EE3CL4: Introduction to Linear Control Systems Section 7: PID - - PowerPoint PPT Presentation

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EE3CL4: Introduction to Linear Control Systems Section 7: PID - - PowerPoint PPT Presentation

EE 3CL4, 7 1 / 17 Tim Davidson PID Control EE3CL4: Introduction to Linear Control Systems Section 7: PID Control Tim Davidson McMaster University Winter 2019 EE 3CL4, 7 2 / 17 Outline Tim Davidson PID Control PID Control 1 EE


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EE 3CL4, §7 1 / 17 Tim Davidson PID Control

EE3CL4: Introduction to Linear Control Systems

Section 7: PID Control Tim Davidson

McMaster University

Winter 2019

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EE 3CL4, §7 2 / 17 Tim Davidson PID Control

Outline

1

PID Control

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EE 3CL4, §7 4 / 17 Tim Davidson PID Control

Cascade compensation

  • Throughout this lecture we consider the case of H(s) = 1.
  • We have looked at using
  • lead compensators to improve the transient

performance of a closed loop

  • lag compensators to improve the steady state error

responses without changing the closed loop transient response too much.

  • What if we wanted to do both? What did we do?
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EE 3CL4, §7 5 / 17 Tim Davidson PID Control

Lead-lag compensation

  • Apply lead design techniques to G(s) to adjust the closed loop

transient response

  • Then apply lag design techniques to GC,lead(s)G(s) to improve

steady state error response without changing the closed loop transient response too much

  • Resulting compensator:

GC(s) = GC,lag(s)GC,lead(s) = KC,lagKC,lead(s + zlag)(s + zlead) (s + plag)(s + plead)

  • How can we gain insight into what the compensator is doing?
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EE 3CL4, §7 6 / 17 Tim Davidson PID Control

Lead-lag approximation

  • GC,lag(s)GC,lead(s) = KC,lagKC,lead(s + zlag)(s + zlead)

(s + plag)(s + plead)

  • Recall that
  • for frequencies between zlead and plead,

lead compensator acts like a differentiator

  • for frequencies between plag and zlag,

lag compensator acts like an integrator

  • Rewrite:

GC,lag(s)GC,lead(s) = ˜ KC,ll(s + zlag)(s + zlead) (s + plag)(1 + s/plead)

  • as plead gets big, and plag gets small this starts to look like

GC,lag(s)GC,lead(s) ≈ ˜ KC,ll(s + zlag)(s + zlead) s for the values of s that are of greatest interest.

  • Not physically realizable (more zeros than poles),

but helpful approximation

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EE 3CL4, §7 7 / 17 Tim Davidson PID Control

Lead-lag to PID

GC,lag(s)GC,lead(s) ≈ ˜ KC,ll(s + zlag)(s + zlead) s

  • Do a partial fraction on RHS and you get

GC,lag(s)GC,lead(s) ≈ KP + KI s + KDs

  • With H(s) = 1, input to the compensator is e(t) = r(t) − y(t)
  • Compensator output: u(t) = L−1

GC,lag(s)GC,lead(s)E(s)

  • =

⇒ u(t) ≈ KPe(t) + KI

  • e(t) dt + KD de(t)

dt

  • That is, (approximately) PID control
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EE 3CL4, §7 8 / 17 Tim Davidson PID Control

Variants of PID control

PID: Gc(s) = KP + KI/s + KDs.

  • With KD = 0 we have a PI controller,

GPI(s) = ˆ KP + ˆ KI/s.

  • With KI = 0 we have a PD controller,

GPD(s) = ¯ KP + ¯ KDs.

  • As implicit in our derivation, a PID controller can be

realized as the cascade of a PI controller and a PD controller; i.e., GPI(s)GPD(s) can be written as KP + KI/s + KDs

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EE 3CL4, §7 9 / 17 Tim Davidson PID Control

PID control and root locus

  • Transfer function of idealized PID controller:

GC(s) = KP + KI/s + KDs = KD(s + z1)(s + z2) s

  • That is, controller adds two zeros and a pole to the open

loop transfer function

  • The pole is at the origin
  • The zeros can be arbitrary real numbers,
  • r an arbitrary complex conjugate pair
  • This provides considerable flexibility in re-shaping the root

locus

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EE 3CL4, §7 10 / 17 Tim Davidson PID Control

PID Tuning

with Gc(s) = KP + KI/s + KDs.

  • How should we choose KP, KI and KD?
  • Can formulate as a optimization problem; e.g.,

Find KP, KI and KD that minimize the settling time, subject to

  • the damping ratio being greater than ζmin,
  • the position and velocity error constants being greater

than Kposn,min and Kv,min,

  • the error constant for a step disturbance being greater

than Kdist,posn,min,

  • and the loop being stable
  • Typically difficult to find the optimal solution
  • Many ad-hoc techniques that usually find “good”

solutions have been proposed.

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EE 3CL4, §7 11 / 17 Tim Davidson PID Control

Zeigler–Nichols Tuning

  • Two well established methods for finding a “good”

solution in some common scenarios

  • Often useful in practice because they can be applied to

cases in which the model has to be measured (no analytic transfer function)

  • We will look at the “ultimate gain” method
  • This is based on the step response of the system
  • However, the method is only suitable for a certain class
  • f systems and a certain class of design goals
  • You need to make sure that the system you wish to

control falls into an appropriate class.

  • You also need to ensure that the ZN tuning goals match

your design goals. The ZN tuning scheme gives considerable weight to the response to disturbances

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EE 3CL4, §7 12 / 17 Tim Davidson PID Control

“Ultimate Gain” Zeigler–Nichols Tuning

1 Set KI and KD to zero. 2 Increase KP until the system is marginally stable

(Poles on the jω-axis)

3 The value of this gain is the “ultimate gain”, KU 4 The period of the sustained oscillations is called the

“ultimate period”, TU (or PU). (The position of the poles on the jω-axis is 2π/TU)

5 The gains are then chosen using the following table

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EE 3CL4, §7 13 / 17 Tim Davidson PID Control

“Ultimate Gain” Zeigler–Nichols Tuning

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EE 3CL4, §7 14 / 17 Tim Davidson PID Control

Manual refinement

  • One way in which the design can be improved, is

searching for “nearby” gains that improve the performance

  • The following table provides guidelines for that local
  • search. These are appropriate for a broad class of

systems

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EE 3CL4, §7 15 / 17 Tim Davidson PID Control

Example

G(s) =

1 s(s+b)(s+2ζωn), with b = 10, ζ = 1/

√ 2 and ωn = 4.

  • Step 2: Plot root locus of G(s) to find KU and TU
  • Step 3: KU = 885.5,
  • Step 4: marginally stable poles: ±j7.5; ⇒ TU = 0.83s
  • Step 5: KP = 521.3, KI = 1280.2, KD = 55.1
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EE 3CL4, §7 16 / 17 Tim Davidson PID Control

Example

Step response of ZN tuned closed loop, KP = 521.3, KI = 1280.2, KD = 55.1

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EE 3CL4, §7 17 / 17 Tim Davidson PID Control

Example

Step response with manually modified gains, KP = 370, KI = 100, KD = 60