Verifying Laplace Analysis: Comparing Equations (1) and (2)

[Saladsamurai]

Homework Statement

I just want to verify that I am dong this correctly and if not, where my misconceptions are. I keep getting a different answer so I am going to go through this step by step:

After a some steps in a controls problem, I end up with a function in the Laplace domain:

$$\Theta(s) = \frac{12}{\frac{6}{5}s^2 +96}\qquad(1)$$

Homework Equations

From table, I know that

$$\mathcal L[Be^{-at}\sin(\omega t)]=\frac{B\omega}{(s+a)^2 + \omega^2}\qquad(2)$$

The Attempt at a Solution

So the idea is to make my equation (1) take the form of the known equation (2):

First, I factored the denominator so that s² has a coefficient of 1:

$$\frac{12}{\frac{6}{5}s^2 +96}=\frac{1}{\frac{6}{5}}\cdot \frac{12}{s^2+80}$$


Comparing the denominators of each we have:

$$(s+a)^2 + \omega^2 = s^2 +80$$

$$\Rightarrow s^2 + 2as + a^2 + \omega^2 =s^2 +80$$

$$\therefore a = 0 \qquad\qquad \omega^2 = 80$$

So we now have to compare the numerators:

$$12 = B*\omega$$

$$\Rightarrow B = \frac{12}{\omega}$$

$$\Rightarrow B=\frac{12}{\sqrt{80}}$$


Thus, we can write:

$$\Theta(s) =\frac{5}{6}\cdot\frac{(\frac{12}{\sqrt{80}})*\sqrt{80}}{(s+0)^2+ \sqrt{80}^2}$$


So, in the time domain, we have:

$$\mathcal L^{-1} [\Theta(s)]=\theta(t) = \frac{10}{\sqrt{80}}\cdot\sin(\sqrt{80}t)$$

or

$$\frac{\sqrt{5}}{2}\sin(4\sqrt5 t)$$

I have gotten this result a couple of times, so I think I am correct, but my friend got something a little different.

Are there any blatant errors I am making?

Thanks!

 

[Dick]

Well, I used Maxima to find the Laplace transform of sqrt(5)*sin(4*sqrt(5)*t)/2. I got 10/(s^2+80). As that is the same function you started out with, I don't think you are doing anything wrong.

 

[Saladsamurai]

Okie dokie! Thanks!