Re: Jointly -WSS?# EE - 电子工程
d*o
1 楼
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(
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(