1. 3F4 Equalisation Dr. I. J. Wassell

2. Introduction When channels are fixed, we have seen that it is possible to design optimum transmit and receive filters, subject to zero ISI In practice, this is not usually possible, –Ideal filters cannot be realised –The channel responses can be unknown and/or time varying –The same transmitter may be used over many different channels

3. Introduction We can improve the situation by including an additional filtering stage at the receiver. This is known as an equalisation filter and usually it is designed to reduce ISI to a minimum Equalisers may be categorised as, –Fixed- The optimal equalisation filter is calculated for a fixed (known) received pulse shape –Adaptive- The filter is adapted continuously to the changing characteristics of the channel

4. Introduction Equalisation may be implemented using, –Analogue filters- A traditional technique mainly confined to fixed channels. Now superseded by, –Digital filters- Have all usual advantage of digital systems, e.g. flexibility, reliability etc. May be either fixed or adaptive. We will consider fixed equalisers implemented as digital filters

5. Digital Filters An analogue signal x(t) is sampled at times t=nT to give a ‘digital’ signal x n The Z-transform of x n is defined analogously to the Laplace transform of a continuous signal as,

6. FIR Filter A Finite Impulse Response (FIR) filter generates a new digital signal y n from x n using delay, multiply and addition operations D X b0b0 D X b1b1 D X b2b2 D X bqbq + xnxn x n-1 x n-2 x n-q ynyn Where b i are known as the filter coefficients and delay D is equal to the sample (symbol) period T

7. FIR Filter Taking the Z transform yields, Where z -n may be taken to mean a delay of n sample periods Now, Hence the transfer function H(z) is,

8. IIR Filters A recursive Infinite Impulse Response Filter generates a new digital signal y n from the input x n as follows, DD X a1a1 D X a2a2 D X apap + xnxn y n-1 y n-2 y n-p ynyn + Where a i are known as the filter coefficients and delay D is equal to the sample (symbol) period T

9. IIR Filters Taking Z transform yields, Rearranging, Now,

10. Zero-Forcing Equalisers Suppose the received pulse in a PAM system is p(t), which suffers ISI This signal is sampled at times t=nT to give a digital signal p n =p(nT) We wish to design a digital filter H E (z) which operates on p n to eliminate ISI Zero ISI implies that the filter output is only non- zero in response to pulse n at sample instant n, i.e. the filter output is the unit pulse  n in response to p n

11. Zero-Forcing Equalisers Note that the Z transform of  n is equal to 1, so, Now, So, Where p i are the sample values of the isolated received pulse

12. Zero-Forcing Equalisers We see that this expression has the form of an IIR filter, If, That is we define the amplitude of the isolated pulse at the optimum sampling point to be unity

13. FIR Approximations to ZFE IIR filters are difficult to deal with in practice –stability is not guaranteed –adaptive methods are difficult to derive –Their recursive nature makes them prone to numerical instability The simplest solution is to use an FIR approximation to the ideal response

14. Truncated Impulse Response A simple way to create an FIR approximation is simply to truncate the ideal impulse response However, this can give rise to significant errors in the filter response

15. Truncated Impulse Response The IIR response has the form, The FIR response has the form, Thus we must perform polynomial division to calculate the coefficients of H(z)

16. Truncated Impulse Response Example –The unequalised pulse response at the receiver in response to a single unit amplitude transmitted pulse at sample times k = 0, 1 and 2 is, p 0 = 1, p 1 = - 0.4 and p 2 = - 0.2 Now, So in this example,

17. Truncated Impulse Response Performing the polynomial division, Truncating to 5 terms gives FIR filter with the coefficients: 1, 0.4, 0.36, 0.224, 0.1616 Now,

18. Direct Zero Forcing The FIR filter equaliser output in the time domain is, In the time domain, the zero forcing constraint is y n = 1 for one value of n and y n = 0 otherwise

19. Direct Zero Forcing This constraint implies an infinite set of simultaneous equations corresponding to, However, we only have q+1 filter coefficients, so we set up q+1 equations in q+1 unknowns and solve for the coefficients

20. Direct Zero Forcing-Example The sampled received pulse in response to a single binary ‘1’ is, Design a 3-tap FIR equaliser to make the response at n=0 equal to 1, and equal to zero for n=1 and n=2 The FIR equaliser filter output is,

21. Direct Zero Forcing-Example Write out previous equation for n=0, 1 and 2, Solving these equations gives, The zero forcing constraint is, y 0 = 1, y 1 = 0, y 2 = 0

22. Error Rates and Noise Equalisation is designed to reduce ISI and hence increase the eye opening However, channel noise also passes through the equaliser and must be handled carefully to predict performance The frequency response of a digital filter may be obtained by substituting,

23. Error Rates and Noise The ideal ZFE has a response, So in the frequency domain, Thus at frequencies where P(e jwT ) is small, large noise amplification will occur.

24. Error Rates and Noise  0 P()P() Received pulse spectrum 0  H E (  ) = 1/P(  ) Equaliser spectral response In this example the low pulse spectrum response near zero will give rise to high gain and noise enhancement by the equaliser in this region.

25. Error Rates and Noise What is the mean-square value (  w ) 2 of the noise at the equaliser output? Suppose the equaliser filter has impulse response b n, (n=0,..,q). Consider the response of the equaliser to noise alone,

26. Error Rates and Noise The mean-squared value is, Assume that v n has a mean-squared value, And that v n is uncorrelated white noise. Then all the terms E[v n v m ] will be zero except when m=n, so,

27. Error Rates and Noise That is, the mean square noise at the filter output is that at the input multiplied by the sum squared of the filter impulse response

28. Error Rates and Noise Hence the worst case BER may be calculated as follows, –Calculate the eye opening h for the equalised pulse –Calculate the mean-squared noise power –Substitute into the BER expression,

29. Error Rates and Noise- Example Returning to the previous example, calculate the worst case BER after equalisation if unipolar line coding with transmit levels of 1V and 0V is employed. The sampled received pulse in response to a single binary ‘1’ is, The direct zero forcing solution is an FIR filter with the following coefficient values,

30. Error Rates and Noise- Example We now need to calculate the worst case eye opening for the equalised pulse. To do this we need to calculate the ‘residual’ values at the output of the equaliser in response to a single received pulse, p n From the earlier equations the FIR filter (equaliser) output is given by,

31. Error Rates and Noise- Example In the example, the input sequence x n is the single pulse p n and q = 2. In this case we have, This direct convoulution yields, Thus the equalised pulse response is,  ‘residuals’

32. Error Rates and Noise- Example So, remembering that for a unipolar scheme only ‘1’s give rise to residuals, the worst case received ‘1’ is, 1 - 0.141 = 0.859 i.e., 1 other ‘1’ contributing The worst case ‘0’ is, 0 + 0.0702 = 0.0702 i.e., 1 other ‘1’ contributing The minimum eye opening h is, h = 0.859 - 0.0702 = 0.789

33. Error Rates and Noise- Example To calculate the rms noise at the output of the equaliser we utilise, Where  v is the rms noise at the equaliser input For our example, Showing that the noise power has been increased

34. Error Rates and Noise- Example The probability of bit error is given by, Substituting for h and  w gives,

35. Error Rates and Noise- Example Note that instead of performing the convolution to give the equaliser output in response to a single received pulse (and hence determine the residuals), an alternative is to multiply the pulse response and equaliser response in the z domain, so

36. Error Rates and Noise- Example Equating this to the expansion for Y(z), Which yields the same expressions for the output sample values y n obtained previously by direct convolution

37. Other Equalisation Methods We have seen that with ZFEs, noise can be amplified leading to poor BER performance Alternative design approaches take into account noise as well as signal propagation through the equaliser The Minimum Mean Squared Error (MMSE) equaliser is one such approach

38. MMSE Equaliser The MMSE explicitly accounts for the presence of noise in the system Assuming a similar model to that used previously, then in Z transform notation, H E (z)+ X(z) V(z) Y(z) Where X(z) is the Z transform of the sampled received signal x n, and V(z) is the Z transform of the noise v n

39. MMSE Equaliser Ideally, the equalised output y n depends only on the transmitted symbols a k. This is not possible owing to the random noise, hence we choose to minimise the total expected mean square error (MSE) between y n and a n with respect to the equaliser H E (z), i.e.,

40. MMSE Equaliser HE(z)HE(z) xnxn ynyn anan - E[(.) 2 ] For a ‘fixed’ equaliser E[(.) 2 ] is minimised by adjusting the coefficients of H E (z). Effectively we have a trade off between noise enhancement and ISI. MMSE equaliser formulation From data source minimise

41. MMSE Equaliser The solution has the form, Where P(z) is the Z transform of the channel pulse response and N o is the noise PSD Note, –The equaliser needs knowledge of the noise PSD –If N o =0, the solution is the same as the ZFE –When noise is present the ZFE solution is modified to make a trade-off between ISI and noise amplification

42. Non-Linear Equalisation The equalisers considered so far are linear, since they simply involve linear filtering operations An alternative we consider now is non- linear equalisation An example is the Decision Feedback Equaliser (DFE)

43. DFE The DFE is a non-linear filter. Again, its purpose is to cancel ISI. The non-linearity allows some of the noise problems associated with linear equalisers to be overcome

44. DFE The structure is, D X a1a1 D X a2a2 D X apap + + - Slicer p n (sampled values of one pulse) xnxn Where a i are known as the filter coefficients (not to be confused with the transmitted symbols a n ) and delay D is equal to the sample (symbol) period T Detected symbols

45. DFE The DFE has almost the same structure as the standard IIR filter based equaliser In the following development this relationship is demonstrated We see that the only significant difference is the position of the data slicer (decision block) A minor difference is the subtract function at the DFE input. Its only effect is to alter the sign of the DFE coefficients compared with those in the IIR filter

46. DFE Development DD X a1a1 D X a2a2 D X apap + xnxn ynyn + Slicer DD X a1a1 D X a2a2 D X apap + xnxn + ynyn IIR Structure IIR

47. DFE Development xnxn + ynyn Slicer DD X a1a1 D X a2a2 D X apap + xnxn + ynyn DD X a1a1 D X a2a2 D X apap + DFE IIR

48. DFE The DFE is almost a standard IIR filter For this structure we know that the ZFE solution is, Where p 0, p 1, etc. is the sampled response at the equaliser input received in response to one transmitted symbol of unity amplitude. Because we define the amplitude of the isolated pulse at the optimum sampling point to be unity, then p o = 1. Comparing with the previous ZF solution where a i = -p i, this time a i = p i owing to the subtract function at the DFE input. The outputs of this filter with no channel noise are unit pulses, weighted by the transmitted symbol amplitudes, a n

49. DFE Example The sampled response to an isolated received pulse p n is given by, p 0 = 1, p 1 = 0.5, p 2 = -0.25 Design a suitable DFE From the earlier work we see that the DFE coefficients are given by a i = p i, so, a 1 = p 1 = 0.5 and a 2 = p 2 = -0.25 Assuming polar data pulses the effect is to – add (subtract) 0.25 (if previous but one bit is a binary one (zero)) to the current input value to remove its effect. –subtract (add) 0.5 (if previous bit is a binary one (zero)) to the current input value to remove its effect. Thus the effect of the previous pulses is eliminated

50. DFE Example D X D X a 2 = p 2 = -0.25 + + - Slicer xnxn a 1 = p 1 = 0.5 1 0.5 -0.5 0 T 2T 3TnT p(nT) p 0 = 1, p 1 = 0.5, p 2 = -0.25 Sampled isolated pulse

51. DFE In noise, we have seen that noise amplification occurs in the IIR filter approach To overcome this, the decision slicer is moved inside the filter loop in the DFE The slicer outputs the symbol estimate which is closest to the value at its input With no noise, this change makes no difference because the ISI is cancelled perfectly by the IIR filter, so the slicer input is a k anyway

52. DFE However, in noise, the slicer acts to ‘clean-up’ the signal, giving a noise free decision at its output. For example, the slicer input becomes a k +v k, where v k is the noise value Provided that v k is small enough the slicer still outputs the correct decision a k So error free cancellation continues without problems of noise amplification

53. DFE-Problems Consider what happens when v k is large enough to cause an error in the slicer decision The error feeds back around the loop and so the ISI is no longer cancelled. Often a long run or ‘burst’ of errors will then be experienced- known as error propagation The length of the burst is of the order of 2 N bits, where N is the number of taps in the feedback filter

54. Automatic Equalisers In practical communication systems the channel is often unknown and/or time varying To overcome this problem so called Automatic Equalisers are employed Two approaches are used –Preset equaliser: The channel is measured periodically by sending some known data. The equaliser coefficients are re-calculated and subsequent data is equalised using the new coefficients.

55. Automatic Equalisers –Adaptive equaliser: The coefficients are adapted continuously based on the received data. A simple approach uses the Least Mean Squares (LMS) algorithm to adjust the coefficient values based on an error criterion. This approach requires that the equaliser is initially trained, so that the coefficients are initialised with approximately the correct values An alternative adaptation algorithm is known as Recursive Least Square (RLS). This has the advantage of faster training but at the cost of higher complexity

56. Summary In this section we have seen –That in practical systems it is difficult to arrange optimum TX and RX filters subject to zero ISI –That additional filters (equalisers) can be added to reduce ISI and improve BER performance –Equalisers can be implemented in an analogue or digital manner and may be fixed or adaptive –Digital implementations are fundamentally of IIR structure but may be approximated by a truncated FIR filter –The zero-forcing criterion may lead to noise enhancement and poor performance –MMSE and non-linear (DFE) approaches can reduce the problem of noise enhancement