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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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