# Step Response of an LTI system

1. Apr 15, 2010

### qwertydump

Hi lads, need help with this question from a signals and systems course.

LTI system is defined as follows

$$H(s) = \frac{1}{ s^2 - 2rs cos(theta) + r^2}$$

where r = 20pi, theta = 1.47.

Find an expression for the step response of the system

I think what I should do is multiply H(s) by 1/s cos it's a step response

and then mix and match it with the laplace tables to get the inverse transform.

So my attempt was:

$$\frac{1}{s^{2}-2rs cos(th) + r^{2}}$$

= $$\frac{1}{s^{2}-2rscos(th)+r^{2}cos^{2}(th) + r^{2}- r^{2}cos^{2}(th)}$$

= $$\frac{1}{(s-rcos(th))^{2}+r^{2}-r^{2}cos^{2}(th)}$$

= $$\frac{1}{(s-rcos(th))^{2}+r^{2}(1-cos^{2}(th))}$$

= $$\frac{1}{(s-rcos(th))^{2}+r^{2}sin^{2}(th) }$$

= $$\frac{1}{(s-rcos(th))^{2}+(rsin(th))^{2}}$$

This was great cos then I could match it to a transform on my laplace transform table

this one:

$$\frac{A(s+a) + Bw}{(s+a)^{2}+w^{2}}$$

= $$e^{-at}[Acos(wt)+Bsin(wt)]$$

but the problem was that I realized that A & B would be zero and i'd get zero as a result and also I hadn't multiplied my equation by 1/s and that screws up everything. I haven't a clue what else to do then.

so ... am I approching this thing all wrong?

any guidance or help very much appreciated

please bear in mind that I don't really know if I should be trying to do the laplace transform or what.

Berty
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