Post on 21-Dec-2015
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