Advanced Control Engineering - Chapter 4 pot

Example 4.1Find the closed-loop transfer function for the system shown in Figure 4.2... Hence the combined transferfunction is G2Gm1ˆ1 ‡ GG2G3 The reduced block diagram is shown in Figur

The elements of a closed-loop control system are represented in block diagramform using the transfer function approach The general form of such a system isshown in Figure 4.1.

The transfer function relating R(s)and C(s)is termed the closed-loop transferfunction

From Figure 4.1

C(s) ˆ G(s)E(s)(4:1)B(s) ˆ H(s)C(s)(4:2)E(s) ˆ R(s) B(s)(4:3)Substituting (4.2)and (4.3)into (4.1)

C(s) ˆ G(s)fR(s) H(s)C(s)gC(s) ˆ G(s)R(s) G(s)H(s)C(s)C(s)f1 ‡ G(s)H(s)g ˆ G(s)R(s)C

4.2 Block diagram reduction 4.2.1 Control systems with multiple loops

A control system may have several feedback control loops For example, with a shipautopilot, the rudder-angle control loop is termed the minor loop, whereas theheading control loop is referred to as the major loop When analysing multiple loopsystems, the minor loops are considered first, until the system is reduced to a singleoverall closed-loop transfer function

To reduce complexity, in the following examples the function of s notation (s)usedfor transfer functions is only included in the final solution

Example 4.1Find the closed-loop transfer function for the system shown in Figure 4.2

Now Gml is multiplied by, or in cascade with G2 Hence the combined transfer

function is

G2Gm1ˆ1 ‡ GG2G3

The reduced block diagram is shown in Figure 4.3

Following a similar process, the second minor loop Gm2may be written

Gm2ˆ

G 2 G 3 1‡G 3 H 3

1 ‡G 2 G 3 H 2 1‡G 3 H 3Multiplying numerator and denominator by 1 ‡ G3H3

Gm2ˆ1 ‡ G G2G3

3H3‡ G2G3H2But Gm2is in cascade with G1, hence

+ –

+ –

The complete, or overall closed-loop transfer function can now be evaluated

C

R(s) ˆ

G 1 G 2 G 3 1‡G 3 H 3 ‡G 2 G 3 H 2

1 ‡ G1 G 2 G 3 H 1 1‡G 3 H 3 ‡G 2 G 3 H 2Multiplying numerator and denominator by 1 ‡ G3H3‡ G2G3H2

+ –

–

Fig 4.4 Second stage of block diagram reduction.

4.2.2 Block diagram manipulation

There are occasions when there is interaction between the control loops and, for the

purpose of analysis, it becomes necessary to re-arrange the block diagram

configur-ation This can be undertaken using Block Diagram Transformation Theorems

1 G

G

G G

G

G G

3 Removing a

block from

a forward path

4 Eliminating

a feedback loop

5 Removing a

block from

a feedback loop

6 Rearranging

summing points

7 Moving a

summing point ahead

of a block

8 Moving a

summing point beyond

a block

9 Moving a

take-off oint ahead

of a block p

10 Moving a

take-off oint beyond

a block p

+

+

+ +

+

+ +

+ –

+ –

+ –

+ – +–

Z

Y

Y Y

X

X

X X

G 1 X

Example 4.2Moving a summing point ahead of a block.

C

R(s) ˆ

G 1 G 2 G 3 G 4 (1‡G 1 G 2 H 1 )(1‡G 3 G 4 H 2 )

1 ‡ 6G 1 G 2 G 3 6G 4 H 3 (6G 1 6G 4 )(1‡G 1 G 2 H 1 )(1‡G 3 G 4 H 2 )

ˆ G1(s)G2(s)G3(s)G4(s)(1 ‡ G1(s)G2(s)H1(s))(1 ‡ G3(s)G4(s)H2(s)) ‡ G2(s)G3(s)H3(s) (4:10)

X

G

+ – Z

Y

X +–

G

1 G

C s ( )

Fig 4.6 Block diagram with interaction.

4.3 Systems with multiple inputs 4.3.1 Principle of superposition

A dynamic system is linear if the Principle of Superposition can be applied This

states that `The response y(t)of a linear system due to several inputs x1(t),

x2(t), , xn(t), acting simultaneously is equal to the sum of the responses of each

input acting alone'

Example 4.4

Find the complete output for the system shown in Figure 4.9 when both inputs act

simultaneously

Solution

The block diagram shown in Figure 4.9 can be reduced and simplified to the form

given in Figure 4.10 Putting R2(s) ˆ 0 and replacing the summing point by ‡1 gives

the block diagram shown in Figure 4.11 In Figure 4.11 note that C1(s)is response to

R1(s)acting alone The closed-loop transfer function is therefore

CI

R1(s) ˆ

G 1 G 2 1‡G 2 H 2

1 ‡G 1 G 2 H 1 1‡G H

– –

1

G 4

H 2

+ –

Fig 4.7 Modified block diagram with no interaction.

Fig 4.8 Reduced block diagram.

CI(s) ˆ1 ‡ G G1(s)G2(s)R1(s)

2(s)H2(s) ‡ G1(s)G2(s)H1(s) (4:11)

Now if R1(s) ˆ 0 and the summing point is replaced by 1, then the response CII(s)

to input R2(s)acting alone is given by Figure 4.12 The choice as to whetherthe summing point is replaced by ‡1 or 1 depends upon the sign at the summingpoint

Note that in Figure 4.12 there is a positive feedback loop Hence the closed-looptransfer function relating R2(s)and CII(s)is

CII

R2(s) ˆ

G 1 G 2 H 1 1‡G 2 H 2

1 G 1 G 2 H 1 1‡G 2 H 2

–

+ +

Fig 4.9 System with multiple inputs.

–

+ +

Fig 4.10 Reduced and simplified block diagram.

CII(s) ˆ1 ‡ G G1(s)G2(s)H1(s)R2(s)

2(s)H2(s) ‡ G1(s)G2(s)H1(s) (4:12)

It should be noticed that the denominators for equations (4.11)and (4.12)are

identical Using the Principle of Superposition, the complete response is given by

One of the most common devices for actuating a control system is the DC

servo-motor shown in Figure 4.13, and can operate under either armature or field control

(a) Armature control: This arrangement is shown in schematic form in Figure 4.14

Now air gap flux  is proportional to if, or

+ –

where Kfd is the field coil constant.

Also, torque developed Tmis proportional to the product of the air gap flux andthe armature current

e t f ( )

Field coil

R ; L a a

e t a ( )

i t a ( ) (a) Physical Arrangement

where Kamis the armature coil constant.

Substituting (4.15)into (4.16)gives

Tm(t) ˆ (KfdKamif)ia(t)(4:17)Since if is constant

where the overall armature constant Kais

When the armature rotates, it behaves like a generator, producing a back emf eb(t)

proportional to the shaft angular velocity

eb(t) ˆ Kbd

where Kbis the back emf constant

The potential difference across the armature winding is therefore

ea(t) eb(t) ˆ Ladia

Taking Laplace transforms of equation (4.21)with zero initial conditions

Ea(s) Eb(s) ˆ (Las ‡ Ra)Ia(s)(4:22)Figure 4.15 combines equations (4.18), (4.20)and (4.22)in block diagram form

Under steady-state conditions, the torque developed by the DC servo-motor is

Fig 4.14 DC servo-motor under armature control e a (t) ˆ Armature excitation voltage; e b (t) ˆ Back emf;

i a (t) ˆ Armature current; R a ˆ Armature resistance; L a ˆ Armature inductance; e f ˆ Constant field

voltage; i f ˆ Constant field current; T m ˆ Torque developed by motor; (t) ˆ Shaft angular displacement;

!(t) ˆ Shaft angular velocity ˆ d dt /

(b) Field control: This arrangement is shown in schematic form in Figure 4.13,with the exception that the armature current iais held at a constant value Equation(4.17)may now be written as

Tm(t) ˆ (KfdKamia)if(t)(4:24)and since ia is a constant, then

E s a ( )

E s b ( )

+ –

Under steady-state conditions, the torque developed by the DC servo-motor is

4.4.2 Linear hydraulic actuators

Hydraulic actuators are employed in such areas as the aerospace industry because

they possess a good power to weight ratio and have a fast response

Figure 4.19 shows a spool-valve controlled linear actuator When the spool-valve

is moved to the right, pressurized hydraulic oil flows into chamber (1)causing the

piston to move to the left, and in so doing forces oil in chamber (2)to be expelled to

the exhaust port

The following analysis will be linearized for small perturbations of the spool-valveand actuator

Increasing e t a ( )

ω( ) (rad/s) t

T t m ( ) (Nm)_

Fig 4.16 Steady-state relationship between T m (t), !(t) and e a (t) for an armature controlled DC

It is assumed that:

the supply pressure Ps is constant the exhaust pressure Peis atmospheric the actuator is in mid-position so that V1 ˆ V2ˆ Vowhich is half the total volume

of hydraulic fluid Vt the hydraulic oil is compressible the piston faces have equal areas A _Q1and _Q2 are the volumetric flow-rates into chamber (1)and out of chamber (2) the average, or load flow-rate _QLhas a value ( _Q1‡ _Q2)/2

P1 and P2 are the fluid pressures in chamber (1)and chamber (2) the load pressure PL has a value (P1 P2)

T t m ( ) (Nm)

Fig 4.18 Steady-state relationship between T m (t), e f (t) and !(t) for a field controlled DC servo-motor.

(a) Actuator analysis: The continuity equation for the chambers may be written

X_Qin

X_Qoutˆ (rate of change of chamber volume)

‡ (rate of change of oil volume)(4:30)

In equation (4.30), the rate of change of chamber volume is due to the piston

movement, i.e dV/dt The rate of change of oil volume is due to compressibility

effects, i.e.:

Bulk Modulus of oil, ˆ Volumetric stress/Volumetric strain

Note that in equation (4.31)the denominator is negative since an increase in pressure

causes a reduction in oil volume

Hence

dV

V ˆ

dP Giving, when differentiated with respect to time

dV

dt ˆ

V

 dP

_QLˆ _Qleak‡1

2

dV1dt

dV2dt

(4:35)

If leakage flow-rate _Qleakis laminar, then

where Cpis the leakage coefficient Also, if V1ˆ V2 ˆ Vo, then

_QLˆ CPPL‡ AdXdto‡4 VtdPdtL (4:39)where

dVo

dt ˆ A

dXodtand

VoˆV2t(b) Linearized spool-valve analysis: Assume that the spool-valve ports are rectan-gular in form, and have area

where W is the width of the port

From orifice theory

From equations (4.41)and (4.42)the load flow-rate may be written as

s

(4:45)Hence

Equation (4.45)can be linearized using the technique described in section 2.7.1 If qL,

xv and pLare small perturbations of parameters QL, XVand PLabout some

operat-ing point `a', then from equation (4.46)

_qLˆ@ _Q@XL

v

axv‡@ _Q@PL

L

... 17

From equations (4. 41)and (4. 42)the load flow-rate may be written as

s

(4: 45)Hence

Equation (4. 45)can be linearized using the technique... Kc To control the level

of damping, it is sometimes necessary to drill small holes through the piston

4. 5 Controllers for closed-loop systems 4. 5.1 The generalized control. ..

A generalized closed-loop control system is shown in Figure 4. 22 The control

problem can be stated as: `The control action u(t)will be such that the controlled

output c(t)will