Unable to get how to solve for mutltivariable function while calculating DTFT

1 次查看(过去 30 天)
Hi ,
I have the frequency response of a system given as:
H(e^(iθ))=1/(1-0.9*e^(-iθ))
and i am struggling to get the response of the above system when the input is
x(n)=0.5*cos((pi*n)/4)
.What i have tried and stucked at is how do i give the x(n) as a input to the function 'H' . . Can anybody help me out with this or other way ?

采纳的回答

Harsh Kumar
Harsh Kumar 2023-7-5
编辑:Harsh Kumar 2023-7-14
Hi Rohit,
I understand that you are trying to find the response of a system for a given input whose frequency response has been given already.
To do this, since the input is sinusoidal, you can use one of its property that the output will be just an amplified and phase differentiated version of itself when passed through the system.
Refer to the below code snippet for better understanding.
clc;
n = -25:1:25;
%system response at w=pi/4
w = pi/4;
H = 1/(1-0.9*exp(-1i*w));
mag = abs(H);
ang = angle(H);
%ouput=mag*sin(wt+ang) for sinosuidal input
output = 1/2*mag*cos((n*pi/4)+ang).*(n>=0);
stem(n,output);
You may refer to these documentation links for better understanding:
  7 个评论
Harsh Kumar
Harsh Kumar 2023-7-14
Yes ,(n>=0) done already in the code and the second was a typing error. Thanks for pointing out and your time .

请先登录,再进行评论。

更多回答(1 个)

William Rose
William Rose 2023-6-28
编辑:William Rose 2023-6-28
[Are you sure that is exactly how the frequency repsonse and the input funtion are defined?]
The frequency response is a function of θ. Let's assume .
The input x is a function of n. Let's assume n=t, so we have . Then we can write x(t) as , where . Now we recall that .
(assuming linearity)
If you continue the algebra and deal with the complex numbers, you should find that Y is real.
  3 个评论
Paul
Paul 2023-6-28
What is the advantage in changing the independent variable from n to t?
This solution is only the steady state solution, which might or might not meet the intent of the question.
William Rose
William Rose 2023-6-29
@Paul,
Good point. I coud have, I should have kept n as n.
ALso a good point about steady state. The question has a transfer function and a sinusoidal input, so I assume that the "steady state" sinusoidal output is what is desired. If a transient solution were desired, then the initial condition of the output would need to be specified, and it is not.

请先登录,再进行评论。

类别

Help CenterFile Exchange 中查找有关 Symbolic Math Toolbox 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by