Preface


This section presents material on complex Fourier series including various examples.

Return to computing page for the first course APMA0330
Return to computing page for the second course APMA0340
Return to Mathematica tutorial for the first course APMA0330
Return to Mathematica tutorial for the second course APMA0340
Return to the main page for the first course APMA0330
Return to the main page for the second course APMA0340
Return to Part V of the course APMA0340
Introduction to Linear Algebra with Mathematica

 

Complex Fourier Series


The complex exponential form of Fourier series is a representation of a periodic function (which is usually a signal) with period \( 2\ell \) as infinite series:
\begin{equation} \label{EqComplex.1} f(x) \,\sim\, \mbox{P.V.} \sum_{k=-\infty}^{\infty} \alpha_k e^{k{\bf j} \pi x/\ell} \qquad ({\bf j}^2 = -1), \end{equation}
where a signal's complex Fourier spectrum is
\begin{equation} \label{EqComplex.2} \alpha_k = \frac{1}{2\ell} \int_{-\ell}^{\ell} f(x)\, e^{-k{\bf j} \pi x/\ell} \,{\text d} x , \qquad k=0, \pm 1, \pm 2, \ldots ; \end{equation}
provided that this series converges in some sense. The abbreviation P.V. means the Cauchy principal value regularization:
\begin{equation} \label{EqComplex.3} \mbox{P.V.} \sum_{k=-\infty}^{\infty} \alpha_k e^{k{\bf j} \pi x/\ell} = \lim_{N\to \infty} \sum_{k=-N}^{N} \alpha_k e^{k{\bf j} \pi x/\ell} \end{equation}
because restoring a function from its Fourier coefficients is an ill-posed problem. Formula \eqref{EqComplex.2} is based on the orthogonality property of exponential functions:
\[ \int_0^T e^{{\bf j}2\pi kt/T} \,e^{-{\bf j}2\pi nt/T} \,{\text d}t = \left\{ \begin{array}{ll} T, & \ \mbox{if $k=n$}, \\ 0 , & \ \mbox{if } k\ne n . \end{array} \right. \]
The conversion of complex Fourier series into standard trigonometric Fourier series is based on Euler's formulas:

\[ \sin \theta = \frac{1}{2{\bf j}} \,e^{{\bf j}\theta} - \frac{1}{2{\bf j}} \,e^{-{\bf j}\theta} = \Im \,e^{{\bf j}\theta} = \mbox{Im} \,e^{{\bf j}\theta}, \qquad \cos \theta = \frac{1}{2} \,e^{{\bf j}\theta} - \frac{1}{2} \,e^{-{\bf j}\theta} = \Re \,e^{{\bf j}\theta} = \mbox{Re} \,e^{{\bf j}\theta}. \]

Here, j is the unit vector in positive vertical direction on the complex plane, so \( {\bf j}^2 =-1. \) For example the Fourier series for the Dirac delta function on a symmetric interval (−&ell, ℓ) is

\begin{equation} \label{EqComplex.4} \delta (x) = \mbox{P.V.} \frac{1}{2\ell} \sum_{k=-\infty}^{\infty} e^{-k{\bf j} \pi x/\ell} =\frac{1}{2\ell} + \frac{1}{\ell}\,\lim_{N\to \infty} \sum_{k=1}^N \cos \left( \frac{k\pi x}{\ell} \right) = \frac{1}{2\ell}\,\lim_{N\to \infty} \frac{\sin \left( 2N+1 \right) \frac{x\pi}{2\ell}}{\sin \frac{x\pi}{2\ell}} . \end{equation}
This means that for every probe function f(x) on the inverval −ℓ < x < ℓ,
\[ \frac{1}{2\ell}\,\lim_{N\to \infty} \int_{-\ell}^{\ell} {\text d}x\,f(x)\,\frac{\sin \left( 2N+1 \right) \frac{x\pi}{2\ell}}{\sin \frac{x\pi}{2\ell}} = f(0) . \]

Mathematica has a default command to calculate complex Fourier series:

FourierSeries[ expr, t, n]        (* gives the n-order (complex) Fourier series expansion of expr in t *)

Mathematica has a special command to find complex Fourier coefficient and to determine its numerical approximation:

FourierCoefficient[ expr, t, n]        (* gives the nth coefficient in the exponential Fourier series expansion of expr in t *)

NFourierCoefficient[ expr, t, n]        (* gives a numerical approximation to the nth coefficient in the Fourier exponential series expansion of expr in t *)

One can easily transfer complex form into trigonometric form vice versa using formulas:

\[ \alpha_k = \begin{cases} \frac{a_0}{2} , & \quad k=0, \\ \frac{1}{2} \left( a_k - {\bf j} b_k \right) , & \quad k=1,2,3,\ldots , \\ \frac{1}{2} \left( a_{-k} + {\bf j} b_{-k} \right) , & \quad k=-1,-2,-3,\ldots \end{cases} \]
and
\[ a_0 = 2\,\alpha_0 , \quad a_k = 2\,\Re \alpha_k = 2\,\mbox{Re} \,\alpha_k , \quad b_k = -2\,\Im \alpha_k = -2\,\mbox{Im} \,\alpha_k , \quad k=1,2,\ldots . \]
Note that \( a_{-n} \quad\mbox{and}\quad b_{-n} \) are only defined when n is negative. Then we get
\[ f(x) \,\sim\, \sum_{k=-\infty}^{\infty} \alpha_k e^{k{\bf j} \pi x/\ell} = \frac{a_0}{2} + \sum_{n\ge 1} \left[ a_n \cos \frac{n\pi x}{\ell} + b_n \sin \frac{n\pi x}{\ell} \right] . \]

The Fourier series command has an option FourierParameters that involves two parameters and when applied, it looks as FourierParameters->{a,b}
This means that complex Fourier coefficient is evaluated according to the formula:

\[ \left\vert \frac{b}{2\,\pi} \right\vert^{(a+1)/2} \, \int_{-\pi/|b|}^{\pi/|b|} f(t)\,e^{-{\bf j}bnt} \,{\text d}t . \]

Example 1: Consider a piecewise constant function on the interval [-2 , 2]:

\[ f(x) = \left\{ \begin{array}{ll} 1, & \ \mbox{on the interval } -2 < x < -1, \\ 0 , & \ \mbox{on the interval } -1 < x < 0 , \\ 2, & \ \mbox{on the interval } 0 < x < 2. \end{array} \right. \]
There is no need to define the function at the points of discontinuity \( x=-2, -1, 0, 2 \) because the corresponding Fourier series will specify the values at these points to be the averages of left and right limit values. Therefore, \( f(-2) = 3/2, \ f(-1) = 1/2, \ f(0) = 1, \ f(2) =3/2 . \) We can find the Fourier coefficients either by evaluating integrals
\[ \alpha_0 = \frac{1}{4} \,\int_{-2}^2 f(x)\,{\text d}x = \frac{5}{4} , \qquad \alpha_k = \frac{1}{4} \,\int_{-2}^2 f(x)\,e^{-k{\bf j} \pi x/2} \,{\text d}x = \frac{\bf j}{2k\pi} \left[ (-1)^k + \cos \frac{k\pi}{2} -2 \right] - \frac{1}{2k\pi} \,\sin \frac{k\pi}{2} , \quad k = \pm 1, \pm 2, \ldots . \]
f[t_] = Piecewise[{{1, -2 < t < -1}, {0, -1 < t < 0}, {2, 0 < t < 2}}]
Integrate[f[t]*Exp[-k*I*Pi*t/2], {t, -2, 2}]/4 // ComplexExpand
Simplify[%]
Out[5]= (I (Cos[(k \[Pi])/2] + Cos[k \[Pi]] + I (2 I + Sin[(k \[Pi])/2] - 3 Sin[k \[Pi]])))/(2 k \[Pi])
or using standard command:
FourierSeries[f[x], x, 5, FourierParameters -> {1, Pi/2}] // ComplexExpand
5/4 - Cos[(\[Pi] x)/2]/\[Pi] + Cos[(3 \[Pi] x)/2]/(3 \[Pi]) - Cos[(5 \[Pi] x)/2]/(5 \[Pi]) + (3 Sin[(\[Pi] x)/2])/\[Pi] + Sin[\[Pi] x]/\[Pi] + Sin[(3 \[Pi] x)/2]/\[Pi] + ( 3 Sin[(5 \[Pi] x)/2])/(5 \[Pi])
The corresponding real trigonometric series is
\[ f(t) = \frac{5}{4} - \frac{1}{\pi} \,\sum_{k\ge 1} \frac{1}{k}\,\sin \frac{k\pi}{2}\, \cos \frac{k\pi t}{2} - \frac{1}{\pi} \,\sum_{k\ge 1} \frac{1}{k}\left[ \cos \frac{k\pi}{2} + (-1)^k -2 \right] \sin \frac{k\pi t}{2} . \]
You can find a single Fourier coefficient (only for complex form) with a command
FourierCoefficient[f[t], t, 5, FourierParameters -> {1, Pi/2}]
-((1/10 + (3 I)/10)/\[Pi])
and when you add to
FourierCoefficient[f[t], t, -5, FourierParameters -> {1, Pi/2}]
-((1/10 - (3 I)/10)/\[Pi])
it will provide you with a simplified answer. Correctness of calculations could be checked numerically:
NFourierCoefficient[f[t], t, 5, FourierParameters -> {1, Pi/2}]
Out[12]= -0.031831 - 0.095493 I
NFourierCoefficient[f[t], t, -5, FourierParameters -> {1, Pi/2}]
Out[13]= -0.031831 + 0.095493 I
N[Sin[5*Pi/2]/10/Pi]
Out[14]= 0.031831
N[3/10/Pi]
Out[15]= 0.095493
because \( (-1)^5 + \cos \frac{5\pi}{2} -2 = -3 . \)

To compare the quality of Fourier approximations, we plot partial sums with 10 and 50 terms, respectively:

curve = 5/4 - (1/Pi)* Sum[Sin[k*Pi/2]*Cos[k*Pi*x/2]/k, {k, 1, 10}] - (1/Pi)* Sum[(Cos[k*Pi/2] + (-1)^k - 2)*Sin[k*Pi*x/2]/k, {k, 1, 10}]
Plot[curve, {x, -3.5, 3.5}, PlotStyle -> Thick]
   
When the complex Fourier series is used to represent a periodic function, then the amplitude spectrum, sketched below, is two-sided. It consists of the points \( \left( \frac{k\pi}{\ell} , \left\vert \alpha_k \right\vert \right) , \quad k= 0, \pm 1, \pm 2, \ldots \) that in our case become
\[ \left( 0, \frac{5}{4} \right) , \quad \left( \frac{k\pi}{2}, \frac{1}{2k\pi} \,\sqrt{\left( (-1)^k + \cos \frac{k\pi}{2} -2 \right)^2 + \sin^2 \frac{k\pi}{2}} \right) , \quad k=\pm 1, \pm 2, \ldots . \]
a[k_] = (-1)^k + Cos[k*Pi/2] - 2
b[k_] = Sin[k*Pi/2]
p0 = Line[{{0, 0}, {0, 5/4}}]; q0 = Graphics[{Thick, p0}]
pm1 = Line[{{-1*Pi/2, 0}, {-1*Pi/2, Sqrt[(a[-1])^2 + (b[-1])^2]/2/Pi}}]
qm1 = Graphics[{Thick, pm1}]
p1 = Line[{{1*Pi/2, 0}, {1*Pi/2, Sqrt[(a[1])^2 + (b[1])^2]/2/Pi}}]
q1 = Graphics[{Thick, p1}]
a = Graphics[Arrow[{{-3*N[Pi]/1.9, 0}, {3*N[Pi]/1.7, 0}}]]
Show[a, q0, q1, q2, q3, qm1, qm2, qm3]
The two-sided amplitude spectrum of the function.

The power spectrum for f is also two-sided, consisting of the points \( \left( \frac{k\pi}{\ell} , \left\vert \alpha_k \right\vert^2 \right) , \quad k= 0, \pm 1, \pm 2, \ldots \) that in our case become
\[ \left( 0, \frac{5}{4} \right) , \quad \left( \frac{k\pi}{2}, \frac{1}{4k^2\pi^2} \,\left[ \left( (-1)^k + \cos \frac{k\pi}{2} -2 \right)^2 + \sin^2 \frac{k\pi}{2} \right] \right) , \quad k=\pm 1, \pm 2, \ldots . \]
f[t_] = Piecewise[{{1, -2 < t < -1}, {0, -1 < t < 0}, {2, 0 < t < 2}}]
DiscretePlot[ Abs[FourierCoefficient[f[x], x, k, FourierParameters -> {1, Pi/2}]]^2, {k, -7, 7}, PlotRange -> All, PlotStyle -> {Thick, PointSize[0.03]}]
Note that here we used Mathematica's command FourierCoefficient that gives the kth coefficient in the Fourier series expansion of f.
The two-sided power spectrum of the function.

   ■

The complex exponential Fourier form has the following advantages compared to the traditional trigonometric form:

A signal's Fourier series spectrum \( \alpha_k \) has interesting properties. The following formula may be helpful for evaluating Fourier coefficients:
\begin{align*} \int p(x)\, e^{ax}\,{\text d}x &= \frac{1}{a} \, e^{ax}\,\sum_{k=0}^n \frac{(-1)^k}{a^k} \,{\texttt D}_x^k p(x) \qquad\quad \left( {\texttt D}_x = \frac{\text d}{{\text d}x} \right) \\ &= \frac{1}{a} \, e^{ax} \left[ p(x) - \frac{p' (x)}{a} + \frac{p'' (x)}{a^2} - \cdots + (-1)^n \,\frac{p^{(n)} (x)}{a^n} \right] , \end{align*}
where \( p(x) = b_n x^n + b_{n-1} x^{n-1} + \cdots + b_1 x + b_0 \) is a polynomial of degree n.

Example 2: Find complex and regular Fourier series expansion of the function \( f(x) = \frac{1-a\,\cos x}{1- 2a\,\cos x + a^2} , \) where real number a has absolute value less than 1: \( |a| <1 . \)

First, we substitute Euler's representation \( \cos x = \frac{1}{2}\,e^{{\bf j}x} + \frac{1}{2}\,e^{-{\bf j}x} . \) instead of cos(x). Then

\[ f(x) = \frac{1 - \frac{a}{2} \left( e^{{\bf j}x} + e^{-{\bf j}x} \right)}{1 - a \left( e^{{\bf j}x} + e^{-{\bf j}x} \right) + a^2} = \frac{1}{2}\,\frac{2- a\, e^{{\bf j}x} - a\,e^{-{\bf j}x}}{\left( 1-a\,e^{{\bf j}x} \right) \left( 1-a\,e^{-{\bf j}x} \right)} . \]
Since the numerator can be written as
\[ 2- a\, e^{{\bf j}x} - a\,e^{-{\bf j}x} = 1- a\, e^{{\bf j}x} + 1 - a\, e^{-{\bf j}x} , \]
we simplify the given function as
\[ f(x) = \frac{1}{2}\,\frac{1- a\, e^{{\bf j}x} + 1 - a\, e^{-{\bf j}x}}{\left( 1-a\,e^{{\bf j}x} \right) \left( 1-a\,e^{-{\bf j}x} \right)} = \frac{1}{2}\,\frac{1}{1-a\,e^{-{\bf j}x}} + \frac{1}{2}\,\frac{1}{1-a\,e^{{\bf j}x}} \]
Using the geometric series formula \( \frac{1}{1-q} = \sum_{n\ge 0} q^n \) twice, first time with \( q= a\,e^{-{\bf j}x} \) and second time with \( q= a\,e^{{\bf j}x} , \) we obtain the required complex Fourier series:
\[ f(x) = \frac{1}{2}\,\sum_{n\ge 0} \left( a\,e^{-{\bf j}x} \right)^n + \frac{1}{2}\,\sum_{n\ge 0} \left( a\,e^{{\bf j}x} \right)^n = \frac{1}{2}\,\sum_{n\ge 0} a^n \, e^{-n{\bf j}x} + \frac{1}{2}\,\sum_{n\ge 0} a^n \, e^{n{\bf j}x} , \]
which we rewrite in a symmetric way
\[ f(x) = 1+ \frac{1}{2}\,\sum_{n\ge 1} a^n \, e^{n{\bf j}x} + \frac{1}{2}\,\sum_{n=-\infty}^{-1} a^{-n} \, e^{n{\bf j}x} . \]
Next, we unite these two sums into one to obtain real trigonometric series:
\[ \frac{1-a\,\cos x}{1- 2a\,\cos x + a^2} = \sum_{n\ge 0} a^n \cos nx . \]
   ■

Example 3: Our next example deals with the periodic rectangular pulse function shown below

\[ \Pi (t,h,T) = \left\{ \begin{array}{ll} 1 , & \ x\in (0, h) \\ 0 , & \ x\in (h,T) \end{array} \right. \qquad\mbox{on the interval } (0,T). \]
The function is a pulse function with amplitude 1, and pulse width h, and period T. Using Mathematica, we can define the pulse function in many ways; however, we demonstrate application of command Which. The Which command provides a logical expression that allows us to evaluate a function in only one statement like the one given in the equation, defining the pulse function. The "Which" command has the general form :
Which[condition1, value1, condition2, value2 ...]
The command will return the value that is true; let us see how this works in practice in an example of the pulse function:
PI[x_, h_, T_] := Which[0 < x < h, 1, h < x < T, 0]
We expand the pulse function into exponential Fourier series:
\[ \Pi (x,h,T) = \sum_{k=-\infty}^{\infty} \alpha_k e^{{\bf j}k2\pi x/T}, \qquad \alpha_k = \frac{1}{T} \,\int_0^h e^{-{\bf j}k2\pi x/T} \,{\text d}x = \frac{\bf j}{2k\pi} \left[ e^{-{\bf j}k2\pi h/T} -1 \right] , \quad k=0, \pm 1, \pm 2, \ldots . \]
Note that the corresponding k = 0 value α0 = h/T does not follow from the general formula directly. The above Fourier series defines the pulse functions at the points of discontinuity as the mean values from left and right:
\[ \Pi (0,h,T) = \Pi (h,h,T) = \frac{1}{2} = \frac{h}{T} + \sum_{k \ge 0} \frac{1}{k\pi} \, \sin \frac{2hk\pi}{T} \qquad\mbox{for any }h, t. \]
We can also convert the complex Fourier form to a regular real trigonometric form:
\[ \Pi (x,h,T) = \frac{h}{T} + \sum_{k \ge 0} \frac{1}{k\pi} \left[ \sin \frac{2hk\pi}{T} \, \cos \frac{2\pi kx}{T} + 2\,\sin^2 \frac{k\pi h}{T} \,\sin \frac{2\pi kx}{T} \right] . \]
To compare the quality of Fourier approximations, we plot partial sums with 10 and 50 terms for particular numerical values h = 1 and T = 3: \)
pulse10 = 1/3 + (1/Pi)*Sum[(1/k)*(Sin[2*k*Pi/3]*Cos[2*Pi*k*x/3] + 2*(Sin[k*Pi/3])^2 *Sin[2*Pi*k*x/3]), {k, 1, 10}]
Plot[pulse10, {x, -5, 5}, PlotStyle -> Thick]
   
We can calculate Fourier coefficients directly:
a0 = 2/3;
an = (2/3)* Integrate[Cos[n*x*2*Pi/3], {x, 0, 1}, Assumptions -> Element[n, Integers]];
bn = (2/3)* Integrate[Sin[n*x*2*Pi/3], {x, 0, 1}, Assumptions -> Element[n, Integers]];
Print[{a0, an, bn}]
fourier[m_] := a0/2 + Sum[an Cos[n x*2*Pi/3] + bn Sin[n x*2*Pi/3], {n, 1, m}]
Plot[fourier[30], {x, -Pi, 2*Pi}, Epilog -> {Red, PointSize[Large], Point[{{0, 0.5}, {1, 0.5}, {3, 0.5}, {-2, 0.5}, {4, 0.5}}]}, PlotStyle -> Thick]
Then we check numerical values of the partial sum with 30 terms at the points of discontinuity:
N[fourier[30] /. x -> 1]
Out[23]= 0.496939
which is closed to the true value \( \Pi (1,1,3)=1/2. \)    ■

Example 4: Consider the piecewise continuous function on the interval [-1,2]:

\[ f(x) = \begin{cases} 1, & \ \mbox {if } -2 < x < -1 , \\ x^2 -1, & \ \mbox {if } -1< x< 2 , \end{cases} \]
Its Fourier coefficients are evaluated with the aid of Mathematica:
f[x_] = Piecewise[{{1, -2 < x < -1}, {x^2 - 1, -1 < x < 2}}, 0]
Integrate[f[x], {x, -2, 2}]/4
1/4
Other coefficients we find by direct integration:
Simplify[Integrate[f[x]*Exp[-n*Pi*I*x/2], {x, -2, 2}]/4 , Assumptions -> {Element[n, Integers], Element[x, Reals]}]
((-1)^n (8 I (-1 + I^(3 n)) + 4 (2 + I^(3 n)) n \[Pi] + I (2 + I^(3 n)) n^2 \[Pi]^2))/(2 n^3 \[Pi]^3)
Now we build partial sums with N = 10, 20, and 100 terms
F10[x_] = 1/4 + Simplify[Sum[ (((-1)^n (8 I (-1 + I^(3 n)) + 4 (2 + I^(3 n)) n \[Pi] + I (2 + I^(3 n)) n^2 \[Pi]^2))/(2 n^3 \[Pi]^3))*Exp[n*Pi*I*x/2], {n, 1, 10}] + Sum[ (((-1)^n (8 I (-1 + I^(3 n)) + 4 (2 + I^(3 n)) n \[Pi] + I (2 + I^(3 n)) n^2 \[Pi]^2))/(2 n^3 \[Pi]^3))*Exp[n*Pi*I*x/2], {n, -10, -1}]]
F20[x_] = 1/4 + Simplify[Sum[ (((-1)^n (8 I (-1 + I^(3 n)) + 4 (2 + I^(3 n)) n \[Pi] + I (2 + I^(3 n)) n^2 \[Pi]^2))/(2 n^3 \[Pi]^3))*Exp[n*Pi*I*x/2], {n, 1, 20}] + Sum[ (((-1)^n (8 I (-1 + I^(3 n)) + 4 (2 + I^(3 n)) n \[Pi] + I (2 + I^(3 n)) n^2 \[Pi]^2))/(2 n^3 \[Pi]^3))*Exp[n*Pi*I*x/2], {n, -20, -1}]]
F100[x_] = 1/4 + Simplify[Sum[ (((-1)^n (8 I (-1 + I^(3 n)) + 4 (2 + I^(3 n)) n \[Pi] + I (2 + I^(3 n)) n^2 \[Pi]^2))/(2 n^3 \[Pi]^3))*Exp[n*Pi*I*x/2], {n, 1, 100}] + Sum[ (((-1)^n (8 I (-1 + I^(3 n)) + 4 (2 + I^(3 n)) n \[Pi] + I (2 + I^(3 n)) n^2 \[Pi]^2))/(2 n^3 \[Pi]^3))*Exp[n*Pi*I*x/2], {n, -100, -1}]]
and plot them
Plot[{f[x], F10[x]}, {x, -2.5, 2.5}, PlotStyle -> {{Thick, Black}, {Thick, Red}}]
Plot[{f[x], F20[x]}, {x, -2.5, 2.5}, PlotStyle -> {{Thick, Black}, {Thick, Orange}}]
Plot[{f[x], F100[x]}, {x, -2.5, 2.5}, PlotStyle -> {{Thick, Black}, {Thick, Blue}}]
   Fourier approximation with 10 terms    Fourier approximation with 20 terms    Fourier approximation with 100 terms
       
   ■

Example 5: Let us consider the Heaviside function on a symmetric interval

\[ H(x) = \begin{cases} 1, & \ \mbox {if } 0 < x < \ell , \\ 0, & \ \mbox {if } -\ell < x< 0 . \end{cases} \]
Expanding the Heaviside function into the Fourier series
\[ H(x) = \mbox{P.V.} \sum_{k=-\infty}^{\infty} \alpha_k e^{k{\bf j} \pi x/\ell} = \lim_{N\to\infty} \sum_{k=-N}^{N} \alpha_k e^{k{\bf j} \pi x/\ell} , \]
where
\[ \alpha_k = \frac{1}{2\ell} \int_{0}^{\ell} e^{-k{\bf j} \pi x/\ell} \,{\text d} x = \frac{\bf j}{2k\pi} \left( 1 - e^{{\bf j}k\pi} \right) \]
Integrate[Exp[k*I*Pi*x/L], {x, 0, L}]/2/L
-((I (-1 + E^(I k \[Pi])))/(2 k \[Pi]))
Since \( e^{{\bf j}k\pi} = (-1)^k , \) , we get
\[ H(x) = \frac{1}{2} + \frac{2}{\pi} \sum_{k\ge 0} \frac{1}{2k+1}\,\sin \frac{\left( 2k+1 \right) \pi x}{\ell} . \]
FourierSeries[HeavisideTheta[t], t, 3]
1/2 + (I E^(-I t))/\[Pi] - (I E^(I t))/\[Pi] + (I E^(-3 I t))/( 3 \[Pi]) - (I E^(3 I t))/(3 \[Pi])
We check the series by plotting partial sums.
S20[x_] = 1/2 + (2/Pi)*Sum[Sin[(2*k + 1)*Pi*x]/(2*k + 1), {k, 0, 20}];
Plot[S20[x], {x, -2, 2}, PlotStyle -> Thick]
However, if you integrate Eq.\eqref{EqComplex.4} term-by-term, you will get another Fourier series expansion
\[ h(x) = \frac{1}{2} + \frac{x\pi}{2\ell} + \frac{1}{\pi} \sum_{k\ge 1} \frac{1}{k}\,\sin \left( \frac{k\pi x}{\ell} \right) . \]
Actually, the function h(x) is a ladder function that coinside with the Heaviside function on the interval (−2ℓ, 2ℓ), as the folloing plots confirm:
SS20[x_] = 1/2 + (x/2) + (1/Pi)*Sum[Sin[k*Pi*x]/(k), {k, 1, 20}];
Plot[S20[x], {x, -3, 5}, PlotStyle -> Thick]
Plot[SS20[x], {x, -3, 5}, PlotStyle -> Thick]
   Fourier approximation with 20 terms of H(x)          Fourier approximation with 20 terms of h(x)
         
   ■

 

 

Return to Mathematica page
Return to the main page (APMA0340)
Return to the Part 1 Matrix Algebra
Return to the Part 2 Linear Systems of Ordinary Differential Equations
Return to the Part 3 Non-linear Systems of Ordinary Differential Equations
Return to the Part 4 Numerical Methods
Return to the Part 5 Fourier Series
Return to the Part 6 Partial Differential Equations
Return to the Part 7 Special Functions