Ftrepcomb mattmiller - Rhea

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.

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