Chapter 17, Problem 19P

Obtain the Fourier series for the periodic waveform in Fig. 17.57.

To determine

Find the Fourier series for the periodic waveform in Figure 17.57.

Answer to Problem 19P

The Fourier series $f\left(t\right)$ for the periodic waveform in Figure 17.57 is $2.5+{\displaystyle \sum _{n=1}^{\infty }\frac{20}{n^2{\pi }^2}\left(\left(2\cos \left(\frac{n\pi }{2}\right)-\cos \left(n\pi \right)-1\right)\cos \left(\frac{n\pi }{2}\right)t+\left(2\sin \left(\frac{n\pi }{2}\right)\right)\sin \left(\frac{n\pi }{2}\right)t\right)}$.

Explanation of Solution

Given data:

Refer to Figure 17.57 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 trigonometric 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.

Write the expression to calculate the dc component of the function $f\left(t\right)$.

$a_0=\frac{1}{T}{\displaystyle \underset{0}{\overset{T}{\int }}f\left(t\right)}dt$ (3)

Write the expression to calculate Fourier coefficients.

$a_n=\frac{2}{T}{\displaystyle \underset{0}{\overset{T}{\int }}f\left(t\right)\cos n{\omega }_{0}t}dt$ (4)

$b_n=\frac{2}{T}{\displaystyle \underset{0}{\overset{T}{\int }}f\left(t\right)\sin n{\omega }_{0}t}dt$ (5)

Calculation:

The given waveform is drawn as Figure 1.

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

$f\left(t\right)=\{\begin{array}{l}10t,0<t<1\\ 20-10t,1<t<2\\ 0,2<t<4\end{array}$

The time period of the function in Figure 1 is,

$T=4$

Substitute $4$ for $T$ in equation (1) to find ${\omega }_0$.

$\begin{array}{c}{\omega }_0=\frac{2\pi }{4}\\ =\frac{\pi }{2}\end{array}$

Substitute $4$ for $T$ in equation (3) to find $a_0$.

$\begin{array}{c}{a}_0=\frac{1}{4}{\displaystyle \underset{0}{\overset{4}{\int }}f\left(t\right)}dt\\ =\frac{1}{4}\left({\displaystyle \underset{0}{\overset{1}{\int }}f\left(t\right)}dt+{\displaystyle \underset{1}{\overset{2}{\int }}f\left(t\right)}dt+{\displaystyle \underset{2}{\overset{4}{\int }}f\left(t\right)}dt\right)\\ =\frac{1}{4}\left({\displaystyle \underset{0}{\overset{1}{\int }}10t}dt+{\displaystyle \underset{1}{\overset{2}{\int }}\left(20-10t\right)}dt+{\displaystyle \underset{2}{\overset{4}{\int }}0}dt\right)\\ =\frac{1}{4}\left(10{\displaystyle \underset{0}{\overset{1}{\int }}t}dt+10{\displaystyle \underset{1}{\overset{2}{\int }}\left(2-t\right)}dt+0\right)\end{array}$

Simplify the above equation to find $a_0$.

$\begin{array}{c}{a}_0=\frac{1}{4}\left(10{\left[\frac{t^2}{2}\right]}_0^1+10{\left[2t-\frac{t^2}{2}\right]}_1^2\right)\\ =\frac{1}{4}\left(\frac{10}{2}\left(1^2-0^2\right)+10\left[\left(2\left(2\right)-\frac{{\left(2\right)}^2}{2}\right)-\left(2\left(1\right)-\frac{{\left(1\right)}^2}{2}\right)\right]\right)\\ =\frac{1}{4}\left(5+10\left[\left(4-\frac{4}{2}\right)-\left(2-\frac{1}{2}\right)\right]\right)\\ =2.5\end{array}$

Simplify the above equation to find $a_0$.

$\begin{array}{c}{a}_0=5\left[\left(1-\frac{1}{2}\right)-0\right]\\ =5\left(0.5\right)\\ =2.5\end{array}$

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

$\begin{array}{c}{a}_n=\frac{2}{4}{\displaystyle \underset{0}{\overset{4}{\int }}f\left(t\right)\cos n\left(\frac{\pi }{2}\right)t}dt\\ =\frac{1}{2}{\displaystyle \underset{0}{\overset{4}{\int }}f\left(t\right)\cos \left(\frac{n\pi }{2}\right)t}dt\\ =\frac{1}{2}\left[{\displaystyle \underset{0}{\overset{1}{\int }}f\left(t\right)\cos \left(\frac{n\pi }{2}\right)t}dt+{\displaystyle \underset{1}{\overset{2}{\int }}f\left(t\right)\cos \left(\frac{n\pi }{2}\right)t}dt+{\displaystyle \underset{2}{\overset{4}{\int }}f\left(t\right)\cos \left(\frac{n\pi }{2}\right)t}dt\right]\\ =\frac{1}{2}\left[{\displaystyle \underset{0}{\overset{1}{\int }}10t\cos \left(\frac{n\pi }{2}\right)t}dt+{\displaystyle \underset{1}{\overset{2}{\int }}\left(20-10t\right)\cos \left(\frac{n\pi }{2}\right)t}dt+{\displaystyle \underset{2}{\overset{4}{\int }}\left(0\right)\cos \left(\frac{n\pi }{2}\right)t}dt\right]\end{array}$

Simplify the above equation to find $a_n$.

$\begin{array}{c}{a}_n=\frac{1}{2}\left[10\left({\displaystyle \underset{0}{\overset{1}{\int }}t\cos \left(\frac{n\pi }{2}\right)t}dt\right)+{\displaystyle \underset{1}{\overset{2}{\int }}10\left(2-t\right)\cos \left(\frac{n\pi }{2}\right)t}dt+0\right]\\ =\frac{1}{2}\left[10\left({\displaystyle \underset{0}{\overset{1}{\int }}t\cos \left(\frac{n\pi }{2}\right)t}dt\right)+10\left({\displaystyle \underset{1}{\overset{2}{\int }}\left(2-t\right)\cos \left(\frac{n\pi }{2}\right)t}dt\right)\right]\\ =\frac{10}{2}\left[{\displaystyle \underset{0}{\overset{1}{\int }}t\cos \left(\frac{n\pi }{2}\right)t}dt+{\displaystyle \underset{1}{\overset{2}{\int }}\left(2-t\right)\cos \left(\frac{n\pi }{2}\right)t}dt\right]\end{array}$

$a_n=5\left[{\displaystyle \underset{0}{\overset{1}{\int }}t\cos \left(\frac{n\pi }{2}\right)t}dt+{\displaystyle \underset{1}{\overset{2}{\int }}\left(2-t\right)\cos \left(\frac{n\pi }{2}\right)t}dt\right]$ (6)

Assume the following to reduce the equation (6).

$x={\displaystyle \underset{0}{\overset{1}{\int }}t\cos \left(\frac{n\pi }{2}\right)t}dt$ (7)

$y={\displaystyle \underset{1}{\overset{2}{\int }}\left(2-t\right)\cos \left(\frac{n\pi }{2}\right)t}dt$ (8)

Substitute the equations (7) and (8) in equation (6) to find $a_n$.

$a_n=5\left[x+y\right]$ (9)

Consider the following integration formula.

${\displaystyle \underset{a}{\overset{b}{\int }}udv}={\left[uv\right]}_a^b-{\displaystyle \underset{a}{\overset{b}{\int }}vdu}$ (10)

Compare the equations (7) and (10) to simplify the equation (7).

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

Using the equation (10), the equation (7) can be reduced as,

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

Simplify the above equation to find $x$.

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

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

$\begin{array}{l}u=2-tdv=\cos \left(\frac{n\pi }{2}\right)tdt\\ du=0-dtv=\frac{\sin \left(\frac{n\pi }{2}\right)t}{\left(\frac{n\pi }{2}\right)}\\ du=-dt\end{array}$

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

$\begin{array}{c}y={\displaystyle \underset{1}{\overset{2}{\int }}\left(2-t\right)\cos \left(\frac{n\pi }{2}\right)t}dt\\ ={\left[\left(2-t\right)\frac{\sin \left(\frac{n\pi }{2}\right)t}{\left(\frac{n\pi }{2}\right)}\right]}_1^2-{\displaystyle \underset{1}{\overset{2}{\int }}\frac{\sin \left(\frac{n\pi }{2}\right)t}{\left(\frac{n\pi }{2}\right)}\left(-dt\right)}\\ =\left[\left(\left(2-2\right)\frac{\sin \left(\frac{n\pi }{2}\right)\left(2\right)}{\left(\frac{n\pi }{2}\right)}\right)-\left(\left(2-1\right)\frac{\sin \left(\frac{n\pi }{2}\right)\left(1\right)}{\left(\frac{n\pi }{2}\right)}\right)\right]+\frac{1}{\left(\frac{n\pi }{2}\right)}{\displaystyle \underset{1}{\overset{2}{\int }}\sin \left(\frac{n\pi }{2}\right)tdt}\\ =\left[0-\left(\frac{\sin \left(\frac{n\pi }{2}\right)}{\left(\frac{n\pi }{2}\right)}\right)\right]+\frac{1}{\left(\frac{n\pi }{2}\right)}{\left[-\frac{\cos \left(\frac{n\pi }{2}\right)t}{\left(\frac{n\pi }{2}\right)}\right]}_1^2\end{array}$

Simplify the above equation to find $y$.

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

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

$\begin{array}{c}{a}_n=5\left[\frac{2}{n\pi }\sin \left(\frac{n\pi }{2}\right)+\frac{4}{n^2{\pi }^2}\left[\cos \left(\frac{n\pi }{2}\right)-1\right]-\frac{2}{n\pi }\sin \left(\frac{n\pi }{2}\right)-\frac{4}{n^2{\pi }^2}\left[\cos \left(n\pi \right)-\cos \left(\frac{n\pi }{2}\right)\right]\right]\\ =5\left[\frac{4}{n^2{\pi }^2}\cos \left(\frac{n\pi }{2}\right)-\frac{4}{n^2{\pi }^2}-\frac{4}{n^2{\pi }^2}\cos \left(n\pi \right)+\frac{4}{n^2{\pi }^2}\cos \left(\frac{n\pi }{2}\right)\right]\\ =5\left[\frac{8}{n^2{\pi }^2}\cos \left(\frac{n\pi }{2}\right)-\frac{4}{n^2{\pi }^2}-\frac{4}{n^2{\pi }^2}\cos \left(n\pi \right)\right]\\ =\frac{20}{{\pi }^2}\left[\frac{2}{n^2}\cos \left(\frac{n\pi }{2}\right)-\frac{1}{n^2}\cos \left(n\pi \right)-\frac{1}{n^2}\right]\end{array}$

Substitute $4$ for $T$ and $\frac{\pi }{2}$ for ${\omega }_0$ in equation (5) to find $b_n$.

$\begin{array}{c}{b}_n=\frac{2}{4}{\displaystyle \underset{0}{\overset{4}{\int }}f\left(t\right)\sin n\left(\frac{\pi }{2}\right)t}dt\\ =\frac{1}{2}{\displaystyle \underset{0}{\overset{4}{\int }}f\left(t\right)\sin \left(\frac{n\pi }{2}\right)t}dt\\ =\frac{1}{2}\left[{\displaystyle \underset{0}{\overset{1}{\int }}f\left(t\right)\sin \left(\frac{n\pi }{2}\right)t}dt+{\displaystyle \underset{1}{\overset{2}{\int }}f\left(t\right)\sin \left(\frac{n\pi }{2}\right)t}dt+{\displaystyle \underset{2}{\overset{4}{\int }}f\left(t\right)\sin \left(\frac{n\pi }{2}\right)t}dt\right]\\ =\frac{1}{2}\left[{\displaystyle \underset{0}{\overset{1}{\int }}10t\sin \left(\frac{n\pi }{2}\right)t}dt+{\displaystyle \underset{1}{\overset{2}{\int }}\left(20-10t\right)\sin \left(\frac{n\pi }{2}\right)t}dt+{\displaystyle \underset{2}{\overset{4}{\int }}\left(0\right)\sin \left(\frac{n\pi }{2}\right)t}dt\right]\end{array}$

Simplify the above equation to find $b_n$.

$\begin{array}{c}{b}_n=\frac{1}{2}\left[10\left({\displaystyle \underset{0}{\overset{1}{\int }}t\sin \left(\frac{n\pi }{2}\right)t}dt\right)+{\displaystyle \underset{1}{\overset{2}{\int }}10\left(2-t\right)\sin \left(\frac{n\pi }{2}\right)t}dt+0\right]\\ =\frac{1}{2}\left[10\left({\displaystyle \underset{0}{\overset{1}{\int }}t\sin \left(\frac{n\pi }{2}\right)t}dt\right)+10\left({\displaystyle \underset{1}{\overset{2}{\int }}\left(2-t\right)\sin \left(\frac{n\pi }{2}\right)t}dt\right)\right]\\ =\frac{10}{2}\left[{\displaystyle \underset{0}{\overset{1}{\int }}t\sin \left(\frac{n\pi }{2}\right)t}dt+{\displaystyle \underset{1}{\overset{2}{\int }}\left(2-t\right)\sin \left(\frac{n\pi }{2}\right)t}dt\right]\end{array}$

$b_n=5\left[{\displaystyle \underset{0}{\overset{1}{\int }}t\sin \left(\frac{n\pi }{2}\right)t}dt+{\displaystyle \underset{1}{\overset{2}{\int }}\left(2-t\right)\sin \left(\frac{n\pi }{2}\right)t}dt\right]$ (11)

Assume the following to reduce the equation (11).

$a={\displaystyle \underset{0}{\overset{1}{\int }}t\sin \left(\frac{n\pi }{2}\right)t}dt$ (12)

$b={\displaystyle \underset{1}{\overset{2}{\int }}\left(2-t\right)\sin \left(\frac{n\pi }{2}\right)t}dt$ (13)

Substitute the equations (12) and (13) in equation (11) to find $a_n$.

$b_n=5\left[a+b\right]$ (14)

Compare the equations (12) and (10) to simplify the equation (12).

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

Using the equation (10), the equation (12) can be reduced as,

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

Simplify the above equation to find $a$.

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

Compare the equations (13) and (10) to simplify the equation (13).

$\begin{array}{l}u=2-tdv=\sin \left(\frac{n\pi }{2}\right)tdt\\ du=0-dtv=-\frac{\cos \left(\frac{n\pi }{2}\right)t}{\left(\frac{n\pi }{2}\right)}\\ du=-dt\end{array}$

Using the equation (10), the equation (13) can be reduced as,

$\begin{array}{c}b={\displaystyle \underset{1}{\overset{2}{\int }}\left(2-t\right)\sin \left(\frac{n\pi }{2}\right)t}dt\\ ={\left[\left(2-t\right)\left(-\frac{\cos \left(\frac{n\pi }{2}\right)t}{\left(\frac{n\pi }{2}\right)}\right)\right]}_1^2-{\displaystyle \underset{1}{\overset{2}{\int }}\left(-\frac{\cos \left(\frac{n\pi }{2}\right)t}{\left(\frac{n\pi }{2}\right)}\right)\left(-dt\right)}\\ =-\left[\left(\left(2-2\right)\frac{\cos \left(\frac{n\pi }{2}\right)\left(2\right)}{\left(\frac{n\pi }{2}\right)}\right)-\left(\left(2-1\right)\frac{\cos \left(\frac{n\pi }{2}\right)\left(1\right)}{\left(\frac{n\pi }{2}\right)}\right)\right]-\frac{1}{\left(\frac{n\pi }{2}\right)}{\displaystyle \underset{1}{\overset{2}{\int }}\cos \left(\frac{n\pi }{2}\right)tdt}\\ =-\left[0-\left(\frac{\cos \left(\frac{n\pi }{2}\right)}{\left(\frac{n\pi }{2}\right)}\right)\right]-\frac{1}{\left(\frac{n\pi }{2}\right)}{\left[\frac{\sin \left(\frac{n\pi }{2}\right)t}{\left(\frac{n\pi }{2}\right)}\right]}_1^2\end{array}$

Simplify the above equation to find $b$.

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

Substitute $-\frac{2}{n\pi }\cos \left(\frac{n\pi }{2}\right)+\frac{4}{n^2{\pi }^2}\sin \left(\frac{n\pi }{2}\right)$ for $a$ and $\frac{2}{n\pi }\cos \left(\frac{n\pi }{2}\right)+\frac{4}{n^2{\pi }^2}\sin \left(\frac{n\pi }{2}\right)$ for $b$ in equation (14) to find $b_n$.

$\begin{array}{c}{b}_n=5\left[-\frac{2}{n\pi }\cos \left(\frac{n\pi }{2}\right)+\frac{4}{n^2{\pi }^2}\sin \left(\frac{n\pi }{2}\right)+\frac{2}{n\pi }\cos \left(\frac{n\pi }{2}\right)+\frac{4}{n^2{\pi }^2}\sin \left(\frac{n\pi }{2}\right)\right]\\ =5\left[\frac{8}{n^2{\pi }^2}\sin \left(\frac{n\pi }{2}\right)\right]\\ =\frac{40}{n^2{\pi }^2}\sin \left(\frac{n\pi }{2}\right)\end{array}$

Substitute $2.5$ for $a_0$, $\frac{20}{{\pi }^2}\left[\frac{2}{n^2}\cos \left(\frac{n\pi }{2}\right)-\frac{1}{n^2}\cos \left(n\pi \right)-\frac{1}{n^2}\right]$ for $a_n$, $\frac{40}{n^2{\pi }^2}\sin \left(\frac{n\pi }{2}\right)$ for $b_n$, and $\frac{\pi }{2}$ for ${\omega }_0$ in equation (2) to find $f\left(t\right)$.

$\begin{array}{c}f\left(t\right)=2.5+{\displaystyle \sum _{n=1}^{\infty }\left(\begin{array}{l}\left(\frac{20}{{\pi }^2}\left[\frac{2}{n^2}\cos \left(\frac{n\pi }{2}\right)-\frac{1}{n^2}\cos \left(n\pi \right)-\frac{1}{n^2}\right]\right)\cos n\left(\frac{\pi }{2}\right)t\\ +\left(\frac{40}{n^2{\pi }^2}\sin \left(\frac{n\pi }{2}\right)\right)\sin n\left(\frac{\pi }{2}\right)t\end{array}\right)}\\ =2.5+{\displaystyle \sum _{n=1}^{\infty }\left(\begin{array}{l}\left(\frac{20}{n^2{\pi }^2}\left[2\cos \left(\frac{n\pi }{2}\right)-\cos \left(n\pi \right)-1\right]\right)\cos n\left(\frac{\pi }{2}\right)t\\ +\left(\frac{40}{n^2{\pi }^2}\sin \left(\frac{n\pi }{2}\right)\right)\sin n\left(\frac{\pi }{2}\right)t\end{array}\right)}\\ =2.5+{\displaystyle \sum _{n=1}^{\infty }\frac{20}{n^2{\pi }^2}\left(\begin{array}{l}\left(2\cos \left(\frac{n\pi }{2}\right)-\cos \left(n\pi \right)-1\right)\cos \left(\frac{n\pi }{2}\right)t\\ +\left(2\sin \left(\frac{n\pi }{2}\right)\right)\sin \left(\frac{n\pi }{2}\right)t\end{array}\right)}\end{array}$

Conclusion:

Thus, the Fourier series $f\left(t\right)$ for the periodic waveform in Figure 17.57 is $2.5+{\displaystyle \sum _{n=1}^{\infty }\frac{20}{n^2{\pi }^2}\left(\left(2\cos \left(\frac{n\pi }{2}\right)-\cos \left(n\pi \right)-1\right)\cos \left(\frac{n\pi }{2}\right)t+\left(2\sin \left(\frac{n\pi }{2}\right)\right)\sin \left(\frac{n\pi }{2}\right)t\right)}$.