Chapter 5
Baseband Digital Transmission
5.1
Preview
In this chapter we consider several baseband digital modulation and demodulation tech­
niques for transmitting digital information through an additive white Gaussian noise
channel. We begin with binary pulse modulation and then we introduce several non­
binary modulation methods. We describe the optimum receivers for these different
signals and consider the evaluation of their performance in terms of the average prob­
ability of error.
5 .2
Binary Signal Transmission
In a binary communication system, binary data consisting of a sequence of O's and
l's are transmitted by means of two signal waveforms, say, 5o(t) and 51 (t). Suppose
that the data rate is specified as R bits per second. Then each bit is mapped into a
corresponding signal waveform according to the rule
where Tb
=
0---+ 5o(t),
0 ::; t ::; Tb
1---+ 51 (t)'
0 ::; t ::; Tb
1/R is defined as the bit time interval. We assume that the data bits 0
and 1 are equally probable-that is, each occurs with probability
� -and are mutually
statistically independent.
The channel through which the signal is transmitted is assumed to corrupt the sig­
nal by the addition of noise, denoted as n(t), which is a sample function of a white
Gaussian process with power spectrum No/2 watts/hertz. Such a channel is called an
additive white Gaussian noise (AWGN) channel. Consequently, the received signal
waveform is expressed as
r(t)
=
5i(t) + n(t),
i
=
0, 1,
(5.2.1)
183
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184
CHAPTER 5. BASEBAND DIGITAL TRANSMISSION
The task of the receiver is to determine whether a 0 or a 1 was transmitted after
observing the received signal r(t) in the interval 0 :-:::: t :-:::: Tb. The receiver is designed
to minimize the probability of error. Such a receiver is called the optimum receiver.
5.2.1
Optimum Receiver for the AWGN Channel
In nearly all basic digital communication texts, it is shown that the optimum receiver for
the AWGN channel consists of two building blocks. One is either a signal correlator
or a matched filter . The other is a detector.
Signal Correlator
The signal correlator cross-correlates the received signal r(t) with the two possible
transmitted signals so(t) and si(t), as illustrated in Figure 5.1. That is, the signal
correlator computes the two outputs
ro(t)
r1 (t)
=
=
J:
f:
r(T)so(T) dT
r(T)S1 (T) dT
in the interval 0 :-:::: t :-:::: Tb, samples the two outputs at t
(5.2.2)
=
Tb, and feeds the sampled
outputs to the detector.
J�O dr:
r(t)
Detector
J�O dr:
Output data
I
Sample
at t
=
1/,
Figure 5.1: Cross-correlation of the received signal r(t) with the two transmitted sig­
nals
---llMlll§jl;§il�Nii;t•]:JMB@I
Illustrative Problem 5.1 [Signal Correlator] Suppose the signal waveforms so(t)
and si(t) are as shown in Figure 5.2, and let so(t) be the transmitted signal. Then
the received signal is
r(t)
=
so(t) + n(t),
(5.2.3)
Determine the correlator outputs at the sampling instants.
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185
5.2. BINARY SIGNAL TRANSMISSION
s0(t)
Ai-----�
A
0
0
-A
Figure 5.2: Signal waveforms so(t) and 51 (t) for a binary communication system
When the signal r (t) is processed by the two signal correlators shown in Figure 5.1,
the outputs ro and r1 at the sampling instant t= Tb are
J:b r (t)so(t) dt
= f:b s6(t) dt + f:b n(t)so(t) dt
ro=
= E+no
(5.2.4)
and
J:b r (t)s1 (t) dt
= J:b so(t)s1 (t) dt + J:b n(t)s1 (t) dt
r1 =
= n1
(5.2.5)
where no and n1 are the noise components at the output of the signal correlators; that
is,
J:b n(t)so(t) dt
n1 = J:b n(t)s1 (t) dt
no=
(5.2.6)
and E= A 2 Tb is the energy of the signals so(t) and s1 (t) . We also note that the two
signal waveforms are orthogonal; that is,
J:b so(t)s1 (t) dt=
0
(5.2.7)
On the other hand, when 51 (t) is the transmitted signal, the received signal is
r (t) = s1 (t) +n(t),
It is easy to show that, in this case, the signal correlator outputs are
(5.2.8)
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186
CHAPTER 5. BASEBAND DIGITAL TRANSMISSION
Output of
Correlator 0
Output of
Correlator 1
Output of
Correlator 0
Output of
Correlator 1
E
E
Tb
0
E
E
2
2
Tb
0
Tb
2
0
0
(a)
(b)
Figure 5.3: Noise-free correlator outputs. (a)
so(t) was transmitted. (b) s1(t) was
transmitted
t .:-::: Tb for
so(t) is transmitted and when s1 (t) is transmitted.
Figure 5.3 illustrates the two noise-free correlator outputs in the interval 0 .:-:::
each of the two cases-that is, when
n(t) is a sample function of a white Gaussian process with power spec­
trum No/2, the noise components no and n1 are Gaussian with zero means-that is,
Because
'Tb
so(t)E [n(t)] dt = 0
Jo
'Tb
s1(t)E [n(t)] dt = 0
E(n1) =
Jo
E(no) =
and variances
(5.2.9)
al, for i = 1, 2, where
al = E(n�)
'Tb 'Tb
= J J Si(t)si(T)E[n(t)n(T)] dtdT
o o
N 'Tb
= J Si(t)si(T)D(t-T)dtdT
o
Tb
= __Q
s? (t)dt
2 0
N,
ENo
i = 0, 1
2
-f
f
i
'
Therefore, when
(5.2.10)
(5.2.11)
so(t) is transmitted, the probability density functions of ro and r1 are
�
- - - - -e- ( ro-E)2/2a2
p(ro I so(t) was transmitted) = J2IT
2
p ( r1 I s0 (t
.
) was transm1tted)
=
These two probability density functions, denoted as
s1(t)
cr
1
rz /2a2
-ffi
- - - -i -e- 1
rr
2 cr
(5.2.12)
p(ro I 0) and p(r1 I 0), are
ro
E
illustrated in Figure 5 .4. Similarly, when
is transmitted,
is zero-mean Gaussian
2
2
with variance cr and
is Gaussian with mean value
and variance cr •
r1
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5.2. BINARY SIGNAL TRANSMISSION
187
0
Figure 5.4: Probability density functions p(ro I 0) and p(r1 I 0) when so(t) is trans­
mitted
--tiMP�ii;f4iif)#lflj;t•]:)l§I·•-----Illustrative Problem 5.2 [Correlation of Signal Waveforms] Sample the signal wave­
forms in Illustrative Problem 5. 1 at a rate F5=20/Tb (sampling interval T5=Tb/20)
and perform the correlation of r(t) with so(t) and s1(t) numerically; that is, compute
and plot
k
ro(kTs)= I r(nTs)so(nT5), k=l,2,...,20
n=l
and
k
r1(kTs)= I r(nT5)si(nT5), k=l,2,...,20
n=l
when (a) so(t) is transmitted signal and (b) s1(t) is the transmitted signal.
Repeat the above computations and plots when the signal samples r(kT5) are cor­
rupted by additive white Gaussian noise samples n(kT5), 1 :-s: k :-s: 20, which have
zero mean and variance a-2 =0.1 and a-2 = 1.
Figure 5.5 illustrates the correlator outputs. We note the effect of the additive noise on
the outputs of the correlator, especially when a-2 = 1. The MATLAB script for these
computations is given below.
----tl®li" -----% MATLAB script for Illustrative
Problem 5 .2
% Initialization:
% Number of samples
K=20;
A=1;
% Signal amplitude
l=O:K;
% Defining signal waveforms:
s_O=A*ones(1,K);
s_l=[A *ones(1,K/2) -A*ones(1,K/2)];
% Initializing output signals:
r_O=zeros(1,K);
r_l=zeros(1,K);
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CHAPTER 5. BASEBAND DIGITAL TRANSMISSION
188
30
20
20
10
10
0
0
lOTb
5Tb
0
0
20Tb
15Tb
(a) cr2= 0 & S0 is transmitted
5Tb
lOTb
15Tb
20Tb
(b) a2= 0 & S1 is transmitted
30
20
20
10
10
0
0
lOTb
5Tb
0
0
20Tb
15Tb
(c) <:P= 0.1 & S0 is transmitted
5Tb
lOTb
15Tb
20Tb
(d) cr2= 0.1 & S1 is transmitted
30
20
20
10
10
....
0
lOTb
5Tb
0
0
....
15Tb
(e) <J2= 1 & S0 is transmitted
--
20Tb
0
5Tb
lOTb
15Tb
20Tb
(f) <J2= 1 & S1 is transmitted
Figure 5.5: Correlator outputs in Illustrative Problem l5.2l Solid and dashed lines rep­
resent the output of correlation with so ( t) and s1 ( t), respectively.
%
Case 1: noise-N(O,O)
noise=random('Normal',0,0,1,K);
%
Sub-case s
s=s_O;
r=s+noise;
for n=1 :K
%
=
s_O:
received signal
r_O(n)=sum(r(1 :n).*s_0(1 :n));
r_l(n)=sum(r(1 :n).*s_l(1 :n));
end
% Plotting the results:
subplot(3,2,1)
plot(l,[O r_O],'-',l,[O r_l],'--')
set(gca, 'XTickLabel',{'0 ','5Tb',' lOTb',' 15Tb','20Tb'})
axis([O 2 0 -5 30])
xlabel('(a) \sigmaA2= 0 & S {O} is transmitted','fontsize',10)
%
Sub-case s
s=s_l;
r=s+noise;
for n=1 :K
%
=
s_J:
received signal
r_O(n)=sum(r(1 :n).*s_0(1 :n));
r_l(n)=sum(r(1 :n).*s_l(1 :n));
end
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5.2. BINARY SIGNAL TRANSMISSION
% Plotting
189
the results:
subplot(3,2,2)
plot(l,[O r_O],'-',l,[O r_l],'--')
set(gca,'XTickLabel',{'0','STb','lOTb','lSTb','20Tb'})
axis([O 20 -5 30])
\sigma"2= 0 & S_{l} is transmitted','fontsize',10)
Case 2: noise-N(0,0.1)
noise=random('Normal',0,0.1,1,K);
% Sub-case s
s_O:
xlabel('(b)
%
=
s=s_O;
r=s+noise;
%
received signal
for n=1 :K
r_O(n)=sum(r(1 :n).*s_Q(1 :n));
r_l(n)=sum(r(1 :n).*s_l(1 :n));
end
% Plotting
the results:
subplot(3,2,3)
plot(l,[O r_O],'-',I,[O r_l],'--')
set(gca,'XTickLabel',{'0','STb','10Tb','lSTb', '20Tb'})
ax.is([O 20 -5 30])
xlabel('(c)
% Sub-case
\sigma"2= 0.1 & S_{O} is transmitted','fontsize',10)
s
s_J:
=
s=s_l;
r=s+noise;
% received signal
for n=1 :K
r_O(n)=sum(r(1 :n).*s_0(1 :n));
r_l(n)=sum(r(1 :n).*s-1(1 :n));
end
% Plotting
the results:
subplot(3,2,4)
plot(l,[O r_O],'-',I,[O r_l],'--')
set(gca,'XTickLabel, ,{'0,',STb','lOTb'',lSTb'',20Tb'})
axis([O 20 -5 30])
\sigma"2= 0.1 & S_{l} is transmitted','fontsize',10)
Case 3: noise-N(O,l)
noise=random('Normal',0,1,1,K);
% Sub-case s
s_O:
xlabel('(d)
%
=
s=s_O;
r=s+noise;
%
received signal
for n=1 :K
r_O(n)=sum(r(1 :n).*s_0(1 :n));
r_l(n)=sum(r(1 :n).*s_l(1 :n));
end
% Plotting
the results:
subplot(3,2,5)
plot(l,[O r_O],'-',l,[O r_l],'--')
set(gca,, XTickLabel, ,{,0,',STb,',lOTb,',lSTb,',20Tb,})
axis([O 20 -5 30])
xlabel('(e)
% Sub-case
\sigma"2= 1 & S_{O} is transmitted','fontsize',10)
s
s_J:
=
s=s_l;
r=s+noise;
% received signal
for n=1 :K
r_O(n)=sum(r(1 :n).*s_0(1 :n));
r_l(n)=sum(r(1 :n).*s_l(1 :n));
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190
CHAPTER 5. BASEBAND DIGITAL TRANSMISSION
end
% Plotting the results:
subplot(3,2,6)
plot(l,[O r_O],' -',l,[O r_l],' -- ')
set(gca, 'XTickLabel',{'0 ','5Tb','lOTb',' 15Tb','20Tb'})
axis([O 20 -5 30])
xlabel('(f) \sigrnaA2= 1 & S_{l} is transrnitted','fontsize',10)
Matched Filter
The matched filter provides an alternative to the signal correlator for demodulating
the received signal r(t). A filter that is matched to the signal waveform s(t), where
0 :s: t :s: Tb, has an impulse response
h(t)
=
s(Tb - t),
(5.2.13)
Consequently, the signal waveform-say, y (t) -at the output of the matched filter
when the input waveform is s(t) is given by the convolution integral
y (t)
=
f:
S(T)h(t - T) dT
(5.2.14)
If we substitute in (5.2.14) for h(t - T) from (5.2.13), we obtain
y (t)
and if we sample y (t) at t
=
=
f:
S(T)s(Tb - t + T) dT
(5.2.15)
Tb, we obtain
(5.2.16)
where E is the energy of the signal s(t). Therefore, the matched filter output at the
sampling instant t
=
Tb is identical to the output of the signal correlator.
Illustrative Problem 5.3 [Matched Filter] Consider the use of matched filters for the
demodulation of the two signal waveforms shown in Figure 5.2 and determine the
outputs.
The impulse responses of the two matched filters are
ho(t)
=
so(Tb - t)
h1(t)
=
s1(Tb - t)
(5.2.17)
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191
5.2. BINARY SIGNAL TRANSMISSION
Ai-----�
A
0
0
-A
Figure 5.6: Impulse responses of matched filters for signals so(t) and s1 (t)
0
(a)
(b)
Figure 5.7: Signal outputs of matched filters when so(t) is transmitted
as illustrated in Figure 5.6. Note that each impulse response is obtained by folding the
signal s(t) to obtain s( -t) and then delaying the folded signal s( -t) by Tb to obtain
s(Tb -t).
Now suppose the signal waveform so(t) is transmitted. Then the received signal
r (t) = so(t) + n(t) is passed through the two matched filters. The response of the
filter with impulse response ho(t) to the signal component so(t) is illustrated in Fig­
ure 5.7(a). Also, the response of the filter with impulse response h1 (t) to the signal
component so(t) is illustrated in Figure 5.7(b). Hence, at the sampling instant t = Tb,
the outputs of the two matched filters with impulse responses ho(t) and hi (t) are
ro = E +no
r1 = n1
(5.2.18)
respectively. Note that these outputs are identical to the outputs obtained from sampling
the signal correlator outputs at t = Tb.
�18!4-il;ful�Tj#lij;l•1:Jl§@I·----Illustrative Problem 5.4 [Match Filtering of Signal Waveforms] Sample the signal
waveform in Illustrative Problem 5 .3 at a rate of F5 = 20 I Tb and perform the matched
filtering of the received signal r (t) with so(t) and s1 (t) numerically; that is, compute
and plot
k
Yo(kTs) =
L r(nTs)so(kTs -nTs),
k= 1,2,... ,20
n=l
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192
CHAPTER 5. BASEBAND DIGITAL TRANSMISSION
and
k
Y1(kTs) =
L r(nTs)si(kTs-nTs),
k= 1,2, ...,20
n=l
when (a) so ( t) is the transmitted signal and (b) s1 ( t) is the transmitted signal.
Repeat the above computations when the signal samples r(kT5) are corrupted by
additive white Gaussian noise samples n(kT5), 1 ::5: k ::5: 20, which have zero mean
and variance cr2 = 0.1 and cr2 = 1.
Figure 5.8 illustrates the matched filter outputs. We note the effect of the additive noise
on the matched filter outputs for the two variances of the noise. Clearly, when cr2 = 1,
the effect of the noise on the matched filter outputs is significant. The MATLAB script
for the computation is given next
2a
20
a
---
-------
-
a
---
--- , ,
_
,
-20
-20
a
2a f-
l
-
a---
1
-
-
------
--
-2a fa
a
5Th lOTh 15Th 20Th a 5Th lOTh 15Th 2aTu a
(a) cr2= 0 & Sa is transmitted
5Th lOTh 15Th 20Th a 5Th lOTh 15Th 20Th a
(b) o2= 0 & S1 is transmitted
20
a
-----
-2a
a
5Th 1aTu 15Th 2aTu a 5Th 1aTu 15Th 2aTu a
(c) cr2= 0.1 & Sa is transmitted
2a
5Th JaTu 15Th 2aTu a STh lOTh ISTh 2aTu a
(d) o2= 0.1 & S1 is transmitted
2a
a
-
-2a
a
-
.,,,,.
,. -
a
- ,,,
5Th lOTh 15Th 20Th a 5Th IaTu 15Th 2aTu a
(e) o2= l & S0 is transmitted
-2a
a
STh IOTh 15Th 20Th a STb IOTb 15Th 2aTu a
(t) o2= 1 & S1 is transmitted
Figure 5.8: Matched filter outputs in Illustrative Problem 5.4. Solid and dashed lines
correspond to the outputs of filters matched to so ( t) and s1 ( t), respectively.
____,,,
_ .. -----% MATLAB script for Illustrative Problem 5.4.
% Initialization:
K=20;
% Number of samples
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5.2. BINARY SIGNAL TRANSMISSION
A=1;
193
% Signal amplitude
l=O:K;
% Defining signal waveforms:
s_O=A*ones(1,K);
s_l=[A*ones(1,K/2) -A*ones(1,K/2)];
% Initializing output signals:
y_O=zeros(1,K);
y_l=zeros(1,K);
% Case 1: noise-N(O,O)
noise=random('Normal',0,0,1,K);
% Sub-case s
=
s_O:
s=s-0;
y=s+noise;
% received signal
y_O=conv(y,wrev(s_O));
y_l=conv(y,wrev(s_ l));
% Plotting the results:
subplot(3,2,1)
plot(l,[O y_0(1:K)],'-k',l,[O y_l(1:K)],'--k')
set(gca,'XTickLabel',{'0 ','STb','lOTb','lSTb','20Tb'})
axis([O 20 -30 30])
xlabel('(a) \sigmaA2= 0 & S {O} is transmitted','fontsize',10)
% Sub-case s
=
s_J:
s=s_l;
y=s+noise;
% received signal
y_O=conv(y,wrev(s_O));
y_l=conv(y,wrev(s_ l));
% Plotting the results:
subplot(3,2,2)
plot(l,[O y_0(1:K)],'-k',l,[O y_l(1:K)],' --k')
set(gca,'XTickLabel',{'0 ','STb','10Tb','15Tb', '20Tb'})
axis([O 20 -30 30])
xlabel('(b) \sigmaA2= 0 & S_{l} is transmitted','fontsize',10)
% Case 2: noise-N(0,0.1)
noise=random('Normal',0,0.1,1,K);
% Sub-case s
=
s_O:
s=s-0;
y=s+noise;
% received signal
y_O=conv(y ,wrev(s_O));
y_l=conv(y,wrev(s_ l));
% Plotting the results:
subplot(3,2,3)
plot(l,[O y_0(1 :K)],' -k' ,l,[O y_l(1 :K)],' --k')
set(gca,'XTickLabel, ,{'0,',STb','10Tb'',lSTb'',20Tb'})
axis([O 20 -30 30])
xlabel('(c) \sigmaA2= 0.1 & S_{O} is transmitted','fontsize',10)
% Sub-case s
=
s_J:
s=s_l;
y=s+noise;
% received signal
y_O=conv(y ,wrev(s_O));
y_l=conv(y,wrev(s_ l));
% Plotting the results:
subplot(3,2,4)
plot(l,[O y_0(1:K)],'-k',l,[O y_l(1:K)],' --k')
set(gca,'XTickLabel',{'0 ','STb','10Tb','lSTb', '20Tb'})
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CHAPTER 5. BASEBAND DIGITAL TRANSMISSION
194
axis([O 20 -30 30])
xlabel('(d) \sigrna"2= 0.1 & S_{l} is transrnitted','fontsize',10}
%
Case 3: noise-N(O,l)
noise=random('Normal',0,1,1,K);
%
Sub-case s
=
s_O:
s=s_O;
y=s+noise;
%
received signal
y_O=conv(y,wrev(s_O));
y_l=conv(y,wrev(s_l)};
%
Plotting the results:
subplot(3,2,5)
plot(l,[O y_0(1 :K}],'-k',l,[O y_l(1 :K}],'--k'}
set(gca,'XTickLabel',{'0 ',' 5Tb',' lOTb',' 15Tb',' 20Tb'})
axis([O 20 -30 30])
xlabel('(e) \sigrna"2= 1 & S {O} is transrnitted','fontsize',10)
%
Sub-case s
=
s_J:
s=s_l;
y=s+noise;
%
received signal
y_O=conv(y,wrev(s_O)};
y_l=conv(y,wrev(s_ l));
%
Plotting the results:
subplot(3,2,6)
plot(l,[O y_0(1 :K)],'-k',l,[O y_l(1 :K)],'--k')
set(gca,'XTickLabel',{'0 ',' 5Tb',' lOTb',' 15Tb',' 20Tb'})
axis([O 20 -30 30])
xlabel('(f) \sigrna"2= 1 & S_{l} is transrnitted','fontsize',10)
The Detector
The detector observes the correlator or matched filter outputs ro and r1 and decides
on whether the transmitted signal waveform is so(t) or s1 (t) , which correspond to the
transmission of either a 0 or a 1, respectively. The optimum detector is defined as the
detector that minimizes the probability of error.
---li!!IJ§ll;fflHtjjj;t•1:Jl@l1
Illustrative Problem 5.5 [Binary Detection] Let us consider the detector for the sig­
nals shown in Figure 5.2, which are equally probable and have equal energies. The
optimum detector for these signals compares ro and r1 and decides that a 0 was trans­
mitted when ro
>
r1
and that a 1 was transmitted when r1
> ro.
Determine the
probability of error.
When s0(t) is the transmitted signal waveform, the probability of error is
Pe
=
P( r1 > ro)
=
P(n1 > E +no)
=
P(n1 - no > E)
(5.2.19)
Because n1 and no are zero-mean Gaussian random variables, their difference x
ni - no is also zero-mean Gaussian. The variance of the random variable xis
(5.2.20)
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195
5.2. BINARY SIGNAL TRANSMISSION
But
E(n1no)
=
0 because the signal waveforms are orthogonal; that is,
E(n1 no)
=
=
=
=
rTb rTb
E J J so(t)s1(T)n(t)n(T) dt dT
o o
No rTb rTb
2 J J So(t)s1(T)O(t - T) dt dT
o o
No rTb
so(t)s1(t) dt
T Jo
(5.2.21)
0
Therefore,
E(x2)
=
2
( E�o ) ENo u�
=
=
(5.2.22)
Hence, the probability of error is
(5.2.23)
The ratio
EI No is called the signal-to-noise ratio (SNR).
The derivation of the detector performance given in the example was based on the
transmission of the signal waveform
so(t). The reader may verify that the probability
of error that is obtained when s1(t) is transmitted is identical to that obtained when
so(t) is transmitted. Because the O's and 1's in the data sequence are equally probable,
the average probability of error is that given by (5.2.23). This expression for the prob­
ability of error is evaluated by using the MATLAB script given next and is plotted in
Figure 5.9 as a function of the SNR, where the SNR is displayed on a logarithmic scale
( 10 log10 EI No). As expected, the probability of error decreases exponentially as the
SNR increases.
-------41111@1il! �---���
% The MATLAB script that generates the probability of error versus the signal-to-noise ratio.
initiaLsnr=O;
finaLsnr=15;
snr_step=0.25;
snr_in_dB=initiaLsnr:snr_step:final_snr;
for i=1 :length(snr_in_dB),
snr=1 DA (snr_in_dB(i)/1 O);
Pe(i)=Qfunct(sqrt(snr));
echo off;
end;
echo on;
semilogy(snr_in_dB,Pe);
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CHAPTER 5. BASEBAND DIGITAL TRANSMISSION
196
°
10
10-
10-
10-
10-
10-
4
5
6
7
10-
10-
2
3
10-
10-
1
8
9
0
Figure 5.9: Probability of error for orthogonal signals
Monte Carlo Simulation of a Binary Communication System
Monte Carlo computer simulations are usually performed in practice to estimate the
probability of error of a digital communication system, especially in cases where the
analysis of the detector performance is difficult to perform. We demonstrate the method
for estimating the probability of error for the binary communication system described
previously.
Illustrative Problem 5.6 [Monte Carlo Simulation] Use Monte Carlo simulation to
estimate and plot Pe versus SNR for a binary communication system that employs
correlators or matched filters. The model of the system is illustrated in Figure 5.10.
We simulate the generation of the random variables ro and r1, which constitute the
input to the detector. We begin by generating a binary sequence of O's and 1's that
occur with equal probability and are mutually statistically independent. To accomplish
this task, we use a random number generator that generates a uniform random number
with a range ( 0, 1). If the number generated is in the range ( 0, 0. 5), the binary source
output is a 0. Otherwise, it is a 1. If a 0 is generated, then ro
a 1 is generated, then ro
=
no and r1
=
=
E + no and r1
=
n1. If
E + n i.
The additive noise components no and n 1 are generated by means of two Gaussian
noise generators. Their means are zero, and their variances are a2
convenience, we may normalize the signal energy E to unity (£
=
=
ENo/2. For
1) and vary a2•
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5.2. BINARY SIGNAL TRANSMISSION
Uniform random
number generator
Binary
data source
197
Gaussian random
number generator
Output
data
0/E
Detector
1/E
'------..i+ 1-------l�I
Gaussian random
number generator
Compare
Error counter
Figure 5.10: Simulation model for Illustrative Problem 5.6
1if �---�--��--�---�
llE
10-s .____....____...._____....._____._____.____,
2
4
12
10
0
8
Figure 5.11: Error probability from Monte Carlo simulation compared with theoretical
error probability for orthogonal signaling
Note that the SNR, which is defined as E/No, is then equal to l/2u2• The detector
output is compared with the binary transmitted sequence, and an error counter is used
to count the number of bit errors.
Figure 5.11 illustrates the results of this simulation for the transmission of N
=
10, 000 bits at several different values of SNR. Note the agreement between the
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CHAPTER 5. BASEBAND DIGITAL TRANSMISSION
198
simulation results and the theoretical value of Pe given by (5.2.23). We should also
note that a simulation of N = 10,000 data bits allows us to estimate the error probabil­
10-3. In other words, with N = 10,000 data bits, we
ity reliably down to about Pe
=
should have at least ten errors for a reliable estimate of Pe. MATLAB scripts for this
problem are given next.
----tl&li" -----% MATLAB script for Illustrative Problem 5.6.
echo on
SNRindB1=0:1 :12;
SNRindB2=0:0.1 :12;
for i=1 :length(SNRindBl),
% simulated error rate
smld_err_prb(i)=smldPe54(SNRindB 1(i));
echo off ;
end;
echo on ;
for i=1 :length(SNRindB2),
SNR=exp(SNRindB2(i)*log(10)/1O);
% theoretical error rate
theo_err_prb(i)=Qfunct(sqrt(SNR));
echo off ;
end;
echo on;
% Plotting commands follow.
semilogy(SNRindBl,smld_err_prb,' * ');
hold
semilogy(SNRindB2,theo_err_prb);
----tl&li" -----function [p]=smldPe54(snr_in_dB)
% [p]=smldPe54(snr_in....dB)
SMWPE54 finds the probability of error for the given
%
%
snr .Jn....dB, signal-to-noise ratio in dB.
E=1;
SNR=exp(snLin_dB*log(10)/10);
sgma=E/sqrt(2*SNR);
% signal-to-noise ratio
% sigma, standard deviation of noise
N=10000;
% generation of the binary data source
for i=1 :N,
temp=rand;
% a uniform random variable over (0,1)
if (temp<0.5),
dsource(i)=O;
% With probability 112, source output is 0.
else
dsource(i)=1;
% With probability 112, source output is 1.
end
end;
% detection, and probability of error calculation
numoferr=O;
for i=1 :N,
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5.2. BINARY SIGNAL TRANSMISSION
% matched filter outputs
if (dsource(i)==O),
rO=E+gngauss(sgma);
rl=gngauss(sgma);
else
rO=gngauss(sgma);
rl=E+gngauss(sgma);
end;
% Detector follows.
if (rO>rl),
decis=O;
else
decis=1;
end;
if (decis-=dsource(i)),
numoferr=numoferr+1;
end;
end;
p=numoferr/N;
199
% if the source output is "O"
% if the source output is "1 "
% Decision is "O".
% Decision is "l ".
% If it is an error, increase the error counter.
% probability of error estimate
---C•1ij§.ilt•Jt•
In Figure 5.11, simulation and theoretical results completely agree at low signal-to­
noise ratios, whereas at higher SNRs they agree less. Can you explain why? How
should we change the simulation process to result in better agreement at higher signal­
to-noise ratios?
5.2.2
Other Binary Signal Transmission Methods
The binary signal transmission method described above was based on the use of orthog­
onal signals. Next, we describe two other methods for transmitting binary information
through a communication channel. One method employs antipodal signals. The other
method employs an on-off-type signal.
Antipodal Signals for Binary Signal Transmission
Two signal waveforms are said to be antipodal if one signal waveform is the negative
of the other. For example, one pair of antipodal signals is illustrated in Figure 5.12(a).
A second pair is illustrated in Figure 5.12(b).
Suppose we use antipodal signal waveforms so(t)
= s(t) and 51 (t) = -s(t) to
transmit binary information, where s(t) is some arbitrary waveform having energy E.
The received signal waveform from an AWGN channel may be expressed as
r(t)
= ±s(t) + n(t),
(5.2.24)
The optimum receiver for recovering the binary information employs a single correlator
or a single matched filter matched to s(t), followed by a detector, as illustrated in
Figure 5.13. Let us suppose that s(t) was transmitted, so that the received signal is
r(t)
= s(t) + n(t)
(5.2.25)
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CHAPTER 5. BASEBAND DIGITAL TRANSMISSION
200
s1(t)
A t------.
0
Tb
0
(a)
A
s0(t)
0
-A
-A
A
Tb
s1(t)
0
(b)
Tb
-A
Figure 5.12: Examples of antipodal signals. (a) A pair of antipodal signals. (b) Another
pair of antipodal signals
Received
s1gna1 r(t)
·
---•�I
Matched filter
s (T.b t)
_
11--- -__,./�
t
Detector
Output
. .
dec1s1on
Sample
at t T,,
=
(a)
Received
signal r(t)
1----------- 1
I
f.o'Odt
I
I
s(t)
1
o
C ���---
Output
i---i11,____, t �
� decision
Sample
at t T,,
=
_____
(b)
Figure 5.13: Optimum receiver for antipodal signals. (a) Matched filter demodulator.
(b) Correlator demodulator
The output of the correlator or matched filter at the sampling instant
r=E+n
where
as
t = Tb is
(5.2.26)
E is the signal energy and n is the additive noise component, which is expressed
r Tb
n= J
o
n(t)s(t) dt
(5.2.27)
Because the additive noise process n(t) is zero mean, it follows that E(n)
The variance of the noise component
n is
rTb fTb
=J
E [n(t)n(T)] s(t)s(T) dt dT
o T T
N f bf b
=T
<5(t - T)S(t)S(T) dt dT
T
NE
N f b 2
= o
s (t) dt = o
=
0.
0
0
2
0
0
2
(5.2.28)
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201
5.2. BINARY SIGNAL TRANSMISSION
-E
E
0
Figure 5.14: Probability density functions for the input to the detector
Consequently, the probability density function of r when s(t) is transmitted is
p(r I s(t) was transmitted) = p(r I 0) =
1
J2IT (J
e-(r-£J2 /20-2
(5.2.29)
Similarly, when the signal waveform -s(t) is transmitted, the input to the detector is
(5.2.30)
r = -E + n
and the probability density function of r is
p(r I -s(t) was transmitted) = p(r I 1) =
1
J2IT
- (r+£)2 /2a2
ue
(5.2.31)
These two probability density functions are illustrated in Figure 5.14.
For equally probable signal waveforms, the optimum detector compares r with the
threshold zero. If r > 0, the decision is made that s(t) was transmitted. If r < 0, the
decision is made that -s(t) was transmitted. The noise that corrupts the signal causes
errors at the detector. The probability of a detector error is easily computed. Suppose
that s(t) was transmitted. Then the probability of error is equal to the probability that
r < 0; that is,
Pe= P(r < 0)
=
=
=
=
1
f
o
E
e-(r- )z 120-2 dr
J2IT (J -00
- E /a
1
e-r2 12
J2IT
--
Q
f
dr
-00
(!)
Q(ffeo)
(5.2.32)
A similar result is obtained when -s(t) is transmitted. Consequently, when the
two signal waveforms are equally probable, the average probability of error is given by
(5.2.32).
When we compare the probability of error for antipodal signals with that for or­
thogonal signals given by (5.2.23), we observe that, for the same transmitted signal
energy E, antipodal signals result in better performance. Alternatively, we may say
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202
CHAPTER 5. BASEBAND DIGITAL TRANSMISSION
that antipodal signals yield the same performance (same error probability) as orthogo­
nal signals by using one-half of the transmitted energy of orthogonal signals. Hence,
antipodal signals are 3 dB more efficient than orthogonal signals.
Illustrative Problem 5.7 [Correlation of Antipodal Signal Waveforms] In antipo­
dal signaling, the signal waveform so(t) is defined as
so(t)
=
A
,
0:::::; t:::::; Tb
and zero otherwise. The received signal is r(t)
signal r(t) and so(t) at the rate F5
=
=
±so(t) + n(t). Sample the received
20/Tb and perform the correlation of r(t) with
so(t) numerically; that is, compute and plot
k
ro(kTs)
=
L r(nT5)so(nT5),
k
=
1,2, ...,20
n=l
when (a) so(t) is the transmitted signal and (b) -so(t) is the transmitted signal. Per­
form the computations when the additive noise is Gaussian having zero mean and vari­
ances a2
=
0, a2
=
0.1, and a2
=
1.
Figure 5.15 illustrates the correlator outputs for different noise variances. The MAT­
LAB script for the computation is given next
--411®'1" 11----% MATLAB script for Illustrative Problem 5.7.
% Initialization:
K=20;
% Number of samples
A=1;
% Signal amplitude
l=O:K;
s_O=A*ones(1,K);% Signal waveform
r_O=zeros(1,K); % Output signal
% Case 1: noise-N(O,O)
noise=random( 'Normal' ,0,0,1,K};
% Sub-case s
s_O:
s=s_O;
r=s+noise; % received signal
for n=1:K
r_O(n)=sum(r(1:n). *s_0(1:n));
end
% Plotting the results:
subplot(3,2,1)
plot(l, [O r_O])
set(gca, 'XTickLabel',{' 0 ',' 5Tb',' lOTb',' 15Tb','20Tb'})
=
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203
5.2. BINARY SIGNAL TRANSMISSION
25
0
20
-5
15
-10
10
-15
5
-20
lOTh
5Th
(a)
2
cr =
15Th
20Th
-25
5Th
0
(b)
0 & S0 is transmitted
2
cr =
lOTh
15Th
20Th
0 & S1 is transmitted
0
25
20
-5
15
-10
10
-15
5
-20
lOTh
5Th
(c)
2
cr =
15Th
20Tb
-25
5Th
0
lOTh
15Th
20Th
(d) cr2= 0.1 & S1 is transmitted
0.1 & S0 is transmitted
25
0
20
15
-10
10
5
0
-20
0
lOTh
5Th
15Th
20Tb
0
(e) cr2= 1 & S0 is transmitted
5Th
lOTh
15Th
20Th
(f) cr2= 1 & S1 is transmitted
Figure 5.15: Correlator outputs in Illustrative Problem 5.7
axis([O 2 0 0 25])
label( '(a) \sigma"2= 0 & S_{O} is transmitted ','fontsize',10)
x
text(l5,3,'\jontsize{IO} r_[O}: - & r..{1}: -','hor','left')
%
%
Sub-case s
=
s_J:
s=-s-0;
r=s+noise; % received signal
for n=1 :K
r_O(n)=sum(r(1 :n).*s_0(1 :n));
end
% Plotting
the results:
subplot(3,2,2)
plot(l,[O r_O])
set( gca,'XTickLabel',{'0 ','5Tb','lOTb',' 15Tb',' 20Tb'})
axis([O 20 -25 OJ)
,10)
xlabel( '(b) \sigma"2= 0 & S_{l} is transmitted ','fontsize'
%
Case 2: noise-N(0,0.1)
noise=random( 'Normal',0,0.1,1,K);
%
Sub-case s
=
s_O:
s=s_O;
r=s+noise; % received signal
for n=1 :K
r_Q(n)=sum(r(1 :n).*s_0(1 :n));
end
% Plotting
the results:
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CHAPTER 5. BASEBAND DIGITAL TRANSMISSION
204
subplot(3,2,3)
plot(l,[O r_O])
set(gca,'XTickLabel',{'0 ',' STb',' lOTb',' lSTb',' 20Tb'})
axis([O 20 0 25])
xlabel('(c) \sigmaA2= 0.1 & S {O} is transmitted ','fontsize',10)
%
Sub-case s
=
s_J:
s=-s_O;
r=s+noise;
%
received signal
for n=1 :K
r_O(n)=sum(r(1 :n).*s_0(1 :n));
end
%
Plotting the results:
subplot(3,2,4)
plot(l,[O r_O])
set(gca,'XTickLabel',{'0 ',' STb',' lOTb',' lSTb',' 20Tb'})
axis([O 20 -25 O])
xlabel(' (d) \sigmaA2= 0.1 & S_{l} is transmitted ','fontsize',10)
%
Case 3: noise-N(O,l)
noise=random('Normal',0,1,1,K);
%
Sub-case s
=
s_O:
s=s_O;
r=s+noise;
%
received signal
for n=1 :K
r_O(n)=sum(r(1 :n).*s_0(1 :n));
end
%
Plotting the results:
subplot(3,2,5)
plot(l,[O r_O])
set(gca,'XTickLabel',{'0 ',' STb',' lOTb',' lSTb',' 20Tb'})
axis([O 20 -5 25])
xlabel(' (e) \sigmaA2= 1 & S {O} is transmitted ','fontsize',10)
%
Sub-case s
=
s_J:
s=-s_O;
r=s+noise;
%
received signal
for n=1 :K
r_O(n)=sum(r(1 :n).*s_0(1 :n));
end
%
Plotting the results:
subplot(3,2,6)
plot(l,[O r_O])
set(gca,'XTickLabel',{'0 ',' STb',' lOTb',' lSTb',' 20Tb'})
axis([O 20 -25 5])
xlabel('(f) \sigmaA2= 1 & S_{l} is transmitted ','fontsize',10)
Illustrative Problem 5.8 [Binary Antipodal Simulation] Use Monte Carlo simula­
tion to estimate and plot the error probability performance of a binary antipodal com­
munication system. The model of the system is illustrated in Figure 5 .16.
As shown, we simulate the generation of the random variable r, which is the input
to the detector. A uniform random number generator is used to generate the binary
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5.2. BINARY SIGNAL TRANSMISSION
Gaussian random
Uniform random
number generator
number generator
Binary
data source
205
±E
Detector
>----- Output
data
Compare
Error counter
Figure 5.16: Model of the binary communication sy stem employing antipodal signals
10-6 �-�-�-�-��-�-�-�-�--��
2
7
10
0
3
8
9
Figure 5.17: Error probability from Monte Carlo simulation compared with theoretical
error probability for antipodal signals
information sequence from the binary data source. The sequence of O's and 1 's is
mapped into a sequence of ±E, where E represents the signal energy. E may be nor­
malized to unity. A Gaussian noise generator is used to generate the sequence of zero­
mean Gaussian random numbers with variance a-2• The detector compares the random
variable r with the threshold of 0. If r > 0, the decision is made that the transmitted
bit is a 0. If r < 0, the decision is made that the transmitted bit is a 1. The output of
the detector is compared with the transmitted sequence of information bits, and the bit
errors are counted. Figure 5.17 illustrates the results of this simulation for the trans­
mission of 10000 bits at several different values of SNR. The theoretical value for Pe
given by (5.2.32) is also plotted in Figure 5.17 for comparison. Note that for 10,000
transmitted bits, the Monte Carlo estimates of Pe below io-3 are not very accurate.
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CHAPTER 5. BASEBAND DIGITAL TRANSMISSION
206
The MATLAB scripts for this problem are given next.
----tl&li" ------% MATLAB script for Illustrative Problem 5.8.
echo on
SNRindB1=0:1 :10;
SNRindB2=0:0.1:1O;
for i=1 :length(SNRindBl),
% simulated error rate
smld_err_prb(i)=smldPe55(SNRindB1(i));
echo off;
end;
echo on;
for i=1 :length(SNRindB2),
SNR=exp(SNRindB2(i)*log(10)/10);
% theoretical error rate
theo_err_prb(i)=Qfunct(sqrt(2*SNR));
echo off;
end;
echo on;
% Plotting commands follow.
sernilogy(SNRindBl,srnld_err_prb,' * ');
hold
sernilogy(SNRindB2,theo_err_prb);
----tl&li" ------function [p]=srnldPe55(snr_in_dB)
% [p]=smldPe55(snr_in_dB)
%
SMWPE55
%
value of snr...in_dB, signal-to-noise ratio in dB.
simulates the probability of error for the particular
E=1;
SNR=exp(snr_in_dB*log(10)/10);
% signal-to-noise ratio
sgma=E/sqrt(2*SNR);
% sigma, standard deviation of noise
N=10000;
% Generation of the binary data source follows.
for i=1 :N,
temp=rand;
% a uniform random variable over
(0,1)
if (temp<0.5),
dsource(i)=O;
% With probability 112, source output is 0.
else
dsource(i)=1;
% With probability
112, source output is 1.
end
end;
% The detection, and probability of error calculation follows.
numoferr=O;
for i=1 :N,
% the matched filter outputs
if (dsource(i)==O),
r=-E+gngauss(sgma);
% if the source output is
"
O
"
else
r=E+gngauss(sgma);
% if the source output is
"1
"
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5.2. BINARY SIGNAL TRANSMISSION
end;
% Detector follows.
if (r<O),
decis=O;
else
decis=1;
end;
if (decis=dsource(i)),
numofe=numoferr+ 1;
end;
end;
p=numoferr/N;
207
% Decision is
"
O
".
% Decision is "1 ".
% If it is an error, increase the error counter.
% probability of error estimate
On-Off Signals for Binary Signal Transmission
A binary information sequence may also be transmitted by use of on-off signals. To
transmit a 0, no signal is transmitted in the time interval of duration Tb. To transmit a
1, a signal waveform s(t) is transmitted. Consequently, the received signal waveform
may be represented as
r(t) =
{n(t),
if a 0 is transmitted
s(t) +n(t),
if a 1 is transmitted
(5.2.33)
where n(t) represents the additive white Gaussian noise.
As in the case of antipodal signals, the optimum receiver consists of a correlator
or a matched filter matched to s(t), whose output is sampled at t = Tb, and followed
by a detector that compares the sampled output to the threshold, denoted as ex.
If
r > ex, a 1 is declared to have been transmitted; otherwise, a 0 is declared to have been
transmitted.
The input to the detector may be expressed as
r-
{n,
E+n,
if a 0 is transmitted
(5.2.34)
if a 1 is transmitted
where n is a zero-mean Gaussian random variable with variance u2 = ENo/2. There­
fore, the conditional probability density functions of the random variable r are
if a 0 is transmitted
if a 1 is transmitted
These probability density functions are illustrated in Figure 5 .18.
When a 0 is transmitted, the probability of error is
Pe o( ex) = P(r > ex) =
1
J2IT (J
f
00
()(
e-r
2 /
2
o-
2
dr
(5.2.35)
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CHAPTER 5. BASEBAND DIGITAL TRANSMISSION
208
Figure 5.18: The probability density functions for the received signal at the output of
the correlator for on-off signals
where ex is the threshold. On the other hand, when a 1 is transmitted, the probability of
error is
Pei (ex)
=
P(r < ex)
=
1
J'Fii a
oe
J
e-(r-E)2 /Za-2
-oo
dr
(5.2.36)
Assuming that the binary information bits are equally probable, we have for the average
probability of error:
(5.2.37)
The value of the threshold ex that minimizes the average probability of error is found
by differentiating Pe(ex) and solving for the optimum threshold. It is easily shown that
the optimum threshold is
exopt
E
=
(5.2.38)
z
Substitution of this optimum value into (5.2.35), (5.2.36), and (5.2.37) yields the prob­
ability of error
Pe(exopt)
=
Q
( f-J;i)
(5.2.39)
We observe that error-rate performance with on-off signals is not as good as with
antipodal signals. It appears to be 6 dB worse than with antipodal signals and 3 dB
worse than with orthogonal signals. However, the average transmitted energy for the
on-off signals is 3 dB less than that for antipodal and orthogonal signals. Consequently,
this difference should be factored in when making comparisons of the performance
with other signal types.
---llMllJ§jl;ful�NiM;l•1:JM§@I
Illustrative Problem 5.9 [Correlation of On-Off Signal Waveforms] The received
signal in on-off signal transmission is given as
r(t)
where
n(t)
=
if a 0 is transmitted
{n(t),
s(t) n(t),
s(t)
t:::;
if a 1 is transmitted
+
represents the additive white Gaussian noise and
=
A
,
0:::;
s(t)
is defined as
Tb
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209
5.2. BINARY SIGNAL TRANSMISSION
and zero otherwise. Suppose the received signal r(t) is sampled at the rate F5
=
30/Yb
and the correlation at the receiver is performed numerically; that is,
k
L r(nT5)s(nTs),
y(kTs)
k
=
1,2, ...,30
n=l
Compute and plot y(kT5) for 1 :-s: k :-s: 30 when the variance of the noise sample is
2
u
=
0, u2
=
0.1, and u2
=
1.
Figure 5.19 illustrates the correlator outputs for the different noise variances.
The
MATLAB script for the computation is given below
10
I
I
I
I
40
I
5
30
0
20
-5
10
-10
0
30Tb
5Tb 10Tb 15Tb 20Tb 25Tb
(a) cl= 0 & 0 is transmitted
10
I
I
I
I
0
30
0
20
-5
10
30Tb
5Tb 10Tb 15Tb 20Tb 25Tb
(c) cl= 0.1 & 0 is transmitted
0
5Tb
10Tb 15Tb 20Tb 25Tb
(b) cl= 0 & 1 is transmitted
30Tb
0
10Tb 15Tb 20Tb 25Tb
5Tb
(d) ef= 0.1 & 1 is transmitted
30Tb
0
5Tb
10Tb 15Tb 20Tb 25Tb
(I) <l= 1 & 1 is transmitted
30Tb
40
I
5
-10
0
0
10
40
5
30
0
20
-5
10
-10
30Tb
5Tb 10Tb 15Tb 20Tb 25Tb
(e) cl= 1 & O is transmitted
0
0
Figure 5.19: Correlator outputs in Illustrative Problem 5.9
----tllli" ---for Illustrative Problem 5 9
% MATLAB script
.
.
% Initialization:
K=30;
% Number
A=1 ;
%
of samples
Signal amplitude
l=O:K;
s=A*ones(1,K);
%
Signal waveform
y=zeros(1,K);
%
Output signal
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210
CHAPTER 5. BASEBAND DIGITAL TRANSMISSION
% Casel: noise-N(O,O)
noise=random( 'Normal',0,0,1,K);
% Sub-case: 0 is transmitted
r=noise;
% received signal
for n=1 :K
y(n)=sum(r(1 :n).*s(1 :n));
end
% Plotting the results:
subplot(3,2,1)
plot(l,[O y])
set(gca, 'XTickLabel',{'0'''5Tb''' lOTb','15Tb'''20Tb','25Tb''' 30Tb'})
axis([O 30 -10 1 OJ)
xlabel('(a) \sigma"2= 0 & 0 is transmitted' ,'fontsize',10)
% Sub-case: 1 is transmitted
r=s+noise; % received signal
for n=1 :K
y(n)=sum(r(1 :n).*s(1 :n));
end
% Plotting the results:
subplot(3,2,2)
plot(l,[O y])
set(gca, 'XTickLabel',{'0, ''5Tb',' lOTb'', 15Tb'', 20Tb'''25Tb'', 30Tb'})
axis([O 30 0 40) )
xlabel(' ( b) \sigma"2= 0 & 1 is transmitted' ,'fontsize',10)
% Case2: noise-N(0,0.1)
noise=random('Normal',0,0.1,1,K);
% Sub-case: 0 is transmitted
r=noise;
% received signal
for n=1 :K
y(n)=sum(r(1 :n).*s(1 :n));
end
% Plotting the results:
subplot(3,2,3)
plot(l,[O y])
set(gca, 'XTickLabel',{'0','5Tb',' lOTb','15Tb','20Tb', '25Tb',' 30Tb'})
axis([O 30 -10 1 OJ)
xlabel('(c) \sigma"2= 0.1 & 0 is transmitted' ,'fontsize',10)
% Sub-case: 1 is transmitted
r=s+noise; % received signal
for n=1 :K
y(n)=sum(r(1 :n).*s(1 :n));
end
% Plotting the results:
subplot(3,2,4)
plot(l,[O y])
set(gca, 'XTickLabel',{'0','5Tb',' lOTb','15Tb','20Tb','25Tb',' 30Tb'})
axis([O 30 0 40])
xlabel(' ( d) \sigma"2= 0.1 & 1 is transmitted' ,'fontsize',10)
% Case3: noise-N(0,1)
noise=random( 'Normal',0,1,1,K);
% Sub-case: 0 is transmitted
r=noise;
% received signal
for n=1 :K
y(n)=sum(r(1 :n).*s(1 :n));
end
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211
5.2. BINARY SIGNAL TRANSMISSION
% Plotting the results:
subplot(3,2,5)
plot(I,[O y])
set(gca, 'XTickLabel',{' 0 '''STb''' lOTb''' 15Tb'''20Tb'''25Tb'''30Tb'})
axis([O 30 -10 1OJ)
xlabel('(e) \sigmaA2= 1 & 0 is transmitted' ,'fontsize',10)
% Sub-case: 1 is transmitted
r=s+noise;
% received signal
for n=1 :K
y(n)=sum(r(1:n).*s(1:n));
end
% Plotting the results:
subplot(3,2,6)
plot(l,[O y])
set(gca, 'XTickLabel',{' 0 '''STb''' lOTb''' 15Tb'''20Tb'''25Tb'' '30Tb'})
axis([O 30 0 40])
xlabel('(f) \sigmaA2= 1 & 1 is transmitted' ,'fontsize',10)
--till!J.il;MlriiiQ;t•]:JM§®·•-----Illustrative Problem 5.10 [On-Off Signaling Simulation] Use Monte Carlo simula­
tion to estimate and plot the performance ofa binary communication system employing
on-off signaling.
The model for the system to be simulated is similar to that shown in Figure 5.16,
except that one ofthe signals is 0. Thus, we generate a sequence ofrandom variables
{rd as given by (5.2.34). The detector compares the random variables {rd to the
optimum threshold E / 2 and makes the appropriate decisions. Figure 5.20 illustrates
the estimated error probability based on 10,000 binary digits. The theoretical error rate
given by (5.2.39) is also illustrated in this figure.
The MATLAB scripts for this problem are given next.
----tllli" -----% MATLAB script for Illustrative Problem 5.10.
echo on
SNRindB1=0:1 :15;
SNRindB2=0:0.1:15;
for i=1:length(SNRindBl),
srnld_err_prb(i)=srnldPe56(SNRindB 1 (i));
% simulated error rate
echo off;
end;
echo on;
for i=1:length(SNRindB2),
SNR=exp(SNRindB2(i)*log(10)/1O);
% signal-to-noise ratio
theo_err_prb(i)=Qfunct(sqrt(SNR/2));
% theoretical error rate
echo off;
end;
echo on;
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212
CHAPTER 5. BASEBAND DIGITAL TRANSMISSION
% Plotting commands follow.
semilogy(SNRindB l ,smld_err_prb, ' * ' );
hold
semilogy(SNRindB2,theo_err_prb);
10� '-------'---'
5
15
10
0
Figure 5 .20: Error probability from Monte Carlo simulation compared with theoretical
error probability for on-off signals
----®'ii" ------function [p]=smldPe56(snr_in_dB)
% [p]=smldPe56(snr_in....dB)
%
SMWPE56 simulates the probability of error for a given
%
snr _in....dB, signal-to-noise ratio in dB.
E=1;
alpha_opt=1/2;
SNR=exp( snr_in_dB*log(10)/10):
sgma=E/sqrt(2*SNR);
% signal-to-noise ratio
% sigma, standard deviation of noise
N=10000;
% Generation of the binary data source follows.
for i=1 :N,
temp=rand;
% a uniform random variable over (0,1)
if (temp<0.5),
dsource(i)=O;
% With probability 112, source output is 0.
else
dsource(i)=1;
% With probability 112, source output is 1.
end
end;
% detection, and probability of error calculation
numoferr=O;
for i=1 :N,
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213
5.2. BINARY SIGNAL TRANSMISSION
% the matched filter outputs
if (dsource(i)==O),
% if the source output is "O"
r=gngauss(sgma);
else
% if the source output is "1 "
r=E+gngauss(sgma);
end;
% Detector follows.
if (r<alpha_opt),
% Decision is "O".
decis=O;
else
% Decision is "1 ".
decis=1;
end;
% If it is an error, increase the error counter.
if (decis-=dsource(i)),
numofe=numoferr+1;
end;
end;
% probability of error estimate
p=numoferr/N;
S.2.3
Signal Constellation Diagrams for Binary Signals
The three types of binary signals-namely, antipodal, on-off, and orthogonal-may
be characterized geometrically as points in "signal space." In the case of antipodal
signals, where the signals are s(t) and -s(t), each having energy E, the two signal
points fall on the real line at ±JE, as shown in Figure 5.21(a). The one-dimensional
geometric representation of antipodal signals follows from the fact that only one signal
waveform or basis function-namely, s(t)-suffices to represent the antipodal signals
in the signal space. On-off signals are also one dimensional. Hence, the two signal
points also fall on the real line at 0 and JE, as shown in Figure 5.21(b).
On the other hand, binary orthogonal signals require a two-dimensional geometric
representation because there are two linearly independent functions, so (t) and s1 (t),
that constitute the two signal waveforms. Consequently, the signal points correspond­
ing to these points are ( JE, 0) and (0, JE), as shown in Figure 5.21(c).
The geometric representations of the binary signals shown in Figure 5.21 are called
signal constellations.
•
-../£
• •
•
•
•
0
0
0
(a)
(b)
(c)
Figure 5.21: Signal constellations for binary signals. (a) Antipodal signals. (b) On-off
signals. (c) Orthogonal signals
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CHAPTER 5. BASEBAND DIGITAL TRANSMISSION
2 14
---li!!�il;tyil�tj#lij;te]:)!#j@I
Illustrative Problem 5.11 [Noise Effect on the Constellation] The effect of noise on
the performance of a binary communication system can be observed from the received
signal plus noise at the input to the detector. For example, let us consider binary orthog­
onal signals, for which the input to the detector consists of the pair of random variables
(ro, r1), where either
or
The noise random variables no and n1 are zero-mean, independent Gaussian random
variables with variance u2• As in Illustrative Problem 5.6, use Monte Carlo simulation
to generate 100 samples of (ro, r1) for each value of u
=
0.1, u
=
0.3, and u
=
0.5,
and plot these 100 samples for each u on different two-dimensional plots. The energy
E of the signal may be normalized to unity.
The results of the Monte Carlo simulation are shown in Figure 5.22. Note that at a
low noise power level (u small) the effect of the noise on performance (error rate)
of the communication system is small. As the noise power level increases, the noise
components increase in size and cause more errors.
The MATLAB script for this problem for u
% MATLAB script for Illustrative Problem 5 11
echo on
n0=.5*randn(100,1);
nl=.5*randn(100,1);
n2=.5*randn(100,1);
n3=.5*randn(100,1);
x1=1.+n0;
yl=nl;
x2=n2;
y2=1.+n3;
plot(xl,y1, 'o' ,x2,y2, ' * ' )
axis(' square')
.
5 .3
=
0.5 is given next.
.
Multiamplitude Signal Transmission
In the preceding section, we treated the transmission of digital information by use of
binary signal waveforms. Thus, each signal waveform conveyed 1 bit of information.
In this section, we use signal waveforms that take on multiple amplitude levels. Thus,
we can transmit multiple bits per signal waveform.
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215
5.3. MULTIAMPLITUDE SIGNAL TRANSMISSION
0
CT= 0.1
..
,.
,.
CT= 0.3
..
..
..
..
,.
,.
0
,.
0
0
0
0
0
0
0
CT= 0.5
Figure 5 .22: Received signal points at input to the detector for orthogonal signals
(Monte Carlo simulation)
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216
CHAPTER 5. BASEBAND DIGITAL TRANSMISSION
s0(t)
Sz(t)
3d
-ff
t------.
d
-ff
T
0
T
0
-d
-ff
-3d
-ff
T
0
0
T
---�
---�
Figure 5.23: Multiamplitude signal waveforms
5.3.1
Signal Waveforms with Four Amplitude Levels
Let us consider a set of signal waveforms of the form
0� t�
Sm(t)=Amg(t),
(5.3.1)
T
where Am is the amplitude of the m th waveform and g(t) is a rectangular pulse de­
fined as
g(t)=
{,Jl/T,
0,
0�t�
T
(5.3.2)
otherwise
where the energy in the pulse g(t) is normalized to unity. In particular, we consider
the case in which the signal amplitude takes on one of four possible equally spaced
values-namely, {Am} = {-3d, -d, d, 3d} or, equivalently,
Am= (2m - 3)d,
m=0,1,2,3
(5.3.3)
where 2d is the Euclidean distance between two adjacent amplitude levels. The four
signal waveforms are illustrated in Figure 5.23. We call this set of waveforms pulse
amplitude modulated (PAM) signals.
The four PAM signal waveforms shown in Figure 5.23 can be used to transmit 2
bits of information per waveform. Thus, we assign the following pairs of information
bits to the four signal waveforms:
00 - so(t)
01
-+
51
(t)
11 - s2(t)
10
-+
53
(t)
Each pair ofinformation bits {00, 01, 10, 11} is called asymbol,and the time duration
is called thesymbol interval. Note that ifthe bit rate is R =1
T T Tb
/Tb
, the symbol interval
. Because all the signal waveforms are scaled versions of the signal basis
function g(t), these signal waveforms may be represented geometrically as points on
is
=2
the real line. Therefore, the geometric representation of the four PAM signals is the
signal constellation diagram shown in Figure 5.24.
As in the case ofbinary signals,we assume that the PAM signal waveforms are trans­
mitted through an AWGN channel. Consequently, the received signal is represented as
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217
5.3. MULTIAMPLITUDE SIGNAL TRANSMISSION
00
01
-3d
-d
11
10
d
3d
0
Figure 5.24: Signal constellation for the four PAM signal waveforms
r(t) = Si(t) +n(t),
i = 0, 1, 2, 3,
Q:::;t:::;T
(5.3.4)
where n(t) is a sample function of a white Gaussian noise process with power spec­
trumNo/ 2 watts/hertz. The task of the receiver is to determine which of the four signal
waveforms was transmitted after observing the received signal r(t) in the interval 0:::;
t:::;T. The optimum receiver is designed to minimize the probability of a symbol error.
5.3.2
Optimum Receiver for the AWGN Channel
The receiver that minimizes the probability of error is implemented by passing the
signal through a signal correlator or matched filter followed by an amplitude detec­
tor. Because the signal correlator and the matched filter yield the same output at the
sampling instant, we consider only the signal correlator in our treatment.
Signal Correlator
The signal correlator cross-correlates the received signal r(t) with the signal pulse
g(t) and its output is sampled at t = T. Thus, the signal correlator output is
r=
=
J:
f:
r(t)g(t)dt
Aig2 (t)dt +
f:
g(t)n(t)dt
=Ai +n
(5.3.5)
where n represents the noise component, defined as
n=
J:
g(t)n(t)dt
(5.3.6)
We note that n is a Gaussian random variable with mean
E(n) =
f:
g(t)E[n(t)] dt = O
(5.3.7)
and variance
uz = E(nz)
=
=
T
g(t)g(T)E [n(t)n(T)] dtdT
J: fo
Tf f
f
T T
N,
No
=
2
No
=2
0
T
0
0
g(t)g(T)8(t - T)dtdT
gZ(t)dt
(5.3.8)
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218
CHAPTER 5. BASEBAND DIGITAL TRANSMISSION
Therefore, the probability density function for the outputr of the signal correlator
is
p(r I Si(t)was transmitted)=
1
J2ri u
e-Cr-Ail2/2u2
(5.3.9)
where Ai is one of the four possible amplitude values.
The Detector
The detector observes the correlator outputr and decides which of the four PAM sig­
nals was transmitted in the signal interval. In the following development of the perfor­
mance of the optimum detector, we assume that the four possible amplitude levels are
equally probable.
Because the received signal amplitude Ai can take the values ± d, ± 3d, as illus­
trated in the signal constellation in Figure 5.24, the optimum amplitude detector com­
pares the correlator outputr with the four possible transmitted amplitude levels and
selects the amplitude level that is closest in Euclidean distance tor. Thus, the opti­
mum amplitude detector computes the distances
i = 0, 1, 2, 3
(5.3.10)
and selects the amplitude corresponding to the smallest distance.
We note that a decision error occurs when the noise variable n exceeds in magni­
tude one-half of the distance between amplitude levels- that is, when In I > d. How­
ever, when the amplitude level + 3d or -3d is transmitted, an error can occur in one
direction only. Because the four amplitude levels are equally probable, the average
probability of a symbol error is
P4 =
3
P(lr - Am i > d)
4
- Joo
1
3
e-xz/2u2 dx
2 d J2ri (}
1
3
- e-xz/2 dx
2 d/u J2ri
=
�
�
oo
J --
� Q (� )
�Q(Na)
(5.3.11)
2 =
We observe that the squared distance between successive amplitude levels is (2d)
c52• Therefore, the average probability of error may be expressed as
P4 =
�Q (\j(82 )
2
4No
(5.3.12)
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5.3. MULTIAMPLITUDE SIGNAL TRANSMISSION
219
10-4
10--0
10-7 �------�--�
0
10
Figure 5.25: Probability of symbol error for four-level PAM
Alternatively, the average probability of error may be expressed in terms of the signal
energy. Because all four amplitude levels are equally probable, the average transmitted
signal energy per symbol is
1
4
Eav = 4 L
k=l
Consequently,
T
f s� (t) dt = 5d2
0
(5.3.13)
d2 = Eav/5 and, hence,
(5.3.14)
Because each transmitted symbol consists of 2 information bits, the transmitted average
energy per bit is
Eav I 2
=
Eavb.
The average probability of error P4 is plotted in Figure 5.25 as a function of the
SNR defined as 10 log1
0
(Eavb I No).
----41114-il;ful�tjii;'•'='•§i•
Illustrative Problem 5.12 [Multiamplitude Signal Simulation] Perform a Monte
Carlo simulation of the four-level (quaternary) PAM communication system that
employs a signal correlator, as described earlier, followed by an amplitude detector.
The model for the system to be simulated is shown in Figure 5.26.
As shown, we simulate the generation of the random variable r, which is the output of
the signal correlator and the input to the detector. We begin by generating a sequence
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CHAPTER 5. BASEBAND DIGITAL TRANSMISSION
220
Error counter
Compare
Am with Am
UniformRNG
r
Mapping to
amplitude level
Detector
Gaussian random
number generator
Figure 5.26: Block diagram of four-level PAM for Monte Carlo simulation
of quaternary symbols that are mapped into corresponding amplitude levels {Am}. To
accomplish this task, we use a random number generator that generates a uniform ran­
dom number in the range (0,1). This range is subdivided into four equal intervals,
(0,0.25), (0.25,0.5), (0.5,0.75), (0.75,1.0), where the subintervals correspond to
the symbols (pairs of information bits) 00, 01, 11, 10, respectively. Thus, the out­
put of the uniform random number generator is mapped into the corresponding signal
amplitude levels -3d, -d, d, 3d, respectively.
The additive noise component having mean 0 and variance cr2 is generated by
means of a Gaussian random number generator. For convenience, we may normalize
the distance parameter d
=
1 and vary u2• The detector observes r
=
Am + n and
computes the distance between r and the four possible transmitted signal amplitudes.
Its output, Am, is the signal amplitude level corresponding to the smallest distance.
Am is compared with the actual transmitted signal amplitude, and an error counter is
used to count the errors made by the detector.
Figure 5 .27 illustrates the results of the simulation for the transmissions of N =
10,000 symbols at different values of the average bit SNR, which is defined as
Eavb
No
=
�
4
( )
d2
u2
(5.3.15)
Note the agreement between the simulation results and the theoretical values of P4
computed from (5.3.14).
The MATLAB scripts for this problem are given next.
----4111&§ii! �----���
% MATLAB script for Illustrated Problem 5 12
.
.
echo on
SNRindB1=0:1 :12;
SNRindB2=0:0.1 :12;
for i=1 :length ( SNRindBl ) ,
%
simulated error rate
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5.3. MULTIAMPLITUDE SIGNAL TRANSMISSION
221
smld_err_prb(i)=smldPe58(SNRindB 1 (i));
echo off;
end;
echo on;
for i=1 :length(SNRindB2),
% signal-to-noise ratio
SNR_per_bit=exp(SNRindB2(i)*log(10)/1O);
% theoretical error rate
theo_err_prb(i)=(3/2)*Qfunct(sqrt((4/5)*SNR_ per_bit));
echo off;
end;
echo on;
% Plotting commands follow.
sernilogy(SNRindBl,smld_ err_prb,' * ');
hold
sernilogy(SNRindB2, theo_ err_prb);
10·4 �---�---�----�---�
2
4
12
0
6
8
10
10 log10 (E!No)
Figure 5.27: Error probability for Monte Carlo simulation compared with theoretical
error probability for M
=
4 PAM
----411Mli" -----function [p]=smldPe58(snr_in_dB)
% [p]=smldPe58(snr_in__dB)
%
SMLDPE58
%
snr ...in--1.lB, signal to noise ratio in dB.
simulates the probability of error for the given
d=1;
SNR=exp(snLin_dB*log( 10)/10);
% signal to noise ratio per bit
sgma=sqrt((5*d"2)/(4*SNR));
% sigma, standard deviation of noise
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222
CHAPTER 5. BASEBAND DIGITAL TRANSMISSION
N=10000;
%
% number of symbols being simulated
Generation of the quaternary data source follows.
for i=1 :N,
temp=rand;
%
a uniform random variable over (0,1)
%
With probability 114, source output is "00."
%
With probability 114, source output is "01."
%
With probability 114, source output is "10."
%
With probability 114, source output is "11."
if (temp<0.25),
dsource(i)=O;
elseif (temp<0.5),
dsource(i)=1;
elseif (temp<0.75),
dsource(i)=2;
else
dsource(i)=3;
end
end;
%
detection, and probability of error calculation
numoferr=O;
for i=1 :N,
%
the matched filter outputs
if (dsource(i)==O),
r=-3*d+gngauss(sgma);
%
if the source output is "00"
%
if the source output is "OJ"
%
if the source output is "10"
%
if the source output is "11"
elseif (dsource(i)==1),
r=-d+gngauss(sgma);
elseif (dsource(i)==2)
r=d+gngauss(sgma);
else
r=3*d+gngauss(sgma);
end;
% Detector
follows.
if (r<-2*d),
decis=O;
% Decision
is "00."
% Decision
is "OJ."
% Decision
is "10."
% Decision
is "11."
elseif (r<O),
decis=1;
elseif (r<2*d),
decis=2;
else
decis=3;
end;
if (decis-=dsource(i)),
% If
it is an error, increase the error counter.
numoferr=numoferr+1;
end;
end;
p=numoferr/N;
5.3.3
%
probability of error estimate
Signal Waveforms with Multiple Amplitude Levels
It is relatively straightforward to construct multiamplitude signals with more than four
levels. In general, a set of M
Sm(t)
=
=
zk multiamplitude signal waveforms is represented as
m
0:::; t:::; T,
Amg(t),
=
0, 1, 2, ... ,M
-
1
where the M amplitude values are equally spaced and given as
Am
=
(2m
-
M + l)d,
m
=
0, 1, ... ,M
-
1
(5.3.16)
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5.3. MULTIAMPLITUDE SIGNAL TRANSMISSION
223
and g(t) is a rectangular pulse, which has been defined in (5.3.2). Each signal wave­
form conveys k = log2M bits of information. When the bit rate is R = 1/Tb , the corre­
sponding symbol rate is l/T = l/kTb. As in the case of four-level PAM, the optimum
receiver consists of a signal correlator (or matched filter) followed by an amplitude de­
tector that computes the Euclidean distances given by (5.3.10) form = 0, 1, ... , M -1.
For equally probable amplitude levels, the decision is made in favor of the amplitude
level that corresponds to the smallest distance.
The probability of error for the optimum detector in an M-level PAM system can
be shown to be
PM=
_.; 1)
2(M
Q
(
6(1ogzM)Eavb
(M2 - l)No
)
(5.3.17)
where Eavb is the average energy for an information bit. Figure 5.28 illustrates the
probability of a symbol error for M = 2, 4, 8, 16.
10·8 �---�--+---�
5
15
25
10
20
0
Figure 5.28: Symbol error probability for M-level PAM for M = 2, 4, 8, 16
---tll!tJaii;MifUli;t•'="3®·•--Illustrative Problem 5.13 [PAM Simulation] Perform a Monte Carlo simulation of a
16-level PAM digital communication system and measure its error-rate performance.
The basic block diagram shown in Figure 5.26 applies in general. A uniform random
number generator is used to generate the sequence of information symbols, which are
viewed as blocks of 4 information bits. The 16-ary symbols may be generated directly
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224
CHAPTER 5. BASEBAND DIGITAL TRANSMISSION
10"8 '--���
5
15
10
25
20
10 log10 (Eavb/No)
Figure 5 .29: Error rate from Monte Carlo simulation compared with the theoretical
error probability for M
=
16 PAM
by subdividing the interval (0, 1) into 16 equal-width subintervals and mapping the
16-ary symbols into the 16-ary signal amplitudes. A white Gaussian noise sequence is
added to the 16-ary information symbol sequence, and the resulting signal plus noise
is fed to the detector. The detector computes the distance metrics given by (5.3.10) and
selects the amplitude corresponding to the smallest metric. The output of the detector
is compared with the transmitted information symbol sequence and errors are counted.
Figure 5 .29 illustrates the measured symbol error rate for 10000 transmitted symbols
and the theoretical symbol error rate given by (5.3.17) with M
=
16.
The MATLAB scripts for this problem are given next.
----tl®li" -----% MATLAB script for Illustrative Problem 5 13
.
.
echo on
SNRindB1=5:1:25;
SNRindB2=5:0.1:25;
M=16;
for i=1:length(SNRindBl),
% simulated error rate
smld_err_prb(i)=smldPe59(SNRindB1(i));
echo off;
end;
echo on ;
for i=1:length(SNRindB2),
SNR_per_bit=exp(SNRindB2(i)*log(10)/1O);
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5.3. MULTIAMPLITUDE SIGNAL TRANSMISSION
225
% theoretical error rate
theo_err_prb(i)=(2*(M-1 )/M)*Qfunct(sqrt((6*log2(M)/(M" 2-1 ) )*SNR_ per_bit));
echo off;
end;
echo on;
% Plotting commands follow.
semilogy(SNRindBl,srnld_err_prb,' * ');
hold
semilogy(SNRindB2,theo_err_prb);
----411Mli" ------function [p]=smldPe59(snr_in_dB)
% [p]=smldPe59(snr_in....dB )
%
SMWPE59
%
snr -1n....dB, signal-to-noise ratio in dB.
simulates the error probability for the given
M=16;
% 16-ary PAM
d=1;
SNR=exp(snr_in_dB*log(10)/1 O);
% signal-to-noise ratio per bit
sgma=sqrt((85*d" 2)/(8*SNR));
% sigma, standard deviation of noise
N=10000;
% number of symbols being simulated
% generation of the data source
for i=1 :N,
temp=rand;
% a uniform random variable over (0,1)
index=floor(M*temp);
% The index is an integer from 0 to M-1, where
% all the possible values are equally likely.
dsource(i)=index;
end;
% detection, and probability of error calculation
numoferr=O;
for i=1 :N,
% matched filter outputs
% (2*dsource(i)-M+l)*d is the mapping to the 16-ary constellation.
r=(2*dsource(i)-M+ 1 )*d+gngauss(sgma);
% the detector
if (r>(M-2)*d),
decis=15;
elseif (r>(M-4)*d),
decis=14;
elseif (r>(M-6)*d),
decis=13;
elseif (r>(M-8)*d),
decis=12;
elseif (r>(M-10}*d),
decis=11;
elseif (r>(M-12)*d),
decis=10;
elseif (r>(M-14)*d),
decis=9;
elseif (r>(M-16)*d),
decis=B;
elseif (r>(M-18}*d),
decis=7;
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226
CHAPTER 5. BASEBAND DIGITAL TRANSMISSION
elseif (r>(M-20)*d),
decis=6;
elseif (r>(M-22)*d),
decis=5;
elseif (r>(M-24)*d),
decis=4;
elseif (r>(M-26)*d),
decis=3;
elseif (r>(M-28)*d),
decis=2;
elseif (r>(M-30)*d),
decis=1;
else
decis=O;
end;
% If it
if (decis-=dsource(i)),
is an error, increase the error counter.
nurnoferr=nurnoferr+1;
end;
end;
% probability
p=nurnoferr/N;
5.4
of error estimate
Multidimensional Signals
In the preceding section we constructed multiamplitude signal waveforms, which al­
lowed us to transmit multiple bits per signal waveform. Thus, with signal waveforms
having M
=
zk amplitude levels, we are able to transmit k
=
log2 M bits of infor­
mation per signal waveform. We also observed that the multiamplitude signals can
be represented geometrically as signal points on the real line (see Figure 5.24). Such
signal waveforms are called one-dimensional signals.
In this section we consider the construction of a class of M
that have a multidimensional representation.
=
zk signal waveforms
That is, the set of signal waveforms
can be represented geometrically as points in N-dimensional space. We have already
observed that binary orthogonal signals are represented geometrically as points in two­
dimensional space.
5.4.1
Multidimensional Orthogonal Signals
There are many ways to construct multidimensional signal waveforms with various
properties. In this section, we consider the construction of a set of M
Si(t), for i
=
=
zk waveforms
0, 1, ... ,M - 1, which have the properties of (a) mutual orthogonality
and (b) equal energy. These two properties may be succinctly expressed as
J:
Si(t)sk(t) dt
=
EDik.
i, k
=
0, 1, ... ,M - 1
(5.4.1)
where Eis the energy of each signal waveform and Dik is called the Kronecker delta,
which is defined as
Dik
=
{l,
0,
(5.4.2)
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227
5.4. MULTIDIMENSIONAL SIGNALS
T
D
T
3T
2
4
, t
)>
T
T
T
4
2
t
T
3T
4
, t
T
Figure 5 .30: An example of four orthogonal, equal-energy signal waveforms
As in our previous discussion, we assume that an information source is providing
a sequence of information bits, which are to be transmitted through a communication
channel. The information bits occur at a uniform rate of R bits per second. The recip­
rocal of R is the bit interval, Tb. The modulator takes k bits at a time and maps them
into one of M = 2k signal waveforms. Each block of k bits is called a symbol. The
time interval available to transmit each symbol is T= kTb. Hence, Tis the symbol
interval.
The simplest way to construct a set of M =2k equal-energy orthogonal waveforms
in the interval (0,T) is to subdivide the interval into M equal subintervals of duration
T/M and to assign a signal waveform for each subinterval. Figure 5 .30 illustrates such
a construction for M = 4 signals. All signal waveforms constructed in this manner
have identical energy, given as
E=
f:
s{(t) dt,
i=0,1,2,... ,M-l
A2T
(5. 4.3)
M
Such a set of orthogonal waveforms can be represented as a set of M-dimensional
orthogonal vectors- that is,
So= (.JE°,0,0,... ,0)
S1 = (O,)E,0,... ,0)
SM = (0,0,... ,0, .JE°)
(5.4.4)
Figure 5 .31 illustrates the signal points (signal constellations) corresponding to M =2
and M = 3 orthogonal signals.
Let us assume that these orthogonal signal waveforms are used to transmit informa­
tion through an AWGN channel. Consequently, if the transmitted waveform is Si(t),
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CHAPTER 5. BASEBAND DIGITAL TRANSMISSION
228
M=2
Figure 5 .31: Signal constellations for M = 2 and M = 3 orthogonal signals
So(t)
f�()dt
f�()dt
Received
Detector
signal r(t)
Output
decision
f�()dt
Figure 5.32: Optimum receiver for multidimensional orthogonal signals
the received waveform is
r(t) = Si(t) + n(t),
0 � t � T,
i = 0, 1, ... ,M - 1
(5.4.5)
where n(t) is a sample function of a white Gaussian noise process with power spec­
trum No/2 watts/hertz. The receiver observes the signal r(t) and decides which of the
M signal waveforms was transmitted.
Optimum Receiver for the AWGN Channel
The receiver that minimizes the probability of error first passes the signal r(t) through
a parallel bank of M matched filters or M correlators. Because the signal correlators
and matched filters yield the same output at the sampling instant, let us consider the
case in which signal correlators are used, as shown in Figure 5.32.
Signal Correlators
The received signal r(t) is cross-correlated with each of the M signal waveforms and
the correlator outputs are sampled at t = T. Thus, the M correlator outputs are
Yi=
f:
r(t)Si(t) dt,
i = 0, 1, ... ,M - 1
(5.4.6)
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5.4. MULTIDIMENSIONAL SIGNALS
229
which may be represented in vector form as r = [Yo, Y1, ... , YM-iJ t. Suppose that
signal waveform so(t) is transmitted. Then
f:
Yo=
f:
s6(t)dt+
n(t)so(t)dt
= E+no
(5.4.7)
and
f:
J:
Yi=
=
f:
so(t)si(t)dt+
n(t)si(t)dt
n(t)si(t)dt = ni,
i= 1,2,3,... ,M-1
(5.4.8)
where
(5.4.9)
Therefore, the output Yo consists of a signal component E and a noise component no.
The other M-1 outputs consist of noise only. Each of the noise components is Gaus­
sian, with mean zero and variance
a2 = E(n?)
t
=
=
f:f:
�Of:f:
Si(t)Si(T)E [n(t)n(T)] dtdT
Si(t)Si(T)8(t - T)dtdT
No rT s?(t)dt
2 Jo 1
NoE
=
(5.4.10)
2
The reader is encouraged to show that E(nmj) = 0, where if= j. Consequently,
the probability density functions for the correlator outputs are
1
p ( Yo I s0 (t) was transm1tte
.
d)= J2IT
p(Yi I so(t) was transmitted)=
a
1
J2Ti a
-(ro-E)2 /Zu2
e
-ri/Zu2
e
'
i= 1,2,... ,M-1
The Detector
The optimum detector observes the M correlator outputs Yi, where i= 0, 1,... ,M-1,
and selects the signal corresponding to the largest correlator output. In the case where
so(t) was transmitted, the probability of a correct decision is simply the probability
that Yo >Yi for i= 1,2,... ,M-1, or
Pc = P (Yo >Y1, Yo >Yz, ... , Yo >YM-1)
(5.4.11)
and the probability of a symbol error is simply
PM=l -Pc
= 1 -P (Yo >Y1, Yo >Yz, ... , Yo >YM-1)
(5.4.12)
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CHAPTER 5. BASEBAND DIGITAL TRANSMISSION
230
It can be shown that PM can be expressed in integral form as
(5.4.13)
For the special case M = 2, the expression in (5.4.13) reduces to
which is the result we obtained in Section 5.2 for binary orthogonal signals.
The same expression for the probability of error is obtained when any one of the
other M
-
1 signals is transmitted. Because all the M signals are equally likely, the
expression for PM given by (5.4.13) is the average probability of a symbol error. This
integral can be evaluated numerically.
Sometimes, it is desirable to convert the probability of a symbol error into an equiv­
alent probability of a binary digit error. For equiprobable orthogonal signals, all symbol
errors are equiprobable and occur with probability
(5.4.14)
Furthermore, there are
(�) ways in which
n bits out ofk may be in error. Hence, the
average number of bit errors perk-bit symbol is
(5.4.15)
and the average bit-error probability is just the result in (5.4.15) divided by k, the
number of bits per symbol. Thus,
Pb=
zk-1
k
2
_
l PM
(5.4.16)
The graphs of the probability of a binary digit error as a function of the SNR per bit,
Eb/No, are shown in Figure 5.33 for M = 2,4, 8, 16, 32,64, where Eb = E/k is the
energy per bit. This figure illustrates that by increasing the number M of waveforms,
one can reduce the SNR per bit required to achieve a given probability of a bit error.
The MATLAB script for the computation of the error probability in (5.4.13) is given
next.
-------4111®1i'' 111---����
% MATLAB script
initial_ snr=O;
that generates the probability of error versus the signal-to-noise ratio
final_snr=15;
snr_step=1;
tolerance=1e-7;
%
minus_inf=-20;
%
tolerance used for the integration
This is practically negative infinity.
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231
5.4. MULTIDIMENSIONAL SIGNALS
%
plus_inf=20;
This is practically infinity.
snr_in_dB=initiaLsnr:snr_step:final_snr;
for i=1 :length(snr_in_dB),
snr=10 (snr_in_dB(i)/1O);
�
Pe_2(i)=Qfunct(sqrt(snr));
Pe_4(i)=(2/3)*quad8(' bdt_int' ,minus_inf,plus_inf,tolerance,[ ],snr,4);
Pe_8(i)=(4/7)*quad8(' bdt_int' ,minus_inf,plus_inf,tolerance,[ ],snr,8);
Pe_l6(i)=(8/15)*quad8(' bdt_int' ,minus_inf,plus_inf,tolerance,[ ],snr,16);
Pe_32(i)=(16/31)*quad8(' bdt_int' ,minus_inf,plus_inf,tolerance,[ ],snr,32);
Pe_64(i)=(32/63)*quad8(' bdt_int' ,minus_inf,plus_inf,tolerance,[ ],snr,64);
end;
%
Plotting commands follow.
10° �----�---�--�
Pb
10-1 �=---
10·6
10·10 .._
0
______
__,________..._...
.
.
_._
....__...__
.
_.____,
5
10
15
Figure 5.33: Bit-error probability for orthogonal signals
Illustrative Problem 5.14 [Orthogonal Signaling Simulation] Perform a Monte Carlo
simulation of a digital communication system that employs M
=
4 orthogonal signals.
The model of the system to be simulated is illustrated in Figure 5 .34.
As shown, we simulate the generation of the random variables ro, r1, rz, r3, which
constitute the input to the detector. We may first generate a binary sequence of O's and
l's that occur with equal probability and are mutually statistically independent, as in
Illustrative Problem 5.6. The binary sequence is grouped into pairs of bits, which are
mapped into the corresponding signal components . An alternative to generating the
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232
CHAPTER 5. BASEBAND DIGITAL TRANSMISSION
Gaussian RNG
0
r1
Output
Detector
Uniform
Mapping to
RNG
signal points
0
rz
0
r3
decision
S;
Error counter
Figure 5.34: Block diagram of system with M
=
4 orthogonal signals for Monte Carlo
simulation
individual bits is to generate the pairs of bits, as in Illustrative Problem 5.12. In any
case, we have the mapping of the four symbols into the signal points:
00
--+
So
=
( JE, 0, 0, 0)
01
--+
S1
=
(0, JE, 0, 0)
10
--+
=
(0, 0, JE, 0)
11
--+
=
(0, 0, 0, vfE)
S2
S3
(5.4.17)
The additive noise components no, n1, nz, n3 are generated by means of four Gaussian
we may normalize the symbol energy to E
follows that Eb
=
=
2
NoE / 2. For convenience,
1 and vary a2. Because E
2Eb, it
noise generators, each having a mean zero and variance a
=
=
�. The detector output is compared with the transmitted sequence of
bits, and an error counter is used to count the number of bit errors.
Figure 5.35 illustrates the results of this simulation for the transmission of 20 ,000
bits at several different values of the SNR Eb/ No. Note the agreement between the
simulation results and the theoretical value of Pb given by (5.4.16).
The MATLAB scripts for this problem are given next.
----Iii" -----% MATLAB script for Illustrative
Problem 5 14
.
.
echo on
SNRindB=0:2:1 O;
for i=1 :length(SNRindB),
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5.4. MULTIDIMENSIONAL SIGNALS
233
% simulated error rate
smld_err_prb(i)=smldP510(SNRindB(i));
echo off;
end;
echo on;
% Plotting commands follow
semilogy(SNRindB,smld_err_prb,
•
* • );
llE
10- s .____.____.____.____,_____._____.____,�__.___.___�
10
0
2
4
7
3
5
6
8
9
Figure 5.35: Bit-error probability for M
=
4 orthogonal signals from a Monte Carlo
simulation compared with theoretical error probability
----tlili" ------function [p]=smldPS 10(snr_in_dB)
% [p]=smldP510(snr_in....dB)
%
SMLDP510
%
snr _in....dB, signal-to-noise ratio in dB.
simulates the probability of error for the given
M=4;
% quaternary orthogonal signaling
E=1;
SNR=exp(snr_in_dB*log(10)/10);
*
sgma=sqrt(E 2/(4 SNR));
% signal-to-noise ratio per bit
N=10000;
% number of symbols being simulated
�
% sigma, standard deviation of noise
% generation of the quaternary data source
for i=1 :N,
temp=rand;
% a uniform random variable over (0,1)
if (temp<0.25),
dsourcel(i)=O;
dsource2(i)=0;
elseif (temp<0.5),
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234
CHAPTER 5. BASEBAND DIGITAL TRANSMISSION
dsourcel(i)=O;
dsource2(i)=1;
elseif (temp<0.75),
dsourcel(i)=1;
dsource2(i)=O;
else
dsourcel(i)=1;
dsource2(i)=1;
end
end;
% detection , and probability of error calculation
numoferr=O;
for i=1 :N,
% matched filter outputs
if ((dsourcel(i)==O) & (dsource2(i)==0)),
rO=sqrt(E)+gngauss(sgma);
rl=gngauss(sgma);
r2=gngauss(sgma);
r3=gngauss(sgma);
elseif ((dsourcel(i)==O) & (dsource2(i)==1)),
rO=gngauss(sgma);
rl=sqrt(E)+gngauss(sgma);
r2=gngauss(sgma);
r3=gngauss(sgma);
elseif ((dsourcel(i)==1) & (dsource2(i)==0)),
rO=gngauss(sgma);
rl=gngauss(sgma);
r2=sqrt(E)+gngauss(sgma);
r3=gngauss(sgma);
else
rO=gngauss(sgma);
rl=gngauss(sgma);
r2=gngauss(sgma);
r3=sqrt(E)+gngauss(sgma);
end;
% the detector
max_r=max([rO rl r2 r3]);
if (rO==max_r),
decisl=O;
decis2=0;
elseif (rl==max_r),
decisl=O;
decis2=1;
elseif (r2==max_r),
decis1=1;
decis2=0;
else
decis1=1;
decis2=1;
end;
% Count the number of bit errors made in this decision.
% If it is an error, increase the error counter.
if (decisr=dsourcel(i)),
numoferr=numoferr+ 1;
end;
if (decisZ-=dsource2(i)),
% If it is an error, increase the error counter.
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2 35
5.4. MULTIDIMENSIONAL SIGNALS
numoferr=numoferr+ 1 ;
end;
end;
p=numoferr/(2*N);
S.4.2
%
bit error probability estimate
Biorthogonal Signals
As we observed in the preceding section, a set of M = 2k equal-energy orthogonal
waveforms can be constructed by subdividing the symbol interval T into M equal
subintervals of duration T /M and assigning a rectangular signal pulse to each subinter­
val. A similar method can be used to construct another set of M= 2k multidimensional
signals that have the property of being biorthogonal. In such a signal set, one-half the
waveforms are orthogonal and the other half are the negative of these orthogonal wave­
forms; that is, so(t), s1(t),... ,5M/2-l (t) are orthogonal signal waveforms. The other
M/2 waveforms are simply Si+M12(t) = -Si(t), for i= 0,1, ... , (M/2) -1. Thus we
obtain M signals, each having M/2 dimensions.
The MI 2 orthogonal waveforms can be easily constructed by subdividing the sym­
bol interval T = k Tb into M/2 nonoverlapping subintervals, each of duration 2 T /M,
and assigning a rectangular pulse to each subinterval. Figure 5.36 illustrates a set of
M = 4 biorthogonal waveforms constructed in this manner.
The geometric repre­
sentation of a set of M signals constructed in this manner is given by the following
(M/2)-dimensional signal points:
so= (.JE,o,o,... ,o)
s1= (O,.JE,0,... ,0)
SM/2-1= (0,0,0,... ,.JE)
sM12= <-.JE,o,o,... ,o)
(5.4.18)
SM-1= (0,0,... , -.JE)
As in the case of orthogonal signals, let us assume that the biorthogonal signal
waveforms are used to transmit information through an AWGN channel.
s0(t)
s1(t)
A
S3(t)
Sz(t)
A
T
T
2
0
Then the
T
2
0
T
2
T
2
0
0
-A
-A
T
Figure 5.36: A set of M= 4 biorthogonal signal waveforms
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236
CHAPTER 5. BASEBAND DIGITAL TRANSMISSION
received signal waveform may be expressed as
r(t) = Si(t) +n(t),
0 �t � T
(5.4.19)
where Si(t) is the transmitted waveform and n(t) is a sample function of a white
Gaussian noise process with power spectrum No/2 watts/hertz.
Optimum Receiver
The optimum receiver may be implemented by cross-correlating the received signal
r(t) with each of the M/2 orthogonal signal waveforms, sampling the correlator out­
puts at t = T, and passing the MI 2 correlator outputs to the detector. Thus, we have
i = 0, 1,...
' � -1
(5.4.20)
Suppose that the transmitted signal waveform is so(t). Then
ri =
=
f: r(t)so(t) dt, = 0, 1,... , � -1
i
E +no,
i= 0
ni,
i * 0
{
(5.4.21)
where
i = 0, 1,...
' � -1
(5.4.22)
and E is the symbol energy for each signal waveform. The noise components are zero­
mean Gaussian and have a variance of cr
2 = ENo/2.
The Detector
The detector observes the M/2 correlator outputs {ri, 0 � i � (M/2) -1} and se­
lects the correlator output whose magnitude lril is largest. Suppose
(5.4.23)
lrjl = max{lril}
t
Then the detector selects the signal s j ( t) if rj > 0 and -s j ( t) if rj < 0.
To determine the probability of error, suppose that so(t) was transmitted. Then the
probability of a correct decision is equal to the probability that ro = E +no > 0 and
lrol > lril fori = 1,2,..., (M/2) -1. Thus,
Pc= r
oo
Jo
1
[ Ff
v LTT
ro/ JENo/2
-r0/)EN0/2
2
12 dx
e-x
]M-1
p(ro) dro
(5.4.24)
where
(5.4.25)
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237
5.4. MULTIDIMENSIONAL SIGNALS
10-4
6
10-
8
10-
10 -
10
10 -
12
1
10 - 4
10 -
16 .__
___
0
_.______.._____._______._____..______,
2
4
8
10
12
Figure 5.37: Symbol-error probability for biorthogonal signals
Finally, the probability of a symbol error is simply
(5.4.26)
Pc and PM may be evaluated numerically for different values of M from (5.4.24)
and (5.4.25). The graph shown in Figure 5.37 illustrates PM as a function of the signal­
to-noise ratio Eb/No, where E
=
kEb for M
=
2, 4, 8, 16, and 32. We observe that
this graph is similar to that for orthogonal signals. However, for biorthogonal signals
we note that P4 > P2. This is due to the fact that we have plotted the symbol-error
probability PM in Figure 5.37. If we plot the equivalent bit-error probability, we would
find that the graphs for M
=
2 and M
=
4 coincide.
The MATLAB script for the computation of the error probability in (5.4.24) and
(5.4.26) is given next.
----tllli" -----% MATLAB script
that generates the probability of error versus the signal-to-noise ratio.
initiaL snr=O;
finaLs=12;
snr_step=0.75;
tolerance=eps;
%
plus_inf=20;
%
tolerance used for the integration
This is practically infinity.
snr_in_dB=initial_snr:snr_step:final_snr;
for i=1:length (snr_in_dB ) ,
snr=1OA (snr_in_dB ( i) /1O ) ;
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CHAPTER 5. BASEBAND DIGITAL TRANSMISSION
238
Pe_2(i)=1-quad8('bdt_int2',O,plus_inf,tolerance,( ],snr,2);
Pe-4(i)=1-quad8('bdt_int2' ,O,plus_inf,tolerance,[ ],snr,4);
Pe_8(i)=1-quad8('bdt_int2',O,plus_inf,tolerance,[ ],snr,8);
Pe_l6(i)=1-quad8('bdt_int2' ,O,plus_inf,tolerance,[ ],snr,16);
Pe_32(i)=1-quad8('bdt_int2' ,O,plus_inf,tolerance,[ ],snr,32);
end;
%
Plotting commands follow.
Illustrative Problem 5.15 [Correlation of Biorthogonal Signal Waveforms] A set
of M
=
4 biorthogonal signal waveforms is shown in Figure 5.36. Note that s2 (t)
-so(t) and s3(t)
=
-s1(t).
=
Therefore, only two correlators are needed at the re­
ceiver to process the received signal, one for correlating r(t) with s1(t) and one for
correlating r(t) with s2 (t).
Suppose the received signal r(t) is sampled at a rate of Fs
=
40/T and the corre­
lations at the receiver are performed numerically, that is,
k
Yo(kTs)
=
L r(nTs)so(nTs).
k
=
1,2,... ,20
n=l
k
Y1(kTs)
=
L r(nTs)si(nTs),
n=21
k
=
21,22, ... , 40
Compute and plot Yo(kTs) and Y1(kTs) when (a) so(t) is transmitted, (b) si(t) is
transmitted, (c) -so(t)
=
s2 (t) is transmitted, and (d) -si(t)
=
s3(t) is transmitted
and the additive noise is zero-mean, white, and Gaussian and the variance of the noise
samples is cr2
=
0, cr2
=
0.1, and cr2
=
1.
Figure 5.38 illustrates the outputs of the two correlators for different noise variances
and transmitted signals. The MATLAB script for the computation is given next.
% MATLAB script for Illustrative
Problem 5.15.
% Initialization:
K=40;
% Number
A=1;
%
of samples
Signal amplitude
m=O:K/2;
n=Kl2:K;
% Defining
signal waveforms:
s_O=[A *ones(1,K/2) zeros(1,K/2)];
s_l=[zeros(1,K/2) A *ones(1,K/2)];
s-2=(-A *ones(1,K/2) zeros(1,K/2)];
s_3=[zeros(1,K/2) -A *ones(1,K/2)];
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239
5.4. MULTIDIMENSIONAL SIGNALS
20
20
10
10
0
,._..__,._-<I>
...
......_.
..
... __,._
....
.........
..
.. ....
....
........
...
0 111>-!1�........
-10
-10
....
- .
-20
-20
0
5Th
lOTh
20Th
15Th
20Tb
25Tb
35Tb
30Th
20
20
� -""
10
10
.... •--*
.......
40Tb
�
r*
...... .... ....--..........
..
o ...��-e--e-<:>-@>-el--&--&-'-<�H'>....'<'J.-��-'f'>'.-4>
0
..
-10
.
·•
. .. .
-10
... .
.....
-20
0
5Th
lOTh
20Th
15Th
-20
20Tb
25Tb
35Tb
30Th
40Tb
20
0
... . ........
-
·
--
-20
-20
0
5Th
lOTh
20Th
15Th
20Tb
25Tb
30Th
35Tb
40Tb
Figure 5.38: Correlator outputs in Illustrative Problem 5.15. Solid, dashed, dotted,
and dash-dotted plots correspond to transmission of so ( t) , s1 ( t) , s2 ( t) , and 53 ( t) ,
respectively.
% Initializing
Outputs:
y_O_O=zeros(1,K);
y_O_l=zeros(1,K);
y_0_2=zeros(1,K);
y_0_3=zeros(1,K);
y_LO=zeros(1,K);
y_Ll=zeros(1,K);
y_L2=zeros(1,K);
y_L3=zeros(1,K);
%
Case I: noise-N(O,O)
noise=random( 'Normal' ,0,0, 1,K);
r_O=s_O+noise; r_l=s_l+noise; %
r_2=s_2+noise; r_3=s_3+noise; %
received signals
received signals
for k=1 :K/2
y_O_O(k)=sum(r-0(1 :k).*s-0(1 :k));
y_O_l(k)=sum(r-1(1 :k).*s_0(1 :k));
y_0_2(k)=sum(r_2(1 :k).*s_0(1 :k));
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CHAPTER 5. BASEBAND DIGITAL TRANSMISSION
240
y_Q_3(k)=sum(r_3(1:k).*s_0(1:k));
l=Kl2+k;
y_LO(l)=sum(r_0(21:l).*s_l(21 :1));
y_Ll(l)=sum(r_I(21:l).*s_l(21 :1));
y_L2(l)=sum(r_2(21:l).*s_l(21:l));
y_L3(l)=sum(r_3(21:l).*s-1(21:l));
end
%
Plotting the results:
subplot(3,2,1)
plot(m,[O y_Q_Q(1:K/2)],' -bo',m,[O y_Q_l(1:K/2)], '--b* ',...
m,[O y_Q_2(1:K/2)],' : b.',m,[O y_Q_3(1:K/2)],' - ')
•
set(gca, 'XTickLabel',{'0','5Tb','lOTb','15Tb','20Tb'})
axis([O 20 -25 25])
xlabel('(a)
\sigmaA2= O & y_{O}(kT_{b})' ,'fontsize',10)
subplot(3,2,2)
plot(n,[O y_LO(K/2+1:K)],' -bo',n,[O y_Ll(K/2+1:K)], '--b* ',...
n,[O y_L2(K/2+1:K)],' : b.',n,[O y_L3(K/2+1:K)],' - ')
•
set(gca, 'XTickLabel',{'20Tb','25Tb',' 30Tb',' 35Tb',' 40Tb'})
axis([20 40 -25 25])
xlabel(' (b) \sigmaA2= O & y_{l}(kT_{b})' , 'fontsize',10)
%
Case 2: noise-N(0,0.1)
noise=random( 'Normal',0,0.1,4,K);
r_O=s_O+noise(1,:); r_}:s_l+noise(2,:); %
r_2=s_2+noise(3,:); r_3=s_3+noise(4,:); %
received signals
received signals
for k=1:K/2
y_Q_Q(k)=sum(r_0(1:k).*s_0(1:k));
y_Q_l(k)=sum(r_l(1:k).*s-0(1:k));
y_Q_2(k)=sum(r_2(1:k).*s_0(1:k));
y_Q_3(k)=sum(r_3(1:k).*s_0(1:k));
l=Kl2+k;
y_LO(l)=sum(r_0(21:l).*s_l(21:l));
y_Ll(l)=sum(r_l(21:l).*s_l(21:l));
y_L2(l)=sum(r_2(21:l).*s-1(21:l));
y_L3(l)=sum(r_3(21:l).*s_l(21 :1));
end
%
Plotting the results:
subplot(3,2,3)
plot(m,[O y_Q_0(1:K/2)],' -bo',m,[O y_Q_l(1:K/2)],' --b* '...
,m,[O y_0_2(1:K/2)],' : b.',m,[O y_0_3(1:K/2)],' - ')
•
set(gca, 'XTickLabel,,{'0,',5Tb'',lOTb'',15Tb'',20Tb'})
axis([O 20 -25 25])
xlabel('(c)
\sigmaA2= 0.1 & y_{O}(kT_{b})' ,'fontsize',10)
subplot(3,2,4)
plot(n,[O y_LQ(K/2+1:K)],' -bo',n,[O y_Ll(K/2+1:K)], '--b* ',...
n,[O y_L2(K/2+1:K)],' : b.',n,[O y_L3(K/2+1:K)],' - ')
•
set(gca, 'XTickLabel',{'2 OTb','25Tb',' 30Tb',' 35Tb',' 40Tb'})
axis([20 40 -25 25])
xlabel('(d)
%
\sigmaA2= 0.1 & y_{l}(kT_{b})' ,'fontsize',10)
Case 3: noise-N(O,l)
noise=random( 'Normal',0,1,4,K);
r_O=s_O+noise(1,:); r_l=s-l+noise(2,:); %
r_2=s_2+noise(3,:); r_3=s_3+noise(4,:); %
received signals
received signals
for k=1:K/2
y_Q_O(k)=sum(r_0(1:k).*s_0(1:k));
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241
5.4. MULTIDIMENSIONAL SIGNALS
y_O_l(k)=sum(r_l(1 :k).*s_0(1 :k));
y_0_2(k)=sum(r_2(1 :k).*s_0(1 :k));
y_0_3(k)=sum(r_3(1 :k).*s_0(1 :k));
l=Kl2+k;
y_LO(l)=sum(r_0(21 :l).*s-1(21 :l));
y_Ll(l)=sum(r_l(21 :l).*s_l(21 :l));
y_L2(l)=sum(r_2(21 :l).*s-1(21 :l));
y_L3(l)=sum(r_3(21 :l).*s_l(21 :l));
end
% Plotting the results:
subplot(3,2,5)
plot(m,(O y_0_0(1 :K/2)],'-bo',m,(O y_O_l(1 :K/2)], '--b* ', ...
m,[O y_0_2(1:K/2)],' : b. ',m,[O y_0_3(1 :K/2)],'- ')
•
set(gca,, XTickLabel, ,{, 0,',5Tb,',lOTb, ,' 15Tb, ,'2OTb, })
axis([O 20 -30 30])
xlabel('(e) \sigrnaA2= 1 & y_{O}(kT_{b})' ,'fontsize',10)
subplot(3,2,6)
plot(n,(O y_LO(K/2+1 :K)], '-bo',n,(O y_Ll(K/2+1 :K)], '--b* ', ...
n,(O y_L2(K/2+1 :K)],' : b. ',n,(O y_L3(K/2+1 :K)],'- ')
set(gca,' XTickLabel, ,{, 2 OTb, ,'25Tb, ,' 30Tb, ,' 35Tb, ,' 40Tb, })
•
axis([20 40 -30 30])
xlabel('(f) \sigrnaA2= 1 & y_{l}(kT_{b})' ,'fontsize',10)
---tll!U-il;fflHTjjj;t•1="§1•
Illustrative Problem 5.16 [Biorthogonal Signaling Simulation] Perform a Monte
Carlo simulation for a digital communication system that employs M
=
4 biorthog­
onal signals. The model of the system to be simulated is illustrated in Figure 5.39.
Gaussian RNG
±-YE
ro
+
Detector
Gaussian RNG
Uniform
RNG
Mapping to
signal points
Output
decision
S;
r1
0
Compare
s; withs;
Symbol
error counter
Figure 5.39: Block diagram of the system with M
=
4 biorthogonal signals for Monte
Carlo simulation
As shown, we simulate the generation of the random variables ro and r1, which con­
stitute the input to the detector. We begin by generating a binary sequence of O's and
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242
CHAPTER 5. BASEBAND DIGITAL TRANSMISSION
10-2
10-6 .__
0
_.____.___,___,____,_____.__..____,____,___,
_
2
3
4
5
6
7
8
9
10
Figure 5.40: Symbol-error probability for M = 4 biorthogonal signals from Monte
Carlo simulation compared with theoretical error probability
1 's that occur with equal probability and are mutually statistically independent, as in
Illustrative Problem 5.6. The binary sequence is grouped into pairs of bits, which are
mapped into the corresponding signal components as follows:
00- so= ()£,O)
01-+ S1= (0, ./£)
10-+ Sz = (0, -JE)
11-+ S3 = (-VE,O)
Alternatively, we may use the method in Illustrative Problem 5.12 to generate the 2-bit
symbols directly.
Because s2 =
- s1 and S3 = - so, the demodulation requires only two correlators
or matched filters, whose outputs are ro and r1. The additive noise components no and
n1 are generated by means of two Gaussian noise generators, each having mean zero
and variance u2 = NoE / 2. For convenience, we may normalize the symbol energy to
E = 1 and vary u2. Because E = 2Eb, it follows that Eb = �. The detector output is
compared with the transmitted sequence of bits, and an error counter is used to count
the number of symbol errors and the number of bit errors.
Figure 5.40 illustrates the results of this simulation for the transmission of 20,000
bits at several different values of the SNR Eb I No. Note the agreement between the
simulation results and the theoretical value of P4 given by (5.4.26) and (5.4.24).
The MATLAB scripts for this problem are given next.
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5.4. MULTIDIMENSIONAL SIGNALS
243
--411ili" ------% MATLAB script for Illustrative Problem 5.16.
echo on
SNRindB=0:2:1O;
for i=1 :length(SNRindB),
% simulated error rate
smld_err_prb(i)=smldP511(SNRindB(i));
echo off;
end;
echo on ;
% Plotting commands follow.
--411ili" ------function [p]=smldP51l(snr_in_dB)
% [p]=smldP5ll(snr_in....dB)
%
SMLDP511
%
snr ...in....dB, signal-to-noise ratio in dB, for the system
%
described in Illustrated Problem 5.11.
simulates the probability of error for the given
M=4;
% quaternary biorthogonal signaling
E=1;
SNR=exp(snr_in_dB*log(10)/1O);
% signal-to-noise ratio per bit
sgma=sqrt(E" 2/(4*SNR));
% sigma, standard deviation of noise
N=10000;
% number of symbols being simulated
% generation of the quaternary data source
for i=1 :N,
temp=rand;
% uniform random variable over (0,1)
if (temp<0.25),
dsource(i)=O;
elseif (temp<0.5),
dsource(i)=1 ;
elseif (temp<0.75),
dsource(i)=2;
else
dsource(i)=3;
end
end;
% detection, and error probability computation
numoferr=O;
for i=1 :N,
% the matched filter outputs
if (dsource(i)==O)
rO=sqrt(E)+gngauss(sgma);
rl=gngauss(sgma);
elseif (dsource(i)==1)
rO=gngauss(sgma);
rl=sqrt(E)+gngauss(sgma);
elseif (dsource(i)==2)
r0=-sqrt(E)+gngauss(sgma);
rl=gngauss(sgma);
else
rO=gngauss(sgma);
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244
CHAPTER 5. BASEBAND DIGITAL TRANSMISSION
rl=-sqrt(E)+gngauss(sgma);
end;
%
detector follows
if (rO>abs(rl)),
decis=O;
elseif (rl>abs(rO)),
decis=1;
elseif (rO<-abs(rl)),
decis=2;
else
decis=3;
end;
if (decis-=dsource(i)),
% If it
is an error, increase the error counter.
numofe=numoferr+1;
end;
end;
p=numoferr/N;
%
bit error probability estimate
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PROBLEMS
245
Problems
5.1 Suppose the two orthogonal signals shown in Figure 5.2 are used to transmit bi­
nary information through an AWGN channel. The received signal in each bit inter­
val of duration Tb is given by (5.2.1). Suppose that the received signal waveform is
sampled at a rate of 10/Tb-that is, at ten samples per bit interval. Hence, in dis­
crete time, the signal waveform so (t) with amplitude A is represented by the ten sam­
ples (A,A, ...,A), and the signal waveform s1 (t) is represented by the ten samples
(A,A,A,A,A, A, A, A, A, A ) . Consequently, the sampled version of the re­
ceived sequence when so(t) is transmitted it is
-
-
-
-
-
k
=
1, 2, ..., 10
and when s1 (t) is transmitted it is
1::::; k::::; 5
6::::; k::::; 10
where the sequence { nd is i.i.d., zero mean, Gaussian with each random variable
having the variance a-2. Write a MATLAB routine that generates the sequence {rk}
for each of the two possible received signals, and perform a discrete-time correlation
of the sequence {rk } with each of the two possible signals so(t) and s1 (t) represented
by their sampled versions for different values of the additive Gaussian noise variance
a-
2
=
to A
0, a-2
=
=
0.1, a-2
=
0.5, and a-2
=
1.0. The signal amplitude may be normalized
1. Plot the correlator outputs at time instants k
1, 2, 3, ..., 10.
=
5.2 Repeat Problem 5.1 for the two signal waveforms so(t) and s1 (t) illustrated in
Figure P5.2. Describe the similarities and differences between this set of two signals
and those illustrated in Figure 5.2. Is one set better than the other from the viewpoint
of transmitting a sequence of binary information signals?
s0(t)
SJ
(t)
A
T
0
0
T
T
2
-A
-A
Figure P5.2
5.3 Repeat the correlation of signal waveforms in Illustrative Problem 5.2 for each of
2
the noise variances a-
2
=
0.1, a-
=
1, a-2
=
3 and plot the correlator outputs for each
2
value of a- .
5.4 Repeat the matched filtering of signal waveforms in Illustrative Problem 5.4 for
each of the noise variances u2
=
0.1, u2
=
1, u2
=
3 and plot the matched filter
2
outputs for each value of a- •
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CHAPTER 5. BASEBAND DIGITAL TRANSMISSION
246
5.5 The antipodal signal waveforms shown in Figure 5.12 (a) are constant over the sig­
nal interval 0 :S: t :S: Tb. In this case, the correlator shown in Figure 5.13 (b) can be
simplified by eliminating the multiplication by s(t). Thus, the correlator is simply an
integrator whose output is reset to zero at the end of each signal interval. As a conse­
quence, the correlator is called an integrate-and-dump (I&D) filter. For the antipodal
signal waveforms in Figure 5.12(a), sketch the I&D filter outputs when so(t) and s1 (t)
are transmitted.
5.6 In this problem, the objective is to substitute two matched filters in place of the two
correlators in Problem 5.1. The condition for generating signals is identical to Problem
5.1.
Write a MATLAB routine that generates the sequence {rk} for each of the two
possible received signals, and perform the discrete-time matched filtering of the se­
quence {rk} with each of the two possible signals so(t) and s1 (t), represented by their
sampled versions, for different values of the additive Gaussian noise variance (]"2
2
(]"
A
0.1, (]"2
=
1.0, and (]"2
=
=
2.0.
1. Plot the correlator outputs at time instants corresponding to k
=
=
0,
The signal amplitude may be normalized to
1, 2, . . . , 10.
=
5.7 Repeat Problem 5.6 for the signal waveforms shown in Figure P5.2.
5.8 Run the MATLAB program that performs a Monte Carlo simulation of the binary
communication system shown in Figure 5.10, based on orthogonal signals. Perform
the simulation for 10,000 bits and measure the error probability for (]"2
2
(]"
2
0.5, and (]"
=
=
=
0, (]"2
=
0.1,
1.0. Plot the theoretical error rate and the error rate measured
from the Monte Carlo simulation and compare the two results. Also plot 1000 received
signal-plus-noise samples at the input to the detector for each value of (]"2•
5.9 Repeat Problem 5.8 for the binary communication system shown in Figure 5.16,
based on antipodal signals.
5.10 Repeat Problem 5.8 for the binary communication system based on on-off sig­
nals.
5.11 Run the MATLAB program that performs a Monte Carlo simulation of a quater­
nary PAM communication system. Perform the simulation for 10,000 symbols (20,000
bits) and measure the symbol-error probability for (]"2
2
and (]"
=
=
0,
2
(]"
=
0.1,
2
(]"
=
0.5,
1.0. Plot the theoretical error rate and the error measured from the Monte
Carlo simulation and compare these results. Also plot 1000 received signal-plus-noise
samples at the input to the detector for each value of (]"2.
5.12 Modify the MATLAB program in Problem 5.11 to simulate M
=
8 PAM signals
and perform the Monte Carlo simulations as specified in Problem 5.11.
5.13 Repeat Illustrative Problem 5.7 for each of the noise variances (]"2
2
(]"
=
=
1, (]"2
=
2,
4.
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PROBLEMS
247
5.14 Repeat Illustrative Problem 5.9 for each of the noise variances u2
(]"
2
=
=
1, u2
=
2,
4.
5.15 Run the MATLAB program that performs the Monte Carlo simulation of a dig­
ital communication system that employs M
4 orthogonal signals, as described in
=
Illustrative Problem 5.14. Perform the simulation for 10,000 symbols (20,000 bits) and
measure the bit-error probability for u2
=
0.1, u2
=
0.5, and u2
=
1.0. Plot the the­
oretical error probability and the error rate measured from the Monte Carlo simulation
and compare these results.
5.16 Consider the four signal waveforms shown in Figure P5.16.
Show that these
four signal waveforms are mutually orthogonal. Will the results of the Monte Carlo
simulation of Problem 5.14 apply to these signals? Why or why not?
Sz(t)
A f------.
A
A
A
T
T
T
T
0
-A
-A
-A
P5.16
5.17 Run the MATLAB program that performs the Monte Carlo simulation of a digital
communication system that employs M
=
4 biorthogonal signals, as described in Il­
lustrative Problem 5.16. Perform the simulations for 10,000 symbols (20,000 bits) and
measure the symbol-error probability for a2
=
0.1, u2
=
1.0, and u2
=
2.0. Plot the
theoretical symbol-error probability and the error rate measured from the Monte Carlo
simulation and compare the results. Also plot 1000 received signal-plus-noise samples
at the input to the detector for each value of u2•
5.18 Repeat Illustrative Problem 5.15 for each of the noise variances u2
(]"
2
=
=
1, u2
=
2,
4.
5.19 Consider the four signal waveforms shown in Figure P5.19. Show that they are
biorthogonal. Will the results of the Monte Carlo simulation in Problem 5.17 apply to
this set of four signal waveforms? Why?
s0(t)
sz(t)
s1(t)
A
A
T
T
0
S3(t)
A
T
2
T
0
0
0
-A
-A
-A
T
2
T
P5.19
5.20 Using the graphs given in this chapter, compare the values of Eb I No required to
achieve a symbol-error probability of 10-6 for M
=
8 PAM, orthogonal, and biorthog­
onal signals.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions l'«tuire it.