetry about 1/(2T ) in the sense that:
* H N (( f )) + H N (1/T − f ) = cons tant for 0 ≤ | f | ≤ 1 T . This means that the spectrum S((ƒ)) of a Nyquist shaped real pulse s(t) satisfies S(( f )) + S * ()1/T − f = cons tant for 0 ≤ | f | ≤ 1 T . Show that this condition guarantees that zero crossings occur in the time-domain at t = ±T, ±2T, ±3T, … as required to eliminate ISI.
Solution: From Section 2, sampling a signal s(t) at t = 0, ±T, ±2T, ±3T, … to produce a sequence of impulses gives a signal referred to as 'sample T (s(t))' whose Fourier transform is (1/T)repeat 1/T (S((f)) where S((f)) is the Fourier transform of s(t).
If s(t) has zero-crossings at t = ±T, ±2T, ±3T, … then all samples of 'sample T (s(t))' will be equal to zero apart from the sample at t=0. Therefore
sample T (s(t)) = s(0)δ(t) i.e. an impulse of strength s(0) at t=0 and zero for t ≠ 0. Its Fourier Transform is equal to the constant s(0) for all f
∞ ∞ since s(0)δ (t)e−2πjft dt = s(0) δ (t)e0 dt = s(0) for all f ∫∫−∞ −∞
But we also know from Section 2 that the FT of 'sample T ( s(t) )' is: 1 ∞ (1/T)repeat 1/T (S((f)) = ∑ S(( f − k /T )) T k=−∞ 1 ∞ therefore ∑ S(( f − k /T )) = s(0) for all f T k=−∞ If S((ƒ)) is band-limited between ƒ = ±1/T Hz, only the terms with k = 0 and k = 1, i.e. S((ƒ)) and S((ƒ-1/T)) , contribute to the expression in the range 0 ≤ ƒ ≤ 1/T. Draw a simple diagram to illustrate this.
S((f)) S((f+1/T)) S((f-1/T)) S((f-2/T))
f -1/T 1/T
CS3282 Section 6 6.8 BMGC/ 25/08/06
Therefore S((f)) +S((f-1/T)) = s(0) for any f in the range 0 ≤ f ≤ 1/T As s(t) is real S(-f) = S*(f) for any f. Therefore S( f −1/T ) = S * (1/T − f ) for any f. Hence, for 0 ≤ ƒ ≤ 1/T, S(( f )) + S * (1/T − f )= s(0) which is constant. Complete the argument for -1/T ≤ ƒ ≤ 0 in a similar way.
Exercise 6.3: Without calculating hrrc(t), i.e. the symbol generated by exciting a 100r % RRC filter with an impulse, would you expect its zero-crossings to occur at t = ±T, ±2T, ±3T, …? Answer: Nope. It may seem strange that this symbol, which is the one actually transmitted and hopefully received, does not have the zero-crossings required for zero ISI. So the transmission along the channel has ISI. It is only when the symbols are received and passed through their matched filter (or correlator), thus squaring the RRC spectrum to make it RC, that the condition for zero ISI is satisfied. Remember that the matched filter performs time-domain convolution, not time-domain multiplication.
Exercise 6.4: If RRC pulses are used, show that if AWGN is added to the received signal, the noise component n0(t) of z(t), i.e. the output of the RRC matched filter at the receiver, has an auto-correlation function which is zero at delays T, 2T, … Why is this a good thing? Answer: ACF is inverse FT of power spectrum. Instead of " t " we have " τ " seconds "delay". At delay τ = T, ACF tells us whether the noise we observe at any time t may be expected to be related to (correlated with) what we observed T seconds ago. If the noise sample was high T seconds ago does this mean that we must expect a high noise level again? If so, a high PB before will lead to a high PB now. Hence errors will tend to occur in "bursts". A lot of errors at once, then few errors. This is not such a good thing. If ACF of n0(t) is zero at τ = ±T, ±2T, etc. there will be no correlation between error prob. at t=nT, & at t=mT for m≠n. Errors more evenly distributed in time. Channel noise n(t) only passes through matched filter. So its magnitude spectrum will be RRC shaped. Its power spectrum will be RRC squared, i.e. RC. Gives required time or "delay" domain zero crossings.
Exercise 6.5: Check from the formula given that an 100r % RC filter satisfies the conditions for being a Nyquist filter. If T= 125 µs, what bandwidth is required for binary signalling with 60%RRC pulses?
Equalisation
As mentioned earlier, the filtering operations that occur when digital information is transmitted and detected may be separated in various ways. The transmitting filter band-limits and shapes the symbols
CS3282 Section 6 6.9 BMGC/ 25/08/06 before they are launched into the channel. The band-limiting is necessary as the channel cannot convey all frequencies, and the shaping is necessary to allow symbols to be detected, eventually, with the minimum of inter-symbol interference (ISI). The channel will distort whatever symbol-shape is launched and noise will be added to the received version. Further band-limiting may be applied at the receiver to reduce out-of-band noise and interference. Matched filtering is then applied to minimise the effect of the noise at the detection instant. This matched filtering also changes the symbol shape. The “equaliser” filter is included at the receiver to cancel out the symbol shaping effect of the channel. Remember that to minimise ISI, we require the product of the frequency responses of the filters just mentioned and the channel itself to be that of a Nyquist filter, HN((ƒ)) say. To make the product of the frequency responses of all the filters and the channel equal to the desired Nyquist frequency-response
HN((ƒ)), generally a “raised cosine” frequency response, we take its square root and construct two “root raised cosine” (RRC) filters. The first RRC acts as the transmitting filter (band-limiting the signal
HN((ƒ))
RRC Channel RRC Equaliser Detector T T Filter Filter launched into the channel) and the second RRC, as well as completing the symbol shaping, also minimises the effect of noise. So each of the RRC filters is doing two useful tasks. The task of the equaliser is now to cancel out the effect of the frequency-response of the channel.
Equalisation of the channel may be achieved using an adaptive FIR “transversal” filter, (or “tapped delay line”) a 4th order example of which is shown below: x(t) Delay Delay T T by T by T
C1 C2 C3 C4 C0 y(t)
Such a filter can be made a “zero forcing” transversal equaliser.
CS3282 Section 6 6.10 BMGC/ 25/08/06
These days, such a filter would probably be implemented digitally, but not necessarily so. It may remind us of the FIR low-pass and band-pass filters encountered in “Digital Signal Processing” courses, but there is an important difference. Note that the delay boxes are not “z-1” (one sample delay) as they were in the DSP course. They are “delay by T seconds” boxes where 1/T is the symbol or Baud rate. If we are receiving 1000 samples/second from the channel and sample the channel output at 8kHz to do the filtering digitally, then each box will be a delay of 8 samples, i.e. a z -8 box rather than a z-1 box. The filter shown above can be made a “zero-forcing” transversal equaliser which looks at the symbols received from the RRC/channel/RRC combination and adapts its coefficients C0, C1, C2, etc. such that, in response to an input waveform x(t) centred on t=0, the output y(t) has exact zero-crossings at t=0, T, 3T & 4T even when, because of the effect of the channel, the input x(t) does not quite have zero crossings at t=±T, and t=±2T. Note that a Nth order transversal filter must delay the centre of the symbol by (N/2)T to do its job.
The transversal filter shown above is of order N = 4, with five “taps” labelled C0, C1, C2, C3, C4.
(Sometimes these are labelled C-2, C-1, C0, C1, C2 ). The output is:
N y(t) = ∑ Cn x(t − nT) n=0 Given a symbol x(t) from the RRC/channel/RRC filtering whose centre is at t = 0. Assume the effect of the channel has been to make x(±T) and x(±2T) non-zero as illustrated. To take an example assume x(-2T) =-0.04, x(-T) = 0.2, x(0) = 1.0, x(T)=0.2 and x(2T)=-0.04.
x(t)
t -2T -T T 2T
To simplify the calculation, use a 3-tap (2nd order) transversal filter. This will allow us only to force zero-crossings at t = +T & -T, the centre of the symbol being delayed by one sample and occurring at t=T.
CS3282 Section 6 6.11 BMGC/ 25/08/06
The full 5-tap filter would have allowed us to force zero-crossings at 0,T, 3T,4T with a 2T delay to the centre of the symbol. The output of the 3-tap transversal filter is:
y(t) = C0 x(t) + C1 x(t −T ) + C2 x(t − 2T ) The output at t=0, T, and 2T is as follows:
y(0) = C0 x(0) + C1 x(−T ) + C2 x(−2T)
y(T ) = C0 x(T) + C1 x(0) + C2 x(−T )
y(2T ) = C0 x(2T) + C1 x(T) + C2 x(0) and we wish to make y(0)=0, y(T)=1, y(2T)=0. Therefore:
y(0) = C0 + 0.2C1 − 0.04C2
y(T ) = 0.2C0 + C1 + 0.2C2
y(2T) = −0.04C0 + 0.2C1 + C2 This is a set of three linear simultaneous equations in three unknowns. In matrix form:
10..2− 004 C 0 0 02..1 02 C1 = 1 − 00..4 02 1 C2 0 i.e. AC = B with the solution C = A-1B You can solve this set of 3 simultaneous equations by several methods. One way is to calculate: DET (A) = 1×(1− 0.04) − 0.2(0.2 + 0.008) − 0.04(0.04 + 0.04) = 0.96 − 0.0416 − 0.0032 = 0.9152 Then, by Cramer’s rule 0 .2 −.04 1 0 −.04 1 .2 0 1 1 1 C = 1 1 .2 C = .2 1 .2 C = .2 1 1 0 det(A) 1 det(A) 2 det(A) 0 .2 1 −.04 0 1 −.04 .2 0
The solution thus obtained is:
C0 = −0.227
C1 = 1.091
C2 = −0.227 A simpler way is to use MATLAB to find A-1, the inverse of A, as follows: A= [1 .2 -0.04; 0.2 1 0.2; -0.04 0.2 1] inv(A) 1.049 -0.227 0.087 -0.227 1.091 -0.227 0.087 -0.227 1.049
CS3282 Section 6 6.12 BMGC/ 25/08/06
Therefore:
C0 1.049 − 0.227 0.087 0 C = − 0.227 1.091 − 0.227 1 1 C2 0.087 − 0.227 1.049 0 and it follows that C0=-0.227, C1=1.091 and C2 = -0.227.
Note that the matrices need not be symmetric in general, and that this second order filter will not attempt to make y(-T), y(2T) or y(3T) equal to zero. This would need a higher order zero-forcing filter. This type of equaliser is often trained by sending a suitable number of individual symbols at the beginning of a transmission and calculating the coefficients which then remain fixed until a break in transmission. An adaptive equaliser adapts the coefficients continuously according to the transmitted data itself.
Exercise 6.6: (Exam 1998) Given x(-2T)= -.05, x(T) = 0.2, x(0) = 1, x(T)= 0.3, x(2T) = -0.07, calculate the coefficients of a 3-term zero-forcing transversal filter and sketch the circuit of the equaliser.
Exercise 6.7: (Exam 2003): (a) What is the purpose of the ‘physical layer’ in a digital communication system? Draw a block-diagram of a digital transmitter and corresponding receiver which is valid for the physical layer transmission of bit-streams over wired or wireless channels. How may the channel and the receiver circuitry be expected to distort the transmitted wave-forms. State the purpose of (i) a pulse shaping filter (ii) a matched filter and (iii) an adaptive equaliser as may be used in this system. Explain why a Nyquist frequency-response is required between appropriate points in the block- diagram and indicate where these points occur. (b) An appropriately shaped symbol with zero-crossings at t = ±T, ±2T, ±3T, etc. relative to the centre is distorted by the frequency-response of a wired channel. It is estimated by averaging over several training symbols that when the received symbol, x(t) say, is normalised to 1 volt at its centre (assumed to occur at t=0), its voltages at t = ±T and ±2T relative to the centre would be as follows in the absence of channel noise: x(−2T) = −0.1, x(−T) = 0.2, x(0) = 1.0, x(T) = −0.3, x(2T) = 0.2 Inter-symbol interference is to be reduced by a 3-term ‘zero-forcing’ transversal equaliser. Explain how this may be achieved and give a diagram of the equaliser. Given that the zero-forcing equaliser has coefficients C0, C1, C2, expressed as a column vector c, show that the required zero-crossings are achieved if Ac = b , where 10..2− 01 A = − 03..1 02 02..− 03 1
CS3282 Section 6 6.13 BMGC/ 25/08/06 and b is a suitably chosen column-vector. Given that the inverse of matrix A is approximately as follows: 09..3 − 015 0.12 −1 A = 03..0 090 − 0.15 − 01..0 030 0.93 calculate the coefficients C0, C1, and C2. Give an expression for the frequency-response of the equaliser and comment on how it will affect a stream of similarly shaped symbols at 1/T baud received with additive white Gaussian noise.
There will be nothing on partial response filtering this year.