Question

Consider the periodic function G(t)=e^t , 0≤ t < c, c a parameter.
G(t+c)=G(t).
Sketch the graph of G(t) and obtain its laplace form

Answer 1

Given $G(t)$ is a periodic function with period $c$, so

$L(G(t)) = \frac{\int_0^c G(t)e^{-st}dt}{1-e^{-cs}}$ .

Now, $\int_0^c G(t)e^{-st}dt = \int_0^c e^te^{-st}dt = \int_0^c e^{(-s+1)t}dt = \frac{e^{(-s+1)t}}{-s+1}$.

$\implies L(G(t))= \frac{e^{(1-s)t}}{(1-s)(1-e^{-cs})}$ .

Graph of given function depends on value of c. To show one plot, we assume c = 2.

So, graph of given function is: