Title: Chapter 10 Stability Analysis and Controller Tuning
1Chapter 10 Stability Analysis and Controller
Tuning
? Bounded-input bounded-output (BIBO) stability
Ex. 10.1 A level process with P control
2(S1) Models (S2) Solution by Laplace transform
where
3- Note
- Stable if Kclt0
- Unstable Kcgt0
- Steady state performance by
4- Ex. 10.3 A level process without control
- Response to a sine flow disturbance
- Response to a step flow disturbance
5? Stability analysis
6Note Assume Gd(s) is stable.
7 Stability of linearized closed-loop
systems Ex. 10.4 The series chemical reactors
with PI controller
8- _at_ Known values
- Process
- Controller
9_at_ Formulation stability
Stable
10? Criterion of stability ? Direct substitution
method
11 The response of controlled output
P1. P2.
12P3.
13?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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15Example A.1 Known transfer functions
16Find (1) Ultimate gain (2) Ultimate
period S1. Characteristic eqn.
17S2. Let at KcKcu
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19 Example A.2
S1.
20S2.
S3.
21Example A.3 Find the following control loop (1)
Ultimate gain (2) Ultimate period
22S1. The characteristic eqn. for
H(s)KT/(?Ts1) S2. Gc-Kc to avoid the
negative gains in the characteristic eqn.
23S3. By direct substitution of at
KcKcu
24 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
25Example A.4 Find the ultimate gain and frequency
of first-order plus dead-time process
S1. Closed-loop system with P control
26S2. Using Pade approximation
27S3. Using direct substitution method
28- Note
- The ultimate gain goes to infinite as the
dead-time approach zero. - The ultimate frequency increases as the dead time
decreases.
29? Root locus
A graphical technique consists of roots of
characteristic equation and control loop
parameter changes.
30Definition Characteristic equation Open-loop
transfer function (OLTF) Generalized OLTF
31Example B.1 a characteristic equation is given
S1. Decide open-loop poles and zeros by OLTF
32S2. Depict by the polynomial (characteristic
equation) Kc1/3
33S3. Analysis
34Example B.2 a characteristic equation is given
S1. Decide poles and zeros
35S2. Depict by the polynomial (characteristic
equation)
36S3. Analysis
37Example B.3 a characteristic equation is given
S1. Decide poles and zeros
S2. Depict by the polynomial (characteristic
equation)
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39S3. Analysis
40_at_ Review of complex number caib
41?Polar notations
42P1. Multiplication for two complex numbers (c,
p) P2. Division for two complex numbers (c, p)
43- _at_ Rules for root locus diagram
- Characteristic equation
- Magnitude and angle conditions
44Since
45- 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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48Example B.4 Depict the root locus of a
characteristic equation (heat exchanger control
loop with P control) S1. OLTF
49- 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. - n3, three branches or loci are indicated.
- Because m0 (zeros), all loci approach infinity
as Kc increases. - Determine CG-0.155 and asymptotes with angles,
?60, 180 , 300 . - Calculate the breakaway point by
50s 0.247 and 0.063
S3. Depict the possible root locus with ?u0.22
(direct substitution method) and Kcu24
51Example B.5 Depict the root locus of a
characteristic equation (heat exchanger control
loop with PI control) S1. OLTF
52S2. Following rules
53S3. Depict root locus
54Exercises
55Ans. 8.1
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57Ans. 8.2
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61 Dynamic responses for various pole locations
62 Which is good method for stability analysis
?
63- ? Bode method
- A brief review
- OLTF
- Frequency response
64? Stability criterion
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66 Frequency response stability criterion Determini
ng the frequency at which the phase angle of OLTF
is 180(p) and AR of OLTF at that frequency
Ex. C.1 Heat exchanger control system (Ex. A.1)
67S1. OLTF
S2. Find MR and ? S3. Bode plot in Fig.
9-2.3 to estimate ?0.219 by ? 180 and
decide MR0.0524
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69S4. Decide Kc as AR1
Stability vs. controller gain In Bode plot,
as ? 180 both ? and MR are determined.
Moreover, ? ?u and Kcu can be obtained.
70Ex. C.2 Analysis of stability for a OTLF S1.
MR and ?
71S2. Show Bode plot (MR vs. ? ? vs. ? )
72S3. Find ?u and Kcu ?u0.16 by ? p
Kcu 12.8
Ex. C.3 The same process with PD
controller and ?0.1 (S1) OLTF
73(S2) By Fig. 9-2.5 ?u0.53 and
MR0.038 Kcu33 and ?u0.53
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75Ex. 10.7
(S1) Bode plot (AR vs. ? ? vs. ? ) for Kc1
76(S2) Stability vs. controller gain Kc
Ex. 10.8 Determine whether this system is stable.
77(S1) Bode plot for Kc15 and TI1
(S2) Since the ARgt1 at , the
system is unstable.
78P1. Bode plot for the first-order system
79P2. Bode plot for the second-order system
80Ex. 10.9 Determine AR and ? of the following
transfer function at
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82 Controller tuning based on Z-N closed-loop
tuning method S1. Calculating ?c by setting
Kc1 and then determine Ku and Pu where
ARc
83S2. Controller tuning constants
Ex. 10.10 Calculate controller tuning constants
for a process, Gp(s)0039/(5s1)3, by uning the
Z-N method S1.
84S2. Bode plot
85S3. Tuning constants
86S4. Closed-loop test
87Ex. 10.14 Integral mode tend to destabilize the
control system
88_at_ 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 GM2 is the same as the Z-N
tuning.
89(2) Phase margin (PM) Typical
specification PMgt45
Ex. D.1 Consider the same heat exchanger to tune
a P controller for specifications (Ex. C.2) (a)
While GM2
90(b) PM 45?? 135. By Fig. in Ex. C.2, we can
find and
91? 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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93Ex. E.2 Consider the amplitude ratio and the
phase angle angle of second-order lag are
given as
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95Ex. E.3 Consider the second-order system with
tuning Kc
96Ex. E.4. Consider the amplitude ratio and the
phase angle angle of pure dead time system
are given as
97? Conformal mapping
98? Nyquist stability criterion (Nyquist plot)
Ex. E.5 Consider a closed-loop system, its OLTF
is given as
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100Unstable stable Kcgt23.8
Marginal stable Kc23.8
stable Kclt23.8
101- Exercises
- Q.10.11
- Q.10.15