Proportional control

Consider forward path gain A

Feedback and Control

VI

A VO VI

A VO

V

V

A

1 A

A

1 AO

I

V

V

A

A

1O

I

If the size of the loop gain is large, that is if |A >> 1, then

or

Consider an amplifier:

Feedback and Control

R1

R2

VI VOVI

R2

R1+R2

A VO

VOR2

R1+R2

V

VA

1AR2

R1 R2

A(R R )

R R ARO

I

1 2

1 2 2

V

V

A(R R )

AR

R R

RO

I

1 2

2

1 2

2

V

V

1 R R

RO

I

1 2

2

If the loop gain AR2/(R1+R2) is large, AR2

>>R1+R2:

In the above, =R2/(R1+R2), so if loop gain is large:

e.g. A=105, R1=6k, R2=4k:

AR2=4x108>>R1+R2=104. VO/VI = 10/4=2.5

Consider an actual amplifier:

Assuming T = 0.001 s

Feedback and Control

VI

R2

R1+R2

A VO1+sT =

Feedback and ControlSecond order systems:

[Unless told otherwise, assume D = 0]

Determine values for C in order to make the system

• Overdamped

• Critically damped

• Underdamped

Higher gain -> faster but more oscillatory response

If too oscillatory - very long time before final value.

We cannot, therefore, just increase controller gain

0.25 Os(1+s0.25)

D

I C

Feedback and ControlSecond order systems, advanced control:

Suppose the system is one with controller C and plant K (at steady state) with no integrator in the plant. {Plant may be of form K/(1+sT)}.

The steady state output is If CK is large, the steady state error is reduced: ~ 1.

But cannot always increase C too much: get oscillatory response

If an integrator is put in so C = C1 + C2/s, steady state errors are removed!

OKII OKC CP

Controllers which cancel part of Plant

I Ocontroller plantE

K

(1+ sT )(1+ sT )1 2Plant

a. Sketch the argand diagram and time response ifT1 is too large for a required fast acting system:

b. Sketch the argand diagram and time response ifT2 is too small for a slow reacting system.

[Assume other time constant was set appropriately]

Controllers which cancel part of Plant

I Ocontroller plantE

a

e

sT+1

sT+1C K

(1+ sT )(1+ sT )1 2

)sT+)(1sT+(1

CK

E

O

1a

loop-1

forward

I

O

1+CK+)T+s(T+TTs

CK

CK)sT+)(1sT+(1

CK

1I

O

1a1a2

1a)sT+)(1sT+(1

CK)sT+)(1sT+(1

CK

1a

1a

Plant

Make Te equal to largest of T1 and T2, say T2, to

speed up system. Then

Gives

Lead Lag Controller

Controllers which cancel part of Plant

The above is an example of pole zero cancellation.A pole is a 1+sT term on denominator, a zero is such a term on numerator.

0 10 20 30 40 500

0.2

0.4

0.6

0.8

1

1.2

1.4

Above shows system where T1=8 and T2 = 15, with and without pole zero cancellation: d.c. gain of controller is 9. System has been speeded up.

Controllers

PI

sC

1+ sT

s

K

(1+ sT )(1+ sT )1 2

O

E

CK

s(1+ sT )1

O

I

CK

s T + s + CK21

Another form of controller is the P+I controller Its transfer function can be written as

Here C = P and T = I/C.P and I can be chosen so the 1+sT term (controller zero) cancels Plant pole.

Suppose Plant

If apply P+I to this Plant, and make T = T2,

then

So

Note, the I term means that the steady state value is 1.

Controllers

P Ds C(1 sT)

K

s(1+ sT)

O

E

CK

s

CKs

CK

I

O

s

TTs)sTT(1C

s

Ds+Ps+I=Ds+

s

I+P 21

221

2

O

E

CK

s

P+D controller

Make 1+s T cancel Plant pole.If Plant is and P+D is applied, then

so

PID controller - can cancel two lags in a plant:

then

In all these examples, by careful arrangement, systems is first/second order.Cancellation may not give best response, but analysis of systems is easier!

The transfer function relating the output position, P, to the armature voltage V of an armature controlled motor is:

where K is 4 Nm/A and T is 0.1s.

The motor forms part of a position control system, as shown below,

where the controller gain C, R required position.

Derive the transfer function relating the R and P.

If R is a unit height step applied at time t = 0, sketch graphs of P and R against time for the following three cases:

a) C = 8 b) C = 10 c) C = 12

In each case show the calculations which you have used to determine the form of the response.

Example Exam Question

sT)s(1

1/K

V

P

R P1/KV

s(1+sT)C

1) For the system below:

a) Sketch the step response when C is 40: label your sketch suitably. Assume D is 0

b) If C is 40, by how much would o change if the plant becomes 0.2/(1+0.25s)

c) If C is 40, and D is a unit step, by how much does o change?

d) Sketch the step response if C = 15.

2) In a central heating system, the transfer function of the heater is

where V is the signal into the heater, T is the temperature.

Sketch variation of T if V is a unit height step.

The heater is put in a feedback system- controller gain 50.

Find the closed loop transfer function, and sketch variation of T if the input is a step.

Comment on the effect on the system of feedback.

Exercises

s101

0.2

V

T

i

0.25 o1+0.25s

D

C

1) For a position control system for which Km = 10 and Tm = 0.2, sketch the response when the input is a unit step, if

a) C = 9 b) C = 12.5 c) C = 15.

2) A field controlled motor has the following parameters:

K = 2 Nm/A, R = 5W, L = 10H, J = 0.06 kgm2 & F = 0.02 Nm per rad/s.

Sketch the open loop and the closed loop step response when the controller has a gain of 1.

3) A motor with transfer function

is connected in a feedback loop with controller gain C.

Find the closed loop transfer function.

Sketch the step response of the closed loop system when:

a) C = 1.25b) C = 2 c) C = 1 d) C = 5

Show all calculations used in producing these sketches.

Exercises

s2)s(1

1/10

V

P

sTm)s(1

1/Km

V

P