3F4 Optimal Transmit and Receive Filtering

Dr. I. J. Wassell.

Transmission System

• The FT of the received pulse is,

Transmit

Filter, HT()

Channel

HC()

Receive

Filter, HR()+

N(),Noise

Weighted impulse train

To slicer

y(t)

)()()()( RCT HHHH

n

sn nTta )(

Transmission System• Where,

– HC() is the channel frequency response, which is often fixed, but is beyond our control

– HT() and HR(), the transmit and receive filters can be designed to give best system performance

• How should we choose HT() and to HR() give optimal performance?

Optimal Filters• Suppose a received pulse shape pR(t) which

satisfies Nyquist’s pulse shaping criterion has been selected, eg, RC spectrum

• The FT of pR(t) is PR(), so the received pulse spectrum is, H()=k PR(), where k is an arbitrary positive gain factor. So, we have the constraint,

)()()()()( RRCT kPHHHH

Optimal Filters• For binary equiprobable symbols,

v

VVQ

2(BER) RateError Bit 01

Where,

•Vo and V1 are the received values of ‘0’ and ‘1’ at the slicer input (in the absence of noise)

• v is the standard deviation of the noise at the slicer• Since Q(.) is a monotonically decreasing function of its arguments, we should,

BER theminimise to2

maximise 01

v

VV

Optimal Filters

x

f(x)

b0

0b

Q(b)

0.5

1.0

Optimal Filters• For binary PAM with transmitted levels A1

and A2 and zero ISI we have,)0()()0()( 121201 RkpAAhAAVV

• Remember we must maximise,

v

R

v

kpAAVV

2

)0(

21201

Now, A1, A2 and pR(0) are fixed, hence we must,

) (or, minimisely equivalentor maximise2

2

kk

k vv

v

Optimal Filters• Noise Power,

– The PSD of the received noise at the slicer is,2

)()()( Rv HNS – Hence the noise power at the slicer is,

dHNdS Rvv

22 )()(2

1)(

2

1

HR()n(t) v(t)

N () Sv()

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

of the energy of the transmitted pulse, hT(t)

dtthE TT2))((

• From Parsevals theorem,

dHE TT2|)(|

2

1

• We know,

)()()()( RRCT kPHHH

Optimal Filters• So,

)()(

)()(

RC

RT HH

kPH

• Giving,

dHE TT2|)(|

2

1

dHH

Pk

RC

R22

22

|)(||)(|

|)(|

2

1

• Rearranging yields,

d

HHP

Ek

RC

R

T

22

2

2

|)(||)(||)(|

21

Optimal Filters• We wish to minimise,

dHH

P

E

dHN

k

RC

R

T

R

v

22

2

2

2

2

|)(||)(||)(|

21

)()(21

d

HH

PdHN

Ek RC

RR

T

v22

22

22

2

|)(||)(|

|)(|)()(

)2(

1

Optimal Filters• Schwartz inequality states that,

2

22 )()(|)(||)(|

dGFdGdF

With equality when,

constantarbitrary an is where)()( * GF Let,

|)(||)(|

|)(|)(

and |)(|)()(

RC

R

R

HH

PG

HNF

Optimal Filters• So we obtain,

d

HH

PdHN

Ek RC

RR

T

v22

22

22

2

|)(||)(|

|)(|)()(

)2(

1

All the terms in the right hand integral are fixed, hence,

ie, equal, areequation theof sides two when theminimised is 2

2

kv

2

2 |)(|

|)(|)(

)2(

1

d

H

PN

E C

R

T

|)(||)(|

|)(| |)(|)(

)()( *

RC

RR HH

PHN

GF

Optimal Filters• Since is arbitrary, let=1, so,

21

)()(

)( |)(|

C

RR

HN

PH

• Utilising,)()()()( RRCT kPHHH

And substituting for HR() gives,2

1

)(

)()( |)(|

C

RT H

NPkH

Receive filter

Transmit filter

Optimal Filters• Looking at the filters

– Dependent on pulse shape PR() selected

– Combination of HT() and HR() act to cancel channel response HC()

– HT() raises transmitted signal power where noise level is high (a kind of pre-emphasis)

– HR() lowers receive gain where noise is high, thereby ‘de-emphasising’ the noise. Note that the signal power has already been raised by HT() to compensate.

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

PSD) sided (2 )( oNN • In this situation,

21

)(

)( |)(|

Co

RR

HN

PH

and2

1

)(

)( |)(|

C

oRT H

NPkH

Optimal Filters• Clearly both |HT()| and |HR()| are proportional

to, 21

)(

)(

C

R

H

P

ie, they have the same ‘shape’ in terms of the magnitude response

• In the expression for |HT()|, 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()|=1, ie an ideal channel, then in

Additive White Gaussian Noise (AWGN),

21

21

)( |)(|

and )( |)(|

RT

RR

PH

PH

• That is, the filters will have an identical RC0.5 (Root Raised Cosine) response (if PR() is RC)

• Any suitable phase responses which satisfy,)()()()( RRCT kPHHH

are appropriate

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 non-causal, 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,

)()exp()()()( RdRCT PtjkHHH

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 = 1s

Delay, td = 10s

Design Example

• Design suitable transmit and receive filters for a binary transmission system with a bit rate of 3kb/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 -3kHz to 3kHz.

• The channel frequency response is,

Design Example

HC(f)

HC()

f (Hz)

rad/s)

03000

23000

-3000

-23000

Design Example• The general RC function is as follows,

PR(f)

f (Hz)

T

0

T2

1 T2

1T2

1

T

1

2121

21 0

)21(4

cos

21

)( 2

TfT

Tf

TfT

TfT

fPR

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

to half the bit rate so, =1/2T=1500 Hz

• Consequently,

0

PR(f)

f (Hz)T2

1

T

1

T

1500 3000(rad/s)21500 200

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

T

x

21

• We have,

10 )(4

cos)( 2 TffTfPR

Where both f and are in Hz

• Alternatively,

)1(20 )(4

cos)( 2 TTPR

Where both and are in rad/s

Design Example• The optimum receive filter is given by,

21

)(

)( |)(|

Co

RR

HN

PH

• Now No and HC() are constant so,

)(4

cos |)(|

)( |)(|

21

2

21

TH

PH

R

RR

Design Example• So,

)(4

cos|)(| aH R

Where a is an arbitrary constant.

• Now,

TT

22

1

• Consequently,

4cos|)(|

TaH R

Design Example

• Similarly we can show that,

21

)( |)(| RT PH

• So that,

4cos|)(|

TaHT

Summary

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

achieve optimum BER performance– A design example