CHAPTER 11: Dynamic Behaviour & Stability of Closed-Loop Control Systems Anis Atikah Ahmad [email protected]

CHAPTER 11: Dynamic Behaviour & Stability of Closed-Loop Control Systems

Anis Atikah Ahmad

[email protected]

Outline Block Diagram Representation Closed-Loop Transfer Function Closed-Loop Response of Simple Control

Systems Stability of Closed-Loop Control Systems

Block Diagram Representation

Figure 11.8: Standard block diagram of a feedback control system



controlled variable

manipulated variable

disturbance variable

controller outputerror signal

measured value of Y


Ysp =

Ysp

set point

internal set point (used by the controller)

Yu

= Yd

= Gc

=

Gv =

change in Y due to U

change in Y due to D

controller transfer

function

transfer function for final control element

(including KIP, if required)Gp = process transfer function

Gd = disturbance transfer function

Gm = transfer function for measuring element andtransmitter

Km = steady-state gain for Gm

Figure 11.9 : Alternative form of the standard block diagram of a feedback control system.


Block Diagram Representation Example:

Figure 11.1 Composition control system for a stirred-tank blending process.

Control Objective : to regulate the tank composition, x, by

adjusting the mass flow rate w2.

Disturbance MV

CV

Flow rate w1

is assumed to be constant & the system is initially operating at the nominal steady rate

1. Process

The approximate dynamic model of a stirred- tank blending system was developed (Example 4.4):

where,

X s

X1 s

W2

s

1 2 (11-1) τs 1 τs 1

K K

K w1 ,1 2 Vρ , 1 xand K (11-2)w w w


Block Diagram of the process

Figure 11.2: Block diagram of the process

2. Sensor-Transmitter (Analyzer) We assume that the dynamic behavior of the

composition sensor-transmitter can be approximated by a first-order transfer function:

Figure 11.2: Block diagram of the process

X m

s

X s τ s

1

(11-3)Km

m

1s

K

m

m

sX sX m

[mass fraction]

[mA

3. Controller Suppose that an electronic proportional plus integral controller

is used.The controller transfer function is (See Chapter 8):

where P s and E(s) are the Laplace transforms of the controller output p(t) and error signal e(t).

Note that pand e are electrical signals that have units of mA, while Kc is dimensionless.

s

KsE

sP

Ic

11

3. Controller The error signal is expressed as:

et xsp t xm t or after taking Laplace transforms,

E s X sp s X m s The symbol xsp denotes the internal set-point composition

expressed as an equivalent electrical current signal. This signal is used internally by the controller xsp t is

related to the actual composition set point by the composition sensor-transmitter gain Km:

(11-5)

(11-6)

xsp t Km xsp

t (11-7)

3. Controller

xsp t Km xsp

t Thus,

Fig 11.4: Block diagram for the controller

(11-7)X sp s

KX

s

(11-8)msp

4. Current to Pressure (I/P) Transducer Because transducers are usually designed to have

linear characteristics and negligible (fast) dynamics, we assume that the transducer transfer function merely consists of a steady-state gain, KIP

IPK sP sPt[mA] [psi]

IP

t KsP

sP

Figure 11.5 Block diagram for the I/P transducer.

5. Control ValveControl valves are usually designed so that theflow

rate through the valve is a nearly linear function of the signal to the valve actuator.

Therefore, a first-order transfer function usually provides an adequate model for operation of an installed valve in the vicinity of a nominal steady state.

Thus, we assume that the control valve can be modelled as

W2

s

(11-10)

Ps τ s 1

Kv

t v

5. Control Valve The block diagram of a control valve is shown in Fig.

11.16

1s

K

v

v

sW2 sPt

[psi] [kg/min]

Block diagram for the entire blending process

Combining the individual block diagram:

Figure 11.7 : Block diagram for the entire blending process (composition control system)

Closed Loop Transfer Function

In deriving closed-loop transfer functions, it is often convenient to combine several blocks into a single block.

For example, consider the three blocks in series in Fig. 11.10.The block diagram indicates the following relations:

Figure 11.10 Three blocks in series.

1. Block diagram reduction

The block diagram indicates the following relations:

X1 G1U

X 2 G2 X1

X 3 G3 X 2

By successive substitution, X3 G3G2G1U

or

X3 GU

where G = G3G2G1

G 3XU

Equivalent block diagram


The closed-loop system behavior for set-point changes is also referred to as the servomechanism (servo) problem in the control literature.

We assume for this case that no disturbance change occurs and thus D = 0.

From Fig. 11.8, it follows that:

Combining gives

Y GpU

2. Set-Point Changes

(11-14)

(11-15)

(11-16)

Y Yd Yu

Yd Gd D 0 (because D 0)

Yu GpU

Block Diagram



Figure 11.8 also indicates the following input/output relations for the individual blocks:


(11-18)

(11-19)

(11-20)

(11-21)

(11-22)

U Gv P

P Gc E

E Ysp Ym

Ysp

KmYsp

Ym GmY


Combining the above equations gives

Y G pGv P G pGvGc E

G pGvGc Ysp Ym G pGvGc KmYsp GmY

Rearranging gives the desired closed-loop transfer function,


(11-26)

1 GcGvGpGm

KmGcGvGp

Ysp

Y

(11-23)

(11-24)

(11-25)


Now consider the case of disturbance changes, which is also referred to as the regulator problem since the process is to be regulated at a constant set point

From Fig. 11.8,

Substituting (11-18) through (11-22) gives,

Y Gd D GpU

Because Ysp = 0 we can arrange (11-28) to give the closed-loop transfer function for disturbance changes:

3. Disturbance Changes

(11-27)Y Yd Yu Gd D G pU

GdD GpGvGc(KmYsp GmY (11.28 )

(11-29)D 1 GcGvGpGm

dY G

A comparison of Eqs. 11-26 and 11-29 indicates that both closed-loop transfer functions have the same denominator, 1 + GcGvGpGm.

The denominator is often written as 1 + GOL where GOL is the

open-loop transfer function; GOL ΔGcGvGpGm

Closed Loop Transfer Function3. Disturbance Changes

(11-26)

(11-29)D 1 GcGvGpGm

dY G

1 GcGvGpGm

KmGcGvGp

Ysp

YServo

Problem

Regulator Problem

Suppose that D ≠ 0 and Ysp ≠ 0, as would be the case if a disturbance occurs during a set-point change. To analyze this situation, we rearrange Eq. 11-28 and substitute the definition of GOL to obtain:

Thus, the response to simultaneous disturbance variable and set-point changes is merely the sum of the individual responses, as can be seen by comparing Eqs. 11-26, 11-29, and 11-30.


(11-30)1 G 1 G

KmGcGvGp

D YGd

spOL OL

Y

Closed-loop transfer functions for more complicated block diagrams can be written in the general form:

Closed Loop Transfer Function4. General Expression for Feedback Control System

Z (11-31)Zi 1 e

f

whereZ is the output variable or any internal variable within

the control loop

Zi is an input variable (e.g., Ysp or D)

f = product of the transfer functions in the forward path from

Zi to Z

= product of every transfer function in the feedback loop

e

Example 11.1 Find the closed-loop transfer function Y/Ysp for the

complex control system in Figure 11.12. Notice that this block diagram has two feedback loops and two disturbance variables

feedback loops and two disturbance variables.

Figure 11.12 Complex control system.

Solution Using the general rule in (11-31), we first reduce the

inner loop to a single block as shown in Fig. 11.13

Figure 11.13 Block diagram for reduced system.

Solution To solve the servo problem, set D1 = D2 = 0

Because Fig. 11.13 contains a single feedback loop, use (11-31) to obtain Fig. 11.14a.

Figure 11.13 Block diagram for reduced system.

Fig. 11.13

Fig. 11.14

Solution The final block diagram is shown in Fig. 11.14b with Y/Ysp = Km1G5.

Substitution for G4 and G5 gives the desired closed-loop transfer function:

Figure 11.14b: Final block diagrams for Example 11.1.

Km1Gc1Gc2G1G2G3

Ysp 1 Gc2G1Gm2 Gc1G2G3Gm1Gc2G1

Y

Closed-Loop Response of Simple Control Systems

Consider the liquid-level control system shown in Fig. 11.15.

Figure 11.15 Liquid-level control system.Figure 11.15 Liquid-level control system.


A second inlet flow rate q1 is the disturbance variable. Assume:

1. The liquid density and the cross-sectional area of the tank A are constant.

2. The flow-head relation is linear, q3 = h/R.

3. The level transmitter, I/P transducer, and control valve have negligible dynamics.

4. An electronic controller with input and output in% is used (full scale = 100%).


Derivation of the process and disturbance transfer functions directly follows Example 4.4.

Consider the unsteady-state mass balance for the tank contents:

Substituting the flow-head relation, q3 = h/R, and introducing deviation variables gives

ρA ρq1 ρq2 ρq3

dt(11-32)dh

A q1 q2 dt

R

(11-33)dh

h


Thus, we obtain the transfer functions

where Kp = R and τ = RA. Note that Gp(s) and Gd(s) are identical because q1 and q2

are both inlet flow rates and thus have the same effect on h.

H s G

s Q2

s

(11-34)

τs 1

K p

K

pp

H s G

Q1

s

s

(11-35)

τs 1d


Figure 11.16: Block diagram for level control system.

Recall:

If a proportional controller is used, then Gc(s) =Kc.

From Fig. 11.16, it follows that the closed-loop transfer function for set-point changes is given by

Closed-Loop Response of Simple Control Systems 1. Proportional Control and Set-Point Changes

H

s

H s 1 KK K K

Kc Kv K p Km / τs

1/ τs

1

(11-36)

sp c v p m

OL

pvcm

sp G

GGGK

Y

Y

1

For a set point change;


This relation can be rearranged in the standard form for a first-order transfer function,

where

H s

H sp

s

K1

τ1s 1(11-37)

K1 1 K

(11-38)KOLOL

1τ

τ (11-39)1 K OL

From Eq. 11-37 it follows that the closed- loop response to a unit step change of magnitude M in set point is given by,


t / τ1

K1M 1 e (11-41)h t

H s K1

τ1s 1

Change to time domain form

M

sX Unit step

change of magnitude M in set point

=


Figure 11.17: Step response for proportional control (set-point change).

Note that a steady-state error or offset exists because the new steady-state value is K1M rather than the desired

value of M.

The offset is defined as

For a step change of magnitude M in set point, h’sp(∞). From (11-41), it is clear that h’(∞)=K1M. Substituting these values and (11-38) into (11-42) gives

(11-38)


OL

OL

K

KK

11

offset M K1M 1 K

(11-43)OL

M

(11-42) h' h'offset Δ sp

The closed-loop transfer function for disturbance changes with proportional control (Gc=Kc) is:

Rearranging gives,

where

Closed-Loop Response of Simple Control Systems 2. Proportional Control and Disturbance Changes

11

2

1

s

K

sQ

sH

11

1

1

1

1

sK

sK

sQ

sH

OL

p

Recall:For a disturbance change;

OL

d

G

G

D

Y

1

OL

p

K

KK

12

OLK

11

Comparing with setpoint changes

(11.37)-(11-39), both are first order and have

same time constant, but have different steady state gains.

The closed loop response to a step change in disturbance of magnitude M is given by:


112teMKth

11

2

1

s

K

sQ

sH

11 1

2

1

2

ss

MK

s

M

s

KsH

Change to time domain form:

The offset can be determined as follows:

Offset =


h0

OL

p

K

MKMK

12


Equal to 0 because this is a regulator

problem (disturbance changes)

Example 11.2Consider the level control system shown in Fig.11.15 implemented with a computer whose inputs & outputs are calibrated in terms of full range(100%).

Example 11.2 The tank is 1m in diameter, while the valve on the exit line acts as a linear

resistance with R = 6.37 min/m2. The level transmitter has a span of 0.5m and an output range of 0 to 100%. The flow characteristics f of the equal percentage control valve is related to

the fraction of lift l by the relation f = (30) l-1. The air-to-open control valve receives a 3 to 15 psi signal from an I/P

transducer, which, in turn, receives a 0 to 100% signal from the computer- implemented proportional-only controller.

When the control valve is fully open (l=1), the flow rate through the valve is 0.2m3/min.

At the nominal operating condition, the control valve is half open (l=0.5).

Using the dynamic model in the block diagram of Fig. 11.16, calculate the closed-loop response to a unit step change in the set point for three values of the controller gain: Kc = 1,2, and 5.


Figure 11.16: Block diagram for level control system.

Example 11.2 The transfer function of the process:

From the question, R = 6.37 min/m2 = Kp

τ = RA= 6.37 min/m2 × 0.052m2 = 5min

H s G

s Q2

s

(11-34)

τs 1

K p

K

pp

H s G

Q1

s

s

(11-35)

τs 1d

where Kp = R and τ =RA.

Example 11.2 The measurement transmitter gain, Km can be

calculated as follows:

Next, calculate gain for the control valve, Kv

The valves relation between flow rate q and fraction of lift,

can be expressed as:

Thus;

rangeinput

rangeoutputKm m

m%200

5.0

%0100

1302.0 q

13030ln2.0

d

dq

Example 11.2 At nominal operating condition,

The control valve gain Kv must include the I/P transducer gain in order to provide a steady-state relation between the computer p (in %) and the flow rate through the control valve, q;

Where pt is the transducer output pressure.

5.0

min124.03030ln2.0 315.0 md

dq

tIPv dp

dqKK

Example 11.2 Using KIP = (15-3) / (100-0) psi/ % =0.12 psi/%, and

the chain rule for differentiation;

If the valve actuator is designed so that the fraction of lift,

varies linearly with the IP transducer output, pt, then

Thus,

ttIPv dp

d

d

dq

dp

dqKK

12.0

10833.0315

01

psipsipdp

d

tt

%min1024.1 33 mKv

Example 11.2 Thus, the closed loop transfer function for a

set-point change is:

Rearranging in a standard form;

Or where

15579.11

15579.1

11

1

sK

sK

sKKKK

sKKKK

sH

sH

c

c

pvcm

pvcm

sp

1579.115

579.11579.1

579.115

579.1

sK

KK

Ks

K

sH

sH

c

cc

c

c

sp

11

1

s

K

sH

sH

sp OL

OL

K

KK

11OLK

11

Example 11.2 For a unit step change in the set point; for Kc=

1,2 & 5;

Kc τ1 (min)

K1

1 1.94 0.612

2 1.20 0.759

5 0.56 0.887

Example 11.2 Set-point responses:

Increasing Kc reduces both offset and time required to reach the new steady state.

Example 11.3 For the liquid level control system and

numerical parameter values of Example 11.2, calculate the closed loop response to a step change in the disturbance variable of 0.05m3/min. Calculate the offset and plot the results for Kc = 1,2 and 5.

Rearranging in a standard form;

15579.11

1537.6

11

1

11

sK

s

sKKKK

sK

G

G

sQ

sH

cpvcm

p

OL

d

1579.115

579.1137.6

579.115

37.6

1

sK

K

KssQ

sH

c

c

c

For a regulator problem (disturbance change), h’sp() = 0 (since there is no change in set point)

Thus, offset = Find using final value theorem (Chapter

3);

Example 11.3


h0

s

s

M

sK

K

c

c

s

1579.115

579.1137.6lim

0

h

ssHthst 0limlim

s

ssK

K

c

c

s

05.0

1579.115

579.1137.6lim

0

cK579.113185.0

Example 11.3 The offsets at Kc = 1,2 & 5;

Kc Offset

1 -0.124

2 -0.077

5 -0.036

Figure 11.19: Disturbance responses

Increasing Kc reduces the amount of offset for disturbance changes.

For PI control, Gc(s) = Kc (1+1/τIs) The closed-loop transfer function for disturbance

changes can be derived as;

Clearing the denominator terms gives;

Closed-Loop Response of Simple Control Systems 3. PI Control and Disturbance Changes

1111

1

11

ssKKKK

sK

G

G

sQ

sH

Icmvp

p

OL

d

1111

1

ssK

sK

IOL

p

sKs

K

sQ

sH

IOL

p

1111

Further rearrangement in standard form of second-order transfer function;

Where;


12 3

223

3

1

ss

sK

sQ

sH

mvcI KKKK 3

I

OL

OL

K

K

1

2

13

OLI K 3

For a unit step change in disturbance,

For 0<ζ<1, the response is a damped oscillation that can be described by;


s

sQ1

1

sss

sKsH

1

12 322

3

3

12 322

3

3

ss

K

3232

33

3 1sin1

33

teK

th t

h’(∞)= 0 Thus, the

addition of integral action

eliminates offset for a step

change in disturbance

Stability of Closed Loop Control Systems 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.

Bounded input: input variable that stays within upper and

lower limits for all values

of time.

tetu 3

ttu

unbounded& unstable

0 1 2 3 4 5 6 70

1

2

3

4

5

6

7

t

u(t

)

0 1 2 3 4 5 6 70E+001E+072E+073E+074E+075E+076E+077E+07

t

u(t

)

Stability of Closed Loop Control Systems General Stability Criterion:

The feedback control system is stable if and only if all roots of characteristic equation are negative or have negative real parts. Otherwise, the system is unstable.

Characteristic equation:

1+ GOL

Example 11.8 Consider a process, that is

open-loop unstable. If Gv= Gm = 1, determine whether proportional controller can stabilize the closed loop system.

The characteristic equation for this system is:

The root, s = 1+ 0.2 Kc

1/2.0 sGp

mvpcOL GGGGG 11

1112.01 sKc

cKs 2.01

Example 11.8

In order for this system to be stable; s must have a negative value (s<0), thus Kc < -5

cKs 2.01

02.01 cK

2.0/1cK

5cK

Stability of Closed Loop Control Systems Routh Stability Criterion

The Routh stability criterion is based on a characteristic equation that has the form of:

All of the coefficients in the characteristic equation must be positive.

If any coefficient is negative or zero, then at least one root of the characteristic equation lies on the right of/on the imaginary axis, and the system is unstable.

0... 011

1 asasasa n

nn

n

Stability of Closed Loop Control Systems If all the coefficients are positive, we next

construct the following Routh array:

1

3211

n

nnnn

a

aaaab

1

3412

n

nnnn

a

aaaab

1

21311 b

baabc nn

1

31512 b

baabc nn

Row

1 an an-2 an-4 …

2 an-1 an-3 an-5 …

3 b1 b2 b3 …

4 c1 c2 …

. .

. .

n + 1 z1

Stability of Closed Loop Control Systems Routh Stability Criterion: A necessary and

sufficient conditions for all roots of the characteristic equation to have negative real parts is that, all of the elements in the left column of the Routh array are positive.

Example 11.10 Find the values of controller gain, Kc that make

the feedback control system with the following transfer function stable:

cc KG 12

1

sGv

15

1

sGG dp 1

1

sGm

Example 11.10-Solution First, find the characteristic equation, 1+GOL

Expanding and make it equal to zero,

mvpcOL GGGGG 11

1

1

12

1

15

111

sssKG cOL

cKsss 11215

0181710 23 cKsss

All of the coefficient is positive. Next, construct the Routh array.

Example 11.10-Solution The Routh array is:

In order for this system to be stable, b1 and c1 must be positive,

Thus, the system will be stable if

0

17

1108171

cK

b

10 8

17 1+Kc

b1 b2

c1

02 b 0

1

1

11

b

Kbc c

6.12cK 1cK

6.121 cK

Example 11.11Consider a feedback control system with Gc= Kc, Gv= 2, Gm =0.25, and Gp = 4e-s/(5s+1)

The characteristic equation is 1+5s+2Kce-s=0

Because this characteristic equation is not a polynomial s,the Routh criterion is not directly applicable.

However, if a polynomial approximation to e-s is introduced, such a Padé approximation, then the Routh criterion can be used to determine approximate stability limits.

For simplicity, use the 1/1 Padé approximation;s

se s

5.01

5.01

Example 11.11Consider a feedback control system with Gc= Kc, Gv= 2, Gm =0.25, and Gp = 4e-s/(5s+1)

The characteristic equation is 1+5s+2Kce-s=0

Because this characteristic equation is not a polynomial s,the Routh criterion is not directly applicable.

However, if a polynomial approximation to e-s is introduced, such a Padé approximation, then the Routh criterion can be used to determine approximate stability limits.

For simplicity, use the 1/1 Padé approximation;

and determine the stability limits for the controller gain.

s

se s

5.01

5.01

Example 11.11-Solution

The characteristic becomes

Rearranging;

The Routh array:

05.01

5.01251

s

sKs c

0215.55.2 2 cc KsKs

2.5 1+2Kc

5.5-Kc 0

b1

0

5.5

215.51

c

cc

K

KKb

5.55.0 cK

Documents

CHAPTER 11: Dynamic Behaviour & Stability of Closed-Loop Control Systems Anis Atikah Ahmad [email protected]