Chapter A7, Problem 1P

To determine

The Laplace transform of the given periodic function.

Answer to Problem 1P

The Laplace transform of the given periodic function is $\frac{8}{s^2+\frac{{\pi }^2}{4}}\left(\frac{s+\frac{\pi }{2}{e}^{-s}+\frac{\pi }{2}{e}^{-3s}-s{e}^{-4s}}{1-e^{-4s}}\right)$.

Explanation of Solution

Given data:

The given periodic waveform is shown in Figure 1.

The time period of this waveform is $T=4\text{s}$.

Calculation:

The function expression calculated from the figure is written as,

$f_1\left(t\right)=\{\begin{array}{l}8\cos \left(\omega t\right)0\le t\le 1\text{s}\\ \text{0}1\le t\le 3\text{s}\\ 8\cos \left(\omega \left(t-4\right)\right)3\le t\le 4\text{s}\end{array}$        (1)

Here,

$\omega$ is the angular frequency.

The angular frequency is given by,

$\omega =\frac{2\pi }{T}$

Substitute $4$ for $T$ in the above equation.

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

Substitute $\frac{\pi }{2}$ for $\omega$ in equation (1).

$f_1\left(t\right)=\{\begin{array}{l}8\cos \left(\left(\frac{\pi }{2}\right)t\right)0\le t\le 1\text{s}\\ \text{0}1\le t\le 3\text{s}\\ 8\cos \left(\frac{\pi }{2}\left(t-4\right)\right)3\le t\le 4\text{s}\end{array}$        (2)

The Laplace transform of a periodic signal is written as,

$L\left\{f\left(t\right)\right\}=\frac{1}{1-e^{-Ts}}L\left\{f_1\left(t\right)\right\}$        (3)

The Laplace transform of the function $f_1\left(t\right)$ can be expressed as,

$L\left\{f_1\left(t\right)\right\}={\displaystyle \underset{0}{\overset{4}{\int }}{f}_1\left(t\right)e^{-st}dt}$

Substitute $\{\begin{array}{l}8\cos \left(\left(\frac{\pi }{2}\right)t\right)0\le t\le 1\text{s}\\ \text{0}1\le t\le 3\text{s}\\ 8\cos \left(\frac{\pi }{2}\left(t-4\right)\right)3\le t\le 4\text{s}\end{array}$ for $f_1\left(t\right)$ in the above equation.

$L\left\{f_1\left(t\right)\right\}={\displaystyle \underset{0}{\overset{1}{\int }}8\cos \left(\frac{\pi }{2}t\right)e^{-st}dt+{\displaystyle \underset{1}{\overset{3}{\int }}0e^{-st}dt+}}{\displaystyle \underset{3}{\overset{4}{\int }}8\cos \left(\frac{\pi }{2}\left(t-4\right)\right)e^{-st}dt}$        (4)

The general form of $\cos \theta$ is expressed as,

$\cos \theta =\frac{e^{j\theta }+e^{-j\theta }}{2}$

Substitute $\frac{\pi }{2}t$ for $\theta$ in the above relation.

$\cos \left(\frac{\pi }{2}t\right)=\frac{e^{j\frac{\pi }{2}t}+e^{-j\frac{\pi }{2}t}}{2}$

Substitute $\frac{e^{j\frac{\pi }{2}t}+e^{-j\frac{\pi }{2}t}}{2}$ for $\cos \left(\frac{\pi }{2}t\right)$ in equation (4).

$\begin{array}{c}L\left\{f_1\left(t\right)\right\}={\displaystyle \underset{0}{\overset{1}{\int }}8\frac{e^{j\frac{\pi }{2}t}+e^{-j\frac{\pi }{2}t}}{2}{e}^{-st}dt}+{\displaystyle \underset{3}{\overset{4}{\int }}8\frac{e^{j\frac{\pi }{2}t}+e^{-j\frac{\pi }{2}t}}{2}{e}^{-st}dt}\\ =4{\displaystyle \underset{0}{\overset{1}{\int }}\left(e^{j\frac{\pi }{2}t}+e^{-j\frac{\pi }{2}t}\right)e^{-st}dt}+4{\displaystyle \underset{3}{\overset{4}{\int }}\left(e^{j\frac{\pi }{2}\left(t-4\right)}+e^{-j\frac{\pi }{2}\left(t-4\right)}\right)e^{-st}dt}\end{array}$

The above equation is divided into two parts as,

$L\left\{f_1\left(t\right)\right\}=I_1+I_2$        (5)

The first part $I_1$ is expressed as,

$I_1=4{\displaystyle \underset{0}{\overset{1}{\int }}\left(e^{j\frac{\pi }{2}t}+e^{-j\frac{\pi }{2}t}\right)e^{-st}dt}$

The second part $I_2$ is expressed as,

$I_2=4{\displaystyle \underset{3}{\overset{4}{\int }}\left(e^{j\frac{\pi }{2}\left(t-4\right)}+e^{-j\frac{\pi }{2}\left(t-4\right)}\right)e^{-st}dt}$        (6)

Solve for $I_1$.

$\begin{array}{c}{I}_1=4{\displaystyle \underset{0}{\overset{1}{\int }}\left(e^{j\frac{\pi }{2}t}+e^{-j\frac{\pi }{2}t}\right)e^{-st}dt}\\ =4{\displaystyle \underset{0}{\overset{1}{\int }}{e}^{j\frac{\pi }{2}t}{e}^{-st}dt}+4{\displaystyle \underset{0}{\overset{1}{\int }}{e}^{-j\frac{\pi }{2}t}{e}^{-st}dt}\\ =4{\displaystyle \underset{0}{\overset{1}{\int }}{e}^{-\left(s-j\frac{\pi }{2}\right)t}dt}+4{\displaystyle \underset{0}{\overset{1}{\int }}{e}^{-\left(s+j\frac{\pi }{2}\right)t}dt}\\ =4{\left[\frac{e^{-\left(s-j\frac{\pi }{2}\right)t}}{-\left(s-j\frac{\pi }{2}\right)}\right]}_0^1+4{\left[\frac{e^{-\left(s+j\frac{\pi }{2}\right)t}}{-\left(s+j\frac{\pi }{2}\right)}\right]}_0^1\end{array}$

Solve further as,

$\begin{array}{c}{I}_1=4\left[\frac{e^{-\left(s-j\frac{\pi }{2}\right)}-1}{-\left(s-j\frac{\pi }{2}\right)}\right]+4\left[\frac{e^{-\left(s+j\frac{\pi }{2}\right)}-1}{-\left(s+j\frac{\pi }{2}\right)}\right]\\ =-4\left[\frac{\left(e^{-\left(s-j\frac{\pi }{2}\right)}-1\right)\left(s+j\frac{\pi }{2}\right)+\left(s-j\frac{\pi }{2}\right)\left(e^{-\left(s+j\frac{\pi }{2}\right)}-1\right)}{\left(s+j\frac{\pi }{2}\right)\left(s-j\frac{\pi }{2}\right)}\right]\\ =-4\left[\frac{\left(e^{-s}{e}^{j\frac{\pi }{2}}-1\right)\left(s+j\frac{\pi }{2}\right)+\left(s-j\frac{\pi }{2}\right)\left(e^{-s}{e}^{-j\frac{\pi }{2}}-1\right)}{\left(s+j\frac{\pi }{2}\right)\left(s-j\frac{\pi }{2}\right)}\right]\\ =-4\left[\frac{\left(j{e}^{-s}-1\right)\left(s+j\frac{\pi }{2}\right)+\left(s-j\frac{\pi }{2}\right)\left(-j{e}^{-s}-1\right)}{\left(s^2+{\left(\frac{\pi }{2}\right)}^2\right)}\right]\end{array}$

Solve further as,

$\begin{array}{c}{I}_1=-4\left[\frac{\text{s}\left(j{e}^{-s}-1-j{e}^{-s}-1\right)+j\frac{\pi }{2}\left(j{e}^{-s}-1-j{e}^{-s}+1\right)}{\left(s^2+{\left(\frac{\pi }{2}\right)}^2\right)}\right]\\ =-4\left[\frac{-\text{2}s+j\frac{\pi }{2}\left(2j{e}^{-s}\right)}{\left(s^2+{\left(\frac{\pi }{2}\right)}^2\right)}\right]\\ =\frac{8\left(s+\frac{\pi }{2}{e}^{-s}\right)}{\left(s^2+{\left(\frac{\pi }{2}\right)}^2\right)}\end{array}$

Solve equation (6) for $I_2$ as,

$\begin{array}{c}{I}_2=4{\displaystyle \underset{3}{\overset{4}{\int }}\left(e^{j\frac{\pi }{2}\left(t-4\right)}+e^{-j\frac{\pi }{2}\left(t-4\right)}\right)e^{-st}dt}\\ =4{\displaystyle \underset{3}{\overset{4}{\int }}{e}^{j\frac{\pi }{2}\left(t-4\right)}{e}^{-st}dt}+4{\displaystyle \underset{3}{\overset{4}{\int }}{e}^{-j\frac{\pi }{2}\left(t-4\right)}{e}^{-st}dt}\\ =4{\displaystyle \underset{3}{\overset{4}{\int }}{e}^{-j2\pi }{e}^{-\left(s-j\frac{\pi }{2}\right)t}dt}+4{\displaystyle \underset{3}{\overset{4}{\int }}{e}^{j2\pi }{e}^{-\left(s+j\frac{\pi }{2}\right)t}dt}\\ =4{\displaystyle \underset{3}{\overset{4}{\int }}{e}^{-\left(s-j\frac{\pi }{2}\right)t}dt}+4{\displaystyle \underset{3}{\overset{4}{\int }}{e}^{-\left(s+j\frac{\pi }{2}\right)t}dt}\end{array}$

Solve further as,

$\begin{array}{c}{I}_2=4{\left[\frac{e^{-\left(s-j\frac{\pi }{2}\right)t}}{-\left(s-j\frac{\pi }{2}\right)}\right]}_3^4+4{\left[\frac{e^{-\left(s+j\frac{\pi }{2}\right)t}}{-\left(s+j\frac{\pi }{2}\right)}\right]}_3^4\\ =4\left[\frac{e^{-\left(s-j\frac{\pi }{2}\right)4}-e^{-\left(s-j\frac{\pi }{2}\right)3}}{-\left(s-j\frac{\pi }{2}\right)}\right]+4\left[\frac{e^{-\left(s+j\frac{\pi }{2}\right)4}-e^{-\left(s+j\frac{\pi }{2}\right)3}}{-\left(s+j\frac{\pi }{2}\right)}\right]\\ =-4\left[\frac{\left(e^{-4s}{e}^{j2\pi }-e^{-3s}{e}^{j\frac{3\pi }{2}}\right)\left(s+j\frac{\pi }{2}\right)+\left(s-j\frac{\pi }{2}\right)\left(e^{-4s}{e}^{-j2\pi }-e^{-3s}{e}^{-j\frac{3\pi }{2}}\right)}{\left(s+j\frac{\pi }{2}\right)\left(s-j\frac{\pi }{2}\right)}\right]\\ =-4\left[\frac{\left(e^{-4s}+j{e}^{-3s}\right)\left(s+j\frac{\pi }{2}\right)+\left(s-j\frac{\pi }{2}\right)\left(e^{-4s}-j{e}^{-3s}\right)}{\left(s^2+{\left(\frac{\pi }{2}\right)}^2\right)}\right]\end{array}$

Solve further as,

$\begin{array}{c}{I}_2=-4\left[\frac{s\left(e^{-4s}+j{e}^{-3s}+e^{-4s}-j{e}^{-3s}\right)+j\frac{\pi }{2}\left(e^{-4s}+j{e}^{-3s}+e^{-4s}-j{e}^{-3s}\right)}{\left(s^2+{\left(\frac{\pi }{2}\right)}^2\right)}\right]\\ =-4\left[\frac{2s{e}^{-4s}-\frac{\pi }{2}2e^{-3s}}{\left(s^2+{\left(\frac{\pi }{2}\right)}^2\right)}\right]\\ =\frac{-8\left(s{e}^{-4s}-\frac{\pi }{2}{e}^{-3s}\right)}{s^2+{\left(\frac{\pi }{2}\right)}^2}\end{array}$

Substitute $\frac{8\left(s+\frac{\pi }{2}{e}^{-s}\right)}{\left(s^2+{\left(\frac{\pi }{2}\right)}^2\right)}$ for $I_1$ and $\frac{-8\left(s{e}^{-4s}-\frac{\pi }{2}{e}^{-3s}\right)}{s^2+{\left(\frac{\pi }{2}\right)}^2}$ for $I_2$ in equation (5).

$L\left\{f_1\left(t\right)\right\}=\frac{8\left(s+\frac{\pi }{2}{e}^{-s}\right)}{\left(s^2+{\left(\frac{\pi }{2}\right)}^2\right)}+\frac{-8\left(s{e}^{-4s}-\frac{\pi }{2}{e}^{-3s}\right)}{s^2+{\left(\frac{\pi }{2}\right)}^2}$

The Laplace transform of $f\left(t\right)$ is $F\left(s\right)$ and the Laplace transform of $f_1\left(t\right)$ is $F_1\left(s\right)$.

Substitute $F_1\left(s\right)$ for $L\left\{f_1\left(t\right)\right\}$ in the above equation.

$\begin{array}{c}{F}_1\left(s\right)=\frac{8\left(s+\frac{\pi }{2}{e}^{-s}-s{e}^{-4s}+\frac{\pi }{2}{e}^{-3s}\right)}{s^2+{\left(\frac{\pi }{2}\right)}^2}\\ =\frac{8}{s^2+\frac{{\pi }^2}{4}}\left(s+\frac{\pi }{2}{e}^{-s}+\frac{\pi }{2}{e}^{-3s}-s{e}^{-4s}\right)\end{array}$

Substitute $F\left(s\right)$ for $L\left\{f\left(t\right)\right\}$ and $F_1\left(s\right)$ for $L\left\{f_1\left(t\right)\right\}$ in equation in equation (3).

$F\left(s\right)=\frac{1}{1-e^{-Ts}}{F}_1\left(s\right)$

Substitute $\frac{8}{s^2+\frac{{\pi }^2}{4}}\left(s+\frac{\pi }{2}{e}^{-s}+\frac{\pi }{2}{e}^{-3s}-s{e}^{-4s}\right)$ for $F_1\left(s\right)$ in the above equation.

$\begin{array}{c}F\left(s\right)=\frac{1}{1-e^{-Ts}}\left(\frac{8}{s^2+\frac{{\pi }^2}{4}}\left(s+\frac{\pi }{2}{e}^{-s}+\frac{\pi }{2}{e}^{-3s}-s{e}^{-4s}\right)\right)\\ =\frac{8}{s^2+\frac{{\pi }^2}{4}}\left(\frac{s+\frac{\pi }{2}{e}^{-s}+\frac{\pi }{2}{e}^{-3s}-s{e}^{-4s}}{1-e^{-Ts}}\right)\end{array}$

Substitute $4$ for $T$ in the above equation.

$F\left(s\right)=\frac{8}{s^2+\frac{{\pi }^2}{4}}\left(\frac{s+\frac{\pi }{2}{e}^{-s}+\frac{\pi }{2}{e}^{-3s}-s{e}^{-4s}}{1-e^{-4s}}\right)$

Conclusion:

Therefore, the Laplace transform of the given periodic function is. $\frac{8}{s^2+\frac{{\pi }^2}{4}}\left(\frac{s+\frac{\pi }{2}{e}^{-s}+\frac{\pi }{2}{e}^{-3s}-s{e}^{-4s}}{1-e^{-4s}}\right)$.