Suppose we observe an output signal Y(t) which consist of a desiredsignal X(t) p
ID: 2953358 • Letter: S
Question
Suppose we observe an output signal Y(t) which consist of a desiredsignal X(t) plus noise N(t), Y(t) = X(t) + N(t). Find the cross-correlation between theoutput signal at 10s and the input signal at 5s, assuming that X(t)and N(t) are independent random signals. Note thatcross-correlation between X and Y at t1 and t2 is Rx,y(t1,t2) = E[ X(t1)Y(t2) ] and auto correlation X at time 5s and 10s is Rx(5,10) = E[ X(5) X(10) ] = 10; while mean of X at time 5s,E[X(5)], is 5 and mean of N at time 10s, E[N(10)], is 0.25 Y(t) = X(t) + N(t). Find the cross-correlation between theoutput signal at 10s and the input signal at 5s, assuming that X(t)and N(t) are independent random signals. Note thatcross-correlation between X and Y at t1 and t2 is Rx,y(t1,t2) = E[ X(t1)Y(t2) ] and auto correlation X at time 5s and 10s is Rx(5,10) = E[ X(5) X(10) ] = 10; while mean of X at time 5s,E[X(5)], is 5 and mean of N at time 10s, E[N(10)], is 0.25Explanation / Answer
We want Rx,y (5,10) = E[X(5)Y(10)]
=E(X(5)(X(10)+N(10))]
= E(X(5)X(10))+E(X(10)N(10))
Given that E(X(5)X(10)) = 10
And we also know that Noise is independent of any signal
So E(X(10)N(10)) = E(X(10)) * E(N(10))
= 5*0.25 = 1.25
Therefore
E(X(5)X(10))+E(X(10)N(10)) = 10+1.25 =11.25
Hope this helps. Feel free to ask for any clarifications