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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