Fourier transforms explained

14 May 2016

The Fourier transform (FT) is as ubiquitous in engineering as light during the day; nevertheless, this topic seems to remain dark for many students. The purpose of this post is to systematically introduce most common forms of the Fourier transform encountered in a standard signal processing course, explain how they relate to each other, and show how to use them.

Definition of the Fourier transform

Let $s(t)$ be a complex-valued function of time, $s: \mathbb{R} \rightarrow \mathbb{C}$. The FT of $s(t)$ is a complex-valued function of frequency, $S: \mathbb{R} \rightarrow \mathbb{C}$, given by the formula

To begin with, let’s not worry too much about smoothness of the functions $s(t)$ and $S(f)$, as the FT can often be defined in a meaningful way even for rather misbehaved functions. Shorthand notation $S(f) = \mathcal{FT} \!\left\{ s(t) \right\}$ is used throughout to denote the integral in \eqref{eq:FT}.

Delta and identity functions

Delta function. The FT of the delta function is the identity function $S(f) \equiv 1$, as immediately follows from the definition:

$$\begin{equation*} \intInf \delta(t) \me^{-j 2\pi f t} \md t = \me^{-j 2\pi f 0} = 1. \end{equation*}$$

Identity function. Nice, so what about computing the FT of the identity function now? Straightforward integration doesn’t work, therefore we have to consider the identity function as the limit of the rect function with width approaching infinity. The FT of the rect function evaluates to

$$\begin{equation} \label{eq:rect} \mathcal{FT} \! \left\{ \rect \left( \frac{t}{P} \right) \right\} = |P|\sinc(fP), \end{equation}$$

where $\sinc(x) = \sin(\pi x) / (\pi x)$ is the normalized sinc function. The figure below shows how these functions look like for $P = 1$.

Now, imagine that $P$ in \eqref{eq:rect}, that controls the width of the rect function, approaches infinity. The rect function becomes wider approaching $1(t)$, while its FT shrinks, in the limit approaching $\delta(f)$. Thus, the following holds:

$$\begin{equation*} \intInf 1 \me^{-j 2\pi f t} \md t = \delta(f). \end{equation*}$$

Delta function again. We’ve used a neat trick to find the FT of the identity function; namely, we’ve approximated our function by another one for which we can relatively easily compute the FT. It is a very general approach. As an exercise, show that the FT of the delta function is indeed the identity function by considering the delta function as the limit of shrinking rectangles with fixed area.

Inverse Fourier transform

One can extract $s(t)$ from the definition \eqref{eq:FT}, as Fourier suggested:

$$\begin{align*} \intInf S(f) \me^{j 2\pi f t} \md f &= \intInf s(\tau) \intInf \me^{-j 2\pi f (\tau-t)} \md f \md \tau \\ &= \intInf s(\tau) \delta(\tau-t) \md \tau = s(t). \end{align*}$$

Leaving out the intermediate steps, we get the following formula for the inverse Fourier transform (IFT):

You can now convince yourself with little effort that the following bidirectional relationships hold:

$$\begin{align*} \delta(t) &\overset{\mathcal{FT}}{\longleftrightarrow} 1(f), \\ \rect \left( \frac{t}{P} \right) &\overset{\mathcal{FT}}{\longleftrightarrow} |P|\sinc(fP). \end{align*}$$

Let’s now check our understanding of the FT and IFT on a real-world example.

Dirac comb

The Dirac comb $\comb_T(t)$ consists of delta functions spaced at intervals of $T$:

$$\begin{equation*} \comb_T(t) = \sumkInf \delta(t - kT) = \frac{1}{T} \comb \left( \frac{t}{T} \right). \end{equation*}$$

Subscript $T$ is customarily omitted when $T = 1$. The figure below gives an idea what this “function” looks like (it is in fact a periodic tempered distribution rather than a function, but let’s not bother).

Its Fourier transform is given by

$$\begin{equation*} \intInf \comb_T(t) \me^{-j 2\pi f t} \md t = \sumnInf \me^{-j 2\pi f n T}. \end{equation*}$$

An infinite sum of complex exponentials is not the most intuitive object to work with, so let’s have a closer look at a symmetrically truncated version of it to get more insight. Assume for simplicity that $N$ is even (the case when $N$ is odd is left as an exercise for the reader), then

$$\begin{align*} \sum_{n=-\frac{N}{2}}^{\frac{N}{2}-1} \me^{-j 2\pi f n T} &= \sum_{n=-\frac{N}{2}}^{\frac{N}{2}-1} q^n = q^{-\frac{N}{2}} \frac{1-q^N}{1-q} = \frac{ q^{-\frac{N}{2}} - q^{\frac{N}{2}} } { q^{\frac{1}{2}} ( q^{-\frac{1}{2}} - q^{\frac{1}{2}} ) } = \me^{j \pi f T} \frac{ \sin(\pi f N T) }{ \sin(\pi f T) }. \end{align*}$$

The absolute value of this truncated series is shown in the following figure.

Pay attention to zeros of the nominator and the denominator. The nominator equals zero at multiples of $1/(NT)$. In our example, these are the points $f = 0, \pm 1/4, \pm 2/4, \pm 3/4,$ and so on. The denominator equals zero less often—at multiples of $1/T$—giving us $f = 0, 1, 2,$ and so on. When $f=0$, the ratio is equal to $N$. Imagine the behavior of this function when $N$ goes to infinity. The peaks become thinner and taller, and the amount of zeros between the peaks increases. Guess what we get in limit? Right—another Dirac comb!

$$\begin{equation} \label{eq:sum_of_exp} \sumnInf \me^{-j 2\pi n f T} = \sumnInf \delta(fT - n). \end{equation}$$

Isn’t that amazing? We’ve just shown that an infinite sum of complex exponentials equals an infinite sum of delta functions. An ancient philosopher would have been infinitely proud of himself for having shown something like this, but we just modestly rejoice and move on.

In \eqref{eq:sum_of_exp}, the relationship is given in the frequency domain. A mere substitution, however, shows that it is equally valid in the time domain as well:

$$\begin{equation*} \sumkInf \me^{j 2\pi k \frac{t}{T}} = \sumkInf \delta(\frac{t}{T} - k). \end{equation*}$$

Summing up the results for the Dirac comb. We have established equality between two representations of a train of delta functions,

and we have computed the FT of a Dirac comb, which turned out to be a Dirac comb again,

$$\begin{align*} \comb_T(t) &\overset{\mathcal{FT}}{\longleftrightarrow} \frac{1}{T} \comb_{\frac{1}{T}}(f). \end{align*}$$

The Dirac delta function is the bridge between the continuous world and the discrete world. We are now ready to enter the realm of digital signal processing.

To be continued…