Chapter 19, Problem 1P

The average values of a function can be determined by

$\overline{f\left(x\right)}=\frac{{\displaystyle {\int }_0^{x}f\left(x\right)dx}}{x}$

Use this relationship to verify the results of Eq. (19.13).

To determine

To prove: The following results,

$\frac{{\displaystyle \sum \sin \left({\omega }_{0}t\right)}}{N}=0,\frac{{\displaystyle \sum \cos \left({\omega }_{0}t\right)}}{N}=0,\frac{{\displaystyle \sum {\sin }^2\left({\omega }_{0}t\right)}}{N}=\frac{1}{2},\frac{{\displaystyle \sum {\cos }^2\left({\omega }_{0}t\right)}}{N}=\frac{1}{2},\frac{{\displaystyle \sum \cos \left({\omega }_{0}t\right)\sin \left({\omega }_{0}t\right)}}{N}=0$

If the average value formula is $\overline{f\left(x\right)}=\frac{{\displaystyle {\int }_0^{x}f\left(x\right)dx}}{x}$.

Explanation of Solution

Given:

The average value formula is $\overline{f\left(x\right)}=\frac{{\displaystyle {\int }_0^{x}f\left(x\right)dx}}{x}$ and the results are,

$\frac{{\displaystyle \sum \sin \left({\omega }_{0}t\right)}}{N}=0,\frac{{\displaystyle \sum \cos \left({\omega }_{0}t\right)}}{N}=0,\frac{{\displaystyle \sum {\sin }^2\left({\omega }_{0}t\right)}}{N}=\frac{1}{2},\frac{{\displaystyle \sum {\cos }^2\left({\omega }_{0}t\right)}}{N}=\frac{1}{2},\frac{{\displaystyle \sum \cos \left({\omega }_{0}t\right)\sin \left({\omega }_{0}t\right)}}{N}=0$

Formula used:

The integral formula are ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx=f\left(b\right)-f\left(a\right),{\displaystyle \int \sin \left(nx\right)dx=-\frac{\cos \left(nx\right)}{n}},{\displaystyle \int \cos xdx=\sin x}$.

The trigonometric identity are,

${\sin }^{2}x=\frac{1-\cos \left(2x\right)}{2},{\cos }^{2}x=\frac{1+\cos \left(2x\right)}{2},\sin \left(2x\right)=2\sin \left(x\right)\cos \left(x\right)$

Proof:

Consider the identity,

$\frac{{\displaystyle \sum \sin \left({\omega }_{0}t\right)}}{N}$

Use average value formula and take $f\left(x\right)=\sin {\omega }_{0}t$ and $T=N$.

As it is known that the angular frequency ${\omega }_0$ is the fraction of $2\pi$ and time period T as shown below:

Therefore,

$\frac{{\displaystyle \sum \sin \left({\omega }_{0}t\right)}}{N}=\frac{{\displaystyle \underset{0}{\overset{T}{\int }}\sin \left({\omega }_{0}t\right)dt}}{T}$

Use integration formula,

$\begin{array}{c}\frac{{\displaystyle \underset{0}{\overset{T}{\int }}\sin \left({\omega }_{0}t\right)}dt}{T}=-\left(\frac{1}{{\omega }_{0}T}\right){\left[\cos \left({\omega }_{0}t\right)\right]}_0^T\\ =-\left(\frac{1}{\frac{2\pi }{T}\cdot T}\right){\left[\cos \left(\frac{2\pi }{T}t\right)\right]}_0^T\\ =-\left(\frac{1}{2\pi }\right)\left[\cos \left(2\pi \right)-\cos \left(0\right)\right]\\ =-\left(\frac{1}{2\pi }\right)\left[1-1\right]\end{array}$

On simplifying further,

$\frac{{\displaystyle \sum \sin \left({\omega }_{0}t\right)}}{N}=0$

Hence, the result is $\frac{{\displaystyle \sum \sin \left({\omega }_{0}t\right)}}{N}=0$.

Now consider next result,

$\frac{{\displaystyle \sum \cos \left({\omega }_{0}t\right)}}{N}$

Use average value formula and take $f\left(x\right)=cos{\omega }_{0}t$ and $T=N$.

As it is known that the angular frequency ${\omega }_0$ is the fraction of $2\pi$ and time period T as shown below:

Therefore,

$\frac{{\displaystyle \sum \cos \left({\omega }_{0}t\right)}}{N}=\frac{{\displaystyle \underset{0}{\overset{T}{\int }}\cos \left({\omega }_{0}t\right)dt}}{T}$

Use the integral formula,

$\begin{array}{c}\frac{{\displaystyle \underset{0}{\overset{T}{\int }}\cos \left({\omega }_{0}t\right)}dt}{T}=\left(\frac{1}{{\omega }_{0}T}\right){\left[\sin \left({\omega }_{0}t\right)\right]}_0^T\\ =\left(\frac{1}{\frac{2\pi }{T}\cdot T}\right)\left[\sin \left(2\pi \right)-\sin \left(0\right)\right]\\ =\left(\frac{1}{\frac{2\pi }{T}\cdot T}\right)\left[0-0\right]\\ =0\end{array}$

Hence, the result is $\frac{{\displaystyle \sum \cos \left({\omega }_{0}t\right)}}{N}=0$.

Now, consider next result,

$\frac{{\displaystyle \sum {\sin }^2\left({\omega }_{0}t\right)}}{N}$

Use average value formula and take $f\left(x\right)={\sin }^2{\omega }_{0}t$ and $T=N$.

As it is known that the angular frequency ${\omega }_0$ is the fraction of $2\pi$ and time period T as shown below:

Therefore,

$\frac{{\displaystyle \sum {\sin }^2\left({\omega }_{0}t\right)}}{N}=\frac{{\displaystyle \underset{0}{\overset{T}{\int }}{\sin }^2\left({\omega }_{0}t\right)dt}}{T}$

Use the trigonometric identity,

$\frac{{\displaystyle \underset{0}{\overset{T}{\int }}{\sin }^2\left({\omega }_{0}t\right)dt}}{T}=\left(\frac{1}{2T}\right){\displaystyle \underset{0}{\overset{T}{\int }}\left(1-\cos \left(2{\omega }_{0}t\right)\right)dt}$

Use the integral formula,

$\begin{array}{c}\frac{{\displaystyle \underset{0}{\overset{T}{\int }}{\sin }^2\left({\omega }_{0}t\right)dt}}{T}=\left(\frac{1}{2T}\right){\left[t-\frac{\sin \left(2{\omega }_{0}t\right)}{2{\omega }_0}\right]}_0^T\\ =\left(\frac{1}{2T}\right)\left[T-\frac{\sin \left(2\left(\frac{2\pi }{T}\right)T\right)}{2\left(\frac{2\pi }{T}\right)}-0+\sin \left(0\right)\right]\\ =\left(\frac{1}{2T}\right)\left[T-0-0+0\right]\\ =\frac{1}{2}\end{array}$

Hence, the result is $\frac{{\displaystyle \sum {\sin }^2\left({\omega }_{0}t\right)}}{N}=\frac{1}{2}$.

Now, consider next result,

$\frac{{\displaystyle \sum {\cos }^2\left({\omega }_{0}t\right)}}{N}$

Use average value formula and take $f\left(x\right)={\cos }^2{\omega }_{0}t$ and $T=N$.

As it is known that the angular frequency ${\omega }_0$ is the fraction of $2\pi$ and time period T as shown below:

Therefore,

$\frac{{\displaystyle \sum {\cos }^2\left({\omega }_{0}t\right)}}{N}=\frac{{\displaystyle \underset{0}{\overset{T}{\int }}{\cos }^2\left({\omega }_{0}t\right)dt}}{T}$

Use the trigonometric identity,

$\frac{{\displaystyle \underset{0}{\overset{T}{\int }}{\cos }^2\left({\omega }_{0}t\right)dt}}{T}=\left(\frac{1}{2T}\right){\displaystyle \underset{0}{\overset{T}{\int }}\left(1+\cos \left(2{\omega }_{0}t\right)\right)dt}$

Use the integral formula,

$\begin{array}{c}\frac{{\displaystyle \underset{0}{\overset{T}{\int }}{\cos }^2\left({\omega }_{0}t\right)dt}}{T}=\left(\frac{1}{2T}\right){\left[t+\frac{\sin \left(2{\omega }_{0}t\right)}{2{\omega }_0}\right]}_0^T\\ =\left(\frac{1}{2T}\right){\left[t+\frac{\sin \left(2\left(\frac{2\pi }{T}\right)T\right)}{2\left(\frac{2\pi }{T}\right)}-0-\sin \left(0\right)\right]}_0^T\\ =\left(\frac{1}{2T}\right)\left[T+\frac{\sin \left(4\pi \right)}{2\left(\frac{2\pi }{T}\right)}-0-0\right]\\ =\left(\frac{1}{2T}\right){\left[T+0-0-0\right]}_0^T\end{array}$

On further simplifying,

$\frac{{\displaystyle \underset{0}{\overset{T}{\int }}{\cos }^2\left({\omega }_{0}t\right)dt}}{T}=\frac{1}{2}$

Hence, the result is $\frac{{\displaystyle \sum {\cos }^2\left({\omega }_{0}t\right)}}{N}=\frac{1}{2}$.

Now, consider next result,

$\frac{{\displaystyle \sum \cos \left({\omega }_{0}t\right)\sin \left({\omega }_{0}t\right)}}{N}$

Use average value formula and take $f\left(x\right)=\left(\cos {\omega }_{0}t\right)\left(\sin {\omega }_{0}t\right)$ and $T=N$.

As it is known that the angular frequency ${\omega }_0$ is the fraction of $2\pi$ and time period T as shown below:

Therefore,

$\frac{{\displaystyle \sum \cos \left({\omega }_{0}t\right)\sin \left({\omega }_{0}t\right)}}{N}=\frac{{\displaystyle \underset{0}{\overset{T}{\int }}\cos \left({\omega }_{0}t\right)\sin \left({\omega }_{0}t\right)}dt}{T}$

Use the trigonometric identity,

$\begin{array}{c}\frac{{\displaystyle \sum \cos \left({\omega }_{0}t\right)\sin \left({\omega }_{0}t\right)}}{N}=\left(\frac{1}{2T}\right){\displaystyle \underset{0}{\overset{T}{\int }}\sin \left(2{\omega }_{0}t\right)}dt\\ =\left(-\frac{1}{4T{\omega }_0}\right){\left[\cos \left(2{\omega }_{0}t\right)\right]}_0^T\\ =\left(-\frac{1}{4T\left(\frac{2\pi }{T}\right)}\right){\left[\cos \left(2\left(\frac{2\pi }{T}\right)T\right)-\cos \left(0\right)\right]}_0^T\\ =\left(-\frac{1}{4T\left(\frac{2\pi }{T}\right)}\right)\left[\cos \left(4\pi \right)-\cos \left(0\right)\right]\end{array}$

On further simplifying,

$\begin{array}{c}\frac{{\displaystyle \sum \cos \left({\omega }_{0}t\right)\sin \left({\omega }_{0}t\right)}}{N}=\left(-\frac{1}{4T\left(\frac{2\pi }{T}\right)}\right)\left[1-1\right]\\ =0\end{array}$

Hence, the result is $\frac{{\displaystyle \sum \cos \left({\omega }_{0}t\right)\sin \left({\omega }_{0}t\right)}}{N}=0$.