Question

Find the inverse Laplace transform of (s^2-4s+7)/(s-1)^4.

Answer 1

Have you considered the Partial Fraction Decomposition?

Answer 2

@Idealist ^^^ Well? Have you?

Answer 3

@Idealist edit your question, please

Answer 4

That's my question.

Answer 5

\[\frac{s^2-4s+7}{(s-1)^4}=\frac{A}{s-1}+\frac{B}{(s-1)^2}+\frac{C}{(s-1)^3}+\frac{D}{(s-1)^4}\]

Answer 6

And how do you find A, B, C, D?

Answer 7

Wait a minute.

Answer 8

\[\begin{align*}s^2-4s+7&=A(s-1)^3+B(s-1)^2+C(s-1)+D\\
&=A(s^3-3s^2+3s-1)+B(s^2-2s+1)+C(s-1)+D\\
&=As^3-3As^2+3As-A+Bs^2-2Bs+B+Cs-C+D\\
&=As^3+(B-3A)s^2+(3A-2B+C)s-A+B-C+D
\end{align*}\]
Matching up coefficients, you have the system
\[\begin{cases}A=0\\B-3A=1\\3A-2B+C=-4\\-A+B-C+D=7\end{cases}\]

Answer 9

Wait a minute, please don't leave.

Answer 10

hahaha, you always ask the helper " don't leave". NoPe!! Whenever you are done, he is back, is it NOT ok?

Answer 11

Okay, so I got A=0, B=1, C=-2, D=4. Now what?

Answer 12

Now you have to find the inverse transform of each term:
\[\begin{align*}\mathcal{L}^{-1}\left\{\frac{s^2-4s+7}{(s-1)^4}\right\}&=\mathcal{L}^{-1}\left\{\frac{1}{(s-1)^2}-\frac{2}{(s-1)^3}+\frac{4}{(s-1)^4}\right\}\\
&=\mathcal{L}^{-1}\left\{\frac{1}{(s-1)^2}\right\}-\mathcal{L}^{-1}\left\{\frac{2}{(s-1)^3}\right\}+\mathcal{L}^{-1}\left\{\frac{4}{(s-1)^4}\right\}
\end{align*}\]
Do you have a table of transforms? It's help a lot here.

The particularly useful formulas I'm referring to are
\[\mathcal{L}^{-1}\left\{F(s-a)\right\}=e^{at}f(t)\\
\mathcal{L}^{-1}\left\{\frac{n!}{s^{n+1}}\right\}=t^n\]

Answer 13

@SithsAndGiggles that means we always use table to solve it, right? or we have to memorize all formula

Answer 14

Let me check the table.

Answer 15

@Loser66, there's an integral formula for the inverse transform, but there's really no need for it if you have a table. Besides, have you seen the formula? lol

http://en.wikipedia.org/wiki/Inverse_laplace_transform#Mellin.27s_inverse_formula

Answer 16

@SithsAndGiggles another question, why don't we use
\[\mathcal{L}^{-1}\left\{\frac{n!}{(s-a)^{n+1}}\right\}=t^n *e^{at}\]It's right for all of them, right?

Answer 17

I hope you mean multiplication, and not convolution. But yeah, that's the same thing I said; your formula combines both.

Answer 18

it's not mine, it's from the table , formula #11

Answer 19

I don't quite get the last one, 4/(s-1)^4, the answer is (2/3)t^3*e^t, how do I get there with the formula t^n*e^at?

Answer 20

\[\frac{4}{(s-1)^4}=\frac{3!}{3!}\cdot\frac{4}{(s-1)^{3+1}}=\frac{4}{6}\cdot\frac{3!}{(s-1)^{3+1}}\]

Answer 21

Thank you!

Answer 22

good , behalf on SithsAndGiggles, ehehehehe

Answer 23

You're welcome!