Recent content by koochiee

Yes I do, thank you! Thank you again for the codes.

:) no worries. So do you think the approximation is right?

Don, I was thinking that the smallest pole is s=-10 is the one that can be ignored, sorry if I'm wrong. But isn't that's the one farthest from imaginary axis so has little effect on the system response? And thank you for the answer & the MATLAB code. I'll check it soon as I get MATLAB.

For a 3rd order system, can I use 2nd order approximation (granted dominant pole is ≈5 Tau away from third pole), when designing a Lead or Lag compensator? Can this OLTF = K/{s(0.1s+1)(s+1)} be approximated to 10K/{s(s+1) which ultimately gives CLTF=10K/{s(s+1)+ K} ? and then design...

Thank you for your response.

Thank you donpacino! Can you explain how this happens? How control bandwidth increases when the gain is increased. From looking at the figure I can say how much I should increase the gain by (40dB). But physically how do I interpret it? I've looked through my notes and textbooks and I can't find...

Actually it was the gain required to improve control BW to 3 rad/s. And I first thought that I'd shift the gain curve 40dBs (according to the figure). But I'm not quite certain. I would really appreciate any help.

The Bode Plot below was given and, I've marked for the Gain Margin & Phase Margin. They're asking how much the gain should be improved to get the Control Bandwidth to 3 rad/s.

Can anyone tell me if Control bandwidth and closed loop bandwidth means the same thing? If not what does control bandwidth mean? Please & Thank you!

Thanks donpacino for the detailed answer.

Thank you very much for your answer Lancelot59! :)

The root locus design method is used to locate poles at desired locations. However, it is not possible to locate poles arbitrarily. Provide reasons for this statement. K = 1/│G(s)H(s)│ K G(s)H(s) = (2k+1)π I've formulated the answer for this as, The root locus gives the path of...

Thank you!

Thanks for the reply! And yes it is, the closed loop T.F is Gc(s)=G(s)/(1+KG(s)). Even after when I convert it to a unity feedback gain system it still is a type 0 system. Which can't have zero steady state error. And they're asking us to show that a gain (possibly in the forward path) of K+1...

Is it possible for a Type 0 Open Loop Transfer function to have 0 steady state error? Context - Control System Gain design to meet a certain steady state error specification. The open loop T.F - G(s)=3/(s^2 +4s+3) (This is type 0 (n=0)) The closed loop T.F is Gc(s)=G(s)/(1+KG(s)), But my problem...