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**Chapter 10 Stability Analysis and Controller Tuning**

※ Bounded-input bounded-output (BIBO) stability ＊ Ex A level process with P control

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

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Note: Stable if Kc<0 Unstable Kc>0 Steady state performance by

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**＊ Ex. 10.3 A level process without control**

Response to a sine flow disturbance Response to a step flow disturbance

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※ Stability analysis

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**Note: Assume Gd(s) is stable.**

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**＊ Stability of linearized closed-loop systems**

Ex The series chemical reactors with PI controller

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@ Known values Process 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:**

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P3.

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**﹪Ultimate gain (Kcu): The controller gain at which this point of marginal instability is reached**

﹪Ultimate period (Tu): 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 Kc=Kcu

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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)=KT/(Ts+1)**

S2. Gc=-Kc to avoid the negative gains in the characteristic eqn.

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**S3. By direct substitution of at Kc=Kcu**

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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: The ultimate gain goes to infinite as the dead-time approach zero. 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)**

Kc: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**

Characteristic equation Magnitude and angle conditions

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Since

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**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 s1 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 From rule 1 where the root locus exists are indicated. From rule 2 indicate that the root locus is originated at the poles of OLTF. n=3, three branches or loci are indicated. Because m=0 (zeros), all loci approach infinity as Kc increases. Determine CG= and asymptotes with angles, =60°, 180 °, 300 °. Calculate the breakaway point by

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s= – and –0.063 S3. Depict the possible root locus with ωu=0.22 (direct substitution method) and Kcu=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: OLTF Frequency response

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◎ Stability criterion

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**＊ Frequency response stability criterion**

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

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

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S4. Decide Kc as AR=1 ＃ ＊ Stability vs. controller gain In Bode plot, as θ= –180° both ω and MR are determined. Moreover, ω = ωu and Kcu 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 ωu and Kcu ωu=0.16 by = –π Kcu =12.8 Ex. C.3 The same process with PD controller and =0.1 (S1) OLTF

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(S2) By Fig u=0.53 and MR=0.038 Kcu=33 and u=0.53

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

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**(S2) Stability vs. controller gain Kc**

Ex Determine whether this system is stable.

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**(S1) Bode plot for Kc=15 and TI=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 Kc=1 and then determine Ku and Pu where ARc=

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**S2. Controller tuning constants**

Ex Calculate controller tuning constants for a process, Gp(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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**@ Effect of modeling errors on stability**

Gain margin (GM): Total loop gain increase to make the system just unstable. The controller gain that yields a gain margin ＊ Typical specification: GM2 If P controller with GM=2 is the same as the Z-N tuning.

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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 Kc**

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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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Unstable stable Kc>23.8 Marginal stable Kc=23.8 stable Kc<23.8

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Exercises: Q.10.11 Q.10.15

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