• 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 DigitalPulse Amplitude Modulation (PAM)
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
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)
tTsT T+Ts 2Ts
p(t)
m(t) s(t) = p(t) m(t)
Pulse shape is rectangular pulse
Ts is symbol period
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 NoiseChannel
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
AHow does the
receiver decide which bit was sent?
13 - 5
Matched Filter• Detection of pulse in presence of additive noise
Receiver knows what pulse shape it is looking forChannel 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 tntgthtwthtgty
+=+=
T is pulse period
13 - 6
power averagepower ousinstantane
)}({|)(|
SNR pulsepeak is where,max
2
20 ==
tnETgη
ηη
Matched Filter Derivation• Design of matched filter
Maximize signal power i.e. power of at t = TMinimize 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 =
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
1x(t)
0 Ts
τ
Rx(τ)
-Ts Ts
Ts
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 powerspectrum in Matlab
{ } 22* )( )( )( )( )( στδσττ =⇔=+= fPtntnER nn
{ } )( )( τxx RFfP =
N = 16384; % number of samplesgaussianNoise = randn(N,1);plot( abs(fft(gaussianNoise)) .^ 2 );
noise floor
{ } �∞
∞−+=+= dttntntntnERn )( )( )( )( )( ** τττ
{ } )(*)( )( )( )( )( )( *** τττττ −=−=−=− �∞
∞−nndttntntntnERn
13 - 92 2 2
0 | )( )(| |)(| �∞
∞−
= dfefGfHTg Tfj π
Matched Filter Derivation
• Noise
• Signal
��∞
∞−
∞
∞−
== dffHN
dffStnE N202 |)(|
2 )(} )( {
f2
0N
Noise power spectrum SW(f)
)()( )(0 fGfHfG =
�∞
∞−
= dfefGfHtg tfj )( )( )( 2 0
π
20 |)(|2
)( )()( fHN
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
13 - 10
�
�∞
∞−
∞
∞−=dffH
N
dfefGfH Tfj
20
2 2
|)(|2
| )( )(| π
η
Matched Filter Derivation• Find h(t) that maximizes pulse peak SNR η
• Schwartz’s inequalityFor vectors:
For functions:
lower bound reached iff
|||| ||||cos |||| |||| | | *
babababa
TT =⇔≤ θ
Rkxkx ∈∀= )( )( 21 φφ
���∞
∞
∞
∞
∞
∞
≤-
22
-
21
2
*2
-1 )( )( )( )( dxxdxxdxxx φφφφ
θ
a
b
13 - 11)( )( Hence,
inequality s' Schwartzby )( )(
whenoccurs which , |)(| 2
|)(| 2
|)(|
2
| )( )(
|)(| |)(| | )( )(
)()( and )()(Let
*
2 *
2
0max
2
020
2 2
222 2
2 *21
tTgkth
kefGkfH
dffGN
dffGN
dffHN
dfefGfH|
dffGdffHdfefGfH|
efGffHf
opt
Tfjopt
-
Tfj
-
Tfj
Tfj
−=∀=
=
≤=
≤
==
−
∞
∞−
∞
∞−∞
∞−
∞
∞
∞
∞−
∞
∞−
∞
∞
−
�
��
�
���
π
π
π
π
η
η
φφ
Matched Filter Derivation
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 kDuration 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
SNR2
|)(| 2
|)(| 2
0
2
0
2
0max ==== ��
∞
∞−
∞
∞− NE
dttgN
dffGN
bη
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 toFirst, integrate for T secSecond, sample at symbol period T secThird, reset integration for next time period
• Integrate and dump circuit
�
Sample and dump
h(t) = ___
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)
CommunicationChannel
input output
x(t) y(t)t
)(0 ty
-A Th
receive ‘0’ bit
tΤh+ΤbΤh
Assume that Th < Tb
t
)(1 ty receive‘1’ bit
Τh+ΤbΤh
A Th
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, …, andthreshold with threshold of zero
• How do we prevent intersymbolinterference (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
Intersymbolinterference
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
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
0Σ
ak∈{-A,A} s(t) x(t) y(t) y(ti)
AWGNw(t)
Decision
Maker
Threshold λ
Sample att=iTb
bits
Clock Tb
pulse shaper
matched filter
���
����
�=
1
00 ln4 p
pATN
boptλ
13 - 18
( ) )( )( )( )(
)(*)()( where)()()(
,i
ikkbkbiii
kbk
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)
intersymbolinterference (ISI)
noise
p(t) is centered at origin
13 - 19
) 2
(rect 21
)(
||,0
, 21
)(
Wf
WfP
Wf
WfWWfP
=
�
�
>
<<−=
Eliminating ISI in PAM• One choice for P(f) is a
rectangular pulseW is the bandwidth of the
systemInverse Fourier transform
of a rectangular pulse isis 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=
13 - 20
�
�
≤≤−
−<≤���
����
����
����
�
−−−
<≤
=
WffW
fWfffW
WfW
ffW
fP
2 || 20
2 || 22
)|(|sin1
41
|| 0 21
)(
1
111
1
π
Eliminating ISI in PAM• Another choice for P(f) is a raised cosine spectrum
• Roll-off factor gives bandwidth in excessof bandwidth W for ideal Nyquist channel
• Raised cosine pulsehas zero ISI whensampled correctly
• Let g(t) and c(t) be square root raised cosines
Wf11−=α
( )222 161
2cos
sinc )(
tWtW
Tt
tps α
απ−��
�
����
�=
ideal Nyquist channel impulse response
dampening adjusted by rolloff factor αααα
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 processMean scaled by H(0)Variance scaled by twice lowpass filter’s bandwidth
• Matched filter’s bandwidth is ½ fb
h(t)Σs(t)
Sample att = nTb
Matched filterv(t)
r(t) r(t) rn � −=k
bk Tktgats ) ( )(
)()()( tvtstr +=
r(t) = h(t) * r(t)
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 dumpSet gain of matched filter to be 1/Tb
Integrate received signal over period, scale, sample
n
Tn
nTb
Tn
nTbn
vA
dttvT
A
dttrT
r
b
b
b
b
+±=
+±=
=
�
�+
+
)(1
)(1
)1(
)1(
0-
Anr
)( nr rPn
A−Probability density function (PDF)
See Slide 13-13
13 - 23
��
���
� >=>=>+−=−=σσAv
PAvPvAPAnTsP nnnb )( )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 variableis Gaussian with
zero mean andvariance of one
��
���
�==��
���
� >=−=−∞
� σπσσσ
AQdve
AvPAnTsP
v
A
n 21
))(|error( 2
2
σ
nv
Q function on next slide
PDF for N(0, 1)
13 - 24
Q Function• Q function
• Complementary errorfunction erfc
• Relationship
�=∞
−
x
y dyexQ 2/2
21
)(π
�=∞
−
x
t dtexerfc22
)(π
��
���
�=22
1)(
xerfcxQ
Erfc[x] in Mathematica
erfc(x) in Matlab
13 - 25
( )
2
2
SNR where,
21
21
))(|error()())(|error()( error)(
σρ
ρ
A
Q�
AQ
�
AQ
�
AQ
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 errordecreases exponentially with SNR
)/())(|error( σAQAnTsP b ==
ρπρ
π
ρ
21
)( )(2
2 −−
≤≤ eQ
xe
xerfcx
ρ, positive largefor x
13 - 26
PAM Symbol Error Probability• Average signal power
GT(ω) is square root of theraised cosine spectrum
Normalization by Tsym willbe removed in lecture 15 slides
• M-level PAM amplitudes
• Assuming each symbol is equally likely
sym
nT
sym
nSignal T
aEdG
TaE
P}{
|)(| 21}{ 2
22
=×= �∞
∞−
ωωπ
[ ]sym
M
i
M
ii
symSignal T
dMid
MTl
TP
3)1( )12(
21
1 22
2
1
2
1
2 −=���
�
�
���
�
�
−=��
���
�= ��==
2, ,0 , ,1
2 ),12(
MMiidli ......+−=−=
2-PAM
d
-d
4-PAMConstellations with decision boundaries
d
-d
3 d
-3 d
13 - 27
symNoise T
Nd
NP
sym
sym2
2
21
0
2/
2/
0 == �−
ωπ
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 constellationError if and only if
where Probablity of error is
• Consider two outer levels in constellation
dnTv symR >|)(|
��
���
�=>σd
QdnTvP symR 2)|)((|
2/02 N=σ
��
���
�=>σd
QdnTvP symR ))((
two-sided power spectral density of AWGN
channel noise filtered by receiver and sampled
0
22
3)1(2
SNRNdM
P
P
Noise
Signal ×−==
13 - 28
��
���
�−=��
���
�+���
����
���
���
�−=σσσd
QM
MdQ
Md
QM
MPe
)1(2
2 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
( )13
SNR since SNR1
3
12 2
2
221
2 −==���
�
�
���
�
���
���
�
−−= M
dP
P
MQ
MM
PNoise
Signale σ
M-2 interior points 2 exterior points
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 overzero crossing
Margin over noise
t + Tsymt
13 - 30
Eye Diagram for 4-PAM
3d
d
-d
-3d
