Question

what is the bilateral laplace transform of the function
\[\Huge x(t)=e^t* \frac{ d }{ dt }[e^{-2t}u(-t)]\]

Answer 1

What's a bilateral laplace transform?

Answer 2

yeah i am also unsure of it. is it just simple laplace transform?

Answer 3

I've never heard of a "bilateral" one. I was hoping you had a definition or something. We can try googling it.

Answer 4

It just means the limits are \(-\infty\to \infty\).

Answer 5

wiki says
http://en.wikipedia.org/wiki/Two-sided_Laplace_transform
so just the limits are -inf to inf

Answer 6

yeah @wio

Answer 7

is \(* \equiv \cdot\)?

Answer 8

yeah .

Answer 9

\[
\int\limits_{-\infty}^{\infty} e^{(1-s)t}\frac{d[e^{-2t}u(-t)]}{dt}
\]

Answer 10

Any idea what \(u\) is?

Answer 11

Is it heaviside?

Answer 12

yeah u(t) means unit step function or heaviside function

Answer 13

It splits into two integrals.

Answer 14

like
\[\int\limits_{-\infty}^{0^-}+\int\limits_{0^-}^{\infty}\]

Answer 15

Basically.

Answer 16

yeah then first one will be zero is'nt it?

Answer 17

And fortunately, one of them should go to \(0\), right?

Answer 18

yeah ok now its just simple laplace

Answer 19

This heaviside function is flipped though, so I think the negative side stays.

Answer 20

how? can we really write that

Answer 21

\[
u(t) = \begin{cases}
0&t<0\\
1&t\geq 0\\
\end{cases}
\]Now \[
1-u(t) =1- \begin{cases}
0&t<0\\
1&t\geq 0\\
\end{cases} =
\begin{cases}
1&t<0\\
0&t\geq 0\\
\end{cases}
= u(-t)
\]I suppose with the exception being \(0\)?

Answer 22

Though in an integral, you can take out any point and it still works...

Answer 23

okay understood :)
i think now i can do it

Answer 24

but one thing
when we are writing
u(-t)=1-u(t)
,does the region of convergence(ROC) will remain unchanged ?

Answer 25

By the way, is \[
\frac{d}{dt}u(t)f(t) = u(t)\frac{d}{dt}f(t)
\]Since \(u(t)\) is two constants?

Answer 26

Eh... now that I think about it... it is a bit risky to use that since they aren't equal at \(0\).

Answer 27

yeah
and for the derivative part i think i will use
\[\large \frac{ d }{ dt }x(t)=s.x(s)\]

Answer 28

Interestingly enough: \[
\lim_{t\to 0^+}u(-t) = 0 = \lim_{t\to 0^+}1-u(t)
\]And \[
\lim_{t\to 0^-}u(-t) = 1 =\lim_{t\to 0^-}1-u(t)
\]

Answer 29

The only issue really is that \(u(-0) \neq 1-u(0)\)

Answer 30

Originally, \(u(0)\) was arbitrary or undefined anyway.

Answer 31

yeah i think u(0) will result a impulse

Answer 32

It's almost sort of a meaningless value.

Answer 33

hey i got something
http://www.southalabama.edu/engineering/ece/faculty/akhan/Courses/EE321-spring-06%29/Lecture%20notes/laplace-transform-chapter%207.pdf
check page 3

Answer 34

\[\begin{split}
\int\limits_{-\infty}^{\infty} e^{(1-s)t}\frac{d[e^{-2t}u(-t)]}{dt}
&=\int\limits_{-\infty}^{0} e^{(1-s)t}\frac{d[e^{-2t}(1)]}{dt}+\int\limits_{0}^{\infty} e^{(1-s)t}\frac{d[e^{-2t}(0)]}{dt}\\
&=\int\limits_{-\infty}^{0} e^{(1-s)t}\frac{d[e^{-2t}]}{dt}\\
&=\int\limits_{-\infty}^{0} e^{(1-s)t}(-2e^{-2t})dt\\
\end{split}
\]\[

\]

Answer 35

\[
=2\int\limits_0^{\infty}e^{(1-s-2)t}dt
\]

Answer 36

I think you can do the rest... unless I made a mistake?

Answer 37

yeah thank u ^_^
i will do it