How can I use the shift property of the Fourier transform to calculate the Fourier transform of an impulse train?

Question

As we all know: $$\mathcal F\{\delta(t)\} = 1$$ and: $$\mathcal F\{x(t-k)\}=X(f)e^{-i2\pi fk}$$

However when I try to use these properties to calculate the Fourier transform (FT) of an impulse train, $x(t) = \sum_{n=-\infty}^{\infty}\delta(t-nT)$, I get: $$\mathcal F\{x(t)\}=\mathcal F\{\sum_{n=-\infty}^{\infty}\delta(t-nT)\} = \sum_{n=-\infty}^{\infty}e^{-i2\pi nTf}$$ As far as I know, the FT of an impulse train should be another impulse train, but the RHS in the expression above doesn't look like it. What am I doing wrong?

Thanks for your time!

Answer 1

The RHS of your equation is the Fourier Series of your desired impulse train.

In other words:

$\Sigma e^{-i2\pi nTf} = \Sigma \delta (f-\frac{n}{T})$

More generally, the Fourier Transform of any periodic function is an impulse train. Here's why:

1. $\mathcal{F}\{e^{i2\pi f_0 t}\}=\delta(f-f_0)$
2. if $f(t)$ is periodic with period $T$, then $f(t)=\Sigma c_n e^{\frac{i2\pi nt}{T}}$

When you take the Fourier Transform of this, you get: $\Sigma c_n \delta(f-\frac{n}{T})$