1. Slides by Prof. Brian L. Evans and Dr. Serene Banerjee
Dept. of Electrical and Computer Engineering
The University of Texas at Austin
EE345S Real-Time Digital Signal Processing Lab Spring 2006
Lecture 13
Matched Filtering and Digital
Pulse Amplitude Modulation (PAM)

2. 13 - 2
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

3. 13 - 3
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)
t
TsT T+Ts 2Ts
p(t)
m(t) s(t) = p(t) m(t)
Pulse shape is
rectangular pulse
Ts is symbol
period

4. 13 - 4
Pulse Amplitude Modulation (PAM)
• Transmission on communication channels is analog
• One way to transmit digital information is called
2-level digital PAM
Τb t
)(1 tx
A
‘1’ bit
Additive Noise
Channel
input output
x(t) y(t)
Τb
)(0 tx
-A
‘0’ bit
t
receive
‘0’ bit
receive
‘1’ bit
)(0 ty
Τb
-A
Τb t
)(1 ty
A
How does the
receiver decide
which bit was sent?

5. 13 - 5
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)
Additive white Gaussian
noise (AWGN) with zero
mean and variance N0 /2
g(t)
Pulse
signal
w(t)
x(t) h(t) y(t)
t = T
y(T)
Matched
filter
)()(
)(*)()(*)()(
0 tntg
thtwthtgty
+=
+=
T is pulse
period

6. 13 - 6
poweraverage
powerousinstantane
)}({
|)(|
SNRpulsepeakiswhere,max
2
2
0
==
tnE
Tg
η
ηη
Matched Filter Derivation
• Design of matched filter
Maximize signal power i.e. power of at t = T
Minimize noise i.e. power of
• Combine design criteria
g(t)
Pulse
signal
w(t)
x(t) h(t) y(t)
t = T
y(T)
Matched
filter
)(*)()( thtwtn =
)(*)()(0 thtgtg =

7. 13 - 7
Power Spectra
• Deterministic signal x(t)
w/ Fourier transform X(f)
Power spectrum is square of
absolute value of magnitude
response (phase is ignored)
Multiplication in Fourier
domain is convolution in
time domain
Conjugation in Fourier domain
is reversal and conjugation
in time
• Autocorrelation of x(t)
Maximum value at Rx(0)
Rx(τ) is even symmetric, i.e.
Rx(τ) = Rx(-τ))()()()( *2
fXfXfXfPx ==
{ })(*)()()( **
ττ −= xxFfXfX
)(*)()( *
τττ −= xxRx
t
1
x(t)
0 Ts
τ
Rx(τ)
-Ts Ts
Ts

8. 13 - 8
Power Spectra
• Power spectrum for signal x(t) is
Autocorrelation of random signal n(t)
For zero-mean Gaussian n(t) with variance σ2
• Estimate noise power
spectrum in Matlab
{ } 22*
)()()()()( στδσττ =⇔=+= fPtntnER nn
{ })()( τxx RFfP =
N = 16384; % number of samples
gaussianNoise = randn(N,1);
plot( abs(fft(gaussianNoise)) .^ 2 );
noise
floor
{ } ∫
∞
∞−
+=+= dttntntntnERn )()()()()( **
τττ
{ } )(*)()()()()()( ***
τττττ −=−=−=− ∫
∞
∞−
nndttntntntnERn

9. 13 - 9
222
0 |)()(||)(| ∫
∞
∞−
= dfefGfHTg Tfj π
Matched Filter Derivation
• Noise
• Signal
∫∫
∞
∞−
∞
∞−
== dffH
N
dffStnE N
202
|)(|
2
)(})({
f
2
0N
Noise power
spectrum SW(f)
)()()(0 fGfHfG =
∫
∞
∞−
= dfefGfHtg tfj
)()()( 2
0
π
20
|)(|
2
)()()( fH
N
fSfSfS HWN ==
g(t)
Pulse
signal w(t)
x(t) h(t) y(t)
t = T
y(T)
Matched filter
)(*)()(0 thtgtg =
)(*)()( thtwtn =
AWGN Matched
filter

10. 13 - 10
∫
∫
∞
∞−
∞
∞−
=
dffH
N
dfefGfH Tfj
20
22
|)(|
2
|)()(| π
η
Matched Filter Derivation
• Find h(t) that maximizes pulse peak SNR η
• Schwartz’s inequality
For vectors:
For functions:
lower bound reached iff
||||||||
cos|||||||||| *
ba
ba
baba
T
T
=⇔≤ θ
Rkxkx ∈∀= )()( 21 φφ
∫∫∫
∞
∞
∞
∞
∞
∞
≤
-
2
2
-
2
1
2
*
2
-
1 )()()()( dxxdxxdxxx φφφφ
θ
a
b

11. 13 - 11
)()(Hence,
inequalitys'Schwartzby)()(
whenoccurswhich,|)(|
2
|)(|
2
|)(|
2
|)()(
|)(||)(||)()(
)()(and)()(Let
*
2*
2
0
max
2
020
22
2222
2*
21
tTgkth
kefGkfH
dffG
N
dffG
N
dffH
N
dfefGfH|
dffGdffHdfefGfH|
efGffHf
opt
Tfj
opt
-
Tfj
-
Tfj
Tfj
−=
∀=
=
≤=
≤
==
−
∞
∞−
∞
∞−
∞
∞−
∞
∞
∞
∞−
∞
∞−
∞
∞
−
∫
∫
∫
∫
∫∫∫
π
π
π
π
η
η
φφ
Matched Filter Derivation

12. 13 - 12
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
Does not depend on pulse shape g(t)
Proportional to signal energy (energy per bit) Eb
Inversely proportional to power spectral density of noise
SNR
2
|)(|
2
|)(|
2
0
2
0
2
0
max ==== ∫∫
∞
∞−
∞
∞−
N
E
dttg
N
dffG
N
b
η

13. 13 - 13
t=kT T
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
Second, sample at symbol period T sec
Third, reset integration for next time period
• Integrate and dump circuit
∫
Sample and dump
h(t) = ___

14. 13 - 14
Transmit One Bit
• Analog transmission over communication channels
• Two-level digital PAM over channel that has
memory but does not add noise
Τh t
)(th
1
Τb t
)(1 tx
A
‘1’ bit
Τb
)(0 tx
-A
‘0’ bit
Model channel as
LTI system with
impulse response
h(t)
Communication
Channel
input output
x(t) y(t)t
)(0 ty
-A Th
receive
‘0’ bit
t
Τh+ΤbΤh
Assume that Th < Tb
t
)(1 ty receive
‘1’ bit
Τh+Τb
Τh
A Th

15. 13 - 15
Transmit Two Bits (Interference)
• Transmitting two bits (pulses) back-to-back
will cause overlap (interference) at the receiver
• Sample y(t) at Tb, 2 Tb, …, and
threshold with threshold of zero
• How do we prevent intersymbol
interference (ISI) at the receiver?
Τh t
)(th
1
Assume that Th < Tb
tΤb
)(tx
A
‘1’ bit ‘0’ bit
2Τb
* =
)(ty
-A Th
tΤb
‘1’ bit ‘0’ bit
Τh+Τb
Intersymbol
interference

16. 13 - 16
Transmit Two Bits (No Interference)
• Prevent intersymbol interference by waiting Th
seconds between pulses (called a guard period)
• Disadvantages?
Τh t
)(th
1
Assume that Th < Tb
* =
tΤb
)(tx
A
‘1’ bit ‘0’ bit
Τh+Τb
t
)(ty
-A Th
Τb
‘1’ bit ‘0’ bit
Τh+Τb
Τh

17. 13 - 17
∑ −=
k
bk Tktgats )()(
Digital 2-level PAM System
• Transmitted signal
• Requires synchronization of clocks between
transmitter and receiver
Transmitter Channel Receiver
bi
Clock Tb
PAM g(t) h(t) c(t)
1
0
Σ
ak∈{-A,A} s(t) x(t) y(t) y(ti)
AWGN
w(t)
Decision
Maker
Threshold λ
Sample at
t=iTb
bits
Clock Tb
pulse
shaper
matched
filter






=
1
00
ln
4 p
p
AT
N
b
optλ

18. 13 - 18
( ) )()()()(
)(*)()(where)()()(
,
i
ikk
bkbiii
k
bk
tnTkipaiTtpaty
tctwtntnkTtpaty
+−+−=⇒
=+−=
∑
∑
≠
µµ
µ
∑ −=
k
bk Tktats )()( δ
Digital PAM Receiver
• Why is g(t) a pulse and not an impulse?
Otherwise, s(t) would require infinite bandwidth
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) = µ p(t) , to eliminate ISI
actual value
(note that ti = i Tb)
intersymbol
interference (ISI)
noise
p(t) is
centered
at origin

19. 13 - 19
)
2
(rect
2
1
)(
||,0
,
2
1
)(
W
f
W
fP
Wf
WfW
WfP
=





>
<<−
=
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
• This is called the Ideal Nyquist Channel
• It is not realizable because the pulse shape is not
causal and is infinite in duration
)2(sinc)( tWtp π=

20. 13 - 20









≤≤−
−<≤













−
−
−
<≤
=
WffW
fWff
fW
Wf
W
ff
W
fP
2||20
2||
22
)|(|
sin1
4
1
||0
2
1
)(
1
11
1
1
π
Eliminating ISI in PAM
• Another choice for P(f) is a raised cosine spectrum
• Roll-off factor gives bandwidth in excess
of bandwidth W for ideal Nyquist channel
• Raised cosine pulse
has zero ISI when
sampled correctly
• Let g(t) and c(t) be square root raised cosines
W
f1
1−=α
( )
222
161
2cos
sinc)(
tW
tW
T
t
tp
s α
απ
−





=
ideal Nyquist channel
impulse response
dampening adjusted by
rolloff factor α

21. 13 - 21
Bit Error Probability for 2-PAM
• Tb is bit period (bit rate is fb = 1/Tb)
v(t) is AWGN with zero mean and variance σ2
• Lowpass filtering a Gaussian random process
produces another Gaussian random process
Mean scaled by H(0)
Variance scaled by twice lowpass filter’s bandwidth
h(t)Σ
s(t)
Sample at
t = nTb
Matched
filterv(t)
r(t) r(t) rn ∑ −=
k
bk Tktgats )()(
)()()( tvtstr +=
r(t) = h(t) * r(t)

22. 13 - 22
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
n
Tn
nTb
Tn
nTb
n
vA
dttv
T
A
dttr
T
r
b
b
b
b
+±=
+±=
=
∫
∫
+
+
)(
1
)(
1
)1(
)1(
0-
A
nr
)( nr rPn
A−
Probability density function (PDF)
See Slide
13-13

23. 13 - 23






>=>=>+−=−=
σσ
Av
PAvPvAPAnTsP n
nnb )()0())(|error(
0 σ/A
σ/nv
Bit Error Probability for 2-PAM
• Probability of error given that the transmitted
pulse has an amplitude of –A
• Random variable
is Gaussian with
zero mean and
variance of one






==





>=−=
−
∞
∫ σπσσ
σ
A
Qdve
Av
PAnTsP
v
A
n
2
1
))(|error( 2
2
σ
nv
Q function on next slide
PDF for
N(0, 1)

24. 13 - 24
Q Function
• Q function
• Complementary error
function erfc
• Relationship
∫=
∞
−
x
y
dyexQ 2/2
2
1
)(
π
∫=
∞
−
x
t
dtexerfc
22
)(
π






=
22
1
)(
x
erfcxQ
Erfc[x] in Mathematica
erfc(x) in Matlab

25. 13 - 25
( )
2
2
SNRwhere,
2
1
2
1
))(|error()())(|error()(error)(
σ
ρ
ρ
A
Q
σ
A
Q
σ
A
Q
σ
A
Q
AnTsPAPAnTsPAPP bb
==
=





=





+





=
−=−+==
Bit Error Probability for 2-PAM
• Probability of error given that the transmitted pulse
has an amplitude of A
• Assume that 0 and 1 are equally likely bits
• Probablity of error
decreases exponentially with SNR
)/())(|error( σAQAnTsP b ==
ρπ
ρ
π
ρ
2
1
)()(
2
2 −
−
≤⇒≤
e
Q
x
e
xerfc
x
ρ,positivelargefor x

26. 13 - 26
PAM Symbol Error Probability
• Average signal power
GT(ω) is square root of the
raised cosine spectrum
Normalization by Tsym will
be removed in lecture 15 slides
• M-level PAM amplitudes
• Assuming each symbol is equally likely
sym
n
T
sym
n
Signal
T
aE
dG
T
aE
P
}{
|)(|
2
1}{ 2
2
2
=×= ∫
∞
∞−
ωω
π
[ ]
sym
M
i
M
i
i
sym
Signal
T
d
Mid
MT
l
T
P
3
)1()12(
211 2
2
2
1
2
1
2
−=










−=





= ∑∑ ==
2
,,0,,1
2
),12(
MM
iidli ......+−=−=
2-PAM
d
-d
4-PAM
Constellations with
decision boundaries
d
-d
3 d
-3 d

27. 13 - 27
sym
Noise
T
N
d
N
P
sym
sym
222
1 0
2/
2/
0
== ∫−
ω
π
ω
ω
)()( symRnsym nTvanTx +=
PAM Symbol Error Probability
• Noise power and SNR
• Assume ideal channel,
i.e. one without ISI
• Consider M-2 inner
levels in constellation
Error if and only if
where
Probablity of error is
• Consider two outer
levels in constellation
dnTv symR >|)(|






=>
σ
d
QdnTvP symR 2)|)((|
2/0
2
N=σ






=>
σ
d
QdnTvP symR ))((
two-sided power spectral
density of AWGN
channel noise filtered
by receiver and sampled
0
22
3
)1(2
SNR
N
dM
P
P
Noise
Signal
×
−
==

28. 13 - 28





−
=





+










−
=
σσσ
d
Q
M
Md
Q
M
d
Q
M
M
Pe
)1(22
2
2
PAM Symbol Error Probability
• Assuming that each symbol is equally likely,
symbol error probability for M-level PAM
• Symbol error probability in terms of SNR
( )1
3
SNRsinceSNR
1
31
2 2
2
2
2
1
2
−==
















−
−
= M
d
P
P
M
Q
M
M
P
Noise
Signal
e
σ
M-2 interior points 2 exterior points

29. 13 - 29
Eye Diagram
• PAM receiver analysis and troubleshooting
• The more open the eye, the better the reception
M=2
t - Tsym
Sampling instant
Interval over which it can be sampled
Slope indicates
sensitivity to
timing error
Distortion over
zero crossing
Margin over noise
t + Tsymt

30. 13 - 30
Eye Diagram for 4-PAM
3d
d
-d
-3d