Question

Find the Inverse Laplace transform of s^2/((s^2+a^2)(s^2+b^2)).

Answer 1

Decompose into partial fractions:
\[\frac{s^2}{(s^2+a^2)(s^2+b^2)}=\frac{Cs+D}{s^2+a^2}+\frac{Es+F}{s^2+b^2}\]
(C,D,E,F are used to avoid confusion with a,b)
\[s^2=(Cs+D)(s^2+b^2)+(Es+F)(s^2+a^2)\\
s^2=(C+E)s^3+(D+F)s^2+(Cb^2+Ea^2)s+(Db^2+Fa^2)\]
yielding the system
\[\begin{cases}C+E=0\\D+F=1\\Cb^2+Ea^2=0\\Db^2+Fa^2=0\end{cases}\]
Need help solving for the constants @cupcake1123 ?

Answer 2

I'll try to solve for the constants, and then will check with you.. Thanks so much!

Answer 3

can we solve this by using the table?

Answer 4

can you help me with the constants please?

Answer 5

From the first equation, you have \(-C=E\). Plug this into the third equation:
\[Cb^2+(-C)a^2=0\\
C(b^2-a^2)=0\]
Assuming \(a\not=b\), you have \(C=0\), and in turn, \(E=0\). (Is this the right assumption? I would think so; otherwise, there'd be no reason to differentiate between the two.)

So you're left with solving the following system:
\[\begin{cases}D+F=1\\Db^2+Fa^2=0\end{cases}\]
Multiply the first equation by \(-a^2\):
\[\begin{cases}-Da^2-Fa^2=-a^2\\Db^2+Fa^2=0\end{cases}\]
Adding the equation together gives you
\[Db^2-Da^2=-a^2\\
D(b^2-a^2)=-a^2\\
D=\frac{a^2}{a^2-b^2}\]
So, you also get \(F=1-\dfrac{a^2}{a^2-b^2}\).

So partial fraction decomp yields
\[\frac{a^2}{a^2-b^2}\cdot\frac{1}{s^2+a^2}+\left(1-\frac{a^2}{a^2-b^2}\right)\cdot \frac{1}{s^2+b^2}\]
Can you take the inverse transform now?

Answer 6

do i solve that using the table?

Answer 7

You could, sure.
\[\mathcal{L}^{-1}\left\{\frac{1}{s^2+a^2}\right\}=\cdots\]
There's some algebraic manipulation you'll have to do first in order to make a proper transformation.

Answer 8

okay.. but is there another approach.. i need to know two ways.. i know the one using the table.

Answer 9

I think there's an actual integral formula for the inverse transform, but I've only ever had to refer to a table of transforms to do these kinds of problems

Answer 10

okay.. i have a question..
can we take individual ones in the multiplication and assign the transform table values.. for instance, if one has laplace inverse of e^at and the one being mutliplied to it has cos(at), can we mutliply the inverses?

Answer 11

Do you mean
\[\mathcal{L}^{-1}\left\{e^{at}\right\}\cdot\mathcal{L}^{-1}\left\{\cos(at)\right\}=\mathcal{L}^{-1}\left\{e^{at}\cos(at)\right\}~~?\]
No, the Laplace transform and its inverse don't work that way with multiplication.

Answer 12

If you meant something else, could you draw it? I'm not sure what you mean.

Answer 13

yes.. that was my question.. so how does it work?

Answer 14

The property you're thinking of only works with addition and scalar multipliers:
\[\mathcal{L}^{-1}\left\{cF(s)+dG(s)\right\}=c\mathcal{L}^{-1}\{F(s)\}+d\mathcal{L}^{-1}\{G(s)\}\]
This means the Laplace transform and its inverse are "linear" operators.

You can derive this property by using the integral definition of the transform.

Answer 15

Just to address your example:
\[\mathcal{L}\left\{e^{at}\cos(bt)\right\}=\frac{s-a}{(s-a)^2+b^2}\]
(I should have removed the "inverse" notation before.)

So basically, if you're given a function multiplied by \(e^{at}\),
\[f(t)=e^{at}g(t)\], then the transform is the transform of the function, shifted by \(a\):
\[\mathcal{L}\{e^{at}g(t)\}=G(s-a)\]
(where \(G(s)\) is the Laplace tranform of \(g(t)\).)