Chapter 17, Problem 11P

Obtain the exponential Fourier series for the signal in Fig. 17.54.

To determine

Find the exponential Fourier series for the signal in Figure 17.54.

Answer to Problem 11P

The exponential Fourier series $y\left(t\right)$ for the signal in Figure 17.54 is,

${\displaystyle \sum _{n=-\infty }^{\infty }\frac{75}{4n^2{\pi }^2}\left(2-2\cos \left(\frac{n\pi }{2}\right)-2j\sin \left(\frac{n\pi }{2}\right)+jn\pi \cos \left(\frac{n\pi }{2}\right)+n\pi \sin \left(\frac{n\pi }{2}\right)\right)e^{j\left(\frac{n\pi }{2}\right)t}}$.

Explanation of Solution

Given data:

Refer to Figure 17.54 in the textbook.

Formula used:

Write the expression to calculate the fundamental angular frequency.

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

Here,

$T$ is the period of the function.

Write the general expression to calculate exponential Fourier series of $y\left(t\right)$.

$y\left(t\right)={\displaystyle \sum _{n=-\infty }^{\infty }{c}_n{e}^{jn{\omega }_{0}t}}$ (2)

Here,

$c_n$ is the exponential Fourier series coefficients,

$n$ is an integer, and

${\omega }_0$ is the angular frequency.

Write the expression to calculate exponential Fourier series coefficients.

$c_n=\frac{1}{T}{\displaystyle \underset{0}{\overset{T}{\int }}y\left(t\right)e^{-jn{\omega }_{0}t}}dt$ (3)

Calculation:

The given waveform is drawn as Figure 1.

Refer to Figure 1. The Fourier series function is defined as,

$f\left(t\right)=\{\begin{array}{l}37.5\left(1+t\right),-1<t<0\\ 37.5,0<t<1\\ 0,1<t<3\end{array}$

The time period of the function in Figure 1 is,

$T=4$

Substitute $4$ for $T$ in equation (1) to find ${\omega }_0$.

$\begin{array}{c}{\omega }_0=\frac{2\pi }{4}\\ =\frac{\pi }{2}\end{array}$

Substitute $4$ for $T$ and $\frac{\pi }{2}$ for ${\omega }_0$ in equation (3) to find $c_n$.

$\begin{array}{c}{c}_n=\frac{1}{4}{\displaystyle \underset{0}{\overset{4}{\int }}y\left(t\right)e^{-jn\left(\frac{\pi }{2}\right)t}}dt\\ =\frac{1}{4}{\displaystyle \underset{0}{\overset{4}{\int }}y\left(t\right)e^{-j\left(\frac{n\pi }{2}\right)t}}dt\\ =\frac{1}{4}\left({\displaystyle \underset{-1}{\overset{0}{\int }}y\left(t\right)e^{-j\left(\frac{n\pi }{2}\right)t}}dt+{\displaystyle \underset{0}{\overset{1}{\int }}y\left(t\right)e^{-j\left(\frac{n\pi }{2}\right)t}}dt+{\displaystyle \underset{1}{\overset{3}{\int }}y\left(t\right)e^{-j\left(\frac{n\pi }{2}\right)t}}dt\right)\\ =\frac{1}{4}\left({\displaystyle \underset{-1}{\overset{0}{\int }}37.5\left(1+t\right)e^{-j\left(\frac{n\pi }{2}\right)t}}dt+{\displaystyle \underset{0}{\overset{1}{\int }}37.5e^{-j\left(\frac{n\pi }{2}\right)t}}dt+{\displaystyle \underset{1}{\overset{3}{\int }}0e^{-j\left(\frac{n\pi }{2}\right)t}}dt\right)\end{array}$

Simplify the above equation to find $c_n$.

$\begin{array}{c}{c}_n=\frac{1}{4}\left({\displaystyle \underset{-1}{\overset{0}{\int }}37.5e^{-j\left(\frac{n\pi }{2}\right)t}}dt+{\displaystyle \underset{-1}{\overset{0}{\int }}37.5t{e}^{-j\left(\frac{n\pi }{2}\right)t}}dt+{\displaystyle \underset{0}{\overset{1}{\int }}37.5e^{-j\left(\frac{n\pi }{2}\right)t}}dt+0\right)\\ =\frac{1}{4}\left(37.5{\left[\frac{e^{-j\left(\frac{n\pi }{2}\right)t}}{-\left(\frac{jn\pi }{2}\right)}\right]}_{-1}^0+37.5{\displaystyle \underset{-1}{\overset{0}{\int }}t{e}^{-j\left(\frac{n\pi }{2}\right)t}}dt+37.5{\left[\frac{e^{-j\left(\frac{n\pi }{2}\right)t}}{-\left(\frac{jn\pi }{2}\right)}\right]}_0^1\right)\\ =\frac{37.5}{4}\left(-\frac{2}{jn\pi }\left[e^{-j\left(\frac{n\pi }{2}\right)\left(0\right)}-e^{-j\left(\frac{n\pi }{2}\right)\left(-1\right)}\right]+{\displaystyle \underset{-1}{\overset{0}{\int }}t{e}^{-j\left(\frac{n\pi }{2}\right)t}}dt-\frac{2}{jn\pi }\left[e^{-j\left(\frac{n\pi }{2}\right)\left(1\right)}-e^{-j\left(\frac{n\pi }{2}\right)\left(0\right)}\right]\right)\\ =\frac{37.5}{4}\left(-\frac{2}{jn\pi }\left[e^0-e^{j\left(\frac{n\pi }{2}\right)}\right]+{\displaystyle \underset{-1}{\overset{0}{\int }}t{e}^{-j\left(\frac{n\pi }{2}\right)t}}dt-\frac{2}{jn\pi }\left[e^{-j\left(\frac{n\pi }{2}\right)}-e^0\right]\right)\end{array}$

Simplify the above equation to find $c_n$.

$\begin{array}{c}{c}_n=\frac{37.5}{4}\left(\begin{array}{l}-\frac{2}{jn\pi }\left[1-\left(\cos \left(\frac{n\pi }{2}\right)+j\sin \left(\frac{n\pi }{2}\right)\right)\right]\\ +{\displaystyle \underset{-1}{\overset{0}{\int }}t{e}^{-j\left(\frac{n\pi }{2}\right)t}}dt\\ -\frac{2}{jn\pi }\left[\left(\cos \left(\frac{n\pi }{2}\right)-j\sin \left(\frac{n\pi }{2}\right)\right)-1\right]\end{array}\right)\left\{\begin{array}{l}\because e^0=1\\ e^{j\left(\frac{n\pi }{2}\right)}=\cos \left(\frac{n\pi }{2}\right)+j\sin \left(\frac{n\pi }{2}\right)\\ e^{-j\left(\frac{n\pi }{2}\right)}=\cos \left(\frac{n\pi }{2}\right)-j\sin \left(\frac{n\pi }{2}\right)\end{array}\right\}\\ \\ =\frac{37.5}{4}\left(\begin{array}{l}-\frac{2}{jn\pi }\left[1-\cos \left(\frac{n\pi }{2}\right)-j\sin \left(\frac{n\pi }{2}\right)\right]\\ +{\displaystyle \underset{-1}{\overset{0}{\int }}t{e}^{-j\left(\frac{n\pi }{2}\right)t}}dt\\ -\frac{2}{jn\pi }\left[\cos \left(\frac{n\pi }{2}\right)-j\sin \left(\frac{n\pi }{2}\right)-1\right]\end{array}\right)\\ \\ =\frac{37.5}{4}\left(\begin{array}{l}-\frac{2}{jn\pi }+\frac{2}{jn\pi }\cos \left(\frac{n\pi }{2}\right)+\frac{2}{n\pi }\sin \left(\frac{n\pi }{2}\right)\\ +{\displaystyle \underset{-1}{\overset{0}{\int }}t{e}^{-j\left(\frac{n\pi }{2}\right)t}}dt-\frac{2}{jn\pi }\cos \left(\frac{n\pi }{2}\right)\\ +\frac{2}{n\pi }\sin \left(\frac{n\pi }{2}\right)+\frac{2}{jn\pi }\end{array}\right)\end{array}$

$c_n=\frac{37.5}{4}\left(\frac{4}{n\pi }\sin \left(\frac{n\pi }{2}\right)+{\displaystyle \underset{-1}{\overset{0}{\int }}t{e}^{-j\left(\frac{n\pi }{2}\right)t}}dt\right)$ (4)

Assume the following to reduce the equation (4).

$x={\displaystyle \underset{-1}{\overset{0}{\int }}t{e}^{-j\left(\frac{n\pi }{2}\right)t}}dt$ (5)

Substitute the equations (5) in equation (4) to find $c_n$.

$c_n=\frac{37.5}{4}\left(\frac{4}{n\pi }\sin \left(\frac{n\pi }{2}\right)+x\right)$ (6)

Consider the following integration formula.

${\displaystyle \underset{a}{\overset{b}{\int }}udv}={\left[uv\right]}_a^b-{\displaystyle \underset{a}{\overset{b}{\int }}vdu}$ (7)

Compare the equations (5) and (7) to simplify the equation (5).

$\begin{array}{l}u=tdv=e^{-j\left(\frac{n\pi }{2}\right)t}dt\\ du=dtv=\frac{e^{-j\left(\frac{n\pi }{2}\right)t}}{-\left(\frac{jn\pi }{2}\right)}\\ v=-\frac{2}{jn\pi }{e}^{-j\left(\frac{n\pi }{2}\right)t}\end{array}$

Using the equation (7), the equation (5) can be reduced as,

$\begin{array}{c}x={\displaystyle \underset{-1}{\overset{0}{\int }}t{e}^{-j\left(\frac{n\pi }{2}\right)t}}dt\\ ={\left[t\left(-\frac{2}{jn\pi }{e}^{-j\left(\frac{n\pi }{2}\right)t}\right)\right]}_{-1}^0-{\displaystyle \underset{-1}{\overset{0}{\int }}\left(-\frac{2}{jn\pi }{e}^{-j\left(\frac{n\pi }{2}\right)t}\right)dt}\\ =-\frac{2}{jn\pi }\left[\left(0\right)e^{-j\left(\frac{n\pi }{2}\right)\left(0\right)}-\left(-1\right)e^{-j\left(\frac{n\pi }{2}\right)\left(-1\right)}\right]+\frac{2}{jn\pi }{\displaystyle \underset{-1}{\overset{0}{\int }}{e}^{-j\left(\frac{n\pi }{2}\right)t}dt}\\ =-\frac{2}{jn\pi }\left[0+e^{j\left(\frac{n\pi }{2}\right)}\right]+\frac{2}{jn\pi }{\left[\frac{e^{-j\left(\frac{n\pi }{2}\right)t}}{-\left(\frac{jn\pi }{2}\right)}\right]}_{-1}^0\end{array}$

Simplify the above equation to find $x$.

$\begin{array}{c}x=-\frac{2}{jn\pi }{e}^{j\left(\frac{n\pi }{2}\right)}-{\left(\frac{2}{jn\pi }\right)}^2\left[e^{-j\left(\frac{n\pi }{2}\right)\left(0\right)}-e^{-j\left(\frac{n\pi }{2}\right)\left(-1\right)}\right]\\ =-\frac{2}{jn\pi }{e}^{j\left(\frac{n\pi }{2}\right)}-\frac{4}{j^2{n}^2{\pi }^2}\left[e^0-e^{j\left(\frac{n\pi }{2}\right)}\right]\\ =\left(\begin{array}{c}-\frac{2}{jn\pi }\left(\cos \left(\frac{n\pi }{2}\right)+j\sin \left(\frac{n\pi }{2}\right)\right)\\ -\frac{4}{\left(-1\right)n^2{\pi }^2}\left[1-\left(\cos \left(\frac{n\pi }{2}\right)+j\sin \left(\frac{n\pi }{2}\right)\right)\right]\end{array}\right)\left\{\begin{array}{l}\because e^0=1,\\ j^2=-1,\\ e^{j\left(\frac{n\pi }{2}\right)}=\cos \left(\frac{n\pi }{2}\right)+j\sin \left(\frac{n\pi }{2}\right)\end{array}\right\}\\ =\left(\begin{array}{c}-\frac{2}{jn\pi }\cos \left(\frac{n\pi }{2}\right)-\frac{2}{n\pi }\sin \left(\frac{n\pi }{2}\right)\\ +\frac{4}{n^2{\pi }^2}\left[1-\cos \left(\frac{n\pi }{2}\right)-j\sin \left(\frac{n\pi }{2}\right)\right]\end{array}\right)\end{array}$

Simplify the above equation to find $x$.

$x=\left(-\frac{2}{jn\pi }\cos \left(\frac{n\pi }{2}\right)-\frac{2}{n\pi }\sin \left(\frac{n\pi }{2}\right)+\frac{4}{n^2{\pi }^2}-\frac{4}{n^2{\pi }^2}\cos \left(\frac{n\pi }{2}\right)-\frac{4}{n^2{\pi }^2}j\sin \left(\frac{n\pi }{2}\right)\right)$

Substitute $\left(-\frac{2}{jn\pi }\cos \left(\frac{n\pi }{2}\right)-\frac{2}{n\pi }\sin \left(\frac{n\pi }{2}\right)+\frac{4}{n^2{\pi }^2}-\frac{4}{n^2{\pi }^2}\cos \left(\frac{n\pi }{2}\right)-\frac{4}{n^2{\pi }^2}j\sin \left(\frac{n\pi }{2}\right)\right)$ for $x$ in equation (6) to find $c_n$.

$\begin{array}{c}{c}_n=\frac{37.5}{4}\left(\begin{array}{l}\frac{4}{n\pi }\sin \left(\frac{n\pi }{2}\right)-\frac{2}{jn\pi }\cos \left(\frac{n\pi }{2}\right)-\frac{2}{n\pi }\sin \left(\frac{n\pi }{2}\right)\\ +\frac{4}{n^2{\pi }^2}-\frac{4}{n^2{\pi }^2}\cos \left(\frac{n\pi }{2}\right)-\frac{4}{n^2{\pi }^2}j\sin \left(\frac{n\pi }{2}\right)\end{array}\right)\\ \\ =\frac{37.5}{4}\left(\begin{array}{l}\frac{2}{n\pi }\sin \left(\frac{n\pi }{2}\right)-\frac{2}{jn\pi }\cos \left(\frac{n\pi }{2}\right)+\frac{4}{n^2{\pi }^2}\\ -\frac{4}{n^2{\pi }^2}\cos \left(\frac{n\pi }{2}\right)-\frac{4}{n^2{\pi }^2}j\sin \left(\frac{n\pi }{2}\right)\end{array}\right)\\ \\ =\frac{37.5}{4}\left(\frac{2}{n^2{\pi }^2}\right)\left(\begin{array}{l}n\pi \sin \left(\frac{n\pi }{2}\right)-\left(-j\right)n\pi \cos \left(\frac{n\pi }{2}\right)+2\\ -2\cos \left(\frac{n\pi }{2}\right)-2j\sin \left(\frac{n\pi }{2}\right)\end{array}\right)\\ \\ =\frac{75}{4n^2{\pi }^2}\left(2-2\cos \left(\frac{n\pi }{2}\right)-2j\sin \left(\frac{n\pi }{2}\right)+jn\pi \cos \left(\frac{n\pi }{2}\right)+n\pi \sin \left(\frac{n\pi }{2}\right)\right)\end{array}$

Substitute $\frac{75}{4n^2{\pi }^2}\left(2-2\cos \left(\frac{n\pi }{2}\right)-2j\sin \left(\frac{n\pi }{2}\right)+jn\pi \cos \left(\frac{n\pi }{2}\right)+n\pi \sin \left(\frac{n\pi }{2}\right)\right)$ for $c_n$ and $\frac{\pi }{2}$ for ${\omega }_0$ in equation (2) to find $y\left(t\right)$.

$\begin{array}{c}y\left(t\right)={\displaystyle \sum _{n=-\infty }^{\infty }\frac{75}{4n^2{\pi }^2}\left(2-2\cos \left(\frac{n\pi }{2}\right)-2j\sin \left(\frac{n\pi }{2}\right)+jn\pi \cos \left(\frac{n\pi }{2}\right)+n\pi \sin \left(\frac{n\pi }{2}\right)\right)e^{jn\left(\frac{\pi }{2}\right)t}}\\ ={\displaystyle \sum _{n=-\infty }^{\infty }\frac{75}{4n^2{\pi }^2}\left(2-2\cos \left(\frac{n\pi }{2}\right)-2j\sin \left(\frac{n\pi }{2}\right)+jn\pi \cos \left(\frac{n\pi }{2}\right)+n\pi \sin \left(\frac{n\pi }{2}\right)\right)e^{j\left(\frac{n\pi }{2}\right)t}}\end{array}$

Conclusion:

Thus, the exponential Fourier series $y\left(t\right)$ for the signal in Figure 17.54 is ${\displaystyle \sum _{n=-\infty }^{\infty }\frac{75}{4n^2{\pi }^2}\left(2-2\cos \left(\frac{n\pi }{2}\right)-2j\sin \left(\frac{n\pi }{2}\right)+jn\pi \cos \left(\frac{n\pi }{2}\right)+n\pi \sin \left(\frac{n\pi }{2}\right)\right)e^{j\left(\frac{n\pi }{2}\right)t}}$.