# The transfer function of a delayed system x(t-T)

1. Nov 12, 2008

Hi, im not looking for the answer, i just want to know how i go about solving it

Ive got to find the transfer function of this equation:
http://img374.imageshack.us/img374/8149/laplaceeq8.jpg [Broken]

I know that the transfer function G(s) is found by:

y(s)=G(s)x(s)

As hard as i try i cannot seem to get x(s) through my workings.

From Laplace i have found the LHS = y(s^2+6s+45) and the RHS to be: e^-sT (not sure about the RHS)

so rearranging i can get y(s) = 1/(s^2 +6s +45) * e^-sT

but i dont know quite what do do here. Ive tried inverse laplace to turn it into a heaviside function, but it didn't seem to help. Any help would be much appreciated, thanks.

Last edited by a moderator: May 3, 2017
2. Nov 12, 2008

That should be LHS $= Y(s)(s^2+6s+45)$ and RHS $= X(s)e^{-sT}$.

3. Nov 12, 2008

Ok thx, so it would be $Y(s) = e^{-sT}/(s^2 + 6s +45)X(s)$
In the form Y(s) = G(s)X(s)

So would the transfer function be:
$G(s) = e^{-sT}/(s^2 + 6s +45)$ ?

4. Nov 13, 2008

$G(s) = e^{-sT}/(s^2 + 6s +45)$