0
\$\begingroup\$

I have an open loop transfer function given by \begin{equation} \frac{(s+3)}{s(s+1)(s+5)} \end{equation}

I'm trying to convert it to a closed loop system with a gain of 0.987 and damping ratio of 0.7071 and find the closed loop transfer function. I know open loop transfer functions are given by KG(s)H(s) and closed loop transfer functions are given by KG(s)/1+KG(s)H(s) but I'm not sure how to relate this to the equation I have here.

Can anyone help me find the closed loop transfer function?

\$\endgroup\$
  • 1
    \$\begingroup\$ It depends on where you're closing the loop. What I would do, for simplicity's sake, is have the entire thing be in the loop, so it'd just be G(s)/(1+G(s)), where G(s) is your open-loop transfer function there. \$\endgroup\$ – Hearth Apr 27 '17 at 21:36
  • \$\begingroup\$ For the previous part of the question I worked out the Gain using matlab so I think I have to use it in this part of the question too, should I just multiply G(s)/(1+G(s)) by K? \$\endgroup\$ – Ca01an Apr 27 '17 at 21:53
  • \$\begingroup\$ Where's the gain in your loop? \$\endgroup\$ – Hearth Apr 27 '17 at 21:57
  • \$\begingroup\$ It's not present in the above loop, I was asked to produce a root locus plot of the above transfer function using Matlab, select a gain for a given closed loop damping ratio and plot the step response of the resulting closed loop system. Because the open loop transfer function I was given doesn't seem to have a gain, I'm not sure if I should add the gain into my closed loop transfer function. \$\endgroup\$ – Ca01an Apr 27 '17 at 22:04
  • \$\begingroup\$ The loop gain is inserted in the OLTF. It may be possible to obtain a 2nd order closed loop factor with \$\zeta=\frac{1}{\sqrt{2}}\$, I haven't checked whether that \$\zeta\$ value is attainable. But it seems that you're stuck with unity closed loop gain as you only have one parameter to play with and that can't satisfy two design criteria. There will also be a 1st order pole, and a zero at s=-3 in the CLTF. \$\endgroup\$ – Chu Apr 28 '17 at 0:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.