3 F 4 Optimal Transmit and Receive Filtering

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3 F 4 Optimal Transmit and Receive Filtering Dr. I. J. Wassell.

3 F 4 Optimal Transmit and Receive Filtering Dr. I. J. Wassell.

Transmission System Transmit Channel Filter, HT(w) H C (w ) Weighted impulse train +

Transmission System Transmit Channel Filter, HT(w) H C (w ) Weighted impulse train + N(w), Noise • The FT of the received pulse is, Receive y(t) Filter, HR(w) To slicer

Transmission System • Where, – HC(w) is the channel frequency response, which is often

Transmission System • Where, – HC(w) is the channel frequency response, which is often fixed, but is beyond our control – HT(w) and HR(w), the transmit and receive filters can be designed to give best system performance • How should we choose HT(w) and to HR(w) give optimal performance?

Optimal Filters • Suppose a received pulse shape p. R(t) which satisfies Nyquist’s pulse

Optimal Filters • Suppose a received pulse shape p. R(t) which satisfies Nyquist’s pulse shaping criterion has been selected, eg, RC spectrum • The FT of p. R(t) is PR(w), so the received pulse spectrum is, H(w)=k PR(w), where k is an arbitrary positive gain factor. So, we have the constraint,

Optimal Filters • For binary equiprobable symbols, Where, • Vo and V 1 are

Optimal Filters • For binary equiprobable symbols, Where, • Vo and V 1 are the received values of ‘ 0’ and ‘ 1’ at the slicer input (in the absence of noise) • sv is the standard deviation of the noise at the slicer • Since Q(. ) is a monotonically decreasing function of its arguments, we should,

Optimal Filters f(x) 0 b x Q(b) 1. 0 0. 5 0 b

Optimal Filters f(x) 0 b x Q(b) 1. 0 0. 5 0 b

Optimal Filters • For binary PAM with transmitted levels A 1 and A 2

Optimal Filters • For binary PAM with transmitted levels A 1 and A 2 and zero ISI we have, • Remember we must maximise, Now, A 1, A 2 and p. R(0) are fixed, hence we must,

Optimal Filters • Noise Power, – The PSD of the received noise at the

Optimal Filters • Noise Power, – The PSD of the received noise at the slicer is, – Hence the noise power at the slicer is, n(t) N (w ) H R (w ) v(t) Sv(w)

Optimal Filters • We now wish to express the gain term, k, in terms

Optimal Filters • We now wish to express the gain term, k, in terms of the energy of the transmitted pulse, h. T(t) • From Parsevals theorem, • We know,

 • So, Optimal Filters • Giving, • Rearranging yields,

• So, Optimal Filters • Giving, • Rearranging yields,

Optimal Filters • We wish to minimise,

Optimal Filters • We wish to minimise,

Optimal Filters • Schwartz inequality states that, With equality when, Let,

Optimal Filters • Schwartz inequality states that, With equality when, Let,

Optimal Filters • So we obtain, All the terms in the right hand integral

Optimal Filters • So we obtain, All the terms in the right hand integral are fixed, hence,

Optimal Filters • Since l is arbitrary, let l=1, so, Receive filter • Utilising,

Optimal Filters • Since l is arbitrary, let l=1, so, Receive filter • Utilising, And substituting for HR(w) gives, Transmit filter

Optimal Filters • Looking at the filters – Dependent on pulse shape PR(w) selected

Optimal Filters • Looking at the filters – Dependent on pulse shape PR(w) selected – Combination of HT(w) and HR(w) act to cancel channel response HC(w) – HT(w) raises transmitted signal power where noise level is high (a kind of pre-emphasis) – HR(w) lowers receive gain where noise is high, thereby ‘de-emphasising’ the noise. Note that the signal power has already been raised by HT(w) to compensate.

Optimal Filters • The usual case is ‘white’ noise where, • In this situation,

Optimal Filters • The usual case is ‘white’ noise where, • In this situation, and

Optimal Filters • Clearly both |HT(w)| and |HR(w)| are proportional to, ie, they have

Optimal Filters • Clearly both |HT(w)| and |HR(w)| are proportional to, ie, they have the same ‘shape’ in terms of the magnitude response • In the expression for |HT(w)|, k is just a scale factor which changes the max amplitude of the transmitted (and hence received) pulses. This will increase the transmit power and consequently improve the BER

Optimal Filters • If |HC(w)|=1, ie an ideal channel, then in Additive White Gaussian

Optimal Filters • If |HC(w)|=1, ie an ideal channel, then in Additive White Gaussian Noise (AWGN), • That is, the filters will have an identical RC 0. 5 (Root Raised Cosine) response (if PR(w) is RC) • Any suitable phase responses which satisfy, are appropriate

Optimal Filters • In practice, the phase responses are chosen to ensure that the

Optimal Filters • In practice, the phase responses are chosen to ensure that the overall system response is linear, ie we have constant group delay with frequency (no phase distortion is introduced) • Filters designed using this method will be noncausal, i. e. , non-zero values before time equals zero. However they can be approximately realised in a practical system by introducing sufficient delays into the TX and RX filters as well as the slicer

Causal Response • Note that this is equivalent to the alternative design constraint, Which

Causal Response • Note that this is equivalent to the alternative design constraint, Which allows for an arbitrary slicer delay td , i. e. , a delay in the time domain is a phase shift in the frequency domain.

Causal Response Non-causal response T=1 s Causal response T = 1 s Delay, td

Causal Response Non-causal response T=1 s Causal response T = 1 s Delay, td = 10 s

Design Example • Design suitable transmit and receive filters for a binary transmission system

Design Example • Design suitable transmit and receive filters for a binary transmission system with a bit rate of 3 kb/s and with a Raised Cosine (RC) received pulse shape and a roll-off factor equal to 1500 Hz. Assume the noise has a uniform power spectral density (psd) and the channel frequency response is flat from 3 k. Hz to 3 k. Hz.

Design Example • The channel frequency response is, HC(f) H C (w ) -3000

Design Example • The channel frequency response is, HC(f) H C (w ) -3000 -2 p 3000 0 3000 f (Hz) 2 p 3000 w (rad/s)

Design Example • The general RC function is as follows, PR(f) T 0 f

Design Example • The general RC function is as follows, PR(f) T 0 f (Hz)

Design Example • For the example system, we see that b is equal to

Design Example • For the example system, we see that b is equal to half the bit rate so, b=1/2 T=1500 Hz • Consequently, PR(f) T 0 f (Hz) 1500 3000 2 p 1500 2 p 3000 w(rad/s)

Design Example • So in this case (also known as x = 1) where,

Design Example • So in this case (also known as x = 1) where, • We have, Where both f and b are in Hz • Alternatively, Where both w and b are in rad/s

Design Example • The optimum receive filter is given by, • Now No and

Design Example • The optimum receive filter is given by, • Now No and HC(w) are constant so,

Design Example • So, Where a is an arbitrary constant. • Now, • Consequently,

Design Example • So, Where a is an arbitrary constant. • Now, • Consequently,

Design Example • Similarly we can show that, • So that,

Design Example • Similarly we can show that, • So that,

Summary • In this section we have seen – How to design transmit and

Summary • In this section we have seen – How to design transmit and receive filters to achieve optimum BER performance – A design example