Lecture 18

Detection of Binary Signal in Gaussian Noise

COMMUNICATION SYSTEMS Lecture # 18

The recovery of signal at the receiver consist of two parts Filter Reduces the received signal to a single variable z(T) z(T) is called the test statistics Detector (or decision circuit) Compares the z(T) to some threshold level 0 , i.e.,

6th Apr 2007 Instructor

WASEEM KHAN

H 1

z (T )

where H1 and H0 are the two possible binary hypothesis

0 H 0

Centre for Advanced Studies in Engineering

Detection of Binary Signal in Gaussian Noise The recovery of signal at the receiver consists of two parts: Waveform-to-sample transformation

Detection of Binary Signal in Gaussian Noise The sample z(T) will be another Gaussian random variable.

Demodulator followed by a sampler Each symbol is sampled at t = T to get a sample z(T).

z(T ) ai (T )

n0 (T ) i 0,1

p( z | s0 )

1 exp 0 2

1 z a0 2 0

p( z | s1 )

1 exp 0 2

1 z a1 2 0

where ai(T) is the desired signal component, and n0(T) is the noise component

2

2

Detection of symbol Assume that input noise is a Gaussian random process, i.e.

1

p (n0 ) 0

2

exp

1 2

n0

2

0

Probabilities Review

1

Choosing the Threshold Maximum a posteriori (MAP) criterion:

P[s0], P[s1] a priori probabilities These probabilities are known before transmission P[z] probability of the received sample p(z|s0), p(z|s1) conditional pdf of received signal z, conditioned on the transmitted symbol si P[s0|z], P[s1|z] a posteriori probabilities

If

p ( s0 | z )

p ( s1 | z )

H0

If

p ( s1 | z )

p ( s0 | z )

H1

Problem is that a posteriori probability are not known. Solution: Use Bay s theorem:

p (s | z) i

p( z | s1 ) P(s1 ) P( z)

H1

H0

p( z | s ) p(s ) i i p(z)

p( z | s0 ) P(s0 ) P( z)

H1

p( z | s1) P(s1)

p( z | s0 ) P(s0 ) H0

1

Choosing the Threshold H1

p ( z | s1 )

L(z)

p( z | s0 )

P (s0 )

likelihood ratio test ( LRT )

P ( s1 )

H0

Substitute the pdfs H0 :

When the two signals, s0(t) and s1(t), are equally likely, i.e., P(s0) = P(s1) = 0.5, then the decision rule becomes

p ( z | s1 )

L( z)

p ( z | s0 )

1

p ( z | s0 )

2

0

2

exp

1 z a0 2 0

exp

1 z a1 2 0

H1

1

max likelihood ratio test

H1 :

1

p ( z | s1 )

H0

2

0

This is known as maximum likelihood ratio test because we are selecting the hypothesis that corresponds to the signal with the maximum likelihood.

H1 p ( z | s1 ) p ( z | s0 )

L(z) In terms of the Bayes criterion, it implies that the cost of both types of error is the same

Solving

1

exp

2

0

1

1

H0

0

exp

2

1 2

2

z

a1

2

z

a0

2

H1

0

1

1 2

H0

0

Hence z ( a1

exp

( a12 a 02 ) 2 02

a0 ) 2 0

1

H1

H1 2 0

z Taking the log of both sides will give

H1 z(a1 a0 )

ln{L( z)}

2 0

(a12 a02 ) 2 02

(a1 a0 )(a1 a0 ) 2 02 (a1 a0 )

H0 0

a0 )

H1 (a

2 1

2 0

a )

2

2 0

( a1

a 0 )( a1 2 02

a0 )

z

Probability of Error

If signals are equally probable

PB

Error will occur if s1 is sent s0 is received

PB

p ( z | s1 ) dz

s0 is sent

s1 is received

P ( H 1 | s0 )

P (e | s0 )

P (e | s0 )

p ( z | s 0 ) dz 0

The total probability of error is the sum of the errors

P ( e, si )

P ( e | s1 ) P ( s1 )

P ( H 1 | s 0 ) dz

P( H1 | s0 )

p ( z | s 0 ) dz

0

P (e | s 0 ) P ( s0 )

0

1

P ( H 1 | s 0 ) P ( s0 )

by Symmetry

Hence, the probability of bit error PB, is the probability that an incorrect hypothesis is made Numerically, PB is the area under the tail of either of the conditional distributions p(z|s1) or p(z|s2)

i 1

P ( H 0 | s1 ) P ( s1 )

P ( H 1 | s 0 ) P ( s0 )

P ( H 1 | s0 )

1 P( H 0 | s1 ) P( H1 | s0 ) 2

PB

2

PB

P ( H 0 | s1 ) P ( s1 ) 1 P ( H 0 | s1 ) 2

P (e | s1 ) 0

P (e | s1 )

0 H0

H0

P ( H 0 | s1 )

0

H0

H1 2 0

a0 ) 2

where z is the minimum error criterion and 0 is optimum threshold For antipodal signal, s1(t) = - s0 (t) a1 = - a0

H0

z ( a1

( a1

z

0

0

2

exp

1 2

z

a0

2

dz

0

2

1

PB 0

exp

2

0

(z

u

1 z 2

a0

2

dz

Q function tabulated

0

a0 ) 0

( a1 a 0 ) 2 0

1 2

u2 du 2

exp

The above equation cannot be evaluated in closed form (Q-function) Hence,

Q( x)

a1 a 0 2 0

PB

Q

1 x 2

exp

x2 2

Problem It is required to transmit an analog signal having 3kHz single-side bandwidth, using ASK. Available bandwidth is 12 kHz and 50% excess bandwidth is recommended for good performance. How many quantization levels are possible, at maximum, while sampling the analog signal?

3

This document was created with Win2PDF available at http://www.daneprairie.com. The unregistered version of Win2PDF is for evaluation or non-commercial use only.