Chapter 17, Problem 15E

To determine

Sketch the line spectrum for the waveform shown in Figure 17.31.

Answer to Problem 15E

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

Explanation of Solution

Given data:

Refer to Figure 17.31 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 17.31, the time period is $T=0.4$.

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

$v\left(t\right)=\{\begin{array}{l}{V}_m\cos 5\pi t,-0.1\le t\le 0.1\\ 0,0.1\le t\le 0.3\end{array}$        (6)

Substitute 0.4 for T in equation (5) to find ${\omega }_0$.

${\omega }_0=\frac{2\pi }{0.4}$

${\omega }_0=5\pi$        (7)

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

$\begin{array}{c}{a}_0=\frac{1}{0.4}{\displaystyle {\int }_0^{0.4}v\left(t\right)}dt\\ =\frac{1}{0.4}\left[{\displaystyle {\int }_{-0.1}^{0.1}{V}_m\cos 5\pi t}dt+{\displaystyle {\int }_{0.1}^{0.3}\left(0\right)}dt+\right]\\ =\frac{V_m}{0.4}{\left[\frac{\sin 5\pi t}{5\pi }\right]}_{-0.1}^{0.1}\\ =\frac{V_m}{0.4}\left[\frac{\sin 5\pi \left(0.1\right)}{5\pi }-\frac{\sin 5\pi \left(-0.1\right)}{5\pi }\right]\\ =\frac{V_m}{0.4}\left[\frac{1+1}{5\pi }\right]\end{array}$

Simplify the above equation as follows,

$a_0=\frac{V_m}{\pi }$

For half wave symmetry and even symmetry,

For all values of ‘n’, $b_n=0$.

Applying equation (6) in equation (3) to find the value of coefficient $a_n$.

$a_n=\frac{2}{0.4}{\displaystyle {\int }_0^{0.4}v\left(t\right)\cos n{\omega }_{0}t}dt$

$a_n=\frac{2V_m}{0.4}\left[{\displaystyle {\int }_{-0.1}^{0.1}\cos \left(5\pi t\right)\cos \left(5\pi nt\right)}dt\right]\left\{\because {\omega }_0=5\pi \right\}$        (8)

Consider the function,

$x={\displaystyle \int \cos \left(5\pi t\right)\cos \left(5\pi nt\right)dt}$

Consider,

$u=5\pi t$

On differentiating the above expression,

$\begin{array}{l}\frac{du}{dt}=5\pi \\ dt=\frac{du}{5\pi }\end{array}$

Equation (8) will be follows,

$x=\frac{1}{5\pi }{\displaystyle \int \cos \left(u\right)\cos \left(nu\right)du}$        (9)

In the above equation, consider,

$\begin{array}{c}y={\displaystyle \int \cos \left(u\right)\cos \left(nu\right)du}\\ ={\displaystyle \int \frac{\cos \left(nu+u\right)+\cos \left(nu-u\right)}{2}}du\left\{\because \cos \left(x\right)\cos \left(y\right)=\frac{1}{2}\left[\cos \left(y+x\right)+\cos \left(y-x\right)\right]\right\}\\ ={\displaystyle \int \frac{\cos \left(\left(n+1\right)u\right)+\cos \left(\left(n-1\right)u\right)}{2}}du\end{array}$

By applying linearity,

$y=\frac{1}{2}{\displaystyle \int \cos \left(\left(n+1\right)u\right)du}+\frac{1}{2}{\displaystyle \int \cos \left(\left(n-1\right)u\right)}du$        (10)

In equation (10),

consider,

$m={\displaystyle \int \cos \left(\left(n+1\right)u\right)du}$        (11)

Let,

$\begin{array}{l}v=\left(n+1\right)u\\ \frac{dv}{du}=\left(n+1\right)\\ du=\frac{dv}{n+1}\end{array}$

Equation (11) will be as follows,

$\begin{array}{c}m=\frac{1}{n+1}{\displaystyle \int \cos \left(v\right)dv}\\ =\frac{\sin \left(v\right)}{n+1}\\ =\frac{\sin \left(\left(n+1\right)u\right)}{n+1}\left\{\because v=\left(n+1\right)u\right\}\end{array}$

Similarly, in equation (10),

consider,

$l={\displaystyle \int \cos \left(\left(n-1\right)u\right)du}$        (12)

Let,

$\begin{array}{l}v=\left(n-1\right)u\\ \frac{dv}{du}=\left(n-1\right)\\ du=\frac{dv}{n-1}\end{array}$

Equation (12) will be as follows,

$\begin{array}{c}l=\frac{1}{n-1}{\displaystyle \int \cos \left(v\right)dv}\\ =\frac{\sin \left(v\right)}{n-1}\\ =\frac{\sin \left(\left(n-1\right)u\right)}{n-1}v=\left(n-1\right)u\end{array}$

Substitute the values of m and l in equation (10) as follows,

$\begin{array}{c}y=\frac{1}{2}\left[\frac{\sin \left(\left(n+1\right)u\right)}{n+1}+\frac{\sin \left(\left(n-1\right)u\right)}{n-1}\right]\\ =\frac{\sin \left(\left(n+1\right)u\right)}{2\left(n+1\right)}+\frac{\sin \left(\left(n-1\right)u\right)}{2\left(n-1\right)}\end{array}$

Substitute the value of y in equation (9) as follows,

$\begin{array}{c}x=\frac{1}{5\pi }\left[\frac{\sin \left(\left(n+1\right)u\right)}{2\left(n+1\right)}+\frac{\sin \left(\left(n-1\right)u\right)}{2\left(n-1\right)}\right]\\ =\frac{\sin \left(\left(n+1\right)u\right)}{10\pi \left(n+1\right)}+\frac{\sin \left(\left(n-1\right)u\right)}{10\pi \left(n-1\right)}\\ =\frac{\sin \left(5\pi \left(n+1\right)t\right)}{10\pi \left(n+1\right)}+\frac{\sin \left(5\pi \left(n-1\right)t\right)}{10\pi \left(n-1\right)}\left\{\because u=5\pi t\right\}\end{array}$

Therefore,

$\begin{array}{c}{\displaystyle \int \cos \left(5\pi t\right)\cos \left(5\pi nt\right)dt}=\frac{\sin \left(5\pi \left(n+1\right)t\right)}{10\pi \left(n+1\right)}+\frac{\sin \left(5\pi \left(n-1\right)t\right)}{10\pi \left(n-1\right)}\\ =\frac{\left(n-1\right)\sin \left(5\pi \left(n+1\right)t\right)+\left(n+1\right)\sin \left(5\pi \left(n-1\right)t\right)}{10\pi \left(n^2-1\right)}\end{array}$

On applying the limits,

$\begin{array}{c}{\displaystyle {\int }_{-0.1}^{0.1}\cos \left(5\pi t\right)\cos \left(5\pi nt\right)}dt={\left[\frac{\left(n-1\right)\sin \left(5\pi \left(n+1\right)t\right)+\left(n+1\right)\sin \left(5\pi \left(n-1\right)t\right)}{10\pi \left(n^2-1\right)}\right]}_{-0.1}^{0.1}\\ =\left[\frac{\left(n-1\right)\sin \left(5\pi \left(n+1\right)\left(0.1\right)\right)+\left(n+1\right)\sin \left(5\pi \left(n-1\right)\left(0.1\right)\right)}{10\pi \left(n^2-1\right)}\right]\\ -\left[\frac{\left(n-1\right)\sin \left(5\pi \left(n+1\right)\left(-0.1\right)\right)+\left(n+1\right)\sin \left(5\pi \left(n-1\right)\left(-0.1\right)\right)}{10\pi \left(n^2-1\right)}\right]\\ =\left[\frac{\left(n-1\right)\sin \left(\frac{n\pi +\pi }{2}\right)+\left(n+1\right)\sin \left(\frac{n\pi -\pi }{2}\right)}{10\pi \left(n^2-1\right)}\right]\\ -\left[\frac{-\left(n-1\right)\sin \left(\frac{n\pi +\pi }{2}\right)-\left(n+1\right)\sin \left(\frac{n\pi -\pi }{2}\right)}{10\pi \left(n^2-1\right)}\right]\\ =\left[\frac{\left(n-1\right)\sin \left(\frac{n\pi +\pi }{2}\right)+\left(n+1\right)\sin \left(\frac{n\pi -\pi }{2}\right)}{10\pi \left(n^2-1\right)}\right]+\\ \left[\frac{\left(n-1\right)\sin \left(\frac{n\pi +\pi }{2}\right)+\left(n+1\right)\sin \left(\frac{n\pi -\pi }{2}\right)}{10\pi \left(n^2-1\right)}\right]\end{array}$

Simplify the equation as follows,

$\begin{array}{c}{\displaystyle {\int }_{-0.1}^{0.1}\cos \left(5\pi t\right)\cos \left(5\pi nt\right)}dt=\frac{2\left[\left(n-1\right)\sin \left(\frac{n\pi +\pi }{2}\right)+\left(n+1\right)\sin \left(\frac{n\pi -\pi }{2}\right)\right]}{10\pi \left(n^2-1\right)}\\ =\frac{2\left[\left(n-1\right)\sin \left(\frac{n\pi }{2}+\frac{\pi }{2}\right)+\left(n+1\right)\sin \left(\frac{n\pi }{2}-\frac{\pi }{2}\right)\right]}{10\pi \left(n^2-1\right)}\\ =\frac{2\left[\left(n-1\right)\cos \left(\frac{n\pi }{2}\right)-\left(n+1\right)\cos \left(-\frac{n\pi }{2}\right)\right]}{10\pi \left(n^2-1\right)}\left\{\begin{array}{l}\because \sin \left(90+\theta \right)=\cos \theta ,\\ \sin \left(-\theta \right)=-\sin \theta \end{array}\right\}\\ =\frac{2\left[n\cos \left(\frac{n\pi }{2}\right)-\cos \left(\frac{n\pi }{2}\right)-\left(n+1\right)\cos \left(-\frac{n\pi }{2}\right)\right]}{10\pi \left(n^2-1\right)}\end{array}$

Simplify the equation as follows,

$\begin{array}{c}{\displaystyle {\int }_{-0.1}^{0.1}\cos \left(5\pi t\right)\cos \left(5\pi nt\right)}dt=\frac{2\left[n\cos \left(\frac{n\pi }{2}\right)-\cos \left(\frac{n\pi }{2}\right)-\left(n+1\right)\cos \left(\frac{n\pi }{2}\right)\right]}{10\pi \left(n^2-1\right)}\left\{\because \cos \left(-\theta \right)=\cos \theta \right\}\\ =\frac{2\left[n\cos \left(\frac{n\pi }{2}\right)-\cos \left(\frac{n\pi }{2}\right)-n\cos \left(\frac{n\pi }{2}\right)-\cos \left(\frac{n\pi }{2}\right)\right]}{10\pi \left(n^2-1\right)}\\ =\frac{2\left[-2\cos \left(\frac{n\pi }{2}\right)\right]}{10\pi \left(n^2-1\right)}\\ =\frac{-2\cos \left(\frac{n\pi }{2}\right)}{5\pi \left(n^2-1\right)}\end{array}$

Substitute the value of ${\displaystyle {\int }_{-0.1}^{0.1}\cos \left(5\pi t\right)\cos \left(5\pi nt\right)}dt$ in equation (8) as follows,

$\begin{array}{c}{a}_n=\frac{2V_m}{0.4}\left[\frac{-2\cos \left(\frac{n\pi }{2}\right)}{5\pi \left(n^2-1\right)}\right]\\ =\frac{-2V_m\cos \left(\frac{n\pi }{2}\right)}{\pi \left(n^2-1\right)}\end{array}$

$a_n=\frac{2V_m}{\pi }\frac{\cos \left(\frac{n\pi }{2}\right)}{\left(1-n^2\right)}\left(n\ne 1\right)$

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

$v\left(t\right)=a_0+{\displaystyle \sum _{n=1}^{\infty }\left(a_n\cos n{f}_{0}t+b_n\sin n{f}_{0}t\right)}$        (13)

Substitute the value of $a_0$, $a_n$, and $b_n$ in equation (13) as follows,

$\begin{array}{c}v\left(t\right)=\frac{V_m}{\pi }+{\displaystyle \sum _{n=1}^{\infty }\left(\left(\frac{2V_m}{\pi }\right)\left(\frac{\cos \left(\frac{n\pi }{2}\right)}{1-n^2}\right)\cos n\left(\frac{5\pi }{2\pi }\right)t+0\right)}\left\{\begin{array}{l}\because f_0=\frac{{\omega }_0}{2\pi },\\ {\omega }_0=5\pi \end{array}\right\}\\ =\frac{V_m}{\pi }+{\displaystyle \sum _{n=1}^{\infty }\left(\left(\frac{2V_m}{\pi }\right)\left(\frac{\cos \left(\frac{n\pi }{2}\right)}{1-n^2}\right)\cos n\left(\frac{5}{2}\right)t\right)}\end{array}$

By considering $V_m=1\text{V}$, the above equation will be as follows,

$v\left(t\right)=\frac{1}{\pi }+{\displaystyle \sum _{n=1}^{\infty }\left(\left(\frac{2}{\pi }\right)\left(\frac{\cos \left(\frac{n\pi }{2}\right)}{1-n^2}\right)\cos n\left(\frac{5}{2}\right)t\right)}$        (14)

For $n=0$, the equation (14) will be as follows,

$v\left(t\right)=\frac{1}{\pi }$

For $n=2$, the equation (14) will be as follows,

$\begin{array}{c}v\left(t\right)=\frac{1}{\pi }+\left(\frac{2}{\pi }\right)\left(\frac{\cos \left(\frac{2\pi }{2}\right)}{1-{\left(2\right)}^2}\right)\cos \left(2\right)\left(\frac{5}{2}\right)t\\ =\frac{1}{\pi }+\left(\frac{2}{\pi }\right)\left(\frac{-1}{-3}\right)\cos \left(5\right)t\\ =\frac{1}{\pi }+\left(\frac{2}{\pi }\right)\left(\frac{-1}{-3}\right)\cos \left(5\right)t\\ =\frac{1}{\pi }+0.2\cos 5t\end{array}$

The sketch for the line spectrum is shown in Figure 1.

Conclusion:

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