Chapter 17, Problem 3P

Given that Fourier coefficients a₀, a_n, and b_n of the waveform in Fig. 17.47. Plot the amplitude and phase spectra.

Figure 17.47

To determine

Find the Fourier coefficients $a_0$, $a_n$ and $b_n$ of the waveform shown in Figure 17.47 and plot the amplitude and phase spectra.

Answer to Problem 3P

The Fourier coefficients $a_0$ is $45$, $a_n$ is $\frac{60}{n\pi }{\left(-1\right)}^{\frac{n+1}{2}}$ for odd $n$ and $b_n$ is $\frac{60}{n\pi }\left(1-2\cos n\pi +\cos \left(\frac{n\pi }{2}\right)\right)$.

Explanation of Solution

Given data:

Refer to Figure 17.47 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 expression to calculate the dc component of the function $g\left(t\right)$.

$a_0=\frac{1}{T}{\displaystyle \underset{0}{\overset{T}{\int }}g\left(t\right)}dt$        (2)

Write the expression to calculate Fourier coefficients $a_n$ and $b_n$.

$a_n=\frac{2}{T}{\displaystyle \underset{0}{\overset{T}{\int }}g\left(t\right)\cos n{\omega }_{0}t}dt$        (3)

$b_n=\frac{2}{T}{\displaystyle \underset{0}{\overset{T}{\int }}g\left(t\right)\sin n{\omega }_{0}t}dt$        (4)

Here,

$n$ is an integer, and

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

Calculation:

The given waveform is drawn as Figure 1.

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

$f\left(t\right)=\{\begin{array}{l}60,0<t<1\\ 120,1<t<2\\ 0,2<t<4\end{array}$

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 (2) to find $a_0$.

$\begin{array}{c}{a}_0=\frac{1}{4}{\displaystyle \underset{0}{\overset{4}{\int }}g\left(t\right)}dt\\ =\frac{1}{4}\left({\displaystyle \underset{0}{\overset{1}{\int }}g\left(t\right)}dt+{\displaystyle \underset{1}{\overset{2}{\int }}g\left(t\right)}dt+{\displaystyle \underset{2}{\overset{4}{\int }}g\left(t\right)}dt\right)\\ =\frac{1}{4}\left({\displaystyle \underset{0}{\overset{1}{\int }}60}dt+{\displaystyle \underset{1}{\overset{2}{\int }}120}dt+{\displaystyle \underset{2}{\overset{4}{\int }}0}dt\right)\\ =\frac{1}{4}\left({\displaystyle \underset{0}{\overset{1}{\int }}60}dt+{\displaystyle \underset{1}{\overset{2}{\int }}120}dt+0\right)\end{array}$

Simplify the above equation to find $a_0$.

$\begin{array}{c}{a}_0=\frac{1}{4}\left(60{\left[t\right]}_0^1+120{\left[t\right]}_1^2\right)\\ =\frac{1}{4}\left(60\left[1-0\right]+120\left[2-1\right]\right)\\ =\frac{1}{4}\left(60\left(1\right)+120\left(1\right)\right)\\ =45\end{array}$

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

$\begin{array}{c}{a}_n=\frac{2}{4}{\displaystyle \underset{0}{\overset{4}{\int }}g\left(t\right)\cos n\left(\frac{\pi }{2}\right)t}dt\\ =\frac{1}{2}\left({\displaystyle \underset{0}{\overset{1}{\int }}g\left(t\right)\cos \left(\frac{n\pi }{2}\right)t}dt+{\displaystyle \underset{1}{\overset{2}{\int }}g\left(t\right)\cos \left(\frac{n\pi }{2}\right)t}dt+{\displaystyle \underset{2}{\overset{4}{\int }}g\left(t\right)\cos \left(\frac{n\pi }{2}\right)t}dt\right)\\ =\frac{1}{2}\left({\displaystyle \underset{0}{\overset{1}{\int }}60\cos \left(\frac{n\pi }{2}\right)t}dt+{\displaystyle \underset{1}{\overset{2}{\int }}120\cos \left(\frac{n\pi }{2}\right)t}dt+{\displaystyle \underset{2}{\overset{4}{\int }}\left(0\right)\cos \left(\frac{n\pi }{2}\right)t}dt\right)\\ =\frac{1}{2}\left({\displaystyle \underset{0}{\overset{1}{\int }}60\cos \left(\frac{n\pi }{2}\right)t}dt+{\displaystyle \underset{1}{\overset{2}{\int }}120\cos \left(\frac{n\pi }{2}\right)t}dt+0\right)\end{array}$

Simplify the above equation to find $a_n$.

$\begin{array}{c}{a}_n=\frac{1}{2}\left(60{\left[\frac{\sin \left(\frac{n\pi }{2}\right)t}{\left(\frac{n\pi }{2}\right)}\right]}_0^1+120{\left[\frac{\sin \left(\frac{n\pi }{2}\right)t}{\left(\frac{n\pi }{2}\right)}\right]}_1^2\right)\\ =\frac{1}{2}\left(60\left(\frac{2}{n\pi }\right)\left[\sin \left(\frac{n\pi }{2}\right)\left(1\right)-\sin \left(\frac{n\pi }{2}\right)\left(0\right)\right]+120\left(\frac{2}{n\pi }\right)\left[\sin \left(\frac{n\pi }{2}\right)\left(2\right)-\sin \left(\frac{n\pi }{2}\right)\left(1\right)\right]\right)\\ =\frac{1}{2}\left(\frac{120}{n\pi }\left[\sin \left(\frac{n\pi }{2}\right)-\sin 0°\right]+\frac{240}{n\pi }\left[\sin n\pi -\sin \left(\frac{n\pi }{2}\right)\right]\right)\\ =\frac{1}{2}\left(\frac{120}{n\pi }\right)\left(\left[\sin \left(\frac{n\pi }{2}\right)-0\right]+2\left[0-\sin \left(\frac{n\pi }{2}\right)\right]\right)\left\{\begin{array}{l}\because \sin n\pi =0\\ \sin 0°=0\end{array}\right\}\end{array}$

Simplify the above equation to find $a_n$.

$\begin{array}{c}{a}_n=\frac{60}{n\pi }\left(\sin \left(\frac{n\pi }{2}\right)-2\sin \left(\frac{n\pi }{2}\right)\right)\\ =\frac{60}{n\pi }\left(-\sin \left(\frac{n\pi }{2}\right)\right)\\ =-\frac{60}{n\pi }\sin \left(\frac{n\pi }{2}\right)\\ =\{\begin{array}{l}\frac{60}{n\pi }{\left(-1\right)}^{\frac{n+1}{2}},n=\text{odd}\\ 0,n=\text{even}\end{array}\end{array}$

From above equation, it is clear that $a_2$, $a_4$, $a_6$ and $a_8$ is equal to zero.

For odd $n$,

$a_n=\frac{60}{n\pi }{\left(-1\right)}^{\frac{n+1}{2}}$        (5)

Substitute $1$ for $n$ in equation (5) to find $a_1$.

$\begin{array}{c}{a}_1=\frac{60}{\left(1\right)\pi }{\left(-1\right)}^{\frac{1+1}{2}}\\ =\frac{60}{\pi }{\left(-1\right)}^1\\ =-19.08\end{array}$

Substitute $3$ for $n$ in equation (5) to find $a_3$.

$\begin{array}{c}{a}_3=\frac{60}{\left(3\right)\pi }{\left(-1\right)}^{\frac{3+1}{2}}\\ =\frac{20}{\pi }{\left(-1\right)}^2\\ =6.36\end{array}$

Substitute $5$ for $n$ in equation (5) to find $a_5$.

$\begin{array}{c}{a}_5=\frac{60}{\left(5\right)\pi }{\left(-1\right)}^{\frac{5+1}{2}}\\ =\frac{12}{\pi }{\left(-1\right)}^3\\ =-3.84\end{array}$

Substitute $7$ for $n$ in equation (5) to find $a_7$.

$\begin{array}{c}{a}_7=\frac{60}{\left(7\right)\pi }{\left(-1\right)}^{\frac{7+1}{2}}\\ =\frac{60}{7\pi }{\left(-1\right)}^4\\ =2.76\end{array}$

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

$\begin{array}{c}{b}_n=\frac{2}{4}{\displaystyle \underset{0}{\overset{4}{\int }}g\left(t\right)\sin n\left(\frac{\pi }{2}\right)t}dt\\ =\frac{1}{2}\left({\displaystyle \underset{0}{\overset{1}{\int }}g\left(t\right)\sin \left(\frac{n\pi }{2}\right)t}dt+{\displaystyle \underset{1}{\overset{2}{\int }}g\left(t\right)\sin \left(\frac{n\pi }{2}\right)t}dt+{\displaystyle \underset{2}{\overset{4}{\int }}g\left(t\right)\sin \left(\frac{n\pi }{2}\right)t}dt\right)\\ =\frac{1}{2}\left({\displaystyle \underset{0}{\overset{1}{\int }}60\sin \left(\frac{n\pi }{2}\right)t}dt+{\displaystyle \underset{1}{\overset{2}{\int }}120\sin \left(\frac{n\pi }{2}\right)t}dt+{\displaystyle \underset{2}{\overset{4}{\int }}\left(0\right)\sin \left(\frac{n\pi }{2}\right)t}dt\right)\\ =\frac{1}{2}\left({\displaystyle \underset{0}{\overset{1}{\int }}60\sin \left(\frac{n\pi }{2}\right)t}dt+{\displaystyle \underset{1}{\overset{2}{\int }}120\sin \left(\frac{n\pi }{2}\right)t}dt+0\right)\end{array}$

Simplify the above equation to find $b_n$.

$\begin{array}{c}{b}_n=\frac{1}{2}\left(60{\left[-\frac{\cos \left(\frac{n\pi }{2}\right)t}{\left(\frac{n\pi }{2}\right)}\right]}_0^1+120{\left[-\frac{\cos \left(\frac{n\pi }{2}\right)t}{\left(\frac{n\pi }{2}\right)}\right]}_1^2\right)\\ =\frac{1}{2}\left(-60\left(\frac{2}{n\pi }\right)\left[\cos \left(\frac{n\pi }{2}\right)\left(1\right)-\cos \left(\frac{n\pi }{2}\right)\left(0\right)\right]-120\left(\frac{2}{n\pi }\right)\left[\cos \left(\frac{n\pi }{2}\right)\left(2\right)-\cos \left(\frac{n\pi }{2}\right)\left(1\right)\right]\right)\\ =\frac{1}{2}\left(-\frac{120}{n\pi }\left[\cos \left(\frac{n\pi }{2}\right)-\cos 0°\right]-\frac{240}{n\pi }\left[\cos n\pi -\cos \left(\frac{n\pi }{2}\right)\right]\right)\\ =\frac{1}{2}\left(-\frac{120}{n\pi }\right)\left(\left[\cos \left(\frac{n\pi }{2}\right)-1\right]+2\left[\cos n\pi -\cos \left(\frac{n\pi }{2}\right)\right]\right)\left\{\because \cos 0°=1\right\}\end{array}$

Simplify the above equation to find $b_n$.

$\begin{array}{c}{b}_n=-\frac{60}{n\pi }\left(\cos \left(\frac{n\pi }{2}\right)-1+2\cos n\pi -2\cos \left(\frac{n\pi }{2}\right)\right)\\ =-\frac{60}{n\pi }\left(-1+2\cos n\pi -\cos \left(\frac{n\pi }{2}\right)\right)\end{array}$

$b_n=\frac{60}{n\pi }\left(1-2\cos n\pi +\cos \left(\frac{n\pi }{2}\right)\right)$        (6)

Substitute $1$ for $n$ in equation (6) to find $b_1$.

$\begin{array}{c}{b}_1=\frac{60}{\left(1\right)\pi }\left(1-2\cos \left(1\right)\pi +\cos \left(\frac{\left(1\right)\pi }{2}\right)\right)\\ =\frac{60}{\pi }\left(1-2\cos \pi +\cos \left(\frac{\pi }{2}\right)\right)\\ =57.29\end{array}$

Substitute $2$ for $n$ in equation (6) to find $b_2$.

$\begin{array}{c}{b}_2=\frac{60}{\left(2\right)\pi }\left(1-2\cos \left(2\right)\pi +\cos \left(\frac{\left(2\right)\pi }{2}\right)\right)\\ =\frac{30}{\pi }\left(1-2\cos 2\pi +\cos \pi \right)\\ =-19.099\end{array}$

Substitute $3$ for $n$ in equation (6) to find $b_3$.

$\begin{array}{c}{b}_3=\frac{60}{\left(3\right)\pi }\left(1-2\cos \left(3\right)\pi +\cos \left(\frac{\left(3\right)\pi }{2}\right)\right)\\ =\frac{20}{\pi }\left(1-2\cos 3\pi +\cos \left(\frac{3\pi }{2}\right)\right)\\ =19.099\end{array}$

Substitute $4$ for $n$ in equation (6) to find $b_4$.

$\begin{array}{c}{b}_4=\frac{60}{\left(4\right)\pi }\left(1-2\cos \left(4\right)\pi +\cos \left(\frac{\left(4\right)\pi }{2}\right)\right)\\ =\frac{60}{4\pi }\left(1-2\cos 4\pi +\cos 2\pi \right)\\ =0\end{array}$

Substitute $5$ for $n$ in equation (6) to find $b_5$.

$\begin{array}{c}{b}_5=\frac{60}{\left(5\right)\pi }\left(1-2\cos \left(5\right)\pi +\cos \left(\frac{\left(5\right)\pi }{2}\right)\right)\\ =\frac{12}{\pi }\left(1-2\cos 5\pi +\cos \left(\frac{5\pi }{2}\right)\right)\\ =11.459\end{array}$

Substitute $6$ for $n$ in equation (6) to find $b_6$.

$\begin{array}{c}{b}_6=\frac{60}{\left(6\right)\pi }\left(1-2\cos \left(6\right)\pi +\cos \left(\frac{\left(6\right)\pi }{2}\right)\right)\\ =\frac{10}{\pi }\left(1-2\cos 6\pi +\cos 3\pi \right)\\ =-6.366\end{array}$

Substitute $7$ for $n$ in equation (6) to find $b_7$.

$\begin{array}{c}{b}_7=\frac{60}{\left(7\right)\pi }\left(1-2\cos \left(7\right)\pi +\cos \left(\frac{\left(7\right)\pi }{2}\right)\right)\\ =\frac{60}{7\pi }\left(1-2\cos 7\pi +\cos \left(\frac{7\pi }{2}\right)\right)\\ =8.185\end{array}$

Substitute $8$ for $n$ in equation (6) to find $b_8$.

$\begin{array}{c}{b}_8=\frac{60}{\left(8\right)\pi }\left(1-2\cos \left(8\right)\pi +\cos \left(\frac{\left(8\right)\pi }{2}\right)\right)\\ =\frac{60}{8\pi }\left(1-2\cos 8\pi +\cos 4\pi \right)\\ =0\end{array}$

Write the expression to calculate the amplitude spectra of the signal.

$A_n=\sqrt{a_n{}^2+b_n{}^2}$        (7)

Substitute $1$ for $n$ in equation (7) to find $A_1$.

$A_1=\sqrt{a_1{}^2+b_1{}^2}$

Substitute $-19.08$ for $a_1$ and $57.29$ for $b_1$ in above equation to find $A_1$.

$\begin{array}{c}{A}_1=\sqrt{{\left(-19.08\right)}^2+{\left(57.29\right)}^2}\\ =60.38\end{array}$

Substitute $2$ for $n$ in equation (7) to find $A_2$.

$A_2=\sqrt{a_2{}^2+b_2{}^2}$

Substitute $0$ for $a_2$ and $-19.099$ for $b_2$ in above equation to find $A_2$.

$\begin{array}{c}{A}_2=\sqrt{{\left(0\right)}^2+{\left(-19.099\right)}^2}\\ =\sqrt{{\left(19.099\right)}^2}\\ =19.099\end{array}$

Substitute $3$ for $n$ in equation (7) to find $A_3$.

$A_3=\sqrt{a_3{}^2+b_3{}^2}$

Substitute $6.36$ for $a_3$ and $19.099$ for $b_3$ in above equation to find $A_3$.

$\begin{array}{c}{A}_3=\sqrt{{\left(6.36\right)}^2+{\left(19.099\right)}^2}\\ =20.13\end{array}$

Substitute $4$ for $n$ in equation (7) to find $A_4$.

$A_4=\sqrt{a_4{}^2+b_4{}^2}$

Substitute $0$ for $a_4$ and $0$ for $b_4$ in above equation to find $A_4$.

$\begin{array}{c}{A}_4=\sqrt{{\left(0\right)}^2+{\left(0\right)}^2}\\ =0\end{array}$

Substitute $5$ for $n$ in equation (7) to find $A_5$.

$A_5=\sqrt{a_5{}^2+b_5{}^2}$

Substitute $-3.84$ for $a_5$ and $11.459$ for $b_5$ in above equation to find $A_5$.

$\begin{array}{c}{A}_5=\sqrt{{\left(-3.84\right)}^2+{\left(11.459\right)}^2}\\ =12.08\end{array}$

Substitute $6$ for $n$ in equation (7) to find $A_6$.

$A_6=\sqrt{a_6{}^2+b_6{}^2}$

Substitute $0$ for $a_6$ and $-6.366$ for $b_6$ in above equation to find $A_6$.

$\begin{array}{c}{A}_6=\sqrt{{\left(0\right)}^2+{\left(-6.366\right)}^2}\\ =6.366\end{array}$

Substitute $7$ for $n$ in equation (7) to find $A_7$.

$A_7=\sqrt{a_7{}^2+b_7{}^2}$

Substitute $2.76$ for $a_7$ and $8.185$ for $b_7$ in above equation to find $A_7$.

$\begin{array}{c}{A}_7=\sqrt{{\left(2.76\right)}^2+{\left(8.185\right)}^2}\\ =8.63\end{array}$

Substitute $8$ for $n$ in equation (7) to find $A_8$.

$A_8=\sqrt{a_8{}^2+b_8{}^2}$

Substitute $0$ for $a_8$ and $0$ for $b_8$ in above equation to find $A_8$.

$\begin{array}{c}{A}_8=\sqrt{{\left(0\right)}^2+{\left(0\right)}^2}\\ =0\end{array}$

From above calculations, the amplitude spectrum of the waveform in Figure 1 is drawn as Figure 2.

Write the expression to calculate the phase spectra of the signal.

${\varphi }_n=-{\tan }^{-1}\left(\frac{b_n}{a_n}\right)$        (8)

Substitute $1$ for $n$ in equation (8) to find ${\varphi }_1$.

${\varphi }_1=-{\tan }^{-1}\left(\frac{b_1}{a_1}\right)$

Substitute $-19.08$ for $a_1$ and $57.29$ for $b_1$ in above equation to find ${\varphi }_1$.

$\begin{array}{c}{\varphi }_1=-{\tan }^{-1}\left(\frac{57.29}{-19.08}\right)\\ =1.25°\end{array}$

Substitute $2$ for $n$ in equation (8) to find ${\varphi }_2$.

${\varphi }_2=-{\tan }^{-1}\left(\frac{b_2}{a_2}\right)$

Substitute $0$ for $a_2$ and $-19.099$ for $b_2$ in above equation to find ${\varphi }_2$.

$\begin{array}{c}{\varphi }_2=-{\tan }^{-1}\left(\frac{-19.099}{0}\right)\\ =-{\tan }^{-1}\left(\infty \right)\\ =-90°\end{array}$

Substitute $3$ for $n$ in equation (8) to find ${\varphi }_3$.

${\varphi }_3=-{\tan }^{-1}\left(\frac{b_3}{a_3}\right)$

Substitute $6.36$ for $a_3$ and $19.099$ for $b_3$ in above equation to find ${\varphi }_3$.

$\begin{array}{c}{\varphi }_3=-{\tan }^{-1}\left(\frac{19.099}{6.36}\right)\\ =-1.25°\end{array}$

Substitute $4$ for $n$ in equation (8) to find ${\varphi }_4$.

${\varphi }_4=-{\tan }^{-1}\left(\frac{b_4}{a_4}\right)$

Substitute $0$ for $a_4$ and $0$ for $b_4$ in above equation to find ${\varphi }_4$.

$\begin{array}{c}{\varphi }_4=-{\tan }^{-1}\left(\frac{0}{0}\right)\\ =-{\tan }^{-1}\left(\infty \right)\\ =-90°\end{array}$

Substitute $5$ for $n$ in equation (8) to find ${\varphi }_5$.

${\varphi }_5=-{\tan }^{-1}\left(\frac{b_5}{a_5}\right)$

Substitute $-3.84$ for $a_5$ and $11.459$ for $b_5$ in above equation to find ${\varphi }_5$.

$\begin{array}{c}{\varphi }_5=-{\tan }^{-1}\left(\frac{11.459}{-3.84}\right)\\ =1.25°\end{array}$

Substitute $6$ for $n$ in equation (8) to find ${\varphi }_6$.

${\varphi }_6=-{\tan }^{-1}\left(\frac{b_6}{a_6}\right)$

Substitute $0$ for $a_6$ and $-6.366$ for $b_6$ in above equation to find ${\varphi }_6$.

$\begin{array}{c}{\varphi }_6=-{\tan }^{-1}\left(\frac{-6.366}{0}\right)\\ =-{\tan }^{-1}\left(\infty \right)\\ =-90°\end{array}$

Substitute $7$ for $n$ in equation (8) to find ${\varphi }_7$.

${\varphi }_7=-{\tan }^{-1}\left(\frac{b_7}{a_7}\right)$

Substitute $2.76$ for $a_7$ and $8.185$ for $b_7$ in above equation to find ${\varphi }_7$.

$\begin{array}{c}{\varphi }_7=-{\tan }^{-1}\left(\frac{8.185}{2.76}\right)\\ =-1.25°\end{array}$

Substitute $8$ for $n$ in equation (8) to find ${\varphi }_8$.

${\varphi }_8=-{\tan }^{-1}\left(\frac{b_8}{a_8}\right)$

Substitute $0$ for $a_8$ and $0$ for $b_8$ in above equation to find ${\varphi }_8$.

$\begin{array}{c}{\varphi }_8=-{\tan }^{-1}\left(\frac{0}{0}\right)\\ =-{\tan }^{-1}\left(\infty \right)\\ =-90°\end{array}$

From above calculations, the phase spectrum of the waveform in Figure 1 is drawn as Figure 3.

Conclusion:

Thus, the Fourier coefficients $a_0$ is $45$, $a_n$ is $\frac{60}{n\pi }{\left(-1\right)}^{\frac{n+1}{2}}$ for odd $n$ and $b_n$ is $\frac{60}{n\pi }\left(1-2\cos n\pi +\cos \left(\frac{n\pi }{2}\right)\right)$ and the amplitude and phase spectra is plotted.