I just have a quick question. In my lecture notes, he defined the PID controller as
And then he said that this is equal to the following
How did he do that? What is k, z1 and z2 equal?
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2\$\begingroup\$ I hate to say it, but simple algebra and perhaps the quadratic formula is all that's needed. Are you unable to see how to work it out? \$\endgroup\$ – jonk Nov 4 '18 at 21:57
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\$\begingroup\$ @jonk Yes. Not sure why I just keep following the same wrong steps and end up nowhere. \$\endgroup\$ – AlfroJang80 Nov 4 '18 at 22:32
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Okay:
$$\begin{align*} k_p + \frac{k_i}{s}+k_d\:s&=\frac{k\left(s+z_1\right)\left(s+z_2\right)}{s}\\\\ k_d\:s^2+k_p\:s + k_i&=k\:s^2 + k\left(z_1+z_2\right)s + k\:z_1\:z_2\\\\\therefore\\\\k&=k_d\\\\ k\left(z_1+z_2\right)&=k_p\\\\ k\:z_1\:z_2&=k_i \end{align*}$$
If you accept that \$k=k_d\$ solves for \$k\$, then that's two remaining equations and two remaining unknowns:
$$\begin{align*} z_1+z_2&=\frac{k_p}{k_d}\\\\ z_1\cdot z_2&=\frac{k_i}{k_d} \end{align*}$$
Can you move forward from here using the quadratic equation?
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\$\begingroup\$ This is a fun answer, one that shows the questioner that simple math and simple deduction skills goes a long way. \$\endgroup\$ – Harry Svensson Nov 4 '18 at 23:03
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\$\begingroup\$ Wow. I'm stupid. I've got it now. Thanks a lot and sorry for the time-wasting question. \$\endgroup\$ – AlfroJang80 Nov 4 '18 at 23:22
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1\$\begingroup\$ @AlfroJang80 All of us go through these kinds of things. You aren't stupid. Just needed a push in the right direction. I'm sure you'll remember the idea now. \$\endgroup\$ – jonk Nov 4 '18 at 23:24
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\$\begingroup\$ Just as a note; it's possible to choose a combination of \$k_p\$, \$k_i\$, and \$k_d\$ that give you unstable or complex zeros. That's usually a clue from the Universe to you that you're about to learn some practical issues of control theory involving buzzes, thumps, or smoke. But sometimes it's actually the right thing to do. \$\endgroup\$ – TimWescott Nov 5 '18 at 1:08
