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RECEIVER IN DIGITALCOMMUNICATION EXERCISE
Exercise 5.11 – 5.12 – 5.13
Textbook: Communication System - Simon Haykin 4th edition
GROUP 1
Nguyễn Văn Nhân 40901817Phạm Ngọc Thoan 40902635
Đoàn Nhựt Tân 40902364
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Problem 5.11Consider the optimum detection of thesinusoidal signal:
in additive white Gaussian noise.a) Determine the correlator output assuming a
noiseless input.b) Determine the corresponding matched filter
output, assuming that the filter includes a
delay T to make it causal.c) Hence show that these two outputs are the
same only at time instant t = T.
T
t t s
8sin)(
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Problem 5.11 (Solution)
For the noiseless case, the received signal
a) The correlator output is
T t t st r 0),()(
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Problem 5.11 (Solution)
b) The matched filter is defined by theimpulse response:
The matched filter output is therefore
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Problem 5.11 (Solution)
Since we have:0 T
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Problem 5.11 (Solution)
c) When the matched filter output issampled at t = T, we get:
which is exactly the same as the coirelator
output determined in part (a)
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Problem 5.12Figure P5.12 shows a pair of signals s
1
(t) and s2
(t) thatare orthogonal to each other over the observationinterval 0 ≤ t ≤ 3T. The received signal is defined by
0 ≤ t ≤ 3T, x(t) = sk{t) + w(t), k = 1, 2
where w(t) is white Gaussian noise of zero mean andpower spectral density N0/2.
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Problem 5.12
(a) Design a receiver that decides in favorof signals Si(f) or $i(t), assuming that thesetwo signals are equiprobable.
(b) Calculate the average probability of symbol error incurred by this receiver forE/No = 4, where E is the signal energy
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Problem 5.12 (Solution)
(a) The matched filter for signal s1(t) isdefined by the impulse response
h1(t) = s
1(T - t)
The matched filter for signal s2(t) is
defined by the impulse responseh2(t) = s2(T - t)
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Problem 5.12 (Solution)The matched filter receiver is as follows
The receiver decides in favor of s2(t) if, for the noisyreceived
We find that x1 > x2. On the other hand, if x2 > x1, itdecides in favor of s2(t). If x1 = x2, the decision is madeby tossing a fair coin.
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Problem 5.12 (Solution)
(b) The energy of s1(t)
= (1)2
0+ (−1)2
2
+ (1)2
2= 3 =
The energy of s2(t)2 = (−1)2/20
+ (1)2/2/2
+ (−1)25/2/2
+ (1)25/2
= 3=
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Problem 5.12 (Solution)
The orthornomal basis functions for the signal-space diagram of these two orthogonal signals
are given by
The signal-space diagram of signals s1 and s2 isas follows:
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Problem 5.12 (Solution)
The distance between the two signal points s1(t)and s2(t) is
The average probability of error is therefore
For E/N0, we therefore have
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Problem 5.13
In the Manchester code, binary symbol 1 isrepresented by the doublet pulse s(t) shown inFigure P5.13, and binary symbol 0 is representedby the negative of this pulse.
Derive the formula for theprobability of error incurred
by the maximum likelihood
detection procedure appliedto this form of signaling over
an AWGN channel.
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Problem 5.13 (Solution)
Energy of binary symbol 1 represented bysignal s1(t) is
Energy of binary symbol 0 represented bysignal s2(t) is the same as shown by
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Problem 5.13 (Solution)
The only basis function of the signal-spacediagram is:
The signal-space diagram of the Manchestercode using the doublet pulse is as follows:
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Problem 5.13 (Solution)
Hence, the distance between the twosignal points is
The average probability of error over an AWGN channel is given by:
T d 2