Fourier Series

Used to decompose periodic non-sinusoidal waveforms into a sum of sinusoidal ones.

Suppose $f(x)$ is a $2\pi$-periodic function.

$$f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \big[ a_n \cos\left(nx\right) + b_n \sin\left(nx\right) \big]$$

Useful trigonometric integrals

$$\int_{\tau}^{\tau + T} \cos(nx)\, \text{d}x = \int_{\tau}^{\tau + T} \sin(nx)\, \text{d}x = 0$$ $$\int_{\tau}^{\tau + T} \sin(nx) \cos(mx) \, \text{d}t = 0$$ $$\int_{\tau}^{\tau + T} \sin(nx) \sin(mx) \, \text{d}x = \begin{cases} \frac{T}{2} & \text{if } n = m \\ 0 & \text{otherwise} \end{cases}$$ $$\int_{\tau}^{\tau + T} \cos(nx) \cos(mx) \, \text{d}x = \begin{cases} \frac{T}{2} & \text{if } n = m \\ 0 & \text{otherwise} \end{cases}$$

Unknown coefficients

$\forall n \geq 0$. $$

$$a_n = \frac{2}{T} \int_{\tau}^{\tau + T} f(x)\cos(nx) \, \text{d}x$$ $$b_n = \frac{2}{T} \int_{\tau}^{\tau + T} f(x)\sin(nx)\, \text{d}x$$

$\cfrac{a_0}{2}$ is the DC offset. $$

Symmetry

For any waveform, subtracting the DC offset results in a symmetrical waveform.

Even Symmetry

When a wave is symmetric about the y-axis.

$$f(x) = f(-t)$$

The fourier series of an even waveform contains only cosine terms.

$$a_n = \frac{4}{T} \int_{\tau}^{\tau + T/2} f(x)\cos(nx) \, \text{d}x \;\;\text{and}\;\; b_n = 0$$

Odd Symmetry

When a wave is symmetric about the origin.

$$f(x) = -f(-t)$$

The fourier series of an odd waveform contains only sine terms.

$$a_0 = a_n = 0 \;\;\text{and}\;\; b_n = \frac{4}{T} \int_{\tau}^{\tau + T/2} f(x)\sin(nx) \, \text{d}x$$

Note

Product of even and odd functions is even or odd function similar to the even and odd numbers.

• Even × Even → Even
• Odd × Odd → Even
• Even × Odd → Odd

Half-Wave Symmetry

When a wave repeats itself with a reversal of sign after half a period.

$$f(x) = -f\left(t + \frac{T}{2}\right) = -f\left(t - \frac{T}{2}\right)$$

The coefficients can be found by:

$$a_n = \begin{cases} \frac{4}{T} \int_{\tau}^{\tau + T/2} f(x)\cos(nx) \, \text{d}x & \text{if } n \text{ is odd} \\ 0 & \text{if } n \text{ is even} \end{cases}$$ $$b_n = \begin{cases} \frac{4}{T} \int_{\tau}^{\tau + T/2} f(x)\sin(nx) \, \text{d}x & \text{if } n \text{ is odd} \\ 0 & \text{if } n \text{ is even} \end{cases}$$

Note

For odd and half-wave symmetry, the Fourier series does not contain a DC offset. Thus, $a_0=0$. $$

Half-wave symmetry can co-exist with odd symmetry or even symmetry. In that case:

$$a_n = \begin{cases} \frac{8}{T} \int_{\tau}^{\tau + T/4} f(x)\cos(nx) \, \text{d}x & \text{if } n \text{ is odd} \\ 0 & \text{if } n \text{ is even} \end{cases}$$ $$b_n = \begin{cases} \frac{8}{T} \int_{\tau}^{\tau + T/4} f(x)\sin(nx) \, \text{d}x & \text{if } n \text{ is odd} \\ 0 & \text{if } n \text{ is even} \end{cases}$$

Dirichlet Conditions

Sufficient conditions for a real-valued, periodic function $f$ to be equal to its Fourier series at a point of continuity.

Suppose $f$ is a periodic function with period $2L$. If:

• Finite number of discontinuities
• Finite number of maxima/minima
• Absolutely integrable over a period

Then:

• At continuity points, Fourier series = f(x)
• At discontinuities, it converges to

$$\frac{f(c^-)+f(c^+)}{2}$$

Piecewise-Defined Periodic Functions

Suppose:

$$f(x) = \begin{cases} \rho(x) & 0\lt x \lt c \\ \theta(x) & c \lt x \lt 2\pi \end{cases}$$

then compute coefficients by splitting integrals:

$$\begin{equation} \nonumber \begin{split} a_0 & =\frac{1}{\pi}\left[\int_0^cρ(x)\,dx+\int_c^{2\pi}θ(x)\,dx\right] \\ a_n & =\frac{1}{\pi}\left[\int_0^cρ(x)\cos nx\,dx+\int_c^{2\pi}θ(x)\cos nx\,dx\right] \\ b_n & =\frac{1}{\pi}\left[\int_0^cρ(x)\sin nx\,dx+\int_c^{2\pi}θ(x)\sin nx\,dx\right] \end{split} \end{equation}$$

Half-Range Series

Used when a function is defined only on (0,π). Extend the function to (−π,π) artificially.

Half-Range Cosine Series

The function is extended as an even function.

$$\begin{equation} \nonumber \begin{split} a_0 & =\frac{2}{\pi}\int_0^\pi f(x)\,dx \\ a_n & =\frac{2}{\pi}\int_0^\pi f(x)\cos nx\,dx \end{split} \end{equation}$$

And $b_n = 0$.

Half-Range Sine Series

The function is extended as an odd function.

$$b_n=\frac{2}{\pi}\int_0^\pi f(x)\sin nx\,dx$$

And $a_0 = a_n = 0$.

Fourier Series of General Period

For period $2l$:

$$f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}\left[a_n\cos\left(\frac{n\pi x}{l}\right)+b_n\sin\left(\frac{n\pi x}{l}\right)\right]$$

Coefficients:

$$a_0=\frac{1}{l}\int_{-l}^{l}f(x)\,dx$$ $$a_n=\frac{1}{l}\int_{-l}^{l}f(x)\cos\left(\frac{n\pi x}{l}\right)\,dx$$ $$b_n=\frac{1}{l}\int_{-l}^{l}f(x)\sin\left(\frac{n\pi x}{l}\right)\,dx$$

Parseval’s Identity

A method to find higher order algebraic series, that is, for $n>2$:

$$\sum_{i=1}^\infty \frac{1}{i^n}$$

For period $2c$:

$$\int_{-c}^{c}[f(x)]^2\,dx = c\left[\frac{a_0^2}{2}+\sum_{n=1}^{\infty}(a_n^2 + b_n^2)\right]$$

Complex Form

For period 2π:

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

where

$$c_n=\frac{1}{2\pi}\int_{-\pi}^{\pi} f(x)e^{inx}\,dx$$

Here:

• $c_0 = a_0/2$
• $c_n = (a_n - ib_n)/2$
• $c_{-n}=(a_n + ib_n)/2$

General period $2l$ also included.