Below figure shows the distribution of noise component.
Pdf of noise signal n(t) is defined as:
Since the area under pdf curve is 1,
h*(2-(-2) = 4h = 1
=> h = 0.25
Symbols s1(t)=a1 and s2(t)=a2 are equally likely to be transmitted, therefore
P(a1) = P(a2) = 0.5
P(Error) = P(a1/a2) + P(a2/a1) [i.e. a2 detected in case of a1 transmitted and vice versa]
= P(a2)*P(a2+n(t)>0) + P(a1)*P(a1+n(t)<0) [ transmitted symbol and noise are independent]
= 0.5*P(-0.8+n(t) >0) + 0.5*P(A+n(t)<0.8) [ a1 = A = 0.8, a2 = -A = -0.8; decision threshold = 0]
= 0.5*P(n(t) >0.8) + 0.5*P(n(t)<-0.8)
= 0.5*((2-0.8)*0.25) + 0.5*((-0.8-(-2)*0.25) [see below figure]
= 0.5*1.2*0.25 + 0.5*1.2*0.25
= 0.3
BER of the system = (no. of bit received with error) / (total no. of received bits)
= 0.3
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