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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.
Thanks for reply. Yes, you are right I made mistake. However, I did not cover Z-transform yet so I cannot use it. This equation is for frequency response
H(e^jΘ) = Σh(m)*e^(-jΘm) where m is from - infinity to +infinity.
What I did is that I converted to exponential and unit vector u is I for 1>= 0. Doing that I got strange solution (probably wrong). Any suggestions
The DTFT is a particular case of Z transform since Z=r*e^(jΘ) , considering that the system is stable , this particular case is when r =1 thus replacing Z by e^(jΘ) in the Z transfer function you wil find the DTFT that you are looking for. This will only be true when the system is stable, the value of the "a" parameters determines the stability of the system , so in order to know in wich values of "a" the system is stable you have to garantee that the poles of the system are inside the unit circle. Thus you must solve the inequation : [1+2*(z/a)^-1+(z/a)^-2]
<1.
With the step above you will get the answer.
Now, considering that you must comput the DTFT directly fro the expression H(e^jΘ) = Σh(m)*e^(-jΘm) you must first understand that the step function is 0 to m<0 thus the expression you have is H(e^jΘ) = Σ(a^n) *cos(n*pi)*e^(-jΘm) where m is from 0 to +infinity. Using euler formula to cos(n*pi) , replacing it in the expression and performing the computation, you will find the answer. Remember to define the values of "a" from wich the sumations converges.
I will not do the computation for you.
Compare with the results from the Z transform explanation and you will see that they are the same.
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