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) is $\underset{\_}{31,415.93\text{rad/s}}$ and $\underset{\_}{3978.87\text{rad/s}}$respectively.

Explanation of Solution

Calculation:

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

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

Substitute $200\mu \text{s}$ for T in equation (1).

$\begin{array}{c}{\omega }_0=\frac{2\pi }{200\mu \text{s}}\\ =31,415.93\text{rad/s}\end{array}$

Substitute $40\mu \text{s}$ for T in equation (1).

$\begin{array}{c}{\omega }_0=\frac{2\pi }{40\mu \text{s}}\\ =3978.87\text{rad/s}\end{array}$

Conclusion:

Thus, the fundamental frequency ${\omega }_0$ for the periodic functions in (a) and (b) is$\underset{\_}{31,415.93\text{rad/s}}$ and $\underset{\_}{3978.87\text{rad/s}}$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) is$\underset{\_}{5000\text{Hz}}$ and $\underset{\_}{25,000\text{Hz}}$respectively.

Explanation of Solution

Calculation:

Consider that the expression for the frequency $f_0$.

$f_0=\frac{1}{T}$ (2)

Substitute $200\mu \text{s}$ for T in equation (2).

$\begin{array}{c}{f}_0=\frac{1}{200\mu \text{s}}\\ =5000\text{Hz}\end{array}$

Substitute $40\mu \text{s}$ for T in equation (2).

$\begin{array}{c}{f}_0=\frac{1}{40\mu \text{s}}\\ =25,000\text{Hz}\end{array}$

Conclusion:

Thus, the fundamental frequency $f_0$ for the periodic functions in (a) and (b) are $\underset{\_}{5000\text{Hz}}\text{and}\underset{\_}{25,000\text{Hz}}$ 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) is $\underset{\_}{0}$ and $\underset{\_}{25V}$respectively.

Explanation of Solution

Calculation:

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

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

$\begin{array}{c}{a}_v=\frac{V_m}{T}\left(10\times {10}^{-6}\right)\\ =\frac{100\left(10\times {10}^{-6}\right)}{40\times {10}^{-6}}\left\{\because V_m=100\text{V}\text{and}T=40\times {10}^{-6}\text{s}\right\}\\ =25\text{V}\end{array}$

Conclusion:

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

d.

To determine

Calculate the Fourier co-efficients $a_k$ and $b_k$for the both periodic functions.

Answer to Problem 1P

The Fourier co-efficients $a_k$ and $b_k$ for the periodic function in (a) is $\underset{\_}{-\frac{80}{\pi k}\sin \frac{\pi k}{2}\left(k\text{odd}\right),0\left(k\text{even}\right)}$ and $\underset{\_}{\frac{240}{\pi k}\left(k\text{odd}\right),0\left(k\text{even}\right)}$respectively. The Fourier co-efficients $a_k$ and $b_k$ for the periodic function in (b) is $\underset{\_}{\frac{200}{\pi k}\sin \frac{\pi k}{4}\left(k\text{odd}\right)}$ and $\underset{\_}{0}$respectively.

Explanation of Solution

Calculation:

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

Calculate the Fourier co-efficient $a_k$ for the function in (a).

$\begin{array}{c}{a}_k=\left[\begin{array}{l}\frac{2}{T}{\displaystyle {\int }_0^{T/4}40\cos \frac{2\pi kt}{T}}dt+\frac{2}{T}{\displaystyle {\int }_{T/4}^{T/2}80\cos \frac{2\pi kt}{T}}dt+\\ \frac{2}{T}{\displaystyle {\int }_{T/2}^{3T/4}-40\cos \frac{2\pi kt}{T}}dt+\frac{2}{T}{\displaystyle {\int }_{3T/4}^T-80\cos \frac{2\pi kt}{T}}dt\end{array}\right]\\ =\left[\begin{array}{l}\frac{80}{T}\frac{T}{2\pi k}{\sin \frac{2\pi kt}{T}|}_0^{T/4}+\frac{160}{T}\frac{T}{2\pi k}{\sin \frac{2\pi kt}{T}|}_{T/4}^{T/2}-\\ \frac{80}{T}\frac{T}{2\pi k}{\sin \frac{2\pi kt}{T}|}_{T/2}^{3T/4}+\frac{-160}{T}\frac{T}{2\pi k}{\sin \frac{2\pi kt}{T}|}_{3T/4}^T\end{array}\right]\\ =-\frac{80}{\pi k}\sin \frac{\pi k}{2},k\text{odd}\end{array}$

And

$a_k=0,k\text{even}$

Calculate the Fourier co-efficient $b_k$ for the function in (a).

$\begin{array}{c}{b}_k=\left[\begin{array}{l}\frac{2}{T}{\displaystyle {\int }_0^{T/4}40\sin \frac{2\pi kt}{T}}dt+\frac{2}{T}{\displaystyle {\int }_{T/4}^{T/2}80\sin \frac{2\pi kt}{T}}dt+\\ \frac{2}{T}{\displaystyle {\int }_{T/2}^{3T/4}-40\sin \frac{2\pi kt}{T}}dt+\frac{2}{T}{\displaystyle {\int }_{3T/4}^T-80\sin \frac{2\pi kt}{T}}dt\end{array}\right]\\ =\left[\begin{array}{l}\frac{-80}{T}\frac{T}{2\pi k}{\cos \frac{2\pi kt}{T}|}_0^{T/4}-\frac{160}{T}\frac{T}{2\pi k}{\cos \frac{2\pi kt}{T}|}_{T/4}^{T/2}+\\ \frac{80}{T}\frac{T}{2\pi k}{\cos \frac{2\pi kt}{T}|}_{T/2}^{3T/4}+\frac{160}{T}\frac{T}{2\pi k}{\cos \frac{2\pi kt}{T}|}_{3T/4}^T\end{array}\right]\\ =\frac{240}{\pi k},k\text{odd}\end{array}$

And

$b_k=0,k\text{even}$

The Fourier co-efficient $a_v$ for the periodic voltage in voltage in part (b) is 25 V.

Calculate the Fourier co-efficient $a_k$ for the function in (b).

$\begin{array}{c}{a}_k=\frac{2}{T}{\displaystyle {\int }_{-T/8}^{T/8}100\cos \frac{2\pi kt}{T}}dt\\ =\frac{200}{T}\frac{T}{2\pi k}{\sin \frac{2\pi k}{T}t|}_{-T/8}^{T/8}\\ =\frac{200}{\pi k}\sin \frac{\pi k}{4}\end{array}$

Calculate the Fourier co-efficient $b_k$ for the function in (b).

$\begin{array}{c}{b}_k=\frac{2}{T}{\displaystyle {\int }_{-T/8}^{T/8}100\sin \frac{2\pi kt}{T}}dt\\ =-\frac{200}{T}\frac{T}{2\pi k}{\cos \frac{2\pi k}{T}t|}_{-T/8}^{T/8}\\ =0\end{array}$

Conclusion:

Thus, the Fourier co-efficients $a_k$ and $b_k$ for the periodic function in (a) is $\underset{\_}{-\frac{80}{\pi k}\sin \frac{\pi k}{2}\left(k\text{odd}\right),0\left(k\text{even}\right)}$ and $\underset{\_}{\frac{240}{\pi k}\left(k\text{odd}\right),0\left(k\text{even}\right)}$respectively. The Fourier co-efficients $a_k$ and $b_k$ for the periodic function in (b) is $\underset{\_}{\frac{200}{\pi k}\sin \frac{\pi k}{4}\left(k\text{odd}\right)}$ and $\underset{\_}{0}$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) is$\underset{\_}{\frac{80}{\pi }{\displaystyle \sum _{n=1,3,5,\cdot \cdot }^{\infty }\left(-\frac{1}{n}\sin \frac{n\pi }{2}\cos n{\omega }_{0}t+\frac{3}{n}\sin n{\omega }_{0}t\right)}\text{V}}$ and $\underset{\_}{25+\frac{200}{\pi }{\displaystyle \sum _{n=1}^{\infty }\left(\frac{1}{n}\sin \frac{n\pi }{4}\cos n{\omega }_{0}t\right)}\text{V}}$respectively.

Explanation of Solution

Calculation:

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

$\begin{array}{c}v\left(t\right)=a_v+{\displaystyle \sum _{n=1,3,5,\cdot \cdot }^{\infty }\left(a_n\cos n{\omega }_{0}t+b_n\sin n{\omega }_{0}t\right)}\text{V}\\ =a_v+{\displaystyle \sum _{n=1,3,5,\cdot \cdot }^{\infty }\left(-\frac{80}{n\pi }\sin \frac{n\pi }{2}\cos n{\omega }_{0}t+\frac{240}{n\pi }\sin n{\omega }_{0}t\right)}\text{V}\\ =0+\frac{80}{\pi }{\displaystyle \sum _{n=1,3,5,\cdot \cdot }^{\infty }\left(-\frac{1}{n}\sin \frac{n\pi }{2}\cos n{\omega }_{0}t+\frac{3}{n}\sin n{\omega }_{0}t\right)}\text{V}\left\{\because a_v=0\right\}\\ =\frac{80}{\pi }{\displaystyle \sum _{n=1,3,5,\cdot \cdot }^{\infty }\left(-\frac{1}{n}\sin \frac{n\pi }{2}\cos n{\omega }_{0}t+\frac{3}{n}\sin n{\omega }_{0}t\right)}\text{V}\end{array}$

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

$\begin{array}{c}v\left(t\right)=a_v+{\displaystyle \sum _{n=1}^{\infty }\left(a_n\cos n{\omega }_{0}t+b_n\sin n{\omega }_{0}t\right)}\text{V}\\ =a_v+{\displaystyle \sum _{n=1}^{\infty }\left(\frac{200}{\pi n}\sin \frac{n\pi }{4}\cos n{\omega }_{0}t+0\right)}\text{V}\\ =25+\frac{200}{\pi }{\displaystyle \sum _{n=1}^{\infty }\left(\frac{1}{n}\sin \frac{n\pi }{4}\cos n{\omega }_{0}t\right)}\text{V}\left\{\because a_v=25\text{V}\right\}\end{array}$

Conclusion:

Thus, the Fourier series expression of voltage $v\left(t\right)$ for the periodic functions in (a) and (b) is$\underset{\_}{\frac{80}{\pi }{\displaystyle \sum _{n=1,3,5,\cdot \cdot }^{\infty }\left(-\frac{1}{n}\sin \frac{n\pi }{2}\cos n{\omega }_{0}t+\frac{3}{n}\sin n{\omega }_{0}t\right)}\text{V}}$and $\underset{\_}{25+\frac{200}{\pi }{\displaystyle \sum _{n=1}^{\infty }\left(\frac{1}{n}\sin \frac{n\pi }{4}\cos n{\omega }_{0}t\right)}\text{V}}$respectively.