2
$\begingroup$

Given the discrete signal $x(n)=\begin{bmatrix} \alpha ^n, n\geq 0 \\0, n<0 \end{bmatrix}$ where $\alpha \in (-1,1)$ and some natural number $N$, we know that the discrete signal $y(n)$ (where $0 \leq n \leq N-1$) holds the following equality: $Y(k)=X(\frac{k}{N})$ where $Y$ is the fourier transform of $y$ and $X$ is the fourier transform of $x$.

Find an expression for $y$.

What I did:

The idea is to perform a fourier transform on $x(n)$ to get $X(k)$, and then find the expression $X(\frac{k}{N})$ and perform an inverse fourier transform on it, to find $y(n)$.

In class we defined the fourier transform of a discrete signal as the fourier transform of the continuous model of the discrete signal. where the continuous model $\tilde x$ is defined as such: $\tilde x(t)=\sum_{n \in \mathbb Z}x(n)\delta(t-nt)$ where $\delta$ is the dirac delta function.

The fourier transform of the discrete signal, is the fourier transform of the continuous model, so:

$X(k)=\int_{\mathbb R}[\sum_{n \in \mathbb Z}x(n)\delta(t-nt)e^{-2\pi i k t}]dt$ but $x(n)=0$ if $n<0$, so the integral is equal to $\int_{\mathbb R}[\sum_{n \in \mathbb N}x(n)\delta(t-nt)e^{-2\pi i k t}]dt=\sum_{n \in \mathbb N}[\alpha ^n\int_{\mathbb R}\delta(t-nt)e^{-2\pi i k t}dt]$

We can see that $\int_{\mathbb R}\delta(t-nt)e^{-2\pi i k t}dt$ is nothing more than the fourier transform of $\delta(t-nt)$, but how can we evaluate that?

I know the fourier transform of $\delta(t-t_0)$ is $e^{-2\pi i k t_0}$ but this is only true if $t_0$ is a scalar. This is not the case in our question...Would appreciate help at evaluating this integral.

$\endgroup$
  • $\begingroup$ You could use $\delta(ax)=\frac{1}{|a|}\delta(x)$ $\endgroup$ – tired Apr 25 '15 at 21:09
  • $\begingroup$ In continuous model shouldn't it be $\delta(t-nT)$ where $T$ is the sampling interval? Then you can use the fact you stated. $\endgroup$ – dioid Apr 25 '15 at 22:13
1
$\begingroup$

The Fourier of a sum is the sum of the Fouriers; that is what makes the Fourier transform a linear operation. Thus, $$\mathcal{F}[\Sigma_{n\in\mathbb{Z}}x(n)\delta(t-nt_0)] = \Sigma_{n\in\mathbb{Z}}x(n)e^{-2\pi iknt_0} $$ In other words, the Fourier transform of a series of modulated Dirac deltas is the sum of the transforms of the individual deltas.

(I changed the notation from $\delta(t-nt)$ to $\delta(t-nt_0)$ since I'm assuming that as $n$ increases we're going forward in time in discrete steps, and not moving around inside a single Dirac delta's zero zone, changing the $n$ of $\delta(t\cdot(1-n))$.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.