Derive exponential Fourier series (CT)

A continuous-time periodic signal $x(t)$ with fundamental period $T$ and fundamental frequency $\omega_0 = 2\pi/T$ can be written as a weighted sum of harmonically related complex exponentials. This is the exponential Fourier series.

Synthesis equation

The set $\phi_k(t) = e^{jk\omega_0 t}$, $k = 0, \pm1, \pm2, \dots$, contains signals that are all periodic with period $T$ (the $k$th one has frequency $k\omega_0$). A linear combination of them is also periodic with period $T$:

$$x(t) = \sum_{k=-\infty}^{\infty} a_k\, e^{jk\omega_0 t} = \sum_{k=-\infty}^{\infty} a_k\, e^{jk(2\pi/T)t}$$

This is the synthesis equation. $a_0$ is the dc term, $k = \pm1$ are the fundamental components, and $k = \pm N$ are the $N$th harmonics.

Analysis equation

To find $a_k$, multiply both sides by $e^{-jn\omega_0 t}$ and integrate over one period $T$:

$$\int_0^T x(t)e^{-jn\omega_0 t}\,dt = \sum_{k=-\infty}^{\infty} a_k \int_0^T e^{j(k-n)\omega_0 t}\,dt$$

Evaluate the integral on the right (orthogonality):

$$\int_0^T e^{j(k-n)\omega_0 t}dt = \int_0^T \cos((k-n)\omega_0 t)\,dt + j\int_0^T \sin((k-n)\omega_0 t)\,dt$$

• For $k \ne n$, both integrals cover a whole number $(k-n)$ of periods, so they are 0.
• For $k = n$, the integrand is 1 and the integral is $T$.

$$\int_0^T e^{j(k-n)\omega_0 t}dt = \begin{cases} T, & k = n \\ 0, & k \ne n \end{cases}$$

So only the $k = n$ term survives:

$$\int_0^T x(t)e^{-jn\omega_0 t}dt = a_n T \quad\Rightarrow\quad a_n = \frac{1}{T}\int_0^T x(t)e^{-jn\omega_0 t}dt$$

Since any interval of length $T$ gives the same result, the analysis equation is

$$a_k = \frac{1}{T}\int_{T} x(t)\,e^{-jk\omega_0 t}\,dt$$

The Fourier series pair

$$x(t) \overset{FS}{\longleftrightarrow} a_k$$

Notes:

• The $a_k$ are the Fourier series coefficients or spectral coefficients; they are complex in general, $a_k = |a_k|e^{j\angle a_k}$, giving the magnitude and phase spectra.
• For real $x(t)$, $a_{-k} = a_k^*$, so $|a_k|$ is even and $\angle a_k$ is odd in $k$.

Example: $x(t) = \cos\omega_0 t = \frac{1}{2}e^{j\omega_0 t} + \frac{1}{2}e^{-j\omega_0 t}$ gives $a_1 = a_{-1} = 1/2$ and all other $a_k = 0$.