Relationship between z-transform and DFT I'm studying for a Signals Processing exam and came across an exercise that I'm finding pretty difficult to solve. It says:


  Asume there is a signal $x[n]$ of length $N$. Its $\mathcal{Z}$-Transform is $X(z)$. Find the expressions of the DFTs of the next signals using a
  correct sampling of $X(z)$.
  
  a) $x_{1}[n]=x[n]$  if  $0\le n \le N-1$ and $x_{1}[n]=0$  if  $N\le n \le 2N-1$
  
  b) $x_{1}[n]=x[n]+x[n-N]$


For a), I thought of expressing  $X[k] = X(z) \bigg|_{z=e^{j\frac{2\pi}{N} k}}$, i.e. the $\mathcal{Z}$-Transform evaluated in $z=j\frac{2\pi}{N} k$ is the DFT of $x[n]$. Due to the zeros added at the end of the original signal, I arrived at the conclusion that  $X[k] = X(z) \bigg|_{z=e^{j\frac{2\pi}{2N} k}}$. Namely, the $\mathcal{Z}$-Transform is now sampled in $2N$ points around the unity circle. Is this reasoning right?

And regarding exercise b), I have absolutely no idea how to solve it.

Thanks for your time!
For a) you're correct. For b), $x_1$ is a length $2N$ signal, and its DFT is given by

$$X_1[k]=\sum_{n=0}^{2N-1}x_1[n]e^{-j2\pi kn/2N}=\sum_{n=0}^{2N-1}x_1[n]e^{-j\pi kn/N}\tag{1}$$

With $x_1=x[n]+x[n-N]$ you get

$$\begin{align}X_1[k]&amp;=\sum_{n=0}^{N-1}x[n]e^{-j\pi kn/N}+\sum_{n=N}^{2N-1}x[n-N]e^{-j\pi kn/N}\\&amp;=X\left(e^{j\pi k/N}\right)+\sum_{n=0}^{N-1}x[n]e^{-jk\pi(n+N)/N}\\&amp;=X\left(e^{j\pi k/N}\right)+e^{-jk\pi}X\left(e^{j\pi k/N}\right)\\&amp;=X\left(e^{j\pi k/N}\right)(1+(-1)^k)=\begin{cases}2X\left(e^{j\pi k/N}\right),&amp;k \text{ even}\\
0&amp;k\text{ odd}\end{cases}\tag{2}\end{align}$$

Eq. $(2)$ shows that decimation in the frequency domain corresponds to aliasing in the time domain.