Chapter 17, Problem 12P

To determine

Find the Fourier series of the given voltage source.

Answer to Problem 12P

The Fourier series $v\left(t\right)$ for the given voltage source is $80{\pi }^2-{\displaystyle \sum _{n=1}^{\infty }\frac{480}{n^2}\cos nt}$.

Explanation of Solution

Given data:

The voltage source of the periodic waveform is,

$v\left(t\right)=120t\left(2\pi -t\right)\text{V,}0<t<2\pi$

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 $v\left(t\right)$.

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

Here,

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

$a_n$ 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 $v\left(t\right)$.

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

Write the expression to calculate Fourier coefficients.

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

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

Calculation:

Given voltage source function is,

$\begin{array}{c}v\left(t\right)=120t\left(2\pi -t\right)\text{V}\\ =\left(240\pi t-120t^2\right)\text{V}\end{array}$

The time period of the given voltage source function is,

$T=2\pi$

Substitute $2\pi$ for $T$ in equation (1) to find ${\omega }_0$.

$\begin{array}{c}{\omega }_0=\frac{2\pi }{2\pi }\\ =1\end{array}$

Substitute $2\pi$ for $T$ in equation (3) to find $a_0$.

$\begin{array}{c}{a}_0=\frac{1}{2\pi }{\displaystyle \underset{0}{\overset{2\pi }{\int }}v\left(t\right)}dt\\ =\frac{1}{2\pi }{\displaystyle \underset{0}{\overset{2\pi }{\int }}\left(240\pi t-120t^2\right)}dt\\ =\frac{1}{2\pi }\left({\displaystyle \underset{0}{\overset{2\pi }{\int }}240\pi t}dt-{\displaystyle \underset{0}{\overset{2\pi }{\int }}120t^2}dt\right)\\ =\frac{1}{2\pi }\left(240\pi {\left[\frac{t^2}{2}\right]}_0^{2\pi }-120{\left[\frac{t^3}{3}\right]}_0^{2\pi }\right)\end{array}$

Simplify the above equation to find $a_0$.

$\begin{array}{c}{a}_0=\frac{1}{2\pi }\left(120\pi \left[{\left(2\pi \right)}^2-{\left(0\right)}^2\right]-40\left[{\left(2\pi \right)}^3-{\left(0\right)}^3\right]\right)\\ =\frac{1}{2\pi }\left(120\pi \left(4{\pi }^2\right)-40\left(8{\pi }^3\right)\right)\\ =\frac{1}{2\pi }\left(480{\pi }^3-320{\pi }^3\right)\\ =80{\pi }^2\end{array}$

Substitute $2\pi$ for $T$ and $1$ for ${\omega }_0$ in equation (4) to find $a_n$.

$\begin{array}{c}{a}_n=\frac{2}{2\pi }{\displaystyle \underset{0}{\overset{2\pi }{\int }}v\left(t\right)\cos n\left(1\right)t}dt\\ =\frac{1}{\pi }{\displaystyle \underset{0}{\overset{2\pi }{\int }}v\left(t\right)\cos nt}dt\\ =\frac{1}{\pi }{\displaystyle \underset{0}{\overset{2\pi }{\int }}\left(240\pi t-120t^2\right)\cos nt}dt\\ =\frac{120}{\pi }{\displaystyle \underset{0}{\overset{2\pi }{\int }}\left(2\pi t-t^2\right)\cos nt}dt\end{array}$

Simplify the above equation to find $a_n$.

$a_n=\frac{120}{\pi }\left({\displaystyle \underset{0}{\overset{2\pi }{\int }}2\pi t\cos nt}dt-{\displaystyle \underset{0}{\overset{2\pi }{\int }}{t}^2\cos nt}dt\right)$

$a_n=\frac{120}{\pi }\left(2\pi {\displaystyle \underset{0}{\overset{2\pi }{\int }}t\cos nt}dt-{\displaystyle \underset{0}{\overset{2\pi }{\int }}{t}^2\cos nt}dt\right)$ (6)

Assume the following to reduce the equation (6).

$x={\displaystyle \underset{0}{\overset{2\pi }{\int }}t\cos nt}dt$ (7)

$y={\displaystyle \underset{0}{\overset{2\pi }{\int }}{t}^2\cos nt}dt$ (8)

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

$a_n=\frac{120}{\pi }\left(2\pi x-y\right)$ (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).

$\begin{array}{l}u=tdv=\cos ntdt\\ du=dtv=\frac{\sin nt}{n}\end{array}$

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

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

Simplify the above equation to find $x$.

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

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

$\begin{array}{l}u=t^{2}dv=\cos ntdt\\ du=2tdtv=\frac{\sin nt}{n}\end{array}$

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

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

Simplify the above equation to find $y$.

$\begin{array}{c}y=\frac{1}{n}\left[4{\pi }^2\sin \left(2n\pi \right)-0\right]-\frac{2}{n}{\displaystyle \underset{0}{\overset{2\pi }{\int }}t\sin ntdt}\\ =\frac{1}{n}\left[4{\pi }^2\left(0\right)\right]-\frac{2}{n}{\displaystyle \underset{0}{\overset{2\pi }{\int }}t\sin ntdt}\left\{\because \sin 2n\pi =0\right\}\\ =0-\frac{2}{n}{\displaystyle \underset{0}{\overset{2\pi }{\int }}t\sin ntdt}\end{array}$

$y=-\frac{2}{n}{\displaystyle \underset{0}{\overset{2\pi }{\int }}t\sin ntdt}$ (11)

Assume the following to reduce the equation (11).

$z={\displaystyle \underset{0}{\overset{2\pi }{\int }}t\sin ntdt}$ (12)

Substitute the equation (12) in equation (11) to find $y$.

$y=-\frac{2}{n}z$ (13)

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

$\begin{array}{l}u=tdv=\sin ntdt\\ du=dtv=-\frac{\cos nt}{n}\end{array}$

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

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

Simplify the above equation to find $z$.

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

Substitute $-\frac{2\pi }{n}$ for $z$ in equation (13) to find $y$.

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

Substitute $0$ for $x$ and $\frac{4\pi }{n^2}$ for $y$ in equation (9) to find $a_n$.

$\begin{array}{c}{a}_n=\frac{120}{\pi }\left(2\pi \left(0\right)-\frac{4\pi }{n^2}\right)\\ =\frac{120}{\pi }\left(0-\frac{4\pi }{n^2}\right)\\ =\frac{120}{\pi }\left(-\frac{4\pi }{n^2}\right)\\ =-\frac{480}{n^2}\end{array}$

Substitute $2\pi$ for $T$ and $1$ for ${\omega }_0$ in equation (5) to find $b_n$.

$\begin{array}{c}{b}_n=\frac{2}{2\pi }{\displaystyle \underset{0}{\overset{2\pi }{\int }}v\left(t\right)\sin n\left(1\right)t}dt\\ =\frac{1}{\pi }{\displaystyle \underset{0}{\overset{2\pi }{\int }}v\left(t\right)\sin nt}dt\\ =\frac{1}{\pi }{\displaystyle \underset{0}{\overset{2\pi }{\int }}\left(240\pi t-120t^2\right)\sin nt}dt\\ =\frac{120}{\pi }{\displaystyle \underset{0}{\overset{2\pi }{\int }}\left(2\pi t-t^2\right)\sin nt}dt\end{array}$

Simplify the above equation to find $b_n$.

$b_n=\frac{120}{\pi }\left({\displaystyle \underset{0}{\overset{2\pi }{\int }}2\pi t\sin nt}dt-{\displaystyle \underset{0}{\overset{2\pi }{\int }}{t}^2\sin nt}dt\right)$

$b_n=\frac{120}{\pi }\left(2\pi {\displaystyle \underset{0}{\overset{2\pi }{\int }}t\sin nt}dt-{\displaystyle \underset{0}{\overset{2\pi }{\int }}{t}^2\sin nt}dt\right)$ (14)

Assume the following to reduce the equation (14).

$a={\displaystyle \underset{0}{\overset{2\pi }{\int }}t\sin nt}dt$ (15)

$b={\displaystyle \underset{0}{\overset{2\pi }{\int }}{t}^2\sin nt}dt$ (16)

Substitute the equations (15) and (16) in equation (14) to find $b_n$.

$b_n=\frac{120}{\pi }\left(2\pi a-b\right)$ (17)

Compare the equations (15) and (10) to simplify the equation (15).

$\begin{array}{l}u=tdv=\sin ntdt\\ du=dtv=-\frac{\cos nt}{n}\end{array}$

Using the equation (10), the equation (15) can be reduced as,

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

Simplify the above equation to find $a$.

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

Compare the equations (16) and (10) to simplify the equation (16).

$\begin{array}{l}u=t^{2}dv=\sin ntdt\\ du=2tdtv=-\frac{\cos nt}{n}\end{array}$

Using the equation (10), the equation (16) can be reduced as,

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

Simplify the above equation to find $b$.

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

$b=-\frac{4{\pi }^2}{n}+\frac{2}{n}{\displaystyle \underset{0}{\overset{2\pi }{\int }}t\cos ntdt}$ (18)

Assume the following to reduce the equation (18).

$c={\displaystyle \underset{0}{\overset{2\pi }{\int }}t\cos ntdt}$ (19)

Substitute the equation (19) in equation (18) to find $b$.

$b=-\frac{4{\pi }^2}{n}+\frac{2}{n}c$ (20)

Compare the equations (19) and (10) to simplify the equation (19).

$\begin{array}{l}u=tdv=\cos ntdt\\ du=dtv=\frac{\sin nt}{n}\end{array}$

Using the equation (10), the equation (19) can be reduced as,

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

Simplify the above equation to find $c$.

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

Substitute $0$ for $c$ in equation (20) to find $b$.

$\begin{array}{c}b=-\frac{4{\pi }^2}{n}+\frac{2}{n}\left(0\right)\\ =-\frac{4{\pi }^2}{n}+0\\ =-\frac{4{\pi }^2}{n}\end{array}$

Substitute $-\frac{2\pi }{n}$ for $a$ and $-\frac{4{\pi }^2}{n}$ for $b$ in equation (17) to find $b_n$.

$\begin{array}{c}{b}_n=\frac{120}{\pi }\left(2\pi \left(-\frac{2\pi }{n}\right)-\left(-\frac{4{\pi }^2}{n}\right)\right)\\ =\frac{120}{\pi }\left(-\frac{4{\pi }^2}{n}+\frac{4{\pi }^2}{n}\right)\\ =\frac{120}{\pi }\left(0\right)\\ =0\end{array}$

Substitute $80{\pi }^2$ for $a_0$, $-\frac{480}{n^2}$ for $a_n$, $0$ for $b_n$ and $1$ for ${\omega }_0$ in equation (2) to find $v\left(t\right)$.

$\begin{array}{c}v\left(t\right)=80{\pi }^2+{\displaystyle \sum _{n=1}^{\infty }\left(-\frac{480}{n^2}\cos n\left(1\right)t+\left(0\right)\sin n\left(1\right)t\right)}\\ =80{\pi }^2+{\displaystyle \sum _{n=1}^{\infty }\left(-\frac{480}{n^2}\cos nt+0\right)}\\ =80{\pi }^2-{\displaystyle \sum _{n=1}^{\infty }\frac{480}{n^2}\cos nt}\end{array}$

Conclusion:

Thus, the Fourier series $v\left(t\right)$ for the given voltage source is $80{\pi }^2-{\displaystyle \sum _{n=1}^{\infty }\frac{480}{n^2}\cos nt}$.