Fourier transform of a modified impulse train

[naimad]

I need to find the Fourier Transform (FT) of:

$$x(t)=\sum^{\infty}_{n=-\infty}((-1)^{n}\delta(t-nT))$$

Not really sure how to solve this problem, so any help will be appreciated.

Also, if you guys know a good reference for non-uniform sampling and reconstruction, please post it.

 

[WolfOfTheSteps]

I need to find the Fourier Transform (FT) of:
$$x(t)=\sum^{\infty}_{n=-\infty}((-1)^{n}\delta(t-nT))$$

Note: I think my answer below is correct, but I've never found a way to check my answers when finding the FT... So, if someone with more experience could verify this, I would appreciate it!

Notice first that all the impulses at odd values of $n$ are being subtracted, while all the impulses at even values of $n$ are being added. This suggests that we write $x(t)$ as follows:

$$x(t)=\sum^{\infty}_{n=-\infty}\delta(t-2nT) - \sum^{\infty}_{n=-\infty}\delta(t-(2n+1)T)$$

$$=\sum^{\infty}_{n=-\infty}\delta(t-2nT) - \sum^{\infty}_{n=-\infty}\delta(t-2nT-T)$$

Now let

$$\hat{x}(t) = \sum^{\infty}_{n=-\infty}\delta(t-2nT)$$

This is a standard impulse train with period 2T. From any table of basic Fourier Transforms:

$$\hat{X}(j\omega) = \frac{2\pi}{2T}\sum^{\infty}_{n=-\infty}\delta\left(\omega-\frac{2\pi n}{2T}\right) = \frac{\pi}{T}\sum^{\infty}_{n=-\infty}\delta\left(\omega-\frac{\pi n}{T}\right)$$

Now notice that $x(t)$, above, can be written as:

[tex]x(t)=\hat{x}(t)-\hat{x}(t-T)[/itex]<br /> <br /> We already know the Fourier transform for $\hat{x}(t)$; now we just need to use the time shifting property to find the transform for $\hat{x}(t-T)$:<br /> <br /> $$FT \{\hat{x}(t-T)\} = e^{-j\omega T}\hat{X}(j\omega)$$<br /> <br /> Now, by linearity:<br /> <br /> $$X(j\omega)=\hat{X}(j\omega)-e^{-j\omega T}\hat{X}(j\omega)<br /> =\hat{X}(j\omega)[1-e^{-j\omega T}]$$<br /> <br /> And since we are dealing with an impulse train, the only values of $\omega$ we have to deal with are those at:<br /> <br /> $$\omega=\frac{\pi n}{T}$$ (since the impulse will be 0 everywhere else).<br /> <br /> From this we can write:<br /> <br /> $$e^{-j\omega T}=e^{-j\pi n }=(-1)^n$$<br /> <br /> So our final answer will be:<br /> <br /> <br /> $$X(j\omega)=\frac{\pi}{T}\sum^{\infty}_{n=-\infty}\delta\left(\omega-\frac{\pi n}{T}\right)[1-(-1)^n]$$[/tex]

 

[naimad]

Thanks WolfOfTheSteps, I checked it out with a friend and it seems to be correct.

Still, if anybody knows a good reference for non-uniform sampling and reconstruction, please post it.