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Okay, so I've been given a system to linearize and it gave me this result $$δ\ddot x+δx=δu$$ I tried state-space modeling this and this is what I got:

$${\dot x}=Ax+Bu=\begin{bmatrix}0&1\\-1&0\end{bmatrix}x+\begin{bmatrix}0\\1\end{bmatrix}u$$ Now, finding the eigen values of this I know that my transfer function will have the following denominator $$(s^2+1)$$ What I'm asked to do is design a compensatorC(s) so that the closed-loop system's dominant poles have a damping ratio of ζ=0,3. enter image description here

I have solved similar problems but in these problems I also had $$y=Cx$$ Now what I'm having trouble with. The eigen values are the poles of the open or the closed loop system? Are they the poles of G(s) in the picture or the poles of the closed loop system? Because finding the closed loop transfer function of the diagram I will also need the numerator of the transfer function and not just the poles.

How do I proceed ?

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  • \$\begingroup\$ What is the output of the system? There may be some info you have that will allow you to determine "C". The eigenvalues are the poles of whatever you are calculating. I don't know what additional info you have. Maybe you have enough into to proceed, and maybe you need to ask your instructor. \$\endgroup\$ – PICyPICyPICy Aug 29 '17 at 16:37
  • \$\begingroup\$ We have another instructor now and I'm solving problems from previous exams(of the previous instructor). This problem has been in 3 of them(identical-without extra info) and no , I haven't left anything out . \$\endgroup\$ – John Katsantas Aug 29 '17 at 16:40
  • \$\begingroup\$ \$\dfrac{C(s)\cdot G(s)}{1+G(s)\cdot C(s)} \$= transfer function of the closed loop and comes something that you want to have a damping ratio of 0.3. Not without knowing the G(s) \$\endgroup\$ – Marko Buršič Aug 29 '17 at 16:46
  • \$\begingroup\$ @MarkoBuršič It states: For the linearized transfer function design a compensator etc etc. G(s) could be the transfer function with the s^2+1 denominator. That or the s^2+1 denominator is for the closed loop system. We could also assume δx as output by ourselves? I'm really upset with some this teacher's problems, half of them are wrong or incomplete. \$\endgroup\$ – John Katsantas Aug 29 '17 at 17:00
  • \$\begingroup\$ Draw the controller canonic block diagram, and see what feedback path plus associated gain term you need to sketch in to give a first derivative term in the TF. Adjust that gain term to get \$\small\zeta=0.3\$. \$\endgroup\$ – Chu Aug 29 '17 at 17:02
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If you assume the plant has no zeros (big assumption, but we don't have anything else to go on) then the controller below will get you pretty close:

\begin{equation} K(s) = \frac{2 \zeta s}{s/\omega_{BIG} + 1} \end{equation}

How close you get depends on where you place the pole in the expression above. For 10 rad/s, I got pretty close.

Some Scilab code:

s = %s

G = syslin('c', 1/(s^2 + 1))

K = syslin('c', 0.6*s/(s/10 + 1))

T=GK/(1+GK)

plzr(T)

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