IFFT of Low Pass Filter function

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Arin Dutta
Arin Dutta 2021 年 8 月 4 日
回答済み: Arin Dutta 2021 年 8 月 4 日
Let, a low pass filter function, R= α^2/(α^2+〖(2πfd)〗^2 ) , where d and α are constant, f is frequency.
If I take the IFFT of R, I should see an exponetial plot (in time domain). But I get it wrong at output. Probably I have done something wrong in my code. Any help would be greately appreciated. I have included my code here:
f = logspace(0,10,1000);
alpha = 0.295/4.343/1000;
d = 1.6693e-12;
A = alpha.^2 ./ (alpha.^2 + (2.*pi.*f.*d).^2);
B =abs(ifftshift(ifft(A)));
B = B(length(B)/2 : end); %
semilogy(B);

回答 (1 件)

Arin Dutta
Arin Dutta 2021 年 8 月 4 日
Thanks for your response. I did the IFFT, because I need to convert the frequency to time. As the given transfer function in frequency domain is a low pass filter, so in the time domain, the output should be exponential.

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