8.11.2019

RTI Vorlesung 8

Recap

2

BIBO

Stability

Poles

Transfer

Functions

2nd-Order

System

Zeros

Static gain

Transfer function

3

Y 𝑠 ⋅ 𝑠2 + 𝑎1 ⋅ 𝑠 + 𝑎0𝑎 𝑠

= 𝑏1 ⋅ 𝑠 + 𝑏0𝑏 𝑠

⋅ 𝑈 𝑠

Origin: differential equation: 𝑦′′ + 𝑎1 ⋅ 𝑦′ + 𝑎0 ⋅ 𝑦 = 𝑏1 ⋅ 𝑢

′ + 𝑏0 ⋅ 𝑢

Y 𝑠 ⋅ 𝑠2 +𝑎1 ⋅ Y 𝑠 ⋅ 𝑠 + 𝑎0 ⋅ Y 𝑠 = 𝑏1 ⋅ 𝑈 𝑠 ⋅ 𝑠 + 𝑏0 ⋅ 𝑈 𝑠

Y 𝑠 =𝑏(𝑠)

𝑎(𝑠)⋅ 𝑈 𝑠

The transfer function is on operator which maps the input 𝑈 𝑠 to the output 𝑌(𝑠) in

the frequency domain

Y 𝑠 = 𝛴(s) ⋅ 𝑈 𝑠

Poles

4

• A pole 𝜋 is a frequency 𝑠 at which Σ 𝑠 is infinite.

• The poles describe the inherent dynamics of the plant.

• The poles are relevant for the stability of the system

The poles are the roots of the denominator polynomial 𝑎(𝑠).

Y 𝑠 ⋅ 𝑠2 + 𝑎1 ⋅ 𝑠 + 𝑎0𝑎 𝑠

= 𝑏1 ⋅ 𝑠 + 𝑏0𝑏 𝑠

⋅ 𝑈 𝑠

Origin: differential equation: 𝑦′′ + 𝑎1 ⋅ 𝑦′ + 𝑎0 ⋅ 𝑦 = 𝑏1 ⋅ 𝑢

′ + 𝑏0 ⋅ 𝑢

Y 𝑠 ⋅ 𝑠2 +𝑎1 ⋅ Y 𝑠 ⋅ 𝑠 + 𝑎0 ⋅ Y 𝑠 = 𝑏1 ⋅ 𝑈 𝑠 ⋅ 𝑠 + 𝑏0 ⋅ 𝑈 𝑠

Zeros

5

• A zero 𝜁 is a frequency 𝑠 at which the transfer function Σ 𝑠 is zero.

• This zero describes the nontrivial dynamics of the system, which for a given initial state

and a given input yield an output of 𝑦 𝑡 = 0 for all times 𝑡 ≥ 0.

The zeros are the roots of the numerator polynomial 𝑏(𝑠).

Y 𝑠 ⋅ 𝑠2 + 𝑎1 ⋅ 𝑠 + 𝑎0𝑎 𝑠

= 𝑏1 ⋅ 𝑠 + 𝑏0𝑏 𝑠

⋅ 𝑈 𝑠

Origin: differential equation: 𝑦′′ + 𝑎1 ⋅ 𝑦′ + 𝑎0 ⋅ 𝑦 = 𝑏1 ⋅ 𝑢

′ + 𝑏0 ⋅ 𝑢

Y 𝑠 ⋅ 𝑠2 +𝑎1 ⋅ Y 𝑠 ⋅ 𝑠 + 𝑎0 ⋅ Y 𝑠 = 𝑏1 ⋅ 𝑈 𝑠 ⋅ 𝑠 + 𝑏0 ⋅ 𝑈 𝑠

Static gain

6

The static gain Σ 0 is the value of the transfer function at 𝑠 = 0: Σ 0 =𝑏0

𝑎0

• The static gain Σ 0 is the asymptotic value of lim𝑡→∞

𝑦(𝑡) in response to a step at the

input 𝑢 𝑡 = ℎ(𝑡).

Y 𝑠 ⋅ 𝑠2 + 𝑎1 ⋅ 𝑠 + 𝑎0𝑎 𝑠

= 𝑏1 ⋅ 𝑠 + 𝑏0𝑏 𝑠

⋅ 𝑈 𝑠

Origin: differential equation: 𝑦′′ + 𝑎1 ⋅ 𝑦′ + 𝑎0 ⋅ 𝑦 = 𝑏1 ⋅ 𝑢

′ + 𝑏0 ⋅ 𝑢

Y 𝑠 ⋅ 𝑠2 +𝑎1 ⋅ Y 𝑠 ⋅ 𝑠 + 𝑎0 ⋅ Y 𝑠 = 𝑏1 ⋅ 𝑈 𝑠 ⋅ 𝑠 + 𝑏0 ⋅ 𝑈 𝑠

BIBO Stability

7

Σ(𝑠)𝑢(𝑡) 𝑦(𝑡)

A system is called bounded-input bounded-output (BIBO) stable iff all

• finite inputs 𝑢 𝑡 < 𝑚 result in

• finite outputs 𝑦 𝑡 < 𝑀

Theorem 1: For linear systems, BIBO stability is given if the following holds

Theorem 2: a system Σ(𝑠) is

• BIBO stable iff all poles 𝜋 have negative real parts, and

• not BIBO stable in all other cases.

𝑢(𝑡) 𝑦(𝑡)

𝜎 𝑡 = Impulse response

2nd Order Systems:

Poles and the Time-Domain Behavior

8

Location of the poles

in the complex plane

Step response

in the time domain

Zeros and the Time-Domain Behavior

9

𝛿 = 0.5

𝛴 𝑠 =−1/𝜁 ⋅ 𝑠 + 1 ⋅ 𝜔0

2

𝑠2 + 2 ⋅ 𝛿 ⋅ 𝜔0 ⋅ 𝑠 + 𝜔02

20.09. Lektion 1 – Einführung

27.09. Lektion 2 – Modellbildung

4.10. Lektion 3 – Systemdarstellung, Normierung, Linearisierung

11.10. Lektion 4 – Analyse I, allg. Lösung, Systeme erster Ordnung, Stabilität

18.10. Lektion 5 – Analyse II, Zustandsraum, Steuerbarkeit/Beobachtbarkeit

25.10. Lektion 6 – Laplace I, Übertragungsfunktionen

1.11. Lektion 7 – Laplace II, Lösung, Pole/Nullstellen, BIBO-Stabilität

8.11. Lektion 8 – Frequenzgänge (RH hält VL)

15.11. Lektion 9 – Systemidentifikation, Modellunsicherheiten

22.11. Lektion 10 – Analyse geschlossener Regelkreise

29.11. Lektion 11 – Randbedingungen

6.12. Lektion 12 – Spezifikationen geregelter Systeme

13.12. Lektion 13 – Reglerentwurf I, PID (RH hält VL)

20.12. Lektion 14 – Reglerentwurf II, „loop shaping“

Modellierung

Systemanalyse im Zeitbereich

Systemanalyse im Frequenzbereich

Reglerauslegung

10

Today: all about one property of asymptotically stable LTI transfer functions 𝛴(𝑠)

12

The mapping 𝛴 𝑗𝜔 ∶ ℝ → ℂ is the frequency response of the system 𝛴(𝑠)

Definition of frequency response (Frequenzgang)

Time Domain Frequency Domain

Given: 𝛴(𝑠)

𝛴 𝑗𝜔 = 𝛼2 + 𝛽2

∠𝛴(𝑗𝜔) = arctan −𝛽

𝛼

𝛴 𝑗𝜔 = 𝛼 − 𝑗𝛽, 𝛼, 𝛽 ∈ ℝ

<

Im

Re

Σ(𝑗ෝ𝜔)

∠𝛴(𝑗ෝ𝜔)

𝛼(𝑗ෝ𝜔)

−𝛽(𝑗ෝ𝜔)

Proof of 𝑦∞(𝑡) = 𝑚 ⋅ cos(𝜔 ⋅ 𝑡 + 𝜑)

13

with

For any asymptotically stable LTI system Σ(𝑠):Laplace Table

𝑥(𝑡) 𝑋(𝑠)

ℎ 𝑡 ⋅ cos(𝜔 ⋅ 𝑡)𝑠

𝑠2 + 𝜔2

ℎ 𝑡 ⋅ sin(𝜔 ⋅ 𝑡)𝜔

𝑠2 + 𝜔2

ℎ 𝑡 ⋅ 𝑡𝑛 ⋅ 𝑒𝛼⋅𝑡𝑛!

𝑠 − 𝛼 𝑛+1

How to get frequency response experimentally

14

• Excite an asymptotically stable LTI

system with a harmonic input

signal of a specific frequency

• Measure the steady-state response

• Compare the amplitude and the

phase of the response with the

amplitude and the phase of the input

signal; save results for chosen

frequency

• Repeat the experiment for various

other excitation frequencies

Example of a Frequency Response

𝑢(𝑡)

𝑦(𝑡)

Linear system Σ 𝑠 :

damped single mass oscillator,

two states → 2nd order system

𝑚𝜔

[−]

Frequency 𝜔𝜑𝜔

[deg]

Frequency 𝜔

General representations of frequency responses

16

The frequency response can be displayed by two different diagrams.

• The frequency-explicit Bode diagram with two separate curves

• The frequency-implicit Nyquist diagram.

«Bode» diagram Nyquist diagram

Real component

Ima

gin

ary

co

mp

on

ent

𝑚𝜔

[−]

𝜑𝜔

[deg]

17

The mapping 𝛴 𝑗𝜔 ∶ ℝ → ℂ is the frequency response of the system 𝛴(𝑠)

Definition of frequency response (Frequenzgang)

Time Domain Frequency Domain

Given: 𝛴(𝑠)

𝛴 𝑗𝜔 = 𝛼2 + 𝛽2

∠𝛴(𝑗𝜔) = arctan −𝛽

𝛼

𝛴 𝑗𝜔 = 𝛼 − 𝑗𝛽, 𝛼, 𝛽 ∈ ℝ

<

Im

Re

Σ(𝑗ෝ𝜔)

∠𝛴(𝑗ෝ𝜔)

𝛼(𝑗ෝ𝜔)

−𝛽(𝑗ෝ𝜔)

18

Time

Domain

Frequency

Domain

𝑡

𝑠 = 𝑎 + 𝑗𝑏

Σ 𝑠 ⋅𝑠

𝑠2 + 𝜔2= 𝑌t 𝑠 +

𝛼 ⋅ 𝑠 + 𝛽 ⋅ 𝜔

s2 + 𝜔2=𝛼 ⋅ 𝑠 + 𝛽 ⋅ 𝜔

s2 + 𝜔2

Σ 𝑠 ⋅𝑠

𝑠2 + 𝜔2= 𝑌t 𝑠 + 𝑌∞ 𝑠 = 𝑌∞(𝑠)

⇒ for lim𝑡→∞

, we must have lim𝑠→𝑗𝜔

:

⇒ Σ 𝑠 ⋅ 𝑠 = (𝑠2 + 𝜔2) ⋅ 𝑌t 𝑠 + 𝛼 ⋅ 𝑠 + 𝛽 ⋅ 𝜔 = (𝛼 ⋅ 𝑠 + 𝛽 ⋅ 𝜔)

⇒ Σ 𝑗𝜔 ⋅ 𝑗𝜔 = 𝛼 ⋅ 𝑗𝜔 + 𝛽 ⋅ 𝜔

How to get frequency response from 𝛴(𝑠) (I)aka “proof: lim

𝑡→∞⇒ lim

𝑠→𝑗𝜔”

𝑌 𝑠 = 𝑌∞ 𝑠 + 𝑌𝑡(𝑠)𝑦 𝑡 = 𝑦∞ 𝑡 + 𝑦𝑡 𝑡

lim𝑡→∞

𝑡 ∈ [0,∞] 𝑠 = 𝑎 + 𝑗 ⋅ 𝑏

𝑦 𝑡 = 𝑦∞ 𝑡 𝑌 𝑠 = 𝑌∞(𝑠)lim𝑠→?

Free variableFree variable

19

Re(Σ 𝑗𝜔 )

Im(Σ 𝑗𝜔 )

How to get frequency response from 𝛴(𝑠) (II)determine 𝑚 and 𝜑

20

The mapping 𝛴 𝑗𝜔 ∶ ℝ → ℂ is the frequency response of the system 𝛴(𝑠)

Definition of frequency response (Frequenzgang)

Time Domain Frequency Domain

Given: 𝛴(𝑠)

𝛴 𝑗𝜔 = 𝛼2 + 𝛽2

∠𝛴(𝑗𝜔) = arctan −𝛽

𝛼

𝛴 𝑗𝜔 = 𝛼 − 𝑗𝛽, 𝛼, 𝛽 ∈ ℝ

<

Im

Re

Σ(𝑗ෝ𝜔)

∠𝛴(𝑗ෝ𝜔)

𝛼(𝑗ෝ𝜔)

−𝛽(𝑗ෝ𝜔)

20.09. Lektion 1 – Einführung

27.09. Lektion 2 – Modellbildung

4.10. Lektion 3 – Systemdarstellung, Normierung, Linearisierung

11.10. Lektion 4 – Analyse I, allg. Lösung, Systeme erster Ordnung, Stabilität

18.10. Lektion 5 – Analyse II, Zustandsraum, Steuerbarkeit/Beobachtbarkeit

25.10. Lektion 6 – Laplace I, Übertragungsfunktionen

1.11. Lektion 7 – Laplace II, Lösung, Pole/Nullstellen, BIBO-Stabilität

8.11. Lektion 8 – Frequenzgänge (RH hält VL)

15.11. Lektion 9 – Systemidentifikation, Modellunsicherheiten

22.11. Lektion 10 – Analyse geschlossener Regelkreise

29.11. Lektion 11 – Randbedingungen

6.12. Lektion 12 – Spezifikationen geregelter Systeme

13.12. Lektion 13 – Reglerentwurf I, PID (RH hält VL)

20.12. Lektion 14 – Reglerentwurf II, „loop shaping“

Modellierung

Systemanalyse im Zeitbereich

Systemanalyse im Frequenzbereich

Reglerauslegung

21

Application of frequency responses

22

Development of nonparametric and parametric

models of the system Σ(𝑠) via measurements

of the frequency response («experiments»).

Lecture 9

Modeling

Controller Design

Using the Nyquist theorem to assess the stability of a closed-loop

system based on the frequency response of the open-loop system.

Lecture 10, 11, …|Σ

𝑗𝜔|[−]

∠Σ𝑗𝜔

[deg]

23

Examples of frequency responses

Bode diagrams of 1st order and 2nd systems

Bode Diagram Convention

24

“Bode diagram” so far Conventional Bode Diagram

• Plot phase in degrees

• Plot frequency axis in log10 base

• Plot magnitude in dB (decibel) – also log10 base

Decibel Scale

25

Conversion Table

Decibel Scale:Decimal Scale Decibel Scale

100 40

10 20

5 13.97…

2 6.02…

1 0

0.1 -20

0.01 -40

0 -Inf

Advantages of log10 and dB (I)

26

• Curved lines become straight asymptotes• Slope of asymptotes change depending on poles/zeros location of Σ(𝑠)

• Poles and zeros of transfer function define “edges”• ⇒ In simple cases, transfer function can be read directly from Bode diagram

“Bode diagram” so far Conventional Bode Diagram

Rules for drawing a Bode diagram

27

TypeMagnitude

Change

Phase

Change

Stable pole @𝜔𝜋 -20 dB/dec -90°

Unstable pole @𝜔𝜋 -20 dB/dec +90°

Minimumphase zero @𝜔𝜁 +20 dB/dec +90°

Non-minimumphase zero @𝜔𝜁 +20 dB/dec -90°

Time delay 0 dB/dec −𝜔 ⋅ 𝑇

Rules for drawing a Bode diagram

𝜔𝜋 = 𝜋 in 𝑟𝑎𝑑

𝑠

𝜔𝜁 = 𝜁 in 𝑟𝑎𝑑

𝑠

Example: reading Σ 𝑠 from bode plot

28

TypeMagnitude

Change

Phase

Change

Stable pole @𝜔𝜋 -20 dB/dec -90°

Rules for drawing a Bode diagram

⇒ 𝜔𝜋1,2= 𝜋1,2 = 𝜔0

⋅ 𝑘

General Bode diagram of a 2nd order system

29

Example: with 𝑘 = 1

𝜔0

𝜔10 ⋅ 𝜔0

𝜔10 ⋅ 𝜔0

0.1 ⋅ 𝜔0

𝑚 𝜔

𝜑(𝜔)

Static gain

Magnitude change

Phase change

Cut-off frequency

30

Calculate the frequency response of

𝜔

|Σ(𝑗𝜔)|

𝜔∠Σ(𝑗𝜔)

Σ 𝑗𝜔 =𝑎 𝑗𝜔

𝑏 𝑗𝜔

∠Σ 𝑗𝜔 = ∠𝑎 𝑗𝜔 − ∠𝑏 𝑗𝜔

Σ 𝑗𝜔 =𝑎 𝑗𝜔

𝑏 𝑗𝜔

Drawing frequency response

Example: 1st order system (II)

Σ 𝑠 =1

𝑠 + 1

Bode diagram of a 1st order system

31

Example: with 𝑘 = 1, 𝜏 = 1

Static gain

Cut-off frequency

Magnitude change

Phase change

TypeMagnitude

Change

Phase

Change

Stable pole @𝜔𝜋 -20 dB/dec -90°

Rules for drawing a Bode diagram

Advantages of log10 and dB (II)

32

• Multiplication of magnitudes becomes addition

Σtot = Σ1 ⋅ |Σ2|

20 ⋅ log10( Σtot ) = 20 ⋅ log10( Σ1 ⋅ Σ1 )

= 20 ⋅ log10 Σ1 + 20 ⋅ log10( Σ2 )

Σ1 𝑗𝜔 = 𝛴1(𝑗𝜔) ⋅ 𝑒𝑗𝜑1(𝑗𝜔)

Σ2 𝑗𝜔 = 𝛴2(𝑗𝜔) ⋅ 𝑒𝑗𝜑2(𝑗𝜔)Σtot 𝑗𝜔 = Σ1 𝑗𝜔 ⋅ Σ2 𝑗𝜔

Sidenote: for phase we have addition property anyway!

⇒ Σtot dB = Σ1 dB + Σ2 dB

∠(Σtot) = ∠(Σ1 + Σ2) = ∠(Σ1) + ∠(Σ2)

Bode Diagram of higher order system

33

Example: Σ 𝑠 = 0.1 ⋅𝑠 + 10

𝑠 + 0.5

Decimal

Scale

Decibel

Scale

2 6.02…

1 0

TypeMagnitude

Change

Phase

Change

Stable pole @𝜔𝜋 -20 dB/dec -90°

Minimumphase

zero @𝜔𝜁

+20 dB/dec +90°

Conversion Table

Rules for drawing a Bode diagram

34

Examples of frequency responses

1st order and 2nd nyquist diagrams

Drawing frequency response

Example: 1st order system (I)

35

Calculate the frequency response of

Re Σ(𝑗𝜔)

Im Σ(𝑗𝜔)

Σ 𝑠 =1

𝑠 + 1

Nyquist diagram of a 1st order system

36

Example: with 𝑘 = 1, 𝜏 = 1 Static gain

Cut-off frequency

Magnitude change

Phase change

Nyquist diagram of a 2nd order

system

37

Example: with 𝑘 = 1

Library of Standard Elements

38

See Appendix A in the script

Why is cosine important? (I)

39

Fourier Series: «any signal can be written as»

𝑢 𝑡 =

𝑘=1

∞

𝐴𝑘 ⋅ cos(𝜔𝑘 ⋅ 𝑡 + 𝜙𝑘)

𝑦∞ 𝑢1 =𝑚 𝜔1 ⋅ 𝐴1⋅ cos 𝜔1 ⋅ 𝑡 + 𝜙1 + 𝜑 𝜔1

𝑦∞ 𝑢2 =𝑚 𝜔2 ⋅ 𝐴2⋅ cos 𝜔2 ⋅ 𝑡 + 𝜙2 + 𝜑 𝜔2

𝑦∞ 𝑢3 = 𝑚 𝜔3 ⋅ 𝐴3 ⋅ cos 𝜔3 ⋅ 𝑡 + 𝜙3 + 𝜑 𝜔3

⋮

Linear System!+

+

+

+

⋮

𝑢1 = 𝐴1 ⋅ cos(𝜔1 ⋅ 𝑡 + 𝜙1)

𝑢2 = 𝐴2 ⋅ cos(𝜔2 ⋅ 𝑡 + 𝜙2)

𝑢3 = 𝐴3 ⋅ cos(𝜔3 ⋅ 𝑡 + 𝜙3)

𝑢4 = 𝐴4 ⋅ cos(𝜔4 ⋅ 𝑡 + 𝜙4)

𝑦∞ 𝑡 = 𝑦∞ 𝑢1 + 𝑦∞ 𝑢2 + 𝑦∞ 𝑢3 +⋯

Why is cosine important? (II)

40

𝛴(𝑠) is a linear system ↔ all linear operations are possible, among which:

• 𝐜𝐨𝐬(𝝎𝒕) or 𝐬𝐢𝐧(𝝎𝒕) are the only periodic signals that keep their shape when

passing through a system Σ 𝑠 (they only change in magnitude and phase)

• All other (periodic) signals change shape in integration or derivation operation

• System output could not be described by scaled and shifted system input

• More than two variables (𝑚,𝜑) would be necessary to describe output signal

Derivative operationd

d𝑡𝑥 𝑡 ↔ 𝑠 ⋅ 𝑋(𝑠)

∫ 𝑥 𝑡 ↔1

𝑠⋅ 𝑋(𝑠)Integral operation

Closed loop stability with Nyquist diagram

41

𝐶 𝑃𝑟 𝑒 𝑢 𝑦

−