# Thread: series RL analysis using Laplace

1. ## series RL analysis using Laplace

Hi Guys,
Could anyone offer some assistance please?

I have a question which states the following:
L[v(t)] = V(s) = 1/(1+s^s)(1-e^(pi*s)
if an inductor (1 H) and a resistor (1 Ohm) are connected in series then show that the resulting current is,

i(t)=sigma(n=0,infinity).f(t-n*pi)
where f(t) =(sin(t)-cos(t)+e^(-t))u(t)

i am able to analysis the circuit to give the following:
L[di/dt] +L[Ri]=L[v(t)]
giving I(s)=1/(s+1)(s^2+1)(1-e^(pi*s))

using partial fractions it becomes:
I(s)=(1/(1-e^(-pi*s)))*0.5*((-s/(s^2+1))+(1/(s^2+1))+(1/(s+1))

I am really not sure where to go next, I am sure that 1/(1-e^(-s*pi)) is some kind of geometric series but other than that I am stumped!<br><br>I would really appreciate it if someone could point me in the right direction.
Regards,

Mckiseldeck

2. ## Re: series RL analysis using Laplace

Hey Mckiseldeck.

Have you considered using repeated applications of the convolution theorem?

3. ## Re: series RL analysis using Laplace

Originally Posted by chiro
Hey Mckiseldeck.

Have you considered using repeated applications of the convolution theorem?

I know L-1{F(s)G(s)}=(f*g)(t) but I don not know what you mean in this instance.

Regards,

Mckiseldeck

4. ## Re: series RL analysis using Laplace

Basically I mean this (note that Laplace transforms satisfy having this theorem too):

Convolution theorem - Wikipedia, the free encyclopedia