industrial control systems - special structures

Industrial Control

Behzad Samadi

Department of Electrical EngineeringAmirkabir University of Technology

Winter 2009Tehran, Iran

Behzad Samadi (Amirkabir University) Industrial Control 1 / 59

Special Control Structures

Outline:

Set-point Weighting

Feedforward Action

Ratio Control

Cascade Control

Override Control

Selective Control

Split Range Control

[Visioli, 2006]

Outline:

Set-point Weighting

Feedforward Action

Ratio Control

Cascade Control

Override Control

Selective Control

Split Range Control

[Visioli, 2006]

Outline:

Set-point Weighting

Feedforward Action

Ratio Control

Cascade Control

Override Control

Selective Control

Split Range Control

[Visioli, 2006]

Outline:

Set-point Weighting

Feedforward Action

Ratio Control

Cascade Control

Override Control

Selective Control

Split Range Control

[Visioli, 2006]

Outline:

Set-point Weighting

Feedforward Action

Ratio Control

Cascade Control

Override Control

Selective Control

Split Range Control

[Visioli, 2006]

Outline:

Set-point Weighting

Feedforward Action

Ratio Control

Cascade Control

Override Control

Selective Control

Split Range Control

[Visioli, 2006]

Outline:

Set-point Weighting

Feedforward Action

Ratio Control

Cascade Control

Override Control

Selective Control

Split Range Control

[Visioli, 2006]

Set-point Weighting :

High load disturbance rejection → too oscillatory set-point stepresponse.

This is actually true especially when the apparent dead time of theprocess is small (with respect to the dominant time constant).

Set-point weighting for PID controllers:

u(t) = Kp(βr(t)− y(t) +1

0e(τ)dτ − Td

u(t): control signal, r(t): reference input,y(t): process output, e(t): tracking error

[Visioli, 2006]

u(t) = Kp(βr(t)− y(t) +1

0e(τ)dτ − Td

[Visioli, 2006]

u(t) = Kp(βr(t)− y(t) +1

0e(τ)dτ − Td

[Visioli, 2006]

Set-point Weighting:

u(t) = Kp(βr(t)− y(t) +1

0e(τ)dτ − Td

Intuitively, given a set of parameters Kp, Ti and Td , the adoption ofa set-point weight β < 1 allows the reduction of the overshoot in theset-point response, since the effect of the proportional action isreduced.

Note that this is achieved without affecting the load disturbancerejection performance.

[Visioli, 2006]

u(t) = Kp(βr(t)− y(t) +1

0e(τ)dτ − Td

Intuitively, given a set of parameters Kp, Ti and Td , the adoption ofa set-point weight β < 1 allows the reduction of the overshoot in theset-point response, since the effect of the proportional action isreduced.

Note that this is achieved without affecting the load disturbancerejection performance.

[Visioli, 2006]

U = Kp(βR − Y +1

Ti sE − TdsY )

U = Kp(β +1

Ti s)R − Kp(1 +

Ti s+ Tds)Y

U = Kp(1 +1

Ti s+ Tds)

βs + 1

1 + Ti s + TiTds2R − Kp(1 +

Ti s+ Tds)Y

U = Kp(1 +1

Ti s+ Tds)(FR − Y )

F =βs + 1

1 + Ti s + TiTds2

U = Kp(βR − Y +1

Ti sE − TdsY )

U = Kp(β +1

Ti s)R − Kp(1 +

Ti s+ Tds)Y

U = Kp(1 +1

Ti s+ Tds)

βs + 1

Ti s+ Tds)Y

U = Kp(1 +1

F =βs + 1

1 + Ti s + TiTds2

U = Kp(βR − Y +1

Ti sE − TdsY )

U = Kp(β +1

Ti s)R − Kp(1 +

Ti s+ Tds)Y

U = Kp(1 +1

Ti s+ Tds)

βs + 1

Ti s+ Tds)Y

U = Kp(1 +1

F =βs + 1

1 + Ti s + TiTds2

U = Kp(βR − Y +1

Ti sE − TdsY )

U = Kp(β +1

Ti s)R − Kp(1 +

Ti s+ Tds)Y

U = Kp(1 +1

Ti s+ Tds)

βs + 1

Ti s+ Tds)Y

U = Kp(1 +1

F =βs + 1

1 + Ti s + TiTds2

u(t) = Kp(βr(t)− y(t) +1

0e(τ)dτ − Td

C (s) = Kp(1 +1

Ti s+ Tds)

F (s) =1 + βTi s

1 + Ti s + TiTds2

[Visioli, 2006]

F (s) =1 + βTi s

1 + Ti s + TiTds2

From another point of view, the overshoot is reduced by smoothingthe set-point signal by means of the filter F .

Example:

P(s) =1

10s + 1e−4s

Ziegler-Nichols: Kp = 3, Ti = 8 and Td = 2

[Visioli, 2006]

F (s) =1 + βTi s

1 + Ti s + TiTds2

From another point of view, the overshoot is reduced by smoothingthe set-point signal by means of the filter F .

Example:

P(s) =1

10s + 1e−4s

Ziegler-Nichols: Kp = 3, Ti = 8 and Td = 2

[Visioli, 2006]

[Visioli, 2006]Behzad Samadi (Amirkabir University) Industrial Control 8 / 59

Feedforward Action:

Set-point tracking:

M(s) represents the desired performance.

G (s) = M(s)

P̃(s)where tildeP(s) is the minimum phase part of P(s).

[Visioli, 2006]

Feedforward Action:

Set-point tracking:

G (s) = M(s)

[Visioli, 2006]

Feedforward Action:

Set-point tracking:

G (s) = M(s)

[Visioli, 2006]

Feedforward Action:

Set-point tracking:

(MC + G )P

1 + PC

Therefore if P̃(s) = P(s) and C (s) = 0 then

R(s)= G (s)P(s) = M(s)

[Visioli, 2006]

Feedforward Action:

Set-point tracking:

(MC + G )P

1 + PC

R(s)= G (s)P(s) = M(s)

[Visioli, 2006]

Feedforward Action:

Set-point tracking:

(MC + G )P

1 + PC

R(s)= G (s)P(s) = M(s)

[Visioli, 2006]

Feedforward Action:

Set-point tracking:

Example:

P(s) =1

10s + 1e−5s

M(s) =1

2s + 1e−5s

G (s) =10s + 1

2s + 1[Visioli, 2006]

Feedforward Action:

Disturbance rejection:

Y = PCE (s) + (H − PG )D(s)

[Visioli, 2006]

Feedforward Action (disturbance rejection):

H − PG = 0⇒ G = HP

Example:

P(s) =K

Ts + 1e−Ls

H(s) =KH

THs + 1e−LH s

If LH = L:

G (s) =H

Ts + 1

THs + 1

[Visioli, 2006]

H − PG = 0⇒ G = HP

Example:

P(s) =K

Ts + 1e−Ls

H(s) =KH

THs + 1e−LH s

If LH = L:

G (s) =H

Ts + 1

THs + 1

[Visioli, 2006]

H − PG = 0⇒ G = HP

Example:

P(s) =K

Ts + 1e−Ls

H(s) =KH

THs + 1e−LH s

If LH = L:

G (s) =H

Ts + 1

THs + 1

[Visioli, 2006]

Example:

P(s) =K

Ts + 1e−Ls

H(s) =KH

THs + 1e−LH s

If LH 6= L:

G (s) =H

P≈ KH

(T − LH + L)s + 1

THs + 1[Visioli, 2006]

Example:

P(s) =K

(T1s + 1)(T2s + 1)

H(s) =KH

THs + 1

G (s) =H

P≈ KH

(T1 + T2)s + 1

THs + 1

[Visioli, 2006]

Example:

P(s) =K

(T1s + 1)(T2s + 1)

H(s) =KH

THs + 1

G (s) =H

P≈ KH

(T1 + T2)s + 1

THs + 1

[Visioli, 2006]

Example:

P(s) =K

(T1s + 1)(T2s + 1)

H(s) =KH

THs + 1

G (s) =H

P≈ KH

(T1 + T2)s + 1

THs + 1

[Visioli, 2006]

Heat Exchanger Process:

heatexdemo - Mathworks.com

t1 = 21.8 (for 28.3% of the final value)

τ = 32(t2− t1)

θ = t2− τ

τ = 32(t2− t1)

θ = t2− τ

τ = 32(t2− t1)

θ = t2− τ

τ = 32(t2− t1)

θ = t2− τheatexdemo - Mathworks.com

Gp(s) =1

23.3s + 1e−14.7s

Gd(s) =1

25s + 1e−35s

PI controller setting using ITAE

Kc = 0.859(θ

τ)−0.997, Ti = (

τ)0.680

Kc = 1.23 (blue) and Kc = 0.9 (red)

Ti = 24.56

F (s) = −Gd

Gp= −21.3s + 1

25s + 1e−20.3s

Ratio Control:

Keep a constant ratio between two (or more) process variables,irrespective of possible set-point changes and load disturbances.

Chemical dosing, water treatment, chlorination, mixing vessels andwaste incinerators

Air-to-fuel ratio for combustion systems

[Chau, 2002]

[Visioli, 2006]

Ratio Control:

[Chau, 2002]

[Visioli, 2006]

Ratio Control:

[Chau, 2002]

[Visioli, 2006]

Ratio Control:Series metered control:

The output y2 is necessarily delayed with respect to y1, due to theclosed-loop dynamics of the second loop.

[Visioli, 2006]

Ratio Control:Parallel metered control:

A disturbance acting on the first process can cause a large error in theratio value.

[Visioli, 2006]

Ratio Control:Cross limited control scheme:

The two loops are interlocked by using a low and a high selectors thatforce the fuel to follow the air flow when the set-point increases andthat force the air to follow the fuel when the set-point decreases.

[Visioli, 2006]

Ratio Control:Blend Station:

r2(t) = a(γr1(t) + (1− γ)y1(t))

Its use is suggested when no disturbances are likely to occur in theprocesses and when the two processes exhibit a different dynamics.

[Visioli, 2006]

Ratio Control:Modified Blend Station:

P1(s) =K1

T1s + 1e−L1s , P2(s) =

T2s + 1e−L2s

u1(t) = Kp1

(βr1(t)− y1(t) +

0(r1(τ)− y1(τ))dτ

)u2(t) = Kp2

(βr2(t)− y2(t) +

0(r2(τ)− y2(τ))dτ

)[Visioli, 2006]

P1(s) =K1

T1s + 1e−L1s , P2(s) =

T2s + 1e−L2s

u1(t) = Kp1

(βr1(t)− y1(t) +

0(r1(τ)− y1(τ))dτ

)u2(t) = Kp2

(βr2(t)− y2(t) +

0(r2(τ)− y2(τ))dτ

)[Visioli, 2006]

{0 if L1 > L2 and t < t0 + L1 − L2

γ? + Kp(er (t) + 1Ti

∫ t0 er (τ)dτ) otherwise

er (t) = ay1(t)− y2(t)

Kp1 Ti1 β1 Kp2 Ti2 β2 γ? Kp Tp0.9T1K1L1

3L1 0 0.9T2K2L2

3L2 0 Ti2Ti1

0.5 L2T2

[Visioli, 2006]

{0 if L1 > L2 and t < t0 + L1 − L2

er (t) = ay1(t)− y2(t)

3L1 0 0.9T2K2L2

3L2 0 Ti2Ti1

0.5 L2T2

[Visioli, 2006]

{0 if L1 > L2 and t < t0 + L1 − L2

er (t) = ay1(t)− y2(t)

3L1 0 0.9T2K2L2

3L2 0 Ti2Ti1

0.5 L2T2

[Visioli, 2006]

Ratio Control:Adaptive Blend Station (Hagglund, 2001):

Ta(ay1 − y2)

Ta = 10 max{T1,T2}

If r1 > max{y1, y2/a}+ eps then S = 1Else if r1 < min{y1, y2/a} − eps then S = −1Else S = 0[Visioli, 2006]

Ta(ay1 − y2)

Ta = 10 max{T1,T2}

Ta(ay1 − y2)

Ta = 10 max{T1,T2}

Cascade Control:

[Chau, 2002]

Cascade Control:

[Chau, 2002]

Cascade Control:

Primary (master) loop: outer loop

Secondary (slave) loop: inner loop

P1: slow dynamics

P2: fast dynamics

The approach can be generalized to more than two loops.

[Visioli, 2006]

Cascade Control:

P1: slow dynamics

P2: fast dynamics

[Visioli, 2006]

Cascade Control:

P1: slow dynamics

P2: fast dynamics

[Visioli, 2006]

Cascade Control:

P1: slow dynamics

P2: fast dynamics

[Visioli, 2006]

Cascade Control:

P1: slow dynamics

P2: fast dynamics

[Visioli, 2006]

Cascade Control:

It appears that the improvement in the cascade control performanceis more significant when disturbances act in the inner loop and whenthe secondary sensor is placed in order to separate as far as possiblethe fast dynamics of the process from the slow dynamics(Krishnaswami et al., 1990).

Additional advantage: The nonlinearities of the process in the innerloop are handled by that loop and therefore they are removed fromthe more important outer loop.

When the secondary process exhibits a significant dead time or thereis an unstable (positive) zero, the use of cascade control is not usefulin general.

[Visioli, 2006]

Cascade Control:

[Visioli, 2006]

Cascade Control:

[Visioli, 2006]

Cascade Control:Design procedure:

First: Design the controller for the secondary loop

Second: Design the controller for the primary loop

[Visioli, 2006]

Cascade Control:Design procedure:

First: Design the controller for the secondary loop

Second: Design the controller for the primary loop

[Visioli, 2006]

Cascade Control:Example:

Gp =0.8

2s + 1, Gv =

s + 1, GL =

[Chau, 2002]

Let us consider a proportional controller Gc2(s) = Kc2 and make thetime constant of the secondary loop equal to 0.1 second.

1 + 0.5Kc2 = 10⇒ Kc2 = 18

G ?v = 9

G ?v Gp = 0.9

0.1s+10.8

G ?v Gp ≈ 0.72

2.05s+1e−0.05s

SIMC tuning rule:

τ1τc + θ

0.15= 18.98

τi = min{τ1, 4(τc + θ)} = min{2.05, 4(0.15)} = 0.6

[Chau, 2002]

1 + 0.5Kc2 = 10⇒ Kc2 = 18

G ?v = 9

G ?v Gp = 0.9

0.1s+10.8

G ?v Gp ≈ 0.72

2.05s+1e−0.05s

SIMC tuning rule:

τ1τc + θ

0.15= 18.98

τi = min{τ1, 4(τc + θ)} = min{2.05, 4(0.15)} = 0.6

[Chau, 2002]

1 + 0.5Kc2 = 10⇒ Kc2 = 18

G ?v = 9

G ?v Gp = 0.9

0.1s+10.8

G ?v Gp ≈ 0.72

2.05s+1e−0.05s

SIMC tuning rule:

τ1τc + θ

0.15= 18.98

τi = min{τ1, 4(τc + θ)} = min{2.05, 4(0.15)} = 0.6

[Chau, 2002]

1 + 0.5Kc2 = 10⇒ Kc2 = 18

G ?v = 9

G ?v Gp = 0.9

0.1s+10.8

G ?v Gp ≈ 0.72

2.05s+1e−0.05s

SIMC tuning rule:

τ1τc + θ

0.15= 18.98

τi = min{τ1, 4(τc + θ)} = min{2.05, 4(0.15)} = 0.6

[Chau, 2002]

1 + 0.5Kc2 = 10⇒ Kc2 = 18

G ?v = 9

G ?v Gp = 0.9

0.1s+10.8

G ?v Gp ≈ 0.72

2.05s+1e−0.05s

SIMC tuning rule:

τ1τc + θ

0.15= 18.98

τi = min{τ1, 4(τc + θ)} = min{2.05, 4(0.15)} = 0.6

[Chau, 2002]

1 + 0.5Kc2 = 10⇒ Kc2 = 18

G ?v = 9

G ?v Gp = 0.9

0.1s+10.8

G ?v Gp ≈ 0.72

2.05s+1e−0.05s

SIMC tuning rule:

τ1τc + θ

0.15= 18.98

τi = min{τ1, 4(τc + θ)} = min{2.05, 4(0.15)} = 0.6

[Chau, 2002]

Cascade Control:Simultaneous tuning of the controllers:

Assume:P2(s) = P2m(s)P2a(s)

P2a(s) is the all-pass portion of the transfer function containing allthe nonminimum phase dynamics (P2a(0) = 1)

[Visioli, 2006]

Cascade Control:Simultaneous tuning of the controllers:Desired inner loop transfer function:

T̄r2y2(s) =P2a(s)

(λ2s + 1)n2

λ2 and n2 are design parameters.

C2(s) =P−1

(λ2s + 1)n2 − P2a(s)

To approximate the controller with a PID controller:

C2(s) =1

sk(s) ≈ 1

[k(0) + k̇(0)s +

k̈(0)

][Visioli, 2006]

Cascade Control:Simultaneous tuning of the controllers:Assume:

P21(s) = P1(s)P2a(s)

(λ2s + 1)n2= P12m(s)P12a(s)

P12a(s) is the nonminimum phase part in all-pass form.

Desired outer loop transfer function

T̄r1y1(s) =P12a(s)

(λ1s + 1)n1

Primary controller transfer function

C1(s) =P−1

12m(s)(λ2s + 1)n2

P2a(s)((λ1s + 1)n1 − P12a(s))

[Visioli, 2006]

P1(s) =K1

T1s + 1e−L1s

P2(s) =K2

T2s + 1e−L2s

T̄r2y2(s) =e−L2s

λ2s + 1

T̄r1y1(s) =e−(L1+L2)s

λ1s + 1[Visioli, 2006]

Kp2 =T2 +

2(λ2+L2)

K2(λ2 + L2)

Ti2 = T2 +L2

2(λ2 + L2)

Td2 =L2

6(λ2 + L2)

3− L2

T2 +L2

22(λ2+L2)

[Visioli, 2006]

Kp1 =T1 + λ2 + (L1+L2)2

2(λ1+L1+L2)

K1(λ1 + L1 + L2)

Ti1 = T1 + λ2 +(L1 + L2)2

2(λ2 + L1 + L2)

Td1 =λ2T1 − (L1+L2)3

6(λ1+L1+L2)

T1 + λ2 + (L1+L2)2

2(λ1+L1+L2)

+(L1 + L2)2

2(λ1 + L1 + L2)

The suggestion is to set:

λ2 = 0.5L2

λ1 = 0.5(L1 + L2)

[Visioli, 2006]

Override Control: There are two modes of operation

Normal operation: One process variable is the controlling variable.

Abnormal operation: Some other process variable becomes thecontrolling variable to prevent it from exceeding a process orequipment limit.

The limiting controller is said to override the normal processcontroller.

http://pse.che.ntu.edu.tw/chencl/Process_Control

Override Control:

[Smith, 2002]

Override Control:

The set point to LC50 is somewhat above h2, as shown in the figure.

The FC50 is a reverse-acting controller, while the LC50 is adirect-acting controller.

The low selector (LS50) selects the lower signal to manipulate thepump speed.

[Smith, 2002]

Override Control:

[Smith, 2002]

Override Control:

[Smith, 2002]

Override Control:Reset Feedback (RFB) or external reset feedback:

This capability allows the controller not selected to override thecontroller selected at the very moment it is necessary.

When FC50 is being selected, its integration is working, but not thatof LC50 (its integration is being forced equal to the output of LS50).

When LC50 is being selected, its integration is working but not thatof FC50 (its integration is being forced equal to the output of LS50).

[Smith, 2002]

Selective Control:

The high selector in this scheme selects the transmitter with thehighest output, and in so doing the controlled variable is always thehighest, or closest to the highest, temperature.

[Smith, 2002]

Split range:A very common control scheme is split range control in which the outputof a controller is split to two or more control valves. For example:

Controller output 0% Valve A is fully open and Valve B fully closed.

Controller output 25% Valve A is 75% open and Valve B 25% open.

Controller output 50% Both valves are 50% open.

Controller output 75% Valve A is 25% open and Valve B 75% open.

Controller output 100% Valve A is fully closed and Valve B fully open.

www.contek-systems.co.uk

Split range:

X: Steam Valve, air-to-open

Y: Cooling Water Valve, air-to-close

TC47: Temperature Controller

TY47: I/P converter

[Love, 2007]

Split range:

ZC47A,B: Valve positioners

TV47A,B: Valves

[Love, 2007]

Split range:Different arrangements are possible. For example, the split can beconfigured as follows:

Controller output 0% Both valves are closed.

Controller output 25% Valve A is 50% open and Valve B still closed.

Controller output 50% Valve A is fully open and Valve B closed.

Controller output 75% Valve A is fully open and Valve B 50% open.

Controller output 100% Both valves are fully open.

Split range:Example:

Split range:

In this application, the flare valve will need to open quickly inresponse to high pressures, but the compressor suction valve will needto move much more slowly to prevent instability in the compressors.The main problem with split range control is that the controller onlyhas one set of tuning parameters.

The solution is to replace the split range controller with twoindependent controllers, both reading the same pressure transmitter,but one controlling the flare valve and the other the suction valve.

Split range:

In this application, the flare valve will need to open quickly inresponse to high pressures, but the compressor suction valve will needto move much more slowly to prevent instability in the compressors.The main problem with split range control is that the controller onlyhas one set of tuning parameters.

The solution is to replace the split range controller with twoindependent controllers, both reading the same pressure transmitter,but one controlling the flare valve and the other the suction valve.

Two controller implementation:

Marriage analogy:

Split - range control is like a good marriage. One partner may bedoing 90of the work, but both partners are occasionally going to sharethe work.

Override control is like a bad marriage. One partner plays a potentiallydominating role, even though the other partner is doing all the work.

Cascade control is more like my marriage. I do the best I can, but mywife Liz constantly and lovingly recalibrates my efforts. She dampensdown the extremes in my behavior so as to promote a stablerelationship and home life.

[Lieberman, 2008]

Marriage analogy:

[Lieberman, 2008]

Marriage analogy:

[Lieberman, 2008]

Smith predictor:

To handle systems with a large dead time

[Chau, 2002]

Smith predictor:

[Chau, 2002]

Smith predictor:

[Chau, 2002]

Chau, P. C. (2002).Process Control: A First Course with MATLAB (Cambridge Series inChemical Engineering).Cambridge University Press, 1 edition.

Lieberman, N. (2008).Troubleshooting Process Plant Control.Wiley.

Love, J. (2007).Process Automation Handbook: A Guide to Theory and Practice.Springer, 1 edition.

Smith, C. A. (2002).Automated Continuous Process Control.Wiley-Interscience, 1 edition.

Visioli, A. (2006).Practical PID Control (Advances in Industrial Control).Springer, 1 edition.

industrial control systems - special structures

Technology

special control structures

td s s

control signal

ti td s2 visioli

ti s r kp1

td s fs

ti s e td sy u

weighting visioli