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I can't find the inverse Laplace of : $$\frac{p_1p_2}{(x+p_1)(x+p_2)}$$

I remove first $p_1,p_2$ and try to $$\mathscr{L}^{-1}\frac{1}{(x+p_1)(x+p_2)}$$ $$\frac{1}{(x+p_1)(x+p_2)}= \frac{A}{(x+p_1)} + \frac{B}{(x+p_2)}$$ $$1 = A(x+p_2) + B(x+p_1)$$

With $x = 0$

$$1 = Ap_2 + Bp_1$$

Then with $x = 1$ and $x = -1$, I get: $$1 = A + B \\ A = 1-B \\ 1 = (1-B)p_2 + Bp_1\\ 1 = p_2 -Bp_2 + Bp_1 \\ B = \frac{1-p_2}{p_1-p_2} \\ A = 1- \frac{1-p_2}{p_1-p_2} = \frac{p_1-1}{p_1-p_2}$$

Which would mean that $$\frac{p_1-1}{p_1-p_2} * \frac{p_1p_2}{x+p_1} + \frac{1-p_2}{p_1-p_2}\frac{p_1p_2}{x+p_2}$$ So I see that we should have $e^{-p_1x}$ and $e^{-p_2x}$ at some point but I don't know what to do with the factors $\frac{p_1-1}{p_1-p_2}$ and $\frac{1-p_2}{p_1-p_2}$. What is wrong and why?

Edit: I am adding a picture of the reply.

Reply from the lecture

Dovendyr
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  • Since $x$ is your variable the terms in $p_1$ and $p_2$ are just multipliers. The inverse will have terms of the form $e^{-p_1 t}$ with variable $t$ not $x$ – Paul Jan 29 '21 at 16:28
  • Sorry I don't understand how do you come to that result? – Dovendyr Jan 29 '21 at 16:46
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    Your A and B are wrong actually. Choose $x=-p_1$ and $x = -p_2$ to find A and B. Then $e^{-p_1 t}$ is the inverse transform of $\frac{1}{x+p_1}$. – Paul Jan 29 '21 at 17:02

1 Answers1

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$$\frac{p_1 p_2}{(p_1+x) (p_2+x)}=\frac{A}{p_1+x}+\frac{B}{p_2+x}=\frac{Ap_2+Ax+Bp_1+Bx}{(p_1+x) (p_2+x)}$$

Therefore it must be $$\frac{(Ap_2+Bp_1)+(A+B)x}{(p_1+x) (p_2+x)}=\frac{(p_1 p_2)+0\cdot x}{(p_1+x) (p_2+x)}$$ this means $$ \begin{cases} Ap_2+Bp_1=p_1p_2\\ A+B=0\\ \end{cases} $$ $$ \begin{cases} Ap_2-Ap_1=p_1p_2\to A(p_2-p_1)=p_1p_2\\ B=-A\\ \end{cases} $$ $$ \begin{cases} A=\frac{p_1p_2}{p_2-p_1}\\ B=-\frac{p_1p_2}{p_2-p_1}\\ \end{cases} $$ therefore $$\frac{p_1 p_2}{(p_1+x) (p_2+x)}=\frac{p_1p_2}{p_2-p_1}\left(\frac{1}{p_1+x}-\frac{1}{p_2+x}\right)$$ Inverse Laplace transform is $$\frac{p_1p_2}{p_2-p_1} \left(e^{-p_1 t}-e^{-p_2 t}\right)$$

Raffaele
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  • Hi again, I see now that $A + B =0$ but why $2−1 = 12$ on line 4 of your explanation? – Dovendyr Jan 30 '21 at 07:14
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    @Dovendyr I added a few lines. Hope it is clearer – Raffaele Jan 30 '21 at 12:04
  • Dear @Raffaele, thank you for your precisions! I get it now, very clear! – Dovendyr Feb 01 '21 at 07:57
  • The reply seems to be = 1 + p2/(p1-p2) * e^(-p1x) - p1/(p1-p2) * e^(-p2x)

    Why is that?

     – Dovendyr Feb 01 '21 at 08:06
  • @Dovendyr You can keep out the constant $\frac{p_1p_2}{p_2-p_1}$. Then look on the tables. Inverse of $\frac{1}{p_1+x}$ is $e^{-p_1t}$ and inverse of $\frac{1}{p_2+x}$ is $e^{-p_2t}$. – Raffaele Feb 01 '21 at 08:26
  • @Dovendyr the answer in your book says that? – Raffaele Feb 01 '21 at 08:26
  • Yes, I will edit my question with a picture of the answer. Where is the 1 coming from? – Dovendyr Feb 01 '21 at 08:31
  • @Dovendyr I don't know. It is wrong. Taking Laplace transform of the answer I get $$\frac{\text{p1} \text{p2}}{s (\text{p1}+s) (\text{p2}+s)}$$ which is not what we started from. They made a mistake. You know, I use Mathematica to compute this stuff and I am quite sure that my result is correct. – Raffaele Feb 01 '21 at 08:33
  • No, indeed. I added a picture now! – Dovendyr Feb 01 '21 at 08:34
  • Hmm the denominators are different, did they do some algebraic transformation there that I do not see? – Dovendyr Feb 01 '21 at 08:58
  • Dear @Raffaele, you were right! And the error is on me, the exercise was not about a Laplace transform (even if it was a big part of it) but about a step response, which consist in multipliying 2 functions, one of them being 1/s. So what you found 3 comments above is absolutely correct. Thank you again for helping me. – Dovendyr Feb 04 '21 at 07:57