Question

Find the Laplace transform

Piecewise function:

2, 0<=x<3
-1, 3<=x 13 years ago

Answer 1

Change it into the Heaviside unit step function.

Answer 2

I guess i thought all you had to do was just treat it like a regular laplace transform problem but instead like for this one, integral from 0 to 3 of e^-sx * 2 dx + integral 3 to infinity of e^-sx * -1 dx.

Answer 3

\[
\begin{split}
F(x)&=\begin{cases}
f(x)&x<c\\
g(x)&x>3
\end{cases}\\
&=
f(x) +
\begin{cases}
0&x<c\\
g(x)-f(x)&x>c
\end{cases}\\
&=
f(x) +
(g(x)-f(x))\begin{cases}
0&x<c\\
1&x>c
\end{cases}\\
&=
f(x) +
(g(x)-f(x))\begin{cases}
0&x-c<0\\
1&x-c>0
\end{cases}\\
F(x)&=f(x)+[g(x)-f(x)]H(x-c)
\end{split}
\]

Answer 4

Yeah, I dunno my first instinct is to use Heaviside for some reason.

Answer 5

well it talks about that in this section, but the examples he uses are really hard to follow... there isnt any specific examples on how to apply it to a problem like the one above, so there isn't much to follow.

Answer 6

Well in this case \(f(x)=2\) and \(g(x)=-1\)
So it would be \(2+(-1-2)H(x-c) = 2-3H(x-c)\)

Answer 7

\[
\mathcal{L}\{2-3H(x-c)\}=\mathcal{L}\{2\}-3\mathcal{L}\{H(x-c)\}
\]

Answer 8

what is H(x-c)??

Answer 9

\(H(x)\) is the standard Heaviside where the step is at \(x=0\). If you put in \(x-c\) then it shifts the step to be a \(x=c\)

Answer 10

In general you want to remember that \[\large
\mathcal{L}\{H(x-c)\} = \frac{e^{-cs}}{s}
\]

Answer 11

Okay I should have written: \[
2+(-1-2)H(x-3) = 2-3H(x-3)
\]Since the step is at \(x=3\)

Answer 12

now when finding L[2] for the first part, are you just taking the integral of e^-sx * 2 and evaluating from 0 to infinity or are you evaluating from 0 to 3

Answer 13

\[
\mathcal{L}\{2\} = 2\mathcal{L}\{1\}
\]It's worth committing to memory that: \[
\mathcal{L}\{1\}=\frac{1}{s}
\]

Answer 14

okay so L[2] = 2/s,

so then H[x-3] = e^-3x / s ??

Answer 15

Yeah. but not \(x\) it's \(s\)

Answer 16

sorry, yes.

so then I will have (2/s) - (3e^-3s)/s =

(2-3e^-3s)/s ?

Answer 17

Yeah, I'm getting \[\Large
\frac{2-3e^{-3s}}{s}
\]

Answer 18

That is what I got the first time as well., I dont have the answer but tried the same thing on a problem where the answer was given.

0, 0<=x<5
2, 5<=x<infinity

what I got for my answer was 5 - [ (2+e^-5s / s]

but the answer they got was 2e^-5s / s

so I am not sure why the same doesnt work?

Answer 19

Okay first of all change it from piece-wise to Heaviside.

Answer 20

Use \[
\begin{split}
F(x)&=\begin{cases}
f(x)&x<c\\
g(x)&x>3
\end{cases}\\
&=
f(x) +
\begin{cases}
0&x<c\\
g(x)-f(x)&x>c
\end{cases}\\
&=
f(x) +
(g(x)-f(x))\begin{cases}
0&x<c\\
1&x>c
\end{cases}\\
&=
f(x) +
(g(x)-f(x))\begin{cases}
0&x-c<0\\
1&x-c>0
\end{cases}\\
F(x)&=f(x)+[g(x)-f(x)]H(x-c)
\end{split}
\]
In this case \(f(x)=0,\;g(x)=5,\;c=5\)

Answer 21

okay, i get it.. it just happened to work out the first time, but using that I get what the book gets

Answer 22

but just for future problem, because I see them coming, if they have say x^2 for f(x) and say (x-5) for g(x), do you still use that same heaviside

Answer 23

Yes, you still want to use the Heaviside.

Answer 24

\[\Large
\mathcal{L}\{H(x-c)f(x)\} = e^{-cs}\mathcal{L}\{f(x)\}
\]It can only make things easier.

Answer 25

thank you for your help... I appreciate all your time.