6. Minimum distance decoding) Suppose that a transmitter sends one of four possi

Question

6. Minimum distance decoding) Suppose that a transmitter sends one of four possible symbols -s- 1,-1). The received signal at a receiver is given by X- S + N, where S is the transmitted symbol, and N (Ni, N2) is additive noise. Assume N; ~ N (0, 22),1-1, 2, and Ni N2, î.?., independent. (a) Assume that every symbol is equally likely to be transmitted, ie., PS-sil-1/4 for all i = 1, 2, 3, 4. Show that the maximurn likelihood (ML) rule will choose the symbol that is closest to the received signal X (according to the Euclidean distance). b) Compute the probability of error under the ML rule

Explanation / Answer

for maximum likehood,if all source probabilities are same

p(x1)=p(x2)=p(x3)=..................p(xk )

yj is detector as xi source

p(xi ) p(yj /xi ) > p(xk ) p(yj /xk)

since p(xi ) = p(xk ) then p(yj /xi ) > p(yj /xk)

which means receiver selects only which has more probable information

but in case let four symbols be s0,s1,s2,s3 have equal euclidean distance

probability of error= 2 Q (sqrt (2Es /No ) * sin^2(pi/M))

where Q is Q function, Es /No =2^2 because all have same distance then Eavg =(r^2 +r^2 +.....) /M = r^2

M= no. of symbols=4 in this case

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