Generality of integral Fourier transforms

As given here, different types of Fourier expressions are defined for different types of input functions.

The most general expression is the Fourier Transform. Given a continuous aperiodic function $x(t)$ , its analysis and synthesis equations are:

$X(\omega) = \int\limits_{-\infty}^{\infty}x(t)e^{-j\omega t}\,dt$
$x(t) = \frac{1}{2\pi}\int\limits_{-\infty}^{\infty}X(\omega)e^{j\omega t}\,d\omega$

In general, the Fourier transform is aperiodic and continuous.

In this article, we will see how the Fourier transform definition above expresses the other related definitions, namely the Discrete-time Fourier transform and the Fourier series.

In this article, we refer to Fourier transform as integral Fourier transform to differentiate it explicitly from other types of Fourier transforms.

The characteristics of the input function that determine the different definitions are periodicity and discreteness.

Firstly, consider an input signal with Dirac delta “functions” separated by a constant time $T$ . Thus,

$x(t) = \sum\limits_{k=-\infty}^{\infty}x(kT)\delta(t-kT)$

Its integral Fourier transform $X(\omega)$ is calculated as:

$\begin{aligned} X(\omega) &= \int\limits_{-\infty}^{\infty}x(t)e^{-j\omega t}\,dt \\ &=\int\limits_{-\infty}^{\infty}\left(\sum\limits_{k=-\infty}^{\infty}x(kT)\delta(t-kT)\right)e^{-j\omega t}\,dt \\ &= \sum\limits_{k=-\infty}^{\infty}x(kT)\left(\int\limits_{-\infty}^{\infty}\delta(t-kT)e^{-j\omega t}\,dt\right) \\ &= \sum\limits_{k=-\infty}^{\infty}x(kT)\left(\int\limits_{-\infty}^{\infty}\delta(t-kT)e^{-j\omega kT}\,dt\right) \mbox{ because }\delta(t-kT) \ne 0 \mbox{ only at } t=kT \\ &= \sum\limits_{k=-\infty}^{\infty}x(kT)e^{-j\omega kT}\left(\int\limits_{-\infty}^{\infty}\delta(t-kT)\,dt\right) \\ &= \sum\limits_{k=-\infty}^{\infty}x(kT)e^{-j\omega kT}\end{aligned}$

We make three observations on the integral Fourier transform of $\sum\limits_{k=-\infty}^{\infty}x(kT)\delta(t-kT)$ :

$X\left(\omega+\frac{2\pi}{T}\right) = X(\omega)$ , so that $X(\omega)$ is periodic with period $\frac{2\pi}{T}$ . Thus, if the original function contains discrete impulses, its integral Fourier transform is periodic.
The integral Fourier transform is periodic only when the input function has Dirac deltas separated by a constant time $T$ . This implies that an input function with (non-zero scaled) Dirac deltas at values that are rational multiples of each other has a periodic integral Fourier transform. On the other hand, if the input function has non-zero scaled Dirac deltas at values that are not rational multiples of each other will not have a periodic integral Fourier transform. Thus, $\pi\delta(t-7)+2\delta(t-9) + \sqrt{2}\,\delta\left(t-\frac{11}{3}\right)$ and $\delta(t-7\sqrt{2}) + \delta(t-23\sqrt{2})$ have periodic integral Fourier transforms, while $\delta(t-2) + \delta(t-\sqrt{2})$ does not have a periodic integral Fourier transform.
If $T = 1$ , then $x(t) = \sum\limits_{n=-\infty}^{\infty}x(n)\delta(t-n)$ . And, $X(\omega) = \sum\limits_{n=-\infty}^{\infty}x(n)e^{-j\omega n}$ , which is nothing but the discrete-time Fourier transform for a discrete function $x[n]$ defined only at integer values of $n$ . (Such a discrete-function takes real-values for each $n$ , rather than Dirac deltas like its continuous-time counterpart $x(t)$ ). Thus, given a discrete-time function, its discrete-time Fourier transform is identical to the integral Fourier Transform of its continuous-time counterpart that contains Dirac deltas appropriately scaled (by $x[n]$ ), separated by unit time.

Next, consider a continuous-time periodic function $x(t)$ .

A continuous-time periodic function has a Fourier series expansion (if $x(t)$ meets certain conditions) as $x(t) = \sum\limits_{k=-\infty}^{\infty}a_ke^{jk\omega_0t}$ , where $\omega_0$ is the fundamental frequency of $x(t)$ .

Now, if $X(\omega)$ is the integral Fourier transform of $x(t)$ ,

Now, $\int\limits_{-\infty}^{\infty}e^{jk\omega_0t}e^{-j\omega t}\,dt$ is the integral Fourier transform of $e^{jk\omega_0t}$ . By properties of Fourier transform, we know that if the Fourier transform of $f(t)$ is $F(\omega)$ , the Fourier transform of $e^{j\omega_0 t}f(t)$ is $F(\omega-\omega_0)$ . Thus, since we know the Fourier transform of the constant function $1$ is $2\pi\delta(\omega)$ , the Fourier transform of $e^{jk\omega_0t}$ would be $2\pi\delta(\omega-k\omega_0)$ .

Thus, continuing the above calculation,

$\begin{aligned} X(\omega) &= \sum\limits_{k=-\infty}^{\infty}a_k\left(\int\limits_{-\infty}^{\infty}e^{jk\omega_0t}e^{j\omega t}\,dt\right) \\ &= 2\pi\sum\limits_{k=-\infty}^{\infty}a_k\delta(\omega-k\omega_0)\end{aligned}$

Thus, the integral Fourier transform of a periodic function consists of appropriately scaled Dirac deltas separated by the fundamental frequency $\omega_0$ . The scale factors used for the Dirac delta at a particular frequency value is just $2\pi$ times the corresponding Fourier series coefficient $a_k$ .

Note that in this case, the (non-zero scaled) Dirac deltas in the integral Fourier transform occur at frequency values that are rational multiples of each other. This observation is “dual” to the observation where for an input signal with (non-zero scaled) Dirac deltas occurring at time values that are rational multiples of each other, the integral Fourier transform is periodic.

In this article, we saw how the discrete-time Fourier transform of a discrete-time function can be regarded as the integral Fourier transform of the counterpart continuous-time signal that contains appropriately scaled Dirac deltas separated by unit time. We also saw how the integral Fourier transform of a continuous-time periodic function is just Dirac deltas separated in frequency by the fundamental frequency value and scaled by $2\pi$ times the corresponding Fourier series coefficients.