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I have the following task:

Assume that $h$ is a filter with finite support $[0..(m−1)]$ for some $m>0$. Let $x$ be a circular extension of $h$.

  1. Prove that $x\in l_\infty$.
  2. When is $x\in𝑙_1 $?
  3. Build a filter $y\in 𝑙_1 $ such that its periodized version $y^m$ equals to $x$ on signals with period $m$.

I answered two first questions but cannot figure out the solution for building filter $y\in l_1$ such that its periodized version $𝑦^𝑚$ equals to $x$ on signals with period $m$.

Could someone help me?

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  • $\begingroup$ so I take it that your notation implies $h$ and $x$ to be discrete. So, for $x$, you can apply the DFT and get one period of the spectrum (which mathematically is periodically extended to infinity). Soooo, your $y*s$ ($*$ means convolution) needs to be identical to $x*s$ for $m$-periodic signals $s$. Hint: convolution property of the DFT. $\endgroup$ – Marcus Müller May 31 at 12:42
  • $\begingroup$ Thanks a lot for the answer! So, am I right that you suggest to construct $y$ so that it is equal to the DFT of $x$? Also, do you mean the convolution theorem (en.wikipedia.org/wiki/Convolution_theorem) by saying "convolution property of the DFT"? I'm afraid I didn't quite get it how this convolution property can be used here $\endgroup$ – darktealeaf May 31 at 13:15
  • $\begingroup$ if two sequences have the same DFT, they are identical, so that's not it. The point is that $s$ being $m$-periodic implied that $s$'s DFT takes a specific shape. $\endgroup$ – Marcus Müller May 31 at 13:20

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