Chapter 19, Problem 4P

Use a continuous Fourier series to approximate the sawtooth wave in Fig. P19.4. Plot the first three terms along with the summation.

FIGURE P19.4

A sawtooth wave.

To determine

To calculate: The Fourier series expansion to approximate the sawtooth wave, as shown in the following figure,

Plot the first three terms along with the summation.

Answer to Problem 4P

Solution:

The Fourier series of the sawtooth curve is $\underset{\_}{f\left(t\right)=\frac{2}{\pi }\left[-\sin \left({\omega }_{0}t\right)+\frac{1}{2}\sin \left(2{\omega }_{0}t\right)-\frac{1}{3}\sin \left(3{\omega }_{0}t\right)+\mathrm{...}\right]}$.

Explanation of Solution

Given Information: The sawtooth wave shown as,

Formula used:

Consider $f\left(t\right)$ is a periodic function with period $T$ defined in the interval $0\le x\le T$ then the Fourier series expansion of the function is,

$f\left(t\right)=a_0+{\displaystyle \sum _{k=1}^{\infty }\left[a_k\cos \left(k{\omega }_{0}t\right)+b_k\sin \left(k{\omega }_{0}t\right)\right]};{\omega }_0=\frac{2\pi }{T}$

And the coefficients are defined by,

$$a_0=\frac{1}{T}{\displaystyle \underset{0}{\overset{T}{\int }}f\left(t\right)dt};a_k=\frac{2}{T}{\displaystyle \underset{0}{\overset{T}{\int }}f\left(t\right)\cos \left(k{\omega }_{0}t\right)dt};b_k=\frac{2}{T}{\displaystyle \underset{0}{\overset{T}{\int }}f\left(t\right)\sin \left(k{\omega }_{0}t\right)dt}$$

Calculation:

Consider the sawtooth wave as shown in the following figure,

Therefore, the sawtooth wave is a periodic function $f\left(t\right)$ with period $T$ in the interval $0\le t\le T$ and from the graph,

$f\left(t\right)$ is a straight line joining the points $\left(0,0\right)$ and $\left(\frac{T}{2},-1\right)$ in $0\le t\le \frac{T}{2}$

; $f\left(t\right)$ is a straight line joining the points $\left(\frac{T}{2},1\right)$ and $\left(T,0\right)$ in $\frac{T}{2}\le t\le T$.

Therefore, the sawtooth wave,

$f\left(t\right)=\{\begin{array}{cc}-\frac{2}{T}t,& 0\le t\le \frac{T}{2}\\ 2-\frac{2}{T}t,& \frac{T}{2}\le t\le T\end{array}$

Therefore, the Fourier series expansion of the function $f\left(t\right)$ is,

$f\left(t\right)=a_0+{\displaystyle \sum _{k=1}^{\infty }\left[a_k\cos \left(k{\omega }_{0}t\right)+b_k\sin \left(k{\omega }_{0}t\right)\right]}$ with ${\omega }_0=\frac{2\pi }{T}$.

Here, the coefficients are defined by,

$a_0=\frac{1}{T}{\displaystyle \underset{0}{\overset{T}{\int }}f\left(t\right)dt};a_k=\frac{2}{T}{\displaystyle \underset{0}{\overset{T}{\int }}f\left(t\right)\cos \left(\frac{2k\pi }{T}t\right)dt};b_k=\frac{2}{T}{\displaystyle \underset{0}{\overset{T}{\int }}f\left(t\right)\sin \left(\frac{2k\pi }{T}t\right)dt}$

Now, find $a_k$.

$\begin{array}{c}{a}_k=\frac{2}{T}{\displaystyle \underset{0}{\overset{T}{\int }}f\left(t\right)\cos \left(\frac{2k\pi }{T}t\right)dt}\\ =\frac{2}{T}{\displaystyle \underset{0}{\overset{\frac{T}{2}}{\int }}\left(-\frac{2}{T}t\right)\cos \left(\frac{2k\pi }{T}t\right)dt}+\frac{2}{T}{\displaystyle \underset{\frac{T}{2}}{\overset{T}{\int }}\left(2-\frac{2}{T}t\right)\cos \left(\frac{2k\pi }{T}t\right)dt}\\ =-\frac{4}{T^2}{\displaystyle \underset{0}{\overset{\frac{T}{2}}{\int }}t\cos \left(\frac{2k\pi }{T}t\right)dt}+\frac{4}{T}{\displaystyle \underset{\frac{T}{2}}{\overset{T}{\int }}\cos \left(\frac{2k\pi }{T}t\right)dt}-\frac{4}{T^2}{\displaystyle \underset{\frac{T}{2}}{\overset{T}{\int }}t\cos \left(\frac{2k\pi }{T}t\right)dt}\end{array}$

Consider,

$\begin{array}{c}I={\displaystyle \int t\cos \left(\frac{2k\pi }{T}t\right)dt}\\ =t{\displaystyle \int \cos \left(\frac{2k\pi }{T}t\right)dt}-{\displaystyle \int \left\{\frac{d}{dt}\left(t\right){\displaystyle \int \cos \left(\frac{2k\pi }{T}t\right)dt}\right\}dt}\\ =t\frac{T}{2k\pi }\sin \left(\frac{2k\pi }{T}t\right)-{\displaystyle \int \frac{T}{2k\pi }\sin \left(\frac{2k\pi }{T}t\right)dt}\\ =\frac{Tt}{2k\pi }\sin \left(\frac{2k\pi }{T}t\right)+\frac{T^2}{4k^2{\pi }^2}\cos \left(\frac{2k\pi }{T}t\right)\end{array}$

Thus,

$\begin{array}{c}{\displaystyle \underset{0}{\overset{\frac{T}{2}}{\int }}t\cos \left(\frac{2k\pi }{T}t\right)dt}={\left[\frac{Tt}{2k\pi }\sin \left(\frac{2k\pi }{T}t\right)+\frac{T^2}{4k^2{\pi }^2}\cos \left(\frac{2k\pi }{T}t\right)\right]}_0^{\frac{T}{2}}\\ =\frac{T}{2k\pi }\frac{T}{2}\sin \left(\frac{2k\pi }{T}\frac{T}{2}\right)+\frac{T^2}{4k^2{\pi }^2}\cos \left(\frac{2k\pi }{T}\frac{T}{2}\right)-\frac{T^2}{4k^2{\pi }^2}\cos \left(0\right)\\ =\frac{T^2}{4k\pi }\sin \left(k\pi \right)+\frac{T^2}{4k^2{\pi }^2}\cos \left(k\pi \right)-\frac{T^2}{4k^2{\pi }^2}\\ =\frac{T^2}{4k^2{\pi }^2}{\left(-1\right)}^k-\frac{T^2}{4k^2{\pi }^2}\end{array}$

$\begin{array}{c}{\displaystyle \underset{\frac{T}{2}}{\overset{T}{\int }}\cos \left(\frac{2k\pi }{T}t\right)dt}=\frac{T}{2k\pi }{\sin \left(\frac{2k\pi }{T}t\right)|}_{\frac{T}{2}}^T\\ =\frac{T}{2k\pi }\left\{\sin \left(2k\pi \right)-\sin \left(k\pi \right)\right\}\\ =0\end{array}$

$\begin{array}{c}{\displaystyle \underset{\frac{T}{2}}{\overset{T}{\int }}t\cos \left(\frac{2k\pi }{T}t\right)dt}={\left[\frac{Tt}{2k\pi }\sin \left(\frac{2k\pi }{T}t\right)+\frac{T^2}{4k^2{\pi }^2}\cos \left(\frac{2k\pi }{T}t\right)\right]}_{\frac{T}{2}}^T\\ =\frac{T^2}{2k\pi }\sin \left(2k\pi \right)+\frac{T^2}{4k^2{\pi }^2}\cos \left(2k\pi \right)-\frac{T^2}{4k\pi }\sin \left(k\pi \right)-\frac{T^2}{4k^2{\pi }^2}\cos \left(k\pi \right)\\ =\frac{T^2}{4k^2{\pi }^2}-\frac{T^2}{4k^2{\pi }^2}{\left(-1\right)}^k\end{array}$

Therefore,

$\begin{array}{c}{a}_k=-\frac{4}{T^2}\left[\frac{T^2}{4k^2{\pi }^2}{\left(-1\right)}^k-\frac{T^2}{4k^2{\pi }^2}\right]+\frac{4}{T}\times 0-\frac{4}{T^2}\left[\frac{T^2}{4k^2{\pi }^2}-\frac{T^2}{4k^2{\pi }^2}{\left(-1\right)}^k\right]\\ =-\frac{1}{k^2{\pi }^2}{\left(-1\right)}^k+\frac{1}{k^2{\pi }^2}-\frac{1}{k^2{\pi }^2}+\frac{1}{k^2{\pi }^2}{\left(-1\right)}^k\\ =0\end{array}$

Now, find $b_k$.

$\begin{array}{c}{b}_k=\frac{2}{T}{\displaystyle \underset{0}{\overset{T}{\int }}f\left(t\right)\sin \left(\frac{2k\pi }{T}t\right)dt}\\ =\frac{2}{T}{\displaystyle \underset{0}{\overset{\frac{T}{2}}{\int }}\left(-\frac{2}{T}t\right)\sin \left(\frac{2k\pi }{T}t\right)dt}+\frac{2}{T}{\displaystyle \underset{\frac{T}{2}}{\overset{T}{\int }}\left(2-\frac{2}{T}t\right)\sin \left(\frac{2k\pi }{T}t\right)dt}\\ =-\frac{4}{T^2}{\displaystyle \underset{0}{\overset{\frac{T}{2}}{\int }}t\sin \left(\frac{2k\pi }{T}t\right)dt}+\frac{4}{T}{\displaystyle \underset{\frac{T}{2}}{\overset{T}{\int }}\sin \left(\frac{2k\pi }{T}t\right)dt}-\frac{4}{T^2}{\displaystyle \underset{\frac{T}{2}}{\overset{T}{\int }}t\sin \left(\frac{2k\pi }{T}t\right)dt}\end{array}$

Consider,

$\begin{array}{c}I={\displaystyle \int t\sin \left(\frac{2k\pi }{T}t\right)dt}\\ =t{\displaystyle \int \sin \left(\frac{2k\pi }{T}t\right)dt}-{\displaystyle \int \left\{\frac{d}{dt}\left(t\right){\displaystyle \int \sin \left(\frac{2k\pi }{T}t\right)dt}\right\}dt}\\ =-t\frac{T}{2k\pi }\cos \left(\frac{2k\pi }{T}t\right)+{\displaystyle \int \frac{T}{2k\pi }\cos \left(\frac{2k\pi }{T}t\right)dt}\\ =-\frac{Tt}{2k\pi }\cos \left(\frac{2k\pi }{T}t\right)+\frac{T^2}{4k^2{\pi }^2}\sin \left(\frac{2k\pi }{T}t\right)\end{array}$

Thus,

$\begin{array}{c}{\displaystyle \underset{0}{\overset{\frac{T}{2}}{\int }}t\sin \left(\frac{2k\pi }{T}t\right)dt}={\left[-\frac{Tt}{2k\pi }\cos \left(\frac{2k\pi }{T}t\right)+\frac{T^2}{4k^2{\pi }^2}\sin \left(\frac{2k\pi }{T}t\right)\right]}_0^{\frac{T}{2}}\\ =-\frac{T^2}{4k\pi }\cos \left(k\pi \right)+\frac{T^2}{4k^2{\pi }^2}\sin \left(k\pi \right)\\ =-\frac{T^2}{4k\pi }{\left(-1\right)}^k\end{array}$

$\begin{array}{c}{\displaystyle \underset{\frac{T}{2}}{\overset{T}{\int }}\sin \left(\frac{2k\pi }{T}t\right)dt}=-\frac{T}{2k\pi }{\cos \left(\frac{2k\pi }{T}t\right)|}_{\frac{T}{2}}^T\\ =-\frac{T}{2k\pi }\cos \left(2k\pi \right)+\frac{T}{2k\pi }\cos \left(k\pi \right)\\ =-\frac{T}{2k\pi }+\frac{T}{2k\pi }{\left(-1\right)}^k\end{array}$

Further,

$\begin{array}{c}{\displaystyle \underset{\frac{T}{2}}{\overset{T}{\int }}t\sin \left(\frac{2k\pi }{T}t\right)dt}={\left[-\frac{Tt}{2k\pi }\cos \left(\frac{2k\pi }{T}t\right)+\frac{T^2}{4k^2{\pi }^2}\sin \left(\frac{2k\pi }{T}t\right)\right]}_{\frac{T}{2}}^T\\ =-\frac{T^2}{2k\pi }\cos \left(2k\pi \right)+\frac{T^2}{4k^2{\pi }^2}\sin \left(2k\pi \right)+\frac{T^2}{4k\pi }\cos \left(k\pi \right)-\frac{T^2}{4k^2{\pi }^2}\sin \left(k\pi \right)\\ =-\frac{T^2}{2k\pi }+\frac{T^2}{4k\pi }{\left(-1\right)}^k\end{array}$

Therefore,

$\begin{array}{c}{b}_k=-\frac{4}{T^2}\left[-\frac{T^2}{4k\pi }{\left(-1\right)}^k\right]+\frac{4}{T}\left[-\frac{T}{2k\pi }+\frac{T}{2k\pi }{\left(-1\right)}^k\right]-\frac{4}{T^2}\left[-\frac{T^2}{2k\pi }+\frac{T^2}{4k\pi }{\left(-1\right)}^k\right]\\ =\frac{1}{k\pi }{\left(-1\right)}^k-\frac{2}{k\pi }+\frac{2}{k\pi }{\left(-1\right)}^k+\frac{2}{k\pi }-\frac{1}{k\pi }{\left(-1\right)}^k\\ =\frac{2}{k\pi }{\left(-1\right)}^k\end{array}$

Therefore, the coefficients of the Fourier series expansions are,

$a_0=0,a_k=0,b_k=\frac{2}{k\pi }{\left(-1\right)}^k$

Therefore, the Fourier series expansion defines ${\omega }_0=\frac{2\pi }{T}$, $\begin{array}{c}f\left(t\right)={\displaystyle \sum _{k=1}^{\infty }\left[\frac{2}{k\pi }{\left(-1\right)}^k\sin \left(k{\omega }_{0}t\right)\right]}\\ =\frac{2}{\pi }{\displaystyle \sum _{k=1}^{\infty }\left[\frac{{\left(-1\right)}^k}{k}\sin \left(k{\omega }_{0}t\right)\right]}\end{array}$

Hence, $f\left(t\right)=\frac{2}{\pi }\left[-\sin \left({\omega }_{0}t\right)+\frac{1}{2}\sin \left(2{\omega }_{0}t\right)-\frac{1}{3}\sin \left(3{\omega }_{0}t\right)+\mathrm{...}\right]$

Graph:

To plot the given sawtooth curve and the approximated Fourier series consider the period, $T=1$.

Therefore, the periodic sawtooth curve is expressed as,

$f\left(t\right)=\{\begin{array}{cc}-2t,& 0\le t\le \frac{1}{2}\\ 2-2t,& \frac{1}{2}\le t\le 1\end{array}$

And, the corresponding Fourier series is,

$f\left(t\right)=\frac{2}{\pi }\left[-\sin \left({\omega }_{0}t\right)+\frac{1}{2}\sin \left(2{\omega }_{0}t\right)-\frac{1}{3}\sin \left(3{\omega }_{0}t\right)+\mathrm{...}\right],{\omega }_0=2\pi$

Hence, $f\left(t\right)=-\frac{2}{\pi }\sin \left(2\pi t\right)+\frac{1}{\pi }\sin \left(4\pi t\right)-\frac{2}{3\pi }\sin \left(6\pi t\right)+\mathrm{...}$

Use the following MATLAB code to plot the first three terms along with the summation.

function Code_97924_19_4P()

% enter the time period

T = 1;

t1 = 0: 0.0001: T/2;

t2 = T/2: 0.0001: T;

% define the function

f1 = @(t) -(2/T)*t;

f2 = @(t) 2-(2/T)*t;

t = [t1, t2];

f = [f1(t1), f2(t2)];

w0 = 2*pi/T;

% define the first three terms in the series

Term1 = @(t) -(2/pi)*sin(w0*t);

Term2 = @(t) (1/pi)*sin(2*w0*t);

Term3 = @(t) -(2/(3*pi))*sin(3*w0*t);

ff = @(t) Term1(t) + Term2(t) + Term3(t);

% plot the results

figure(); hold on

plot(t,f,'–k','Linewidth',2);

plot(t, Term1(t),'-.m','Linewidth',2);

plot(t, Term2(t),'-.r','Linewidth',2);

plot(t, Term3(t),'-.g','Linewidth',2);

plot(t, ff(t),'b','Linewidth',2);

% define the graphical properties

ylim([-2 2]); grid on

legend('Original','1^{st} Term','2^{nd} Term','3^{rd} Term','Sum')

set(gca,'XTick',[0 T/4 T/2 3*T/4 T]);

set(gca,'XTickLabel',{'0' 'T/4' 'T/2' '3T/4' 'T'});

set(gca,'Fontname','Times New Roman','FontSize',12);

end

Execute the above code to obtain the plot as,

Interpretation: The above plot shows the comparison between the variation in the first three terms of the series along with the variation in summation.