Chapter 17, Problem 14E

To determine

Sketch the line spectrum for the waveform shown in Figure 17.4c (limited to the five largest terms).

Answer to Problem 14E

The line spectrum for the waveform shown in Figure 17.4c is sketched as shown in Figure 1.

Explanation of Solution

Given data:

Refer to Figure 17.4c in the textbook.

Formula used:

Write the general expression for Fourier series expansion.

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

Write the general expression for Fourier series coefficient $a_0$.

$a_0=\frac{1}{T}{\displaystyle {\int }_0^{T}f\left(t\right)}dt$        (2)

Write the general expression for Fourier series coefficient $a_n$.

$a_n=\frac{2}{T}{\displaystyle {\int }_0^{T}f\left(t\right)\cos n{\omega }_{0}t}dt$        (3)

Write the general expression for Fourier series coefficient $b_n$.

$b_n=\frac{2}{T}{\displaystyle {\int }_0^{T}f\left(t\right)\sin n{\omega }_{0}t}dt$        (4)

Write the expression to calculate the fundamental angular frequency.

${\omega }_0=\frac{2\pi }{T}=2\pi f_0$        (5)

Here,

$T$ is the period of the function.

Calculation:

In the given Figure 17.4a, the time period is $T=2$.

The function $v\left(t\right)$ for the given waveform is,

$v\left(t\right)=\{\begin{array}{l}2t,-0.5\le t\le 0.5\\ 2-2t,0.5\le t\le 1.5\end{array}$        (6)

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

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

${\omega }_0=\pi$        (7)

Applying equation (6) in equation (2) to find $a_0$ as follows,

$\begin{array}{c}{a}_0=\frac{1}{2}{\displaystyle {\int }_0^{2}f\left(t\right)}dt\\ =\frac{1}{2}\left[{\displaystyle {\int }_{-0.5}^{0.5}\left(2t\right)}dt+{\displaystyle {\int }_{0.5}^{1.5}\left(2-2t\right)}dt\right]\\ =\frac{1}{2}\left[2{\left[\frac{t^2}{2}\right]}_{-0.5}^{0.5}-2{\left[t-\frac{t^2}{2}\right]}_{0.5}^{1.5}\right]\\ =\frac{1}{2}\left[2\left[\frac{{\left(0.5\right)}^2}{2}-\frac{{\left(-0.5\right)}^2}{2}\right]-2\left[\left(1.5-\frac{{\left(1.5\right)}^2}{2}\right)-\left(0.5-\frac{{\left(0.5\right)}^2}{2}\right)\right]\right]\end{array}$

Simplify the above equation as follows,

$\begin{array}{c}{a}_0=\frac{1}{2}\left[2\left[\frac{0.25}{2}-\frac{0.25}{2}\right]-2\left[\left(1.5-\frac{2.25}{2}\right)-\left(0.5-\frac{0.25}{2}\right)\right]\right]\\ =\frac{1}{2}\left[\left(0\right)-2\left[\frac{0.75}{2}-\frac{0.75}{2}\right]\right]\\ =\frac{1}{2}\left[0-0\right]\\ =0\end{array}$

As the given function is an odd symmetry. Therefore,

$a_n=0$

Applying equation (6) in equation (4) to finding the Fourier coefficient $b_n$.

$b_n=\frac{2}{2}{\displaystyle {\int }_0^{2}f\left(t\right)\sin n{\omega }_{0}t}dt$

$b_n={\displaystyle {\int }_{-0.5}^{0.5}\left(2t\right)\sin n{\omega }_{0}t}dt+{\displaystyle {\int }_{0.5}^{1.5}\left(2-2t\right)\sin n{\omega }_{0}t}dt$        (8)

Equation (8) is simplified as,

$x={\displaystyle {\int }_{-0.5}^{0.5}\left(2t\right)\sin n{\omega }_{0}t}dt$

$y={\displaystyle {\int }_{0.5}^{1.5}\left(2-2t\right)\sin n{\omega }_{0}t}dt$

Therefore, equation (8) becomes,

$b_n=x+y$        (9)

Consider the function,

$x=2{\displaystyle {\int }_{-0.5}^{0.5}t\sin n{\omega }_{0}t}dt$        (10)

Consider the following integration formula.

${\displaystyle \underset{a}{\overset{b}{\int }}udv}={\left[uv\right]}_a^b-{\displaystyle \underset{a}{\overset{b}{\int }}vdu}$        (11)

Compare the equations (10) and (11) to simplify the equation (10).

$\begin{array}{l}u=tdv=\sin n\pi tdt\left\{\because {\omega }_0=\pi \right\}\\ du=dtv=\frac{-\cos n\pi t}{n\pi }\end{array}$

Using the equation (11), the equation (10) can be reduced as,

$\begin{array}{c}x=2\left\{{\left[t\left(\frac{-\cos n\pi t}{n\pi }\right)\right]}_{-0.5}^{0.5}-{\displaystyle {\int }_{-0.5}^{0.5}\left(\frac{-\cos n\pi t}{n\pi }\right)dt}\right\}\\ =2\left\{\left[\left(0.5\right)\left(\frac{-\cos n\pi \left(0.5\right)}{n\pi }\right)+\left(-0.5\right)\left(\frac{-\cos n\pi \left(-0.5\right)}{n\pi }\right)\right]-{\left[\frac{-\sin n\pi t}{n^2{\pi }^2}\right]}_{-0.5}^{0.5}\right\}\\ =2\left\{\left[\frac{-\cos 0.25n\pi }{n\pi }-\frac{\cos 0.25n\pi }{n\pi }\right]-\left[\frac{-\sin 0.5n\pi }{n^2{\pi }^2}-\frac{-\sin \left(-0.5\right)n\pi }{n^2{\pi }^2}\right]\right\}\\ =2\left\{\left[\frac{-\cos 0.5n\pi }{n\pi }\right]-\left[-\frac{\sin 0.5n\pi }{n^2{\pi }^2}-\frac{\sin 0.5n\pi }{n^2{\pi }^2}\right]\right\}\end{array}$

Simplify the above equation as follows,

$x=\frac{-2\cos 0.5n\pi }{n\pi }+\frac{4\sin 0.5n\pi }{n^2{\pi }^2}$

Consider the function,

$y={\displaystyle {\int }_{0.5}^{1.5}\left(2-2t\right)\sin n{\omega }_{0}t}dt$        (12)

Compare the equations (12) and (11) to simplify the equation (12).

$\begin{array}{l}u=2-2tdv=\sin n\pi tdt\left\{\because {\omega }_0=\pi \right\}\\ du=-2dtv=\frac{-\cos n\pi t}{n\pi }\end{array}$

Using the equation (11), the equation (12) can be reduced as,

$\begin{array}{c}y={\left[\left(2-2t\right)\left(\frac{-\cos n\pi t}{n\pi }\right)\right]}_{0.5}^{1.5}-2{\displaystyle {\int }_{0.5}^{1.5}\left(\frac{\cos n\pi t}{n\pi }\right)dt}\\ ={\left[\left(2-2\left(1.5\right)\right)\left(\frac{-\cos n\pi \left(1.5\right)}{n\pi }\right)-\left(2-2\left(0.5\right)\right)\left(\frac{-\cos n\pi \left(0.5\right)}{n\pi }\right)\right]}_{0.5}^{1.5}-2{\left[\frac{\sin n\pi t}{n^2{\pi }^2}\right]}_{0.5}^{1.5}\\ =\left[\left(-1\right)\left(\frac{-\cos 1.5n\pi }{n\pi }\right)-\left(1\right)\left(\frac{-\cos 0.5n\pi }{n\pi }\right)\right]-2\left[\frac{\sin n\pi \left(1.5\right)}{n^2{\pi }^2}-\frac{\sin n\pi \left(0.5\right)}{n^2{\pi }^2}\right]\\ =\frac{\cos 1.5n\pi }{n\pi }+\frac{\cos 0.5n\pi }{n\pi }-\frac{2\sin 1.5n\pi }{n^2{\pi }^2}+\frac{2\sin 0.5n\pi }{n^2{\pi }^2}\end{array}$

Simplify the above equation as follows,

$y=\frac{n\pi \cos 1.5n\pi }{n^2{\pi }^2}+\frac{n\pi \cos 0.5n\pi }{n^2{\pi }^2}-\frac{2\sin 1.5n\pi }{n^2{\pi }^2}+\frac{2\sin 0.5n\pi }{n^2{\pi }^2}$

Substituting the values of x and y in equation (9) as follows,

$b_n=\frac{-2\cos 0.5n\pi }{n\pi }+\frac{4\sin 0.5n\pi }{n^2{\pi }^2}+\frac{n\pi \cos 1.5n\pi }{n^2{\pi }^2}+\frac{n\pi \cos 0.5n\pi }{n^2{\pi }^2}-\frac{2\sin 1.5n\pi }{n^2{\pi }^2}+\frac{2\sin 0.5n\pi }{n^2{\pi }^2}$

$b_n=\frac{-2\cos \left(\frac{n\pi }{2}\right)}{n\pi }+\frac{4\sin \left(\frac{n\pi }{2}\right)}{n^2{\pi }^2}+\frac{n\pi \cos 1.5n\pi }{n^2{\pi }^2}+\frac{n\pi \cos \left(\frac{n\pi }{2}\right)}{n^2{\pi }^2}-\frac{2\sin 1.5n\pi }{n^2{\pi }^2}+\frac{2\sin \left(\frac{n\pi }{2}\right)}{n^2{\pi }^2}$

Converting the equation (1) which is in angular frequency into frequency.

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

Substitute the values of $a_0$, $a_n$, and $b_n$ in the above equation as follows,

$\begin{array}{c}v\left(t\right)=0+0+{\displaystyle \sum _{n=1}^{\infty }\left(\begin{array}{l}\frac{-2\cos \left(\frac{n\pi }{2}\right)}{n\pi }+\frac{4\sin \left(\frac{n\pi }{2}\right)}{n^2{\pi }^2}+\frac{n\pi \cos 1.5n\pi }{n^2{\pi }^2}\\ +\frac{n\pi \cos \left(\frac{n\pi }{2}\right)}{n^2{\pi }^2}-\frac{2\sin 1.5n\pi }{n^2{\pi }^2}+\frac{2\sin \left(\frac{n\pi }{2}\right)}{n^2{\pi }^2}\end{array}\right)\sin n{f}_{0}t}\\ ={\displaystyle \sum _{n=1}^{\infty }\left(\begin{array}{l}\frac{-2\cos \left(\frac{n\pi }{2}\right)}{n\pi }+\frac{4\sin \left(\frac{n\pi }{2}\right)}{n^2{\pi }^2}+\frac{n\pi \cos 1.5n\pi }{n^2{\pi }^2}\\ +\frac{n\pi \cos \left(\frac{n\pi }{2}\right)}{n^2{\pi }^2}-\frac{2\sin 1.5n\pi }{n^2{\pi }^2}+\frac{2\sin \left(\frac{n\pi }{2}\right)}{n^2{\pi }^2}\end{array}\right)\sin n\left(\frac{{\omega }_0}{2\pi }\right)t}\left\{\because f_0=\frac{{\omega }_0}{2\pi }\right\}\\ ={\displaystyle \sum _{n=1}^{\infty }\left(\begin{array}{l}\frac{-2\cos \left(\frac{n\pi }{2}\right)}{n\pi }+\frac{4\sin \left(\frac{n\pi }{2}\right)}{n^2{\pi }^2}+\frac{n\pi \cos 1.5n\pi }{n^2{\pi }^2}\\ +\frac{n\pi \cos \left(\frac{n\pi }{2}\right)}{n^2{\pi }^2}-\frac{2\sin 1.5n\pi }{n^2{\pi }^2}+\frac{2\sin \left(\frac{n\pi }{2}\right)}{n^2{\pi }^2}\end{array}\right)\sin n\left(\frac{\pi }{2\pi }\right)t\left\{\because {\omega }_0=\pi \right\}}\end{array}$

$v\left(t\right)={\displaystyle \sum _{n=1}^{\infty }\left(\begin{array}{l}\frac{-2\cos \left(\frac{n\pi }{2}\right)}{n\pi }+\frac{4\sin \left(\frac{n\pi }{2}\right)}{n^2{\pi }^2}+\frac{n\pi \cos 1.5n\pi }{n^2{\pi }^2}\\ +\frac{n\pi \cos \left(\frac{n\pi }{2}\right)}{n^2{\pi }^2}-\frac{2\sin 1.5n\pi }{n^2{\pi }^2}+\frac{2\sin \left(\frac{n\pi }{2}\right)}{n^2{\pi }^2}\end{array}\right)\sin n\left(\frac{1}{2}\right)t}$        (13)

For $n=1$, equation (13) will be as follows,

$\begin{array}{c}v\left(t\right)=\left(\begin{array}{l}\frac{-2\cos \left(\frac{\left(1\right)\pi }{2}\right)}{\left(1\right)\pi }+\frac{4\sin \left(\frac{\left(1\right)\pi }{2}\right)}{{\left(1\right)}^2{\pi }^2}+\frac{\left(1\right)\pi \cos 1.5\left(1\right)\pi }{{\left(1\right)}^2{\pi }^2}\\ +\frac{\left(1\right)\pi \cos \left(\frac{\left(1\right)\pi }{2}\right)}{{\left(1\right)}^2{\pi }^2}-\frac{2\sin 1.5\left(1\right)\pi }{{\left(1\right)}^2{\pi }^2}+\frac{2\sin \left(\frac{\left(1\right)\pi }{2}\right)}{{\left(1\right)}^2{\pi }^2}\end{array}\right)\sin \left(1\right)\left(\frac{1}{2}\right)t\\ =\left(\begin{array}{l}\frac{-2\cos \left(\frac{\pi }{2}\right)}{\pi }+\frac{4\sin \left(\frac{\pi }{2}\right)}{{\pi }^2}+\frac{\pi \cos 1.5\pi }{{\pi }^2}+\frac{\pi \cos \left(\frac{\pi }{2}\right)}{{\pi }^2}\\ -\frac{2\sin 1.5\pi }{{\pi }^2}+\frac{2\sin \left(\frac{\pi }{2}\right)}{{\pi }^2}\end{array}\right)\sin \left(\frac{1}{2}\right)t\\ =\left(0+\frac{4\left(1\right)}{{\pi }^2}+0+0+\frac{2}{{\pi }^2}+\frac{2}{{\pi }^2}\right)\sin \left(\frac{1}{2}\right)t\left\{\begin{array}{l}\because \cos 90=0,\\ \sin 90=1\end{array}\right\}\\ =\frac{8}{{\pi }^2}\sin \left(\frac{1}{2}\right)t\end{array}$

For $n=3$, equation (13) will be as follows,

$\begin{array}{c}v\left(t\right)=\left(\begin{array}{l}\frac{-2\cos \left(\frac{\left(3\right)\pi }{2}\right)}{\left(3\right)\pi }+\frac{4\sin \left(\frac{\left(3\right)\pi }{2}\right)}{{\left(3\right)}^2{\pi }^2}+\frac{\left(3\right)\pi \cos 1.5\left(3\right)\pi }{{\left(3\right)}^2{\pi }^2}\\ +\frac{\left(3\right)\pi \cos \left(\frac{\left(3\right)\pi }{2}\right)}{{\left(3\right)}^2{\pi }^2}-\frac{2\sin 1.5\left(3\right)\pi }{{\left(3\right)}^2{\pi }^2}+\frac{2\sin \left(\frac{\left(3\right)\pi }{2}\right)}{{\left(3\right)}^2{\pi }^2}\end{array}\right)\sin \left(3\right)\left(\frac{1}{2}\right)t\\ =\left(\begin{array}{l}\frac{-2\cos \left(\frac{3\pi }{2}\right)}{3\pi }+\frac{4\sin \left(\frac{3\pi }{2}\right)}{9{\pi }^2}+\frac{3\pi \cos 4.5\pi }{9{\pi }^2}+\frac{3\pi \cos \left(\frac{\pi }{2}\right)}{9{\pi }^2}\\ -\frac{2\sin 4.5\pi }{9{\pi }^2}+\frac{2\sin \left(\frac{3\pi }{2}\right)}{9{\pi }^2}\end{array}\right)\sin \left(\frac{3}{2}\right)t\\ =\left(0+\frac{-4}{9{\pi }^2}+0+0+\frac{-2}{9{\pi }^2}+\frac{-2}{9{\pi }^2}\right)\sin \left(\frac{3}{2}\right)t\\ =\frac{-8}{9{\pi }^2}\sin \left(\frac{3}{2}\right)t\end{array}$

The sketch for the line spectrum of $v\left(t\right)$ is shown in Figure 1.

Conclusion:

Thus, the line spectrum for the waveform shown in Figure 17.4c is sketched as shown in Figure 1.