two level control of

TWO-LEVEL CONTROL OF

Processes with Dead Time and Input Constraints

Qing-Chang Zhong∗ and Chang-Chieh Hang∗∗

[email protected], [email protected]

∗School of Electronics

University of Glamorgan

United Kingdom

∗∗Dept. of Elec. & Comp. Eng.

National Univ. of Singapore

Singapore

Outline

Background information

Motivation

Two-level control signal

Controller design

Implementation of the controller

Simulation examples

Q.-C. ZHONG & C.-C. HANG: TWO-LEVEL CONTROL OF PROCESSES WITH DEAD TIME AND INPUT CONSTRAINTS – p. 2/25

Background informationregulationv.s. set-point responses

regulation: the major taskset-point responses: often necessary2DOF controller

fast set-point response

fast but without overshootdead timeinput constraint

In general, this is difficult. However, for some pro-cesses, this can be done.


Systems under considerationThe most common chemical processes:the first-order plus dead time (FOPDT)

G(s) =Ke−τs

Ts + 1,

whereK is the static gain,τ is the dead time andT isthe apparent time constant.

G(s)ZOHC(z)F (z) i i- - - - - - -?

��

6

d(s)y(s)r(z) r′ e u

−

Ts


MotivationA typical control signal using a PI controller withF (z) = 1

Three stages:I: due to the integratingeffect, the control signalincreases until the actua-tor saturates;II : the integrator “windsup” and the actuator sat-urates;III : the control signalsettles down.

I II III

u

u

o Time (sec)


Why is the response slow?Stage I: The proportional gain cannot be too largeotherwise the actuator saturates very quickly.This means that the potential of the controller isoften not fully used to speed up the systemresponse; see the shaded area in the figure.

Stage II: The integrator windup requires the errorsignal to go opposite for a long period to drag theintegrator back to normal. This causes a largeovershoot and long settling time.

Stage III: The oscillation is not desirable either,which causes a long settling time.

=⇒The desired control signal


Two-level control signal

Time (sec) O

u

u

r/K

(n + 1)Ts lTs

The desired control signal when the set-point changeis the boundr of all step changes


Two-level control signal (cont’d)The signal can be expressed as:

1 − an+1z−n−1

K(1 − an+1)· r,

which gives the desired transfer func-tion from r to u:

T dur(z) =

1 − an+1z−n−1

K(1 − an+1).

Time (sec) O

u

u

r/K

(n + 1)Ts lTs

The first part should be under the saturation boundu:

r

K(1 − an+1)≤ u ⇒ n ≥

T

Tsln

Ku

Ku − r− 1.


Controller design

G(z) = K1 − a

z − az−l, C(z) =

(1 − az−1)N(z)

D(z)

l = τ/Ts is a positive integer anda = e−Ts/T . Theorder of polynomialsN(z) andD(z) in z−1 is n andm, respectively.

Tyr(z) = F (z)K(1 − a)N(z)z−(l+1)

D(z) + K(1 − a)N(z)z−(l+1)

Tur(z) = F (z)(1 − az−1)N(z)

D(z) + K(1 − a)N(z)z−(l+1).


Controller design: F (z)

Since the closed-loop system is stable,F (z) can besimply chosen to cancel the closed-loop poles.

F (z) =D(z)

K(1 − a)+ N(z)z−(l+1).

Then,Tyr(z) = N(z)z−(l+1)

Tur(z) = N(z)1 − az−1

K(1 − a).

The outputy is expected to start just after the deadtime,N(0) 6= 0. Hence,D(0) 6= 0.


Controller design: N(z)

The desired transfer function:

T dur(z) =

1 − an+1z−n−1

K(1 − an+1).

The actual transfer function:

Tur(z) = N(z)1 − az−1

K(1 − a).

⇓

N(z) =

∑ni=0 aiz−i

∑ni=0 ai

.


Controller design: D(z)

D(z) is designed to guarantee the stability of theclosed-loop system. One possibility is to choose

D(z) =1 − z−1

KIN(z)

to offer a PI controller:

C(z) =(1 − az−1)N(z)

D(z)= KI

1 − az−1

1 − z−1.

The corresponding open-loop transfer function is

L(z) = C(z)G(z) =KIK(1 − a)

(z − 1)zl.


A typical root-locus diagram

O

1 Re

Im


Tuning of the controller: KI

Theorem The closed-loop system is stable if

0 < KI <2

K(1 − a)sin

π

4l + 2.

To obtain a phase margin ofφm, KI can be chosen as

KI =2

K(1 − a)sin

π − 2φm

4l + 2.

To obtain a gain margin ofgm, KI can be chosen as

KI =2

K(1 − a)gmsin

π

4l + 2.


Some commentsthe settling time is approximately

(l + n + 1)Ts ≈ τ + T lnKu

Ku − r.

It is independent of the control parameterKI and the sampling period. It depends on the

saturation boundu and is hence an inherent property of the system. There is no way to

make the response any faster.

the static error is0 becauseN(1) = 1.

There is no braking control.

there is no need for such a brake because the response reachesthe steady state in

finite time and there is no overshoot;

the benefit of a large negative action is very small whenu is not very large, which

is the common case in practice,

the control strategy is more sensitive when there is a large negative control action.


An alternative implementation

G(s)ZOHFu(z)

Gm(z)

C(z) j

j j6

-

-

- - - - -?

��

6

d(s)y(s)r(z)

ym

uuo

uc

−Ts

Fu(z) =1 − an+1z−n−1

K(1 − an+1), Gm(z) = K

1 − a

1 − az−1z−(l+1).

This structure appeared in [Wallén and Åström, 2002].


Advantages of this implementationThe control signalu is split into two parts:

u = uo + uc,

with the desired (open-loop) control signaluo and the contributionuc of the feedback

controllerC resulted from disturbances and model uncertainties.

There is strong connection with the input-shaping technique.

It is clearer that the desired control signal can be designedin an open-loop way if the

plant is stable. What’s extra is to inject this desired control signal into the modelGm of

the process and to obtain the error between the model outputym and the process outputy

for error feedback.

The feedback controllerC does not affect the shape of the control signal, which is not

explicit in the case discussed before (whereN(z) is a part of the controller). This means

that the controller may not be limited to a PI controller as designed above. In other words,

the proposed technique can be regarded as a “bolt-on” to any standard well-tuned PID

controllers.

The sampling periods for the feedforward controllerFu(z) and the feedback loop can be

different to give more freedom to the design of the feedback controller.Q.-C. ZHONG & C.-C. HANG: TWO-LEVEL CONTROL OF PROCESSES WITH DEAD TIME AND INPUT CONSTRAINTS – p. 17/25

Internal model controlIf C(z) is designed to be

C(z) =Fu(z)

1 − Gm(z)Fu(z),

then the system is actually the well-known IMC.

G(s)ZOHFu(z)

Gm(z)

m m m

C(z)

6

�

6

- - - - - - -?

��

d(s)

y(s)r(z) u

−

Ts


An example

G(s) =e−5s

s + 1

Ts = 0.25s ⇒ a = 0.7788, l = 20

u = 1.45, r = 1 ⇒ n ≥ 3.68

N(z) = 0.31+0.2414z−1 +0.1881z−2 +0.1464z−3 +0.1141z−4.

φm = 45◦ ⇒ KI = 0.173.


The converted set point r′

0r

Con

vert

ed s

et p

oint

Time (sec)

effect of D(z)

effect of N(z)

r’

mTs lTs

m = n + 1 = 5, l = 20, Ts = 0.25


The system response

0 1 2 3 4 5 6 7 8 9 10−1

0

1

2

3

4

5

6

7

8

9

e

Time (sec)

The error signale = r′ − y

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

Time (sec)

u

y

The output y and thecontrol signalu


Another example

G(s) =e−0.5s

s + 1


Comparative studies

(a) the system outputs (b) the control signals


Robustness

(a)T increased by20% (b) τ increased by20%


SummaryA controller is designed to obtain atwo-level controlsignal and adeadbeatset-point response. The con-troller is tuned to obtain the desired stability margin.

Time (sec) O

u

u

r/K

(n + 1)Ts lTs


two level control of

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