Fourier Transform of Rep and Comb Functions

A slecture by ECE student Matt Miller

Partly based on the ECE438 Fall 2014 lecture material of Prof. Mireille Boutin.

In this slecture, we are going to look at the fourier transforms of the $comb_T()$ and $rep_T()$ functions.

First, we will define the a function $p_T(t)$ as pulse train function, or series of time-shifted impulses:

\begin{align} p_T(t) = \sum_{n=-\infty}^{\infty}\delta(t-kT) \end{align}

Now let's take a look at the $comb_T()$ function. By definition:

\begin{align}comb_T(x(t)) :&= x(t)p_T(t) \\ &= x(t)\sum_{k=-\infty}^{\infty}\delta(t-kT) \\ &= \sum_{k=-\infty}^{\infty}x(t)\delta(t-kT) \\ &= \sum_{k=-\infty}^{\infty}x(kt)\delta(t-kT) \end{align}

The result of this function is a set of time-shifted impulses whose amplitudes match those of the input signal x(t) at each given point. This signal can be referred to as a sampling of x(t) with a sampling rate of 1/T.

Before we get started with the fourier transform of this function, lets take a look at the $rep_T()$ function, which is very similar to the $comb_T()$ function:

\begin{align}rep_T(x(t)) :&= x(t)*p_T(t) \\ &= x(t)*\sum_{k=-\infty}^{\infty}\delta(t-kT) \\ &= \sum_{k=-\infty}^{\infty}x(t)*\delta(t-kT) \\ &= \sum_{k=-\infty}^{\infty}x(t-kT)\end{align}

As shown above, the $rep_T()$ function differs from the $comb_T()$ function in that it convolves the input signal with the impulse train rather than simply multiplying the two.

Now that we've seen the $rep_T()$ function, we can go back to the fourier transform of the $comb_T()$ function:

${\mathcal F}(comb_T(x(t))) = {\mathcal F}(x(t)p_T(t))$

Using the multiplication property:

\begin{align} &= {\mathcal X}(f)*{\mathcal F}(p_T(t)) \\ &= {\mathcal X}(f)*{\mathcal F}(\sum_{n=-\infty}^{\infty}\frac{1}{T}e^{j{\frac{2 \pi}{T}}nt}) \\ &= {\mathcal X}(f)*\sum_{n=-\infty}^{\infty}\frac{1}{T}{\mathcal F}(e^{j{\frac{2 \pi}{T}}nt}) \\ &= {\mathcal X}(f)*\frac{1}{T}\sum_{n=-\infty}^{\infty}\delta(f - \frac{n}{T}) \\ &= \frac{1}{T}{\mathcal X}(f)*{\mathcal P}_{\frac{1}{T}}(f)\end{align}

Now, knowing that a signal convolved with a pulse train results in a $rep_T()$ function, we can simplify this as:

${\mathcal F}(comb_T(x(t))) = \frac{1}{T}rep_{\frac{1}{T}}({\mathcal X}(f))$

Simply put, the fourier transform of a comb function is a rep function. Now let's see if the inverse is true:

${\mathcal F}(rep_T(x(t))) = {\mathcal F}(x(t)*p_T(t))$

Using the convolution property:

\begin{align} &= {\mathcal X}(f){\mathcal F}(p_T(t)) \\ &= {\mathcal X}(f){\mathcal F}(\sum_{n=-\infty}^{\infty}\frac{1}{T}e^{j{\frac{2 \pi}{T}}nt}) \\ &= {\mathcal X}(f)\sum_{n=-\infty}^{\infty}\frac{1}{T}{\mathcal F}(e^{j{\frac{2 \pi}{T}}nt}) \\ &= {\mathcal X}(f)\frac{1}{T}\sum_{n=-\infty}^{\infty}\delta(f - \frac{n}{T}) \\ &= {\mathcal X}(f)\frac{1}{T}{\mathcal P}_{\frac{1}{T}}(f) \\ &= \frac{1}{T}comb_{\frac{1}{T}}({\mathcal X}(f))\end{align}

As shown above, we now know that the fourier transform of a rep is comb. This the duality of the two functions between the time and frequency domains; one in the time domain is the other in the frequency domain, or vise-versa. One difference that it is important to remember, however, is that the resulting fourier transform's impulse-train period is the inverse of the original in the time domain. The frequency domain signal is also amplitude scaled by the inverse of the period T.