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I want to generate the following periodic sequence: h[n]={1,2,3,4,0,0,0,0,1,2,3,4,0,0,0,0,1,2...}. I am planning to generate this as an impulse response of a filter with an order no more than 8.

I start by finding the z transform of h[n] but this is infinite sequence and can't achieve an order less than 8.

Any hint will be appreciated.

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You need a system with order $8$, because the period of the desired sequence is $8$. Obviously, you'll need a recursive system to generate such an infinite sequence. Since this looks like another homework problem, I won't write out the complete solution but I'll give you a few hints:

Let $g[n]=\{1,2,3,4,0,0,0,0\}$ be the first period of $h[n]$. Now note that

$$g[n]=h[n]-h[n-8]\tag{1}$$

The $\mathcal{Z}$-transform of (1) is

$$G(z)=H(z)(1-z^{-8})\tag{2}$$

from which

$$H(z)=\frac{G(z)}{1-z^{-8}}\tag{3}$$

follows. Now it only remains to find the $\mathcal{Z}$-transform $G(z)$ of $g[n]$. Then you can use Eq. (3) to compute $H(z)$ from $G(z)$. If you realize a filter with transfer function $H(z)$ (which is a recursive filter of order $8$), then its response to a unit impulse will equal the desired sequence $h[n]$.

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