9.2: Complex Exponential Fourier Series

Before deriving the Fourier transform, we will need to rewrite the trigonometric Fourier series representation as a complex exponential Fourier series. We first recall from Chapter ?? the trigonometric Fourier series representation of a function defined on \([-\pi, \pi]\) with period \(2 \pi\). The Fourier series is given by

\[f(x) \sim \frac{a_{0}}{2}+\sum_{n=1}^{\infty}\left(a_{n} \cos n x+b_{n} \sin n x\right),\label{eq:1} \]

where the Fourier coefficients were found as

\[\begin{align} &a_{n}=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos n x d x, \quad n=0,1, \ldots,\nonumber \\ &b_{n}=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin n x d x, \quad n=1,2, \ldots\label{eq:2} \end{align} \]

In order to derive the exponential Fourier series, we replace the trigonometric functions with exponential functions and collect like exponential terms. This gives

\[\begin{align} f(x) & \sim \frac{a_{0}}{2}+\sum_{n=1}^{\infty}\left[a_{n}\left(\frac{e^{i n x}+e^{-i n x}}{2}\right)+b_{n}\left(\frac{e^{i n x}-e^{-i n x}}{2 i}\right)\right]\nonumber \\ &=\frac{a_{0}}{2}+\sum_{n=1}^{\infty}\left(\frac{a_{n}-i b_{n}}{2}\right) e^{i n x}+\sum_{n=1}^{\infty}\left(\frac{a_{n}+i b_{n}}{2}\right) e^{-i n x} .\label{eq:3} \end{align} \]

The coefficients of the complex exponentials can be rewritten by defining

\[c_{n}=\frac{1}{2}\left(a_{n}+i b_{n}\right), \quad n=1,2, \ldots .\label{eq:4} \]

This implies that

\[\bar{c}_{n}=\frac{1}{2}\left(a_{n}-i b_{n}\right), \quad n=1,2, \ldots\label{eq:5} \]

So far the representation is rewritten as

\[f(x) \sim \frac{a_{0}}{2}+\sum_{n=1}^{\infty} \bar{c}_{n} e^{i n x}+\sum_{n=1}^{\infty} c_{n} e^{-i n x} .\nonumber \]

Re-indexing the first sum, by introducing \(k=-n\), we can write

\[f(x) \sim \frac{a_{0}}{2}+\sum_{k=-1}^{-\infty} \bar{c}_{-k} e^{-i k x}+\sum_{n=1}^{\infty} c_{n} e^{-i n x}\nonumber \]

Since \(k\) is a dummy index, we replace it with a new \(n\) as

\[f(x) \sim \frac{a_{0}}{2}+\sum_{n=-1}^{-\infty} \bar{c}_{-n} e^{-i n x}+\sum_{n=1}^{\infty} c_{n} e^{-i n x} .\nonumber \]

We can now combine all of the terms into a simple sum. We first define \(c_{n}\) for negative \(n^{\prime}\)s by

\[c_{n}=\bar{c}_{-n}, \quad n=-1,-2, \ldots\nonumber \]

Letting \(c_{0}=\frac{a_{0}}{2}\), we can write the complex exponential Fourier series representation as

\[f(x) \sim \sum_{n=-\infty}^{\infty} c_{n} e^{-i n x},\label{eq:6} \]

where

\[\begin{align} &c_{n}=\frac{1}{2}\left(a_{n}+i b_{n}\right), \quad n=1,2, \ldots\nonumber \\ &c_{n}=\frac{1}{2}\left(a_{-n}-i b_{-n}\right), \quad n=-1,-2, \ldots\nonumber \\ &c_{0}=\frac{a_{0}}{2}\label{eq:7} \end{align} \]

Given such a representation, we would like to write out the integral forms of the coefficients, \(c_{n}\). So, we replace the \(a_{n}\) ’s and \(b_{n}\) ’s with their integral representations and replace the trigonometric functions with complex exponential functions. Doing this, we have for \(n=1,2, \ldots\).

\[\begin{align} c_{n} &=\frac{1}{2}\left(a_{n}+i b_{n}\right)\nonumber \\ &=\frac{1}{2}\left[\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos n x d x+\frac{i}{\pi} \int_{-\pi}^{\pi} f(x) \sin n x d x\right]\nonumber \\ &=\frac{1}{2 \pi} \int_{-\pi}^{\pi} f(x)\left(\frac{e^{i n x}+e^{-i n x}}{2}\right) d x+\frac{i}{2 \pi} \int_{-\pi}^{\pi} f(x)\left(\frac{e^{i n x}-e^{-i n x}}{2 i}\right) d x\nonumber \\ &=\frac{1}{2 \pi} \int_{-\pi}^{\pi} f(x) e^{i n x} d x .\label{eq:8} \end{align} \]

It is a simple matter to determine the \(c_{n}\) ’s for other values of \(n\). For \(n=0\), we have that

\[c_{0}=\frac{a_{0}}{2}=\frac{1}{2 \pi} \int_{-\pi}^{\pi} f(x) d x .\nonumber \]

For \(n=-1,-2, \ldots\), we find that

\[c_{n}=\bar{c}_{n}=\frac{1}{2 \pi} \int_{-\pi}^{\pi} f(x) \overline{e^{-i n x}} d x=\frac{1}{2 \pi} \int_{-\pi}^{\pi} f(x) e^{i n x} d x .\nonumber \]

Therefore, we have obtained the complex exponential Fourier series coefficients for all \(n\). Now we can define the complex exponential Fourier series for the function \(f(x)\) defined on \([-\pi, \pi]\) as shown below.

Complex Exponential Series for \(f(x)\) defined on \([-\pi ,\pi ]\)

\[f(x) \sim \sum_{n=-\infty}^{\infty} c_{n} e^{-i n x},\label{eq:9} \]

\[c_{n}=\frac{1}{2 \pi} \int_{-\pi}^{\pi} f(x) e^{i n x} d x .\label{eq:10} \]

We can easily extend the above analysis to other intervals. For example, for \(x \in[-L, L]\) the Fourier trigonometric series is

\[f(x) \sim \frac{a_{0}}{2}+\sum_{n=1}^{\infty}\left(a_{n} \cos \frac{n \pi x}{L}+b_{n} \sin \frac{n \pi x}{L}\right)\nonumber \]

with Fourier coefficients

\[\begin{aligned} &a_{n}=\frac{1}{L} \int_{-L}^{L} f(x) \cos \frac{n \pi x}{L} d x, \quad n=0,1, \ldots, \\ &b_{n}=\frac{1}{L} \int_{-L}^{L} f(x) \sin \frac{n \pi x}{L} d x, \quad n=1,2, \ldots . \end{aligned} \nonumber \]

This can be rewritten as an exponential Fourier series of the form

Complex Exponential Series for \(f(x)\) defined on \([-L,L]\).

\[f(x) \sim \sum_{n=-\infty}^{\infty} c_{n} e^{-i n \pi x / L},\label{eq:11} \]

\[c_{n}=\frac{1}{2 L} \int_{-L}^{L} f(x) e^{i n \pi x / L} d x .\label{eq:12} \]

We can now use this complex exponential Fourier series for function defined on \([-L, L]\) to derive the Fourier transform by letting \(L\) get large. This will lead to a sum over a continuous set of frequencies, as opposed to the sum over discrete frequencies, which Fourier series represent.