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$B(z)+B(-z) = 2c$, explain the structure of $b[n]$ and find the constraint of its length given that $c$ cannot be $0$.

This is a homework problem. "Explain the structure" means that $b[n]$ is zero for certain values of $n$, and has a certain shape.

I'm trying to take the inverse $\mathcal Z$-transform of $B(z)$ and $B(-z)$, but I'm not sure how the inverse $\mathcal Z$-transform of $B(-z)$ is related to $b[n]$, so I'm stuck...can anyone give me some advice?

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Just use the definition of the $\mathcal{Z}$-transform:

$$B(z)=\sum_{n=-\infty}^{\infty}b[n]z^{-n}\tag{1}$$

from which it follows that

$$B(-z)=\sum_{n=-\infty}^{\infty}b[n](-z)^{-n}=\sum_{n=-\infty}^{\infty}b[n](-1)^nz^{-n}\tag{2}$$

From $(2)$, the sequence that corresponds to $B(-z)$ is $b[n](-1)^n$. Now taking the inverse $\mathcal{Z}$-transform of

$$B(z)+B(-z)=2c\tag{3}$$

gives

$$b[n]+b[n](-1)^n=2c\delta[n]\tag{4}$$

For odd $n$, the left side of $(4)$ equals zero. For even $n$ it equals $2b[n]$. Consequently, equation $(4)$ can only be satisfied if $b[n]=0$ for even $n$ except for $n=0$, where we require $b[0]=c$:

$$b[n]=\begin{cases}c,&n=0\\ 0,&n\text{ even}\end{cases}$$

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  • $\begingroup$ Thanks a lot! The question also asks if there's any constraint on the length of $b[n]$. From your answer, I think there is no constraint on the length, since $b[n]$ is not $0$ when $n$ is odd? $\endgroup$ – q85ts Jul 3 '16 at 21:37
  • $\begingroup$ @qwt185: that's right, no length constraint. $\endgroup$ – Matt L. Jul 4 '16 at 7:30

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