Question:-
Statement for Linked Answer Questions:

Consider a baseband binary PAM receiver shown below. The additive channel noise n(t) is whit with power spectral density SN(f)=N0/2=10-20 W/Hz.
The low-pass filter is ideal with unity gain and cutoff frequency 1MHz. Let Yk represent the random variable y(tk).
Yk=Nk if transmitted bit bk=0
Yk=a+Nk if transmitted bit bk=1
Where Nk represents the noise sample value. The noise sample has a probability density function, PNK(n)=0.5ae-a|n| (This has mean zero and variance 2/a2).
Assume transmitted bits to be equiprobable and threshold z is set to a/2=10-6V.


The value of the parameter a (in V-1) is
Option (A)
1010
Option (B)
107
Option(C)
1.414x10-10
Option(D)
2x10-20
Correct Option:
(B)
Question Solution:
exist in fourier transform pair
R(0) = (N0/2), Power = 2 (N0 /2) x 1x 10-16 = 2 x 10-14
Variance = (2/d2) which is equal to power since mean is zero Þ a = 107
question-answer-faq-2193