Workbook V 19 · Control Systems I Prof. Dr. Bayerlein WB file Workbook CS I_19_2.docx from...

English Version of my ‘Ergänzungsblätter’ Only supplementary information to my lecture V18.1=V17.1: Sign ≙

Additional available documentation in English:

All Blocks p.16 WB with Screens Regcsharp V3.pdf RTskript_rev4US complete.pdf

Workbook V 19.2

Control Systems for ISE/MSOE

by

Prof. Dr. Bayerlein

WS 2019/20

Control Systems I Prof. Dr. Bayerlein

WB file Workbook CS I_19_2.docx from 04.09.2019 2

Contents of Workbook Contents of Workbook ............................................................................................................... 2 1. Content of the lecture Control Systems .................................................................................. 3 References .................................................................................................................................. 4 Software ..................................................................................................................................... 4 RegCSharp.exe ........................................................................................................................... 4 WindfC# ..................................................................................................................................... 4

Mechanical Elements ............................................................................................................ 6 Conversion rules for Block diagrams ..................................................................................... 7

3. Overview of Dynamic Relationships ..................................................................................... 8 Laplace-Transform Table (1.Nb.=degree num., 2. Nb. = degree denom.) ............................ 9 To Chap 3.5. Graphical display of Transfer Functions ........................................................ 12 Example to chap 3.5.3 mesh plot of a transfer function ....................................................... 14 Mesh plot with matlab .......................................................................................................... 15 Table with standard blocks (p. 16) ....................................................................................... 16

Calculation 3 RC- chain with OpAmp ..................................................................................... 17 Control System Lab exp. 1 part 3 ......................................................................................... 17 Further pages to 3.7 : DT2-block ......................................................................................... 18 Further pages to 3.7 : PIDT1 - Block ................................................................................... 20 Further pages to 3.7: PT2 - Block ........................................................................................ 22 Project 2: Temperature control for a hand drying fan with PIDT1 ...................................... 25

1. Step: Determine the necessary hardware in addition to the fan. ................................. 25 2. Step: Identification of process dynamic. ...................................................................... 26 3. Step: Design of the controller ....................................................................................... 28 4. Step: Realization of the controller ................................................................................ 31

Chap 3.8 ‘Flash-design’ by formula ..................................................................................... 33 Chap 3.9. Nichols - Formulas ............................................................................................... 34

4. Chap. : Stability Criteria ....................................................................................................... 35 Poles ................................................................................................................................. 35 Hurwitz Criterion ............................................................................................................. 35

5. Chap FRA-Design with the 4PT1- example ........................................................................ 36 Chap 5.2 Step-by-step-procedure FRA-design of PIDT1 .................................................... 37

FRA- Design 4 PT1- Example with new plots ......................................................................... 38 P- Controller (process 4 PT1) ............................................................................................... 38 I-Controller (process 4 PT1) ................................................................................................ 39 PI- Controller (process 4 PT1) with polecompensation ....................................................... 39 PI- Controller (process 4 PT1) with Symmetrical Optimum .............................................. 40 PDT1- controller with polecompensation (process 4 PT1) .................................................. 40 PIDT1- controller PI with sym. Opt. and Tv with polecomp. ............................................. 41 PIDT1- controller PI with polecomp. and Tv with polecomp. (process 4 PT1) .................. 41 Conclusion results of FRA- design with 4 PT1- process ..................................................... 42

FRA- Design 3 PT1-I - Example with new plots .................................................................... 42 P- Controller (process 3 PT1-I) ............................................................................................ 43 PI- Controller (process 3 PT1-I) with symmetrical optimum .............................................. 43 PDT1- Controller (process 3 PT1-I) ..................................................................................... 44 PIDT1- Controller (process 3 PT1-I) with sym. Opt. .......................................................... 44 Conclusion results of FRA- design with 3 PT1 + I- process ................................................ 45 Simulation results of 4 PT1- process with RegCSharp. ....................................................... 46 Simulation results process with 3PT1 with Integrator. ........................................................ 47 Improvement of control performance, chap. 5.3 .................................................................. 50

Design of PIDT1- controller with FRA .................................................................................... 52



Table for PI controller with symmetrical optimum .................................................................. 53 -12 dB method .......................................................................................................................... 53 -30 dB method .......................................................................................................................... 53

1. Content of the lecture Control Systems

1. Introduction 1.1 Fundamentals 1.2 Terms, designations 2. Block diagram 2.1 Construction of a block diagram 2.2 Reduction rules 2.3 General strategy 3. Characterization of linear control loops 3.1 Description in the time domain Differential equation, step response, impulse response 3.2 Description in the frequency domain Laplace-transform Transform pairs Characteristics 3.3 Procedure to calculate system responses Examples 3.4 Complex transfer function Laplace-transform of a differential equation 3.5 Graphical display of the complex transfer function Polar plot Pole - Zero - diagram Mesh plot Amplitude - and phase diagram 3.6 Bode plots PC - Demo of some programs

3.7 Elementary transfer blocks with step response, bode plot, symbol, characteristic values

3.7.1 P- block 3.7.2 I- block 3.7.3 D- block 3.7.4 PT1- block 3.7.5,6 PD- block and PDT1- block 3.7.7 PI- block Project 1: 1. Project Cruise control Loop with PI + PT1 3.7 Elementary transfer blocks continued 3.7.8 PID- block 3.7.9 PIDT1- block 3.7.10 Delay time - block 3.7.13 PT2-block 3.8 Project 2: Temperature controller 2PT1- process with PI- and PIDT1- controller 3.9 Bode plots for closed control loops 3.9.1 Graphic approximation design 3.9.2 Nichols-Formulas



4. Stability 4.1 Definition of stability 4.2 Condition for poles (roots) 4.3 Hurwitz-Criterion 4.4 Special Nyquist Criterion 5. Design of Control Loops 5.1 General 5.1.1 Four objectives 5.1.2 Performance criteria 5.2 The Frequency-Response-Approach (FRA) design 5.2.1 Without I - block (Projekt 3: Different controller wih 4 PT1) 5.2.2 With I – block (Projekt 4: Different controller wih 3 PT1 + I) 5.3 Improvements of the control loop performance Diverse exercises with cascaded controllers

References 1. Phillips, C.L., Feedback Control Systems, Prentice Hall 1991

(€ 70.- not available in Germany) 2. Ogata, K. Modern Control Engineering. Prentice Hall, 1991 3. Saadat, Hadi, Computational Aids in Control Systems Using MATLAB. McGraw

Hill, 1993. (available, about e 30.-).

Software

RegCSharp.exe Simulation program with signal block diagram editor. Calculation of bode plots, step responses and FRA- design of PIDT1- controllers. Download: moodle course “Toolprogramme Bayerlein” there block RegCsharp There you can find a file: REGC_-Manual US V1_3.pdf

WindfC# Design of digital filters, bode plots of analog and digital filters, some identification modules, some controller design modules, some hardware interfaces (National Instruments, Meilhaus), real time controllers. Download: moodle course “Toolprogramme Bayerlein” block WindfCsharp Document: WindfC#Manual.pdf in the same path. Source code developed in c# with MS VStudio2017



2. Block Diagrams electrical elements



Mechanical Elements



Conversion rules for Block diagrams



3. Overview of Dynamic Relationships

The possible transfers are shown as arrows. It should be mentioned that the transfers from top to bottom can be exactly computed, but the transfer from bottom to top can only be estimated and it is very complex to do so. (Bottom – up process is called identification, see CS II. Possible controller design discussed in my lecture: 1: DE/state space not discussed 2. F(p) e.g. flash design, analytical formulas 3. step response Indentify F(p), then goto 2 4. Bode plot FRA-design, RegCSharp

time domain frequency domain

1 2

3 4

general

description

for any xe xa can

be computed

diff. equ.

state spacedescription

F(p)

special testsig-

nals at input

measurable

easily

step response bode plot

solvingofdiff. equ.

p=jw

identification

? identification

Laplacetransformnumerical integration

FFT, DFT

identicalcoefficients



Laplace-Transform Table (1.Nb.=degree num., 2. Nb. = degree denom.)

0.0.a 1 (t) 0.1.a 1 p 1 t

0.1.b T Tp( )1 exp( ) t T

0.2.a 1 2p t

0.2.b 1 1p Tp 1 exp /t T

0.2.c

1

12 Tp

1

2Tt

t

T

exp

0.2.d

1

1 11 2 T p T p 1

1 2

1 2

T Te e

t T t T

0.2.e 1

1 2 2 T p

T

t

Tsinh

1

0.2.f 1

1 2 2 T p

1

T

t

Tsin

0.2.g 1

1 2 2 2 dTp T p for d2<1 1

1

11

2

2

T de

Td tdt T

/ sin with o=1/T

1.2.a 1 2 Sp p t S

1.2.b

1

1

Sp

p Tp 1

S T

Te t T

1.2.c

1

12

Sp

Tp

T S

Tt

S

Te t T

3 2

1.2.d

1

1 11 2

Sp

T p T p

T S

T T Te

T S

T T Te

t T t T1

1 1 2

2

2 1 2

1 2

1.2.e 1

1 2 2

Sp

T p

12T

t

T

S

T

t

Tsinh cosh

1.2.f 1

1 2 2

Sp

T p 1

112

2TS

T

t

T

sin with

1.2.g 1

1 2 2 2

Sp

dTp T p for d2<1 1 2

1

12

2 2

2

2

T

T dTS S

de

d

Tt

dtT

sin ,

1.2.h

p

Tp12

13T

T t t T exp

1.2.i

p

T p T p1 11 2 1

1 2 1 21 2 2 1T T T T

T t T T t T( )

exp exp

1.2.j p T p1 2 2 1 12T Tcosh

1.2.k p T p1 2 2 1 2T t Tcos

1.2.l p

dTp T p1 2 2 2 for d2<1 exp

sin

dt T

T d

d

Tt

2 2

2

1

1 , tan

1 2d

dand o=1/T

F p f t

tanS

T

tan

S d

T dS

1 2



0.n.a 1

pn

1

11

( )!( )

nt n

for n>1

0.3.a 1 12p Tp T et

Tt T

1

0.3.b 1 12

p Tp 1 T t

Te t T

0.3.c

1

1 11 2p T p T p 1

1

2 11 2

1 2

T TT e T e

t T t T

0.3.d 1 1 2 2p T p 1 cosh t T

0.3.e 1 1 2 2p T p 1 cos t T

0.3.f

1

1 2 2 2p dTp T p for d2<1 1

1

1 12

2 2

exp( )sin , tan ,

dt T

d

d

Tt

d

d = 1/ T o

0.3.g

1

13 Tp

1

2 32

Tt e t T

0.3.h

1

1 11 2

2 T p T p

T

T Tt T

T T t T T

T T Tt T1

2 1

2 12 1 1 2

2 2 1

2 2

exp( / ) exp( / )

0.3.i

1

1 1 11 2 3 T p T p T p

T t T

T T T T

T t T

T T T T

T t T

T T T T1 1

1 2 1 3

2 2

2 1 2 3

3 3

3 1 3 2

exp( ) exp( ) exp( )

0.3.j

1

1 1 2

1

1 2 22 2

2

T p dT p T p

for d and o = 1 T2

T e

T dT T T

ed

Tt

d T dT T T

T d

T dT

t Tdt T

1

22

1 2 12

2

2

222

1 2 12

12

2 1

1

2

2

1

1 2

1

sin

, tan

1.3.a

p

Tp13

t

T

t

Te t T

3

2

42

1.3.b

p

T p T p1 11 2

2

e

T T T T

t

T T Te

t Tt T1

2

1 2

2

2 1

222

2 1

1

1.3.c

p

T p T p T p1 1 11 2 3

T T e T T e T T e

T T T T T T

t T t T t T2 3 3 1 1 2

1 2 2 3 3 1

1 2 3

1.3.d

p

T p T p1 11 22 2

1 1

22

12

2 22

12

2

1

2

1

T Te

T T T

t

Twith

T

Tt T

cos tan

1.3.e

1

12

Sp

p Tp S T e tt T 1

1.3.f

1

12

Sp

p Tp 1 1

2

S T

Tt e t T

1.3.g

1

1 11 2

Sp

p T p T p 1 1

2 1

2

2 1

1 2

T S

T Te

T S

T Te

t T t T

1.3.h

1

1 2 2

Sp

p T p 1 1

2

S

T

t

Twith

S

Tcos tan



1.3.i

1

1 2 2 2 2 1

Sp

p dTp T p d

o = 1 T

))(arctan()arctan(,

,1

,1

,sin)(

1 22

22

aaTda

abT

bS

teab at

1.3.j

1

1 11 2

2

Sp

T p T p

T S

T Te

T S

T T Tt

S T

T Te

t T t T1

1 2

22

22

2 1

1

2 1

21 2

1.3.k

1

1 1 11 2 3

Sp

T p T p T p

( ) exp( ) ( ) exp( ) ( ) exp( )T S t T

T T T T

T S t T

T T T T

T S t T

T T T T1 1

1 2 1 3

2 2

2 1 2 3

3 3

3 1 3 2

2.3.a

1

1

2

2

ap bp

p Tp t a T a T

b

Tt T

exp( / )

2.3.b

1

1

2

2

ap bp

p Tp 1 1

2

2

3

b

T

b aT T

Tt t Texp( / )

2.3.c

1

1 1

2

1 2

ap bp

p T p T p

1 1 12

1 2 11

2 22

2 2 12

b aT T

T T Tt T

b aT T

T T Tt Texp exp

2.3.d

1

1

2

3

ap bp

Tp b

T

aT b

Tt

T aT b

Tt t T

34

2

522

2

exp

2.3.e

1

1

2

2 2

ap bp

p T p 1

12

2 2 2 2

T

T b a Tt

Tcos with

2.3.f

p

Tp

2

31

1

22 4

52 2

TT Tt t t T exp( / )

2.3.g

p

T p T p

2

1 2

21 1

exp( / )exp( / )

t T

T T T

T T

T T T

t

T T Tt T1

1 1 2

21 2

22

1 2

223

1 22

2

2.3.h

p

T p T p T p

2

1 2 31 1 1

exp( / ) exp( / ) exp( / )

t T

T T T T T

t T

T T T T T

t T

T T T T T1

1 1 2 1 3

2

2 2 1 2 3

3

3 3 1 3 2

2.3.i

1

1 2 1

1

2

1 12 2

2

2

ap bp

dT p T p T p

for d and = 1 / To 1

DA

Tt T E

dt

T

d

Tt

DT dT T T

ET

T B dT BC C

d

22

1

2

1

12

1 2 22

12

12 2

12

2

1

1

2

1 2

1

exp( / ) exp sin

,

with

arctan ,C d

T B dC

1 2

1

A T aT b B T dT T aT b

C T b T T aT dbT

22

2 12

1 2 2

2 12

2 12

1

2

2

;

tan

aT

T b2



To Chap 3.5. Graphical display of Transfer Functions In the lecture the different types of display of a complex transfer function are discussed in the chapter 3.5. There the following example is used:

F pp

p p p pDEMOF UFK Eq( )

. . . .( . ) ( . . )

4 3 20 8 8 74 3 432 14 40414 1

With the program WindfC# behind key *ufk with key Factorize the complex zeroes can be numerically calculated from the denominator and the result is: p1,2= -0.2 j 1.5 and p3,4 = - 0.2 j 2.5.

With these terms F(p) can also be written in the pole / zero form like

F pp p

p p p p p p p pEqo( )

( )( )( )( )( . . )

1

1 2 3 4

4 2

By multiplying the terms with the conjugate complex pole pairs, you get the form

F pp

p p p pDEMOF ZK Eq( )

( . . )( . . )( . ) ( . . )

2 20 4 2 29 0 4 6 294 3

which can be interpreted as cascaded blocks with a D block and two PT2 blocks. The parameters d und 0 of the PT2 blocks are: (e.g. calculated with WindfC#)

d1=0.132, 01=1.513 and d1=0.0797, 02=2.508

and K=0.06943.

In the next picture the pole / zero plot is presented.

On the following pages the polar plot is presented (you can use Program RegCSharp to draw polar plots) and you can see the magnitude of F in the bode plot, here derived with the program RegCSharp .In the last picture a mesh plot is shown (generated with the program WindfC#).



Bodeplot of function DEMOF with RegCsharp, sized with 0.5, polar plot with frequency zoom 1-10. Blue curve is unit circle. Frequency marks with right mouse click.



Example to chap 3.5.3 mesh plot of a transfer function Program Windfc’; File Demof.zk.

First buttom „Load Filter“. Dann „Lade zk-file“.

Content of window:

Then and you get



Mesh plot with matlab



Table with standard blocks (p. 16)

Unit step response



Calculation 3 RC- chain with OpAmp

Control System Lab exp. 1 part 3 Circuit

𝐹 𝑝 𝑈𝑈

𝑏𝑎 𝑎 𝑝 𝑎 𝑝 𝑎 𝑝

With b0= R5 and 𝑎 𝑅 𝑅 𝑅 𝑅 𝑎 𝑅 𝑅 𝐶 𝑅 𝑅 𝑅 𝐶 𝑅 𝐶 𝑅 𝐶 𝑅 𝑅 𝑅 𝐶 𝑅 𝑅 𝐶 𝑎 𝑅 𝑅 𝐶 𝑅 𝐶 𝑅 𝐶 𝑅 𝐶 𝑅 𝑅 𝑅 𝑅 𝐶 𝐶 𝑅 𝑅 𝑅 𝐶 𝐶 𝑎 𝑅 𝑅 𝑅 𝑅 𝐶 𝐶 𝐶 With measured elements R1=9.99 kOhm, R2=9.82 kOhm, R3=9.807 kOhm, R4=21.47 kOhm, R5=50.94 kOhm, C1=4.288 µF, C2=4.303 µF, C3=4.258 µF we get with new Tool in Windfc#

(since V7.6.11): Shifted into the UFK- Window gives

After Button „Factorize“: So we get three new time constants T1=102.6 ms, T2=24.26 ms and T3=12.76 ms.



Further pages to 3.7 : DT2-block With the following circuit a DT2 - block can be set up:

Then we obtain the transfer function:

F pU

U

Kpd

p po o

( )

2

12

212 1

.

The parameter K,d and o can be

calculated to:

KR R C

R R

R R

R R R C C

2 3 1

1 20

2 1 2

1 2 3 1 2

, and

2

0

1 2 1 2

1 2

d R R C C

R R

( ).

The bode plot (magnitude curve) looks like this:

Numerical example: R1=R2=1K, R3=82K, C1=C2=0.22F gives K=9.02 ms, o =709.88

1/s and d=0.078087. (You have to consider the unit of K !!! [K]=s ). The unit step response (like in the figure on the next page for the above-mentioned example)

has the equation. h tK

de td t

E( ) sin

0

210 with E d 0

21 .

This function has the zeroes t ii E0 * / and the first maximum at

td

ddm

E E

1 1 12

arctan arccos( ) the amount of

U K d t Km m 0 0 0 2exp( ) exp( ( / ) tan ) with

arctand

d1 2.

Also here there exists like the PT2-block, the logarithmic decrement with the relationship:

U

U

d

d

m

min

exp( tan ) exp( )

1 2,

working backwards d again can be calculated to d

ln( )

ln ( )

2 2.

The DT2-unit-step response for the above-mentioned numerical example looks like this:

1/2d in dB

0

log

|F(j)|

1/K

in dB

00K 0

Attention!! Here phases in rad !!

-

+U1 U2

R1

C2

C1

R3

R2



Number values for the above-mentioned DT2 - block: K=9.02 ms, o=709.88 1/s and

d=0.078087. The characteristic values of this example are then: t0=4.44 ms

tm=2.11 ms

Um=5.696

Resonance step-up over asymptote intersection: 16.12 dB =4.479 =exp(tan())=1.279. The damping for optimal disturbance behavior (tolerance range:10% of the peak value) emerges with

=10 to .

d ln( ) ln ( ) . 2 2 10 0 5912

Um

tmt0

2t0

Umin

t



Further pages to 3.7 : PIDT1 - Block A PID can not be realized because of the ideal D - behaviour, an actual PID - block is always a PIDT1 - block. It always consists of a PID and PT1 - block, as the right figure shows.

Usually there are two forms of the transfer function: The parallel form (Eq.7.1) and the bodeform (Eq.7.2). With the table on the next page the parameters can be converted into each other (page 21 Eq. 7.4):

PIDT1 parallel form: (Eq.7.1) PIDT1 bodeform: (Eq. 7.2) The stepdepth st = Tv /T1 returns thereby the "quality" of the D-part quite well. If for

example st=1, the PIDT1 - block passes over into a PI - block, for st -> the PIDT1 becomes an ideal PID - block. If st=10 the ratio between the corner frequencies 1/Tv and 1/T1 is one

decade, the PDT1- HF-gain-increase is just 20dB. Unit step response:

h t Kt T

T

T

T

T

Tt T Eq

I

D

I

( ) exp ( . . )

1

1

111 1 7 3

Important: Especially the beginning impulse is determined from st. In most cases of control loops this is the maximum value of the control output, if a reference step input is applied.

F p KpT pT

pTI D( )

1 1

1 1

F p KpT pT

pT pTRN V

N

( )( )( )

( )

1 1

1 1



A circuit of the PIDT1 is shown in the next figure. We obtain (with v):

Another circuit looks like: We obtain:

The following relationships are valid for the possible block diagrams:

Bodeform (serial form) Summingform (parallel form) Wiki- P+I+DT1- form The conversion of the bodeform in the parallel form can be done with following equations: Conversion formula: Bodeform Parallel form Wiki- Form(Equ 7.4) ParallelBode BodeParallel Wiki Parallel

K K T T

T T T T

T T T T

R D I

N I D I

V I D I

05 1 1 4

05 1 1 4

05 1 1 4

.

.

.

K KT T

T

T T T

TT T

T T

RN V

N

I N V

DN V

N V

F p KpT pT

pT pT

U

URN V

N

( )( )( )

( )

1 1

1 1

2

1

TD

TI

F p KpT pT

pTI D( )

1 1

1 1

F p KpT pT

pT pTRN V

N

( )( )( )

( )

1 1

1 1

KR

RT R C T C R R T C RR N V 2

12 2 1 3 1 1 1 3, , ( ),

KR

R RT R C T C R T C

R R

R RR N V

3

1 23 2 1 2 1 1

1 2

1 2

, , ,

11

11)(pT

pTpTKpF b

aa

ba

aD

I

DIb

aIIa

aa

Ia

TTTT

TTundT

TT

TTT

TTTundTTT

TTKKundT

TKK

11

11

11

11 11



Further pages to 3.7: PT2 - Block

22

1

2

121

)(pp

dK

U

UpF

oo

The most important values of the magnitude plot can be seen in the left figure. The most important characteristic value is the value at the natural frequency 0 . The frequency

0 is located at the phase displacement

of exactly -90. With it the values 0

and d can be determined immediately from a measured course of the plot.

2

0

22

0

21

|)(|

d

KF and for the phase

2

0

0

1

2

arctan

d

.

Unit step responses :

You can see the display of the unit step responses for several different values of d on the overnext page. Some characteristic values should still be indicated: The first maximum of the unit step function has the time of tm and the value hmax1 with

)1(/

2dt

o

Em

and h Kd

dEqmax exp

( )( . . )1 2

11

6 4

If now the final value K is known, overshoot ü can be defined: (with U1=hmax1-K)

If the damping factor d is very small (ü1), it is better to compute d with the logarithmic decrement (where hmax2 is the second maximum of the step response and U3=hmax2-K):

for K=1!!

Bode plot of PT2

d h t Ke

dt Eq

d t

E

o

1 11

6 22

: ( )( )

sin( ) ( . . )

with d and d dE o ( ) tan ( )1 12 2

d h t KT

T Te

T

T Te for PT Blocks Eqt T t T

1 1 2 1 6 31

1 2

2

2 1

1 2: ( ) ( . . )

üU

K

d

dor d

ü

üEq

1

2 2 2165exp

( )

ln( )

(ln )( . . )



U

U

d

dü or d Eq1

32

2

2 2

2

1 46 6exp

( )

ln

(ln )( . . )

The left figure shows the typical course of the PT2 - step response for d < 0.2. If K is not known, K can be determined in such way, that the intersections with the final value K have the equal distance tm. If only U2 is known, but not U3, the relation

U

Uü2

1

applies or in general

and with Eq.6.5 d can also

be determined. Try to prove these relations yourselve with above curve, it has been drawn with d=0.1.

Conversion PT2 in 2 PT1 _____________________________________________________ For d >= 1 the PT2 block can be converted into 2 PT1 - blocks. If d<1, this conversion is not allowed / not possible, because in this case the time constants become complex.

Then Eq. 6.1 becomes

The conversion formulae of Eq. 6.1 (PT2) to Eq. 6.7 (2PT1) is:

Td d

and Td d

Eq1

2

02

2

0

1 168

( . . )

and vice versa from Eq. 6.7 (2PT1) to Eq. 6.1 (PT2) :

0

1 2

01 21

26 9

T T

dT T

Eqand ( . . )

U

Uün n

1

1 tm

tm tm

U1

U2K

U3

K K T1 1 T2o,d



Passive PT2-blocks Oscillating circuit:

Spring-mass-system:

MC

Dd

M

C 1

2,0

Several different unit-step-functions of a PT2 with K=1, o=1:

0

1

2

LCd

R C

L,

C M D

y(t) x(t)

F pU

U dp p

o o

( )

2

12

2

1

12 1

F pX

Y dp p

o o

( )

1

12 1

22

d 0

0.1

0.2

0.320.4

0.7071

1

2



Project 2: Temperature control for a hand drying fan with PIDT1

On the next pages I want to describe the design of the control system for the air temperature of a fan (ventilator). You can easily transfer this procedure to other control tasks with similar process behaviour. The hardware is set up on a board realized with standard - components and is demonstrated in my lecture.

.

Goal:The air temperature should be controlled between room temperature and 100°C arbitrarily adjustable and as exactly as possible (errors < 0.25°C). Overshoots of the temperature should be avoided, because the decreasing of temperature by the cooling of the process via room temperature is very time-consuming. Goal: 0.25°C overshoot in a 10° step from 35°C to 45 °C, that is overshoot ü=2.5% .

1. Step: Determine the necessary hardware in addition to the fan.

A two point controller (mostly installed) is excluded, because the temperature oscillation amplitude is too large (measured: 5°C). Therefore an actuator is used, which can change the

heating power continuously. The heating resistor has a power of 1.7kW at 230V AC. The cheapest possibility is a triac - circuit to cut the phase at adjustable times. This is available at Conrad (electronic supplier) and is called a light control with a maximum power of 2kVA with a 0 - 10V -

input for about 50€. The DC gain is calculated to Kp=1.7kW / 10V=170W/V, the time constant of the control block is neglected, because its time constant of about 20ms is small compared with them of the heater. These lie within many seconds (s.b.)

Furthermore a temperature sensor is necessary, which outputs the measured temperature as a voltage value. I have chosen an instrument with a digital display to check the measured temperature. It has a waterproof sensor (Greisner GTH175, approx. 35€). This instrument had

still no fitting voltage output. But the sensor voltage can be contacted and amplified with an

Up P

Kp

Ut

Kt



Op - circuit to get the desired range. At the output I finally get a voltage Ut of 0-10V at a temperature between 0 -100°C. The gain totals in Kt=10V / 100°C=0.1V/°C. The dynamic is discussed later.

+15V

V

Us

Ut-Ut

Up

-(Us-Ut)

100K

100K

100K

100K

100KR2

R3

R1

C1

C2

100n0.2%

0.2%

0.2%

0.2%

S1

V

S2

OP's: TL081,TL071 o.ä.

D1

The still missing controller is realized with an operational amplifier circuit described on PIDT1- page. The desired value is applied as a voltage and can be changed with a potentiometer. The error signal = (desired value - actual value) is formed by an adder. The negative sign is realized with an inverting amplifier. The total circuit can be seen in the above figure. With S1 the integration capacitor of the PIDT1 - controller can be unloaded, with S2 you can open the feedback of the adder and increase the loop gain of several decades. This caused a two point controller behaviour without hysteresis. Negative control values and wrong polarity at the electrolyte capacitor C2 should be prevented by the diode D1. The four 100k - resistors have a very small tolerance to avoid too large errors in the steady state. Actually the values Us and Ut should not differ more than 1% (1% is 1°C! ).

2. Step: Identification of process dynamic.

To design the controller a model of the process is required. Often you can get these model through a measurement of the step response. Therefore I have applied a step with the step

amplitude of 4.5V to the actuator block (corresponds to 4.5*170W). The curve is measured with program WindfC#.

You can recognize the exponential increase, the final value is not reached yet. In the beginning course you can see with

corresponding zoom a horizontal tangent. Now you can identify a model with 2PT1 blocks with methods described in chap. 6 of the lecture. Practically this is done easily with my program WindfC#, which accomplishes a parameter optimization. The result is a process model with a step response, which differs hardly compared with the measured one. The curve is offset- cleaned and converted by division of 4.5 to a unit step response. :



The model is a 2PT1- model and for simply use I rounded the time constants. Final model curve looks like this:

You can find this identification method in WindfC# menu 'Identification/Regression f(t)' . The parameters are Ks=0.73, T1=10s and T2=1 s. After 5T150s the final value is reached within 1% .

Now the control loop is defined except of the controller. It has the following signal block diagram

The entire loop gain is Ks = Kp*K*Kt = 0.73 without any unit. So you can state: e.g. at a control value of 1V the temperature increases 7.3°C according to 0.73V. The disturbance input presents the room temperature, because

the temperature increase versus the environment is directly proportional to the heating power P.

Ks T1 1 T2

Kp K T1 1 T2 KtUs

UtUp P

u

Regler

Stellglied MessgliedStrecke Gebläse

s



3. Step: Design of the controller

Now still the controller is missing. The design can be done with diverse methods (RegCSharp with FRA - design etc), but the fastest method is here the flash controller design surely, described on p. 33. For a PIDT1 -controller with 2PT1 - process the design formulas are:

with

TN=T1 and Tv=T2. FR describes the controller and FS the process. Here the compensation of the largest time constants is accomplished, the stepdepth st and the damping d can be freely chosen within limits. The closed control loop behaves exactly like a PT2 in this special case

with the parameters Kw=1, d used in the above formulae and o as calculated. With this knowledge you can discuss very quickly the reference behaviour, since the PT2 block has been comprehensively described on the pages before. Especially the values tm and ü of the reference step response

are immediately calculated with

t dm o ( )1 2 and üd

d

exp( )

1 2.

With the program WindfC# you can quickly calculate these numbers. We want to compare three different controller designs, which are designed altogether with a TN=T1=10 s and a damping of d=0.7613. The damping corresponds to an overshoot of 2.5%, i.e. at a 10°C - step this causes an overshoot of just 0.25°C. The first design should be a PI - controller (st=1), the second design is a PIDT1 with st=10 and the third controller becomes also a PIDT1 with st=2. See all numbers in the overview (presets in bold letters):

F KpT pT

pT pT

st

FK

pT pTK

stT

T d Kst

dTR RN v

Nv

S R o

( )( )

( );

( )( );

1 1

1 1 1 4 21 2

1

22

2

Kw d,0

Us Ut



Above a screen shot of window design PIDT1- controller with WindfC#.

KR st col Krst Pmax(1V) tm Ü 0 Taus 2.5%

PI 5.9 1 Red 5.9 1003 W 7.4 s 2.5% 0.657 5.2s*

PIDT1 59 10 Blue 590 100 kW 0.74 s 2.5% 6.57 0.52s*

dto begrenzt " 10 Yellow " 1.7 kW 6.2 s* 14 %* " >20 s*

PIDT1 11.8 2 Lgreen 23.6 4.0 kW 3.7 s 2.5% 1.31 2.6s*

dto begrenzt " 2 Lblue " 1.7 kW 6 s* 7%* " 18s*

(* measured with simulation)

The first controller is relatively slow. The second design needs at a 1V -step (=10K) a gigantic control output impulse by 100 kilowatt, which of course can not be realized. If you limit this controller output to 1.7 kilowatt, this design becomes also useless, since it has a strong overshoot with a long decreasing time caused by Wind-up-effect. The third design however becomes not bad. With limitation of the control output the overshoot remains at 3%, the settling time has been simulated about 18s. The values are all computed with RegCSharp, you can find the reference steps of all controllers with and without limitation in the next figure. You can recognize that a stepdepth decreases the performance, if the control output is strongly limited. Often the increase of the stepdepth from 1 to 2 is very successful already, but in this application a PI or a PIDT1 with antiwindup-procedure is OK. With this step (e.g. 1V=10°C), the PIDT1 with st=2 is optimal. This last design has then been realized.



Simulation with RegCSharp: red: PI, lgreen: PIDT1 st=2 without limitation, blue: PIDT1 st=10 without limitation. without AWU, yellow: PIDT1 st=10 with limitation. without AWU, lblue: PIDT1 st=2 with limitation without AWU.

The situation will be improved, if you add an anti- wind-up-mechanism. Here the integrator of the controller is disabled in the case of limitation of controller output, the PIDT1 works then as a PDT1. Resultsa can be seen in the next diagram (limitation to 10 V).

lgreen: PI as above

Red: PIDT1 st=2 as above

Blue: PIDT1, st=10 as above

lblue: PIDT1, st=2 with AWU

yellow: PIDT1, st=10 with AWU

Is there a realisation of this antiwindup- mechanism (onla possible in a digital controller) , then the st=10 Controller is the best, otherwise the PI.



4. Step: Realization of the controller

The controller should have the parameters TN=10s, KR=5.9 / st=1 and the PIDT1- controller the parameters TN=10s, Tv=1s, KR=11.8 / st=2 with the left circiut. The equations for the components are :

KR

RT R C T C R R T C RR N V 2

12 2 1 3 1 1 1 3, , ( ),

with R1 =220 kOhm is C2 =7.7uF, R2=1300 kOhm, fort he PIDT1 : R1 =110 kOhm results in C2 =7.7uF, R2=1300 kOhm C1=4.54 uF and R3=110 kOhm. If you realize the Caps with electrolytes, this could be a problem. Because UP is always positive the polarity with C2 is no problem, but with C1 it is. Furthermore unfortunately the el.-caps have a parasitic parallel resistance however,

which in our case has a very annoying side effect. It could destroy the desired indefinite large DC gain of the I-behaviour of the controller and the PIDT1 behaves as a PDT1 for low frequencies.

Up

R2

R3

R1

C1

C2

S1

Op: TL081,TL071 o.ä.

D1

D2

D2 is AWU -mech.

ideal I-part

influence of parasitic resistance

Bode plot PI-controller



Results of real measurements. Two curves are results with onboard- hardware controller realized with OPamp. The other are measured with digital controller with PC and AD card sampled with 100 ms:

black: PI digital To=10ms Design 1 free KR=5.9 TN=10s blue: PI RCOP Kr, TN dito red: PIDT1 digital To=10ms Design II free st=2 KR =11.9, TN =10 s, Tv=1 s violet: PTDT1 RCOP st=2 KR, TN dito The oversoots are rather great caused by wind up. Only with anti wind up- mechanism, which is only possible in the digital controller, you get acceptable results:

red: above digital PIDT1, Design I with AWU blue: above digital PIDT1, Design II with AWU



Chap 3.8 ‘Flash-design’ by formula

Under certain circumstances the following controller design can be used, which evades the bode-plot (FRA) - or other troublesome designs. The controller parameters can then be determined directly per formula with the choice of the desired characteristics of the closed loop, if the process parameter are known. Requirement: 1. No delay time available (does not work therefore at digital controllers!)

2. Fo =Fc FP has the form 𝐹 .

This is the case at following controller / process -combinations: 2.1 P - controllers and IT1 - process 2.2 I - controllers and PT1 - process 2.3 PI - controllers and 2PT1 - process (special case of 2.5 for st=1) with polcompensation 2.4 PDT1 - controllers and IT1 - process with polcompensation 2.5 PIDT1 - controllers and 2PT1 - process with polcompensations The closed control loops of these combinations always yield exactly PT2 - behaviour with the PT2 - parameters K=1, d und o. The design happens to such an extent, that the

damping as well as the stepdepth are given and KR und o then can be computed:

2.1 : 1

211

21;

4

1

)1(; dTKdT

KpTp

KFKF o

ic

ipcc

2.2: 1

211

21;

4

1

)1(; dTKdT

KpT

KF

p

KF ocp

cc

2.3: look at 2.5 for st=1

2.4:1

21

11

2;4

;)1(

;1

1dT

stKdT

stKTT

pTp

KF

st

Tp

pTKF o

icv

ip

v

vcc

2.5:2

22

1

212;

4)1)(1(;

)1(

)1)(1(dT

stKdT

stTK

pTpT

KF

st

TppT

pTpTKF ocp

vN

vNcc

with TN=T1 and TV=T2, if T1 is the largest process time constant. You find the formula 2.5 programmed in WindfC#, menu 'controller design', then flash..... If one does not want to choose the damping d, but the phase margin R , the damping d is calculated by using the relationship

, see WindfC#.

There you can type in d, ü or R =phir, both other numbers then be calculated.

d R 0 5 90 0 5 0 252 2. (tan ( ) . ) .



Chap 3.9. Nichols - Formulas The Nichols - Formulas can be used to calculate the bode plot of the closed control loop from the data of the bode plot of the open loop, commonly used to design cascaded controllers.

Under the requirement, that no transfer block exists in the feedback, F0 is the complex transfer function of the open control loop and F then is the function of the closed control loop, as indicated in the left figure. The conversion below is well known

FF

FEq

0

0151( . )

Exact calculation with the Nichols - formulas:

Attention: In this equation both the F and Fo-values are not in dB !!!

This function must be calculated for each frequency point of the bode plot. However if a transfer block is available in the feedback branch, then these formulas can not be used directly. Then the block diagram has to be transformed into a new one without feedback

block before using Nichols- equations. The procedure is illustrated in the above

figure: G and H are multiplied to equal Fo, and then with the

Nichols-Formulas F is calculated. Then the inverse function H-1 is added. A PC - program, which evaluates 5.2 and 5.3 is found in WindfC#.

w andith F F j F F j exp( ) exp( ). 0 0 0

Approximation consideration: If |Fo| < -20 dB, then FFo is approximately valid. If |Fo| > 20 dB, then F 1 is

approximately valid, i.e. |F| = 0 dB and =0 . (Colloquially: In the bode plots |F0| becomes |F| by cutting off |F0 | with the 0-dB-line).

)3.5.(cos

sinarctan

)2.5.(cos21

00

0

2

000

0

EqF

EqFF

FF



4. Chap. : Stability Criteria

Poles

Disadvantage of this method: The calculation of the denominator zeroes for degrees higher than 2 is only

numerically possible and Fw(p) must be known. Systems with variable parameters can not be investigated in this way with higher degrees than 2.

Hurwitz Criterion

Fw is defined by Fnumerator

c p c p c p cwn

nn

n

11

1 0.... If all ci are negative , Fw must be

expanded with (-1)!

If only one condition is hurt the system becomes instable, and the investigation can be aborted. The definition of the n * n - Hurwitz determinant is:

H

c c

c c

c c

c c

c

n

n n

n n

n n

n n

1 3

2

1 3

2

0

0

0

0 0

0 0

0 0

0

. . .

. . .

. .

. .

. . . .

. . . .

, H cn1 1 , Hc c

c cn n

n n2

1 3

2

etc.

For general denominator polynoms with n=2 to n=4 you can find the stability conditions easily in other form without the computation of determinants:

Disadvantage of the Hurwitz criterion: Fw must be known and delay times are not allowed. Advantage: Well suited for calculation of stability boundaries and at unstable processes.

Fow x

A system with given Fw=Fo/(1+Fo) is stable if and only if all poles of Fw(p) have a negative real part.

A system with given Fw=Fo/(1+Fo) is stable if and only if the denominator of Fw has following characteristics: 1. All ci must be positive and unequal to zero and 2. The Hurwitz determinant Hn must be positive and unequal to zero and 3. All ‘north-west’ sub-determinants H1 to Hn-1 must be also positive and unequal to zero.

A system with a degree of 2 is stable if and only if all c0, c1 and c2 have equal signs and are unequal to zero.

A system with a degree of 3 is stable if and only if all c0, c1, c2 and c3 have equal signs, are unequal to zero and the condition c2c1 > c0c3 in addition is valid (product of the inside coefficients > product of the outside coefficients ).

A system with a degree of 4 is stable if and only if the following conditions are valid: 1. All c0, c1, c2, c3 and c4 positive and unequal to zero and 2. c2c3 > c1 c4 and 3. c1c2c3 > c1

2 c4 + c0c32

.



5. Chap FRA-Design with the 4PT1- example The following pages will discuss the design of PID-controllers with the FRA-design (Frequency-Response Approach). This application example comes from the area of power electronics, but is easily transferred to other areas. 1. Project 3: Block diagram with 4 PT1-blocks

Results of FRA- design with 4 PT1- process (values during lecture) Controller d Kc TN contr(d) Tv st Ref.error remarks P I PI polecomp PI Sym. Opt. PDT1 PIDT1 sym. Opt. PIDT1 polecomp

2. Project 4: Block diagram with I-block and 3 PT1-blocks

Results of FRA- design with 3 PT1 + I- process (values during lecture) Controller d Kc TN contr(d) Tv st Ref.error remarks P I PI polecomp PI sym. Opt. PDT1 PIDT1 sym. Opt. PIDT1 polecomp

1 0.01 1 1 2 0.05 1 0.33

1

Z

1 0.01 1 0.11 1.6 0.05 0.1

1.25

Z1 Z2



Chap 5.2 Step-by-step-procedure FRA-design of PIDT1 abbrev: FRA: frequency response approach R: phase margin =180° + 0 at d.

contr: phase of PI-part of controller at d so: contr:=PI(d) 0 : open loop phase = p +PDT1 + PI

Design of a PI or PIDT1 - controller with PI with Symmetrical Optimum 1. Generate process magnitude | Fp |and phase p of Fp as bode plot (drawing or

measurement). 2. If st is not 1 (st=1 : PI – case) add the phase of the PDT1- part of the controller. Choose

Tv (e.g. with polecompensation or -30 dB method) and st and the phase is

PDT1=arctan(Tv)-arctan(*Tv/st), this gives new = p +PDT1

3. The intersection of the new phase curve new with -180 +R -contr determines the crossover frequency d . (e.g. R=60 and contr= -15 gives new(d) = -105). 4. With known d and phase contr you can calculate corner frequency of the PI -

controller N= 1/TN with equation contr= -90 + arctan(d * TN ) and determine TN. (It

works also with factor A, s.b.in table). 5. Kc now can be calculated at frequency d with Nyquist |Fo(d )| = |Fc(d )| |Fp(d )| =1.

2

22

)/(1(

)(1()(1()(

stT

T

T

TKF

Vd

Vd

Nd

NdCdC

.

Get now |Fp(d)| = value from Bodeplot. Note that bode plot value is in dB, the equations need factors. This can be solved to the single last unknown Kc.

Design of a PI or PIDT1- controller with PI with Polecompensation 1. identical to Sym. Opt. (see above) point 1 2. identical to Sym. Opt. (see above) point 2 3. Choose TN (e.g. with polecompensation or -12 dB method) and add the phase of the PI –

part of the controller: PI()= -90°+arctan(TN); this gives 0 =new +PI 4. The intersection of the new phase curve 0 with -180 +R determines the crossover

frequency d . (e.g. R=60 gives 0(d)= -120). 5. identical to Sym. Opt. (see above) point 5

Table for symmetrical optimum at ITx - process and PI -controllers R contr p+PDT1 A= N/ d dTN = 1/A

30 -30 -120 0.5774 1.732 40 -24 -116 0.4452 2.246 50 -20 -110 0.3640 2.747 60 -15 -105 0.2679 3.732



FRA- Design 4 PT1- Example with new plots

All examples use phase margin value R=60 and both Nyquist equations 1. 0(d) = -180 +R = -120° 2. |Fo(d )| = |Fc(d )| |Fp(d )| =1.

All radian frequencies have the unit 1/s = 1/sec. Note, that I use the symbol „≙ “ for „is equivalent to“ to differentiate between dB- value and factor. All diagrams could easily be reproduced with the toolprogramm Windfc#, the filterfilenames are given.

P- Controller (process 4 PT1) Process magnitude (blue) and phase (red). Filter file 4PT1.zk

P- Controller: open loop phase φ0 = process phase φp (P has no phase), at -120° the ωd =2.68. Process magnitude at ωd =2.68 is -5.98 dB ≙ 0.502. To get |Fo (ωd)| = 1 Kc = 1.99.



I-Controller (process 4 PT1) Process magnitude (blue) and phase (red), open loop phase φ0 = φp -90° green. Filter file: 4PT1 and I-controller.zk

I- Controller: open loop phase φ0 = process phase φp -90°, at -120° the ωd =0.388. Process magnitude at ωd =0.388 is 5.47 dB ≙ 1.877. To get |Fo (ωd)| = 1 |Fo (ωd)| = |Fc (ωd)| |Fp (ωd)| = Kc/ ωd *1.877 = 1. Result Kc=0.207.

PI- Controller (process 4 PT1) with polecompensation open loop phase φ0 = φp -90° + arctan (ω* TN) with TN=1 s, green curve. Filter file: 4PT1 with PI polecomp.zk

At -120° the ωd =1.4. Magnitude at ωd =1.4 is 0.646 dB ≙ 1.077. To get |Fo (ωd)| = 1

|Fo (ωd)| = |Fc (ωd)| |Fp (ωd)| = ∗ ∗

∗ *1.077 = 1.

Result Kc=0.757



PI- Controller (process 4 PT1) with Symmetrical Optimum Open loop phase φ0 = φp + φc. At ωd open loop phase φ0 = φp + φc = -120°. With φc (ωd) = -15° is φp (ωd) = -105°. Filter file: 4PT1.zk

At -105° the ωd =2.03. Process magnitude at ωd =2.03 is -2.72 dB ≙ 0.731. TN with PI- phase equation: φc(ωd) = -15° = -90° + arctan(ωd TN) ωd TN =tan (75°)=3.732. TN=1.84s. To get Kc use |Fo (ωd)| = 1

|Fo (ωd)| = |Fc (ωd)| |Fp (ωd)| = ∗ ∗

∗ *0.731 = 1.

Result Kc=.

. . 1.32.

PDT1- controller with polecompensation (process 4 PT1) TV=second largest process time constant = 0.33, stepdepth is given st=10. Add the PDT1- phase to get open loop phase φ0 = φp + arctan(ω*0.33)-arctan(ω*0.033) (green curve), at -120° the ωd =7.31. Filter file: 4PT1 PDT1 with polecomp.zk



Process magnitude at ωd =7.31is -20.2 dB ≙ 0.0977. To get |Fo (ωd)| = 1 Kc .

|Fc (ωd)| |Fp (ωd)| = ∗ ∗

∗ / *0.0977 = 1.

Result Kc= . ∗ .

. . ∗ . = 4.03.

PIDT1- controller PI with sym. Opt. and Tv with polecomp. TV=second largest process time constant = 0.33, stepdepth is given st=10. Add first the PDT1- phase to get new phase φnew = φp + arctan(ω*0.33)-arctan(ω*0.033) (green curve, same as PDT1- controller). Filter file: 4PT1 PDT1 with polecomp.zk

Open loop phase φ0 = φnew + φc. At ωd open loop phase φ0 (ωd) = φnew + φc = -120°. With φc (ωd) = -15° is φnew (ωd) = -105°. ωd=5.01. TN with PI- phase equation: φc(ωd) =-15° = -90° + arctan(ωd TN) ωd TN =tan (75°)=3.732. TN=0.745s. Process magnitude is -14.2 dB ≙ 0.195. To get Kc use |Fo (ωd)|=1.

|Fo (ωd)| = |Fc (ωd)| |Fp (ωd)| = ∗ ∗

∗

∗

∗ / *0.195 = 1.

Result Kc= . . ∗ .

. . . ∗ . =

. .

. . = 2.6.

PIDT1- controller PI with polecomp. and Tv with polecomp. (process 4 PT1) TN=largest process time constant (p.t.c.)= 1s, TV=second largest p.t.c. = 0.33s, stepdepth is given st=10. Add first the PIDT1- phase to get open loop phase φo = φp -90 + arctan(ω *1) + arctan(ω*0.33)-arctan(ω*0.033) brown curve. Filter file: 4PT1 PIDT1 with polecomp.zk



φo (ωd) = -120° results in ωd =5.76. Process mag. is -16.2 dB ≙ 0.155. To get Kc use |Fo (ωd)|=1.

|Fo (ωd)| = |Fc (ωd)| |Fp (ωd)| = ∗ ∗

∗

∗

∗ / *0.155 = 1.


. . . ∗ . =

. .

. . = 3.01.

Conclusion results of FRA- design with 4 PT1- process Controller d Kc TN contr(d) Tv st Ref. Error* remarks P 2.68 1.99 - 0° - 1 0.2 I 0.388 0.207 - -90° - 1 0 PI polecomp 1.4 0.757 1 -35.5° - 1 0 PI Sym. Opt. 2.03 1.32 1.84 -15° - 1 0 PDT1 7.31 4.03 - 53.9° 0.33 10 0.11 PIDT1 sym. Opt. 5.01 2.6 0.745 34.4° 0.33 10 0 PIDT1 polecomp 5.76 3.01 1 41.6° 0.33 10 0

Ref. error *: Error at reference step 𝑥 𝑡 → ∞

FRA- Design 3 PT1-I - Example with new plots

Bode plots with WindfC# (Version higher than 7.6.8). Filter file process: 3PT1 und I lecture.zk.



P- Controller (process 3 PT1-I) Blue: Mag process, red: Phase process, filter file 3PT1 und I lecture.zk

P- Controller: open loop phase φ0 = process phase φp (P has no phase), at -120° the ωd =3.17. Process magnitude at ωd =3.17 is -24.4 dB ≙ 0.0603. To get |Fo (ωd)| = 1 Kc = 16.6.

PI- Controller (process 3 PT1-I) with symmetrical optimum Blue: Mag process, red: Phase process, filter file 3PT1 und I lecture.zk

PI- Controller: At ωd open loop phase φ0 = φp + φc = -120°. With φc (ωd) = -15° is φp (ωd) = -105°. Result: the ωd =1.55. TN with PI- phase equation: φc(ωd) = -15° = -90° + arctan(ωd TN) ωd TN =tan (75°)=3.732. TN=2.41s. Process magnitude at ωd =1.55 is -17.9 dB ≙ 0.127. To get Kc use |Fo (ωd)|=1.

|Fo (ωd)| = |Fc (ωd)| |Fp (ωd)| = ∗ ∗

∗ *0.127 = 1.

Result Kc= .

. . = 7.61.



PDT1- Controller (process 3 PT1-I) Blue: Mag process, red: Phase process, green: Phase process + PDT1 with Tv = 0.11 (polecompensation) and st=10. Filter file green curve: 3PT1 und I lecture with PDT1 st10.zk

PDT1- Controller: open loop phase φ0 = φp + arctan(ω*0.11)-arctan(ω*0.011), at -120° the ωd =7.63. Process magnitude at ωd =7.63 is -34.4 dB ≙ 0.0191. To get |Fo (ωd)| = 1 Kc .

|Fc (ωd)| |Fp (ωd)| = ∗ ∗

∗ / *0.0191 = 1.

Result Kc= . ∗ .

. . ∗ . = 40.2.

PIDT1- Controller (process 3 PT1-I) with sym. Opt. Blue: Mag process, red: Phase process, green: Phase process + PDT1 with Tv = 0.11 (polecompensation) and st=10, same curve as PDT1. Now add PI with Sym. Opt. Filer file: 3PT1 und I lecture with PDT1 st10.zk

First add PDT1- Phase to process: φnew = φp + arctan(ω*0.11)-arctan(ω*0.011) green curve PIDT1- Controller: phase φ0 = φnew + φPI , ωd at -120°.



With φPI (ωd) =-15° is φnew (ωd) = -105°. Result: ωd =3.76. TN with PI- phase equation: φc(ωd) = -15° = -90° + arctan(ωd TN) ωd TN =tan (75°)=3.732. TN=0.99s. Process magnitude at ωd =3.76 is -26.4 dB ≙ 0.0479. To get Kc use |Fo (ωd)|=1.

|Fo (ωd)| = |Fc (ωd)| |Fp (ωd)| = ∗ ∗

∗

∗

∗ / *0.0479 = 1.


. . . ∗ . = 18.7.

Conclusion results of FRA- design with 3 PT1 + I- process Controller d Kc TN contr(d) Tv st Ref. error remarks P 3.17 16.6 - 0° - 1 0 PI polecomp 0.11 unstable PI Sym. Opt. 1.55 7.61 2.41 -15° - 1 0 PDT1 7.63 40.2 - 35.2° 0.11 10 0 PIDT1 sym. Opt. 3.76 18.7 0.99 5.1° 0.11 10 0 PIDT1 polecomp 0.11 unstable



Simulation results of 4 PT1- process with RegCSharp.

Reference step responses Red: P Green: PI, Polcomp. lblue: PDT1, st=10 blue: PIDT1, st=10



Reference step responses All PIDT1 with 60° Green: PI (st=1) ymax=0.8 Red: PIDT1 st=2, ymax=2.6 Blue: PIDT1 st=5, ymax=11 lblue: PIDT1 st=10, ymax=30

Reference step responses All controller with 60° Red: PI Polcompensation Green: PI, symmetrical optimum Blue: PI, TN too large, TN=10s lblue: P controller to compare



Simulation results process with 3PT1 with Integrator.

Reference step responses All controller with 60° Red: P, ymax=17 Green: PI, symm. Opt. ymax=7,6 Blue: PDT1, st=10 ymax=417 lblue: PIDT1, st=10 ymax=242 vio: PIDT1, st=10, ymax=10, AWU



Disturbance Step responses process 3PT1 with Integrator Variation of controller type

Disturbance step responses All controller 60° Red: P, error 2,8% Green: PI, symm. opt. error 0 % Blue: PDT1, st=10 error 1,3% lblue: PIDT1, st=10 error 0 %



Reference- and Disturbance Step responses Process with 3PT1 and Integrator, Variation of phase margin

Reference step responses All PI-controller Red: 30° Green: 40° blue: 50° lblue: 60°

Disturbance step responses All PI- controller Red: 30° Green: 40° blue: 50° lblue: 60°



Improvement of control performance, chap. 5.3



Design of PIDT1- controller with FRA You have to know phase margin R and st and bode plot of process |Fp| and p. If st= 1 then no PDT1- part (pure PI). With PDT1- part (PIDT1 or PDT1- controller)

Without PDT1- part (PI or P- contoller, st=1)

Tv with pole compensation or -30-dB- method. Then add phase of PDT1- part with PDT1=arctan(Tv)-arctan(*Tv/st) to process phase.

Nothing, PDT1=0.

Continue with

PI with pole compensation

PI with Symmetrical Optimum

Only P

Get TN: Either TN =known largest process time constant or with -12 dB- method. Then add controller phase PI()= -90°+arctan(TN) to phase curve. Then get d at crossing point of phase curve p + PDT1 @ -180 +R.

Depending on given phase margin R get controller phasecontr at d out of table next page. Then get d at crossing point of phase curve p + PDT1

@ -180 +R -contr.

Get d at crossing point of phase curve p + PDT1 @ -180 +R.

Calculate corner frequency of PI-controller N= 1/TN with equation Reg= -90 + arctan(d * TN ) and solve to TN.

Finally calculate Kc with Nyquist equation |Fo(d)| = |Fc(d)| |Fp(d)| = 1.

Get |Fp(d)| from bode magnitude curve. |Fc(d)| =

With PDT1 2

22

)/(1

)(1)(1

stT

T

T

TK

Vd

Vd

Nd

Ndc

2

22

)/(1

)(1)(1

stT

T

T

TK

Vd

Vd

Nd

Ndc

2

2

)/(1

)(1

stT

TK

Vd

Vdc

Without PDT1 Nd

Ndc T

TK

2)(1

Nd

Ndc T

TK

2)(1 cK



Table for PI controller with symmetrical optimum

-12 dB method Draw tangent with gradient -12 dB/ decade, read radian freq. at tangent point (green arrow), here 1/sec= 12

A good approximation of TN is TN= 1 /12 Result: TN= 1 sec Doesn`t work, if process has I- part

-30 dB method Draw tangent with gradient -30 dB/ decade read radian freq at at tangent point (green arrow), here 3.4/sec= 30

A good approximation of Tv is Tv= 1 /30 Result: Tv= 0.294 sec Works in any case.

R contr @ d p+PDT1 @ d

30 -30 -120 40 -24 -116 50 -20 -110 60 -15 -105

R d ü 30 0.2686 0.416 40 0.3672 0.289 50 0.4777 0.181 60 0.6124 0.0877

Damping values for different phase margins in ideal case (closed loop is PT2), but approximately usable also for all other loops.

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