# Fourier transform of a modified impulse train

1. Jul 26, 2007

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.

2. Jul 26, 2007

### WolfOfTheSteps

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:

$$x(t)=\hat{x}(t)-\hat{x}(t-T)[/itex] We already know the fourier transform for $\hat{x}(t)$; now we just need to use the time shifting property to find the transform for $\hat{x}(t-T)$: [tex]FT \{\hat{x}(t-T)\} = e^{-j\omega T}\hat{X}(j\omega)$$

Now, by linearity:

$$X(j\omega)=\hat{X}(j\omega)-e^{-j\omega T}\hat{X}(j\omega) =\hat{X}(j\omega)[1-e^{-j\omega T}]$$

And since we are dealing with an impulse train, the only values of $\omega$ we have to deal with are those at:

$$\omega=\frac{\pi n}{T}$$ (since the impulse will be 0 everywhere else).

From this we can write:

$$e^{-j\omega T}=e^{-j\pi n }=(-1)^n$$

So our final answer will be:

$$X(j\omega)=\frac{\pi}{T}\sum^{\infty}_{n=-\infty}\delta\left(\omega-\frac{\pi n}{T}\right)[1-(-1)^n]$$

Last edited: Jul 26, 2007
3. Jul 27, 2007