naimad said:
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 />
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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 />
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FT \{\hat{x}(t-T)\} = e^{-j\omega T}\hat{X}(j\omega)<br />
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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 />
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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 />
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From this we can write:<br />
<br />
e^{-j\omega T}=e^{-j\pi n }=(-1)^n<br />
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So our final answer will be:<br />
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<br />
X(j\omega)=\frac{\pi}{T}\sum^{\infty}_{n=-\infty}\delta\left(\omega-\frac{\pi n}{T}\right)[1-(-1)^n]
