# MATLAB TUTORIAL for the Second Cource. Part 6: Complex Fourier series

Complex Fourier Series

# 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:
$f(x) \,\sim\, \sum_{k=-\infty}^{\infty} \alpha_k e^{k{\bf j} \pi x/\ell}$
where a signal's complex Fourier spectrum is
$\alpha_k = \frac{1}{2\ell} \int_{-\ell}^{\ell} f(x)\, e^{-k{\bf j} \pi x/\ell} \,{\text d} x , \qquad k=0, \pm 1, \pm 2, \ldots ;$
provided that this series converges in some sense. The above formula is based on the orthogonality property of exponential functions:
$\int_0^T e^{{\bf j}2\pi kt/T} \,e^{-{\bf j}2\pi nt/T} \,{\text d}t = \left\{ \begin{array}{ll} T, & \ \mbox{if k=n}, \\ 0 , & \ \mbox{if } k\ne n . \end{array} \right.$
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 $${\bf j}^2 =-1.$$ matlab has a default command to calculate complex Fourier series:

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

matlab has a special command to find complex Fourier coefficent 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 trigonomtric 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:

$\left\vert \frac{b}{2\,\pi} \right\vert^{(a+1)/2} \, \int_{-\pi/|b|}^{\pi/|b|} f(t)\,e^{-{\bf j}bnt} \,{\text d}t .$
The complex exponential Fourier form has the following advantages compaired to traditional trigonometric form:
• only need to perform one integration;
• a single exponential can be manipulated more easily than a sum of sinusoids, and
• it provides a logical transition into a further discussion of the Fourier Transform.
A signal's Fourier series spectrum $$\alpha_k$$ has interesting properties.
• If f(t) is a real function, then $$\alpha_{-k} = \overline{\alpha_k} ;$$
• if f(t) is an even function, then $$\alpha_{-k} = \alpha_k ;$$ if f(t) is an odd function, then $$\alpha_{-k} = -\alpha_k ,$$ , and
• the Fourier series obeys Parseval's Theorem
$\int_0^T f^2 (t) \,{\text d}t = T\,\sum_k \left\vert \alpha_k \right\vert^2 .$

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

$f(x) = \left\{ \begin{array}{ll} 1, & \ \mbox{on the interval (-2 , -1)}, \\ 0 , & \ \mbox{on the interval } (-1, 0) , \\ 2, & \ \mbox{on the interval } (0, 2). \end{array} \right.$
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 comamnd: FourierSeries[f[x], x, 20, FourierParameters -> {1, Pi/2}] // ComplexExpand 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} .$
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 $$(-1)^5 + \cos \frac{5\pi}{2} -2 = -3 .$$

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]
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.

Example: 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 instead of $$\cos x$$ its Euler representation: $$\cos x = \frac{1}{2}\,e^{{\bf j}x} + \frac{1}{2}\,e^{-{\bf j}x} .$$ 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 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 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: Our next example deals with the periodic 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, definding 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' s 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, 1]
We expand the pulse function into exponential Fourier series:
$\Pi (t,h,T) = \sum_{k=-\infty}^{\infty} \alpha_k e^{{\bf j}k2\pi x/T}, \qquad \alpha_k = \frac{1}{T} \,\int_0^h e^{-{\bf j}k2\pi x/T} \,{\text d}x = \frac{\bf j}{2k\pi} \left[ e^{-{\bf j}k2\pi h/T} -1 \right] , \quad k=0, \pm 1, \pm 2, \ldots .$
Note that the value of $$\alpha_0 = h/T$$ 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: $$\Pi (0,h,T) = \Pi (h,h,T) = 1/2 .$$ We can also convert the complex Fourier form to a regular real trigonometric form:
$\Pi (t,h,T) = \frac{h}{T} + \sum_{k \ge 0} \frac{1}{k\pi} \left[ \sin \frac{2hk\pi}{T} \, \cos \frac{2\pi kx}{T} + 2\,\sin^2 \frac{k\pi h}{T} \,\sin \frac{2\pi kx}{T} \right] .$
Then we plot partial sums with 10 and 50 terms for particular numerical values $$h=1 \quad\mbox{and}\quad T=3:$$
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 $$\Pi (1,1,3)=1/2.$$

# Motivated examples

Complete


Complete

#### II. Plotting

Direction fields

Since Matlab has no friendly subroutine to plot direction fileds for ODEs, we present several codes that allow to plot these fields directly. Many others can found on the Internet.

Separable equations

Enter text here

Equations reducible to separable equations
Exact equations
Integrating Factors
Linear and Bernoulli equations
Riccati equation
Existence and Uniqueness
Qualitative analysis
Applications