# Matched Filtering and Digital Pulse Amplitude Modulation (PAM) - PDF

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```					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
Pulse shape is                                        period
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 )                                                       y0 (t )        receive
‘0’ bit                                                                    ‘0’ bit
Τb           input                    output                  Τb
t                 Channel
x(t)                      y(t)
-A                                                            -A

x1 (t )                                                       y1 (t )        receive
How does the                                      ‘1’ bit
A                                                             A
Τb       t
which bit was sent?                           Τb     t

‘1’ bit                                                                    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)          x(t)    h(t)      y(t)         y(T)       T is pulse
Pulse                                    t=T               period
signal                 Matched
w(t)         filter
Additive white Gaussian        y (t ) = g (t ) * h(t ) + w(t ) * h(t )
noise (AWGN) with zero                = g 0 (t ) + n(t )
mean and variance N0 /2
13 - 5
Matched Filter Derivation
• Design of matched filter
Maximize signal power i.e. power of g 0 (t ) = g (t ) * h(t ) at t = T
Minimize noise i.e. power of n(t ) = w(t ) * h(t )
• Combine design criteria
max η , where η is peak pulse SNR
| g 0 (T ) |2 instantane ous power
η=        2
=
E{n (t )}        average power

g(t)             x(t)    h(t)     y(t)         y(T)

Pulse                                       t=T
signal                     Matched
w(t)           filter                            13 - 6
Power Spectra
• Deterministic signal x(t)      • Autocorrelation of x(t)
w/ Fourier transform X(f)          Rx (τ ) = x(τ ) * x* (−τ )
Power spectrum is square of      Maximum value at Rx(0)
absolute value of magnitude
Rx(τ) is even symmetric, i.e.
response (phase is ignored)
2                  Rx(τ) = Rx(-τ)
Px ( f ) = X ( f ) = X ( f ) X ( f )
*

x(t)
Multiplication in Fourier                                              1
domain is convolution in
time domain                                         0           Ts        t
Conjugation in Fourier domain                        Rx(τ)
is reversal and conjugation                                Ts
in time
{
X ( f ) X * ( f ) = F x(τ ) * x* (−τ )   }   -Ts                Ts            τ
13 - 7
Power Spectra
• Power spectrum for signal x(t) is Px ( f ) = F { Rx (τ ) }
Autocorrelation of random signal n(t)
{
Rn (τ ) = E n(t ) n (t + τ ) =
*
}        ∞

−∞
n(t ) n* (t + τ ) dt
{
Rn (−τ ) = E n(t ) n* (t − τ ) =     }      ∞

−∞
n(t ) n* (t − τ ) dt = n(τ ) * n* (−τ )
For zero-mean Gaussian n(t) with variance σ2
{                     }
Rn (τ ) = E n(t ) n* (t + τ ) = σ 2 δ (τ ) ⇔ Pn ( f ) = σ 2
• Estimate noise power
spectrum in Matlab
noise
N = 16384; % number of samples
floor
gaussianNoise = randn(N,1);
plot( abs(fft(gaussianNoise)) .^ 2 );                                            13 - 8
Matched Filter Derivation
Noise power
g(t)                       x(t)      h(t)           y(t)            y(T)      spectrum SW(f)
N0
Pulse                                                          t=T                                   2
signal         w(t)             Matched filter                                                       f
N0
• Noise n(t ) = w(t ) * h(t )                       S N ( f ) = SW ( f ) S H ( f ) =      | H ( f ) |2
2
AWGN Matched
∞                       ∞
N0                                         filter
E{ n (t ) } = S N ( f ) df =
2
| H ( f ) |2 df
−∞
2             −∞

• Signal g 0 (t ) = g (t ) * h(t )                    G 0 ( f ) = H ( f )G ( f )
∞
g 0 (t ) =        H ( f ) G( f ) e j 2π      f t
df
−∞
∞
| g 0 (T ) |2 = |          H ( f ) G( f ) e j 2π        fT
df |2                         13 - 9
−∞
Matched Filter Derivation
• Find h(t) that maximizes pulse peak SNR η
∞
|        H ( f ) G( f ) e j 2π     f T
df |2                                         a
η=       −∞
∞
N0
| H ( f ) |2 df
2    −∞                                                    θ
b
• Schwartz’s inequality
aT b
For vectors:                      | aT b* | ≤ || a || || b || ⇔ cosθ =
|| a || || b ||
∞                   2     ∞                  ∞
2                    2
For functions:                         φ1 ( x) φ ( x) dx ≤
*
2               φ1 ( x) dx         φ2 ( x) dx
-∞                         -∞                 -∞

lower bound reached iff φ1 ( x) = k φ2 ( x) ∀k ∈ R
13 - 10
Matched Filter Derivation
Let φ1 ( f ) = H ( f ) and φ2 ( f ) = G* ( f ) e− j 2 π f T
∞                                     ∞                   ∞
| H ( f ) G( f ) e j 2 π f T df |2 ≤ | H ( f ) |2 df          | G( f ) |2 df
-∞                                    −∞                 −∞
∞
| H ( f ) G( f ) e j 2 π f T df |2           ∞
2
η=   -∞
∞
≤           | G( f ) |2 df
N0                                N0
| H ( f ) |2 df               −∞
2   −∞
∞
2
ηmax =         | G( f ) |2 df , which occurs when
N0 −∞
Hopt ( f ) = k G* ( f ) e− j 2 π f T ∀ k by Schwartz ' inequality
s
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
∞                     ∞
2                     2                     2Eb
ηmax =       | G( f ) | df =
2
| g (t ) | dt =
2
= SNR
N0 −∞                 N0 −∞                 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
Second, sample at symbol period T sec         Sample and dump

Third, reset integration for next time period
• Integrate and dump circuit

t=kT                  T

h(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 )                                                                   y0 (t )   receive
‘0’ bit                                                                           ‘0’ bit
Τb           input                         output                 Τh       Τh+Τb
Communication                                                   t
t                  Channel
x(t)                          y(t)        -A Th
-A

x1 (t )                                          h(t )                    y1 (t )
Model channel as                                              ‘1’ bit
A                            LTI system with                   1       A Th
impulse response
t
Τb       t          h(t)                   Τh       t             Τh       Τh+Τb
‘1’ bit                     Assume that Th < Tb                                    13 - 14
Transmit Two Bits (Interference)
• Transmitting two bits (pulses) back-to-back
will cause overlap (interference) at the receiver
x(t )                           h(t )
y (t )
A
*               1       =
2Τb                                                         Τh+Τb

Τb               t              Τh       t                        Τb                   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                                    Intersymbol
interference
• How do we prevent intersymbol
interference (ISI) at the receiver?                                                       13 - 15
Transmit Two Bits (No Interference)
• Prevent intersymbol interference by waiting Th
seconds between pulses (called a guard period)
x(t )                                 h(t )
y (t )
A
*               1       =
Τh+Τb                                                            Τh+Τb

Τb               t              Τh       t                 Τh Τb                        t
-A Th
‘1’ bit          ‘0’ bit       Assume that Th < Tb                    ‘1’ bit       ‘0’ bit

13 - 16
Digital 2-level PAM System
ak∈{-A,A} s(t)                   x(t)          y(t) y(ti)

bi                                                             Decision
1
PAM        g(t)     h(t)     Σ          c(t)
bits                                                   Sample at Maker
t=iTb
0
pulse          AWGN matched
Clock Tb                                        Threshold λ
shaper           w(t) filter Clock T
b          N    p
λopt = 0 ln 0
4 ATb      p1

• Transmitted signal s (t ) =                 ak g (t − k Tb )
k
• Requires synchronization of clocks between
13 - 17
• Why is g(t) a pulse and not an impulse?
Otherwise, s(t) would require infinite bandwidth
s (t ) =        ak δ (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) = µ p(t) , to eliminate ISI
y (t ) = µ       a k p (t − kTb ) + n (t ) where n (t ) = w(t ) * c (t )                p(t) is
k
centered
y (t i ) = µ ai p (t i − iTb ) + µ                 a k p ((i − k )Tb ) + n (t i )   at origin
k ,k ≠i

actual value                     intersymbol                  noise
(note that ti = i Tb)            interference (ISI)
13 - 18
Eliminating ISI in PAM
• One choice for P(f) is a                 1
rectangular pulse                           ,−W < f < W
P( f ) = 2 W
W is the bandwidth of the
0      , | f |> W
system
Inverse Fourier transform
of a rectangular pulse is             1         f
P( f ) =    rect (    )
is a sinc function                   2W        2W
p(t ) = sinc(2 π 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
13 - 19
Eliminating ISI in PAM
• Another choice for P(f) is a raised cosine spectrum
1
0 ≤ | f | < f1
2W
1         π (| f | −W )
P( f ) =    1 − sin                   f1 ≤ | f | < 2W − f1
4W           2W − 2 f1
0              2W − f1 ≤ | f | ≤ 2W

• Roll-off factor gives bandwidth in excess                            α = 1−
f1
of bandwidth W for ideal Nyquist channel                                      W
• Raised cosine pulse p(t ) = sinc t cos(2π 2α W t )
Ts 1 − 16 α W 2 t 2
has zero ISI when
sampled correctly ideal Nyquist channel dampening adjusted by
impulse response rolloff factor α
• 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 )
Σ       h(t)                               k

Sample at      r (t ) = s (t ) + v(t )
Matched    t = nTb
v(t)    filter                  r(t) = h(t) * r(t)

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
• 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
See Slide
Set gain of matched filter to be 1/Tb                                         13-13
Integrate received signal over period, scale, sample
( n + 1 ) Tb
1                                                      Prn ( rn )
rn =                r ( t ) dt
Tb      nT b
( n + 1 ) Tb
1                                                                      rn
= ±A+                            v ( t ) dt           -       0
Tb                  nT b                     −A                  A
= ± A + vn                                    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                      >
σ         σ
• Random variable                                          vn / σ
vn
σ  is Gaussian with                                                             PDF for
zero mean and                                                                   N(0, 1)
variance of one                                      0 A/σ
∞              v2
vn       A             1      −                   A
P(error | s (nT ) = − A) = P        >       =            e     2
dv = Q
σ        σ       A     2π                     σ
σ

Q function on next slide
13 - 23
Q Function
• Q function
1 ∞ − y2 / 2
Q( x) =      e       dy
2π x
• Complementary error
function erfc
2 ∞ −t 2
erfc( x) =    e dt
π   x

• Relationship               Erfc[x] in Mathematica
1            x
Q( x) = erfc                     erfc(x) in Matlab
2             2
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 / σ )
• 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
= Q
2
A     1
+ Q
2
A
=Q
A
=Q ρ( )
A2                                                           −
ρ
where, ρ = SNR = 2                              e−x
2
1 e 2
σ         erfc( x) ≤                 Q( ρ ) ≤
• Probablity of error          πx                                             2π        ρ
decreases exponentially with SNR                            for large positive x, ρ
13 - 25
PAM Symbol Error Probability
• Average signal power                                                                           3d
2                 ∞                        2
E{a n } 1                                  E{a n }
PSignal    =        ×                  | GT (ω ) | dω =
2

Tsym    2π            −∞
Tsym                       d         d

GT(ω) is square root of the                                                         -d       -d
raised cosine spectrum
Normalization by Tsym will                                                                   -3 d
be removed in lecture 15 slides                                           2-PAM       4-PAM
• M-level PAM amplitudes                                                        Constellations with
M                         M
li = d ( 2i − 1),             i=−      + 1, . . . , 0, . . . ,                decision boundaries
2                         2
• Assuming each symbol is equally likely
M
1     M
1 2    2
d2
PSignal =                 l   i
2
=               [d (2i − 1)]
2
= ( M − 1)
2

Tsym   i =1              T M    i =1                             3Tsym          13 - 26
PAM Symbol Error Probability
• Noise power and SNR                                  • Consider M-2 inner
1
ω sym / 2
N0       N0                 levels in constellation
PNoise   =                      dω =
2π   −ω sym / 2
2       2Tsym               Error if and only if
| vR (nTsym ) |> d
two-sided power spectral
density of AWGN                       where σ 2 = N 0 / 2
PSignal     2 ( M 2 − 1) d 2                  Probablity of error is
SNR =                =             ×
PNoise           3        N0                                                  d
P(| vR (nTsym ) |> d ) = 2 Q
σ
• Assume ideal channel,                                • Consider two outer
i.e. one without ISI                                 levels in constellation
x ( nT sym ) = a n + v R ( nT sym )                                            d
P(vR (nTsym ) > d ) = Q
channel noise filtered
σ
13 - 27
PAM Symbol Error Probability
• Assuming that each symbol is equally likely,
symbol error probability for M-level PAM
M −2    d                2   d            2( M − 1)   d
Pe =      2Q                +   Q            =           Q
M      σ                M   σ               M        σ

M-2 interior points     2 exterior points

• Symbol error probability in terms of SNR
1
M −1                                            PSignal
Pe = 2
M
Q
M −1
3
2
SNR
2
since SNR =
PNoise
=
d2
3σ 2
(M 2 −1 )

13 - 28
Eye Diagram
• PAM receiver analysis and troubleshooting
Sampling instant

M=2
Margin over noise
Distortion over
Slope indicates                                            zero crossing
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

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