Transcript 投影片 1 - YunTech

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:
(1) Stable if Kc<0
(2) Unstable Kc>0
(3) Steady state performance by
＊ Ex. 10.3 A level process without control
(1) Response to a sine flow disturbance
(2) 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
1. Process
2. Controller
@ Formulation& stability
Stable
◎ Criterion of stability
※ Direct substitution method
The response of controlled output:
P1.
P2.
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 s  iu in the
characteristic equation
Example A.1 Known transfer functions
Find: (1) Ultimate gain
(2) Ultimate period
S1. Characteristic eqn.
S2. Let s  iu 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 s  i u 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 firstorder plus dead-time process
S1. Closed-loop system with P control
S2. Using Pade approximation
S3. Using direct substitution method
Note:
1. The ultimate gain goes to infinite as the dead-time
approach zero.
2. 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
(1) Characteristic equation
(2) Magnitude and angle conditions
Since
(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 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
(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 Kc
increases.
(v) Determine CG=-0.155 and asymptotes with angles, =60°,
180 °, 300 °.
(vi) 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:
(1) OLTF
(2) 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 c  1.31 , 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   0.4
＊ 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
(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.
(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
  0  AR  K and   0o

1

o



AR

0
.
707
K
and



45



    AR  0. and   90o

Ex. E.2 Consider the amplitude ratio and the phase angle angle of
second-order lag are given as
  0  AR  K and   0o


1
K
o



AR

and



90


2


o




AR

0
.
and



180

Ex. E.3 Consider the second-order system with tuning Kc
Kc
GM ( s)Gc ( s) 
( s  1)(5s  1)
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:
(1) Q.10.11
(2) Q.10.15