single input single output system

8/18/2019 Single Input Single Output System

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054461 PROCESS CONTROL LABLECTURE ONE

Daniel R. Lewin, Technion1

Course IntroductionPROCESS CONTROL LAB - (c) Daniel R. Lewin1-1

054461 Process Control Laboratory

LECTURE 1:

COURSE INTRODUCTION

Daniel R. Lewin

Department of Chemical EngineeringTechnion, Haifa, Israel


Lecture Objectives

Be familiar with the course structure andobjectives.

Have recalled all of the material you learned inthe introductory control course, and inparticular, be able to:

a. Formulate a linear process model

b. Sketch the response of a linear system

c. Design a simple (PID) feedback controllerusing the Root Locus method

On completing this section, you should:


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Course Objectives

Derive an empirical linear model for a realprocess by generating experimental data and itsanalysis.

Design, tune and implement SISO controllers fora real process. The types of controllers that youshould be able to implement are: simple feedback(e.g. PID), cascade controllers, IMC and FF.

Design and implement control system for MIMOprocesses.

On completing this course, you should:


Linear Systems – Review + More

Modeling process transient behavior

Linearization

Laplace transforms

Linear system response Root locus design of FB controllers


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Modeling - Dead Sea Pond (1)

Prepare a model describing an evaporating pond in the DeadSea Works.

Feed brine [T/h]

Brine conc. [kg salt/kg]

Evap. Rate [T/h]

Solution.

rate of accumu- rate of input rate of out- rate of lossby evaporationlation w ith feed put w ith effluent

f0 q q E

= − − = − −

Overall mass balance:

fThus, q q E= −


Modeling - Dead Sea Pond (2)

( )

rate of accumu- rate of salt rate of saltout-

lation of salt input with feed put with effluent

f f

dVc q c qc

dt

= −

ρ = −

The balance for salt gives:

fUsing q q E, and noting that and V are constant:= − ρ

( )1 , 0ff

dc E 1 c c c 0 cqdt

+

= − =τ

( )ff f c q Eq cdcV Vdt

−−

=ρ ρ

fDefining V q :τ = ρ What is the sign of

(E/qf – 1) ?


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Linearization (1) Consider the general SISO nonlinear process:

( )dx

f x,udt

=

Linearization around the stationary point (x0, u0) givesthe linear equation:

( ) ( ) ( )0 0 0 0

0 0 0 0x ,u x ,u

dx f ff x ,u x x u u

dt x u

∂ ∂+ − + −

∂ ∂

But: ( )0 0 0dx

f x , udt

=

Hence: ( ) ( ) ( )

0 0 0 0

0 0 0 0

00 0

x ,u x ,u

x ,u x ,u

d x x f fx x u udt x u

dx f fx u

dt x u

− ∂ ∂− + −∂ ∂

∂ ∂+

∂ ∂


Linearization (2) Extension to MIMO nonlinear process:

( )

( )

dxf x,u

dt y g x,u

=

=

Linearization around the stationary point (x*, u*) gives

the linear system:

dxAx Bu

dt y Cx Du

= +

= +

ii,j

fA a

x

∂≡ =

∂

The matrices are Jacobian matrices, e.g.A,B,C and D

*

*

x x x

u u u

= −

= −


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Linearization – Example (1)

( ) ( )1 , 0f f ff

dc Ef c;q ,c ,E 1 c c c 0 cqdt

+

= = − =τ

Continuing the Dead Sea Ponds example, we have:

Data: qf = 10 T/h, E = 5 T/h, cf = 0.1 kg salt/kg brine, ρV = 100 T

At steady state:

( )= ⇒ = − =f fdc

0 c c 1 E q 0.2 kg salt/kg brinedt

Stationary point: c* = 0.2; qf* = 10; cf* = 0.1; E* = 5; τ* = 10 h.

Linearization:

ffss ssf fss ss

dc f f f fc q c E

dt c q c E

∂ ∂ ∂ ∂= + + +

∂ ∂ ∂ ∂


Linearization – Example (2)

-1*

*ss f*

Ef 11 0.05 h

c q

∂= − = −

∂ τ -1*f*

f ss

c cf0.001 T

q V

−∂= = −

∂ ρ

-1

*f ss

f 10.1 h

c

∂= =

∂ τ-1*

ss

cf0.002 T

E V

∂= =

∂ ρ

( )ffdc

0.05c 0.001q 0.1c 0.002E, c 0 0dt

= − − + + =

ff

ss ssf fss ss

dc f f f fc q c E

dt c q c E

∂ ∂ ∂ ∂= + + +

∂ ∂ ∂ ∂


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Laplace Transforms General MIMO linear process is:

dxAx Bu

dt y Cx Du

= +

= +

*

*

x x x

u u u

= −

= −

Example: [ ]

( )f

f

qdc

0.05 c 0.001 0.1 0.002 c ,c 0 0dt

E

= − + − =

Taking Laplace Transforms (around steady state):

( ) ( ) ( )

( ) ( ) ( )

sX s AX s BU s

Y s CX s DU s

= +

= +

Hence: ( ) ( ) ( )1

Y s C sI A B D U s− = − +

P(s) –transferfunction matrixrelating all inputsto outputs


Laplace Transforms

A strictly proper system has n > m (physical system).A proper system has n = m (e.g. PI controller) Roots of numerator are zeros and roots of denominatorare poles:

Analysis of transfer functions: ( ) ( )1

P s C sI A B D−

= − +

The transfer function matrix is composed of elements:

( )m m 1

m m 1 1 0i,j n n 1

n n 1 1 0

b s b s b s bp s

a s a s a s a

−−

−−

+ + + +=

+ + + +…

…

zi are zeros

( ) ( ) ( ) ( )

( )( ) ( )1 2 m

i,j1 2 n

s z s z s zp s

s p s p s p

− − −=

− − −

pi are poles

For stability ALL poles must have a negative real part. Complex poles give oscillatory response. Zeros shape the response (see later)


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Response to step

in OP: --OP —PVClassification

Transfer

Function

Stable self-

regulatings

p

p e1s

K θ−+τ

Non self-regulating

sp es

K θ−

Unstable sp

p e1s

K θ−+τ−

Linear System Response


Identification by Step Test First order response of process, y, to a step change in input, u:

Dynamics are approximated by the FOPTD model:p s

Kp(s) e

s 1−θ=

τ +

pwith K y u, and θ and estimated from the trajectory= ∆ ∆ τ

u(t)

y(t)

∆u

0.623∆ y ∆ y


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Higher Order Responses (1)

p

2 2

Kp(s)

s 2 s 1=

τ + τξ +

Response of 2nd order transfer function to a unit step:

( ) ( )( )= =

− −τ + τξ +p p 1 2

2 21 2

K K p p y(s)

s s p s ps s 2 s 1ξ −ξ

= − ±τ τ

2

1,2

1p

Need to differentiate between three cases:

static gain

damping coefficient

ξ >ξ =ξ <

A) 1B) 1C) 1

natural period


Higher Order Responses (2)ξ > 1 2A) For 1, p and p are real negative roots:

( )− −= − −1 2p t p tp 1 2 y(t) K 1 A e A e

ξ = = = − τ1 2B) For 1, p p 1/ :

( )( )− τ

= − + τt

p y(t) K 1 1 t e


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Higher Order Responses (3)ξ < ξ −2 1 2C) For 1, 1 is imaginary, and p and p

are complex roots:

( )tp 21

y(t) K 1 e sin t1

−ξ τ

= − ω − φ − ξ

= − ξ τ ± − ξ τ21,2p i 1

− ξω =

τ

21

( )−

φ = − ξ ξ

1 2

tan 1


Higher Order Responses (4) Example 1: A processing system with its controller

( ) ( )( ) ( )

p

1 2

KCp s s

F s 1 s 1= =

τ + τ +

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )( )s

C s P s B s e s

P s B s C s C s

=

= −

( ) ( ) ( )

( ) ( )

c p

c p

1 2 1 2

c p c p

s 2

K K

K K 1

K K 1 K K 1

P s B sCs

C P s B s 1s s 1

+

τ τ τ + τ

+ +

= =+

+ +


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Root Locus for Control Design

1. The open-loop transfer function is expressed in the form: The Rules:

2. The root loci start (at K = 0) at the poles of pc(s) and end(K = ∞) at the zeros of pc(s) or at ± ∞

3. On the real axis of the complex plane, the RL isconstructed from right to left, according to the total polesand zeros (Σpz) met along the way:

If K > 0, draw the RL if Σpz is odd

If K < 0, draw the RL if Σpz is even

4. The number of loci = order of the system = number ofpoles of pc(s). The complex loci always appear as complexconjugates (mirror image either side of the real axis).

( ) ( ) ( ) ( )

( ) ( ) ( )1 2 m

1 2 n

s z s z s zpc s K

s p s p s p

− − −=

− − −

…

…



5. The angle of the asymptotes of the loci (as K →∞) is:

The Rules (Cont’d):

( )

( )

180 2k 1 k 0,1, ,n m 1 (K 0)

n m180 2k

k 0,1, ,n m 1 (K 0)n m

−= − − >

−

= − − <−

…

…

where n and m are the number of poles and zeros of pc(s).

6. The asymptotes are centered atwhere pi and zi are the locations of poles and zeros of

pc(s), respectively.

( ) ( )i ip z n mσ = − −∑ ∑


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Root Locus for Control Design( )

( ) ( ) ( ) c

1 p s ,c s K

s 1 s 3= =

+ +Example 3

1 2-1-2-3-4

-1

-2

1

2

Re(s)

Im(s)3

-3

-5 ××××××××



( )( )( )

( ) ( )c i21

p s ,c s K 1 1 Tss 1 s 3

= = ++ +

Example 4

1 2-1-2-3-4

-1

-2

1

2

Re(s)

Im(s)3

-3

-5 ××××××××

2 ××××


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Root Locus for Control Design( )

( ) ( ) ( )s c i

1 p s e ,c s K 1 1 Ts

s 1−= = +

+

1 2-1-2-3-4

-1

-2

1

2

Re(s)

Im(s)3

-3

-5

Example 5

×××××××× ××××


Design Specifications for RL The desired closed loop response is:

2 2s

y 1(s)

y s 2 s 1=

τ + τξ +

( )t

2

1

y(t) 1 e sin t1

−ξ τ

= − ω − φ− ξ

The response to a unit step in ys is:

The analytical response is used to estimate desired overshootand settling time.

Overshoot: 2OS exp 1 = −πξ − ξ

Thus: ( )

( )

2

e2 2

e

log OS

log OSξ ≥

+ π


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Summary

Linearization

Laplace transforms

Linear system response

Root locus design of FB controllers Use of approximate specifications in design

( ) ( )= =x f x,u , y g x,u

Modeling process transient behavior Generating models in standard form:

single input single output system

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