Control Theory

D action – Tuning

When there’s too much oscillation, this can sometimes be solved by adding a derivative action. This action will take into account how fast the error signal is changing.

In the time domain this leads to:- the PD controller:

- and the PID controller:

dt

tdeTKteKtu Dcc

)()()(

dt

tdeTKdtte

T

KteKtu Dc

t

I

cc

)()()()(

0

Derivative Action

)1()( sTKsT DcR PD in the s domain:

PID in the s domain:

sT

sTsTTK

sTsTKsT

I

IDIc

IDcR

1)

11()(

2

Derivative Action

Group Task

P or PD? 2/(s2+2s)r(t) z(t)

ProcessController

Compare a P controller with Kc=10 with a PD controller with Kc=10 and TD=0,2

(step response for the servo problem)

Group Task

Tuning = choosing appropriate parameters for the controller

For PID control: choice of Kc, TI and TD

2 possibilities:- Based on (open/closed loop) model- Based on experiments

2 pretty much used methods:- Ziegler-Nichols- Cohen-Coon (not discussed)

Tuning

(one type of) ZIEGLER-NICHOLS:1) Start with a P controller with low Kc and increase this gain

slowly until y oscillates with a constant amplitude. (Remark: every time Kc is increased, one should bring the closed loop system out of equilibrium, e.g. by slightly changing the reference!)

2) The gain we have now is called the ultimate gain Ku and the period of the oscillation is called the ultimate period Tu

3) Choose the PID parameters according to the following table:

Kc TI TD

P Ku/2 - -

PI 0.45 Ku Tu/1.2 -

PID 0.6 Ku Tu/2 Tu/8

Tuning

Example:

A process with the following step response:

Tuning

Example:

Ku = 1.45Tu = 4.5s

bv.PID:Kc = 0.87TI = 2.25sTD = 0.65s

Tuning

Example:

Ku = 1.45Tu = 4.5s

bv.PID:Kc = 0.87TI = 2.25sTD = 0.65s

Remark: ZN typically leads to pretty much overshoot...(there are revised ZN methods that solve this problem)

5 10 15

1

Tuning

Example:

Suppose a model of the process was given by

2.023

4)(

23

ssssT

Can you now tune according to ZN based on this model?

Tuning

Example:

2.023

4)(

23

ssssT

-3.227dB

1.4rad/s

Tuning

Problem(s) introduced by D-control?

A. Control action in servo problem becomes “infinitely” high.

B. High frequency behavior/noise is amplified.C. None of the above.D. Both A and B

Bode plots!Bode plot PD:

We introduce an extra pole to solve the problem at high frequencies:

1)(,

s

sTsT

D

DtamD

This means we first filter the error with a low pass filter.Off course the time constant of the filtering << TD

(Remark: to solve the problem of suddenly changing reference values, often another solution is presented: the derivative action is put on the MEASUREMENT instead of on the ERROR!)

Derivative Action with filtering

Difference in Bode plot:

Remark: The positive effect on the phase shift is now lost at higher frequencies!

Blue: D actionGreen: D action + prefiltering

Derivative Action with filtering

1) What is a time delay?

Typical example: measurement comes too late, e.g.:

TR TR,mTR

TR,mtd

Influence of a time delay

In other words:

Dead time td

f(t) f(t-td)

This is of course a pure phase shift. In Laplace (see table):

Dead time td

F(s) e-td s F(s)

A dead time is given as e-td s in the s domain: it’s non-linear!On the Bodeplot?

AR = |e-td jω|=1 φ = (e-td jω) = -tdω (in rad)

Pure phase shift

Influence of a time delay

A- Bode plot of a time delay:

2) What is the influence on a feedback system?

Influence of dead timeCHAPTER 3. PID CONTROL

φ = (e-td jω) = -tdω * 180/π (in degrees)

This can have severe impact on the stability:The information comes too late.

How can we see this in our analysis?- Based on TF: difficult: time delay = non-linear thing- … but it is a pure phase shift… on Bode plot?

Influence of a time delay2) What is the influence on a feedback system?