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As in the picture , how can I determine ki , kp , kd I tried (using provided data) by finding ζ and ωn But I end up with third order transfer function could not subsitute my results to find PID parameters (ki kp kd ) values

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    \$\begingroup\$ Hi and welcome. Please edit the post so the graphic appears correctly oriented. \$\endgroup\$ – mike65535 Nov 27 '18 at 16:22
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    \$\begingroup\$ Look up "dominant pole" in your book. If you have one pole or pole pair that's much slower than the rest, that pole or pole pair tends to dominate. From the math provided you've got the freedom to select all three poles. Choose the real pole to be enough higher than the resonant pair to be considered "dominant" by your book, and proceed. \$\endgroup\$ – TimWescott Nov 27 '18 at 16:24
  • \$\begingroup\$ BTW: your provided transfer function doesn't look right for a PID in the forward path. You may want to put your block diagram into a separate question along with your derivation; ask if you got it right. \$\endgroup\$ – TimWescott Nov 27 '18 at 16:25
  • \$\begingroup\$ Hi in fact this is answer sheet in page 5 they used "standered transfer function .. Why they equate alpha to 10 .. studocu.com/en-us/document/university-of-the-west-of-england/… \$\endgroup\$ – Freeman Nov 27 '18 at 21:40
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Well, we know that a second order system can be written in the following form:

$$\mathcal{H}\left(\text{s}\right):=\frac{\text{Y}\left(\text{s}\right)}{\text{X}\left(\text{s}\right)}=\frac{1}{\frac{1}{\omega_0^2}\cdot\text{s}^2+\frac{2\beta}{\omega_0}\cdot\text{s}+1}\tag1$$

When there is overshoot we know that:

  • Overshoot (in \$\text{%}\$): $$\text{D}=100\cdot\exp\left(-\pi\cdot\frac{\beta}{\sqrt{1-\beta^2}}\right)\tag2$$
  • Rise time: $$\text{t}_\text{p}=\frac{\pi}{\omega}\tag3$$

And we have overshoot so we know the relation between \$\omega\$, \$\omega_0\$ and \$\beta\$:

$$\omega=\omega_0\cdot\sqrt{1-\beta^2}\tag4$$

So, we get (using your biggest values):

  • $$10=100\cdot\exp\left(-\pi\cdot\frac{\beta}{\sqrt{1-\beta^2}}\right)\space\Longleftrightarrow\space\beta=\frac{\ln\left(10\right)}{\sqrt{\pi^2+\ln^2\left(10\right)}}\tag5$$
  • $$1=\frac{\pi}{\omega}\space\Longleftrightarrow\space\omega=\pi\tag6$$
  • $$\omega=\omega_0\cdot\sqrt{1-\beta^2}\space\Longleftrightarrow\space\omega_0=\sqrt{\pi^2+\ln^2\left(10\right)}\tag7$$

So:

$$\mathcal{H}\left(\text{s}\right):=\frac{\text{Y}\left(\text{s}\right)}{\text{X}\left(\text{s}\right)}=\frac{1}{\frac{1}{\pi^2+\ln^2\left(10\right)}\cdot\text{s}^2+\frac{\ln\left(100\right)}{\pi^2+\ln^2\left(10\right)}\cdot\text{s}+1}\tag8$$

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