Inverse laplace transform

**Greengoblin** · April 13th 2010, 11:31 AM

Hi, I'm trying to do an inverse laplace transform of the following:

$H(s)=\frac{1}{(1+as)^2}$

I know from a laplace transform table:

$L^{-1}\left\{ \frac{1}{s-a} \right\} = e^{at}$

I'm not sure if this helps me, but I can split the transfer function into partial fractions:

$H(s)=\frac{1}{(1+as)^2}=\frac{A}{(1+as)}+\frac{B}{ (1+as)^2}$

but I'm not sure how I can actually get the time domain function $h(t)=L^{-1}\{H(s))\}$ from these things

Thanks

**chisigma** · April 13th 2010, 11:57 AM

One of the properties of the Laplace Transform is ...

$\mathcal {L}^{-1} \{H(s-\alpha)\} = e^{\alpha t}\cdot \mathcal {L}^{-1} \{H(s)\}$ (1)

... so that is...

$\mathcal {L}^{-1} \{\frac{1}{(1+ a s)^{2}}\} = \frac{1}{a^{2}}\cdot e^{-\frac{t}{a}}\cdot \mathcal {L}^{-1} \{\frac{1}{s^{2}}\} = \frac{t\cdot e^{-\frac{t}{a}}}{a^{2}}\cdot \mathcal {U} (t)$ (2)

Kind regards

$\chi$ $\sigma$

**Greengoblin** · April 13th 2010, 12:39 PM

Thanks alot, what other laplace identities have you used there? For instance how did you get the 1/a^2 and 1/s^2? Also is u(t) simply the step function (heaviside function)?

edit: by the way, in my first post, a is a real constant not a complex frequency

**drumist** · April 13th 2010, 12:56 PM

chisigma is using some algebra:

$\frac{1}{(1+as)^2} = \frac{1}{a^2\left(\frac{1}{a}+s \right)^2} = \frac{1}{a^2} \cdot \frac{1}{\left(\frac{1}{a}+s \right)^2}$

Thus we can simply pull out the constant term $\frac{1}{a^2}$ in front. Then we notice that

$\mathcal{L}^{-1} \left\{ \frac{1}{\left(\frac{1}{a}+s \right)^2} \right\} = e^{-t/a} \cdot \mathcal{L}^{-1} \left\{ \frac{1}{s^2} \right\}$

from the property that he gave.

**Greengoblin** · April 13th 2010, 01:24 PM

Thanks, I can see where the 1/a^2 comes from now, but still don't understand how:

$\frac{1}{\left(\frac{1}{a}+s\right)^2}$

can be written in the form H(s-a), so that the rule can be applied, because:

$H(s)=\frac{1}{(1+as)^2}$

not:

$H(s)=\frac{1}{s^2}$

**drumist** · April 13th 2010, 02:08 PM

First, don't mix the $a$ and $\alpha$ (alpha). To make it more clear, I'll change the name of the variables. So, the general rule is like this:

$\mathcal {L}^{-1} \{H(s-k)\} = e^{k t}\cdot \mathcal {L}^{-1} \{H(s)\}$

(where k is a constant with respect to s)

By the way, an equivalent way to write this rule is like this:

$\mathcal {L}^{-1} \{H(s)\} = e^{k t}\cdot \mathcal {L}^{-1} \{H(s+k)\}$

The basic importance is that the argument of the right side is the argument of the left side plus the constant k.

So, we have that $H(s)=\frac{1}{a^2} \cdot \frac{1}{\left(\frac{1}{a}+s \right)^2}$

Let $k=-\tfrac{1}{a}$ . Then:

$H(s+k)=H\left(s-\tfrac{1}{a}\right)=\frac{1}{a^2} \cdot \frac{1}{\left(\frac{1}{a}+s-\frac{1}{a} \right)^2}= \frac{1}{a^2} \cdot \frac{1}{s^2}$

So we get:

$\mathcal {L}^{-1} \{H(s)\} = e^{k t}\cdot \mathcal {L}^{-1} \{H(s+k)\} = e^{-t/a}\cdot \mathcal {L}^{-1} \left\{H\left(s-\tfrac{1}{a}\right)\right\}$

$= e^{-t/a}\cdot \mathcal {L}^{-1} \left\{ \frac{1}{a^2} \cdot \frac{1}{s^2} \right\} = \frac{e^{-t/a}}{a^2} \cdot \mathcal {L}^{-1} \left\{ \frac{1}{s^2} \right\}$

**Greengoblin** · April 13th 2010, 03:44 PM

Thanks alot, thats really helpful. How can I calculate the laplace transform of 1/s^2? my reference table gives three different solutions, 1, U(t), and H(t), and I'm not sure which one to use. In fact I'm not sure what H(t) and U(t) are even supposed to represent, but I think U(t) might be the step function. Is it different depending on context?

**drumist** · April 13th 2010, 06:30 PM

A Laplace transform table should give you:

$\mathcal{L}^{-1}\left\{ \frac{1}{s^2} \right\} = t \cdot u(t)$

where $u(t)$ is the unit step function. The unit step function is also sometimes called the Heaviside step function and notated as $H(t)$ . So, both of those are actually the same thing. You should just stick with whatever notation your teacher is using.

However, 1 is definitely not the same thing. I'm not sure where you are getting that.

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