Question

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

Answer 1

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,
\(F(s)=L\left\{f(t)\right\} =L\left\{f_1(t)\right\} \cdot \frac{1}{1-e^{-sp}} = \frac{\int_0^p f(t)e^{-st}dt}{1-e^{-sp}}\)
\(=\frac{16s-1}{2s^2(1-e^{-4s})}-\frac{(12-1)e^{-4s}}{s^2(1-e^{-4s})}\)
\(\Rightarrow F(s)=\frac{16s-1}{2s^2(1-e^{-4s})}-\frac{(12-1)e^{-4s}}{s^2(1-e^{-4s})}\)