- #1

- 1

- 0

LTI system is defined as follows

[tex] H(s) = \frac{1}{ s^2 - 2rs cos(theta) + r^2}[/tex]

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:

[tex]\frac{1}{s^{2}-2rs cos(th) + r^{2}}[/tex]

= [tex]\frac{1}{s^{2}-2rscos(th)+r^{2}cos^{2}(th) + r^{2}- r^{2}cos^{2}(th)}[/tex]

= [tex]\frac{1}{(s-rcos(th))^{2}+r^{2}-r^{2}cos^{2}(th)}[/tex]

= [tex]\frac{1}{(s-rcos(th))^{2}+r^{2}(1-cos^{2}(th))}[/tex]

= [tex]\frac{1}{(s-rcos(th))^{2}+r^{2}sin^{2}(th) }[/tex]

= [tex]\frac{1}{(s-rcos(th))^{2}+(rsin(th))^{2}}[/tex]

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

this one:

[tex]\frac{A(s+a) + Bw}{(s+a)^{2}+w^{2}}[/tex]

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

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