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.

  • $\begingroup$ 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

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 $\delta(t\cdot(1-n))$.)


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