1. Matched Filtering and Digital
Pulse Amplitude Modulation (PAM)
• By
• Ravi Kumar Thakur
• Nrusingha CH. Pradhan
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. 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)
13 - 3
m(t) s(t) = p(t) m(t)
t
p(t)
Ts T T+Ts 2Ts
Pulse shape is
rectangular pulse
Ts is symbol
period
4. Pulse Amplitude Modulation (PAM)
• Transmission on communication channels is analog
• One way to transmit digital information is called
2-level digital PAM
13 - 4
Tb t
( ) 1 x t
A
‘1’ bit
input output
Additive Noise
x(t) Channel
y(t)
Tb
( ) 0 x t
-A
‘0’ bit
t
receive
‘0’ bit
receive
‘1’ bit
( ) 0 y t
Tb
-A
Tb t
( ) 1 y t
A
How does the
receiver decide
which bit was sent?
5. y t = g t h t +
w t h t
( ) ( )* ( ) ( )* ( )
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)
g(t)
Pulse
signal
w(t)
x(t) h(t) y(t)
Additive white Gaussian
noise (AWGN) with zero
mean and variance N0 /2
t = T
y(T)
Matched
filter
0 g t n t
= +
( ) ( )
T is pulse
period
6. Maximize signal power i.e. power of at t = T
Minimize noise i.e. power of
13 - 6
Matched Filter Derivation
• Design of matched filter
• Combine design criteria
h h
n(t) = w(t)*h(t)
max , where is peak pulse SNR
instantaneous power
average power
h g T
| ( ) |
= 0 =
E n 2
t
2
{ ( )}
g(t)
Pulse
signal
w(t)
x(t) h(t) y(t)
t = T
y(T)
Matched
filter
( ) ( )* ( ) 0 g t = g t h t
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(t) is even symmetric, i.e.
( ) ( ) 2 ( ) *( ) Rx(t) = Rx(-t) P f X f X f X f x = =
X ( f ) X * ( f ) = F{ x(t )* x*(-t ) }
R (t ) = x(t )* x* (-t ) x
t
1
x(t)
0 Ts
t
Rx(t)
Ts
-Ts Ts
8. R = E n t n t + = n t n t + dt n (t ) ( ) * ( t ) ( ) * ( t )
(-t ) = { ( ) * ( -t ) } = ò ( ) * ( -t ) = (t )* * (-t ) ¥
13 - 8
Power Spectra
• Power spectrum for signal x(t) is
Autocorrelation of random signal n(t)
R E n t n t n t n t dt n n n
For zero-mean Gaussian n(t) with variance s2
R (t ) = E{ n(t) n* (t +t ) } =s 2 d (t ) Û P ( f ) =s 2 n n
• Estimate noise power
spectrum in Matlab
( ) { (t ) } x x P f = F R
N = 16384; % number of samples
gaussianNoise = randn(N,1);
plot( abs(fft(gaussianNoise)) .^ 2 );
noise
floor
{ } ò¥
-¥
-¥
9. S ( f ) S ( f ) S ( f ) N H f N W H = =
13 - 9
Matched Filter Derivation
n(t) = w(t)*h(t)
E n 2 t = S f df = N 0 | H ( f ) |2
df N
g (t) = H( f ) G( f ) e j 2 f t df
0
¥
¥
¥
| g ( T ) | = | ò H ( f ) G ( f ) e j p f T df
|
0 2 2 2
-¥
• Noise
• Signal
ò ò
-¥
¥
-¥
2
{ ( ) } ( )
0 N
2
f
Noise power
spectrum SW(f)
( ) ( ) ( ) 0 G f = H f G f
ò
-¥
p
0 | ( ) |2
2
g(t)
Pulse
signal w(t)
x(t) h(t) y(t)
t = T
y(T)
Matched filter
( ) ( )* ( ) 0 g t = g t h t
AWGN Matched
filter
10. a
b
13 - 10
H f G f e j f T df
| ( ) ( ) p |
ò
ò
¥
-¥
¥
= -¥
2 2
N 0 | H ( f ) |
2
df
2
h
Matched Filter Derivation
• Find h(t) that maximizes pulse peak SNR h
• Schwartz’s inequality
For vectors:
T
q
a b a b a b
¥
¥
ò ò ò
For functions:
lower bound reached iff
|| || || ||
| * | || || || || cos
a b
T £ Û q =
¥
¥
(x) = k (x) "k ÎR 1 2 f f
¥
¥
£
-
2
2
-
2
1
2
*
2
-
1 f (x) f (x) dx f (x) dx f (x) dx
11. Matched Filter Derivation
* 2
f = H f f =
G f e
Let ( ) ( ) and ( ) ( )
1 2
ò ò ò
2 2 2 2
| H ( f ) G ( f ) e df | | H ( f ) | df | G ( f ) |
df
2 2
| H f G f e df
( ) ( ) |
= £
2 | G ( f ) | 2
df
, which occurs when
* 2
N
h
H f = k G f e "
k
( ) ( ) by Schwartz 's inequality
13 - 11
Hence, ( ) ( )
2 | ( ) |
| ( ) |
2
*
0
max
2
0 2 0
h t k g T t
G f df
N
N H f df
opt
j f T
opt
-
j f T
-
j f T
j f T
= -
=
£
-
¥
-¥
¥
-¥
¥
-¥
¥
¥
¥
-¥
¥
-¥
¥
¥
-
ò
ò
ò
ò
p
p
p
p
h
f f
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
¥
¥
g t dt E
2 | ( ) | 2 | ( ) | 2 SNR
max = ò = ò = =
h b
0
2
G f df
N
-¥ 0
2
0
-¥
N
N
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. 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
13 - 13
First, integrate for T sec
Second, sample at symbol period T sec
Third, reset integration for next time period
• Integrate and dump circuit
t=kT T
ò
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
input output
h(t)
1
Th t
Tb
Tb t
( ) 0 x t
( ) 1 x t
A
‘1’ bit
-A
‘0’ bit
Communication
Model channel as
LTI system with
impulse response
h(t)
Channel
t x(t) y(t)
( ) 0 y t
-A Th
receive
‘0’ bit
t
Th+Tb Th
Assume that Th < Tb
t
( ) 1 y t receive
‘1’ bit
Th+Tb Th
A Th
15. Transmit Two Bits (Interference)
• Transmitting two bits (pulses) back-to-back
will cause overlap (interference) at the receiver
13 - 15
h(t)
* =
1
Th t
Assume that Th < Tb
x(t)
2Tb
T t b
A
‘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?
y(t)
-A Th
Th+Tb
T t b
‘1’ bit ‘0’ bit
Intersymbol
interference
16. Transmit Two Bits (No Interference)
• Prevent intersymbol interference by waiting Th
seconds between pulses (called a guard period)
13 - 16
* =
Th+Tb
• Disadvantages?
h(t)
1
Th t
Assume that Th < Tb
T t b
x(t)
A
‘1’ bit ‘0’ bit
t
y(t)
-A Th
Tb
Th+Tb
Th
‘1’ bit ‘0’ bit
17. æ
p
N
0 ln 0
4 p
13 - 17
Digital 2-level PAM System
akÎ{-A,A} s(t) x(t) y(t) y(ti)
PAM g(t) h(t) c(t)
Sample at
t=iTb
Transmitter Channel Receiver
=å -
k b s(t) a g(t k T )
• Transmitted signal
k
• Requires synchronization of clocks between
transmitter and receiver
bi
Clock Tb
1
0
S
AWGN
w(t)
Decision
Maker
Threshold l
bits
Clock Tb
pulse
shaper
matched
filter
ö
÷ ÷ø
ç çè
=
1
AT
b
opt l
18. 13 - 18
• Why is g(t) a pulse and not an impulse?
Otherwise, s(t) would require infinite bandwidth
å
Digital PAM Receiver
k b s(t) a d (t k T )
y t = a p t - kT + n t n t =
w t c t
Þ = - + - +
( ) ( ) ( ) where ( ) ( )* ( )
( ) ( ) (( ) ) ( )
å
i i i b k b
,
i
k k i
k
k b
y t a p t iT a p i k T n t
¹
m m
m
=å -
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) = m p(t) , to eliminate ISI
actual value
(note that ti = i Tb)
intersymbol
interference (ISI)
noise
p(t) is
centered
at origin
19. Eliminating ISI in PAM
W f W
- < <
13 - 19
)
2
,
rect(
ì
( ) 1
2
0 ,| |
2
1
( )
f
W
W
P f
f W
P f W
=
ïî
ïí
>
=
• 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 p W t)
• This is called the Ideal Nyquist Channel
• It is not realizable because the pulse shape is not
causal and is infinite in duration
20. • Another choice for P(f) is a raised cosine spectrum
fa =1- 1
13 - 20
Eliminating ISI in PAM
ì
ï ï ï
í
ï ï ï
î
- < £ ÷ ÷ø
f | f | 2
W f
1 1
- £ £
ö
æ
ç çè
ö
÷ ÷ø
æ
f W
- -
1 sin (| | )
ç çè
-
£ <
=
W f f W
W f
1
W
f f
W
P f
2 2
0 2 | | 2
4
0 | |
2
1
( )
1
1
1
p
• 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
( )
æ
p t t
p a
ö
( ) sinc cos 2
W t
- ÷ ÷ø
1 16 2 W 2 t
2
T
s a
ç çè
=
ideal Nyquist channel
impulse response
dampening adjusted by
rolloff factor a
21. Bit Error Probability for 2-PAM
k b s(t) a g(t k T )
r(t) = s(t) + v(t)
r(t) = h(t) * r(t)
13 - 21
• Tb is bit period (bit rate is fb = 1/Tb)
s(t)
r(t) r(t) rn =å -
S h(t)
Sample at
Matched
t = nTv(t) filter
b • Matched filter’s bandwidth is ½ fk
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
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
See Slide
13-13
13 - 22
Set gain of matched filter to be 1/Tb
Integrate received signal over period, scale, sample
n T
( 1)
1 ( )
n
n T
( 1)
b nT
b nT
n
=
= ± +
A v
v t dt
T
A
r t dt
T
r
b
b
b
b
=± +
ò
ò
+
+
1 ( )
- 0
A
n r
( ) r n P r n
- A
Probability density function (PDF)
23. Bit Error Probability for 2-PAM
• Probability of error given that the transmitted
pulse has an amplitude of –A
P s nT A P A v P v A P vn A
ö çè
÷ø
PDF for
N(0, 1)
13 - 23
= - = - + > = > = æ >
s s
b n n (error | ( ) ) ( 0) ( )
n /s v
0 A/s
• Random variable
v
is Gaussian with
zero mean and
variance of one
P s nT A P v A e dv Q A
ö çè
÷ø
= - = æ > ¥ -
(error | ( ) ) 1 2
ö çè
æ = = ÷ø
s s ò p s
s
v
A
n
2
2
s n
Q function on next slide
24. 13 - 24
Q Function
• Q function
¥
( ) 1
Q x e y2 / 2dy
= ò
p
• Complementary error
function erfc
erfc x e t dt ( ) 2 2
• Relationship
-
x
2
¥
= ò
-
x
p
Q(x) 1 erfc x
ö çè
÷ø
= æ
2 2
Erfc[x] in Mathematica
erfc(x) in Matlab
25. Bit Error Probability for 2-PAM
• Probability of error given that the transmitted pulse
has an amplitude of A
P(error | s(nT ) A) Q(A/s ) b = =
• Assume that 0 and 1 are equally likely bits
P = P A P s nT = A + P - A P s nT = -
A b b
(error) ( ) (error | ( ) ) ( ) (error | ( ) )
2 - 2 -
£ Þ Q £ e
( ) ( ) 1
13 - 25
( )
1
2
2
1
Q A
= æ
where, SNR
2
2
s
r
r
A
Q
Q A
ö σ
çè
Q A
ö σ
çè
σ
= =
= ÷ø
æ = ÷ø
æ + ÷øö çè
• Probablity of error
decreases exponentially with SNR
p r
r
p
r
2
x
erfc x e
x
for large positive x,r
26. PAM Symbol Error Probability
3 d
d
-d
-3 d
13 - 26
• Average signal power
¥
{ 2
} 1 2
P E a | G ( ) | d E { a
}
= ´ ò w 2
w
=
Signal T
T
n
T
sym
p
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
n
sym
l d(2i 1), i M M i = - = - + . . . . . .
d
-d
• Assuming each symbol is equally likely
ö
÷ ÷ ÷
æ
1 1 2 (2 1) ( 1) 2
[ ]
= æå å
- = ÷ø
Signal T
sym
M
i
M
i
i
sym
d i M d
T M
l
T
P
3
2
2
1
2
1
2 = -
ø
ç ç ç
è
ö çè
= =
2
1, , 0, ,
2
2-PAM
4-PAM
Constellations with
decision boundaries
27. PAM Symbol Error Probability
P v nT d Q d R sym (| ( ) | ) 2
ö çè
P v nT d Q d R sym ( ( ) )
ö çè
13 - 27
• Noise power and SNR
/ 2
P N d N
1 0
= ò 0 =
Noise T
sym
sym
2
2
2
sym / 2
-
w
p
w
w
2 2
3
0
M d
P
= Signal = - ´
SNR 2( 1)
N
• Assume ideal channel,
i.e. one without ISI
( ) ( ) sym n R sym x nT =a +v nT
• Consider M-2 inner
levels in constellation
Error if and only if
v nT d R sym | ( ) |>
where
Probablity of error is
• Consider two outer
levels in constellation
÷ø
> = æ
s
/ 2 0
s 2 = N
÷ø
> = æ
s
two-sided power spectral
density of AWGN
channel noise filtered
by receiver and sampled
P
Noise
28. PAM Symbol Error Probability
13 - 28
• Assuming that each symbol is equally likely,
symbol error probability for M-level PAM
Q d
ö çè
÷ø
Q d M
M
æ - = ÷ø
+ æ ÷ ÷ø
ö çè
ö
æ
ç çè
Q d
ö çè
÷ø
P M e
= - æ
M
s s s
M
2 2 2 2( 1)
• Symbol error probability in terms of SNR
ö
P
2 2
æ
P M
= - d M
2 1 3 2
( 1)
e s
3
÷ ÷ ÷
ö çè
SNR since SNR
1
2
1
Signal
2 = = -
ø
ç ç ç
è
÷ø
æ
-
P
M
Q
M
Noise
M-2 interior points 2 exterior points
29. Distortion over
zero crossing
13 - 29
Eye Diagram
• PAM receiver analysis and troubleshooting
M=2
Margin over noise
Interval over which it can be sampled
t - Tsym
Sampling instant
Slope indicates
sensitivity to
timing error
t + Tsym t
• The more open the eye, the better the reception
