Chapter 16, Problem 1P

a.

To determine

Calculate the fundamental frequency ${\omega }_0$ for the both periodic functions.

Answer to Problem 1P

The fundamental frequency ${\omega }_0$ for the periodic functions in (a) and (b) are $\underset{\_}{785.4\text{rad/s}}\text{and}\underset{\_}{78.54\text{krad/s}}$ respectively.

Explanation of Solution

Calculation:

Consider that the expression for the fundamental frequency ${\omega }_0$.

${\omega }_0=\frac{2\pi }{T}$        (1)

Substitute $8\text{ms}$ for T in equation (1).

$\begin{array}{c}{\omega }_0=\frac{2\pi }{8\text{ms}}\\ =785.4\text{rad/s}\end{array}$

Substitute $80\text{ms}$ for T in equation (1).

$\begin{array}{c}{\omega }_0=\frac{2\pi }{80\text{ms}}\\ =78.54\text{krad/s}\end{array}$

Conclusion:

Thus, the fundamental frequency ${\omega }_0$ for the periodic functions in (a) and (b) are $\underset{\_}{785.4\text{rad/s}}\text{and}\underset{\_}{78.54\text{krad/s}}$ respectively.

b.

To determine

Calculate the frequency $f_0$ for the both periodic functions.

Answer to Problem 1P

The fundamental frequency $f_0$ for the periodic functions in (a) and (b) are $\underset{\_}{125\text{Hz}}\text{and}\underset{\_}{12.5\text{Hz}}$ respectively.

Explanation of Solution

Calculation:

Consider that the expression for the frequency $f_0$.

$f_0=\frac{1}{T}$        (2)

Substitute $8\text{ms}$ for T in equation (2).

$\begin{array}{c}{f}_0=\frac{1}{8\text{ms}}\\ =125\text{Hz}\end{array}$

Substitute $80\text{ms}$ for T in equation (2).

$\begin{array}{c}{f}_0=\frac{1}{80\text{ms}}\\ =12.5\text{Hz}\end{array}$

Conclusion:

Thus, the fundamental frequency $f_0$ for the periodic functions in (a) and (b) are $\underset{\_}{125\text{Hz}}\text{and}\underset{\_}{12.5\text{Hz}}$ respectively.

c.

To determine

Calculate the Fourier co-efficient $a_v$ for the both periodic functions.

Answer to Problem 1P

The Fourier co-efficient $a_v$ for the periodic functions in (a) and (b) are $\underset{\_}{25\text{V}\text{and}0}$ respectively.

Explanation of Solution

Calculation:

Calculate the Fourier co-efficient $a_v$ for the periodic voltage in voltage in part (a).

$\begin{array}{c}{a}_v=\frac{V_m}{T}\left(4\times {10}^{-3}\right)\\ =\frac{50\left(4\times {10}^{-3}\right)}{8\times {10}^{-3}}\left\{\because V_m=50\text{V}\text{and}T=8\times {10}^{-3}\text{s}\right\}\\ =25\text{V}\end{array}$

For the periodic voltage in part (b), the Fourier co-efficient $a_v$ is 0 since it is the odd function with half-wave symmetry.

Conclusion:

Thus, the Fourier co-efficient $a_v$ for the periodic functions in (a) and (b) are $\underset{\_}{25\text{V}\text{and}0}$ respectively.

d.

To determine

Calculate the Fourier co-efficients $a_k\text{and}{b}_k$ for the both periodic functions.

Answer to Problem 1P

The Fourier co-efficients $a_k\text{and}{b}_k$ for the periodic function in (a) are $\underset{\_}{\frac{100}{\pi k}\sin \frac{\pi k}{2}\text{and}0}$ respectively. The Fourier co-efficients $a_k\text{and}{b}_k$ for the periodic function in (b) are $\underset{\_}{\frac{120}{\pi k}\sin \frac{\pi k}{4}\text{and}\frac{120}{\pi k}\left[1-\cos \left(\pi k\right)\right]}$ respectively.

Explanation of Solution

Calculation:

Consider that the periodic function in Figure P16.1(a). The Fourier co-efficient $a_v$ is 25 V.

Calculate the Fourier co-efficient $a_k$ using Figure P16.1(a).

$\begin{array}{c}{a}_k=\frac{2}{T}{\displaystyle {\int }_{-T/4}^{T/4}50\cos \frac{2\pi kt}{T}}dt\\ =\frac{100}{T}\frac{T}{2\pi k}{\sin \frac{2\pi kt}{T}|}_{-T/4}^{T/4}\\ =\frac{100}{\pi k}\sin \frac{\pi k}{2}\end{array}$

Calculate the Fourier co-efficient $b_k$ using Figure P16.1(a).

$\begin{array}{c}{b}_k=\frac{2}{T}{\displaystyle {\int }_{-T/4}^{T/4}50\sin \frac{2\pi kt}{T}}dt\\ =-\frac{100}{T}\frac{T}{2\pi k}{\cos \frac{2\pi kt}{T}|}_{-T/4}^{T/4}\\ =0\end{array}$

Consider that the periodic function in Figure P16.1(b). The Fourier co-efficient $a_v$ is 0.

Calculate the Fourier co-efficient $a_k$ using Figure P16.1(b).

$\begin{array}{c}{a}_k=\left\{\begin{array}{l}\frac{2}{T}\left[{\displaystyle {\int }_0^{T/4}90\cos \frac{2\pi kt}{T}}dt+{\displaystyle {\int }_{T/4}^{T/2}30\cos \frac{2\pi kt}{T}}dt\right]-\\ \frac{2}{T}\left[{\displaystyle {\int }_{T/2}^{3T/4}90\cos \frac{2\pi kt}{T}}dt+{\displaystyle {\int }_{3T/4}^{T}30\cos \frac{2\pi kt}{T}}dt\right]\end{array}\right\}\\ =\left\{\begin{array}{l}\frac{60}{T}\frac{T}{2\pi k}\left[{3\sin \frac{2\pi kt}{T}|}_0^{T/4}+{\sin \frac{2\pi kt}{T}|}_{T/4}^{T/2}\right]\\ -\frac{60}{T}\frac{T}{2\pi k}\left[{3\sin \frac{2\pi kt}{T}|}_{T/2}^{3T/4}+{\sin \frac{2\pi kt}{T}|}_{3T/4}^T\right]\end{array}\right\}\\ =\frac{30}{\pi k}\left[2\sin \frac{\pi k}{2}-2\sin \frac{3\pi k}{2}\right]\\ =\frac{120}{\pi k}\sin \frac{\pi k}{2}\end{array}$

Calculate the Fourier co-efficient $b_k$ using Figure P16.1(b).

$\begin{array}{c}{b}_k=\left\{\begin{array}{l}\frac{2}{T}\left[{\displaystyle {\int }_0^{T/4}90\sin \frac{2\pi kt}{T}}dt+{\displaystyle {\int }_{T/4}^{T/2}30\sin \frac{2\pi kt}{T}}dt\right]-\\ \frac{2}{T}\left[{\displaystyle {\int }_{T/2}^{3T/4}90\sin \frac{2\pi kt}{T}}dt+{\displaystyle {\int }_{3T/4}^{T}30\sin \frac{2\pi kt}{T}}dt\right]\end{array}\right\}\\ =\left\{\begin{array}{l}-\frac{60}{T}\frac{T}{2\pi k}\left[{3\cos \frac{2\pi kt}{T}|}_0^{T/4}+{\cos \frac{2\pi kt}{T}|}_{T/4}^{T/2}\right]\\ +\frac{60}{T}\frac{T}{2\pi k}\left[{3\cos \frac{2\pi kt}{T}|}_{T/2}^{3T/4}+{\cos \frac{2\pi kt}{T}|}_{3T/4}^T\right]\end{array}\right\}\\ =\frac{120}{\pi k}\left[1-\cos \left(k\pi \right)\right]\end{array}$

The value of $b_k$ is 0 for even values of k and the value of $b_k$ is $\frac{120\left(2\right)}{\pi k}$ for off values of k.

Conclusion:

Thus, the Fourier co-efficients $a_k\text{and}{b}_k$ for the periodic function in (a) are $\underset{\_}{\frac{100}{\pi k}\sin \frac{\pi k}{2}\text{and}0}$ respectively. The Fourier co-efficients $a_k\text{and}{b}_k$ for the periodic function in (b) are $\underset{\_}{\frac{120}{\pi k}\sin \frac{\pi k}{4}\text{and}\frac{120}{\pi k}\left[1-\cos \left(\pi k\right)\right]}$ respectively.

e.

To determine

Derive the Fourier series expression for the voltage $v\left(t\right)$.

Answer to Problem 1P

The Fourier series expression of voltage $v\left(t\right)$ for the periodic functions in (a) and (b) are $\underset{\_}{25+\frac{100}{\pi }{\displaystyle \sum _{n=1}^{\infty }\left(\frac{1}{n}\sin \frac{n\pi }{2}\cos n{\omega }_{0}t\right)}\text{V}\text{and}\frac{120}{\pi }{\displaystyle \sum _{n=1,3,5\cdot \cdot \cdot }^{\infty }\frac{1}{n}\left(\sin \frac{n\pi }{2}\cos n{\omega }_{0}t+2\sin n{\omega }_{0}t\right)}\text{V}}$ respectively.

Explanation of Solution

Calculation:

Write the Fourier series expression of voltage $v\left(t\right)$ for the periodic function in (a).

$\begin{array}{c}v\left(t\right)=a_v+{\displaystyle \sum _{n=1}^{\infty }\left(a_n\cos n{\omega }_{0}t\right)}\text{V}\\ =25+{\displaystyle \sum _{n=1}^{\infty }\left(\frac{100}{n\pi }\sin \frac{n\pi }{2}\cos n{\omega }_{0}t\right)}\text{V}\left\{\begin{array}{l}\because a_v=25,\\ a_n=\frac{100}{n\pi }\sin \frac{n\pi }{2}\end{array}\right\}\\ =25+\frac{100}{\pi }{\displaystyle \sum _{n=1}^{\infty }\left(\frac{1}{n}\sin \frac{n\pi }{2}\cos n{\omega }_{0}t\right)}\text{V}\end{array}$

Write the Fourier series expression of voltage $v\left(t\right)$ for the periodic function in (b).

$\begin{array}{c}v\left(t\right)=a_v+{\displaystyle \sum _{n=1,2,3,\cdot \cdot \cdot }^{\infty }\left(a_n\cos n{\omega }_{0}t+b_n\sin n{\omega }_{0}t\right)}\text{V}\\ =0+{\displaystyle \sum _{n=1,2,3,\cdot \cdot \cdot }^{\infty }\left(\frac{120}{n\pi }\sin \frac{n\pi }{2}\cos n{\omega }_{0}t+\frac{120\left(2\right)}{\pi n}\sin n{\omega }_{0}t\right)}\text{V}\left\{\begin{array}{l}\because a_v=0,\\ a_n=\frac{120}{n\pi }\sin \frac{n\pi }{2},\\ b_n=\frac{120\left(2\right)}{\pi n}\end{array}\right\}\\ =0+\frac{120}{\pi }{\displaystyle \sum _{n=1,3,5\cdot \cdot \cdot }^{\infty }\frac{1}{n}\left(\sin \frac{n\pi }{2}\cos n{\omega }_{0}t+2\sin n{\omega }_{0}t\right)}\text{V}\\ =\frac{120}{\pi }{\displaystyle \sum _{n=1,3,5\cdot \cdot \cdot }^{\infty }\frac{1}{n}\left(\sin \frac{n\pi }{2}\cos n{\omega }_{0}t+2\sin n{\omega }_{0}t\right)}\text{V}\end{array}$

Conclusion:

Thus, the Fourier series expression of voltage $v\left(t\right)$ for the periodic functions in (a) and (b) are $\underset{\_}{25+\frac{100}{\pi }{\displaystyle \sum _{n=1}^{\infty }\left(\frac{1}{n}\sin \frac{n\pi }{2}\cos n{\omega }_{0}t\right)}\text{V}\text{and}\frac{120}{\pi }{\displaystyle \sum _{n=1,3,5\cdot \cdot \cdot }^{\infty }\frac{1}{n}\left(\sin \frac{n\pi }{2}\cos n{\omega }_{0}t+2\sin n{\omega }_{0}t\right)}\text{V}}$ respectively.