Chapter 17, Problem 50CP

(a)

To determine

To show: The left hand side of the given equation is $0$ if $m$ is even and $\frac{4A}{m\omega }$ if $m$ is odd.

Answer to Problem 50CP

The left hand side of the given equation is $0$ if $m$ is even and $\frac{4A}{m\omega }$ if $m$ is odd.

Explanation of Solution

Given info: The frequency of wave is $f$, the amplitude of wave is $A$. The period of wave is described by $y\left(t\right)=\{\begin{array}{l}A0<t<\frac{T}{2}\\ -A\frac{T}{2}<t<T\end{array}$. The given equation is $y\left(t\right)={\displaystyle \sum _{\text{n}}\left(A_{\text{n}}\sin n\omega t+B_{\text{n}}\cos n\omega t\right)}$.

Write the expression of the $y\left(t\right)$.

$y\left(t\right)={\displaystyle \sum _{\text{n}}\left(A_{\text{n}}\sin n\omega t+B_{\text{n}}\cos n\omega t\right)}$

Multiply $\sin m\omega t$ both sides in above equation.

$y\left(t\right)\sin m\omega t={\displaystyle \sum _{}\left(A_{\text{n}}\sin n\omega t+B_{\text{n}}\cos n\omega t\right)}\left(\sin m\omega t\right)$

Integrate the above equation over on period of time $T$.

${\displaystyle {\int }_0^{T}y\left(t\right)\sin m\omega t}=\left[\begin{array}{l}{\displaystyle \sum _{\text{n}}{\displaystyle \underset{0}{\overset{T}{\int }}{A}_{\text{n}}\left(\sin n\omega t\right)\left(\sin m\omega t\right)dt}}\\ +{\displaystyle \sum {\displaystyle \underset{0}{\overset{T}{\int }}{B}_{\text{n}}\left(\cos n\omega T\right)\left(\sin n\omega T\right)dt}}\end{array}\right]$ (1)

The term $y\left(t\right)$ is a positive constant $A$ for half of the period $T$ and $-A$ for next half period.

The left side of the equation is,

$\begin{array}{c}{\displaystyle {\int }_0^{T}y\left(t\right)\sin m\omega t}={\displaystyle {\int }_0^{\frac{T}{2}}A\sin m\omega t}+{\displaystyle {\int }_0^{\frac{T}{2}}-A\sin m\omega t}\\ =-\frac{A}{m\omega }\left[\cos m\omega \left(\frac{T}{2}\right)-\cos 0\right]+\frac{A}{m\omega }\left[\cos m\omega T-\cos m\omega \left(\frac{T}{2}\right)\right]\\ =\left\{\begin{array}{l}\frac{2A}{m\omega }m\text{odd}\\ \text{0}m\text{even}\end{array}\right\}+\left\{\begin{array}{l}\frac{2A}{m\omega }m\text{odd}\\ \text{0}m\text{even}\end{array}\right\}\\ =\left\{\begin{array}{l}\frac{4A}{m\omega }m\text{odd}\\ \text{0}m\text{even}\end{array}\right\}\end{array}$

Conclusion:

Therefore, the left hand side of the given equation is $0$ if $m$ is even and $\frac{4A}{m\omega }$ if $m$ is odd.

(b)

To determine

To show: The terms of the right hand side of the given equation involving $B_{\text{n}}$ are equal to zero.

Answer to Problem 50CP

The terms of the right hand side of the given equation involving $B_{\text{n}}$ are equal to zero.

Explanation of Solution

Given info: The frequency of wave is $f$, the amplitude of wave is $A$. The period of wave is described by $y\left(t\right)=\{\begin{array}{l}A0<t<\frac{T}{2}\\ -A\frac{T}{2}<t<T\end{array}$. The given equation is $y\left(t\right)={\displaystyle \sum _{\text{n}}\left(A_{\text{n}}\sin n\omega t+B_{\text{n}}\cos n\omega t\right)}$.

The part of the right hand of the equation (1) involving $B_n$ is,

$\begin{array}{c}{\displaystyle \sum {\displaystyle \underset{0}{\overset{T}{\int }}{B}_{\text{n}}\left(\cos n\omega T\right)\left(\sin n\omega T\right)dt}}={\displaystyle \sum \frac{1}{2}{B}_n{\displaystyle \underset{0}{\overset{T}{\int }}\left[\sin \left(n\omega t+m\omega t\right)-\sin \left(n\omega t-m\omega t\right)\right]dt}}\\ =0+0\\ =0\end{array}$

Conclusion:

Therefore, the terms of the right hand side of the given equation involving $B_{\text{n}}$ are equal to zero.

(c)

To determine

To show: The terms of the right hand side of the given equation involving $A_{\text{n}}$ are equal to zero.

Answer to Problem 50CP

The terms of the right hand side of the given equation involving $A_{\text{n}}$ are equal to zero.

Explanation of Solution

Given info: The frequency of wave is $f$, the amplitude of wave is $A$. The period of wave is described by $y\left(t\right)=\{\begin{array}{l}A0<t<\frac{T}{2}\\ -A\frac{T}{2}<t<T\end{array}$. The given equation is $y\left(t\right)={\displaystyle \sum _{\text{n}}\left(A_{\text{n}}\sin n\omega t+B_{\text{n}}\cos n\omega t\right)}$.

The part of the right hand of the equation (1) involving $A_n$ is,

$\begin{array}{c}{\displaystyle \sum {\displaystyle \underset{0}{\overset{T}{\int }}{B}_{\text{n}}\left(\cos n\omega T\right)\left(\sin n\omega T\right)dt}}={\displaystyle \sum \frac{1}{2}{A}_{\text{n}}{\displaystyle \underset{0}{\overset{T}{\int }}\left[\cos \left(n\omega t-m\omega t\right)-\cos \left(n\omega t+m\omega t\right)\right]dt}}\\ ={\displaystyle \sum \frac{1}{2}{A}_{\text{n}}{\displaystyle \underset{0}{\overset{T}{\int }}\left[\cos \left(n-m\right)\omega t-\cos \left(n+m\right)\omega t\right]dt}}\\ ={{\displaystyle \sum \frac{1}{2}{A}_{\text{n}}\left[\begin{array}{l}\frac{1}{\left(n-m\right)\omega }\sin \left(n-m\right)\omega t\\ -\frac{1}{\left(n+m\right)\omega }\sin \left(n+m\right)\omega t\end{array}\right]}}_0^T\\ =0\end{array}$

The integration of the $\sin \theta$ over the period of cycle is zero.

Conclusion:

Therefore, the terms of the right hand side of the given equation involving $A_{\text{n}}$ are equal to zero.

(d)

To determine

To show: The entire right hand side of the equation reduces to $\frac{1}{2}{A}_{\text{m}}T$.

Answer to Problem 50CP

The entire right hand side of the equation reduces to $\frac{1}{2}{A}_{\text{m}}T$.

Explanation of Solution

Given info: The frequency of wave is $f$, the amplitude of wave is $A$. The period of wave is described by $y\left(t\right)=\{\begin{array}{l}A0<t<\frac{T}{2}\\ -A\frac{T}{2}<t<T\end{array}$. The given equation is $y\left(t\right)={\displaystyle \sum _{\text{n}}\left(A_{\text{n}}\sin n\omega t+B_{\text{n}}\cos n\omega t\right)}$.

The right hand side of the equation (1) is,

$\begin{array}{c}\left[\begin{array}{l}{\displaystyle \sum {\displaystyle \underset{0}{\overset{T}{\int }}{A}_{\text{n}}\left(\sin n\omega t\right)\left(\sin m\omega t\right)dt}}\\ +{\displaystyle \sum {\displaystyle \underset{0}{\overset{T}{\int }}{B}_{\text{n}}\left(\cos n\omega T\right)\left(\sin n\omega T\right)dt}}\end{array}\right]=\left[\begin{array}{l}{\displaystyle \sum {\displaystyle \underset{0}{\overset{T}{\int }}{A}_{\text{n}}\left(\sin n\omega t\right)\left(\sin m\omega t\right)dt}}\\ +{\displaystyle \sum \frac{1}{2}{\displaystyle \underset{0}{\overset{T}{\int }}{B}_{\text{n}}\left(\sin 2n\omega T\right)dt}}\end{array}\right]\\ ={\displaystyle \sum {\displaystyle \underset{0}{\overset{T}{\int }}{A}_{\text{n}}\left(\sin n\omega t\right)\left(\sin m\omega t\right)dt}}+0\end{array}$

The integer $n=m$.

Substitute $m$ for $n$ in above equation.

$\begin{array}{c}{\displaystyle \sum {\displaystyle \underset{0}{\overset{T}{\int }}{A}_{\text{n}}\left(\sin n\omega t\right)\left(\sin m\omega t\right)dt}}={\displaystyle \sum {\displaystyle \underset{0}{\overset{T}{\int }}{A}_{\text{m}}\left(\sin m\omega t\right)\left(\sin m\omega t\right)dt}}\\ =\frac{1}{2}{A}_{\text{m}}{\displaystyle \underset{0}{\overset{T}{\int }}\left[\cos \left(m-m\right)\omega T\right]dt}\\ =\frac{1}{2}{A}_{\text{m}}{\displaystyle \underset{0}{\overset{T}{\int }}1}dt\\ =\frac{1}{2}{A}_{\text{m}}T\end{array}$

Conclusion:

Therefore, the entire right hand side of the equation reduces to $\frac{1}{2}{A}_{\text{m}}T$.

(e)

To determine

To show: The Fourier series expansion for a square wave is $y\left(t\right)={\displaystyle \sum _{\text{n}}\frac{4A}{n\pi }\sin n\omega t}$.

Answer to Problem 50CP

The Fourier series expansion for a square wave is $y\left(t\right)={\displaystyle \sum _{\text{n}}\frac{4A}{n\pi }\sin n\omega t}$.

Explanation of Solution

Given info: The frequency of wave is $f$, the amplitude of wave is $A$. The period of wave is described by $y\left(t\right)=\{\begin{array}{l}A0<t<\frac{T}{2}\\ -A\frac{T}{2}<t<T\end{array}$. The given equation is $y\left(t\right)={\displaystyle \sum _{\text{n}}\left(A_{\text{n}}\sin n\omega t+B_{\text{n}}\cos n\omega t\right)}$.

The general equation of $y\left(t\right)$ is,

$y\left(t\right)={\displaystyle \sum A_{\text{n}}\sin n\omega t}$ (2)

From the part (a), The value of ${\displaystyle \underset{0}{\overset{T}{\int }}y\left(t\right)\sin m\omega tdt}=\frac{4A}{m\omega }$ when $m$ is odd.

From the part (d), the value of ${\displaystyle \underset{0}{\overset{T}{\int }}y\left(t\right)\sin m\omega tdt}=\frac{1}{2}{A}_{\text{m}}T$, when $n=m$.

Therefore,

$\begin{array}{c}\frac{4A}{n\omega }=\frac{1}{2}{A}_{\text{m}}T\\ A_{\text{m}}=\frac{8A}{n\omega T}\end{array}$

Substitute $\frac{2\pi }{T}$ for $\omega$ in above equation.

$\begin{array}{c}{A}_{\text{m}}=\frac{8A}{n\left(\frac{2\pi }{T}\right)T}\\ =\frac{4A}{n\pi }\end{array}$

Substitute $A_{\text{m}}$ for $A_{\text{n}}$ in equation (2).

$y\left(t\right)={\displaystyle \sum A_{\text{m}}\sin n\omega t}$

Substitute $\frac{4A}{n\pi }$ for $A_m$ in above equation.

$y\left(t\right)={\displaystyle \sum \frac{4A}{n\pi }\sin n\omega t}$

Conclusion:

Therefore, the Fourier series expansion for a square wave is $y\left(t\right)={\displaystyle \sum _{\text{n}}\frac{4A}{n\pi }\sin n\omega t}$.