Matched Filter & PAM - University of Texas at Austin
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Transcript Matched Filter & PAM - University of Texas at Austin
EE345S Real-Time Digital Signal Processing Lab
Spring 2006
Matched Filtering and Digital
Pulse Amplitude Modulation (PAM)
Slides by Prof. Brian L. Evans and Dr. Serene Banerjee
Dept. of Electrical and Computer Engineering
The University of Texas at Austin
Lecture 13
Outline
• PAM
• Matched Filtering
• PAM System
• Transmit Bits
• Intersymbol Interference (ISI)
– Bit error probability for binary signals
– Bit error probability for M-ary (multilevel) signals
• Eye Diagram
13 - 2
Pulse Amplitude Modulation (PAM)
• Amplitude of periodic pulse train is varied with a
sampled message signal m
– Digital PAM: coded pulses of the sampled and quantized
message signal are transmitted (next slide)
– Analog PAM: periodic pulse train with period Ts is the
carrier (below)
m(t)
s(t) = p(t) m(t)
p(t)
Ts is symbol
period
Pulse shape is
rectangular pulse
t
T
Ts T+Ts 2Ts
13 - 3
Pulse Amplitude Modulation (PAM)
• Transmission on communication channels is analog
• One way to transmit digital information is called
2-level digital PAM
x0 (t )
receive
‘0’ bit
y0 (t )
‘0’ bit
input
Tb
t
x(t)
-A
x1 (t )
A
Tb
‘1’ bit
t
Additive Noise
Channel
output
y(t)
How does the
receiver decide
which bit was sent?
Tb
-A
receive
‘1’ bit
y1 (t )
A
Tb
t
13 - 4
Matched Filter
• Detection of pulse in presence of additive noise
Receiver knows what pulse shape it is looking for
Channel memory ignored (assumed compensated by other
means, e.g. channel equalizer in receiver)
g(t)
Pulse
signal
x(t)
h(t)
y(T)
y(t)
t=T
T is pulse
period
Matched
filter
w(t)
Additive white Gaussian
noise (AWGN) with zero
mean and variance N0 /2
y (t ) g (t ) * h(t ) w(t ) * h(t )
g 0 (t ) n(t )
13 - 5
Matched Filter Derivation
• Design of matched filter
Maximize signal power i.e. power of g0 (t ) g (t ) * h(t ) at t = T
Minimize noise i.e. power of n(t ) w(t ) * h(t )
• Combine design criteria
max , where is peak pulse SNR
| g 0 (T ) |2 inst ant aneous power
2
E{n (t )}
averagepower
g(t)
x(t)
h(t)
y(T)
y(t)
t=T
Pulse
signal
w(t)
Matched
filter
13 - 6
Power Spectra
• Deterministic signal x(t)
• Autocorrelation of x(t)
w/ Fourier transform X(f)
Rx (t ) x(t ) * x* (t )
Power spectrum is square of
Maximum value at Rx(0)
absolute value of magnitude
Rx(t) is even symmetric, i.e.
response (phase is ignored)
Rx(t) = Rx(-t)
2
*
Px ( f ) X ( f ) X ( f ) X ( f )
x(t)
Multiplication in Fourier
domain is convolution in
time domain
Conjugation in Fourier domain
is reversal and conjugation
in time
X ( f ) X * ( f ) F x(t ) * x* (t )
1
0
Ts
t
Rx(t)
Ts
-Ts
Ts
13 - 7
t
Power Spectra
• Power spectrum for signal x(t) is Px ( f ) F Rx (t )
Autocorrelation of random signal n(t)
Rn (t ) E n(t ) n (t t ) n(t ) n* (t t ) dt
*
Rn (t ) E n(t ) n* (t t ) n(t ) n* (t t ) dt n(t ) * n* (t )
For zero-mean Gaussian n(t) with variance s2
Rn (t ) E n(t ) n* (t t ) s 2 (t ) Pn ( f ) s 2
• Estimate noise power
spectrum in Matlab
N = 16384; % number of samples
gaussianNoise = randn(N,1);
plot( abs(fft(gaussianNoise)) .^ 2 );
noise
floor
13 - 8
Matched Filter Derivation
g(t)
Pulse
signal
x(t)
N0
2
t=T
w(t)
Matched filter
E{ n2 (t ) } S N ( f ) df
AWGN
N0
2
|
H
(
f
)
|
df
2
• Signal g0 (t ) g (t ) * h(t )
j 2
H
(
f
)
G
(
f
)
e
f t
Matched
filter
G0 ( f ) H ( f )G( f )
f
N0
S N ( f ) SW ( f ) S H ( f )
| H ( f ) |2
2
• Noise n(t ) w(t ) * h(t )
g 0 (t )
y(T)
y(t)
h(t)
Noise power
spectrum SW(f)
df
| g0 (T ) |2 | H ( f ) G( f ) e j 2
fT
df |2
13 - 9
Matched Filter Derivation
• Find h(t) that maximizes pulse peak SNR
|
j 2
H
(
f
)
G
(
f
)
e
N0
2
fT
df |2
a
2
|
H
(
f
)
|
df
b
• Schwartz’s inequality
For vectors:
T
a
b
T
*
| a b | || a || || b || cos
|| a || || b ||
2
For functions:
( x)
1
-
*
2
( x) dx
( x)
1
-
2
dx
( x)
2
2
dx
-
lower bound reached iff 1 ( x) k 2 ( x) k R
13 - 10
Matched Filter Derivation
Let 1 ( f ) H ( f ) and 2 ( f ) G * ( f ) e j 2 f T
| H ( f ) G ( f ) e j 2 f T df |2 | H ( f ) |2 df
-
| G ( f ) |2 df
| H ( f ) G ( f ) e j 2 f T df |2
-
N0
2
| H ( f ) |2 df
2
N0
| G ( f ) |2 df
2
2
max
|
G
(
f
)
|
df , which occurs when
N 0
H opt ( f ) k G * ( f ) e j 2 f T k by Schwartz ' s inequality
Hence, hopt (t ) k g * (T t )
13 - 11
Matched Filter
• Given transmitter pulse shape g(t) of duration T,
matched filter is given by hopt(t) = k g*(T-t) for all k
Duration and shape of impulse response of the optimal filter is
determined by pulse shape g(t)
hopt(t) is scaled, time-reversed, and shifted version of g(t)
• Optimal filter maximizes peak pulse SNR
2Eb
2
2
2
2
max
| G( f ) | df
| g (t ) | dt
SNR
N 0
N 0
N0
Does not depend on pulse shape g(t)
Proportional to signal energy (energy per bit) Eb
Inversely proportional to power spectral density of noise
13 - 12
Matched Filter for Rectangular Pulse
• Matched filter for causal rectangular pulse has an
impulse response that is a causal rectangular pulse
• Convolve input with rectangular pulse of duration
T sec and sample result at T sec is same as to
First, integrate for T sec
Sample and dump
Second, sample at symbol period T sec
Third, reset integration for next time period
• Integrate and dump circuit
h(t) = ___
t=kT
T
13 - 13
Transmit One Bit
• Analog transmission over communication channels
• Two-level digital PAM over channel that has
memory but does not add noise
x0 (t )
‘0’ bit
input
Tb
t
x(t)
-A
x1 (t )
A
Tb
‘1’ bit
t
output
Communication
Channel
Model channel as
LTI system with
impulse response
h(t)
y(t)
Assume that Th < Tb
receive
‘0’ bit
Th
Th+Tb
y1 (t )
receive
‘1’ bit
t
-A Th
h(t )
A Th
1
Th
y0 (t )
t
Th
Th+Tb
13 - 14
t
Transmit Two Bits (Interference)
• Transmitting two bits (pulses) back-to-back
will cause overlap (interference) at the receiver
h(t )
x(t )
*
A
y(t )
=
1
Th+Tb
2Tb
Tb
t
Th
Tb
t
t
-A Th
‘1’ bit
‘0’ bit
Assume that Th < Tb
‘1’ bit
‘0’ bit
• Sample y(t) at Tb, 2 Tb, …, and
threshold with threshold of zero
• How do we prevent intersymbol
interference (ISI) at the receiver?
Intersymbol
interference
13 - 15
Transmit Two Bits (No Interference)
• Prevent intersymbol interference by waiting Th
seconds between pulses (called a guard period)
h(t )
x(t )
*
A
y(t )
=
1
Th+Tb
Tb
Th+Tb
t
Th
Th Tb
t
t
-A Th
‘1’ bit
‘0’ bit
Assume that Th < Tb
‘1’ bit
‘0’ bit
• Disadvantages?
13 - 16
Digital 2-level PAM System
ak{-A,A} s(t)
bi
bits
PAM
Clock Tb
g(t)
pulse
shaper
x(t)
y(t) y(ti)
h(t)
c(t)
Sample at Maker
t=iTb
0
AWGN matched
Threshold l
filter Clock T
w(t)
b
lopt
Transmitter
1
Decision
Channel
• Transmitted signal s(t )
Receiver
a
k
p
N0
ln 0
4 ATb p1
g (t k Tb )
k
• Requires synchronization of clocks between
transmitter and receiver
13 - 17
Digital PAM Receiver
• Why is g(t) a pulse and not an impulse?
Otherwise, s(t) would require infinite bandwidth
s(t ) ak (t k Tb )
k
Since we cannot send an signal of infinite bandwidth, we limit
its bandwidth by using a pulse shaping filter
• Neglecting noise, would like y(t) = g(t) * h(t) * c(t)
to be a pulse, i.e. y(t) = p(t) , to eliminate ISI
y(t ) ak p(t kTb ) n(t ) where n(t ) w(t ) * c(t )
k
y(ti ) ai p(ti iTb ) ak p(i k )Tb n(ti )
k , k i
actual value
(note that ti = i Tb)
intersymbol
interference (ISI)
p(t) is
centered
at origin
noise
13 - 18
Eliminating ISI in PAM
• One choice for P(f) is a
rectangular pulse
W is the bandwidth of the
system
Inverse Fourier transform
of a rectangular pulse is
is a sinc function
p(t ) sinc(2 W t )
1
2 W ,W f W
P( f )
0
, | f | W
1
f
P( f )
rect(
)
2W
2W
• This is called the Ideal Nyquist Channel
• It is not realizable because the pulse shape is not
causal and is infinite in duration
13 - 19
Eliminating ISI in PAM
• Another choice for P(f) is a raised cosine spectrum
1
P( f )
4W
1
2W
1 sin (| f | W )
2W 2 f
1
0
0 | f | f1
f1 | f | 2W f1
2W f1 | f | 2W
• Roll-off factor gives bandwidth in excess
f1
1
W
of bandwidth W for ideal Nyquist channel
• Raised cosine pulse p(t ) sinc t cos2 2 W2 t 2
T 1 16 W t
s
has zero ISI when
sampled correctly ideal Nyquist channel dampening adjusted by
impulse response rolloff factor
• Let g(t) and c(t) be square root raised cosines
13 - 20
Bit Error Probability for 2-PAM
• Tb is bit period (bit rate is fb = 1/Tb)
s(t)
r(t)
r(t)
rn
s(t ) ak g (t k Tb )
k
h(t)
Sample at
Matched
filter
r (t ) s(t ) v(t )
t = nTb
r(t) = h(t) * r(t)
v(t)
v(t) is AWGN with zero mean and variance s2
• Lowpass filtering a Gaussian random process
produces another Gaussian random process
Mean scaled by H(0)
Variance scaled by twice lowpass filter’s bandwidth
• Matched filter’s bandwidth is ½ fb
13 - 21
Bit Error Probability for 2-PAM
• Binary waveform (rectangular pulse shape) is A
over nth bit period nTb < t < (n+1)Tb
• Matched filtering by integrate and dump
Set gain of matched filter to be 1/Tb
Integrate received signal over period, scale, sample
1
rn
Tb
( n 1)Tb
See Slide
13-13
Prn (rn )
r (t ) dt
nTb
1
A
Tb
A vn
( n 1)Tb
v(t ) dt
nTb
-
A
rn
0
A
Probability density function (PDF)
13 - 22
Bit Error Probability for 2-PAM
• Probability of error given that the transmitted
pulse has an amplitude of –A
vn A
P(error| s(nTb ) A) P( A vn 0) P(vn A) P
s s
vn / s
• Random variable
vn
is Gaussian with
zero mean and
variance of one
PDF for
N(0, 1)
s
0 A /s
1
vn A
P(error | s (nT ) A) P
e
s s A 2
v2
2
A
dv Q
s
s
Q function on next slide
13 - 23
Q Function
• Q function
1 y2 / 2
Q( x)
dy
e
2 x
• Complementary error
function erfc
2 t 2
erfc( x)
e dt
• Relationship
x
1
x
Q( x) erfc
2
2
Erfc[x] in Mathematica
erfc(x) in Matlab
13 - 24
Bit Error Probability for 2-PAM
• Probability of error given that the transmitted pulse
has an amplitude of A
P(error| s(nTb ) A) Q( A / s )
• Assume that 0 and 1 are equally likely bits
P(error) P( A) P(error| s (nTb ) A) P( A) P(error| s (nTb ) A)
1 A 1 A
A
Q Q Q Q
2 σ 2 σ
σ
A2
x2
where, SNR 2
1 e 2
erfc( x)
Q( )
x
2
• Probablity of error
s
e
decreases exponentially with SNR
for large positivex,
13 - 25
PAM Symbol Error Probability
• Average signal power
PSignal
E{an2 } 1
Tsym
2
3d
E{an2 }
| GT (w ) | dw Tsym
2
GT(w) is square root of the
raised cosine spectrum
Normalization by Tsym will
be removed in lecture 15 slides
• M-level PAM amplitudes
li d (2i 1),
i
M
M
1, . . . , 0, . . . ,
2
2
d
d
-d
-d
-3 d
2-PAM
4-PAM
Constellations with
decision boundaries
• Assuming each symbol is equally likely
PSignal
1 M 2 1 2
li
Tsym i 1 T M
d2
2
2
d (2i 1) (M 1)
3Tsym
i 1
M
2
13 - 26
PAM Symbol Error Probability
• Noise power and SNR
PNoise
1
2
w sym / 2
w
sym / 2
N0
N0
dw
2
2Tsym
two-sided power spectral
density of AWGN
SNR
PSignal
PNoise
2( M 2 1) d 2
3
N0
• Assume ideal channel,
i.e. one without ISI
x(nTsym ) an vR (nTsym )
channel noise filtered
by receiver and sampled
• Consider M-2 inner
levels in constellation
Error if and only if
| vR (nTsym ) | d
where s 2 N0 / 2
Probablity of error is
d
P(| vR (nTsym ) | d ) 2 Q
s
• Consider two outer
levels in constellation
d
P(vR (nTsym ) d ) Q
s
13 - 27
PAM Symbol Error Probability
• Assuming that each symbol is equally likely,
symbol error probability for M-level PAM
M 2
2 d 2(M 1) d
d
Pe
2 Q
Q
Q
M
M
s M s
s
M-2 interior points
2 exterior points
• Symbol error probability in terms of SNR
1
PSignal
M 1 3
d2
2
2
Pe 2
Q 2
SNR since SNR
M
1
2
M
PNoise 3s
M 1
13 - 28
Eye Diagram
• PAM receiver analysis and troubleshooting
Sampling instant
M=2
Margin over noise
Distortion over
zero crossing
Slope indicates
sensitivity to
timing error
Interval over which it can be sampled
t - Tsym
t
t + Tsym
• The more open the eye, the better the reception
13 - 29
Eye Diagram for 4-PAM
3d
d
-d
-3d
13 - 30