# Build a filter such that its periodized version equals to x on signals with period m

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?

• 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. – Marcus Müller May 31 at 12:42
• 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 – darktealeaf May 31 at 13:15
• 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. – Marcus Müller May 31 at 13:20