Dynamic Behavior of Closed-Loop Control Systems Chapter 11.

Dynamic Behavior of Closed-Loop Control Systems

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Next, we develop a transfer function for each of the five elements in the feedback control loop. For the sake of simplicity, flow rate w1 is assumed to be constant, and the system is initially operating at the nominal steady rate.

Process

In section 4.1 the approximate dynamic model of a stirred-tank blending system was developed:

1 21 2 (11-1)

τ 1 τ 1

K KX s X s W s

s s

where

11 2

ρ 1, , and (11-2)

wV xK K

w w w

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The symbol denotes the internal set-point composition expressed as an equivalent electrical current signal. is related to the actual composition set point by the composition sensor-transmitter gain Km:

spx t spx t

spx t

(11-7)sp m spx t K x t

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Current-to-Pressure (I/P) Transducer

The transducer transfer function merely consists of a steady-state gain KIP:

(11-9)tIP

P sK

P s

Control Valve

As discussed in Section 9.2, control valves are usually designed so that the flow rate through the valve is a nearly linear function of the signal to the valve actuator. Therefore, a first-order transfer function is an adequate model

2 (11-10)τ 1

v

t v

W s K

P s s

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Composition Sensor-Transmitter (Analyzer)

We assume that the dynamic behavior of the composition sensor-transmitter can be approximated by a first-order transfer function, but τm is small so it can be neglected.

mm

X sK

X s

Controller

Suppose that an electronic proportional plus integral controller is used.

11 (11-4)

τcI

P sK

E s s

where and E(s) are the Laplace transforms of the controller output and the error signal e(t). Kc is dimensionless.

P s p t

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1. Summer

2. Comparator

3. Block

•Blocks in Series

are equivalent to...

G(s)X(s)Y(s) Ch

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“Closed-Loop” Transfer Functions

•Indicate dynamic behavior of the controlled process (i.e., process plus controller, transmitter, valve etc.)

•Set-point Changes (“Servo Problem”)

Assume Ysp 0 and D = 0 (set-point change while disturbance change is zero)

(11-26)

•Disturbance Changes (“Regulator Problem”)

Assume D 0 and Ysp = 0 (constant set-point)

(11-29)

*Note same denominator for Y/D, Y/Ysp.

( )

( ) 1m c v p

sp c v p m

K G G GY s

Y s G G G G

( )

( ) 1d

c v p m

GY s

D s G G G G

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Figure 11.16 Block diagram for level control system.

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EXAMPLE 1:EXAMPLE 1: P.I. control of liquid level

Block Diagram:

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Assumptions

1. q1, varies with time; q2 is constant.

2. Constant density and x-sectional area of tank, A.

3. (for uncontrolled process)

4. The transmitter and control valve have negligible dynamics (compared with dynamics of tank).

5. Ideal PI controller is used (direct-acting).

)h(fq3

0KAs

1)s(G

As

1)s(G

K)s(G

K)s(Gs

11K)s(G

CL

P

VV

MMI

CC

For these assumptions, the transfer functions are:

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1 1d

C V P M

GY H

D Q G G G G

MVI

C KAs

Ks

K

AsD

Y

11

11

1

MPCIMVCI

I

KKKsKKKsA

s

D

Y

2

02 MPCIMVCI KKKsKKKsA

1s2s

K)s(G

22

The closed-loop transfer function is:

Substitute,

Simplify,

Characteristic Equation:

Recall the standard 2nd Order Transfer Function:

(11-68)

(2)

(3)

(4)

(5)

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For 0 < < 1 , closed-loop response is oscillatory. Thus decreased degree of oscillation by increasing Kc or I (for constantKv, KM, and A).

To place Eqn. (4) in the same form as the denominator of the T.F. in Eqn. (5), divide by Kc, KV, KM :

01ssKKK

AI

2

MVC

I

A

KKK

2

1 IMVC

10

Comparing coefficients (5) and (6) gives:

Substitute,

22

KKK

A

KKK

A

II

MVC

I

MVC

I2

•unusual property of PI control of integrating system•better to use P only

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Stability of Closed-Loop Control Systems

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Proportional Control of First-Order Process

Set-point change:

MPVCOLOLOL

OL

sp

MPVC

MPVC

sp

KKKKKKK

KK

s

K

Y

Ys

KKKKs

KKKK

Y

Y

11

1

11

1

11

1

1

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1

s

KG P

P

),,(K

gainsconstant ,,

V mC

MCV

KK

GGG

Set-point change = M

Offset =

See Section 11.3 for tank example

11( ) 1 ty t K M e

1sp

OL

My y

K

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Closed-Loop Transfer function approach:

First-order behavior closed-loop time constant

(faster, depends on Kc)

11

1

1

sKK

KK

KK

KKs

KK

Y

Y

C

C

C

C

C

sp

CKK

1

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General Stability Criterion

Most industrial processes are stable without feedback control. Thus, they are said to be open-loop stable or self-regulating. An open-loop stable process will return to the original steady state after a transient disturbance (one that is not sustained) occurs. By contrast there are a few processes, such as exothermic chemical reactors, that can be open-loop unstable.

Definition of Stability. An unconstrained linear system is said to be stable if the output response is bounded for all bounded inputs. Otherwise it is said to be unstable.C

hap

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11

Effect of PID Control on a Disturbance Change

For a regulator (disturbance change), we want the disturbance effects to attenuate when control is applied.

Consider the closed-loop transfer function for proportional control of a third-order system (disturbance change).

Kc is the controller function, i.e., .

3 2

8( ) ( )

6 12 8 8 C

Y s D ss s s K

is unspecified)(sD

3

81 1

2V M P dG G G G

s

CC K)s(G

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Let

If Kc = 1,

Since all of the factors are positive, ,

the step response will be the sum of negative

exponentials, but will exhibit oscillation.

CKssss 88126)( 23

jsjssssss 31314424)( 2

ate

as

1

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If Kc = 8,3 2 2( ) 6 12 72 ( 6)( 12)s s s s s s

Corresponds to sine wave (undamped), so this case is marginally stable.

If Kc = 27

Since the sign of the real part of the root is negative, we obtain a positive exponential for the response. Inverse transformation shows how the controller gain affects the roots of the system.

Offset with proportional control (disturbance step-response; D(s) =1/s )

2828224126)( 223 sssssss

CCs

C

KKssYty

sKssssY

1

1

88

8)(lim)(

1

88126

8)(

0

23

j331sj331s8s

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Therefore, if Kc is made very large, y(t) approaches 0, but does not equal zero. There is some offset with proportional control, and it can be rather large when large values of Kc create instability.

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Integral Control:

s

K)s(G)s(E

s

K)s(Ptdte

KP

I

CC

I

Ct

0I

C

For a unit step load-change and Kc=1,

)(y0)s(sYlim

s

182ss

s8)s(Y

0s

I

3

no offset

(note 4th order polynomial)

adjust Kc and I to obtain satisfactory response (roots of equation which is 4th order).

PI Control:

no offset

PID Control: (pure PID)

0)s(sYlims

1

sK8K8)2s(s

s8)s(Y

s

11K)s(G

0s

CI

C3

ICC

s

s

11K)s(G D

ICC

No offset, adjust Kc, I , D to obtain satisfactory result (requires solving for roots of 4th order characteristic equation).

Analysis of roots of characteristic equation is one way toanalyze controller behavior 0GGGG1 MPVC

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Rule of Thumb:Closed-loop response becomes less oscillatory and more stable bydecreasing Kc or increasing .

General Stability CriterionConsider the “characteristic equation,”

Note that the left-hand side is merely the denominator of theclosed-loop transfer function.

The roots (poles) of the characteristic equation (s - pi) determinethe type of response that occurs:

Complex roots oscillatory responseAll real roots no oscillations

***All roots in left half of complex plane = stable system

0GGGG1 MPVC

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Figure 11.25 Stability regions in the complex plane for roots of the characteristic equation.

Stability Considerations

• Feedback control can result in oscillatory or even unstable closed-loop responses.

• Typical behavior (for different values of controller gain, Kc).

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Roots of 1 + GcGvGpGm

(Each test is for differentvalue of Kc)

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(Note complex roots always occur in pairs)

Figure 11.26 Contributions of characteristic equation roots to closed-loop response.

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)3)(2)(1(

2)(

sss

KsG C

OL

Routh Stability Criterion

Characteristic equation

Where an . According to the Routh criterion, if any of the coefficients a0, a1, …, an-1 are negative or zero, then at least one root of the characteristic equation lies in the RHP, and thus the system is unstable. On the other hand, if all of the coefficients are positive, then one must construct the Routh Array shown below:

(11-93)0011

1 asasasa n

nn

n

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For stability, all elements in the first column must be positive.

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The first two rows of the Routh Array are comprised of thecoefficients in the characteristic equation. The elements in theremaining rows are calculated from coefficients by using theformulas:

(n+1 rows must be constructed; n = order of the characteristic eqn.)

n-1 n-2 n n-31

n-1

a a a ab

a

1n

5nn4n1n2 a

aaaab

1

21n3n11 b

baabc

1

31n5n12 b

baabc

.

.

(11-94)

(11-95)

(11-96)

(11-97)

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The important constraint is Kc<8. Any Kc 8 will cause instability.

Application of the Routh Array:

Characteristic Eqn is

We want to know what value of Kc causes instability, I.e., at leastone root of the above equation is positive. Using the Routh array,

Conditions for Stability

CCMV3LP KG1GG)2s(

8GG

01 MPVC GGGG

0K88s12s6s

0K8)2s(0)2s(

K81

C23

C3

3C

0K88

06

K881)12(6K886

3n121

C

C

C

1K0K88

8K0K8872

CC

CC

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Figure 11.29 Flowchart for performing a

stability analysis.

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1. Bode Stability Criterion• Ch. 14 - can handle time delays

2. Nyquist Stability Criterion• Ch. 14

Additional Stability Criteria

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Direct Substitution Method

Imaginary axis is the dividing line between stable and unstable systems.

1. Substitute s = j into characteristic equation

2. Solve for Kcm and c

(a) one equation for real part(b) one equation for imaginary part

Example (cf. Example 11.11)

characteristic equation: 1 + 5s + 2Kce-s = 0 (11-101)

set s = j 1 + 5j + 2Kce-j = 0

1 + 5j + 2Kc (cos( – j sin() = 0

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Direct Substitution Method (continued)

Re: 1 + 2Kc cos = 0 (1)

Im: 5 – 2Kc sin = 0 (2)

solve for Kc in (1) and substitute into (2):

sin

5 5 tan 0cos

Solve for : c = 1.69 rad/min (96.87°/min)

from (1) Kcm = 4.25

(vs. 5.5 using Pade approximation in Example 11.11)

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