1. Pid controller pole/zero cancellation
By Frank Owen, PhD, PE
polyXengineering, Inc.
San Luis Obispo, California

2. This is just one strategy for tuning a pid
There are many strategies to follow to tune a PID
The most common aims for using a PID are:
Reduce or eliminate steady-state error
Make response faster without causing more overshoot
The strategy described here is completely different
We have a 2nd-order system whose dynamics we do not like
We cancel the system’s dynamics out and replace them with different dynamics
Let’s see how to do this…

3. 2nd–order with complex poles
Open-loop poles: 𝑠=−3±𝑗∙2
System has two asymptotes
No matter the value of KP , the system oscillates
Besides this, system has ess when given a step input
To be complete, let’s specify that the open-loop system has a gain of 1 (KOL = 1)
Aim:
Make system non-oscillatory with a time constant of 1 second
Eliminate or reduce ess

4. 2nd–order with complex poles
A PID controller has:
A pole at the origin
Two zeros, which can be placed at any location
A gain, which can be used to move the closed-loop poles along the revised root locus
(See the video “Beyond P-only control via root locus”)
Put the pole at the origin
Place the two zeros on top of the OL poles to cancel them
The revised RL is just a path running from the origin along an asymptote at 180°…
…so the open-loop system with the PID controller is now just an integrator and a gain

5. 2nd–order with complex poles
Now adjust the gain to move the closed-loop pole out to the left so that T = 1 second.
Adjust gain to move closed-loop pole to T = 1
(pole at -1/T)
-1/1
The final step is to calculate KP , KI , and KD that give us these results…

6. 2nd–order with complex poles
For the graphical solution, we draw the vectors from all poles and zeros to the desired closed-loop pole.
…but Mp1 = Mz1
Recall that the formula for the evaluation of GOL at the desired location is:
𝐾 𝑃𝐼𝐷 𝐺 𝑂𝐿 = 𝐾 𝑃𝐼𝐷 𝐾 𝑂𝐿 Π 𝑀 𝑧 Π 𝑀 𝑝 ∠ Σ 𝜃 𝑧 −Σ 𝜃 𝑝
Mp2 , Mz2
Mp3
Mp1 , Mz1
-1/1
…also 𝜃 𝑝1 = 𝜃 𝑧1 and 𝜃 𝑝2 = 𝜃 𝑧2 . So these all cancel in the calculation. (This just shows that the zeros cancel the poles.) So we are left with:
𝐾 𝑃𝐼𝐷 𝐺 𝑂𝐿 = 𝐾 𝑃𝐼𝐷 𝐾 𝑂𝐿 1 𝑀 𝑝3 ∠ 0− 𝜃 𝑝3 = 𝐾 𝑃𝐼𝐷 𝐾 𝑂𝐿 ∠ −180°
…and Mp2 = Mz2

7. 2nd–order with complex poles
𝐾 𝑃𝐼𝐷 𝐺 𝑂𝐿 = 𝐾 𝑃𝐼𝐷 𝐾 𝑂𝐿 ∠−180°
To meet the magnitude criterion, 𝐾 𝑃𝐼𝐷 = 1 𝐾 𝑂𝐿 . But the question is, what is 𝐾 𝑃𝐼𝐷 ?
Let’s put the PID controller together. We added
𝑠+3+𝑗∙2 (𝑠+3−𝑗∙2) 𝑠 = 𝑠 2 +6𝑠+13 𝑠
-1/1
For this part of the controller to have a gain of 1, we need to divide it by 13. Note that this does not change the location of the zeros.

8. 2nd–order with complex poles
Thus the PID without its gain (or, otherwise stated, with a gain of 1) is
1 13 𝑠 𝑠+1 𝑠
From the last slide, 𝐾 𝑃𝐼𝐷 = 1 𝐾 𝑂𝐿 , but the original problem statement was that KOL = 1. Thus the overall PID is 1.
-1/1
The complete PID controller is thus:
𝐺 𝑃𝐼𝐷 = 𝑠 𝑠+1 𝑠

9. 2nd–order with complex poles
We match this with the standard form:
𝐺 𝑃𝐼𝐷 = 𝑠 𝑠+1 𝑠 = 𝐾 𝐷 ∙ 𝑠 2 + 𝐾 𝑃 ∙𝑠+ 𝐾 𝐼 𝑠
Thus
𝐾 𝑃 = 6 13
𝐾 𝐼 =1
𝐾 𝐷 = 1 13

10. The Problem with Pole/zero cancellation
The system’s open-loop poles can drift during operation (for example, hydraulic fluid warming up during operation) or over time (kinematic or friction changes). Thus the cancellation of the poles with the controller zeros is not complete. Part of your study, if pole/zero cancellation is used, may be to assess the effect of this pole drift.

11. © polyXengineering, Inc. San Luis Obispo, California
That’s all folks!
Fin
© polyXengineering, Inc.
San Luis Obispo, California