0
$\begingroup$

I studied various signal processing materials for a long time, and I have a question. Considering an LTI filter, one can define its frequency response by evaluating its transfer function $H(z)$ on the unit circle $H(e^{j\omega})$. Is this definition correct?

I also came across the second definition, which says that the filter's frequency response is the Fourier Transform of its impulse response. Are these definitions equivalent? And if so, can DFT be used as the Fourier transform?

Thank you.

$\endgroup$

1 Answer 1

2
$\begingroup$

Considering an LTI filter, one can define its frequency response by evaluating its transfer function H(z) on the unit circle H(ejω). Is this definition correct?

This is only correct if the LTI filter discrete in time. But if it is: yes.

I also came across the second definition, which says that the filter's frequency response is the Fourier Transform of its impulse response.

Correct

Are these definitions equivalent?

Depends on what you mean by equivalent, but mostly yes: they are closely related.

And if so, can DFT be used as the Fourier transform?

Sort of. This is actually a somewhat complicated questions. There are four different types for Fourier Transform which are applicable to all 4 possible combinations of the time domain and/or the frequency domain signals being discrete or continuous. http://fourier.eng.hmc.edu/e101/lectures/handout4/node3.html

A signal that's discrete in one domain is periodic in the other. The DFT applies to both signals being discrete. That means that both signals are also periodic as well. However, the impulse response of a causal LTI systems can NOT be periodic so there is an inherent conflict here.

In practice you can work around this fairly easily by making sure that your time domain window for the DFT is long enough to capture the entire impulse response. However, this can never be done "exactly" so you always will end up with some residual amount of time- and/frequency- domain aliasing.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.