Lecture 11 Controller Specifications CL-417 Process Control Prof. - - PowerPoint PPT Presentation

Lecture 11 Controller Specifications

CL-417 Process Control

• Prof. Kannan M. Moudgalya

IIT Bombay Wednesday, 14 August 2013

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Outline

• 1. Closed loop as a second order system
• 2. Expressions for performance requirement
• 3. Desired region

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• 1. Closed loop as a second order system

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Recall Feedback Control of Mixing

AC Control Pure A x2 = 1 w2 =? x, w AT Mixture A, B x1, w1 Valve

We will see a schematic of it in the next page

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Closed loop feedback diagram

Ysp Gm Gc = Kc

• 1 + 1

τis

• KIp

Gv = Kv τvs + 1 Gp Gd

−

Km

Y

U Uin P E Ym

It has transfer functions for the measurement system, valve and the controller.

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Block diagram of transfer function model

∆h(s) = K1 τs + 1∆Qi(s) − K2 τs + 1∆x(s)

− K2 τs + 1 K1 τs + 1 ∆Qi ∆x Disturbance Variable Manipulated Variable ∆h Controlled Variable +

K2 block replaces 3 blocks in full schematic!

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Work with Simplified Feedback Control System: Flow Control System

− −K2 τs + 1 K1 τs + 1 ∆Qi ∆x ∆h e Setpoint Gc

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Closed loop system

v Gc u G −

e

y

r

◮ Gc: controller, G: plant ◮ Derive the closed loop transfer function

between r and y: Gcl = GGc 1 + GGc , y(s) = Gclr(s)

◮ To do: derive the transfer function

between v and y

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Gcl as a Second Order System

◮ Gcl =

GGc 1 + GGc

◮ When will Gcl be a second order system ◮ If Gc = K a constant and G a second

• rder system

◮ If Gc and G both are first order systems

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Identify the false statement

We model the closed loop system as second

• rder, because
• 1. Need a simple model that has oscillations
• 2. Second order underdampled system is the

simplest

• 3. All closed loop systems can be modelled

as a second order system Answer: 3

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Why Model a Second Order System

◮ Need a simple model that has oscillations ◮ Second order underdampled system is the

simplest

◮ Would want the closed loop system to be

fast

◮ So we want the closed loop system to be

underdamped second order

◮ Instead of writing Gcl, we will denote the

closed loop transfer function as simply G from now on

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Recall: Step response of second order system

Y(s) = K τ 2s2 + 2ζτs + 1U(s)

◮ Let U(s) = 1/s, calculate Y(s) and invert

it.

◮ For ζ values of < 1, = 1 and > 1 ◮ How would you calculate?

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Underdamped system

We studied second order systems, G(s) = K τ 2s2 + 2ζτs + 1 Also consider an equivalent and popular representation with gain K = 1: G(s) = ω2

n

s2 + 2ζωns + ω2

n

◮ ωn: natural frequency, ζ: damping factor. ◮ For ζ < 1, get an underdamped system ◮ Calculate the roots for ζ < 1

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Solution to underdamped system

Underdamped second order system: G(s) = ω2

n

s2 + 2ζωns + ω2

n

For ζ < 1, rts of den. = −ζωn ± jωn

• 1 − ζ2

The step response is, y(t) = 1 − e−ζωnt

• 1 − ζ2 sin
• ωn
• 1 − ζ2t + tan−1
• 1 − ζ2

ζ

• We will consider the underdamped case only

in the rest of this lecture

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Step response of underdamped system

y(t) tr tp ts

Mp

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• 2. Expressions for performance

requirement

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Performance Requirements

◮ Rise time tr is the time required for y(t)

to reach the final value (1 in this case) the first time

◮ Peak time tp is when the first peak occurs ◮ Mp is the corresponding overshoot ◮ ts is the settling time at which the

response y(t) enters a tube around the steady state and stays within it Want tr, Mp, ts to be small

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Step response of underdamped system

Want (a) tr, (b) Mp, (c) ts to be small

y(t) tr tp ts

Mp

Want to translate this to conditions on G(s) = ω2

n

s2 + 2ζωns + ω2

n

• r on the poles: = −ζωn ± jωn
• 1 − ζ2

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Translation of performance requirement to pole locations

We want to translate performance conditions

• n to pole locations, because
• 1. We want to get simple expressions for

performance requirements

• 2. This could help us back calculate the

required controller

• 3. This could help picture the situation

clearly and makes it easy to understand Answer: 2, may help calculate the controller

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• a. Rise time, tr

Recall the solution: y(t) = 1 − e−ζωnt

• 1 − ζ2 sin
• ωn
• 1 − ζ2t + tan−1
• 1 − ζ2

ζ

• y(∞) = 1. At t = tr, y(t) = 1.

sin

• ωn
• 1 − ζ2tr + tan−1
• 1 − ζ2

ζ

• = 0

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Expression for rise time

◮ Soln: argument = nπ, n = 0, 1, . . . ◮ For n = 0, tr < 0, hence not valid. First

valid solution is for n = 1. tr =

• π − tan−1
• 1 − ζ2

ζ

• /ωn
• 1 − ζ2

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Approximate expression for rise time

tr =

• π − tan−1
• 1 − ζ2

ζ

• /ωn
• 1 − ζ2

◮ Want a simple expression ◮ For intermediate value of ζ = 0.5,

tr = 2.42/ωn.

◮ Small tr is obtained by large ωn ◮ ωn is the absolute value of

= −ζωn ± jωn

• 1 − ζ2

◮ i.e. ωn is the distance of the pole from

the origin

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Poles of second order system

Recall the expression for poles: −ζωn ± jωn

• 1 − ζ2

θ × Im(s) Re(s) ζωn ωn ωn

• 1 − ζ2

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How do we get a small rise time?

ωn Im(s) Re(s)

Desired region where poles should lie is shaded

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• b. Want small overshoot in step response

y(t) tr tp ts

Mp

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Peak overshoot, Mp

Recall the solution: y(t) = 1 − e−ζωnt

• 1 − ζ2 sin
• ωn
• 1 − ζ2t + tan−1
• 1 − ζ2

ζ

• Differentiate and equate to zero

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Expression for Mp

tp = π ωn

• 1 − ζ2

The corresponding peak expression is: Mp = exp

• −

πζ

• 1 − ζ2
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CL-417 Process Control Control Specification

Towards an approximate expression for Mp

Scilab code:

1

zeta = 0 : 0 . 0 1 : 0 . 6 ;

2 Mp = exp(−%pi∗ zeta

. / s q r t (1− zeta ˆ2) ) ;

3 L = gca ( ) ; 4 L . t h i c k n e s s = 2 ; 5

p l o t 2 d ( zeta ,Mp)

6

x t i t l e ( ’ Plot

• f Mp vs .

zeta ’ , ’ zeta ’ , ’Mp ’ ) ;

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Scilab output plot

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 0.0 0.1 0.2 0.3 0.4 0.5 0.6 Plot of Mp vs. zeta zeta Mp

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Approximate expression for Mp

Mp ≃ 1 − ζ 0.6 for 0 ≤ ζ < 0.6 Want ζ large for small Mp

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What is ζ?

Recall the expression for poles: −ζωn ± jωn

• 1 − ζ2

θ × Im(s) Re(s) ζωn ωn ωn

• 1 − ζ2

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Want small overshoot in step response

Re(s) Im(s)

Shaded region is desired

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• c. Want small settling time

y(t) tr tp ts

Mp

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Small settling time

Recall the solution: y(t) = 1 − e−ζωnt

• 1 − ζ2 sin
• ωn
• 1 − ζ2t + tan−1
• 1 − ζ2

ζ

• e−ζωnts = ε

For ε = 0.02, ts = 4 ζωn For ε = 0.01, ts = 4.6 ζωn . ts small ⇒ ζωn large

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Poles of second order system

θ × Im(s) Re(s) ζωn ωn ωn

• 1 − ζ2

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Small settling time

Re(s) Im(s)

Desired region where poles should lie is shaded

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• 3. Desired region

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• c. Small settling time

Re(s) Im(s)

Desired region where poles should lie is shaded

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• a. Small rise time

ωn Im(s) Re(s)

Desired region where poles should lie is shaded

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• b. Small overshoot

Re(s) Im(s)

Desired region where poles should lie is shaded

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Desired region for all three conditions

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Desired region for all three conditions

Intersection of the three regions

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What we learnt today

◮ Closed loop as a second order system ◮ Expressions for performance requirement ◮ Desired region

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Thank you

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