d****o
发帖数: 7 | 1 I think so --- and I presume that you also have the condition that X(t) and
W(t) are independent.
First verify that Y(t) is WSS:
1) clearly E[Y(t)]=0;
2) By expanding E[Y(t)Y(t+\tau)], you can also see that this term is
independent of t. There is an intuitive way to
see this: remember that when W(t) is a deterministic function, then
E[Y(t)Y(t+\tau)] is indepdent of t for WSS X(t);
now W(t) is a random process, this means that for every fixed realization of
W(t) (or conditioned on W(t)), E[Y(t)Y( | d****o
发帖数: 7 | 2
on
To show E{Y(t)Y(t+\tau)] depends only on \tau, I guess the following will
work:
E[Y(t)Y(t+\tau)]= E[ \int \int X(t-u)W(u)X(t+\tau-v)W(v) du dv ]
=\int \int E[X(t-u) X (t+\tau-v)] E[W(u) W(v)] dudv
=\int \int R_X(\tau-v+u) R_W(u, v) dudv
where R_X is the autocorrelation function of X(t) (expressed as single
variable function since X(t) is WSS) and
R_W is the autocorrelation function of W(t); and the second equality above
follows from that W(t) and X(t) are independent.
Then clearly, this fin |
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