Fourier Series and Fourier Transform - III

This is the third part of the series on Fourier Series and Fourier Transform. In the first two parts, we have discussed the Fourier Series. From this part onwards, we will forcus on the Fourier Transform. In a nutshell, Fourier transform could be thought of as the extension of Fourier Series to the continuous domain, which means that we could extend the Fourier Series to the functions which are not periodic.

Review of Fourier Series

For a periodic function $f(x)$ with period $1$ (you can normalize the period to $1$ by scaling the $x$-axis), we can write it as a Fourier Series:

$f(x) = \sum_{n=-\infty}^{\infty} c_n e^{2\pi i n x}$

where $c_n$ is the Fourier coefficient, which is the projection of $f(x)$ onto the basis $e^{2\pi i n x}$:

$c_n =\langle f(x), e^{2\pi i n x} \rangle = \int_0^1 f(x) e^{-2\pi i n x} dx$

where $c_n$ is the Fourier coefficient, which is the projection of $f(x)$ onto the basis $e^{2\pi i n x}$.

Conside a square wave of period $1$, which is defined as:

$f(x) = \begin{cases} 1 & 0 \leq x < \frac{1}{2} \\ -1 & \frac{1}{2} \leq x < 1 \end{cases}$

Figure 1. Illustration of square wave.

We have calculated the Fourier coefficients of the square wave in the previous part:

$\begin{aligned} f(x) & = \sum_{n \text{ is odd}} \frac{2}{\pi i n}e^{2\pi i n x} \\ & = \sum_{-\infty}^{\infty} \frac{2}{\pi i (2k+1)}e^{2\pi i (2k+1) x} \\ & = \sum_{-\infty}^{\infty} \frac{2}{\pi i (2k+1)}(e^{2\pi i (2k+1) x} - e^{-2\pi i (2k+1) x}) \\ & = \sum_{k=0}^{\infty} \frac{4}{\pi (2k+1)} \sin[2\pi (2k+1) x] \\ & = \frac{4}{\pi} \sum_{k=0}^{\infty} \frac{1}{2k+1} \sin[2\pi (2k+1) x] \end{aligned}$

The above formula shows the fourier coefficients of the square wave is zero for even $n$ and non-zero for odd $n$. And each component of the Fourier Series is a sine wave with frequency $2k+1$. The following figure shows how the square wave is constructed by the Fourier Series.

Figure 2. Illustration of Fourier Series (this figure was taken from TikZ.net by Izaak Neutelings).

This gives us a hint that we can use Fourier Series to decompose a periodic function into a series of sine waves with different frequencies. Now, we want to extend this idea to the non-periodic functions.

Fourier Transform

We now define the Fourier Transform of a function $f(t)$ as:

$\hat{f}(s) = \int_{-\infty}^{\infty} f(t) e^{-2\pi i s t} dt$

For any $s \in \mathbb{R}$, we can define a function $g_s(t) = e^{-2\pi i s t}$. Then, we can write the Fourier Transform as:

$\hat{f}(s) = \langle f(t), g_s(t) \rangle$

which is the projection of $f(t)$ onto the basis $g_s(t)$. Moreover, for any $s \in \mathbb{R}$, integrating $f(t)$ against the complex-valued function $g_s(t)$ gives us a complex-valued function $\hat{f}(s)$ of $s$.

The inverse Fourier Transform is defined as:

$f(t) = \int_{-\infty}^{\infty} \hat{f}(s) e^{2\pi i s t} ds$

Now, let’s check an example. Consider the triangle function, defined by

$\wedge (t) = \begin{cases} 1 - |t| & |t| \leq 1 \\ 0 & |t| > 1 \end{cases}$

For the Fourier transform,

$\begin{aligned} \mathcal{F}[\wedge(t)] & = \int_{-\infty}^{\infty} \wedge(t) e^{-2\pi i s t} dt \\ & = \int_{-1}^{1} (1 - |t|) e^{-2\pi i s t} dt \\ & = \int_{-1}^{0} (1 + t) e^{-2\pi i s t} dt + \int_{0}^{1} (1 - t) e^{-2\pi i s t} dt \end{aligned}$

Now, let’s consider the first integral:

$A(s) = \int_{-1}^{0} (1 + t) e^{-2\pi i s t} dt$

Let $u = -t$, then $du = -dt$. Then, we have:

$\begin{aligned} A(s) & = \int_{1}^{0} (1 - u) e^{2\pi i s u} (-du) \\ & = \int_{0}^{1} (1 - u) e^{2\pi i s u} du \\ \end{aligned}$

This gives us:

$A(-s) = \int_{0}^{1} (1 - u) e^{-2\pi i s u} du$

Thus,

$\mathcal{F}[\wedge(t)] = A(s) + A(-s)$

Now, let’s consider the first integral:

$\begin{aligned} A(s) & = \int_{-1}^{0} (1 + t) e^{-2\pi i s t} dt \\ & = [(1+t) \frac{e^{-2\pi i s t}}{-2\pi i s}]_{-1}^{0} - \int_{-1}^{0} \frac{e^{-2\pi i s t}}{-2\pi i s} dt \\ & = \frac{1}{2\pi i s} - \frac{1}{-2\pi i s} \int_{-1}^{0} e^{-2\pi i s t} dt \\ & = \frac{1}{2\pi i s} - [ \frac{1}{-4\pi^2 s^2} e^{-2\pi i s t}]_{-1}^{0} \\ & = \frac{1}{2\pi i s} - \frac{1}{4\pi^2 s^2} + \frac{1}{4\pi^2 s^2} e^{2\pi i s} \\ & = - \frac{1}{2 \pi is } + \frac{1}{4\pi^2 s^2} (1 - e^{2\pi i s}) \end{aligned}$

Then

$\begin{aligned} \mathcal{F}[\wedge(t)] & = A(s) + A(-s) \\ & = - \frac{1}{2 \pi is } + \frac{1}{4\pi^2 s^2} (1 - e^{2\pi i s}) - \frac{1}{2 \pi i (-s) } + \frac{1}{4\pi^2 (-s)^2} (1 - e^{-2\pi i s}) \\ & = \big ( \frac{\sin \pi s}{\pi s} \big )^2 \\ & = \text{sinc}^2(\pi s) \end{aligned}$

Here is the graph of the Fourier transform of the triangle function:

As you can see, we are transofrming a function from the time domain to the frequency domain. Both functions are continuous.

After introducing the Fourier transform, we will list some properties of the Fourier transform.

• linearity: $\mathcal{F}[af(t) + bg(t)] = a\mathcal{F}[f(t)] + b\mathcal{F}[g(t)]$
• time shift: $\mathcal{F}[f(t - t_0)] = e^{-2\pi i s t_0} \mathcal{F}[f(t)]$
• stretch: $\mathcal{F}[f(at)] = \frac{1}{|a|} \mathcal{F}[f(t)]$
• derivative: $\mathcal{F}[f'(t)] = 2\pi i s \mathcal{F}[f(t)]$

Convolution