1
Chapter 7
Performance of
Communication Systems
Corrupted by Noise
2
Introduction
(chapter objectives)
? Bit error rate for binary systems (unipolar,
polar,bipolar,OOK,BPSK,FSK,and MSK)
? Output signal-to-noise ratio for analog systems(AM,
SSB,PM,and FM)
T r a n s m i t t i n g
e n d
I n f o r m a t i o n
s o u r c e
m o d u l a t o r
c h a n n e l
D e m o d u -
l a t o r
S i n k
O r ( a c c e p t a n t )
R e c e i v i n g
e n d n o i s e
? How much bandwidth occupied in the air?
How many AM stations can be?
? What is the listening-performance?
Is it very noisy to the listener?
? Equipment cost and complexity?
Bandwidth of Hz
SNR
Money and time
3
Introduction
Eample 2,A digital communication case,
? How much bandwidth occupied in the air?
How many AM stations can be?
? What is the listening-performance?
Is it very noisy to the listener?
? Equipment cost and complexity?
Data-rate of bsp or
Bandwidth of Hz
Error rate (probability of
error) or SNR
Money and time
4
Introduction
(chapter objectives)
? Channel,not a,free lunch”
– Always having a limited bandwidth
– With noise,and even interference
– Often,with distortion
We mainly concern the most common AWGN channel,
– Additive noise
– Noise is White and Gaussian
– No distortion,maybe some amplifying of A,maybe
some delay of t0
– Bandwidth is enough for a certain transmission
5
Introduction
In this chapter,we discuss the,performance-and-cost” of
communication systems
– SNR and/or Pe (probability of error),bit-error-rate
– Bandwidth and/or data-rate
For analog system,
Signal-to-
Noise-Ratio of,
AM
SSB
PM
FM
For digital system,
bit-error-rate
of,
OOK
BPSK
FSK
QPSK
MSK
For PCM
system,
Signal-to-
Noise-
Ratio
6
Introduction
Error probability for binary signaling
1,General process and bit error rate
2,Results for Gaussian Noise
3,Optimal reception with Match-filters
Performance of baseband binary systems
Unipolar/Polar/Bipolar signaling
Coherent detection of bandpass binary system
Non-coherent detection of bandpass binary system
QPSK and MSK
Comparison of digital signaling systems
Output SNR for PCM systems
Output SNR for analog systems
Comparison of analog systems
7
7.1 Error Probabilities for Binary Signaling
(General Results)
– Baseband signaling in Ch3,such as line-code,
processing = LPF+AMP
– Bandpass signaling in Ch4&5,as OOK,PSK,FSK,MSK,and
etc,processing = superheterodyne receiver (mixer + IF-amp +
detector)
8
7.1 Error Probabilities for Binary Signaling
(General Results)
data Transmitter
output signal
At the input
of receiver
After sampling Data
received
1 s1(t) r(t)=s1(t)+n(t)
Noise is
added
r01(t0) ? r01
r01(t) sampled at t0,
most likely close to
“1”
1,
Most
likely
0 s2(t) r(t)=s2(t)+n(t)
Noise is
added
r02(t0) ? r02
r02(t) sampled at t0,
most likely close to
“2”
0,
Most
likely
9
7.1 Error Probabilities for Binary
Signaling
1,Source data,
2,Transmitted signal
3,Input of receiver
4,After sampling
5,Data received
0
1
,0
,1
?
?
??
??
d a ta
d a tam
0
1
),(
),()(
2
1
?
?
??
??
d a t a
d a t a
ts
tsts
??
?
?
?
?
??
0
1
),()(
),()()(
2
1
d a ta
d a ta
tnts
tntstr
??
?
?
???
0
1
),(
),()(
002
001
000 d a ta
d a ta
tr
trtrr
??
?
?
?
?
??
0
1
"0"
"1"
,0
,1~
0
0
d a ta
d a ta
r
rm
Src data of 0 and 1 at
prob,of ?
randomly
A signal,up to src
data is in the
interval (0,T)
Noise added
r(t) is then processed
and r0(t) output,
Sampled at t0,the
sync-timing; noise
is inside of r0
r01 and r02 are diff,
then which data
could be told
10
7.1 Error Probabilities for Binary
Signaling
? Receiver,detection + decision
In the receiver,detect-processing is used to convert the waveforms r(t)
into r0
– With no noise,s1(t) and s2(t) are mapped to the apart,clean
centers”
– The noise corrupts the waveform and then the r0 is,dirty” and
away from the center,
Decision is to obtain 0 or 1 from the r0
– The rule is,go to the nearest neighbor”
– Or equivalent to,use a threshold VT to seperate,
11
7.1 Error Probabilities for Binary
Signaling
? Errors occur when r0 falls in the wrong areas due to noise,
? The r0 is a random variable,basically from the
randomness of n(t) in r(t),The distribution could be as
following,
? f(r0|s1) and f(r0|s2) are generated from n(t) by detect-
processing
– VT is a proper-defined threshold
12
7.1 Error Probabilities for Binary
Signaling
? When signal plus noise is present at the receiver
input,Error can occur in two ways,an error occurs when
r0<VT if a binary 1 is sent,and an error occurs when r0>VT
if a binary 0 is sent
(7, 6 ) )|() |(
5.7 )|() |(
0202
0101
??
??
??
??
T
T
V
V
drsrfs e n tser r o rP
drsrfs e n tser r o rP ）（
13
7.1 Error Probabilities for Binary
Signaling
? The BER is
????
??
????
TT VV
e
drsrfs en tsPdrsrfs en tsP
s en tsPs en tser r o rPs en tsPs en tser r o rPP
02020101
2211
)|() ()|() (
7.7 ) () |() () |( ）（
– P(s1 sent) and P(s2 sent) are source statistics and are
known before transmission,so called prior statistics;
P(s1 sent) and P(s2 sent) are normally considered to be
equally likely,
? Pe is a standard measure of error rate of the system,
– It is on a bit-basis,so called as (Average) bit error rate
? For given n(t),good processing and opt-VT yield good Pe
? We assume that the source statistics are equally likely,
14
7.1 Error Probabilities for Binary Signaling
Results for Gaussian Noise
? Two assumptions make things easy,
– 1,linear detection except for the threshold device
– 2,The AWGN channel ? white and Gaussian n(t)
with zero mean and E{| n(t) |2}=σ02 ? Notes,
? For baseband signaling,the processing circuits
consisting of linear filter with some gain,
? For bandpass signaling,a superheterodyne circuit
consisting of a mixer,IF stage,and product detector
being also a linear circuit,
? If automatic gain control (AGC) or limiters or a
nonlinear detector such as envelope detector is used,
the results of this section will not be applicable,
15
7.1 Error Probabilities for Binary Signaling
Results for Gaussian Noise
? For the case of a linear-processing receiver circuit with a
binary signal plus noise at the input,the sampled output is,
( 7, 1 0 ) 000 nsr ??
? s0 is a constant that depends on the signal being sent
(7, 1 1 ) s en t 0b i n ary af o r s en t 1b i n ary af o r
02
01
0 ??
??
s
ss
? Since the output noise n0 is a zero-mean Gaussian random
variable,the total output sample r0 is a Gaussian random
variable with a mean value of either s01 or s02,depending on
whether a binary 1 or a binary 0 was sent,That is to say,
s en t 0b i n ar y af o r
s en t 1b i n ar y af o r
02
01
0 ??
??
s
sm
r
16
7.1 Error Probabilities for Binary Signaling
Results for Gaussian Noise
? Then the two conditional PDFs are
? ? ( 7, 1 2 )
2
1)|( )2/(
0
10
2
0
2
010 ?
??
sresrf ???
? ? ( 7, 1 3 )
2
1)|( )2/(
0
20
2
0
2
020 ?
??
sresrf ???
? Using equally likely source statistics,the BER becomes,
? ? ? ?
????
????
????
??
??
??
??
????
????
TT
TT
TT
V
sr
V
sr
VV
VV
e
dredre
drsrfdrsrf
drsrfs e n tsPdrsrfs e n tsPP
0
)2/(
0
0
)2/(
0
020010
02020101
2
0
2
020
2
0
2
010
2
1
2
1
2
1
2
1
)|(
2
1
)|(
2
1
)|() ()|() (
??
????
17
7.1 Error Probabilities for Binary Signaling
Results for Gaussian Noise
? thus
? ? ? ?
?
?
?
?
?
?
?
? ?
???
?
?
?
?
?
? ??
?
????
?
?
?
?
?
??
0
01
0
01
/
2/2/
2
1
2
1
2
1
2
1
2
1
2
1
002
2
0/01
2
??
?
?
?
? ?
??
?
sV
Q
sV
Q
dedeP
TT
sV
e
T
sTV
? To find the VT that minimizes Pe we need to solve
dPe/dVT=0
? ? ? ? 0
2
1
2
1
2
1
2
1 )2/(
0
)2/(
0
2
0
2
020
2
0
2
010 ??? ???? ??
????
srsr
T
e ee
dV
dP
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
2
0
2
0
2
0201
0201
44
)(
( m i n )
2
)(
??
d
e
T
E
Q
ss
QP
ss
b e s tV
18
7.1 Error Probabilities for Binary Signaling
Results for Gaussian Noise and Matched-Filter Reception
? If the receiving filter is optimized the BER can be
reduced,To minimize Pe,we need to maximize the
argument of Q,thus we need to find the linear filter
that maximizes,
? ? )())()((
2
0
2
0
2
0
2
002001
??
tststs d??
? [sd(t0)]2 is the instantaneous power of the difference
output signal at t=t0
19
7.1 Error Probabilities for Binary Signaling
Results for Gaussian Noise and Matched-Filter Reception
? The linear filter that maximizes the instantaneous
output signal power at the sampling t=t0 when
compared with the average output noise power σ02=n02(t)
is the matched filter,For the case of noise white noise at
the receiver input,the matched filter needs to be
matched to the difference signal sd(t)=s1(t)-s2(t),Thus
the impulse response of the matched filter for binary
signaling is,
）（ 18.7 )]()([)( 0201 ttsttsCth ????
? The output peak signal to average noise ratio that is
obtained from the matched filter is,
? ?
(t )]-(t )[
2)(
0
2
21
0
2
0
2
0
dtssE
N
Ets
T
d
dd
?
?
?
?
?
??
?
?
???
?
???
??
02
)( NEQo p tP de
20
7.1 Error Probabilities for Binary Signaling
Results for Gaussian Noise and Matched-Filter Reception
? The performance is measured by
Pe=P(err|s1) P(s1) + P(err|s2) P(s2) => [P(err|s1) + P(err|s2) ]/2
? For AWGN channel,when linear detector used,
???
?
???
??
???
?
???
? ????
2
0
2
0
2
02010201
44
)(( m in ),
2)( ??
d
eT
EQssQPandssb e s tV
? More,optimal linear receiver or match filter C[s1(t0-
t)-s2(t0-t)] can maximize the SNR of r0 and Pe even
smaller
???
?
???
??
02
)( NEQo p tP de
21
7.2 Performance of Baseband Binary Systems
(Unipolar Signaling)
22
7.2 Performance of Baseband Binary Systems
(Unipolar Signaling)
? Source data,
0
1
,0
,1
?
?
??
??
d a ta
d a tam
0
1
,0
,)(
?
?
??
? ??
d a t a
d a t aAts
??
?
?
???
0
1
),(
),()(
d a t a
d a t a
tn
tnAtr
??
?
?
???
0
1
,
,
0
001
0 d a t a
d a t a
n
nsr
??
?
?
?
?
??
0
1
0,0
,1~
0
010
d a ta
d a ta
r
srm
? Transmitted
signal
? Input of
receiver
? After
sampling
? Data received
23
7.2 Performance of Baseband Binary Systems
(Unipolar Signaling)
● Receive with (linear) LPF
● Detect with LPF of BHz to limit the noise and pass the
signal
● Sampling at any time in the pulses,best at middle of them
● If gain=1,s01=A and s02=0,
Var of n0(t) = (2B) (N0/2) = N0B
● Decision with VT=A/2
???
?
???
??
???
?
???
??
BN
AQEQP d
e
0
2
2
0 44 ?
For Rect-pulse,or for sin(x)/x pulse [B=1/(2T)] =A/2
???
?
???
??
0
2
4 N
TAQP
e ??
?
?
???
??
0
2
2 N
TAQP
e
24
7.2 Performance of Baseband Binary Systems
(Unipolar Signaling)
Receive with (linear) Match filter
– Detect with filter,h(t) = s1(T-t)-s2(T-t) = A; In fact,
this is a integrator! (Ed=A2T)
– Sampling at T (the end of pulses)
– s01=AT and s02=0;
Var of n0(t),do not care
– Decision with VT=AT/2
? Results
???
?
???
??
???
?
???
??
???
?
???
??
00
2
0 22 N
EQ
N
TAQ
N
EQP bd
e
? Eb is the average bit-energe,
? No matter what pulse-shape,Pe is the same
? The best Pe,is always smaller than or equal to that of
LPF-Pe.。
2/)( 2 TAE b ?
25
7.2 Performance of Baseband Binary Systems
(Unipolar Signaling)
? For LPF,smaller BW introduces less noise and then
good Pe,But,BW must be bigger enough to let the
signal to pass,
? However,for MF the BW might be very large,For
Example,the rect-pulse need rect-MF h(t),Its H(f) is a
sinc-function and the absolute BW is infinitive! Then
it would introduce much noise,why it could make the
best Pe?
26
7.2 Performance of Baseband Binary Systems
(Polar Signaling)
Bandwidth is BHz and the signal is low freq,in (0-B) Hz,
For Rect-pulses,B = 1/T Hz
27
7.2 Performance of Baseband Binary Systems
(Polar Signaling)
? Source data,
0
1
,0
,1
?
?
??
??
d a ta
d a tam
? Input of
receiver
? After
sampling
? Data received
? Transmitted
signal 01,,)( ????? ??? d a t ad a t aAAts
??
?
?
?
??
??
0
1
),(
),()(
d a t a
d a t a
tnA
tnAtr
??
?
?
?
??
??
0
1
,
,
001
001
0 d a ta
d a ta
ns
nsr
??
?
?
?
??
??
0
1
,0
,1~
010
010
d a ta
d a ta
sr
srm
28
7.2 Performance of Baseband Binary Systems
(Polar Signaling)
● Receive with (linear) LPF
● Detect with LPF of BHz to limit the noise and pass the
signal
● Sampling at any time in the pulses,best at middle of
them
● If gain=1,s01=A and s02=-A,
Var of n0(t) = (2B) (N0/2) = N0B
● Decision with VT=0
???
?
???
?
???
?
?
???
??
BN
AQEQP d
e
0
2
2
04 ?For Rect-pulse,or for sin(x)/x pulse [B=1/(2T)]
=A/2
???
?
???
??
0
2
N
TAQP
e ??
?
?
???
??
0
22
N
TAQP
e
29
7.2 Performance of Baseband Binary Systems
(Polar Signaling)
Receive with (linear) Match filter
– Detect with filter,h(t) = [s1(T-t)-s2(T-t)]/2 = A; In
fact,this is a integrator! (Ed=4A2T)
– Sampling at T (the end of pulses)
– s01=AT and s02=-AT;
Var of n0(t),do not care
– Decision with VT=0
? Results
? Eb is the average bit-energe,
? No matter what pulse-shape,Pe is the same
? The best Pe,is always smaller than or equal to that of
LPF-Pe.。
TAEb 2?
???
?
???
??
???
?
???
??
???
?
???
??
00
2
0
222 NEQN TAQNEQP bde
30
7.2 Performance of Baseband Binary Systems
(Bipolar Signaling)
31
7.2 Performance of Baseband Binary Systems
(Bipolar Signaling)
Read it youself with the following hints,
1,3-levels
1,Calculate the conditional PDFs
2,Get the best thresholds,VT1 and VT2,with the limitation,
VT1=-VT2=VT
Answer,
1,What is thresholds,VT1 and VT2?
2,What is Pe for normal LPF and match filter?
3,What is the match filter h(t)?
5.0
25.0
25.0
0
1
1
0
,
,
)(
?
?
?
?
?
?
??
?
?
?
?
?
?
P
P
P
d a t a
d a t a
d a t a
A
A
ts
32
7.2 Performance of Baseband Binary Systems
(Bipolar Signaling)
? Source data,
0
1
,0
,1
?
?
??
??
d a ta
d a tam
? Input of
receiver
? After
sampling
? Data received
? Transmitted
signal
??
?
?
?????
0
1
),(
)(,),()(
d a t a
d a t a
tn
tnAortnAtr
??
?
?
?????
0
1
,
,,
0
001001
0 d a t a
d a t a
n
nsornsr
?
?
?
?
?
?
????
0
1
0
,,
,0
,1~
0
010010
d a t a
d a t a
r
srorsrm
5.0
25.0
25.0
0
1
1
0
,
,
)(
?
?
?
?
?
?
??
?
?
?
?
?
?
P
P
P
d a t a
d a t a
d a t a
A
A
ts
33
7.2 Performance of Baseband Binary Systems
(Bipolar Signaling)
For the case of additive Gaussian noise,the BER is,
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
??????
00
22
2
2
1
2
4
1
) () |(
) () |() () |(
??
TT
e
V
Q
VA
Q
s e n tsPs e n tse r r o rP
s e n tAPs e n tAe r r o rPs e n tAPs e n tAe r r o rPP
Where the optimum value of VT is
2ln)/(2/ 20 AAV T ???
When σ0<<A,thus
2/AV T ?
? ?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
0
2
0
2
0
2
22
3
N
EA
A
QP
d
outN
S
e
?
?
MFf o r
2
3
L P Ff o r
42
3
0
0
2
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
N
E
QP
BN
A
QP
b
e
e
34
7.3 Coherent Detection of Bandpass Binary
Signals(On-Off Keying )
·The detection is
mixer+H(f),
which is linear
processing
· Note the LO is
2cos(·) to keep
gain of signal
being 1
· LPF is to filter
out the 2fc
component
produced by
mixer
35
7.3 Coherent Detection of Bandpass Binary
Signals (On-Off Keying )
? Source signal,
0
1
,0
,1
?
?
??
??
b i n a r y
b i n a r ym
0
1
,0
),c o s ()(
?
?
??
? ??
b i n a r y
b i n a r ytAts cc ??
??
? ???
0
1
),(
),()c o s ()(
b i n a r y
b i n a r y
tn
tntAtr cc ??
??
?
?
???
0
1
,
,
0
001
0 b i n a r y
b i n a r y
n
nsr
?
?
?
?
?
?
??
0
1
0,0
,1~
0
010
b i n a r y
b i n a r y
r
srm
? Input of
receiver
? After
sampling
? Data received
? Transmitted
signal
36
7.3 Coherent Detection of Bandpass Binary
Signals (On-Off Keying )
Noise goes through mixer,
加：备课内容 lxf_s4 (1 slides)
37
7.3 Coherent Detection of Bandpass Binary
Signals (On-Off Keying )
Receive with Mixer+LPF,(like baseband-unipolar case)
– Detect with LPF of BHz to filter off the high-freq,
then unpolar-like pulses produced,
– Sampling at any time in the pulses,best at middle of
them
– If gain=1,s01=A and s02=0;
Var of n0(t) = 2N0B
– Decision with VT=A/2
? Results
???
?
???
??
???
?
???
??
BN
AQEQP d
e
0
2
2
0 84 ?For Rect-pulse,or for sin(x)/x pulse [B=1/(2T)]
=A/2
???
?
???
??
0
2
8 N
TAQP
e ??
?
?
???
??
0
2
4 N
TAQP
e
38
7.3 Coherent Detection of Bandpass Binary
Signals (On-Off Keying )
Receive with (linear) Match filter
– Detect with filter
Ed=A2T/2
– Sampling at T (the end of pulses)
– s01=AT and s02=0; Var of n0(t),do not care
– Decision with VT=AT/2
? Eb is the average bit-energe,
? No matter what pulse-shape,Pe is the same
? The best Pe,is always smaller than or equal to that of
LPF-Pe.。
? ? ))(c o s ()()()( 21 cc tTAtTstTsCth ?? ???????
???
?
???
??
???
?
???
??
???
?
???
??
00
2
0 42 N
EQ
N
TAQ
N
EQP bd
e
4/)( 2 TAE b ?
39
7.3 Coherent Detection of Bandpass Binary
Signals
Think about it,
? What is the similarity and difference
between the OOK and Baseband unipolar
signal?
40
7.3 Coherent Detection of Bandpass Binary
Signals
Different structures of MF
加：备课内容 lxf_s4 (1 slides)
41
7.3 Coherent Detection of Bandpass Binary
Signals (Binary-Phase-Shift Keying)
Read it youself with the following hints,
1,Similar to that of Baseband Polar Signaling
2,Note the bandpass (cos) characteristic
3,Note the results
Answer,
1,Pe for LPF and MF sturcture?
2,What is the match filter h(t)?
3,What makes the difference on Pe between
PSK(or Polar) and OOK(or Unipolar),noise or
Ed?
42
7.3 Coherent Detection of Bandpass Binary
Signals (Binary-Phase-Shift Keying)
? Source data
0
1
,0
,1
?
?
??
??
b i n a r y
b i n a r ym
0
1
),co s (
),co s ()(
?
?
??
?
??
??
b i n a r y
b i n a r y
tA
tAts
cc
cc
??
??
??
?
???
???
0
1
),()co s (
),()co s ()(
b i n a r y
b i n a r y
tntA
tntAtr
cc
cc
??
??
??
?
?
?
??
??
0
1
,
,
0
0
0 b i n a r y
b i n a r y
nA
nAr
?
?
?
?
?
??
??
0
1
,0
,1~
0
0
b i n a r y
b i n a r y
Ar
Arm
? Input of
receiver
? After
sampling
? Data received
? Transmitted
signal
43
7.3 Coherent Detection of Bandpass Binary
Signals
Receive with Mixer+LPF,(like baseband-polar case)
– Detect with BPF of BHz,then polar-like pulses
produced,
– Sampling at any time in the pulses,best at middle
of them
– If gain=1,s01=A and s02=-A;
Var of n0(t) = 2N0B
– Decision with VT=A/2
? Results
???
?
???
?
???
?
?
???
??
BN
AQEQP d
e
0
2
2
0 24 ?For Rect-pulse,or for sin(x)/x pulse [B=1/(2T)]
=A/2
???
?
???
??
0
2
2 N
TAQP
e ??
?
?
?
?
?
?
?
0
2
N
TAQP
e
44
7.3 Coherent Detection of Bandpass Binary
Signals
Receive with (linear) Match filter
– Detect with filter
Ed=2A2T
– Sampling at T (the end of pulses)
– s01=AT and s02=-AT; Var of n0(t),do not care
– Decision with VT=AT/2
? Eb is the average bit-energe,
? No matter what pulse-shape,Pe is the same
? The best Pe,is always smaller than or equal to that of
LPF-Pe.。
? ? ))(c o s (2)()()( 21 cc tTAtTstTsCth ?? ???????
?
?
?
?
?
?
?
?
???
?
???
??
?
?
?
?
?
?
?
?
?
00
2
22 NEQN TAQP be
45
7.3 Coherent Detection of Bandpass Binary
Signals (Frequency-Shift Keying)
46
7.3 Coherent Detection of Bandpass Binary
Signals (Frequency-Shift Keying)
? Source data,
0
1
,0
,1
?
?
??
??
d a ta
d a tam
0
1
),co s (
),co s ()(
22
11
?
?
??
?
?
??
d a t a
d a t a
tA
tAts
??
??
??
?
?
?
??
???
0
1
),()c o s (
),()c o s ()(
222
111
d a ta
d a ta
tntA
tntAtr
??
??
??
?
?
?
???
????
0
1
),()0(
),()0(
020102
020101
0 d a ta
d a ta
nns
nnsr
??
?
?
?
??
??
0
1
,0
,1~
020
010
d a ta
d a ta
sr
srm
? Input of
receiver
? After
sampling
? Data received
? Transmitted
signal
47
7.3 Coherent Detection of Bandpass Binary
Signals (Frequency-Shift Keying)
Receive with Mixer+LPF,(like baseband-polar case)
– Detect with LPF of BHz to filter off the high-freqs,then
polar-like pulses produced,
– Sampling at any time in the pulses,best at middle of them
– If gain=1,s01=+A and -s02=-A;
Var of n0(t) = 2x2N0B = 4N0B (doubled)
– Decision with VT=0
? Results
? For Rect-pulse,or for sin(x)/x pulse [B=1/(2T)]
???
?
???
??
0
2
4 N
TAQP
e ?
?
?
?
???
??
0
2
2 N
TAQP
e
???
?
???
??
???
?
???
??
BN
AQEQP d
e
0
2
2
0 16
4
4 ?
48
7.3 Coherent Detection of Bandpass Binary
Signals (Frequency-Shift Keying)
Receive with (linear) Match filter
– Detect with filter
– Sampling at T (the end of pulses)
– s01=AT and -s02=-AT;
Var of n0(t),do not care
– Decision with VT=0
?
?
???
????
T
T
b
td tffATA
dttTstTsE
0
21
22
0
21
)(2c o s
)]()([
?
? ?
)])(c o s ())([ c o s (2
)()()(
2211
21
???? ??????
????
tTtTA
tTstTsCth
49
7.3 Coherent Detection of Bandpass Binary
Signals (Frequency-Shift Keying)
? Results
– Ed (max) = A2T,when f1-f2=n/(2T)=n(bit-rate/2); Then
Pe is min as
? f1 and f2 is called orthogonal,when configured as above
condition
Or,when f1-f2>>bit-rate,approximately orthogonal
– Eb is the average bit-energe,
– No matter what pulse-shape,Pe is the same
– This is best Pe,smaller than or equal to LPF-Pe
???
?
???
??
???
?
???
??
???
?
???
??
00
2
0 22 N
EQ
N
TAQ
N
EQP bd
e
2/)( 2 TAE b ?
? Unlike OOK,FSK signal is constant-enveloped,Why Pe of
FSK is not the same as PSK?
50
Summary
? When modulated signal is passed thru the AWGN channel,receiver
with linear detector is often used
– The mixer+LPF provides a basic result,
Best Pe when sinc-pulse (smallest BW of LPF)
3dB worse when rect-pulse； even-worse if wider-BW pulses used
– The match filter=correlator,C[s1(T-t)-s2(T-t)] can minimize Pe,so
best result
? Different modulations yield different Pe,With the same Eb/N0,means,
Modulations A Pe of MF Comment (same N0)
OOK = unipolar 1.414 3dB worse and larger
signal peak
PSK = polar 1 Best
FSK (keep f1 and f2
orthogonal)
1 3dB worse; noise
doubled due to 2 bands
? ?0/ NEQ b
? ?)/(2 0NEQ b
? ?0/ NEQ b
51
7.4 Noncoherent Detection of Bandpass Binary
Singals
52
Homework
Chapter 7
Performance of
Communication Systems
Corrupted by Noise
2
Introduction
(chapter objectives)
? Bit error rate for binary systems (unipolar,
polar,bipolar,OOK,BPSK,FSK,and MSK)
? Output signal-to-noise ratio for analog systems(AM,
SSB,PM,and FM)
T r a n s m i t t i n g
e n d
I n f o r m a t i o n
s o u r c e
m o d u l a t o r
c h a n n e l
D e m o d u -
l a t o r
S i n k
O r ( a c c e p t a n t )
R e c e i v i n g
e n d n o i s e
? How much bandwidth occupied in the air?
How many AM stations can be?
? What is the listening-performance?
Is it very noisy to the listener?
? Equipment cost and complexity?
Bandwidth of Hz
SNR
Money and time
3
Introduction
Eample 2,A digital communication case,
? How much bandwidth occupied in the air?
How many AM stations can be?
? What is the listening-performance?
Is it very noisy to the listener?
? Equipment cost and complexity?
Data-rate of bsp or
Bandwidth of Hz
Error rate (probability of
error) or SNR
Money and time
4
Introduction
(chapter objectives)
? Channel,not a,free lunch”
– Always having a limited bandwidth
– With noise,and even interference
– Often,with distortion
We mainly concern the most common AWGN channel,
– Additive noise
– Noise is White and Gaussian
– No distortion,maybe some amplifying of A,maybe
some delay of t0
– Bandwidth is enough for a certain transmission
5
Introduction
In this chapter,we discuss the,performance-and-cost” of
communication systems
– SNR and/or Pe (probability of error),bit-error-rate
– Bandwidth and/or data-rate
For analog system,
Signal-to-
Noise-Ratio of,
AM
SSB
PM
FM
For digital system,
bit-error-rate
of,
OOK
BPSK
FSK
QPSK
MSK
For PCM
system,
Signal-to-
Noise-
Ratio
6
Introduction
Error probability for binary signaling
1,General process and bit error rate
2,Results for Gaussian Noise
3,Optimal reception with Match-filters
Performance of baseband binary systems
Unipolar/Polar/Bipolar signaling
Coherent detection of bandpass binary system
Non-coherent detection of bandpass binary system
QPSK and MSK
Comparison of digital signaling systems
Output SNR for PCM systems
Output SNR for analog systems
Comparison of analog systems
7
7.1 Error Probabilities for Binary Signaling
(General Results)
– Baseband signaling in Ch3,such as line-code,
processing = LPF+AMP
– Bandpass signaling in Ch4&5,as OOK,PSK,FSK,MSK,and
etc,processing = superheterodyne receiver (mixer + IF-amp +
detector)
8
7.1 Error Probabilities for Binary Signaling
(General Results)
data Transmitter
output signal
At the input
of receiver
After sampling Data
received
1 s1(t) r(t)=s1(t)+n(t)
Noise is
added
r01(t0) ? r01
r01(t) sampled at t0,
most likely close to
“1”
1,
Most
likely
0 s2(t) r(t)=s2(t)+n(t)
Noise is
added
r02(t0) ? r02
r02(t) sampled at t0,
most likely close to
“2”
0,
Most
likely
9
7.1 Error Probabilities for Binary
Signaling
1,Source data,
2,Transmitted signal
3,Input of receiver
4,After sampling
5,Data received
0
1
,0
,1
?
?
??
??
d a ta
d a tam
0
1
),(
),()(
2
1
?
?
??
??
d a t a
d a t a
ts
tsts
??
?
?
?
?
??
0
1
),()(
),()()(
2
1
d a ta
d a ta
tnts
tntstr
??
?
?
???
0
1
),(
),()(
002
001
000 d a ta
d a ta
tr
trtrr
??
?
?
?
?
??
0
1
"0"
"1"
,0
,1~
0
0
d a ta
d a ta
r
rm
Src data of 0 and 1 at
prob,of ?
randomly
A signal,up to src
data is in the
interval (0,T)
Noise added
r(t) is then processed
and r0(t) output,
Sampled at t0,the
sync-timing; noise
is inside of r0
r01 and r02 are diff,
then which data
could be told
10
7.1 Error Probabilities for Binary
Signaling
? Receiver,detection + decision
In the receiver,detect-processing is used to convert the waveforms r(t)
into r0
– With no noise,s1(t) and s2(t) are mapped to the apart,clean
centers”
– The noise corrupts the waveform and then the r0 is,dirty” and
away from the center,
Decision is to obtain 0 or 1 from the r0
– The rule is,go to the nearest neighbor”
– Or equivalent to,use a threshold VT to seperate,
11
7.1 Error Probabilities for Binary
Signaling
? Errors occur when r0 falls in the wrong areas due to noise,
? The r0 is a random variable,basically from the
randomness of n(t) in r(t),The distribution could be as
following,
? f(r0|s1) and f(r0|s2) are generated from n(t) by detect-
processing
– VT is a proper-defined threshold
12
7.1 Error Probabilities for Binary
Signaling
? When signal plus noise is present at the receiver
input,Error can occur in two ways,an error occurs when
r0<VT if a binary 1 is sent,and an error occurs when r0>VT
if a binary 0 is sent
(7, 6 ) )|() |(
5.7 )|() |(
0202
0101
??
??
??
??
T
T
V
V
drsrfs e n tser r o rP
drsrfs e n tser r o rP ）（
13
7.1 Error Probabilities for Binary
Signaling
? The BER is
????
??
????
TT VV
e
drsrfs en tsPdrsrfs en tsP
s en tsPs en tser r o rPs en tsPs en tser r o rPP
02020101
2211
)|() ()|() (
7.7 ) () |() () |( ）（
– P(s1 sent) and P(s2 sent) are source statistics and are
known before transmission,so called prior statistics;
P(s1 sent) and P(s2 sent) are normally considered to be
equally likely,
? Pe is a standard measure of error rate of the system,
– It is on a bit-basis,so called as (Average) bit error rate
? For given n(t),good processing and opt-VT yield good Pe
? We assume that the source statistics are equally likely,
14
7.1 Error Probabilities for Binary Signaling
Results for Gaussian Noise
? Two assumptions make things easy,
– 1,linear detection except for the threshold device
– 2,The AWGN channel ? white and Gaussian n(t)
with zero mean and E{| n(t) |2}=σ02 ? Notes,
? For baseband signaling,the processing circuits
consisting of linear filter with some gain,
? For bandpass signaling,a superheterodyne circuit
consisting of a mixer,IF stage,and product detector
being also a linear circuit,
? If automatic gain control (AGC) or limiters or a
nonlinear detector such as envelope detector is used,
the results of this section will not be applicable,
15
7.1 Error Probabilities for Binary Signaling
Results for Gaussian Noise
? For the case of a linear-processing receiver circuit with a
binary signal plus noise at the input,the sampled output is,
( 7, 1 0 ) 000 nsr ??
? s0 is a constant that depends on the signal being sent
(7, 1 1 ) s en t 0b i n ary af o r s en t 1b i n ary af o r
02
01
0 ??
??
s
ss
? Since the output noise n0 is a zero-mean Gaussian random
variable,the total output sample r0 is a Gaussian random
variable with a mean value of either s01 or s02,depending on
whether a binary 1 or a binary 0 was sent,That is to say,
s en t 0b i n ar y af o r
s en t 1b i n ar y af o r
02
01
0 ??
??
s
sm
r
16
7.1 Error Probabilities for Binary Signaling
Results for Gaussian Noise
? Then the two conditional PDFs are
? ? ( 7, 1 2 )
2
1)|( )2/(
0
10
2
0
2
010 ?
??
sresrf ???
? ? ( 7, 1 3 )
2
1)|( )2/(
0
20
2
0
2
020 ?
??
sresrf ???
? Using equally likely source statistics,the BER becomes,
? ? ? ?
????
????
????
??
??
??
??
????
????
TT
TT
TT
V
sr
V
sr
VV
VV
e
dredre
drsrfdrsrf
drsrfs e n tsPdrsrfs e n tsPP
0
)2/(
0
0
)2/(
0
020010
02020101
2
0
2
020
2
0
2
010
2
1
2
1
2
1
2
1
)|(
2
1
)|(
2
1
)|() ()|() (
??
????
17
7.1 Error Probabilities for Binary Signaling
Results for Gaussian Noise
? thus
? ? ? ?
?
?
?
?
?
?
?
? ?
???
?
?
?
?
?
? ??
?
????
?
?
?
?
?
??
0
01
0
01
/
2/2/
2
1
2
1
2
1
2
1
2
1
2
1
002
2
0/01
2
??
?
?
?
? ?
??
?
sV
Q
sV
Q
dedeP
TT
sV
e
T
sTV
? To find the VT that minimizes Pe we need to solve
dPe/dVT=0
? ? ? ? 0
2
1
2
1
2
1
2
1 )2/(
0
)2/(
0
2
0
2
020
2
0
2
010 ??? ???? ??
????
srsr
T
e ee
dV
dP
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
2
0
2
0
2
0201
0201
44
)(
( m i n )
2
)(
??
d
e
T
E
Q
ss
QP
ss
b e s tV
18
7.1 Error Probabilities for Binary Signaling
Results for Gaussian Noise and Matched-Filter Reception
? If the receiving filter is optimized the BER can be
reduced,To minimize Pe,we need to maximize the
argument of Q,thus we need to find the linear filter
that maximizes,
? ? )())()((
2
0
2
0
2
0
2
002001
??
tststs d??
? [sd(t0)]2 is the instantaneous power of the difference
output signal at t=t0
19
7.1 Error Probabilities for Binary Signaling
Results for Gaussian Noise and Matched-Filter Reception
? The linear filter that maximizes the instantaneous
output signal power at the sampling t=t0 when
compared with the average output noise power σ02=n02(t)
is the matched filter,For the case of noise white noise at
the receiver input,the matched filter needs to be
matched to the difference signal sd(t)=s1(t)-s2(t),Thus
the impulse response of the matched filter for binary
signaling is,
）（ 18.7 )]()([)( 0201 ttsttsCth ????
? The output peak signal to average noise ratio that is
obtained from the matched filter is,
? ?
(t )]-(t )[
2)(
0
2
21
0
2
0
2
0
dtssE
N
Ets
T
d
dd
?
?
?
?
?
??
?
?
???
?
???
??
02
)( NEQo p tP de
20
7.1 Error Probabilities for Binary Signaling
Results for Gaussian Noise and Matched-Filter Reception
? The performance is measured by
Pe=P(err|s1) P(s1) + P(err|s2) P(s2) => [P(err|s1) + P(err|s2) ]/2
? For AWGN channel,when linear detector used,
???
?
???
??
???
?
???
? ????
2
0
2
0
2
02010201
44
)(( m in ),
2)( ??
d
eT
EQssQPandssb e s tV
? More,optimal linear receiver or match filter C[s1(t0-
t)-s2(t0-t)] can maximize the SNR of r0 and Pe even
smaller
???
?
???
??
02
)( NEQo p tP de
21
7.2 Performance of Baseband Binary Systems
(Unipolar Signaling)
22
7.2 Performance of Baseband Binary Systems
(Unipolar Signaling)
? Source data,
0
1
,0
,1
?
?
??
??
d a ta
d a tam
0
1
,0
,)(
?
?
??
? ??
d a t a
d a t aAts
??
?
?
???
0
1
),(
),()(
d a t a
d a t a
tn
tnAtr
??
?
?
???
0
1
,
,
0
001
0 d a t a
d a t a
n
nsr
??
?
?
?
?
??
0
1
0,0
,1~
0
010
d a ta
d a ta
r
srm
? Transmitted
signal
? Input of
receiver
? After
sampling
? Data received
23
7.2 Performance of Baseband Binary Systems
(Unipolar Signaling)
● Receive with (linear) LPF
● Detect with LPF of BHz to limit the noise and pass the
signal
● Sampling at any time in the pulses,best at middle of them
● If gain=1,s01=A and s02=0,
Var of n0(t) = (2B) (N0/2) = N0B
● Decision with VT=A/2
???
?
???
??
???
?
???
??
BN
AQEQP d
e
0
2
2
0 44 ?
For Rect-pulse,or for sin(x)/x pulse [B=1/(2T)] =A/2
???
?
???
??
0
2
4 N
TAQP
e ??
?
?
???
??
0
2
2 N
TAQP
e
24
7.2 Performance of Baseband Binary Systems
(Unipolar Signaling)
Receive with (linear) Match filter
– Detect with filter,h(t) = s1(T-t)-s2(T-t) = A; In fact,
this is a integrator! (Ed=A2T)
– Sampling at T (the end of pulses)
– s01=AT and s02=0;
Var of n0(t),do not care
– Decision with VT=AT/2
? Results
???
?
???
??
???
?
???
??
???
?
???
??
00
2
0 22 N
EQ
N
TAQ
N
EQP bd
e
? Eb is the average bit-energe,
? No matter what pulse-shape,Pe is the same
? The best Pe,is always smaller than or equal to that of
LPF-Pe.。
2/)( 2 TAE b ?
25
7.2 Performance of Baseband Binary Systems
(Unipolar Signaling)
? For LPF,smaller BW introduces less noise and then
good Pe,But,BW must be bigger enough to let the
signal to pass,
? However,for MF the BW might be very large,For
Example,the rect-pulse need rect-MF h(t),Its H(f) is a
sinc-function and the absolute BW is infinitive! Then
it would introduce much noise,why it could make the
best Pe?
26
7.2 Performance of Baseband Binary Systems
(Polar Signaling)
Bandwidth is BHz and the signal is low freq,in (0-B) Hz,
For Rect-pulses,B = 1/T Hz
27
7.2 Performance of Baseband Binary Systems
(Polar Signaling)
? Source data,
0
1
,0
,1
?
?
??
??
d a ta
d a tam
? Input of
receiver
? After
sampling
? Data received
? Transmitted
signal 01,,)( ????? ??? d a t ad a t aAAts
??
?
?
?
??
??
0
1
),(
),()(
d a t a
d a t a
tnA
tnAtr
??
?
?
?
??
??
0
1
,
,
001
001
0 d a ta
d a ta
ns
nsr
??
?
?
?
??
??
0
1
,0
,1~
010
010
d a ta
d a ta
sr
srm
28
7.2 Performance of Baseband Binary Systems
(Polar Signaling)
● Receive with (linear) LPF
● Detect with LPF of BHz to limit the noise and pass the
signal
● Sampling at any time in the pulses,best at middle of
them
● If gain=1,s01=A and s02=-A,
Var of n0(t) = (2B) (N0/2) = N0B
● Decision with VT=0
???
?
???
?
???
?
?
???
??
BN
AQEQP d
e
0
2
2
04 ?For Rect-pulse,or for sin(x)/x pulse [B=1/(2T)]
=A/2
???
?
???
??
0
2
N
TAQP
e ??
?
?
???
??
0
22
N
TAQP
e
29
7.2 Performance of Baseband Binary Systems
(Polar Signaling)
Receive with (linear) Match filter
– Detect with filter,h(t) = [s1(T-t)-s2(T-t)]/2 = A; In
fact,this is a integrator! (Ed=4A2T)
– Sampling at T (the end of pulses)
– s01=AT and s02=-AT;
Var of n0(t),do not care
– Decision with VT=0
? Results
? Eb is the average bit-energe,
? No matter what pulse-shape,Pe is the same
? The best Pe,is always smaller than or equal to that of
LPF-Pe.。
TAEb 2?
???
?
???
??
???
?
???
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?
???
??
00
2
0
222 NEQN TAQNEQP bde
30
7.2 Performance of Baseband Binary Systems
(Bipolar Signaling)
31
7.2 Performance of Baseband Binary Systems
(Bipolar Signaling)
Read it youself with the following hints,
1,3-levels
1,Calculate the conditional PDFs
2,Get the best thresholds,VT1 and VT2,with the limitation,
VT1=-VT2=VT
Answer,
1,What is thresholds,VT1 and VT2?
2,What is Pe for normal LPF and match filter?
3,What is the match filter h(t)?
5.0
25.0
25.0
0
1
1
0
,
,
)(
?
?
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?
?
?
??
?
?
?
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P
P
P
d a t a
d a t a
d a t a
A
A
ts
32
7.2 Performance of Baseband Binary Systems
(Bipolar Signaling)
? Source data,
0
1
,0
,1
?
?
??
??
d a ta
d a tam
? Input of
receiver
? After
sampling
? Data received
? Transmitted
signal
??
?
?
?????
0
1
),(
)(,),()(
d a t a
d a t a
tn
tnAortnAtr
??
?
?
?????
0
1
,
,,
0
001001
0 d a t a
d a t a
n
nsornsr
?
?
?
?
?
?
????
0
1
0
,,
,0
,1~
0
010010
d a t a
d a t a
r
srorsrm
5.0
25.0
25.0
0
1
1
0
,
,
)(
?
?
?
?
?
?
??
?
?
?
?
?
?
P
P
P
d a t a
d a t a
d a t a
A
A
ts
33
7.2 Performance of Baseband Binary Systems
(Bipolar Signaling)
For the case of additive Gaussian noise,the BER is,
?
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00
22
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1
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) () |() () |(
??
TT
e
V
Q
VA
Q
s e n tsPs e n tse r r o rP
s e n tAPs e n tAe r r o rPs e n tAPs e n tAe r r o rPP
Where the optimum value of VT is
2ln)/(2/ 20 AAV T ???
When σ0<<A,thus
2/AV T ?
? ?
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2
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2
22
3
N
EA
A
QP
d
outN
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MFf o r
2
3
L P Ff o r
42
3
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0
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34
7.3 Coherent Detection of Bandpass Binary
Signals(On-Off Keying )
·The detection is
mixer+H(f),
which is linear
processing
· Note the LO is
2cos(·) to keep
gain of signal
being 1
· LPF is to filter
out the 2fc
component
produced by
mixer
35
7.3 Coherent Detection of Bandpass Binary
Signals (On-Off Keying )
? Source signal,
0
1
,0
,1
?
?
??
??
b i n a r y
b i n a r ym
0
1
,0
),c o s ()(
?
?
??
? ??
b i n a r y
b i n a r ytAts cc ??
??
? ???
0
1
),(
),()c o s ()(
b i n a r y
b i n a r y
tn
tntAtr cc ??
??
?
?
???
0
1
,
,
0
001
0 b i n a r y
b i n a r y
n
nsr
?
?
?
?
?
?
??
0
1
0,0
,1~
0
010
b i n a r y
b i n a r y
r
srm
? Input of
receiver
? After
sampling
? Data received
? Transmitted
signal
36
7.3 Coherent Detection of Bandpass Binary
Signals (On-Off Keying )
Noise goes through mixer,
加：备课内容 lxf_s4 (1 slides)
37
7.3 Coherent Detection of Bandpass Binary
Signals (On-Off Keying )
Receive with Mixer+LPF,(like baseband-unipolar case)
– Detect with LPF of BHz to filter off the high-freq,
then unpolar-like pulses produced,
– Sampling at any time in the pulses,best at middle of
them
– If gain=1,s01=A and s02=0;
Var of n0(t) = 2N0B
– Decision with VT=A/2
? Results
???
?
???
??
???
?
???
??
BN
AQEQP d
e
0
2
2
0 84 ?For Rect-pulse,or for sin(x)/x pulse [B=1/(2T)]
=A/2
???
?
???
??
0
2
8 N
TAQP
e ??
?
?
???
??
0
2
4 N
TAQP
e
38
7.3 Coherent Detection of Bandpass Binary
Signals (On-Off Keying )
Receive with (linear) Match filter
– Detect with filter
Ed=A2T/2
– Sampling at T (the end of pulses)
– s01=AT and s02=0; Var of n0(t),do not care
– Decision with VT=AT/2
? Eb is the average bit-energe,
? No matter what pulse-shape,Pe is the same
? The best Pe,is always smaller than or equal to that of
LPF-Pe.。
? ? ))(c o s ()()()( 21 cc tTAtTstTsCth ?? ???????
???
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???
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???
??
00
2
0 42 N
EQ
N
TAQ
N
EQP bd
e
4/)( 2 TAE b ?
39
7.3 Coherent Detection of Bandpass Binary
Signals
Think about it,
? What is the similarity and difference
between the OOK and Baseband unipolar
signal?
40
7.3 Coherent Detection of Bandpass Binary
Signals
Different structures of MF
加：备课内容 lxf_s4 (1 slides)
41
7.3 Coherent Detection of Bandpass Binary
Signals (Binary-Phase-Shift Keying)
Read it youself with the following hints,
1,Similar to that of Baseband Polar Signaling
2,Note the bandpass (cos) characteristic
3,Note the results
Answer,
1,Pe for LPF and MF sturcture?
2,What is the match filter h(t)?
3,What makes the difference on Pe between
PSK(or Polar) and OOK(or Unipolar),noise or
Ed?
42
7.3 Coherent Detection of Bandpass Binary
Signals (Binary-Phase-Shift Keying)
? Source data
0
1
,0
,1
?
?
??
??
b i n a r y
b i n a r ym
0
1
),co s (
),co s ()(
?
?
??
?
??
??
b i n a r y
b i n a r y
tA
tAts
cc
cc
??
??
??
?
???
???
0
1
),()co s (
),()co s ()(
b i n a r y
b i n a r y
tntA
tntAtr
cc
cc
??
??
??
?
?
?
??
??
0
1
,
,
0
0
0 b i n a r y
b i n a r y
nA
nAr
?
?
?
?
?
??
??
0
1
,0
,1~
0
0
b i n a r y
b i n a r y
Ar
Arm
? Input of
receiver
? After
sampling
? Data received
? Transmitted
signal
43
7.3 Coherent Detection of Bandpass Binary
Signals
Receive with Mixer+LPF,(like baseband-polar case)
– Detect with BPF of BHz,then polar-like pulses
produced,
– Sampling at any time in the pulses,best at middle
of them
– If gain=1,s01=A and s02=-A;
Var of n0(t) = 2N0B
– Decision with VT=A/2
? Results
???
?
???
?
???
?
?
???
??
BN
AQEQP d
e
0
2
2
0 24 ?For Rect-pulse,or for sin(x)/x pulse [B=1/(2T)]
=A/2
???
?
???
??
0
2
2 N
TAQP
e ??
?
?
?
?
?
?
?
0
2
N
TAQP
e
44
7.3 Coherent Detection of Bandpass Binary
Signals
Receive with (linear) Match filter
– Detect with filter
Ed=2A2T
– Sampling at T (the end of pulses)
– s01=AT and s02=-AT; Var of n0(t),do not care
– Decision with VT=AT/2
? Eb is the average bit-energe,
? No matter what pulse-shape,Pe is the same
? The best Pe,is always smaller than or equal to that of
LPF-Pe.。
? ? ))(c o s (2)()()( 21 cc tTAtTstTsCth ?? ???????
?
?
?
?
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00
2
22 NEQN TAQP be
45
7.3 Coherent Detection of Bandpass Binary
Signals (Frequency-Shift Keying)
46
7.3 Coherent Detection of Bandpass Binary
Signals (Frequency-Shift Keying)
? Source data,
0
1
,0
,1
?
?
??
??
d a ta
d a tam
0
1
),co s (
),co s ()(
22
11
?
?
??
?
?
??
d a t a
d a t a
tA
tAts
??
??
??
?
?
?
??
???
0
1
),()c o s (
),()c o s ()(
222
111
d a ta
d a ta
tntA
tntAtr
??
??
??
?
?
?
???
????
0
1
),()0(
),()0(
020102
020101
0 d a ta
d a ta
nns
nnsr
??
?
?
?
??
??
0
1
,0
,1~
020
010
d a ta
d a ta
sr
srm
? Input of
receiver
? After
sampling
? Data received
? Transmitted
signal
47
7.3 Coherent Detection of Bandpass Binary
Signals (Frequency-Shift Keying)
Receive with Mixer+LPF,(like baseband-polar case)
– Detect with LPF of BHz to filter off the high-freqs,then
polar-like pulses produced,
– Sampling at any time in the pulses,best at middle of them
– If gain=1,s01=+A and -s02=-A;
Var of n0(t) = 2x2N0B = 4N0B (doubled)
– Decision with VT=0
? Results
? For Rect-pulse,or for sin(x)/x pulse [B=1/(2T)]
???
?
???
??
0
2
4 N
TAQP
e ?
?
?
?
???
??
0
2
2 N
TAQP
e
???
?
???
??
???
?
???
??
BN
AQEQP d
e
0
2
2
0 16
4
4 ?
48
7.3 Coherent Detection of Bandpass Binary
Signals (Frequency-Shift Keying)
Receive with (linear) Match filter
– Detect with filter
– Sampling at T (the end of pulses)
– s01=AT and -s02=-AT;
Var of n0(t),do not care
– Decision with VT=0
?
?
???
????
T
T
b
td tffATA
dttTstTsE
0
21
22
0
21
)(2c o s
)]()([
?
? ?
)])(c o s ())([ c o s (2
)()()(
2211
21
???? ??????
????
tTtTA
tTstTsCth
49
7.3 Coherent Detection of Bandpass Binary
Signals (Frequency-Shift Keying)
? Results
– Ed (max) = A2T,when f1-f2=n/(2T)=n(bit-rate/2); Then
Pe is min as
? f1 and f2 is called orthogonal,when configured as above
condition
Or,when f1-f2>>bit-rate,approximately orthogonal
– Eb is the average bit-energe,
– No matter what pulse-shape,Pe is the same
– This is best Pe,smaller than or equal to LPF-Pe
???
?
???
??
???
?
???
??
???
?
???
??
00
2
0 22 N
EQ
N
TAQ
N
EQP bd
e
2/)( 2 TAE b ?
? Unlike OOK,FSK signal is constant-enveloped,Why Pe of
FSK is not the same as PSK?
50
Summary
? When modulated signal is passed thru the AWGN channel,receiver
with linear detector is often used
– The mixer+LPF provides a basic result,
Best Pe when sinc-pulse (smallest BW of LPF)
3dB worse when rect-pulse； even-worse if wider-BW pulses used
– The match filter=correlator,C[s1(T-t)-s2(T-t)] can minimize Pe,so
best result
? Different modulations yield different Pe,With the same Eb/N0,means,
Modulations A Pe of MF Comment (same N0)
OOK = unipolar 1.414 3dB worse and larger
signal peak
PSK = polar 1 Best
FSK (keep f1 and f2
orthogonal)
1 3dB worse; noise
doubled due to 2 bands
? ?0/ NEQ b
? ?)/(2 0NEQ b
? ?0/ NEQ b
51
7.4 Noncoherent Detection of Bandpass Binary
Singals
52
Homework
