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Chapter 10 Stability Analysis and Controller Tuning ※ Bounded-input bounded-output (BIBO) stability ＊ Ex. 10.1 A level process with P control

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(S1) Models (S2) Solution by Laplace transform where

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Note: (1)Stable if K c <0 (2)Unstable K c >0 (3)Steady state performance by

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＊ Ex. 10.3 A level process without control (1)Response to a sine flow disturbance (2)Response to a step flow disturbance

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※ Stability analysis

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Note: Assume G d (s) is stable.

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＊ Stability of linearized closed-loop systems Ex. 10.4 The series chemical reactors with PI controller

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@ Known values 1.Process 2.Controller

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@ Formulation& stability Stable

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◎ Criterion of stability ※ Direct substitution method

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The response of controlled output: P1. P2.

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P3.

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﹪ Ultimate gain (K cu ): The controller gain at which this point of marginal instability is reached ﹪ Ultimate period (T u ): It shows the period of the oscillation at the ultimate gain ＊ Using the direct substitution method by in the characteristic equation

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Example A.1 Known transfer functions

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Find: (1) Ultimate gain (2) Ultimate period S1. Characteristic eqn.

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S2. Let at K c =K cu

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Example A.2 S1.

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S2. S3.

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Example A.3 Find the following control loop: (1) Ultimate gain (2) Ultimate period

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S1. The characteristic eqn. for H(s)=K T /( T s+1) S2. G c =-K c to avoid the negative gains in the characteristic eqn.

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S3. By direct substitution of at K c =K cu

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＊ Dead-time Since the direct substitution method fails when any of blocks on the loop contains deadt-ime term, an approximation to the dead-time transfer function is used. First-order Padé approximation:

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Example A.4 Find the ultimate gain and frequency of first- order plus dead-time process S1. Closed-loop system with P control

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S2. Using Pade approximation

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S3. Using direct substitution method

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Note: 1.The ultimate gain goes to infinite as the dead-time approach zero. 2.The ultimate frequency increases as the dead time decreases.

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※ Root locus A graphical technique consists of roots of characteristic equation and control loop parameter changes.

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＊ Definition: Characteristic equation: Open-loop transfer function (OLTF): Generalized OLTF:

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Example B.1: a characteristic equation is given S1. Decide open-loop poles and zeros by OLTF

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S2. Depict by the polynomial (characteristic equation) K c :1/3

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S3. Analysis

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Example B.2: a characteristic equation is given S1. Decide poles and zeros

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S2. Depict by the polynomial (characteristic equation)

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S3. Analysis

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Example B.3: a characteristic equation is given S1. Decide poles and zeros S2. Depict by the polynomial (characteristic equation)

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S3. Analysis

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@ Review of complex number c=a+ib

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Polar notations

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P1. Multiplication for two complex numbers (c, p) P2. Division for two complex numbers (c, p)

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@ Rules for root locus diagram (1)Characteristic equation (2)Magnitude and angle conditions

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Since

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(3)Rule for searching roots of characteristic equation Ex. A system have two OLTF poles (x) and one OLTF zero (o) Note: If the angle condition is satisfied, then the point s 1 is the part of the root locus

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Example B.4 Depict the root locus of a characteristic equation (heat exchanger control loop with P control) S1. OLTF

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S2. Rule for root locus (i)From rule 1 where the root locus exists are indicated. (ii)From rule 2 indicate that the root locus is originated at the poles of OLTF. (iii)n=3, three branches or loci are indicated. (iv)Because m=0 (zeros), all loci approach infinity as K c increases. (v)Determine CG=-0.155 and asymptotes with angles, =60°, 180 °, 300 °. (vi)Calculate the breakaway point by

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s= – 0.247 and –0.063 S3. Depict the possible root locus with ω u =0.22 (direct substitution method) and K cu =24

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Example B.5 Depict the root locus of a characteristic equation (heat exchanger control loop with PI control) S1. OLTF

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S2. Following rules

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S3. Depict root locus

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＊ Exercises

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Ans. 8.1

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Ans. 8.2

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＊ Dynamic responses for various pole locations

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◎ ＊ Which is good method for stability analysis

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※ Bode method A brief review: (1)OLTF (2)Frequency response

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◎ Stability criterion

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Ex. C.1 Heat exchanger control system (Ex. A.1) ＊ Frequency response stability criterion Determining the frequency at which the phase angle of OLTF is –180°(–π) and AR of OLTF at that frequency

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S2. Find MR and θ S3. Bode plot in Fig. 9-2.3 to estimate ω=0.219 by θ= –180° and decide MR=0.0524 S1. OLTF

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S4. Decide K c as AR=1 ＃ ＊ Stability vs. controller gain In Bode plot, as θ= –180° both ω and MR are determined. Moreover, ω = ω u and K cu can be obtained.

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Ex. C.2 Analysis of stability for a OTLF S1. MR and θ

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S2. Show Bode plot (MR vs. & vs. )

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S3. Find ω u and K cu ω u =0.16 by = –π K cu =12.8 Ex. C.3 The same process with PD controller and =0.1 (S1) OLTF

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(S2) By Fig. 9-2.5 u =0.53 and MR=0.038 K cu =33 and u =0.53

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Ex. 10.7 (S1) Bode plot (AR vs. & vs. ) for K c =1

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(S2) Stability vs. controller gain K c Ex. 10.8 Determine whether this system is stable.

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(S1) Bode plot for K c =15 and T I =1 (S2) Since the AR>1 at, the system is unstable.

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P1. Bode plot for the first-order system

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P2. Bode plot for the second-order system

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Ex. 10.9 Determine AR and of the following transfer function at

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＊ Controller tuning based on Z-N closed-loop tuning method S1. Calculating c by setting K c =1 and then determine K u and P u where AR c =

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S2. Controller tuning constants Ex. 10.10 Calculate controller tuning constants for a process, G p (s)=0039/(5s+1) 3, by uning the Z-N method S1.

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S2. Bode plot

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S3. Tuning constants

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S4. Closed-loop test

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Ex. 10.14 Integral mode tend to destabilize the control system

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(1)Gain margin (GM): Total loop gain increase to make the system just unstable. The controller gain that yields a gain margin ＊ Typical specification: GM 2 If P controller with GM=2 is the same as the Z-N tuning. @ Effect of modeling errors on stability

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(2) Phase margin (PM): ＊ Typical specification: PM>45° Ex. D.1 Consider the same heat exchanger to tune a P controller for specifications (Ex. C.2) (a) While GM=2

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(b) PM= 45° θ= –135°. By Fig. in Ex. C.2, we can find and

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※ Polar plot The polar plot is a graph of the complex-valued function G(i ) as goes from 0 to . Ex. E.1 Consider the amplitude ratio and the phase angle angle of first-order lag are given as

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Ex. E.2 Consider the amplitude ratio and the phase angle angle of second-order lag are given as

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Ex. E.3 Consider the second-order system with tuning K c

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Ex. E.4. Consider the amplitude ratio and the phase angle angle of pure dead time system are given as

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※ Conformal mapping

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※ Nyquist stability criterion (Nyquist plot) Ex. E.5 Consider a closed-loop system, its OLTF is given as

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Marginal stable K c =23.8 stable K c <23.8 Unstable stable K c >23.8

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Exercises: (1)Q.10.11 (2)Q.10.15

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