Help with inverse Laplace transform.

[yungman]

I want to perform the inverse of

\frac s { [(s+α)^2-β^2](s^2+ω^2)}

I know the conventional way is

\frac s { [(s+α)^2-β^2](s^2+ω^2)}= \frac{As+B}{[(s+α)^2-β^2]}+\frac{Ds+E}{(s^2+ω^2)}

s= (As+B)(s^2+ω^2)+(Ds+E)[(s+α)^2-β^2]

\Rightarrow\; A+D=0,\; B+E+2\alpha A=0,\;\alpha^2A-\beta^2A+2\alpha B +D\omega^2=1,\; B\alpha^2-B\beta^2+E\omega^2=0

You then expand and find A, B, D and E.

But this is very complicated and long. on top, α and β is a function of equations of constants. So the algebra really blow up. Is there any easy way to solve this?

Is there any site that you can enter the equation in s domain and it will transform back to the original domain.

 

[LCKurtz]

I want to perform the inverse of

\frac s { [(s+α)^2-β^2](s^2+ω^2)}

I know the conventional way is

\frac s { [(s+α)^2-β^2](s^2+ω^2)}= \frac{As+B}{[(s+α)^2-β^2]}+\frac{Ds+E}{(s^2+ω^2)}

It would be quite a bit simpler if you factor:$$
(s+\alpha)^2-\beta^2 = ((s+\alpha)+\beta)((s+\alpha)-\beta)$$giving two linear factors.

 

[Ray Vickson]

I want to perform the inverse of

\frac s { [(s+α)^2-β^2](s^2+ω^2)}

I know the conventional way is

\frac s { [(s+α)^2-β^2](s^2+ω^2)}= \frac{As+B}{[(s+α)^2-β^2]}+\frac{Ds+E}{(s^2+ω^2)}

s= (As+B)(s^2+ω^2)+(Ds+E)[(s+α)^2-β^2]

\Rightarrow\; A+D=0,\; B+E+2\alpha A=0,\;\alpha^2A-\beta^2A+2\alpha B +D\omega^2=1,\; B\alpha^2-B\beta^2+E\omega^2=0

You then expand and find A, B, D and E.

But this is very complicated and long. on top, α and β is a function of equations of constants. So the algebra really blow up. Is there any easy way to solve this?

Is there any site that you can enter the equation in s domain and it will transform back to the original domain.

Welcome to the real world, in which sometimes things are just long and complicated and there is nothing you can do about it. You can, however, define a number of auxiliary quantities, and express the answer in terms of them. For example, you could solve for the A,B,C and D, but leave your final answer in terms of them (without substituting in their final values in terms of α, β and ω).

Alternatively, you can try to find your f(s) in a table of transforms, and read off the answer from there, or you can use a symbolic computer package.

RGV

 

[vela]

I'd use the residue theorem to evaluate the Bromwich integral.

 

[yungman]

It would be quite a bit simpler if you factor:$$
(s+\alpha)^2-\beta^2 = ((s+\alpha)+\beta)((s+\alpha)-\beta)$$giving two linear factors.

Thanks for the reply

Can you check whether I did this correct, I solve for A and B. D and E is easier.

\frac s { [(s+α)^2-β^2](s^2+ω^2)} \;=\; \frac s {(s^2+ω^2)[(s+\alpha)+\beta][(s+\alpha)-\beta]}\;=\; \frac {As+B}{(s^2+\omega^2)}\;+\;\frac D {(s+\alpha)+\beta}\;+\;\frac E {(s+\alpha)-\beta}

\Rightarrow\;\frac s {[(s+\alpha)+\beta][(s+\alpha)-\beta]}\;=\; (As+B)\;+\;\frac {D(s^2+ω^2)} {(s+\alpha)+\beta}\;+\;\frac {E(s^2+ω^2)} {(s+\alpha)-\beta}

We let s^2=-\omega^2\;\Rightarrow\;s=j\omega.

\Rightarrow\;\frac {j\omega} {[(j\omega+\alpha)+\beta][(j\omega+\alpha)-\beta]}=j\omega A + B

\frac {j\omega} {[(j\omega+\alpha)+\beta][(j\omega+\alpha)-\beta]}\;=\;\frac {j\omega}{(\alpha^2-\beta^2-\omega^2)+j2\alpha\omega}\left[\frac {(\alpha^2-\beta^2-\omega^2)-j2\alpha\omega} {(\alpha^2-\beta^2-\omega^2)-j2\alpha\omega}\right]\;= \frac {j\omega (\alpha^2-\beta^2-\omega^2)+2\alpha \omega^2}{(\alpha^2-\beta^2-\omega^2)^2+4\alpha^2\omega^2}\;=\;j\omega A + B

\Rightarrow A\;=\;\frac {(\alpha^2-\beta^2-\omega^2)}{(\alpha^2-\beta^2-\omega^2)^2+4\alpha^2\omega^2}\;\hbox {and }\;B\;= \frac {2\alpha \omega^2}{(\alpha^2-\beta^2-\omega^2)^2+4\alpha^2\omega^2}

 

[yungman]

Anyone can help verify I did it right?

 

[Ray Vickson]

Anyone can help verify I did it right?

You did it incorrectly. You need to solve for A, B, D and E all at the same time. Your expressions for A and B are not correct.

RGV

 

[yungman]

You did it incorrectly. You need to solve for A, B, D and E all at the same time. Your expressions for A and B are not correct.

RGV

But I got A and B represented by all constants independent of D and E! What did I do wrong? That's the reason I posted the steps in detail, it looked to simple to me. But I cannot see any problem.

 

[Ray Vickson]

But I got A and B represented by all constants independent of D and E! What did I do wrong? That's the reason I posted the steps in detail, it looked to simple to me. But I cannot see any problem.

You have
\frac{As+B}{s^2+\omega^2}+\frac{D}{s+\alpha+\beta}+\frac{E}{s+\alpha-\beta}<br /> = \frac{N}{(s^2+\omega^2)(s+\alpha+\beta)(s+\alpha-\beta)},
where
N = (As+B)(s+\alpha+\beta)(s+\alpha-\beta)+ D(s^2+\omega^2)(s+\alpha-\beta)<br /> + E(s^2+\omega^2)(s+\alpha+\beta).
The numerator N must be = s for all values of s, so that gives 4 equations in the 4 parameters A, B, D and E (that is, the N-coefficients of s³, s² and s⁰ must vanish, while the coefficient of s¹ must = 1).

However, that is not the way I would do it. I would write your Laplace transform f(s) in the form
f(s) = \frac{A(s+\alpha) + B}{(s + \alpha)^2 -\beta^2} + \frac{Cs + D}{s^2 + \omega^2} \equiv f_1(s+\alpha) + f_2(s),
then note that there is an easy way to get the inverse of f₁(s+α) from that of f₁(s). Both f₁(s) and f₂(s) have standard, textbook inverses. Getting A, B, C and D would be done in the manner indicated above.

RGV

 

[yungman]

Thanks for the reply, why can't I multiply both side just with s^2+\omega^2\; and let s^2=-\omega^2\; as shown in post #5 to cancel out both D and E and just concentrate on A and B?

This is pretty standard method of partial fraction also to multiply with part of the fraction and then set the value of s to make said fraction to zero. anything about the s^2+\omega^2\; nature that I cannot do this?

I don't quite understand the last part of your comment:

f(s) = \frac{A(s+\alpha) + B}{(s + \alpha)^2 -\beta^2} + \frac{Cs + D}{s^2 + \omega^2} \equiv f_1(s+\alpha) + f_2(s),

I understand the last part of your comment. I actually gone this route and find it very tedious as in my post #1, that's the reason I come here to look for an easier way.

 

[Ray Vickson]

You asked before what is wrong with getting A and B independently of D and E. Ehat is wrong is that they are not independent, and you got a wrong answer. I have said all I want on this topic, so I am quitting now.

RGV

 

[yungman]

Anyone can help?

 

[vela]

Your results for A and B are fine. They match what Mathematica comes up with.

 

[yungman]

Your results for A and B are fine. They match what Mathematica comes up with.

Thanks, I was thinking about this, I kind of think the answer is correct except the sign, It can be either +v or -ve. So I have to look into this a little more.

Thanks for your help.