Question

# DIFFERENTIAL EQUATIONGiven f(t)=-frac{1}{2t}+8 , 0leq t<4 , f(t+4)=f(t)Find F(s)=Lleft{f(t)right} of the Periodic Function

Laplace transform

DIFFERENTIAL EQUATION
Given $$f(t)=-\frac{1}{2t}+8 , 0\leq t<4 , f(t+4)=f(t)$$
Find $$F(s)=L\left\{f(t)\right\}$$ of the Periodic Function

2020-10-27

Step 1
According to the given information it is required to find
$$F(s)=L\left\{f(t)\right\}$$ Where, L represents Laplace Transformation and given:
$$f(t)=-\frac{1}{2t}+8 , 0\leq t<4 , f(t+4)=f(t)$$
Step 2
Laplace transformation of a periodic function with period $$p>0$$ is given by
$$L\left\{f(t)\right\}=L\left\{f_1(t)\right\} \cdot \frac{1}{1-e^{-sp}}$$
$$L\left\{f(t)\right\}=\frac{\int_0^p f(t)e^{-st}dt}{1-e^{-sp}}$$
here, $$f_1(t)$$ is one period of the function.
Step 3
Calculate the required integral to find the Laplace transformation of given periodic function, as
$$\int_0^pf(t)e^{-st}dt=\int_0^4(-\frac{1}{2}t+8)e^{-st}dt$$
$$=-\frac{1}{2}\int_0^4(t-16)e^{-st}dt$$
$$=-\frac{1}{2}\left[(t-16)\int e^{-st}dt-\int\left(\frac{d}{dt}(t-16)\int e^{-st}dt\right)dt\right]_0^4$$
$$=-\frac{1}{2}\left[\frac{(t-16)e^{-st}}{-s}-\int\left(\frac{e^{-st}}{-s}\right)dt\right]_0^4=-\frac{1}{2}\left[-\frac{t-16}{se^{-st}}-\frac{1}{e^{st}s^2}\right]_0^4$$
$$=-\frac{1}{2}\left[-\frac{4-16}{se^{4s}}+\frac{0-16}{se^{0-s}}-\frac{1}{e^{4s}s^2}+\frac{1}{e^{0-s}s^2}\right]=-\frac{1}{2}\left[-\frac{12}{se^{4s}}-\frac{16}{s}-\frac{1}{e^{4s}s^2}+\frac{1}{s^2}\right]$$
$$=\frac{16s-1}{2s^2}-\frac{(12-1)e^{-4s}}{s^2}$$
Step 4
Thus,
$$F(s)=L\left\{f(t)\right\} =L\left\{f_1(t)\right\} \cdot \frac{1}{1-e^{-sp}} = \frac{\int_0^p f(t)e^{-st}dt}{1-e^{-sp}}$$
$$=\frac{16s-1}{2s^2(1-e^{-4s})}-\frac{(12-1)e^{-4s}}{s^2(1-e^{-4s})}$$
$$\Rightarrow F(s)=\frac{16s-1}{2s^2(1-e^{-4s})}-\frac{(12-1)e^{-4s}}{s^2(1-e^{-4s})}$$