slides by prof. brian l. evans and dr. serene banerjee dept. of electrical and computer engineering...
TRANSCRIPT
Slides by Prof. Brian L. Evans and Dr. Serene Banerjee
Dept. of Electrical and Computer Engineering
The University of Texas at Austin
EE445S Real-Time Digital Signal Processing Lab Spring 2014
Lecture 14
Matched Filtering and DigitalPulse Amplitude Modulation (PAM)
14 - 2
Outline
• Transmitting one bit at a time
• Matched filtering
• PAM system
• Intersymbol interference
• Communication performanceBit error probability for binary signals
Symbol error probability for M-ary (multilevel) signals
• Eye diagram
14 - 3
Transmitting One Bit
• Transmission on communication channels is analog• One way to transmit digital information is called
2-level digital pulse amplitude modulation (PAM)
b t
)(1 tx
A
‘1’ bit
Additive NoiseChannel
input output
x(t) y(t)
b
)(0 tx
-A
‘0’ bit
t
How does the receiver decide
which bit was sent?
receive‘1’ bit
b t
)(1 ty
A
receive ‘0’ bit
)(0 ty
b
-A
t
14 - 4
Transmitting One Bit
• Two-level digital pulse amplitude modulation over channel that has memory but does not add noise
h t
)(tc
1
b t
)(1 tx
A
‘1’ bit
b
)(0 tx
-A
‘0’ bit
Model channel as LTI system with impulse response
c(t)
LTIChannel
input output
x(t) y(t)t
)(0 ty
-A Th
receive ‘0’ bit
th+bh
Assume that Th < Tb
t
)(1 ty receive‘1’ bit
h+bh
A Th
14 - 5
Transmitting 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?
h t
)(tc
1
Assume that Th < Tb
tb
)(tx
A
‘1’ bit ‘0’ bit
b
* =)(ty
-A Th
tb
‘1’ bit ‘0’ bit
h+b
Intersymbol interference
14 - 6
Preventing ISI at Receiver
• Option #1: wait Th seconds between pulses in transmitter (called guard period or guard interval)
Disadvantages?
• Option #2: use channel equalizer in receiverFIR filter designed via training sequences sent by transmitterDesign goal: cascade of channel memory and channel
equalizer should give all-pass frequency response
h t
)(tc
1
Assume that Th < Tb
* =
tb
)(tx
A
‘1’ bit ‘0’ bit
h+b
t
)(ty
-A Th
b
‘1’ bit ‘0’ bit
h+b
h
14 - 7
k
bk Tktgats ) ( )(
Digital 2-level PAM System
• Transmitted signal
• Requires synchronization of clocksbetween transmitter and receiver
Transmitter Channel Receiver
bi
Clock Tb
PAM g(t) c(t) h(t)10
ak{-A,A} s(t) x(t) y(t) y(ti)
AWGNw(t)
Decision
Maker
Threshold
Sample at
t=iTb
bits
Clock Tb
pulse shaper
matched filter
1
00 ln4 p
p
AT
N
boptN(0, N0/2)
p0 is the probability bit ‘0’
sent
bits
14 - 8
Matched Filter
• Detection of pulse in presence of additive noiseReceiver 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 tntgthtwthtgty
T is the symbol period
14 - 9
power average
power ousinstantane
)}({
|)(|
SNR pulsepeak is where,max
2
20
tnE
Tg
Matched Filter Derivation
• Design of matched filterMaximize 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
T is the symbol period
14 - 10
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 & conjugation in time
• Autocorrelation of x(t)
Maximum value (when it exists) is at Rx(0)
Rx() is even symmetric,i.e. Rx() = Rx(-) )( )()()( *2
fXfXfXfPx
)(*)( )( )( ** xxFfXfX
)(*)()( * xxRx
t
1x(t)
0 Ts
Rx()
-Ts Ts
Ts
Power Spectra
• Two-sided random signal n(t)Fourier transform may not exist, but power spectrum exists
For zero-mean Gaussian random process n(t) with variance 2
• Estimate noise powerspectrum in Matlab
)( )( )( )( 2* tntnERn
)( )( nn RFfP
N = 16384; % finite no. of samplesgaussianNoise = randn(N,1);plot( abs(fft(gaussianNoise)) .^ 2 );
approximate noise floor
dttntntntnERn )( )( )( )( )( **
)(*)( )( )( )( )( )( ***
nndttntntntnERn
0 when 0 )( )( )( * tntnERn2)( fPn
14 - 11
14 - 122 2 2
0 | )( )(| |)(|
dfefGfHTg Tfj
Matched Filter Derivation
• Noise
• Signal
dffHN
dffStnE N202 |)(|
2 )(} )( {
f
20N
Noise power spectrum SW(f)
)()( )(0 fGfHfG
dfefGfHtg tfj )( )( )( 2 0
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
T is the symbol period
14 - 13
dffH
N
dfefGfH Tfj
20
2 2
|)(|2
| )( )(|
Matched Filter Derivation• Find h(t) that maximizes pulse peak SNR
• Schwartz’s inequality
For vectors:
For functions:
upper bound reached iff
|||| ||||cos |||| |||| | | *
ba
bababa
TT
Rkxkx )( )( 21
-
2
2
-
2
1
2
*2
-
1 )( )( )( )( dxxdxxdxxx
a
b
14 - 14)( )( 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
Matched Filter Derivation
T is the symbol period
14 - 15
Matched Filter
• Impulse response is hopt(t) = k g*(T - t)
Symbol period T, transmitter pulse shape g(t) and gain k
Scaled, conjugated, time-reversed, and shifted version of g(t)
Duration and shape determined by pulse shape g(t)
• 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
N
Edttg
NdffG
Nb
14 - 16
t=nT T
Matched Filter for Rectangular Pulse
• Matched filter for causal rectangular pulse shapeImpulse response is causal rectangular pulse of same duration
• Convolve input with rectangular pulse of duration T sec and sample result at T sec is same asFirst, 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 - 17
k
bk Tktgats ) ( )(
Digital 2-level PAM System
• Transmitted signal
• Requires synchronization of clocksbetween transmitter and receiver
Transmitter Channel Receiver
bi
Clock Tb
PAM g(t) c(t) h(t)10
ak{-A,A} s(t) x(t) y(t) y(ti)
AWGNw(t)
Decision
Maker
Threshold
Sample at
t=iTb
bits
Clock Tb
pulse shaper
matched filter
1
00 ln4 p
p
AT
N
boptN(0, N0/2)
p0 is the probability bit ‘0’
sent
bits
14 - 18
)( )( )( )(
)(*)()( where)()()(
,i
ikkbkbiii
kbk
tnTkipaiTtpaty
thtwtntnkTtpaty
k
bk Tktats ) ()(
Digital 2-level PAM System
• Why is g(t) a pulse and not an impulse?Otherwise, s(t) would require infinite bandwidth
We limit its bandwidth by using a pulse shaping filter
• Neglecting noise, would like y(t) = g(t) * c(t) * h(t) to be a pulse, i.e. y(t) = p(t) , to eliminate ISI
actual value(note that ti = i Tb)
intersymbolinterference (ISI)
noise
p(t) is centered at origin
14 - 19
) 2
(rect 2
1 )(
||,0
, 2
1
)(
W
f
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 pulse shape is not
causal and is infinite in duration
) 2(sinc)( tWtp
14 - 20
WffW
fWfffW
Wf
W
ffW
fP
2 || 20
2 || 22
)|(|sin1
4
1
|| 0 2
1
)(
1
111
1
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 h(t) be square root raised cosine pulses
W
f11
222 161
2cos
sinc )(
tW
tW
T
ttp
s
ideal Nyquist channel impulse response
dampening adjusted by rolloff factor
14 - 21
Bit Error Probability for 2-PAM
• Tb is bit period (bit rate is fb = 1/Tb)
w(t) is AWGN with zero mean and variance 2
• 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)s(t)
Sample att = nTb
Matched filterw(t)
r(t) r(t) rn k
bk Tktgats ) ( )(
)()()( twtstr
r(t) = h(t) * r(t)
Bit Error Probability for 2-PAM
• Noise power at matched filter output
14 - 22
dnTwhnTv
)()()(
22 )()()( dnTwhEnTvE
})()()()({ 212211
ddnTwgnTwgE TT
212121 )}()({)()( ddnTwnTwEgg TT
TdHdh
sym
sym
22/
2/
2222 )(2
1)(
Noise power
T = TsymFiltered noise
2 (1–2)
14 - 23
Bit Error Probability for 2-PAM
• Symbol amplitudes of +A and -A• Rectangular pulse shape with amplitude 1
• Bit duration (Tb) of 1 second
• Matched filtering with gain of one (see slide 14-15)Integrate received signal over nth bit period and sample
n
n
n
n
n
n
vA
dttwA
dttrr
)(
)(
1
1
0-
Anr
)( nr rPn
AProbability density function (PDF)
14 - 24
Av
PAvPvAPAnTsP nnnb )( )0())(|error(
Bit Error Probability for 2-PAM
• Probability of error given thattransmitted pulse has amplitude –A
• Random variable is Gaussian withzero mean andvariance of one
AQdve
AvPAnTsP
v
A
n 2
1))(|error( 2
2
nv
Q function on next slide
PDF for N(0, 1)
0 /A
/ nvforPDF
Tb = 1
14 - 25
Q Function
• Q function
• Complementary error function erfc
• Relationship
x
y dyexQ 2/2
21
)(
x
t dtexerfc22
)(
22
1)(
xerfcxQ
Erfc[x] in Mathematica
erfc(x) in Matlab
14 - 26
2
2
SNR where,
2
1
2
1
))(|error()())(|error()( error)(
A
Qσ
AQ
σ
AQ
σ
AQ
AnTsPAPAnTsPAPP bb
Bit Error Probability for 2-PAM• Probability of error given that
transmitted pulse has amplitude A
• Assume that 0 and 1 are equally likely bits
• Probability of error exponentiallydecreases with SNR (see slide 8-16)
)/())(|error( AQAnTsP b
2
1)(
2
e
Q
positive largefor
Tb = 1
14 - 27
PAM Symbol Error Probability
• Set symbol time (Tsym) to 1 second
• Average transmitted signal power
GT() square root raised cosine spectrum
• M-level PAM symbol amplitudes
• With each symbol equally likely
}{|)(| 2
1 }{ 222
nTnSignal aEdGaEP
3
)1( )12(2
1 2
22
1
22
12
2 dMid
Ml
MP
M
i
M
Mi
iSignal
2, ,0 , ,1
2 ),12(
MMiidli ......
2-PAM
d
-d
4-PAM
Constellation points with receiver
decision boundaries
d
d
3 d
3 d
14 - 28
2
2
2
1 0
2/
2/
0 Nd
NP
sym
sym
Noise
nnn var
PAM Symbol Error Probability
• Noise power and SNR
• Assume ideal channel,i.e. one without ISI
• Consider M-2 inner levels in constellationError only if
where
Probability of error is
• Consider two outer levels in constellation
dvn ||
d
QdvP n 2)|(|
2/02 N
d
QdvP n )(
two-sided power spectral density of AWGN
channel noise after matched filtering and sampling
0
22
3
)1(2 SNR
N
dM
P
P
Noise
Signal
14 - 29
d
QM
MdQ
M
dQ
M
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
22
1
2
Md
P
P
MQ
M
MP
Noise
Signale
M-2 interior points 2 exterior points
14 - 30
Visualizing ISI
• Eye diagram is empirical measure of signal quality
• Intersymbol interference (ISI):
Raised cosine filter has zeroISI when correctly sampled
knk
k
symsymknsymsymk g
kTnTgaagTkTnganx
)0(
)( )0() ( )(
nkk
sym
nkk
symsym
g
kTgdM
g
kTnTgdMD
,, )0(
)( )1(
)0(
)( )1(
14 - 31
Eye Diagram for 2-PAM• Useful for PAM transmitter and 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
14 - 32
Eye Diagram for 4-PAM
3d
d
-d
-3d
Due to startup transients.
Fix is to discard first few symbols equal to number of symbol periods in pulse shape.