Statement of Fourier's theorem

Given a function ${\displaystyle y(t)}$

1. periodic with period T
2. bounded
3. integrable on [0,T]

then ${\displaystyle y(t)}$ can be expanded in a Fourier series, that is

${\displaystyle y(t)=a_0+{\displaystyle \sum _{n=1}^{\mathrm{\infty }}}[a_n\cos ({\omega }_{n}t)+b_n\sin ({\omega }_{n}t)]}$

(1)

where a_n and b_n are suitable coefficients that depend on the analytic form of the periodic function ${\displaystyle y(t)}$, and

${\displaystyle {\omega }_n=\frac{2\pi n}{T}}$.

The sinusoids of the series therefore have frequencies ${\displaystyle f_n={\omega }_n/(2\pi )=n/T}$, integer multiples of the fundamental frequency ${\displaystyle f_0=1/T}$.

The sinusoid whose frequency equals the fundamental frequency ${\displaystyle f_0}$ is also called the fundamental harmonic or first harmonic; the sinusoids whose frequencies are multiples of the fundamental frequency are called higher harmonics (second, third, etc.) or also partials. An oscillation (wave) made up of several harmonics will be called a complex wave. Each complex wave will have its own harmonic content, characterized by the different values of the amplitudes and phases of the various harmonics into which it can be decomposed.

Calculation of the coefficients of the Fourier expansion

To calculate the coefficients ${\displaystyle a_n}$ and ${\displaystyle b_n}$, keep in mind that:

• if the function ${\displaystyle y(t)}$ is odd, as the sine function is, a series of sines only will be obtained (that is, the coefficients ${\displaystyle a_n}$ are all zero);

• if the function ${\displaystyle y(t)}$ is even, as the cosine function is, a series of cosines only will be obtained (that is, the coefficients ${\displaystyle b_n}$ are all zero).

In general, the coefficients are calculated by means of the following integrals:

${\displaystyle a_0=\frac{1}{T}{\displaystyle \underset{0}{\overset{T}{\int }}}y(t)dt}$ which is nothing but the mean value of the function ${\displaystyle y(t)}$ over the periodicity interval [0,T].

${\displaystyle a_n=\frac{2}{T}{\displaystyle \underset{-\frac{T}{2}}{\overset{\frac{T}{2}}{\int }}}y(t)\cos ({\omega }_{n}t)dt}$

(2)

${\displaystyle b_n=\frac{2}{T}{\displaystyle \underset{-\frac{T}{2}}{\overset{\frac{T}{2}}{\int }}}y(t)\sin ({\omega }_{n}t)dt}$

(3)

Examples of Fourier series

We present some notable expansions that you will also find used in our virtual laboratory to build the default waveforms.

In what follows we will always assume that:

• the waveforms have period T (their analytic expression will be given only over one periodicity interval which, to exploit any symmetries of the function, has been "centred" at the origin, that is, it will go from -T/2 to T/2);

• the mean value of the function is always zero, so that the coefficient ${\displaystyle a_0}$ is always zero;

• the amplitude of the function is 1.

Triangle wave

With the previous assumptions, the triangle wave can be represented analytically by the expression

${\displaystyle y(t)=\{\begin{array}{ll}1+\frac{4}{T}t& \mathrm{i}\mathrm{f}-\frac{T}{2}\le t<0\\ 1-\frac{4}{T}t& \mathrm{i}\mathrm{f}0\le t\le \frac{T}{2}\end{array}}$

Since the function is even, a series of cosines only will be obtained. Calculating the integrals (2) and (3) indeed leads to the result

${\displaystyle b_n=0}$

${\displaystyle a_n=\{\begin{array}{ll}\frac{8}{n^2{\pi }^2}& n\mathrm{o}\mathrm{d}\mathrm{d}\\ 0& n\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}\end{array}}$

The result obtained therefore shows that the series contains only cosines and that only odd harmonics are present. You can check these results directly with the app (selecting forme d'onda:triangolo (waveforms: triangle)):

• note that only odd harmonics are present;
• by moving the mouse over the slider of each harmonic you can check that the amplitude of the n-th harmonic is indeed given (apart from rounding errors) by

${\displaystyle a_n=\frac{8}{n^2{\pi }^2}}$

• that the phase of each harmonic is zero (the app works precisely with cosine functions)

waveform in time	amplitudes of the Fourier components (cosine)

Square wave

With the previous assumptions, the square wave can be represented analytically by the expression

${\displaystyle y(t)=\{\begin{array}{ll}-1& \mathrm{i}\mathrm{f}-\frac{T}{2}\le t<0\\ 1& \mathrm{i}\mathrm{f}0\le t\le \frac{T}{2}\end{array}}$

Since the function is odd, a series of sines only will be obtained. Calculating the integrals (2) and (3) indeed leads to the result

${\displaystyle a_n=0}$

${\displaystyle b_n=\{\begin{array}{ll}\frac{4}{n\pi }& n\mathrm{o}\mathrm{d}\mathrm{d}\\ 0& n\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}\end{array}}$

The result obtained therefore shows that the series contains only sines and that only odd harmonics are present. You can check these results directly with the app (selecting forme d'onda:Onda quadra (waveforms: square wave)):

• note that only odd harmonics are present;
• by moving the mouse over the slider of each harmonic you can check that the amplitude of the n-th harmonic is indeed given (apart from rounding errors) by

${\displaystyle a_n=\frac{4}{n\pi }}$.

• It is curious to note that the amplitude of the fundamental is larger than the amplitude of the whole square wave;
• the phase of each harmonic is equal to ${\displaystyle -\frac{\pi }{2}}$. This is correct, remembering
  • that the app expands in series using only cosine functions;
  • the trigonometric identity ${\displaystyle \cos (\alpha -\frac{\pi }{2})=\sin (\alpha )}$

waveform in time	amplitudes of the Fourier components (cosine)

Sawtooth wave

With the previous assumptions, the "sawtooth" wave can be represented analytically by the expression

${\displaystyle y(t)=\{\begin{array}{ll}\frac{2}{T}t& \mathrm{i}\mathrm{f}-\frac{T}{2}\le t<\frac{T}{2}\end{array}}$

Since the function is odd, a series of sines only will be obtained. Calculating the integrals (2) and (3) indeed leads to the result

${\displaystyle a_n=0}$

${\displaystyle b_n=\{\begin{array}{ll}\frac{2}{n\pi }& n\mathrm{o}\mathrm{d}\mathrm{d}\\ -\frac{2}{n\pi }& n\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}\end{array}}$

The result obtained therefore shows that the series contains only sines and that all harmonics, even and odd, are present.

waveform in time	amplitudes of the Fourier components (sine)

You can check these results directly with the app (selecting forme d'onda:Dente di sega (waveforms: sawtooth)):

• note that now all harmonics are present;
• by moving the mouse over the slider of each harmonic you can check that the amplitude of the n-th harmonic is indeed given (apart from rounding errors) by

${\displaystyle a_n=\frac{2}{n\pi }}$.

• that the phase of the harmonics is alternately ${\displaystyle -\frac{\pi }{2}}$ and ${\displaystyle \frac{\pi }{2}}$ . This is correct, remembering
  • that the app expands in series using only cosine functions;
  • the trigonometric identities

${\displaystyle \cos (\alpha -\frac{\pi }{2})=\sin (\alpha )}$

${\displaystyle \cos (\alpha +\frac{\pi }{2})=-\sin (\alpha )}$

Taking the amplitudes as positive, the app "shifts" the minus sign of the ${\displaystyle b_n}$ with even n onto the phase of the cosine wave.

Check your understanding