Control outputi.e. Output = Input, despite disturbancesThis is achieved by feedback.

Open loop systems - i.e. without feedback

Process with transfer function P perturbed by a disturbance D.

Suppose P is 10 and disturbance D is 0.If the output O is to be 1, just make input I = 0.1.

But, if P changes by 10% to 11 then O changes by 10% to 1.1.

If disturbance D is 0.1, then O will also change by 0.1.

Feedback control

I OP

DProcessDisturbance

OutputInput

Closed loop system: feedback added!Now consider the ‘closed-loop’ system below: P represents the device being controlled; C is the controller.

Ignoring disturbances (D = 0), by forward over 1 minus loop rule

Let P = 10, as before, and C = 10;

If I is 1, then O is 0.99 i.e. it is within 1% of being 1

If P changes to 11; O = I * 110/111; if I is 1, O is still about 0.99.

Feedback control

I OP

D

C

Controller ProcessDisturbance

OutputInput

O =C * P

1 + C * PI

I 0.99I011

100=I

01* 10 + 1

10 * 10=O

To see the effect of disturbances, assume I is 0. Then

If C = 10, P = 10 and D = 0.1

Negative Feedback•reduces effects on output of disturbances•reduces effects on output of parameter changes

if |closed loop gain| < |open loop gain|

Disturbances control

I OP

D

C

Controller ProcessDisturbance

OutputInput

P * C + 1

1=

Loop-1

Forward

D

O

0.00099011

0.1=0.1

01* 10 + 1

1=O

If CP large: O ~ I + 0 = I

So Feedback • makes output almost same as input, • minimises effects of disturbances and • reduces effect of change in device.This is true because the ‘loop gain’, C * P, is high.

Note, can’t just keep increasing the ‘gain’ of C.Also, need to consider the dynamics of the blocksAlso, there can be a block in the feedback path which we must consider

Principle of Superposition

I OP

D

C

Controller ProcessDisturbance

OutputInput

DP * C + 1

1I

P * C + 1

P*C=O

Specify a motor and resistor for R2-D2?

Assume we walk at 2 m/s & R2-D2’s wheel diameter is 4cm. Therefore, required angular velocity is: 16 revs/s -> 100 radians/s

Input voltage is 4 D cells giving 6V input.

Ke = 6/100 = 0.06 V/rad/s

Weight of R2-D2 gives an inertia torque (J):

J = 0.05 kgm2

Assume current is 1 amp -> R = 1

Want T = 0.2s

KT = 25 Nm/A

We need to form a relationship between input voltage and output velocity:

R2-D2 Motor System

sT

K

I

O

11

te

a

KK

JRT 1

Has a time response to a unit step input:

Unit Step Response of System

Time

Output Output when K = 1, T= 0.01

Input

Output Output when K = 1.6, T= 0.02

Input

Now include armature inductance:

Armature resistor:

Back emf of motor:

Torque proportional to armature current:

Torque is opposed by the inertia torque:

Hint: apply Kirchhoff’s voltage law to the armature circuit

We need to form a relationship between input voltage and output velocity:

R2-D2 Motor System

Rba vvv

sVa sIa

sVb

T

aTiKT eb Kv aaR Riv

dt

dJT

dt

diLv aaL

Include new component:Components of Motor System

sIRsV aR

sKsV eb

sJssT

bV

RV

eK

RI

Js

T

+_

bV

aV RVRba vvv

aTiKT

eb Kv

aaR Riv

dt

dJT

sIKsT TTI TK

ssILsV aL

LVLasIdt

diLv aaL

Reduce block diagram:

bV

aV+_

1

1

2 sKKJR

sKKJL

K

te

a

te

a

e

1.

2.

3.

eKbV

TK Js1

bV

aV+_

eKbV

2sJLsJR

K

aa

T

aV

aIba VV +_ 1/R

sLaaILV

sLR aa 1

Output linked to input:

Previously (inductor = 0)

Can be expressed much more simply!:

Previously (inductor = 0)

Transfer Function of System

1

1

sKKJRK

V

te

a

e

a

11 21

sTsT

K

Va

1

1

2

sKKJR

sKKJL

K

V

te

a

te

a

e

a

11

sT

K

Va

Any system of the form:

Has a time response (depending on input):

Time Response of System

sTsT

K

I

O

21 11

Time

Output Varies!

Has a time response to a unit step input:

Over Damping

Time

Output Output is over damped

Input

Output

Output changes with k, T1 & T2

Input

Has a time response to a unit step input:

Critical Damping

Time

Output

Output is critically damped

Input

Output

Output from unique T1, T2 & k

Input

Has a time response to a unit step input:

Under Damping

Time

Output

Output is under damped

Input

Output

Output changes with k, T1 & T2

Input

d0 = tf(1.0,[1 0 1]) %undamped

d1 = tf(1.0,[1 2 1]) %critically damped

d2 = tf(4.0,[1 2 4]) %under damped

d3 = tf(0.5,[1 2 0.5]) %over damped

T= [0: 0.01: 20];%set up the time increments

[y0,t]=step(d0,T);%step response over one second

[y1,t]=step(d1,T);%step response over one second

[y2,t]=step(d2,T);%step response over one second

[y3,t]=step(d3,T);%step response over one second

stept = 1 + 0*t; %graph to show step response

clf; %clear all graphs

hold on % put each graph on top of each other

plot(t,y0, 'r');

plot(t,y1,'k');

plot(t,y2,'g');

plot(t,y3,'b');

plot(t,stept,'m');

Matlab code

1) Write down the transfer function for the RC circuit when

R= 2kΩ and C = 5mF.

Sketch the response of the system to a unit step input, marking the time constant's position on the time axis and the final value on the other axis.

2) Find the transfer function of the thermal system for which

R = 4 KW-1 and C = 2 JK-1.

Sketch the response of the output if the input is a step change of 2 W

3) For each of the following work out O/I and sketch response if I is a step.

Exercises

I

5s

2

2

5/s OI

3

2/s OO I