Chapter 17, Problem 25P

Determine the Fourier series representation of the function in Fig. 17.63.

Figure 17.63

To determine

Find the Fourier series representation of the function given in Figure 17.63.

Answer to Problem 25P

The Fourier series representation $f\left(t\right)$ of the function given in Figure 17.63 is ${\displaystyle \sum _{\begin{array}{l}n=1\\ n=\text{odd}\end{array}}^{\infty }\left(\left[\frac{4}{n\pi }\sin \left(\frac{2n\pi }{3}\right)+\frac{6}{n^2{\pi }^2}\left(\cos \left(\frac{2n\pi }{3}\right)-1\right)\right]\cos \left(\frac{2n\pi }{3}\right)t+\left[\frac{6}{n^2{\pi }^2}\sin \left(\frac{2n\pi }{3}\right)-\frac{4}{n\pi }\cos \left(\frac{2n\pi }{3}\right)\right]\sin \left(\frac{2n\pi }{3}\right)t\right)}$.

Explanation of Solution

Given data:

Refer to Figure 17.63 in the textbook.

Formula used:

Write the expression to calculate the fundamental angular frequency.

${\omega }_0=\frac{2\pi }{T}$ (1)

Here,

$T$ is the period of the function.

Write the general expression to calculate the Fourier series of $f\left(t\right)$.

$f\left(t\right)=a_0+{\displaystyle \sum _{n=1}^{\infty }\left(a_n\cos n{\omega }_{0}t+b_n\sin n{\omega }_{0}t\right)}$ (2)

Here,

$a_0$ is the dc component of $f\left(t\right)$,

$a_n$ and $b_n$ are the Fourier coefficients,

$n$ is an integer, and

${\omega }_0$ is the angular frequency.

Calculation:

The given waveform is drawn as Figure 1.

Refer to Figure 1, each half cycle is the mirror image of the next half cycle. Therefore, the waveform is called as half-wave symmetry and it is expressed as,

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

Write the expressions for Fourier coefficients for the half-wave odd symmetric function.

$a_0=0$

$a_n=\{\begin{array}{l}\frac{4}{T}{\displaystyle \underset{0}{\overset{\frac{T}{2}}{\int }}f\left(t\right)\cos n{\omega }_{0}tdt,n=\text{odd}}\\ 0,n=\text{even}\end{array}$ (3)

$b_n=\{\begin{array}{l}\frac{4}{T}{\displaystyle \underset{0}{\overset{\frac{T}{2}}{\int }}f\left(t\right)\sin n{\omega }_{0}tdt,n=\text{odd}}\\ 0,n=\text{even}\end{array}$ (4)

Refer to the Figure 1. The Fourier series function of the waveform is defined as,

$f\left(t\right)=\{\begin{array}{l}2t,0<t<1\\ 0,1<t<1.5\\ -2t+3,1.5<t<2.5\\ 0,2.5<t<3\end{array}$

The time period of the function in Figure 1 is,

$T=3$

Substitute $3$ for $T$ in equation (1) to find ${\omega }_0$.

${\omega }_0=\frac{2\pi }{3}$

Substitute $\frac{2\pi }{3}$ for ${\omega }_0$ and $3$ for $T$ for odd n in equation (3) to find $a_n$.

$\begin{array}{c}{a}_n=\frac{4}{3}{\displaystyle \underset{0}{\overset{\frac{3}{2}}{\int }}f\left(t\right)\cos n\left(\frac{2\pi }{3}\right)tdt}\\ =\frac{4}{3}{\displaystyle \underset{0}{\overset{1.5}{\int }}f\left(t\right)\cos \left(\frac{2n\pi }{3}\right)tdt}\\ =\frac{4}{3}\left({\displaystyle \underset{0}{\overset{1}{\int }}f\left(t\right)\cos \left(\frac{2n\pi }{3}\right)tdt}+\frac{4}{3}{\displaystyle \underset{1}{\overset{1.5}{\int }}f\left(t\right)\cos \left(\frac{2n\pi }{3}\right)tdt}\right)\\ =\frac{4}{3}\left({\displaystyle \underset{0}{\overset{1}{\int }}2t\cos \left(\frac{2n\pi }{3}\right)tdt}+\frac{4}{3}{\displaystyle \underset{1}{\overset{1.5}{\int }}\left(0\right)\cos \left(\frac{2n\pi }{3}\right)tdt}\right)\end{array}$

Simplify the above equation to find $a_n$.

$\begin{array}{c}{a}_n=\frac{4}{3}\left({\displaystyle \underset{0}{\overset{1}{\int }}2t\cos \left(\frac{2n\pi }{3}\right)tdt}+0\right)\\ =\frac{4}{3}\left({\displaystyle \underset{0}{\overset{1}{\int }}2t\cos \left(\frac{2n\pi }{3}\right)tdt}\right)\end{array}$

$a_n=\frac{8}{3}\left({\displaystyle \underset{0}{\overset{1}{\int }}t\cos \left(\frac{2n\pi }{3}\right)tdt}\right)$ (5)

Assume the following to reduce the equation (5).

$x={\displaystyle \underset{0}{\overset{1}{\int }}t\cos \left(\frac{2n\pi }{3}\right)tdt}$ (6)

Substitute the equation (6) in equation (5) to find $a_n$.

$a_n=\frac{8}{3}x$ (7)

Consider the following integration formula.

${\displaystyle \underset{a}{\overset{b}{\int }}udv}={\left[uv\right]}_a^b-{\displaystyle \underset{a}{\overset{b}{\int }}vdu}$ (8)

Compare the equations (6) and (8) to simplify the equation (6).

$\begin{array}{l}u=tdv=\cos \left(\frac{2n\pi }{3}\right)tdt\\ du=dtv=\frac{\sin \left(\frac{2n\pi }{3}\right)t}{\left(\frac{2n\pi }{3}\right)}\end{array}$

Using the equation (8), the equation (6) can be reduced as,

$\begin{array}{c}x={\displaystyle \underset{0}{\overset{1}{\int }}t\cos \left(\frac{2n\pi }{3}\right)tdt}\\ ={\left[\left(t\right)\frac{\sin \left(\frac{2n\pi }{3}\right)t}{\left(\frac{2n\pi }{3}\right)}\right]}_0^1-{\displaystyle \underset{0}{\overset{1}{\int }}\frac{\sin \left(\frac{2n\pi }{3}\right)t}{\left(\frac{2n\pi }{3}\right)}dt}\\ =\left[\left(1\right)\frac{\sin \left(\frac{2n\pi }{3}\right)\left(1\right)}{\left(\frac{2n\pi }{3}\right)}-\left(0\right)\frac{\sin \left(\frac{2n\pi }{3}\right)\left(0\right)}{\left(\frac{2n\pi }{3}\right)}\right]-\left(\frac{1}{\left(\frac{2n\pi }{3}\right)}\right){\left[-\frac{\cos \left(\frac{2n\pi }{3}\right)t}{\left(\frac{2n\pi }{3}\right)}\right]}_0^1\\ =\left[\frac{\sin \left(\frac{2n\pi }{3}\right)}{\left(\frac{2n\pi }{3}\right)}-0\right]+\frac{1}{{\left(\frac{2n\pi }{3}\right)}^2}\left[\cos \left(\frac{2n\pi }{3}\right)\left(1\right)-\cos \left(\frac{2n\pi }{3}\right)\left(0\right)\right]\end{array}$

Simplify the above equation to find $x$.

$\begin{array}{c}x=\frac{\sin \left(\frac{2n\pi }{3}\right)}{\left(\frac{2n\pi }{3}\right)}+\frac{1}{{\left(\frac{2n\pi }{3}\right)}^2}\left[\cos \left(\frac{2n\pi }{3}\right)-\cos 0°\right]\\ =\frac{3}{2n\pi }\sin \left(\frac{2n\pi }{3}\right)+\frac{1}{\left(\frac{4n^2{\pi }^2}{9}\right)}\left[\cos \left(\frac{2n\pi }{3}\right)-1\right]\left\{\because \cos \left(0°\right)=1\right\}\\ =\frac{3}{2n\pi }\sin \left(\frac{2n\pi }{3}\right)+\frac{9}{4n^2{\pi }^2}\left(\cos \left(\frac{2n\pi }{3}\right)-1\right)\end{array}$

Substitute $\frac{3}{2n\pi }\sin \left(\frac{2n\pi }{3}\right)+\frac{9}{4n^2{\pi }^2}\left(\cos \left(\frac{2n\pi }{3}\right)-1\right)$ for $x$ in equation (7) to find $a_n$.

$\begin{array}{c}{a}_n=\frac{8}{3}\left(\frac{3}{2n\pi }\sin \left(\frac{2n\pi }{3}\right)+\frac{9}{4n^2{\pi }^2}\left(\cos \left(\frac{2n\pi }{3}\right)-1\right)\right)\\ =\frac{24}{6n\pi }\sin \left(\frac{2n\pi }{3}\right)+\frac{72}{12n^2{\pi }^2}\left(\cos \left(\frac{2n\pi }{3}\right)-1\right)\\ =\frac{4}{n\pi }\sin \left(\frac{2n\pi }{3}\right)+\frac{6}{n^2{\pi }^2}\left(\cos \left(\frac{2n\pi }{3}\right)-1\right)\end{array}$

Substitute $\frac{2\pi }{3}$ for ${\omega }_0$ and $3$ for $T$ for odd n in equation (4) to find $b_n$.

$\begin{array}{c}{b}_n=\frac{4}{3}{\displaystyle \underset{0}{\overset{\frac{3}{2}}{\int }}f\left(t\right)\sin n\left(\frac{2\pi }{3}\right)tdt}\\ =\frac{4}{3}{\displaystyle \underset{0}{\overset{1.5}{\int }}f\left(t\right)\sin \left(\frac{2n\pi }{3}\right)tdt}\\ =\frac{4}{3}\left({\displaystyle \underset{0}{\overset{1}{\int }}f\left(t\right)\sin \left(\frac{2n\pi }{3}\right)tdt}+{\displaystyle \underset{1}{\overset{1.5}{\int }}f\left(t\right)\sin \left(\frac{2n\pi }{3}\right)tdt}\right)\\ =\frac{4}{3}\left({\displaystyle \underset{0}{\overset{1}{\int }}2t\sin \left(\frac{2n\pi }{3}\right)tdt}+{\displaystyle \underset{1}{\overset{1.5}{\int }}\left(0\right)\sin \left(\frac{2n\pi }{3}\right)tdt}\right)\end{array}$

Simplify the above equation to find $b_n$.

$\begin{array}{c}{b}_n=\frac{4}{3}\left({\displaystyle \underset{0}{\overset{1}{\int }}2t\sin \left(\frac{2n\pi }{3}\right)tdt}+0\right)\\ =\frac{4}{3}\left({\displaystyle \underset{0}{\overset{1}{\int }}2t\sin \left(\frac{2n\pi }{3}\right)tdt}\right)\end{array}$

$b_n=\frac{8}{3}\left({\displaystyle \underset{0}{\overset{1}{\int }}t\sin \left(\frac{2n\pi }{3}\right)tdt}\right)$ (9)

Assume the following to reduce the equation (9).

$y={\displaystyle \underset{0}{\overset{1}{\int }}t\sin \left(\frac{2n\pi }{3}\right)tdt}$ (10)

Substitute the equation (10) in equation (9) to find $b_n$.

$b_n=\frac{8}{3}y$ (11)

Compare the equations (8) and (10) to simplify the equation (10).

$\begin{array}{l}u=tdv=\sin \left(\frac{2n\pi }{3}\right)tdt\\ du=dtv=-\frac{\cos \left(\frac{2n\pi }{3}\right)t}{\left(\frac{2n\pi }{3}\right)}\end{array}$

Using the equation (8), the equation (10) can be reduced as,

$\begin{array}{c}y={\displaystyle \underset{0}{\overset{1}{\int }}t\sin \left(\frac{2n\pi }{3}\right)tdt}\\ ={\left[\left(t\right)\left(-\frac{\cos \left(\frac{2n\pi }{3}\right)t}{\left(\frac{2n\pi }{3}\right)}\right)\right]}_0^1-{\displaystyle \underset{0}{\overset{1}{\int }}\left(-\frac{\cos \left(\frac{2n\pi }{3}\right)t}{\left(\frac{2n\pi }{3}\right)}\right)dt}\\ =-\left[\left(1\right)\frac{\cos \left(\frac{2n\pi }{3}\right)\left(1\right)}{\left(\frac{2n\pi }{3}\right)}-\left(0\right)\frac{\cos \left(\frac{2n\pi }{3}\right)\left(0\right)}{\left(\frac{2n\pi }{3}\right)}\right]+\frac{1}{\left(\frac{2n\pi }{3}\right)}{\left[\frac{\sin \left(\frac{2n\pi }{3}\right)t}{\left(\frac{2n\pi }{3}\right)}\right]}_0^1\\ =-\left[\frac{\cos \left(\frac{2n\pi }{3}\right)}{\left(\frac{2n\pi }{3}\right)}-0\right]+\frac{1}{{\left(\frac{2n\pi }{3}\right)}^2}\left[\sin \left(\frac{2n\pi }{3}\right)\left(1\right)-\sin \left(\frac{2n\pi }{3}\right)\left(0\right)\right]\end{array}$

Simplify the above equation to find $y$.

$\begin{array}{c}y=-\frac{\cos \left(\frac{2n\pi }{3}\right)}{\left(\frac{2n\pi }{3}\right)}+\frac{1}{{\left(\frac{2n\pi }{3}\right)}^2}\left[\sin \left(\frac{2n\pi }{3}\right)-\sin \left(0°\right)\right]\\ =-\frac{3}{2n\pi }\cos \left(\frac{2n\pi }{3}\right)+\frac{1}{\left(\frac{4n^2{\pi }^2}{9}\right)}\left[\sin \left(\frac{2n\pi }{3}\right)-0\right]\left\{\because \sin \left(0°\right)=0\right\}\\ =-\frac{3}{2n\pi }\cos \left(\frac{2n\pi }{3}\right)+\frac{9}{4n^2{\pi }^2}\sin \left(\frac{2n\pi }{3}\right)\\ =\frac{9}{4n^2{\pi }^2}\sin \left(\frac{2n\pi }{3}\right)-\frac{3}{2n\pi }\cos \left(\frac{2n\pi }{3}\right)\end{array}$

Substitute $\frac{9}{4n^2{\pi }^2}\sin \left(\frac{2n\pi }{3}\right)-\frac{3}{2n\pi }\cos \left(\frac{2n\pi }{3}\right)$ for $y$ in equation (11) to find $b_n$.

$\begin{array}{c}{b}_n=\frac{8}{3}\left(\frac{9}{4n^2{\pi }^2}\sin \left(\frac{2n\pi }{3}\right)-\frac{3}{2n\pi }\cos \left(\frac{2n\pi }{3}\right)\right)\\ =\frac{72}{12n^2{\pi }^2}\sin \left(\frac{2n\pi }{3}\right)-\frac{24}{6n\pi }\cos \left(\frac{2n\pi }{3}\right)\\ =\frac{6}{n^2{\pi }^2}\sin \left(\frac{2n\pi }{3}\right)-\frac{4}{n\pi }\cos \left(\frac{2n\pi }{3}\right)\end{array}$

Substitute $\frac{2\pi }{3}$ for ${\omega }_0$, $0$ for $a_0$, $\frac{4}{n\pi }\sin \left(\frac{2n\pi }{3}\right)+\frac{6}{n^2{\pi }^2}\left(\cos \left(\frac{2n\pi }{3}\right)-1\right)$ for odd n for $a_n$ and $\frac{6}{n^2{\pi }^2}\sin \left(\frac{2n\pi }{3}\right)-\frac{4}{n\pi }\cos \left(\frac{2n\pi }{3}\right)$ for odd n for $b_n$ in equation (2) to find $f\left(t\right)$.

$\begin{array}{c}f\left(t\right)=0+{\displaystyle \sum _{\begin{array}{l}n=1\\ n=\text{odd}\end{array}}^{\infty }\left(\begin{array}{l}\left[\frac{4}{n\pi }\sin \left(\frac{2n\pi }{3}\right)+\frac{6}{n^2{\pi }^2}\left(\cos \left(\frac{2n\pi }{3}\right)-1\right)\right]\cos n\left(\frac{2\pi }{3}\right)t\\ +\left[\frac{6}{n^2{\pi }^2}\sin \left(\frac{2n\pi }{3}\right)-\frac{4}{n\pi }\cos \left(\frac{2n\pi }{3}\right)\right]\sin n\left(\frac{2\pi }{3}\right)t\end{array}\right)}\\ ={\displaystyle \sum _{\begin{array}{l}n=1\\ n=\text{odd}\end{array}}^{\infty }\left(\begin{array}{l}\left[\frac{4}{n\pi }\sin \left(\frac{2n\pi }{3}\right)+\frac{6}{n^2{\pi }^2}\left(\cos \left(\frac{2n\pi }{3}\right)-1\right)\right]\cos \left(\frac{2n\pi }{3}\right)t\\ +\left[\frac{6}{n^2{\pi }^2}\sin \left(\frac{2n\pi }{3}\right)-\frac{4}{n\pi }\cos \left(\frac{2n\pi }{3}\right)\right]\sin \left(\frac{2n\pi }{3}\right)t\end{array}\right)}\end{array}$

Conclusion:

Thus, the Fourier series representation $f\left(t\right)$ of the function given in Figure 17.63 is ${\displaystyle \sum _{\begin{array}{l}n=1\\ n=\text{odd}\end{array}}^{\infty }\left(\left[\frac{4}{n\pi }\sin \left(\frac{2n\pi }{3}\right)+\frac{6}{n^2{\pi }^2}\left(\cos \left(\frac{2n\pi }{3}\right)-1\right)\right]\cos \left(\frac{2n\pi }{3}\right)t+\left[\frac{6}{n^2{\pi }^2}\sin \left(\frac{2n\pi }{3}\right)-\frac{4}{n\pi }\cos \left(\frac{2n\pi }{3}\right)\right]\sin \left(\frac{2n\pi }{3}\right)t\right)}$.