Constructing and Verifying Cyber Physical...

Marcus Völp

Constructing and Verifying Cyber Physical Systems

Effect of Poles and Zeros / PID Control

Summer 2015 Constructing and Verifying Cyber Physical Systems - Marcus Völp 109

Overview

Math

FeedbackControl

RTOS

Verification

Physics

Introduction

Mathematical Foundations (Differential Equations and Laplace Transformation)

Control and Feedback

Transfer Functions and State Space Models

Poles, Zeros / PID Control

Stability, Root Locust Method, Digital Control

Mixed-Criticality Scheduling and Real-Time Operating Systems (RTOS)

Program Verification

Differential Dynamic Logic and KeYmaera X

Differential Invariants


Overview

Control

Regulating Control

Open Loop Control

Tracking Control

Closed Loop Control

Noise

Observability

Sensor Fusion

Block Diagrams / Simulink

Robustness

Stability / Final Value Theorem

K. Åström, R. Murray“Feedback Systems –An Introduction for Scientists and Engineers”Princeton University Press

G. Franklin, J.D. Powell, A. Emami-Naeini“Feedback Control of Dynamic Systems”7th ed. Pearson

Control Equations


Overview

Observability

Sensor Fusion

Robustness

Stability



PID Control

Rise Time / Overshoot / Settling

Effect of Poles and Zeros



𝐻 𝑠 =𝑎𝑛𝑠

𝑛 + 𝑎𝑛−1𝑠𝑛−1 + …+ 𝑎1𝑠 + 𝑎0

𝑎𝑚𝑠𝑚 + 𝑏𝑚−1𝑠𝑚−1 + …+ 𝑏1𝑠 + 𝑏0

= 𝐾 𝑖=1

𝑛 𝑠 + 𝑧𝑖

𝑗=1𝑚 𝑠 + 𝑝𝑗

Gain

Transfer function is Laplace transform of impulse responsewe can directly compute the impulse response from the

poles and zeroes of the transfer function call impulse response the natural response of the system

Modes


Effect of Real-Valued Poles

𝐻 𝑠 =1

(𝑠 + 𝜎)

ℎ 𝑡 = 𝑒−𝜎𝑡1(𝑡)

1.0

1.0

1

𝑒

impulse response

step response

𝐻 𝑠 =2𝑠 + 1

𝑠2 + 3𝑠 + 2

= 2𝑠 +

12

𝑠 + 1 𝑠 + 2

partial fraction expansion

xx o

Im(s)

Re(s)1-1-2


Partial Fraction Expansion

real-valued function where 𝑄 𝑥 = 𝑥 − 𝑥0 … 𝑥 − 𝑥𝑚 , degree 𝑃 𝑥 < 𝑚

• for each 𝑟𝑖-fold real pole 𝑥 − 𝑥𝑖𝑟𝑖, add the following polynomial to 𝐴 𝑥 , using fresh constants:

𝑅 𝑥 =𝑃(𝑥)

𝑄(𝑥)

𝑎𝑖1𝑥 − 𝑥𝑖

+𝑎𝑖2

𝑥 − 𝑥𝑖2+ …+

𝑎𝑖𝑟𝑖𝑥 − 𝑥𝑖

𝑟𝑖

Suppose 𝑅 𝑥 = 𝐴 𝑥 where 𝐴 𝑥 is constructed as follows:

• because 𝑅(𝑥) is a real-valued function, both 𝑧𝑖 and 𝑧𝑖 appear as poles;for each 𝑠𝑖-fold complex pole 𝑥 − 𝑧𝑖

𝑠𝑖 , add the following to 𝐴 𝑥 :

𝑏𝑖1𝑥 + 𝑐𝑖1𝑥2 + 𝑝𝑖𝑥 + 𝑞𝑖

+𝑏𝑖2𝑥 + 𝑐𝑖2

𝑥2 + 𝑝𝑖𝑥 + 𝑞𝑖2+ …+

𝑏𝑖𝑠𝑖𝑥 + 𝑐𝑖𝑠𝑖𝑥2 + 𝑝𝑖𝑥 + 𝑞𝑖

𝑠𝑖

where 𝑥2+𝑝𝑖𝑥 + 𝑞𝑖 = 𝑥 − 𝑧𝑖 𝑥 − 𝑧𝑖

• solve by comparison of the coefficients



𝐻 𝑠 =1

(𝑠 + 𝜎)


1.0

1.0

1

𝑒

impulse response

step response

𝐻 𝑠 =2𝑠 + 1

𝑠2 + 3𝑠 + 2

= 2𝑠 +

12

𝑠 + 1 𝑠 + 2

𝐻(𝑠) = −1

𝑠 + 1+

3

𝑠 + 2


𝐻 𝑠 = 2𝑠 +

12

𝑠 + 1 𝑠 + 2=

𝑎

𝑠 + 1+

𝑏

𝑠 + 2

2𝑠 + 1 = 𝑎𝑠 + 2𝑎 + 𝑏𝑠 + 𝑏

2𝑠 = 𝑎𝑠 + 𝑏𝑠 1 = 2𝑎 + 𝑏

𝑎 = −1, 𝑏 = 3

ℎ 𝑡 = −𝑒−𝑡 + 3𝑒−2𝑡 for 𝑡 ≥ 0



𝐻 𝑠 =1

(𝑠 + 𝜎)


1.0

1.0

1

𝑒

impulse response

step response

𝐻 𝑠 =2𝑠 + 1

𝑠2 + 3𝑠 + 2

= 2𝑠 +

12

𝑠 + 1 𝑠 + 2

𝐻(𝑠) = −1

𝑠 + 1+

3

𝑠 + 2


ℎ 𝑡 = −𝑒−𝑡 + 3𝑒−2𝑡 𝑓𝑜𝑟 𝑡 ≥ 00 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

decays faster than 𝑒−𝑡

=> say: “Pole is faster”

impulse step


Effect of Compex Poles

𝐻 𝑠 =𝜔𝑛

2

𝑠2 + 2𝜉𝜔𝑛𝑠 + 𝜔𝑛2

Because H(s) is real valued, complex poles always occur as conjugate pairs 𝑝𝑖 and 𝑝𝑖

Characterize complex poles in terms of their real and imaginary parts: 𝑠 = −𝜎 ± 𝑗𝜔𝑑

denominator: 𝑏(𝑠) = (𝑠 + 𝜎 − 𝑗𝜔𝑑) (𝑠 + 𝜎 + 𝑗𝜔𝑑) in 𝐻 𝑠 =𝑎(𝑠)

𝑏(𝑠)

= 𝑠 + 𝜎 2 + 𝜔𝑑2

= 𝑠2 + 2𝜎𝑠 + 𝜎2 + 𝜔𝑑2

Write result when finding transfer function of second order differential equation as:

𝜎 = 𝜉𝜔𝑛 𝜔𝑑 = 𝜔𝑛 1 − 𝜉2

damping ratio undamped natural frequency


Effect of Compex Poles

𝐻 𝑠 =𝜔𝑛

2


(𝑠 + 𝜎 − 𝑗𝜔𝑑) (𝑠 + 𝜎 + 𝑗𝜔𝑑)


x

x

𝜔𝑛

𝜃 = sin−1 𝜉

𝜎

𝜔𝑑

=𝜔𝑛

2

(𝑠 + 𝜉𝜔𝑛)2+𝜔𝑛

2(1 − 𝜉2)

ℎ 𝑡 =𝜔𝑛

1 − 𝜉2𝑒−𝜎𝑡 sin𝜔𝑑𝑡 1(𝑡)

𝑒−𝜎𝑡

−𝑒−𝜎𝑡


Effect of Zeros

𝐻 𝑠 =𝑠 + 𝑎

𝑏(𝑠)where a, is not a pole of 𝑏(𝑠)

plant𝑢(𝑡)

𝐻 𝑠ℒ 𝑈(𝑠) ∙

𝑢 𝑡 = 𝑒−𝑎𝑡 ℒ 𝑒−𝑎𝑡 =1

𝑠+𝑎𝑌 𝑠 =

1

(𝑠 + 𝑎)

𝑠 + 𝑎

𝑏(𝑠)=

1

𝑏(𝑠)

Zeros cancel frequencies; they are transmission blocking(unless there is also a pole)


Effect of Real-Valued Zeros

𝐻1 𝑠 =2

(𝑠 + 1)(𝑠 + 2)

=2

(𝑠 + 1)−

2

(𝑠 + 2)

𝐻1 𝑠 =2(𝑠 + 1.1)

1.1(𝑠 + 1)(𝑠 + 2)

=2

1.1

0.1

(𝑠 + 1)+

0.9

(𝑠 + 2)=

0.18

(𝑠 + 1)+

1.64

(𝑠 + 2)

xx o-1-2

Zeros at pole location cancel the mode defined by the pole.Zeros near a pole damp the effect.

impulse

step

impulse

step


Effect of Complex Zeros

𝐻 𝑠 =

1𝑎𝜉

𝑠 + 1

𝑠2 + 2𝜉𝑠 + 1Pair of complex poles at 𝜔𝑛 = 1, 𝜃 = sin−1 𝜉

e.g., 𝜉 = 0.5 and 𝜃 = 30°

x

x

-1

𝐻 𝑠 =1

𝑎𝜉

𝑠 + 𝑎𝜉

𝑠2 + 2𝜉𝑠 + 1

o

𝑎 = 1

𝑎 = 2𝑎 = 4

Nearby zeros cause a faster rise at the cost of higher overshoot.

Im(s)

Re(s)


Effect of Complex Zeros

𝐻 𝑠 =

1𝑎𝜉

𝑠 + 1

𝑠2 + 2𝜉𝑠 + 1

=1

𝑠2 + 2𝜉𝑠 + 1+

1

𝑎𝜉

𝑠

𝑠2 + 2𝜉𝑠 + 1Recall: ℒ

𝜕𝑦

𝜕𝑡= 𝑠𝐻 𝑠 , where ℒ 𝑦 = 𝐻 𝑠

= 𝐻0(𝑠) +1

𝑎𝜉𝑠𝐻0(𝑠)

𝑦(𝑡) = 𝑦0(𝑡) +1

𝑎𝜉

𝜕𝑦0𝜕𝑡

𝑦(𝑡)

𝑦0(𝑡)

1

𝑎𝜉

𝜕𝑦0𝜕𝑡


Time Domain Specification

0,9

0,1

settling time: 𝑡𝑠time after which signal remains within epsilon distance around reference

peak time: 𝑡𝑝 - time of maximum overshoot

overshoot: 𝑀𝑝 - maximum over reference signal

rise time: 𝑡𝑟 - time to reach vicinity of settling point

Second order system

rise time 𝜔𝑛 ≥1.8

𝑡𝑟

overshoot 𝑀𝑝 = 𝑒

−𝜋𝜉

1−𝜉2

=> 𝜉 ≥ 𝜉(𝑀𝑝)

settling time 𝜎 ≥4.6

𝑡𝑠(decay below 1%)

𝐻 𝑠 =𝜔𝑛

2




Time Domain Specification

Second order system

rise time 𝜔𝑛 ≥1.8

𝑡𝑟

overshoot 𝑀𝑝 = 𝑒

−𝜋𝜉

1−𝜉2

=> 𝜉 ≥ 𝜉(𝑀𝑝)

settling time 𝜎 ≥4.6

𝑡𝑠(decay below 1%)

𝐻 𝑠 =𝜔𝑛

2



Im(s)

Re(s)

Im(s)

Re(s)

Im(s)

Re(s)x

x

x

sin−1 𝜉

𝜔𝑛

𝜎

rise time overshoot settling timeIm(s)

Re(s)

x

x

x

x

So far:

• by adding zeros,

• we can block / damp certain frequencies

• we can eliminate / damp modes if the zeros are near poles

• we can influence rise time at cost of overshoot

• by adding roots,

• we can add sinusoids and exponentials to the response

• we can influence overshoot, settling time and rise time


Control

Control: add roots and zeros to obtain desired plant behavior.


Exercise: Tracking and Pole Placement

Compute the coefficients 𝑐𝑖 , 𝑑𝑖 such that the closed loop has the characteristic equation 𝑠 + 6 𝑠 + 3 𝑠2 + 3𝑠 + 9 = 0.

(characteristic equation is the denominator of the closed-loop transfer function set to zero)

Controller𝑐2𝑠

2 + 𝑐1𝑠 + 𝑐0𝑠(𝑠 + 𝑑1)

Plant1

𝑠2 + 3𝑠 + 9

𝑅(𝑠) 𝑌(𝑠)Σ+

+

𝑊(𝑠)

disturbance

reference signal output

𝑈(𝑠)Σ

Σ

+

-

++

𝑉(𝑠)

sensor noise

𝐸(𝑠)

𝑌(𝑠)

𝑅(𝑠)= 𝑇(𝑠) = 𝐺𝐷𝑆 =

𝐺𝐷

1 + 𝐺𝐷

1 + 𝐺𝐷 = 0

1 +𝑐2𝑠

2 + 𝑐1𝑠 + 𝑐0𝑠(𝑠 + 𝑑1)

1

𝑠2 + 3𝑠 + 9= 0


Exercise: Tracking and Pole Placement

𝑠 + 6 𝑠 + 3 𝑠2 + 3𝑠 + 9 = 0

Controller𝑐2𝑠

2 + 𝑐1𝑠 + 𝑐0𝑠(𝑠 + 𝑑1)

Plant1

𝑠2 + 3𝑠 + 9


+

𝑊(𝑠)

disturbance


𝑈(𝑠)Σ

Σ

+

-

++

𝑉(𝑠)

sensor noise

𝐸(𝑠)

1 +𝑐2𝑠

2 + 𝑐1𝑠 + 𝑐0𝑠(𝑠 + 𝑑1)

1

𝑠2 + 3𝑠 + 9= 0

𝑠 𝑠 + 𝑑1 𝑠2 + 3𝑠 + 9 + 𝑐2𝑠2 + 𝑐1𝑠 + 𝑐0 = 0

𝑠4 + 12𝑠3 + 54𝑠2 + 135 𝑠 + 162 = 0

𝑠4 + (𝑑1 +3)𝑠3 + 𝑐2 + 3𝑑1 + 9 𝑠2 + 9𝑑1 + 𝑐1 𝑠 + 𝑐0 = 0

𝑑1 + 3 = 12

𝑐2 + 3𝑑1 + 9 = 54

9𝑑1 + 𝑐1 = 135

𝑐0 = 162

𝑑1 = 9, 𝑐2 = 18, 𝑐1 = 54, 𝑐0 = 162


PID Control

Controller𝐷𝑐𝑙𝑜𝑠𝑒𝑑(𝑠)

Plant𝐺(𝑠)


+

𝑊(𝑠)

disturbance


𝑈(𝑠)Σ

Σ

+

-

++

𝑉(𝑠)

sensor noise

𝑌 = 𝑇𝑅 = 𝐺𝐷𝑆𝑅 = 𝐺𝐷1

1 + 𝐺𝐷R

𝐸 = 𝑅 − 𝑌 = 𝑆𝑅 =1

1 + 𝐺𝐷𝑅

P: feedback proportional to error 𝑢 𝑡 = 𝑘𝑝𝑒(𝑡)

𝐷𝑐𝑙𝑜𝑠𝑒𝑑 𝑠 =𝑈(𝑠)

𝐸(𝑠)= 𝑘𝑝

purely algebraic; no dynamics proportional gain

𝐸(𝑠)


PID Control

Controller𝑘𝑝

Plant𝐺(𝑠)


+

𝑊(𝑠)

disturbance


𝑈(𝑠)Σ

Σ

+

-

++

𝑉(𝑠)

sensor noise

𝑌(𝑠) =𝑘𝑝𝐺(𝑠)

1 + 𝑘𝑝𝐺(𝑠)R(s)

P: feedback proportional to error 𝑢 𝑡 = 𝑘𝑝𝑒(𝑡)

e.g. second order plant: 𝐺 𝑠 =𝐴

𝑠2+𝑎1𝑠+ 𝑎2

𝐸(𝑠)

characteristic equation: 1 + 𝑘𝑝𝐺 𝑠 = 0

𝑠2 + 𝑎1𝑠 + 𝑎2 + 𝑘𝑝𝐴 = 0

control natural frequencyno control over damping

𝑎1 = 1.4, 𝑎2 = 𝐴 = 1

𝑘𝑝 = 1.5

𝑘𝑝 = 6

improve tracking and rise time,but decrease damping (2nd order only)

higher order: increase damping for some poles; increase for others


PID Control

Controller𝑘𝐼𝑠

Plant𝐺(𝑠)


+

𝑊(𝑠)

disturbance


𝑈(𝑠)Σ

Σ

+

-

++

𝑉(𝑠)

sensor noise

I: integral feedback 𝑢 𝑡 = 𝑘𝑖 𝑡0𝑡𝑒(𝑡)𝜕𝑡

𝐸(𝑠)

Infinite value of control with zero system errorConsider all past values

=> cancel constant disturbances zero error


𝐸(𝑠)=

𝑘𝐼𝑠

integral gain𝑎1 = 1.4, 𝑎2 = 𝐴 = 1, 𝑘𝐼 = 0.5


PID Control

Plant𝐺(𝑠)


+

𝑊(𝑠)

disturbance


𝑈(𝑠)Σ

Σ

+

-

++

𝑉(𝑠)

sensor noise

𝐸(𝑠)

𝐸(𝑠)

𝑅(𝑠)=

1

1 +𝑘𝐼𝑠 𝐺(𝑠)

=𝑠

𝑠 + 𝑘𝐼𝐺(𝑠)

𝑈(𝑠)

𝑅(𝑠)=

𝑘𝐼𝑠


=𝑘𝐼


𝑌(𝑠)

𝑅(𝑠)=

𝑘𝐼𝑠 𝐺(𝑠)


=𝑘𝐼𝐺(𝑠)


Controller𝑘𝐼𝑠

error

control

response

𝑒 ∞ =0

0 + 𝑘𝐼𝐺(0)= 0

𝑢 ∞ =𝑘𝐼

0 + 𝑘𝐼𝐺(0)= 𝐺−1 0 = 1

𝑦 ∞ =𝑘𝐼𝐺(0)

0 + 𝑘𝐼𝐺(0)= 1

𝑟 𝑡 = 1 𝑡

𝑅 𝑠 =1

𝑠

final value thm.

2nd order 𝑎1 = 𝐴 = 0

Integral control is robustagainst plant variation


PID Control

Plant𝐺(𝑠)


+

𝑊(𝑠)

disturbance


𝑈(𝑠)Σ

Σ

+

-

++

𝑉(𝑠)

sensor noise

𝐸(𝑠) Controller…+ 𝑘𝐷 𝑠

D: derivative feedback 𝑢 𝑡 = …+ 𝑘𝐷𝜕𝑒(𝑡)

𝜕𝑡


𝐸(𝑠)= ⋯+ 𝑘𝐷𝑠

Derivative supplies no information about desired end state=> use in combination with other controllers (e.g., P, PI)

Add derivative control for increased stability


PID Control

Plant𝐺(𝑠)


+

𝑊(𝑠)

disturbance


𝑈(𝑠)Σ

Σ

+

-

++

𝑉(𝑠)

sensor noise

𝐸(𝑠) Controller

𝑘𝑝 +𝑘𝐼𝑠+ 𝑘𝐷𝑠

PID: derivative feedback 𝑢 𝑡 = 𝑘𝑝𝑒 𝑡 + 𝑘𝐼 𝑡0𝑡𝑒(𝜏)𝜕𝜏 + 𝑘𝐷

𝜕𝑒(𝑡)

𝜕𝑡


𝐸(𝑠)= 𝑘𝑝 +

𝑘𝐼𝑠+ 𝑘𝐷𝑠

𝐽𝑚 = 1.13 × 10−2 𝑁 ∙ 𝑚 ∙ 𝑠𝑒𝑐2/𝑟𝑎𝑑

DC Motor: Θ𝑚(𝑠)

𝑉𝑎(𝑠)=

𝐾

𝑠(𝜏𝑠+1)with 𝐾 =

𝐾𝑡

𝑏𝑅𝑎+𝐾𝑡𝐾𝑒, 𝜏 =

𝑅𝑎𝐽𝑚

𝑏𝑅𝑎+𝐾𝑡𝐾𝑒

𝑏 = 0.028 𝑁 ∙ 𝑚 ∙ 𝑠𝑒𝑐/𝑟𝑎𝑑

𝑅𝑎 = 0.45Ω

𝐾𝑡 = 0.067 𝑁 ∙ 𝑚/amp 𝐾𝑒 = 0.067 𝑉 ∙ 𝑠𝑒𝑐/rad


PID Control

P

PI

PID

step input step disturbance𝑘𝑝 = 3

𝑘𝐼 = 9

𝑘𝐷 = 0.3


Summary

Observability

Sensor Fusion

Robustness

Stability



PID Control

Rise Time / Overshoot / Settling



Leftovers

Stability• How to derive stability from location of zeros and poles in s-plane?• How to move zeros and poles by adding control?

Sensitivity• Feedback control systems are much less error prone

to plants changing their gain as open loop systems.


Overview

Math

FeedbackControl

RTOS

Verification

Physics

Introduction

Mathematical Foundations (Differential Equations and Laplace Transformation)

Control and Feedback

Transfer Functions and State Space Models

Poles, Zeros / PID Control

Stability, Root Locust Method, Digital Control

Mixed-Criticality Scheduling and Real-Time Operating Systems (RTOS)

Coordinating Networked Cyber-Physical Systems

Program Verification

Differential Dynamic Logic and KeYmaera X

Differential Invariants

CPS

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