- Find $\mathscr{Z}$ transform of the following discrete signal: $$x[n]=\begin{cases}\left(\displaystyle\frac{1}{2}\right)^{-n}&\text{if $n$ is a multiple of 3},\\1&\text{otherwise}.\end{cases}$$
- Find radius and graph radius of convergence of the $\mathscr{Z}$ transform of the previous signal.
I know that $$\mathscr{Z}(x[n])=\sum_{n=0}^\infty x[n]z^{-n}.$$ and $$x[n]=\begin{cases}\left(\displaystyle\frac{1}{2}\right)^{-n}&n=3k,\\1&n\neq3k.\end{cases}$$ However, the "$1$" is bothering me.
If I had this signal: $$y[n]=\begin{cases}\left(\displaystyle\frac{1}{2}\right)^{-n}&n=3k,\\\color{red}0&n\neq3k.\end{cases}$$ then, as $\{y[n]\}_{n\in\Bbb N}=\{1,0,0,2^3,0,0,2^6,0,0,2^9,\ldots\}$ we can say: $$\mathscr{Z}(y[n])=\sum_{n=0}^\infty y[3n]z^{-3n}=\sum_{n=0}^\infty\left(\frac{1}{2}\right)^{-3n}z^{-3n}=\sum_{n=0}^\infty\left(\frac{8}{z^3}\right)^{n}=\frac{1}{1-\frac{8}{z^3}}=\boxed{\frac{z^3}{z^3-8}},$$ and $|z|<2$.
But here we have $\{x[n]\}_{n\in\Bbb N}=\{1,1,1,2^3,1,1,2^6,1,1,2^9,\ldots\}$ and I do not know how to interpret the $1$'s. Maybe the transform is $\fbox{$(\sum_{n=0}^\infty x[3n]z^{-3n})+1=z^3/(z^3+8)+1$}$?
Thanks!!
You are almost there. You can write the $\mathscr{Z}$-transform of $x[n]$ as $$ \mathscr{Z}(x[n]) = \sum_{n= 0}^{\infty} x[3n] z^{-3n} + \sum_{n= 0}^{\infty} x[3n + 1] z^{-(3n+1)} + \sum_{n= 0}^{\infty} x[3n+2] z^{-(3n+2)}$$
You can deal with the first part as you have done for $y[n]$ and the second and the third part are simply the $\mathscr{Z}$-transform of the constant sequence $u[n] = 1 \ \forall \ n$ evaluated at $z^3$ and offset by $z^{-1}$ and $z^{-2}$ respectively. Just evaluate them all and add them up. I believe the final answer should be $\displaystyle \mathscr{Z}(x[n]) = \frac{z^3}{z^3 - 8} + \frac{z^2 + z}{z^3 - 1}$ and the region of convergence would be $|z| > 2$. Note that you have flipped the sign in the region of convergence in the question.
Hope this helps.