0
$\begingroup$

Question :

Find inverse Laplace of :

$$\dfrac{1}{(s^2+a^2)^2}$$

My try :

$$\dfrac{1}{(s^2+a^2)^2}=-\frac{1}{2s}\frac{\mathrm d}{\mathrm ds}\left( \frac{1}{s^2+a^2}\right)$$

I need use this identity :

$tf(t)\stackrel{\mathcal{L}}\longleftrightarrow-\frac{\mathrm d}{\mathrm ds}F(s)$

And :

$\int_{0}^{t}g(\tau)d\tau\stackrel{\mathcal{L}}\longleftrightarrow\frac{1}{s}G(s)$

but I don't understand how I applied ?

$\endgroup$
2
$\begingroup$

Hint: Other method, which isn't the last one is using convolution $${\cal L}^{-1}\left(\dfrac{1}{(s^2+a^2)^2}\right)=\dfrac{1}{a^2}\int_0^x\sin a(x-t)\sin at\ dt$$

$\endgroup$
2
$\begingroup$

As

$$ F(s) = \int_0^{\infty}e^{-st}f(t)dt $$

we have

$$ \frac{d}{ds}F(s) = \frac{d}{ds}\int_0^{\infty}e^{-st}f(t)dt = -\int_0^{\infty}e^{-st}t f(t)dt $$

then knowing that $\mathcal{L}^{-1}\left(\frac{1}{s^2+a^2}\right) = \frac 1a\sin(a t)$

$$ -\frac{d}{ds}\left(\frac{1}{s^2+a^2}\right) = \int_0^{\infty}e^{-st} \frac ta\sin(a t)dt $$

then

$$ \frac{2s}{(s^2+a^2)^2}\Leftrightarrow \frac ta\sin(a t) $$

and

$$ \frac{1}{2s}\frac{2s}{(s^2+a^2)^2}\Leftrightarrow \frac 12\int_0^t\frac {\tau}{a}\sin(a \tau)d\tau = \frac{\sin (a t)-a t \cos (a t)}{2 a^3} $$

$\endgroup$
2
$\begingroup$

The second rule says$$\mathcal L^{-1}\left[\frac1sG(s)\right]=\int_0^t g(\tau)\mathrm d\tau$$where $g(\tau)=\mathcal L^{-1}[G](\tau)$. In this example, our $g(t)$ is $$g(t)=\mathcal L^{-1}\left[-\frac d{ds}\left(\frac1{s^2+a^2}\right)\right]=tf(t)$$ where $$f(t)=\mathcal L^{-1}\left[\frac{1}{s^2+a^2}\right]$$So combining these together: $$\mathcal L^{-1}\left[\frac1{(s^2+a^2)^2}\right]=\int_0^t\tau\mathcal L^{-1}\left[\frac1{s^2+a^2}\right](\tau)\mathrm d\tau$$

From here you can integrate this to get the final result $$\frac1{2a^3}(\sin at - at \cos at)$$

$\endgroup$
  • 2
    $\begingroup$ You end abruptly with a rather complicated formula ; it would be good to indicate the result to be reached, and the method (integration by parts) : $\tfrac1{2a^3}(\sin at - at \cos at)$. $\endgroup$ – Jean Marie May 18 at 17:21
  • 1
    $\begingroup$ Perhaps, but I assumed OP would know the result of the inverse Laplace transform in the integral, and would've then known what to do. (if not they could ask). I didn't want to give away the method for the next bit since they hadn't indicated they'd struggled with that, or even tried it yet. The question seemed to me to ask how to combine those two rules for Laplace transforms in this particular case. I guess including the final answer wouldn't have hurt though. $\endgroup$ – John Doe May 18 at 18:50
2
$\begingroup$

Using the fact that $(s^2+a^2)^2=(s-ai)^2(s+ai)^2$, we can write the given expression under the form :

$$\dfrac{1}{4ai}\dfrac{1}{s}\left(\dfrac{1}{(s-ai)^2}-\dfrac{1}{(s+ai)^2}\right)$$

which gives, reading backwards the classical LT table :

$$\dfrac{1}{4ai}\text{Prim}\left(te^{ait}-te^{-ait}\right)=\dfrac{1}{4ai}\text{Prim}\left(t(e^{ait}-e^{-ait})\right)$$

where "Prim" means "take the primitive function" of what follows.

Otherwise said :

$$\dfrac{1}{2a}\dfrac{1}{\require{cancel}\cancel{2i}}\text{Prim}\left(t \cancel{2i} \sin(at)\right)$$

Let us integrate by parts :

$$\dfrac{1}{2a}\left( [-t \tfrac1{a} \cos(at)]- \text{Prim} ((-\tfrac1{a})\cos(at))\right)$$

$$\dfrac{1}{2a^2}\left( -t \cos(at)+\tfrac1{a}\sin(at)\right)$$

$$\dfrac{1}{2a^3}\left(-at \cos(at)+\sin(at)\right)$$

$\endgroup$
1
$\begingroup$

Other method might be useful is complex method, the function $f(s)=\dfrac{e^{st}}{(s^2+a^2)^2}$ has two poles $s=\pm ia$ of order $2$, with residues $$\operatorname{Res}_{s=ia}f(s)= \lim_{s\to ia}\dfrac{d}{ds}\left((s-ia)^2f(s)\right)=e^{ iat}\dfrac{iat-1}{-4ia^3}$$ $$\operatorname{Res}_{s=-ia}f(s)= \lim_{s\to-ia}\dfrac{d}{ds}\left((s+ia)^2f(s)\right)=e^{-iat}\dfrac{iat+1}{-4ia^3}$$ thus the sum of residues is $$\dfrac{at\cos at-\sin at}{-2a^3}$$

$\endgroup$
0
$\begingroup$

$$ \frac{1}{(s^2+a^2)^2}= \frac{1}{(s+ia)^2(s-ia)^2} \\ = -\frac{1}{4a^2}\left[\frac{1}{s+ia}-\frac{1}{s-ia}\right]^2 \\ = -\frac{1}{4a^2}\left[\frac{1}{(s+ia)^2}-2\frac{1}{(s+ia)(s-ia)}+\frac{1}{(s-ia)^2}\right] \\ = -\frac{1}{4a^2}\left[\frac{1}{(s+ia)^2}-\frac{1}{ia}\left(\frac{1}{s-ia}-\frac{1}{s+ia}\right)+\frac{1}{(s-ia)^2}\right] \\ = \frac{1}{4a^2}\frac{d}{ds}\left[\frac{1}{s+ia}+\frac{1}{s-ia}\right]+\frac{1}{4ia^3}\left(\frac{1}{s-ia}-\frac{1}{s+ia}\right) \\ = \frac{1}{4a^2}\frac{d}{ds}\int_0^{\infty}e^{-(s+ia)t}+e^{-(s-ia)t}dt \\ +\frac{1}{4ia^3}\int_0^{\infty}e^{-(s-ia)t}-e^{-(s+ia)t}dt \\ = \frac{1}{4a^2}\int_0^{\infty}-t(e^{-(s+ia)t}+e^{-(s-ia)t})dt \\ +\frac{1}{4ia^3}\int_0^{\infty}e^{-(s-ia)t}-e^{-(s+ia)t}dt \\ = \int_0^{\infty}e^{-st}\left[\frac{1}{4a^2}\left(-te^{-iat}-te^{iat}\right)+\frac{1}{4ia^3}(e^{iat}-e^{-iat})\right]dt \\ = \int_0^{\infty}e^{-st}\left[-\frac{1}{2a^2}t\cos(at)+\frac{1}{2a^3}\sin(at)\right]dt $$ So the inverse Laplace transform is $$ -\frac{1}{2a^2}t\cos(at)+\frac{1}{2a^3}\sin(at). $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.