I was wondering if I could get some help with this question:
Consider the function:
$$f(t) = \begin{cases} te^{2t} \quad \,0 \leq t < 3\\
0 \quad \, 3 \leq t
\end{cases}$$(a) Find the laplace transform of $f$ by directly using the integral definition of a laplace transform.
(b) Find $f$ in terms of step functions and use the t-shifting theorem to find the laplace transform of $f$
.
For part (a):
By integral definition,
$\mathcal{L}$$f(t)$ = $\int_0^\infty te^{2t} . e^{-st}$ $dt$
= $\int_0^3 te^{(2-s)t} $$dt$
I let $u$=$t$ and $v'$= $e^{(2-s)t}$ and thus $du=dt$ and $v$= $\frac {1}{2-s}$$e^{(2-s)t}$.
Using integration by parts,
F(s)= $\bigg[\frac{t}{(2-s)}e^{(2-s)t}-\frac{1}{2-s}\int e^{(2-s)t} dt\bigg]_{t=0}^{t=3}$
=$\bigg[\frac{t}{(2-s)}e^{(2-s)t}-\frac{1}{(2-s)^2}e^{(2-s)t} \bigg]_{t=0}^{t=3}$
= $\bigg[\frac{3}{(2-s)}e^{(6-3s)}-\frac{1}{(2-s)^2}e^{(6-3s)} -[0-\frac{1}{(2-s)^2}]\bigg]_{t=0}^{t=3}$
I managed to get the answer $\frac {3e^{6-3s}}{2-s}$–$\frac {e^{6-3s}}{(2-s)^{2}}$+$\frac {1}{(2-s)^2}$ by using integration by parts.
For part (b):
I manage to write $f$ in terms of the unit step function and got $f(t) = te^{2t} u(t) – te^{2t} u(t-3)$.
I know that
$\mathcal{L}$$f(t)$= $\mathcal{L}$ [$te^{2t} u(t)$] – $\mathcal{L}$
[$te^{2t} u(t-3)$]
and
$\mathcal{L}$[$te^{2t} u(t)$] = $\frac {1}{(s-2)^2}$
I'm having trouble finding the laplace transform of $te^{2t} u(t-3)$ using the unit step function. Here is what I have tried so far:
$\mathcal{L}$[$te^{2t} u(t-3)$]
= $\mathcal{L}$ $[(t-3)e^{2(t-3)} u(t-3)]$
= $\mathcal{L}$ $[(t-3 +3)e^{2t-6+6} u(t-3)]$
Up to this step, I do not know how I should continue from here… Any advice?
Thanks in advance!
Best Answer
We want to find the Laplace Transform of
$$\tag 1 te^{2t} u(t-3)$$
We will make use of two properties from LTs
$$\mathcal{L}(u(t − a)f(t − a)) = e^{−as}F(s)~~ \text{and}~~ \mathcal{L}(t f(t)) = -\dfrac{d}{ds}(F(s))$$
To make use of the properties, we need to get our expression to the same forms , we can write $(1)$ as
$$\mathcal{L}(te^{2t} u(t-3)) = -\dfrac{d}{ds}\mathcal{L}\left(e^{(2(t-3))} e^6 u(t-3)\right) = -e^6\dfrac{d}{ds}\mathcal{L}\left(e^{(2(t-3))} u(t-3)\right)$$
Now we find $$-e^6\dfrac{d}{ds}\left(\dfrac{e^{-3s}}{s-2}\right) = -e^6\left(\frac{e^{-3 s} (5-3 s)}{(s-2)^2}\right) = \frac{e^{6-3 s} (3 s-5)}{(s-2)^2}$$
As an alternate approach, solve it using the definition of the LT
$$\mathcal{L}\left\{ {f\left( t \right)} \right\} = \int_{{\,0}}^{{\,\infty }}{{{{\bf{e}}^{ - s\,t}}f\left( t \right)\,dt}}$$