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Lecture Outline Regeltechniek Previous lecture: representation of dynamic models, transfer func- Lecture 3 Stability analysis and transient response tions and state-space models. Robert Babu ska Today: Delft Center for Systems and

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Regeltechniek

Lecture 3 – Stability analysis and transient response

Robert Babuˇ ska Delft Center for Systems and Control Faculty of Mechanical Engineering Delft University of Technology The Netherlands e-mail: r.babuska@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: representation of dynamic models, transfer func- tions and state-space models. Today:

  • Assignment for the first computer session.
  • Stability analysis.
  • Transient response.

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

Information on Computer Sessions

  • Compulsory for all students who have not successfully completed

it in previous year(s).

  • First session in week 4 (calendar week 39)

– Wednesday 1+2 or 3+4 or Thursday 1+2 or 3+4 depending on in which group you are - check Blackboard from Monday next week.

  • Location: ’Computer room 020’ at Civil Engineering.
  • Homework preparation required!

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

Homework for Computer Session

  • Read thoroughly the entire handout “Matlab and Simulink for

Modeling and Control”.

  • Work out items a) through e) of Section 5 (by hand).
  • If you have never used Matlab before, familiarize yourself with

this tool (type ‘demo’ to start).

  • Bring the handout “Matlab and Simulink for Modeling and Con-

trol” with you to the computer lab.

  • Bring your own laptop with Matlab, if you have one.

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

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Purpose of Analysis

Analyze the available model in order to:

  • Understand the behavior of the process under study.
  • Define meaningful specification for the controlled system.
  • Give basis for control design choices (controller structure, pa-

rameters). We are mainly interested in:

  • Stability of the open-loop process.
  • Transient response (impulse, step, ramp).
  • Steady-state response (constant or sinusoidal input).

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

Stability – General Notions

  • Nonlinear systems – stability of a trajectory (solutions of differ-

ential equations).

  • Mostly we consider stability of equilibria, i.e., solutions of 0 =

f(x0, u0).

  • One system may have many equilibria, some stable, some un-

stable.

  • Linear systems – stability of an equilibrium implies stability of

the whole system.

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

Stability of LTI Systems

G(s) = Y (s) U(s) = b0sm + b1sm−1 + · · · + bm sn + a1sn−1 + · · · + an = K m

j=1(s − zj)

n

i=1(s − pi)

Response to initial conditions: y(t) =

n

  • i=1

Kiepit pi are real or complex poles of the system The exponential terms decay iff Re{pi} < 0 = the system is stable

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

Real Poles

y(t) = Kept y t ( ) t y t ( ) t p=0 p < y t ( ) t p>0 A single pole p = 0 . . . system marginally stable Multiple poles at p = 0 . . . system unstable

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

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Complex Poles

y(t) = K′eσt sin(ωt + ϕ) y t ( ) t >0 y t ( ) t =0 y t ( ) t <0

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

Transient and Steady-State Response

  • Responses to an arbitrary input signal cannot be computed an-

alytically (we have to resort to simulation).

  • However, some specific input signals are useful:

– step response – impulse response – response to a ramp input (ramp response) – response to a sinusoidal input (frequency response)

  • Importance both for analysis and for identification of model

parameters from measured data.

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

First-Order Systems: Step Response

Y (s) = G(s)U(s) = 1 τs + 1 · 1 s Expand in partial fractions: Y (s) = 1 s − τ τs + 1 = 1 s − 1 s + 1/τ Corresponding time signal: y(t) = 1 − e−t/τ

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

First-Order Systems: Step Response

y t

63%

1

3τ . . . 95%, 4τ . . . 98%, 5τ . . . 99% of steady state value

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

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Estimate Parameters From Step Response

  • Assume we have a stable first-order process with unknown gain

and time constant: G(s) = K τs + 1

  • Apply a step input to the process (choose a suitable amplitude,

not necessarily the unit step).

  • Plot the corresponding output and read the parameters from

the graph.

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

Demo: Rotational Pendulum

m1 m2 l2 l1 m2g m1g

motor

M(θ)¨ θ + C(θ, ˙ θ) ˙ θ + G(θ) =    kmu   

M(θ) =    P1 + P2 + 2P3 cos θ2 P2 + P3 cos θ2 P2 + P3 cos θ2 P2    C(θ, ˙ θ) =    b1 − P3 ˙ θ2 sin θ2 −P3( ˙ θ1 + ˙ θ2) sin θ2 P3 ˙ θ1 sin θ2 b2    G(θ) =    −g1 sin θ1 − g2 sin(θ1 + θ2) −g2 sin(θ1 + θ2)   

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

Linearized Model

G(s) = θ2(s) U(s) = 661.2903(s2 + 49.05) s(s + 33.06)(s2 + 0.6783s + 98.11)

  • one pole in origin (pure integration)
  • one fast real pole (motor mechanical time constant)
  • a pair of poorly damped complex poles
  • a pair of complex zeros

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

First-Order Systems: Ramp Response

Y (s) = 1 τs + 1 · 1 s2 Expand in partial fractions: Y (s) = 1 s2 − τ s + τ2 τs + 1 Corresponding time signal: y(t) = t − τ + τe−t/τ

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

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First-Order Systems: Impulse Response

Y (s) = G(s)U(s) = 1 τs + 1 · 1 Expand in partial fractions: Y (s) = 1 τs + 1 = 1/τ s + 1/τ Corresponding time signal: y(t) = 1 τ e−t/τ

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

Relationship Between the Responses

  • Ramp response:

G(s) · 1

s2

  • Step response:

G(s) · 1

s

(derivative of ramp response)

  • Impulse response: G(s) · 1

(derivative of step response) So far we considered τ > 0 (asymptotically stable first-order sys- tem). Work out the impulse, step and ramp response for τ < 0 (unstable system) and for an integrator (G(s) = 1/s).

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

Second-Order System

H(s) = ω2

n

s2 + 2ζωns + ω2

n

= k (s + σ)2 + ω2

d

ωn : undamped natural frequency ζ : relative damping σ : attenuation (damping) ωd : damped natural frequency

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

Second-Order System

H s s s

n n n

b g =

+ + ω ζω ω

2 2 2

2

σ ωd Re Im θ ωn

σ ζω ω ω ζ θ ζ = = − =

− n d n 1 2 1

sin

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

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SLIDE 6

Second-Order System: Time Response

y(t) = 1 − e−σt cos ωdt + σ ωd sin ωdt

  • t

t+T y

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

Influence of Damping

2 4 6 8 10 12 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 t y(t) z=0.0 z=0.1 z=0.3 z=0.5 z=0.7 z=0.9 z=1.0

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

Step Response Characteristics

0.2 0.4 0.6 0.8 1 1.2 1.4

time y(t) tr ±1% ts Mp tp

90% 10%

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

Relation to Natural Frequency and Damping

t t t M e

r n p n s n p

= = − = ± =

− −

18 1 4 6 1%

2 1

2

. . ω π ω ζ ζω

πζ ζ

for

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

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SLIDE 7

Performance Specifications for Closed-Loop

t t M

r s p

ω ζ ζ σ

n r p s

t M t

≥ ≥ ≥

1 8 4 6 . .

✁ ✂

ω n

Re Re Re Im Im Im

θ σ

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

Performance Specifications: Example

ω ζ ζ σ

n r p s

t M t ≥ ≥ ≥ 18 4 6 . .

d i

Re Im

t s M t s

r p s

≤ ≤ ≤ 06 10 3 . . % .

  • 5-4-3-2-1

01 3 2 1

  • 1

1 1

ω ζ σ

n

rad s ≥ ≥ ≥ 28 0 6 15 . . . .

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

Additional Pole in the System

H(s) = ω2

nωnγ

  • s2 + 2ζωns + ω2

n

  • s + γωn
  • The 3rd order system can be accurately approximated by a second
  • rder system if

γ ≥ 10 In such a case, the two complex poles are dominant.

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 27

Zero in a Second-Order System

H(s) = ω2

n(bs + 1)

s2 + 2ζωns + ω2

n

H(s) = ω2

n

s2 + 2ζωns + ω2

n

  • H0

+ b · ω2

ns

s2 + 2ζωns + ω2

n

  • Hd

Time response of Hd is the derivative of response of H0.

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 28

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SLIDE 8

Step Response With an Extra Zero

1 2 3 4 5 6 7 8 9 10 −0.2 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 Time (sec) y(t)

H(s) H0(s) Hd(s)

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 29

Consequences of Extra Zeros

H(s) = ω2

n

s2 + 2ζωns + ω2

n

  • H0

+ b · ω2

ns

s2 + 2ζωns + ω2

n

  • Hd

For large b → overshoot Mp increases If b < 0 (zero in RHP) initial response is negative (such systems are called nonminimum-phase systems or inverse- response systems)

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 30

Nonminimum-Phase Response

1 2 3 4 5 6 7 8 9 10 −1.5 −1 −0.5 0.5 1 1.5 Time (sec) y(t)

H(s) H0(s) Hd(s)

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 31