Lecture Outline Systeem- en Regeltechniek II Previous lecture: PID - - PowerPoint PPT Presentation

Systeem- en Regeltechniek II

Lecture 10 – Nyquist plot and stability criterium

Robert Babuˇ ska Delft Center for Systems and Control Faculty of Mechanical Engineering Delft University of Technology The Netherlands e-mail: r.babuska@dcsc.tudelft.nl www.dcsc.tudelft.nl/˜babuska tel: 015-27 85117

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 1

Lecture Outline

Previous lecture: PID controller design, lead and lag compen- sators. Today:

• Nyquist plot.
• Nyquist stability criterion.

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 2

Frequency Domain Methods

frequency response Nyquist plot transfer function

• 40
• 30
• 20
• 10

• 1

• 1 5 0
• 1 0 0
• 5 0

data (experiment) Bode plot

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 3

Frequency Domain Methods

The key of frequency domain design: provide sufficient phase at the crossover frequency (= get the closed-loop far enough from the point of becoming un- stable) ⇒ Bode plots are well suited as a design and analysis tool. . . . , so, do we need yet another kind plot? In fact, we do, let’s have a look why . . .

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 4

Motivating Example

G(s) = 10 s − 1 sketch the bode plot, indicate PM

1 20

• 10
• 180
• 90

magnitude [dB] phase [deg] frequency [rad/s] 1 frequency [rad/s]

PM

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 5

Deficiency of Bode Plots

For systems with poles in right half-plane, the Bode plot alone does not provide any good indication of stability / instability. → In the above example, the phase will never cross −180◦, and yet, for K < 0.1, the closed loop becomes unstable (check the root locus!). Is there a method for frequency domain design, considering stabil- ity for all kinds of systems? Yes, the Nyquist plot and stability criterion.

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 6

Complex Numbers as Vectors

Im

s

Re

p s - p

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 7

Euler Representation

Im

s

Re

p s - p

s − p = |s − p| ejφ

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 8

Transfer Function

G(s) = s − z1 (s − p1)(s − p2) = |s − z1| |s − p1| · |s − p2| ej(ψ1−φ1−φ2)

Im

s

Re

p1

x

p2 z1

x

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 9

Nyquist (Polar) Plot

Let s = jω for ω ∈ [0, ∞) and plot Im[G(jω)] against Re[G(jω)].

−12 −10 −8 −6 −4 −2 2 −6 −5 −4 −3 −2 −1 1 2 3 4 5 Nyquist Diagram Real Axis Imaginary Axis

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 10

Relation Bode Plot – Nyquist Plot

G(s) = 1 (s + 1)(s + 5)

• 180

magnitude [dB] phase [deg] Re Im

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 11

Argument Principle (I)

Im

G s ( ) G s -plane ( )

Re Im

s s-plane

Re

x x

No zero/pole encirclement = ⇒ no origin encirclement

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 12

Argument Principle (II)

Im

G s ( ) G s -plane ( )

Re Im

s s-plane

Re

x x

Encirclement of one zero or pole = ⇒ one encirclement of the origin

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 13

Argument Principle in General

For a clockwise contour in the s-plane, denote: P number of poles encircled in the s-plane Z number of zeros encircled in the s-plane N number of clockwise encirclements of the origin by G(s) N = Z - P Recall: ∠G(s) =

• i

∠(s − zi) −

• j

∠(s − pj) =

• i

ψi −

• j

φj

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 14

Argument Principle for Stability Analysis

Given the Nyquist plot of KG(s), we want to determine whether the closed loop: Gcl(s) = Y (s) R(s) = KG(s) 1 + KG(s) is stable.

• Closed-loop stability ⇐

⇒ Gcl(s) has no poles in RHP.

• Poles are given by 1 + KG(s), so let us study 1 + KG(s).

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 15

Argument Principle for Stability Analysis

Gcl(s) = Y (s) R(s) = KG(s) 1 + KG(s) Poles of Gcl(s) are the solutions of 1 + KG(s) = 0, i.e.: poles of Gcl(s) are the zeros of (1 + KG(s)) in addition, as: 1 + KG(s) = 0 → 1 + K b(s) a(s) = 0 → a(s) + Kb(s) a(s) = 0 poles of G(s) are the poles of (1 + KG(s))

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 16

Argument Principle for Stability Analysis

So, we want to find out, if 1 + KG(s) has no RHP zeros. Im s-plane Re encircle the entire RHP = draw the Nyquist diagram for frequencies ω ∈ (−∞, ∞)

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 17

Argument Principle for Stability Analysis

Im KG s ( )

• 1

Re Im 1 + ( ) KG s Re

Draw the Nyquist diagram of the loop TF L(s) = KG(s), count clockwise encirclements of −1 : N = Z − P

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 18

Argument Principle for Stability Analysis

N = Z - P Z = number of RHP zeros of (1 + KG(s)) P = number of RHP poles of (1 + KG(s)) Given that: 1 + KG(s) = 0 → a(s) + Kb(s) a(s) = 0 we have: Z = number of RHP poles of Gcl(s) . . . CL poles P = number of RHP poles of G(s) . . . OL poles

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 19

Nyquist Stability Criterion

Z = N + P In words: number of RHP closed-loop poles = clockwise encirclements + number of RHP open-loop poles

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 20

Stability Margins in Nyquist Plot

• 1

Re Im

1/GM PM

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 21

Why Don’t We Ride These Bikes?

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 22

Simple Bicycle Models

Front steering: G(s) = φ(s) δ(s) = K s + v

a

s2 − g

h

Rear steering: G(s) = φ(s) δ(s) = K −s + v

a

s2 − g

h

a – distance of COM to fixed wheel center h – height of COM above ground

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 23

Front Steering: Nyquist Plot

−1.4 −1.2 −1 −0.8 −0.6 −0.4 −0.2 −1.5 −1 −0.5 0.5 1 1.5 Nyquist Diagram Real Axis Imaginary Axis

5G(s) G(s)

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 24

Rear Steering: Nyquist Plot

−1.4 −1.2 −1 −0.8 −0.6 −0.4 −0.2 −1.5 −1 −0.5 0.5 1 1.5 Nyquist Diagram Real Axis Imaginary Axis

5G(s) G(s)

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 25

Nyquist: Homework Assignments

• Read Section 6.3 (Nyquist stability criterion).
• Work out examples in this section and verify the results by using

Matlab.

• Work out problems 6.18 – 6.22 and verify your results by using

Matlab.

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 26