Lecture 13: PID controllers - ZTMiR

Three-term controller Ziegler-Nichols tuning rules Summary

Basics of Automation and Control I

Lecture 13: PID controllers

Paweł Malczyk

Division of Theory of Machines and RobotsInstitute of Aeronautics and Applied MechanicsFaculty of Power and Aeronautical Engineering

Warsaw University of Technology

© Paweł Malczyk. Basics of Automation and Control I Lecture 13: PID controllers 1 / 35


Outline

1 Three-term controller

2 Ziegler-Nichols tuning rules

3 Summary



Three-term controller

1 Three-term controllerIntroductionBasic control functionsP controllerPI controllerPD controllerPID controller


3 Summary



Introduction

Fig. 1: Block diagram of a process with feedback controller

1 The three-term controller, i.e. proportional-integral-derivative (PID)controller, is a control loop feedback mechanism widely used in manyindustrial control systems.

2 PID controllers appear in many different forms: as stand-alonecontrollers, as part of hierarchical, distributed control systems and builtinto embedded components.

3 By tuning the three parameters in the PID controller algorithm, thecontroller can provide control action designed for specific processrequirements (benign process dynamics, moderate performance).



Basic control functions

Fig. 2: Block diagram of feedback control systemwith ideal PID controller

The input/output relation for an ideal PID controller*:

u(t) = kp[e(t) +

1Ti

∫ t

0e(τ)dτ + Td

de(t)dt

]=

= kpe(t) + ki∫ t

0e(τ)dτ + kd

de(t)dt

(1)

e(t) = r(t) − y(t) – error signal, u(t) – control signal,kp – proportional gain, Ti – integral time, Td – derivative time,

ki =kpTi– integral gain, kd = kpTd – derivative gain.

*Modified versions of the PID controller are used in practical applications.



Basic control functionsIdeal PID controller:

u(t) = kp

[e(t) +

1Ti

∫ t

0

e(τ)dτ + Tdde(t)dt

]→ C(s) =

U(s)E(s)

= kp(1+

1Tis

+ Tds) (2)

Fig. 3: Block diagram of ideal PID controller• P controller (Ti = ∞, Td = 0)

C(s) = kp (3)

• PI controller (Td = 0)

C(s) = kp(1+

1Tis

)(4)

• PD controller (Ti = ∞)C(s) = kp(1+ Tds) (5)

Fig. 4: PID interpretation→Gentle introduction toPID controllers: https://www.youtube.com/watch?v=4Y7zG48uHRo.


https://www.youtube.com/watch?v=4Y7zG48uHRo


Mechanical system

Fig. 5: Układmechaniczny. Stałe fizyczne: m = 1 kg, b = 2 Nsm , k = 5 N

m

Goal: choose a control force f(t) (in N) such that the mass will stop at thedesired position (say 1 m from equilibrium).Plant tranfer function G(s) = Y(s)

F(s) =1

ms2+bs+k .Reference signal (desired position): r(t) = yd · 1(t) (yd = 1 m).In steady-state the force f(t) need to balance the spring force, i.e.:f = k · yd · 1(t), then

yss = limt→∞

y(t) = lims→0

sY(s) = lims→0

sG(s)F(s) = lims→0

sG(s)kyd1s= G(0)kyd = yd



Mechanical systemwith disturbance

Fig. 6: Mechanical system. Physical constants: m = 1 kg, b = 2 Nsm , k = 5 N

m , and g = 10 ms2

Plant transfer function G(s) = Y(s)F(s) =

1ms2+bs+k .

Disturbance – gravity force: d(t) = −mg · 1(t).Reference signal (desired position): r(t) = yd · 1(t) (yd = 1 m).In steady-state the force f(t) need to balance the sum of spring force andthe gravity force, i.e.: u = (k · yd +mg) · 1(t).Assumption: perfect knowledge of parameters and disturbance...



Mechanical systemwith parametric uncertainty

True values: m, b, k.Estimated params: m, b, k.Real parameters anddisturbance signals areusually unknown. Fig. 7: Mechanical systemwith parametric uncertainty

Control signal: u = (k · yd +mg) · 1(t).Input signal: f = (kyd + (m − m)g) · 1(t).Steady-state response:

yss = limt→∞

y(t) = lims→0

sY(s) = lims→0

sG(s)(kyd+(m−m)g)1s=

1k(kyd+(m−m)g)

Error signal:

ess = yss − yss = yd − 1k(kyd + (m − m)g) =

1k((k − k)yd + (m − m)g)

The greater the uncertainty is, the greater the steady-state error is.Transient-response is not controlled. Open-loop control is sensitive toparametric uncertainty.



Proportional control

Fig. 8: Feedback control system

Assumption: u(t) = kp(r(t) − y(t)) – force proportional to the error.Open-loop transfer function: L(s) = C(s)G(s).

Closed-loop transfer function: T(s) = Y(s)R(s) =

L(s)1+L(s) =

kpms2+bs+k+kp

.

kp – virtual spring coefficient.Let us calculate the response y(t) and steady-state error e(t).





The response: Y(s) = T(s)R(s) + G(s)S(s)D(s).Sensitivity function:

S(s) =1

1+ L(s)=

ms2 + bs+ kms2 + bs+ k+ kp

Load disturbance sensitivity function:

W(s) = G(s)S(s) =1

ms2 + bs+ k+ kpReference signal: r(t) = yd · 1(t) → R(s) = yd

sDisturbance: d(t) = −mg · 1(t) → D(s) = −mg

s .© Paweł Malczyk. Basics of Automation and Control I Lecture 13: PID controllers 11 / 35




Steady-state response:

yss = limt→∞

y(t) = lims→0

sY(s) = lims→0

s(T(s)yds

+ W(s)(−mgs

)) =

= ydT(0) − mgW(0) = ydkp

k+ kp− mg

1k+ kp

If kp → ∞, then kpk+kp

→ 1 and 1k+kp

→ 0. In consequence: yss = yd.

Conclusion:→ disturbance rejection (even unknown) and trajectory tracking.



P controller

Fig. 11: P controller, C(s) = kp

Effects of P controllerProportional term, P, causes a corrective control actuation proportionalto the error.The systemwithP controllerwill usually havenonzero steady-state errors.As kp increases, then the static position error decreases.As kp increases, the stability margins decrease and the system maybecome unstable.As kp increases, the BW (as well as overshoot and settling time) increases.It is difficult to ensure both good transient response accuracy and steady-state performance.



P controller

0 1 2 3 4 5 6 7 8 9 10

t

0

0.2

0.4

0.6

0.8

1

1.2

1.4

y(t)

kp=15kp=20kp=25kp=30

r(t)

0 1 2 3 4 5 6 7 8 9 10

t

-15-10-505

1015202530

u(t)

kp=15kp=20kp=25kp=30

10-2 10-1 100 101 102

Frequency (rad/sec)

-60-50-40-30-20-10

010203040

Mag

nitu

de (

dB)

kp=30

10-2 10-1 100 101 102

Frequency (rad/sec)

-180

-150

-120

-90

-60

-30

0

Pha

se (

deg)

kp=30

GCLST

Fig. 12: P controller characteristics

Comment 1If kp ↗, thenMp ↗, tp ↘, ts ↗, ess ↘, stability issues.



P controller

10-2 10-1 100 101 102

Frequency (rad/sec)

-60

-50

-40

-30

-20

-10

0

10

20

30

40

Mag

nitu

de (

dB)

kp=30

GCLST

0 1 2 3 4 5 6 7 8 9 10

t

0

0.2

0.4

0.6

0.8

1

1.2

1.4

y(t)

kp=15

kp=20kp=25kp=30

r(t)

Fig. 13: P controller characteristics



PI controller

Fig. 14: PI controller, C(s) = kp(1 + 1

Tis

)Effects of PI controller

The integral term, I, gives a correction proportional to the integral of theerror.PI largely reduces the steady-state errors (compared to P controller).It may cause the closed loop system less stable (or even unstable).PI may slow down the transient response.Filtering out high-frequency noise.Windup - interaction of integral action and saturations.



PI controller

0 1 2 3 4 5 6 7 8 9 10

t

-0.6-0.4-0.2

00.20.40.60.8

11.21.41.6

y(t)

kp=30

Ti=1Ti=5Ti=10Ti=100

0 1 2 3 4 5 6 7 8 9 10

t

-20-15-10-505

1015202530

u(t)

kp=30

Ti=1Ti=5Ti=10Ti=100

10-2 10-1 100 101 102

Frequency (rad/sec)

-80

-60

-40

-20

0

20

40

60

80

Mag

nitu

de (

dB)

kp=30, Ti=1

G C L S T

10-2 10-1 100 101 102

Frequency (rad/sec)

-180

-150

-120

-90

-60

-30

0

Pha

se (

deg)

kp=30, Ti=1

GCLST

Fig. 15: PI controller characteristics

Comment 2If Ti ↘, thenMp ↗, ts ↗, ↘, ess ↘.



PI controller

10-2 10-1 100 101 102

Frequency (rad/sec)

-80

-60

-40

-20

0

20

40

60

80

Mag

nitu

de (

dB)

kp=30, Ti=1

GCLST

0 1 2 3 4 5 6 7 8 9 10

t

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

y(t)

kp=30

Ti=1Ti=5Ti=10Ti=100

Fig. 16: PI controller characteristics



PD controller

Fig. 17: PD controller, C(s) = kp(1 + Tds)

Effects of PD controllerThe derivative term, D, gives a predictive capability yielding a controlaction proportional to the rate of change of the error.Derivative action tends to have a stabilizing effect.PD controller adds damping and reduces maximum overshoot.It improves the transient response (decreases rise and settling time).PD controller improves relative stability (increases gain and phasemargins).Possibly accentuating noise at higher frequencies.Filtered version of the PD (or PID) controller is used due to derivative kick,then, C(s) = kp(1+ Tds 1

TdN s+1

). Typical values of N are 8 to 20.



PD controller

0 1 2 3 4 5 6 7 8 9 10

t

-0.6-0.4-0.2

00.20.40.60.8

11.21.4

y(t)

kp=30

Td=0Td=0.01Td=0.05Td=0.1

0 1 2 3 4 5 6 7 8 9 10

t

-15-10-505

1015202530

u(t)

kp=30

Td=0Td=0.01Td=0.05Td=0.1

10-2 10-1 100 101 102 103

Frequency (rad/sec)

-60

-40

-20

0

20

40

60

80

Mag

nitu

de (

dB)

kp=30, Td=0.1

G C L S T

10-2 10-1 100 101 102

Frequency (rad/sec)

-180-150-120

-90-60-30

0306090

120

Pha

se (

deg)

kp=30, Td=0.1

GCLST

Fig. 18: PD controller characteristics

Comment 3If Td ↗, thenMp ↘, tp ∼↗, ts ↘.



PD controller

10-2 10-1 100 101 102 103

Frequency (rad/sec)

-60

-40

-20

0

20

40

60

80

Mag

nitu

de (

dB)

kp=30, Td=0.1

GCLST

0 1 2 3 4 5 6 7 8 9 10

t

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

y(t)

kp=30

Td=0Td=0.01Td=0.05Td=0.1

Fig. 19: PD controller characteristics



PID controller

Fig. 20: PID controller, C(s) = kp(1 + 1

Tis+ Tds

)Effects of PID controller

The features of each of the PI and PD controllers are utilized.Improves both steady-state errors as well as transient responsespecifications.The PID controller algorithm involves three separate constant parametersthat need to be tuned.



PID controller

0 1 2 3 4 5 6 7 8 9 10

t

0

0.2

0.4

0.6

0.8

1

1.2

1.4

y(t)

kp=30, Td=0.1, Ti=1

y(t)r(t)

0 1 2 3 4 5 6 7 8 9 10

t

-10

0

10

20

30

40

50

u(t)

kp=30, Td=0.1, Ti=1

u(t)

10-3 10-2 10-1 100 101 102 103

Frequency (rad/sec)

-80

-60

-40

-20

0

20

40

60

80M

agni

tude

(dB

)kp=30, Td=0.1, Ti=1

G C L S T

10-3 10-2 10-1 100 101 102 103

Frequency (rad/sec)

-180-150-120

-90-60-30

0306090

120

Pha

se (

deg)

kp=30, Td=0.1, Ti=1

G C L S T

Fig. 21: PID controller characteristics



PID controller

10-3 10-2 10-1 100 101 102 103

Frequency (rad/sec)

-80

-60

-40

-20

0

20

40

60

80

Mag

nitu

de (

dB)

kp=30, Td=0.1, Ti=1

G C L S T

0 1 2 3 4 5 6 7 8 9 10

t

0

0.2

0.4

0.6

0.8

1

1.2

1.4

y(t)

kp=30, Td=0.1, Ti=1

y(t)r(t)

Fig. 22: PID controller characteristics



PID controllers

Fig. 23: Ideal P, PI, PD and PID controllers1 PID controllers are applicable to many control problems, and often

perform satisfactorily.2 Ideal PID are linear controllers with constant coefficients.3 There are many variations of PID controllers associated with practical

control applications (e.g. preventing from integrator windup).© Paweł Malczyk. Basics of Automation and Control I Lecture 13: PID controllers 25 / 35


Ziegler-Nichols tuning rules


2 Ziegler-Nichols tuning rulesIntroductionFirst methodSecondmethodCommentsExample

3 Summary



Introduction

Fig. 24: Ideal PID controller and plant

If a mathematical model of the plant is known, then it is possible todesign the PID controller that will meet the transient and steady-statespecifications of the closed-loop system.If the mathematical model cannot be easily obtained, one should selectthe controller parameters experimentally, which known as a process ofcontroller tuning.Experimental Ziegler and Nichols tuning rules for PID controllers as astarting point for fine tuning.



First method

Experimental response of the plant due toa unit-step input. The transfer function Y(s)

U(s)may be approximated by

G(s) =Y(s)U(s)

=Ke−τs

Ts+ 1(6)

T – time constant, τ – delay, K – gain. Comment 4The controller parameters are designed toresult in 25% amplitude decay ratio in oneperiod.

Fig. 25: S-shaped response of a closed-loopsystem due to a unit-step input. The plantinvolves neither integrators nor dominantcomplex conjugate closed-loop poles

Ziegler-Nichols tuning rule based on a step response of a plant

Type kp Ti TdP T

Kτ ∞ 0

PI 0.9 TKτ

τ0.3 0

PID 1.2 TKτ 2τ 0.5τ



Secondmethod

Fig. 26: Closed-loop systemwith P controller and sustained oscillation with period Tcr

1. Set Ti = ∞, Td = 0 and increase kpfrom0 to a critical value kcr, where theoutput exhibits sustained oscillations.

2. Find the critical period Tcr.3. Select the parameters kp, Ti and Td

according to the table.

Ziegler-Nichols tuning rule basedon a critical period Tcr

Type kp Ti TdP 0.5kcr ∞ 0

PI 0.45kcr 11.2Tcr 0

PID 0.6kcr 0.5Tcr 0.125Tcr

Comment 5 The controller parameters are designed to result in 25%amplitude decay ratio in one period. Comment 6 One can use Nyquist plot to find kcr, and Tcr.



Comments

1 Ziegler-Nichols tuning rules (and other tuning rules) have been widelyused in process control systems.

2 The plant dynamics may be known or may be not precisely known.3 Experience has shown that the Z-N rules provide acceptable closed-loop

response for many systems.4 The empirical PID tuning strategies are only starting points in the process

to obtain the right controller.5 Tuning method (by Astrom & Hagglund) based on relay feedback is

available.



Example

Example 1 Consider the control system shown in the Fig. 27. The processhas transfer function G(s) = Y(s)

U(s) = 1(Ts+1)2 . Apply the Ziegler-Nichols tuning

rule to determine the constants of the P, PI, and PID controller.

Fig. 27: Liquid-level control system



ExampleProcess model G(s) = Ke−τs

Ts+1 .K = 1τ = t1 = 12.72secT = t3 − t1 = 122.737sec

Type kp Ti TdP T

Kτ ∞ 0

PI 0.9 TKτ

τ0.3 0

PID 1.2 TKτ 2τ 0.5τ

Type kp Ti TdP 9.6491 ∞ 0

PI 8.6842 42.3576 0

PID 11.579 25.44 6.36

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300−0.2

−0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1.1

1.2

Fig. 28: Process reaction curve

CP(s) = 9.6491, CPI(s) = 8.6842(1+

142.3576s

),

CPID(s) = 11.579(1+

125.44s

+ 6.36s)



Example

0 50 100 150 200 250 3000

0.20.40.60.81

1.21.41.61.8

PPIPID

0 50 100 150 200 250 300−6−4−202468101214161820

PPIPID

0 50 100 150 200 250 300−0.05

0

0.05

0.1

0.15

0.2PPIPID

0 50 100 150 200 250 3000

0.025

0.05

0.075

0.1PPIPID

Fig. 29: Closed loop system responses for various controllers



Review questions

1 What is themain objective of introducing proportional control, integral control or derivativecontrol?

2 What is a P, PI, PD, PID controller? Write the input-output relation, unit-step response andBode plots.

3 Give the effects of P, PI, PD and PID controllers on the feedback control systemperformance.4 Formulate and comment on the twoZiegler-Nichols tuning rules for a PID controller. Discuss

the range of applicability of both methods.



Summary



3 Summary


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Lecture 13: PID controllers - ZTMiR