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How do you find the inverse Laplace Transform of the following,

$$\frac{2 (s^2+4 s+5)^2+40}{(s^2+4 s+5)^2}$$

Separating them into complex coefficients is to long.

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3 Answers

up vote 2 down vote accepted

Let $F(s)$ denote the fraction in the post, hence $F(s)=2+40\frac1{(s^2+4s+5)^2}$. The $2$ part of $F(s)$ is the Laplace transform of twice the Dirac measure at $0$. The fraction $\frac1{s^2+4s+5}$ is a linear combination of $\frac1{s+2\pm\mathrm i}$ hence it is the Laplace transform of a linear combination of the functions $t\mapsto\exp(-(2\pm\mathrm i)t)$ on $t\geqslant0$, namely, the Laplace transform of $f$ where $f(t)=\mathrm e^{-2t}\sin(t)$ for every $t\geqslant0$. Thus, the fraction $\frac1{(s^2+4s+5)^2}$ is the Laplace transform of $f\ast f$, which might (or might not, check for yourself) be such that $(f\ast f)(t)=\frac12\mathrm e^{-2t}(\sin(t)-t\cos(t))$ for every $t\geqslant0$.

Finally, the function $F$ is the Laplace transform of the measure $\nu=2\delta_0+40\mu$ where $\mu$ has density $f\ast f$ with respect to the Lebesgue measure on $t\geqslant0$. Note that $F$ is the Laplace transform of no function, only a measure can do, and one has $$ F(s)=\int_0^{+\infty}\mathrm e^{-st}\mathrm d\nu(t). $$

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Although I didn't quite get how L(f*f(t)) is equal to that, thanks for the answer :) –  Eddy May 13 '13 at 17:30
    
It is a general fact that L(f*g)=(Lf)(Lg) hence for f=g, one gets... –  Did May 13 '13 at 19:16
    
Are you certain? –  Eddy May 14 '13 at 13:25
    
Certain about what? (For what it is worth, as explained (and acknowledged by the OP) in the comments, the answer you accepted does not answer your question. Can you explain?) –  Did May 14 '13 at 16:56
    
L(f*g)=(Lf)(Lg) that this true. My book states: L(f(t)*g(t)) does not equal to L(f(t))*L(g(t)) Are you saying the accepted answer is wrong? It's like yours. –  Eddy May 14 '13 at 17:16
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The problem reduces to $$2+\frac{40}{(s^2+4s+5)^2}$$

Let us denote the Laplace Transform of $f(t)$ as $L\left(f(t)\right)=F(s)$

and the inverse as $L^{-1}\left(F(s)\right)=f(t)$

For $L^{-1}\frac1{(s^2+4s+5)^2},$

$$\text{If }L\left(f(t)\right)=\frac{s-a}{\{(s-a)^2+b^2\}^2},$$

$$L\left(\frac {f(t)}t\right)=\int_s^\infty \frac{s-a}{\{(s-a)^2+b^2\}^2} ds$$ $$=\frac12\int_s^\infty \frac{2(s-a)}{\{(s-a)^2+b^2\}^2} ds=-\frac12\left(\frac1{(s-a)^2+b^2}\right)_s^\infty=\frac12\frac1{(s-a)^2+b^2}$$

$$\implies \frac {2f(t)}t=L^{-1}\left(\frac1{(s-a)^2+b^2}\right)=\frac{e^{at}\sin bt}{b} $$

$$\implies L^{-1}\frac{s-a}{\{(s-a)^2+b^2\}^2} =f(t)=\frac{te^{at}\sin bt}{2b}$$

$$a=0\implies L^{-1}\frac s {\{s^2+b^2\}^2} =\frac{t\sin bt}{2b}$$

As $L\{g(t)\}=G(s)\implies \int_0^tg(t)dt=L^{-1}\left(\frac{G(s)}s\right)$

$$L^{-1}\frac1{(s^2+b^2)^2}=L^{-1}\left(\frac1s\cdot \frac s{(s^2+b^2)^2}\right)=\int_0^t\frac{t\sin bt}{2b}dt=\frac{\sin bt-bt\cos bt}{2b^3}$$ (Integrating by parts)

Now using shifting property $L\{g(t)\}=G(s)\implies L(e^{at}g(t))=G(s-a)$

So, $$L^{-1}\frac1{\{(s-a)^2+b^2\}^2}=\frac{e^{at}(\sin bt-bt\cos bt)}{2b^3}$$

Here $s^2+4s+5=(s+2)^2+1^2\implies a=-2,b=1$

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Sorry but what does $L^{-1}(2)=\frac 2s$ mean? –  Did May 13 '13 at 19:17
    
    
Quite unrelated, I would say. –  Did May 13 '13 at 19:22
    
@Did, not sure what you mean? How do you handle the 1st part? –  lab bhattacharjee May 13 '13 at 19:24
    
Sorry but this is not/should not be the question: the question is what you mean by $L^{-1}(2)=\frac 2s$. –  Did May 13 '13 at 19:26
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I have done it, you can see if it's useful to you, I.m not sure so if you look out the mistake please fix it..

$\begin{array}{l} F(s) = \frac{{2{{({s^2} + 4s + 5)}^2} + 4}}{{{{({s^2} + 4s + 5)}^2}}} = 2 + \frac{{40}}{{{{{\rm{[}}{{(s + 2)}^2} + 1{\rm{]}}}^2}}} \\ {L^{ - 1}}{\rm{[2] = }}2\delta (t) \\ {\rm{\{ }}\int_0^\infty {2.{e^{ - st}}\delta (t)dt = 2} {\rm{.}}{{\rm{e}}^{ - s0}}{\rm{ = 2\} }} \\ G(s) = \frac{{40}}{{{{{\rm{[}}{{(s + 2)}^2} + 1{\rm{]}}}^2}}}{\rm{ = 40}}{\rm{.}}\frac{1}{{{\rm{[}}{{(s + 2)}^2} + 1{\rm{]}}}}.\frac{1}{{{\rm{[}}{{(s + 2)}^2} + 1{\rm{]}}}} = 40L\left[ {{e^{ - 2t}}.\sin t} \right].{L^{ - 1}}\left[ {{e^{ - 2t}}.\sin t} \right] \\ = > g(t) = {e^{ - 2t}}.\sin t*{e^{ - 2t}}.\sin t = \int_0^t {{e^{ - 2(t - u)}}.\sin (t - u).{e^{ - 2u}}.\sin u.du} \\ = {e^{ - 2t}}\int_0^t {(\sin t.\cos u - \cos t.\sin u)\sin u.du} \\ = > g(t) = {e^{ - 2t}}\left[ {\sin t\underbrace {\int_0^t {\cos u.\sin u.du} }_{{I_1}} - \cos t\underbrace {\int_0^t {{{\sin }^2}udu} }_{{I_2}}} \right] \\ = > f(t) = 2\delta (t) + g(t) \\ \end{array}$

You can evalue the integrals I1 and I2 yourself

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