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3F4 Equalisation

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

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3F4 Equalisation

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  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 Now superseded by, • Digital filters- Have all usual advantage of digital systems, e.g. flexibility, reliability etc.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 xn • The Z-transform of xn is defined analogously to the Laplace transform of a continuous signal as,

  6. xn xn-1 xn-2 xn-q D D D D b0 b1 b2 bq X X X X yn + FIR Filter • A Finite Impulse Response (FIR) filter generates a new digital signal yn from xn using delay, multiply and addition operations Where bi 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. yn xn yn-1 yn-2 yn-p + D D D D a1 a2 ap X X X + IIR Filters • A recursive Infinite Impulse Response Filter generates a new digital signal yn from the input xn as follows, Where ai 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 pn=p(nT) • We wish to design a digital filter HE(z) which operates on pn 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 dn in response to pn

  11. Zero-Forcing Equalisers • Note that the Z transform of dn is equal to 1, so, • Now, Where piare the sample values of the isolated received pulse • So,

  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, p0= 1, p1= - 0.4 and p2= - 0.2 Now, So in this example,

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

  18. Direct Zero Forcing • The FIR filter equaliser output in the time domain is, • In the time domain, the zero forcing constraint is yn= 1 for one value of n and yn= 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 • The zero forcing constraint is, y0 = 1, y1 = 0, y2 = 0 • Write out previous equation for n=0, 1 and 2, • Solving these equations gives,

  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(ejwT) is small, large noise amplification will occur.

  24. HE(w) = 1/P(w) P(w) 0 0 w w Error Rates and Noise Received pulse spectrum 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 (sw)2 of the noise at the equaliser output? • Suppose the equaliser filter has impulse response bn, (n=0,..,q). • Consider the response of the equaliser to noise alone,

  26. Error Rates and Noise • The mean-squared value is, • Assume that vn has a mean-squared value, And that vn is uncorrelated white noise. Then all the terms E[vnvm] 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, pn • From the earlier equations the FIR filter (equaliser) output is given by,

  31. Error Rates and Noise- Example • In the example, the input sequence xn is the single pulse pn and q = 2.In this case we have, • This direct convoulution yields, • Thus the equalised pulse response is, 1, 0, 0, -0.141, 0.0702 ‘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 sv 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 sw 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 yn 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. X(z) Y(z) + HE(z) V(z) 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, Where X(z) is the Z transform of the sampled received signal xn, and V(z) is the Z transform of the noise vn

  39. MMSE Equaliser • Ideally, the equalised output yn depends only on the transmitted symbols ak. This is not possible owing to the random noise, hence we choose to minimise the total expected mean square error (MSE) between yn and an with respect to the equaliser HE(z), i.e.,

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

  41. MMSE Equaliser • The solution has the form, Where P(z) is the Z transform of the channel pulse response and No is the noise PSD • Note, • The equaliser needs knowledge of the noise PSD • If No=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. xn + Slicer - pn (sampled values of one pulse) + ap a2 a1 X X X D D D DFE • The structure is, Detected symbols Where ai are known as the filter coefficients (not to be confused with the transmitted symbols an) and delay D is equal to the sample (symbol) period T

  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 yn IIR Structure Slicer xn + D D D D ap a1 a2 X X X + xn yn Slicer + D D D D ap a1 a2 IIR X X X +

  47. DFE Development xn yn Slicer + + IIR ap a1 a2 X X X D D D D yn xn + Slicer DFE + ap a1 a2 X X X D D D D

  48. DFE • The DFE is almost a standard IIR filter • For this structure we know that the ZFE solution is, Where p0, p1, 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 po = 1. Comparing with the previous ZF solution where ai = -pi, this time ai = pi 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, an

  49. DFE Example • The sampled response to an isolated received pulse pn is given by, p0 = 1, p1 = 0.5, p2 = -0.25 • Design a suitable DFE • From the earlier work we see that the DFE coefficients are given by ai= pi, so, a1 = p1 = 0.5 and a2 = p2 = -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. xn + Slicer - + a1= p1 = 0.5 a2= p2 = -0.25 X X D D DFE Example p(nT) 1 Sampled isolated pulse 0.5 0 2T 3T nT T -0.5 p0 = 1, p1 = 0.5, p2 = -0.25

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