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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!

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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

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

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

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  • $\begingroup$ Thank you so much! One last question: what do you mean by "decimation in the frequency domain corresponds to aliasing in the time domain"? I mean, I know what every word in that sentence means, I just don't see it in Eq (2). For the second signal, its DFT is clearly a decimation of the original one, that's clear. But why does that correspond to aliasing in the time domain? $\endgroup$ – Tendero Jan 26 '16 at 13:50
  • $\begingroup$ @M.S.: I mean that the DFT values of $X[k]$ are decimated in the sense that every other sample is omitted (and then replaced by a zero), according to Eq. (2). In the time domain this corresponds to an addition of a shifted version of $x[n]$, i.e., $x[n]+x[n-N]$, which is basically aliasing in the time domain, even though the shifted versions don't overlap. $\endgroup$ – Matt L. Jan 26 '16 at 15:37
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here is the relationship in a quick and concise (or terse) manner:

discrete-time signal: $x[n]$ where $n$ is an integer.

Z transform:

$$ X(z) \triangleq \sum\limits_{n=-\infty}^{+\infty} x[n] \ z^{-n} $$

Discrete-time Fourier transform (DTFT):

$$\begin{align} X\left(e^{j\omega}\right) \ &= \ X(z) \Big|_{z=e^{j\omega}} \\ &= \ \sum\limits_{n=-\infty}^{+\infty} x[n] \ e^{-j \omega n} \\ \end{align}$$

that is how the DTFT is related to the Z transform.

DFT:

$$\begin{align} X[k] &= X\left(e^{j\omega}\right)\Big|_{\omega = \frac{2 \pi k}{N}} \quad \text{where } x[n]=0 \text{ for } n<0 \text{ or }n\ge N \\ &= \ \sum\limits_{n=-\infty}^{+\infty} x[n] (u[n]-u[n-N]) \ e^{-j \frac{2 \pi k}{N} n} \\ &= \ \sum\limits_{n=0}^{N-1} x[n] \ e^{-j \frac{2 \pi k}{N} n} \\ \end{align}$$

here $k$ is an integer and $u[n]$ is the discrete unit step function.

$$ u[n] \triangleq \begin{cases} 1 \quad \text{for } n \ge 0 \\ 0 \quad \text{for } n < 0 \end{cases} $$

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  • $\begingroup$ Thanks for the answer, but that was not what I was looking for, I'm sorry :/ I was trying to get an answer for the specific problem I wrote above, as it has some zero-padding involved and it's not a standard DFT-ZT relationship question $\endgroup$ – Tendero Jan 26 '16 at 5:08

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