Recent content by bexter

(d) Let y(t) be defined as: y(t) = (B(q)/A(q))x(t) Can this signal be used as a ONE STEP ahead prediction of v(t)? Justify your answer I got that can you give me hints for part c

this is the final equation I got for : Need to compute E[x(t-k)e(t-k)] x(t) = (SUM(ak*e(t-k)) - (SUM(bk*e(t-k) - (SUM(ak*x(t-k)) multiply x(t) by e(t): x(t)e(t)= (SUM(ak*e(t-k)e(t)) - (SUM(bk*e(t-k)e(t)) - (SUM(ak*x(t-k)x(t)) then shift by j for both (x) and (e) x(t-j)e(t-j)=...

so I get for part c ---> answer is zero

Perfect I see it now after expanding A(q) I would get σe2 What in general will E[e(t-k)e(t-j)] evaluate to? zero for all j not equal to k but if j=k then σe2

Same it would equal to zero as E[e(t-k) A(q) e(t-k)] = A(q)E[e(t-k)]E[(e(t-k)] ----> and E[e(t-k)] is zero.

I am not sure if I got that right or not. can you give me more hints if its not correct

Okay, can you evaluate E[e(t) q-1 e(t)]? with your hint : e(t) and e(t-1) are independent then E[e(t)]E[e(t-1)] ---> and that is zero since the mean is zero.

That sounds about right Steff. Now let's move to part c: (c) Compute the expected value E[x(t − k)e(t − k)], for k = 0, 1, 2, . . .. x(t) = ((A(q) − B(q))/B(q)) v(t) where v(t) = (B(q)/A(q) )e(t) substituting v(t) into x(t) we get x(t) = ((A(q) − B(q))/B(q))*(B(q)/A(q) )e(t) --->...

This is what i am doing for this question : (b) Write the relationship between x(t) and v(t) as a difference equation. What is the coefficient of v(t) (i.e. at zero delay)? Is this filter stable? x(t) = ((A(q) − B(q))/B(q)) v(t) x(t) * B(q) = (A(q) - B(q) ) * v(t) bk *x(t-k) = ak*v(t-k) -...