Suppose that the fourier transform of a signal $x(t)$ is $\hat x(u)=\frac{1}{2u_m}(1+\cos (\frac{\pi u}{u_m}))$ where $u_m \geq |u|$.$t$ here stands for time so $t \geq 0$

We sample the original signal such that the interval between two samples is $T=\frac{1}{u_m}$.

We are asked to find the sampled signal.

I have an idea of how a solution would work, but I'm having difficulties at the first step.

My idea is to find the original $x(t)$ by performing an inverse fourier transform, and then the sampled signal would be $s(n)=x(\frac{n}{u_m})$ for $n=0,1,2,3,...$.

But finding the inverse is somewhat difficult.

$x(t)=\int_{\mathbb R}\hat x(u)e^{2\pi i u t}du=\int_{\mathbb R}\frac{1}{2u_m}(1+\cos (\frac{\pi u}{u_m}))e^{2\pi i u t}du$

so $2u_m x(t)=\int_{\mathbb R}e^{2\pi i u t}du+\int_{\mathbb R}\cos(\frac{ \pi u}{u_m})e^{2\pi i u t}du$. This is supposed to be some function of $t$. but in reality, it isnt. just looking at the first integral we can see that this is going to be infinity regardless of the value of $t$.

So given that the fourier transform of some function $x(t)$ is $\hat x(u)=\frac{1}{2u_m}(1+\cos (\frac{\pi u}{u_m}))$, how do I find $x(t)$?

  • $\begingroup$ Take care that $\int e^{2\pi ut}\mathrm{d}u\neq+\infty$! You must understand these integrals as distributions. See this good article en.wikipedia.org/wiki/… at "Tables of important Fourier transforms" to get your answer. $\endgroup$ – Nicolas Apr 25 '15 at 15:58

$$\frac{1}{2u_{m}}\left(1+\cos\left( \frac{\pi u}{u_{m}}\right)\right) = \frac{1}{2u_{m}} + \frac{1}{4u_{m}}\left(e^{i\pi u/u_{m}}+e^{-i\pi u/u_{m}} \right) = \frac{e^{-i\pi u/u_{m}}}{4u_{m}} - \frac{e^{i\pi 0/u_{m}}}{2u_{m}} + \frac{e^{i\pi u /u_{m}}}{4u_{m}}$$

View this is as a Fourier Series with the majority of the Fourier coefficients equal to zero.


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