# Laplace transform integral equation

• Oct 4th 2012, 10:45 AM
Skruven
Laplace transform integral equation
Determine i(t)

$\displaystyle {i}'(t) + 4i(t) + 5\int_{0}^{t}i(\tau )d\tau = \delta (t-2)$

$\displaystyle t>0, i(0)=0$ where $\displaystyle \delta (t-2)$ is a dirac delta function

i(t) is the current, with laplace transformation i got $\displaystyle i(t) = \frac{1}{2}e^{((-2-i)(t-2)))}((1+2i)e^{(2i(t-2))}+(1-2i))\theta (t-2)$
where $\displaystyle \theta (t-2)$ is a heaviside function

then they ask of $\displaystyle \lim_{t \to \infty }\int_{0}^{t}i(\tau )d\tau$ i think it will be 0 but how do i see that? do i really need to integrate my i(t)?

regards!
• Oct 4th 2012, 12:17 PM
MaxJasper
Re: Laplace transform integral equation
$\displaystyle \lim_{t \to \infty }\int_{0}^{t}i(\tau )d\tau \to 0$

http://mathhelpforum.com/attachment....1&d=1349385167
• Oct 4th 2012, 06:39 PM
TheEmptySet
Re: Laplace transform integral equation
Quote:

Originally Posted by Skruven
Determine i(t)

$\displaystyle {i}'(t) + 4i(t) + 5\int_{0}^{t}i(\tau )d\tau = \delta (t-2)$

$\displaystyle t>0, i(0)=0$ where $\displaystyle \delta (t-2)$ is a dirac delta function

i(t) is the current, with laplace transformation i got $\displaystyle i(t) = \frac{1}{2}e^{((-2-i)(t-2)))}((1+2i)e^{(2i(t-2))}+(1-2i))\theta (t-2)$
where $\displaystyle \theta (t-2)$ is a heaviside function

then they ask of $\displaystyle \lim_{t \to \infty }\int_{0}^{t}i(\tau )d\tau$ i think it will be 0 but how do i see that? do i really need to integrate my i(t)?

regards!

Note: It is really bad style or notation to use i as a function name and then use i as the imaginary unit.

I don't recall the name of the theorem but If all of the poles of the laplace transform are in the left half plane, then

$\displaystyle \lim_{t \to \infty}f(t) =\lim_{s \to 0^{+}}sF(s)$

So in your case if you define $\displaystyle g(t) =\int_{0}^{t}i(\tau)d\tau$ then

$\displaystyle \lim_{t \to \infty}g(t)=\lim_{s \to 0^+}s\mathcal{L} \{\int_{0}^{t}i(\tau)d\tau\}=\lim_{s \to 0^+}I(s)$

Now if you take the laplace transform you get

If you take the laplace transform you get

$\displaystyle sI+4I+\frac{5}{s}I=e^{2s} \iff (s^2+4s+5)I=se^{2s}$

$\displaystyle I = \frac{s}{(s+2)^2+1}e^{2s}$

So

$\displaystyle \lim_{s \to 0^+}\frac{s}{(s+2)^2+1}e^{2s}=0$