frequency response
Dear rkk76,
The form that expression is presented is doubty.
Wouldn't you mean:
h
= (a^n) *cos(n*pi)*u
, If so, it is easy.
You just find the z transform of h
making x
= cos(n*pi)*u
, thus we have h
= (a^n)x
.
Using the property:
(a^n)x
---Z----> X(Z/a)
Since from a Z transform table.
cos(n*pi)*u
---Z----> [1 +z^-1] / [1+2*z^-1+z^-2]
We find:
(a^n) *cos(n*pi)*u
---Z----> [1 +(z/a)^-1] / [1+2*(z/a)^-1+(z/a)^-2]
Depending of the purposeof the question, this is the answer you are looking for. However, this is actually the Z transform from the given expression and not its frequency response.
In order to find the frequency respond you have to determine a specific value to "a" and choose a sample frequency, with these data at hand, you can use the freqz comand in matlab and find the frequency response.
Well, this is correct if the expression is indeed h
= (a^n) *cos(n*pi)*u
as I presumed it is and probably it is,Nevertheless, If it is h
= (a)^n *cos(n*pi)*u
, considering n *cos(n*pi)*u
as expoent of a like h
= (a)^(n *cos(n*pi)*u
) the situation is tough and I am sorry I can't help.
Regards