3
$\begingroup$

Consider the LTI system given by: $H(z) = 1 - \frac{1}{2}z^{-1}+\frac{3}{4}z^{-2}$

Let $x[n] = (\frac{1}{2})^nu[n]$ be the input to the system. We want to find the output for $n = 0,1,...,N_a$, using the DFT of a relevant portion of the input and samples of $H(z)$ on the unit circle, at $z = e^{j\frac{2\pi}{N_b}k}$, for $k = 0,...,Nb-1$.

Now, because $H(z)$ is FIR, the samples of $H(z)$ on the unit circle are actually the values of the DFT of length $N_b$ of $h[n]$. I can work out the appropriate length of the DFT of $x[n]$ to use and it would work and using the circular convolution property, find the actual values of the output.

  • What if instead of a FIR, I'm given a causal IIR such as $H(z) = \frac{1}{1-\frac{1}{2}z^{-1}}$?

    Can I use the samples of this new $H(z)$ on the unit circle just as in the first case? My guess is yes, because the system is stable (because it has a pole at $\frac{1}{2}$ and it is causal) and then the Fourier transform exists, so I can obtain the samples of the Fourier transform and use them as DFT of $H(z)$, (I believe) it would be like truncating $h[n]$ and then using its DFT.

  • Now, what if the system was unstable and causal, for example:

    $H(z) = \frac{1}{1-2z^{-1}}$?

    The Fourier transform clearly doesn't exist, and I can't truncate $h[n]$

$\endgroup$
0
$\begingroup$

I believe that you already have an answer in your question. Clearly, Fourier and DFT are applicable under limited set of circumstances.

For unstable system, Fourier transform does not exist for all types of systems, as you already noticed from your example. If you are interested in more, look at the following question that is related: Discrete Time Fourier Transform (DTFT) for an unstable system (Ideal Low Pass Filter)

It is not a trivial question, and requires careful analysis of the assumptions under which the Fourier and Discrete Fourier are applied.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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