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

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

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

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

※ Stability analysis

Note: Assume Gd(s) is stable.

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

@ Known values Process Controller

@ Formulation& stability Stable

◎ Criterion of stability ※ Direct substitution method

The response of controlled output:

P3.

﹪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

Example A.1 Known transfer functions

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

S2. Let at Kc=Kcu

Example A.2 S1.

S2. S3.

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

S1. The characteristic eqn. for H(s)=KT/(Ts+1) S2. Gc=-Kc to avoid the negative gains in the characteristic eqn.

S3. By direct substitution of at Kc=Kcu

＊ 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:

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

S2. Using Pade approximation

S3. Using direct substitution method

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

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

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

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

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

S3. Analysis

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

S2. Depict by the polynomial (characteristic equation)

S3. Analysis

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

S3. Analysis

@ Review of complex number c=a+ib

Polar notations

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

@ Rules for root locus diagram Characteristic equation Magnitude and angle conditions

Since

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

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

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=-0.155 and asymptotes with angles, =60°, 180 °, 300 °. Calculate the breakaway point by

s= – 0.247 and –0.063 S3. Depict the possible root locus with ωu=0.22 (direct substitution method) and Kcu=24

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

S2. Following rules

S3. Depict root locus

＊Exercises

Ans. 8.1

Ans. 8.2

＊ Dynamic responses for various pole locations

＊ Which is good method for stability analysis ◎

※ Bode method A brief review: OLTF Frequency response

◎ Stability criterion

＊ 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)

S1. OLTF 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

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.

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

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

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

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

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

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

(S1) Bode plot for Kc=15 and TI=1 (S2) Since the AR>1 at , the system is unstable.

P1. Bode plot for the first-order system

P2. Bode plot for the second-order system

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

＊ 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=

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

S2. Bode plot

S3. Tuning constants

S4. Closed-loop test

Ex. 10.14 Integral mode tend to destabilize the control system

@ 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: GM2 If P controller with GM=2 is the same as the Z-N tuning.

(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

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

※ 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

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

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

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

※ Conformal mapping

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

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

Exercises: Q.10.11 Q.10.15