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:

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)&lt;br /&gt; =\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]

 

[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.