458.308 Process Control & Design

458.308 Process Control & Design

Lecture 7: Dynamic Behavior and Stability of Closed-Loop Control Systems

Jong Min Lee

Chemical & Biomolecular Engineering Seoul National University

Overview: Closed-loop

To simplify the notation, the primes andsdependence have been omitted; thus,Yis used rather thanY^′(s)

Y(s)

Ysp(s) = KmGcGvGp 1 +GcGvGpGm

= KmGcGvGp 1 +G_OL Y(s)

D(s) = Gd

1 +GcGvGpGm

= Gd 1 +GOL

GOL∆

=GcGvGpGm Different from open-loop!

Depends onGc

Analysis and Design Problems

Analysis: Given particularG's andG_c Are the closed-loop dynamics stable?

Speed of response? Damping?

Gains forY/YspandY/D

Design: Given particularG's, choose (``design")G_cso that the closed-loop dynamics are stable

Y

Ysp has a gain of∼1and ^Y_D has a gain of∼0

the dynamics are sufficiently fast (but not too fast) and smooth (without excessive oscillation).

Model Used for Analysis and Design

Case I (Less Frequent)

From a fundamental model, perform linearization and Laplace transform of the linearized ODEs to findGp(s)andGd(s) Find actuator and measurement dynamicsGvandGm

Case II (More Frequent)

The composite modelG(=GmGpGv)is fitted to data ofym

obtained by perturbingp(e.g., by making a step change).

PID Controller

p(t) = ¯p+K_c (

e(t) + 1 τI

∫ _t

0

e(t^∗)dt^∗+τD

de dt

)

⇒ P^′(s) = K_c

(

E(s) + 1

τIsE(s) +τDsE(s) )

⇒ P^′(s)

E(s) = K_c (

1 + 1

τ_Is+τDs )

Y⁰

sp E

K_c

1 + 1 τ_Is+τ_Ds

P⁰

G

Y⁰

G_d

D⁰

Calculation of Closed-Loop Functions

++

+− ++

Y’_sp+− E

D’

G

G

G

Y^′ = G_dD^′+GP^′ (P^′ =G_c(Y^′_sp−Y^′)) Y^′ = G_dD^′+GG_c(Y^′_sp−Y^′)

.Convenience of L.T.

. . . ...

Y^′(s)

D^′(s) = G_d

1 +GG_c and Y^′(s)

Y^′_sp(s) = GG_c 1 +GG_c

Calculation of Closed-Loop Functions:

Generalization

++

+− ++

+− Y

G₁ G_i G_i+1 G_j X

Y(s)

X(s) = G_i+1G_i+2· · ·G_j 1 +G₁G₂· · ·G_j =

∏

f

1 +∏

e Assumenegativefeedback

∏

f: Product of the transfer functions in the forward path fromXtoY.

∏

e: Product of every transfer function in the feedback loop.

Analysis of P-only Control

Y0

sp E

K_c

P0 K_p

τ s+ 1

Y⁰ K_d

τ s+ 1 D⁰

Y^′(s) Y^′_sp(s) =

KcKp τs+1

1 +^K_τs+1^cKp = KcKp

τs+ 1 +KcKp

=

KcKp 1+KcKp

τ 1+KcKps+ 1

Gain is not 1 unlessK_c=∞, Time constant decreases with increasingK_c.

Y^′(s) D^′(s) =

Kd τs+1

1 +^K_τs+1^cKp = Kd

τs+ 1 +KcKp

=

Kd 1+KcKp

τ 1+KcKps+ 1

Gain is not 0 unlessKc=∞, Time constant decreases with increasingKc.

Analysis of PI Control

Ysp⁰ E

Kc

1 + ¹

τ_Is

P⁰ Kp

τ s+1

Y⁰

Kd

τ s+1

D⁰

Y^′(s) Y^′_sp(s) =

KcKp(τIs+1) τIs(τs+1)

1+^KcKp(τ_τ ^Is+1)

Is(τs+1)

= _τ ^KcKp(τIs+1)

Is(τs+1)+KcKp(τIs+1)

= _τIτ ^τIs+1

KcKps²+^1+KcKp_KcKp τ_Is+1

Gain = 1always! No offset 2nd order dynamics

Underdamped dynamics for very small

Closed-Loop Stability

.Characteristic Equation .

. . ...

1 + G

= 0

Roots of the above equation are thepoles of the closed-loop functions(important information for analyzing closed-loop dynamics)

For stability, make sure all the roots are in the Left-Half-Plane (negative real parts)

Can be checked byRouth's test Or bydirect substitution

Example: Routh's Test

Main Idea: Form a Routh array to see if any roots are in the RHP

1 + 6K_c

(2s+ 1)(4s+ 1)(6s+ 1) = 0

48s³+ 44s²+ 12s+ (1 + 6K_c) = 0

48 12

44 1 + 6K_c

44×12−48(1+6Kc)

44 0

(¹²⁰₁₁−⁷²₁₁Kc)×(1+6Kc)−44×0

120 11−⁷²₁₁Kc

Must be allpositivefor closed-loop stability!

120

11 −⁷²₁₁K_c>0, 1 + 6K_c>0

Example: Direct Substitution

Main Idea: At the limits of instability, the closed-loop poles will be on theimaginary axis(between LHP and RHP)

48s³+ 44s²+ 12s+ (1 + 6K_c) = 0^s=jω−→

−48jω³−44ω²+ 12jω+ (1 + 6K_c) = 0 (−48ω³+ 12ω)j+(

−44ω²+ (1 + 6K_c))

= 0

−48ω³+ 12ω= 0 −44ω²+ (1 + 6K_c) = 0 ω= 0, K_c =−1/6 ω=±1/2, K_c= 5/3 .Note

. . . ...

This method works with a system with time delay. Routh's method does not.