Control
Control Systems
Systems Design
Design
2007年1月31日
1
課程用書
iTextbook:
Richard C. Dorf and Robert H. Bishop, “MODERN CONTROL SYSTEMS” 10th Edition
iReference:
Benjamin C. Kuo, “AUTOMATIC CONTROL SYSTEMS” 7th Edition
Katsuhiko Ogata, ”Modern Control Engineering" 2th Edition
iMATLAB and SIMULINK:
Mathworks Inc., “Matlab User’s Guide” and “Simulink User’s Guide”
授課方式
i以投影片為主
i模擬軟體 MATLAB、SIMULINK
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2
課程內容
i Chapter
1 Introduction to Control Systems
i Chapter 2 Mathematical Models of Systems
i Chapter 3 State Variable Models
i Chapter 4 Feedback Control System Characteristics
i Chapter 5 The Performance of Feedback Control Systems
i Chapter 6 The Stability of Linear Feedback Systems
i Chapter 7 The Root Locus Method
i Chapter 8 Frequency Response Methods
i Chapter 9 Stability in The Frequency Domain
i Chapter 10 The Design of Feedback Control Systems
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CHAPTER 1
Introduction to Control Systems
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1
FIGURE 1.2
Open-loop control system (without feedback).
An open-loop control system utilizes an actuating
device to control the process directly without using
feedback
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Control :
Output
Results
Input
Reference
Plant or
Process
• Manual Control
• Automatic Control
−
−
−
−
−
M
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Automobile Cruise-Control
Airplane Flight Control
Robot Arm Control
Inkjet Printer Head Control
Disk Drive Read/Write Head Positioning Control
3
FIGURE 1.3
Closed-loop feedback control
system (with feedback).
A closed-loop control system uses a measurement of
the output and feedback of this signal to compare it
with the desired output (negative feedback).
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4
FIGURE 1.4
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Multivariable control system.
5
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FIGURE 1.5
Watt’s flyball governor.
6
Advantages of Feedback:
• Increased accuracy (reduced the steady-state error)
• Reduced sensitivity to parameter variations
• Reduced effects of disturbances
• Increased speed of response and bandwidth
Computer Aided Control System Design (CACSD)
• Matlab & Simulink
• Matrixx
• Simnon
• Program CC
M
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FIGURE 1.19 The control system design process.
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CHAPTER 2
Mathematical Models of Systems
► Differential Equations of Physical System
► Linear Approximations of Physical System
► The Laplace Transform
► The Transfer Function Of Linear System
► Block Diagram Models
► Signal-Flow Graph Model
► Computer Analysis of control system
► Design Examples
► The Simulation of System Using Matlab
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Differential Equations of Physical System
The approach to dynamic system problem can be listed as follow:
1. Define the system and its components
2. Formulate the mathematical model and list the necessary assumptions.
3. Write the differential equations describing the model.
4. Solve the equations for the desired output variables.
5. Examine the solutions and the assumptions.
6. If necessary, reanalyze or redesign the system
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FIGURE 2.1 (a) Torsional spring-mass
system.(b) Spring element.
3
d 2 y (t )
dy (t )
M
+
b
+ ky (t ) = r (t )
dt 2
dt
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; Linear Approximations of Physical Systems
A linear system satisfied the properties of superposition and homogeneity.
r(t)
y(t)
system
r (t ) → y (t ) ⇔ y (t ) = g [ r (t )]
r1 (t ) → y1 (t ) ⎫
⎬ ⇒ r (t ) = αr1 (t ) + βr2 (t ) → y (t ) = αy1 (t ) + βy2 (t )
r2 (t ) → y2 (t ) ⎭
g [αr1 (t ) + βr2 (t )] = αg [ r1 (t )] + βg [ r2 (t )]
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Linearization of a nonlinear systems:
y (t ) = g [ x (t )]
Expanding the nonlinear equation into a Taylor series about the operation point,
then we have
( x − xo ) 2
( x − xo ) d 2
d
+ 2 g ( x)
+L
y = g ( x ) = g ( xo ) + g ( x )
2!
dx
1!
dx
x= x
x= x
o
o
Neglecting all the high order terms, to yield
y = g ( xo ) +
d
( x − xo )
g ( x)
= y0 + m ⋅ ( x − xo )
dx
1!
x= x
o
⇒ y − y 0 = m ⋅ ( x − xo )
or Δy = m ⋅ Δx
y = g ( x1 ,L, xn )
= g ( x1o ,L, xno ) +
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∂g
∂x1
x = xo
( x1 − x1o ) +
∂g
∂x2
x = xo
( x2 − x2 o ) + L +
∂g
∂xn
x = xo
( xn − xno )
8
Example 2.1 Pendulum oscillator model
T = Mgl sin θ
∂ sin θ
|θ =θ 0 (θ − θ 0 )
∂θ
where T0 = 0.
T − T0 ≅ Mgl
T = Mgl (cos 00 )(θ − 0)
= Mglθ
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; Rotational Motion
1. Inertia
Newton’s Law for rotation motion :
d
d2
∑ T ( t ) = Jα ( t ) = J ω ( t ) = J 2 θ ( t )
dt
dt
θ (t )
J
T (t )
T : Torque J : Inertia α : Angular acceleration
ω : Angular velocity θ : Angular displacement
θ (t )
2. Torsional Spring
T ( t ) = kθ ( t )
k
k : Torsional spring constant
T (t )
3. Friction for rotational motion : viscous, static, coulomb friction
d
T (t ) = B θ (t )
θ (t )
dt
B : Viscous friction coefficient
B
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T (t )
10
t
dv(t ) 1
v(t )
+C
+ ∫ v(t ) dt = r (t )
dt
L0
R
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; The Transfer Function of Linear Systems
Definition:
The ratio of the Laplace transform of the output variable to the
Laplace transform of the input variable, with all initial conditions
assumed to be zero.
The Laplace transform of the impulse response, with all the initial
conditions set to zero.
r(t)
y(t)
system
Y ( s)
= G( s)
R( s )
or
G ( s ) = L[ y (t )]
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Example : Mass-Spring-Friction System
y (t )
k
m
f (t )
B
d
d2
f1 (t ) = ma (t ) = m v (t ) = m 2 y (t )
dt
dt
f 2 (t ) = ky (t )
f 3 (t ) = B
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d
y (t )
dt
13
∑ f (t ) = f1 + f 2 + f 3
d2
d
= m 2 y (t ) + B y (t ) + ky (t )
dt
dt
&y&(t ) +
B
1
k
y& (t ) + y (t ) = f (t )
m
m
m
Taking the Laplace transform with zero initial conditions, we have
s 2Y ( s ) + s
B
1
k
Y ( s) + Y ( s) = F ( s)
m
m
m
Then the transfer function between Y(s) and F(s) is obtained
Y ( s)
1
= 2
F ( s ) s m + sB + k
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1. Inertia
Newton’s Law for rotation motion :
d
d2
∑ T ( t ) = Jα ( t ) = J ω ( t ) = J 2 θ ( t )
dt
dt
θ (t )
J
T (t )
T : Torque J : Inertia α : Angular acceleration
ω : Angular velocity θ : Angular displacement
θ (t )
2. Torsional Spring
T ( t ) = kθ ( t )
k
k : Torsional spring constant
T (t )
3. Friction for rotational motion : viscous, static, coulomb friction
d
T (t ) = B θ (t )
θ (t )
dt
B : Viscous friction coefficient
B
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T (t )
15
• Field-current controlled DC motor (Homework #1)
• Armature-controlled DC motor
Mathematical modeling of PM DC motor:
Ra
La
+
ia (t )
va (t )
eb (t )
−
J,B
TLoad
ω ,θ
The air-gap flux of the dc motor is
Ra
v a (t )
eb (t )
km
T (t )
J
θ (t )
Armature resistance La
Applied voltage
ia (t )
Back-emf
kb
Torque constant
Tm (t )
Load torque
Td
Rotor inertia
B
Rotor displacement ω (t )
Armature inductance
Armature current
Back-emf constant
Motor torque
Disturbance torque
Viscous friction coefficient
Rotor angular velocity
φ (t ) = k f i f (t )
and the torque developed by by the dc motor is
Tm (t ) = k1φ (t )ia (t ) = k1k f i f (t )ia (t )
= k1k f I f ia (t ), i f (t ) = constant
= k mia (t )
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If La/Ra≈ 0, the approximate model of the dc motor is obtained as
ω ( s)
Va ( s )
=
km
JRa s + ( BRa + kb k m )
Δ
k
sτ + 1
where
τ=
JRa
km
, k=
BRa + kb km
BRa + kb km
Relation between km and kb
The power input to the rotor = The power delivered to the shaft
eb (t )ia (t ) = T (t )ω (t )
⇔ kbω (t )ia (t ) = kmia (t )ω (t )
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Õ
km = kb
17
The cause and effect equation for the motor circuit are
d
ia (t ) + eb (t )
dt
eb (t ) = kbω (t ) = kbθ (t )
va (t ) = Ra ia (t ) + La
d
ω (t ) + Bω (t ) + Td
dt
d2
d
= J 2 θ (t ) + B θ (t ) + Td
dt
dt
T (t ) = J
The motor torque is equal to the torque delivered to the load, that is
Tm (t ) = T (t )
Hence the transfer function of the dc motor, with Td=0, is
ω ( s)
Va ( s )
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=
km
( Ra + sLa )( Js + B ) + kb km
18
• Tachometer
et (t ) = ktω (t ) = kt
d
θ (t )
dt
• Potentiometer
e p (t ) = k pθ (t ), k p =
Vapp
2πN
Vapp
ep
• Incremental encoder (Homework #2)
1. Measuring angular displacement
2. Measuring angular velocity
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DC motor Filed-controlled
Home work 2:
ω( s)
V f ( s)
=?
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Homework 3
V2 ( s )
=?
V1 ( s )
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Homework 4:
θ L = nθ m
Tm
=?
TL
T
m
T
L
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The block diagram
FIGURE 2.26 Multiple-loop feedback control
system.
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FIGURE 2.27
Fig.2.26
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Block diagram reduction of the system of
25
Signal-flow Graph Modes
V f (s )
G (s )
θ (s )
FIGURE 2.28 Signal-flow graph of the dc
motor.
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Signal-flow Graph Modes
•
•
•
•
•
Branch:
Nodes:
Forward-path gain:
Loop gain:
Non-touching:
FIGURE 2.31 Two-path
interacting system.
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Mason’s signal-flow gain formula:
Tij
∑
=
k
Pij Δ ijk
Δ
input node
=
output node
Pijk :
Δ:
Δ ijk :
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Example 2.11
G7
G8
R
G1
G2
G3
G4
H4
H2
G5
G6
Y
H1
H3
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The Simulation of System
Assuming that a mode and the simulation are reliably accurate,computer
simulation has the following advantages:
1.
System performance can be observed under all conceivable conditions.
2.
Results of field-system performance can be extrapolated with a simulation
model for prediction purposes.
3.
Decisions concerning future system presently in a conceptual stage can be
examined.
4.
Trials of system under test can be accomplished in a much-reduceed period
of time.
5.
Simulation results can be obtained at lower cost than real experimentation.
6.
Study of hypothetical situation can be achieved even when the hypothetical
situation would be unrealizable in actual life at the present time.
7.
Computer modeling and simulation is often the only feasible or safe
technique to analyze and evaluate a system.
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FIGURE 2.34 Analysis and design
using a system model.
31
Example 2.12 Electric traction motor control
ω( s)
vin ( s )
=?
FIGURE 2.35 Speed control of an electric traction
motor.
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FIGURE 2.35 Speed control of an electric traction motor.
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The simulation of system using Matlab
M&y& + by& + ky = r (t )
The unforced dynamic response,y(t), of the spring-mass-damper mechanical system is
y (t ) =
y ( 0)
1−ξ 2
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e −ξωnt sin(ωn 1 − ξ 2 t + θ )
34
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FIGURE 2.40 Script to analyze the spring-mass35
damper.
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FIGURE 2.41 Spring-mass-damper unforced response.
36
Exercises
E2.8 E2.9
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CHAPTER 4
Feedback Control System Characteristics
•
•
•
•
•
•
Open and Closed-Loop Control.
Sensitivity of Control System to Parameter Variations.
Control of the Transient Response of Control system.
Disturbance Signals in a Feedback Control system.
Steady-state Error.
The cost of Feedback
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; Open-Loop and Closed-Loop Control Systems
• Open-loop
R(s)
C(s)
G(s)
Transfer function:
G( s) =
C ( s)
R( s)
Error signal: E ( s ) = R ( s ) − C ( s )
Output: C ( s ) = G ( s ) E ( s ) = G ( s )[ R ( s ) − C ( s )]
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3
• Closed-loop
R (s )
+
E (s )
C(s)
G (s )
m
H (s )
B(s)
Error signal: E ( s ) = R ( s ) − B ( s )
Feedback signal: B ( s ) = H ( s )C ( s )
Output: C ( s ) = G ( s ) E ( s ) = G ( s )[ R ( s ) − B ( s )]
Transfer function:
T ( s) =
Error transfer function:
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C ( s)
G( s)
=
R( s ) 1 ± G ( s ) H ( s )
1
E ( s)
=
R( s) 1 ± G ( s) H ( s)
4
Sensitivity of Control Systems to Parameter variations
Definite: System sensitivity is the ratio of the change in the system transfer
function to the change of a process transfer function( or parameter) for a
small incremental change
∫
T (s)
G(s)
ΔT ( s ) / T ( s ) ΔT ( s ) G ( s )
×
=
ΔG ( s ) / G ( s ) ΔG ( s ) T ( s )
ΔT ( s ) G ( s ) ∂T ( s ) G ( s )
= lim
×
=
×
Δ →0 ΔG ( s )
T ( s ) ∂G ( s ) T ( s )
∂ ln(T ( s ))
=
∂ ln(G ( s ))
=
The chain rule:
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∫
T (s)
α
=
∫ ∫
T (s) G(s)
G(s) α
5
• Open-loop
Transfer function:
∫
T (s)
=
G(s)
T ( s) =
C ( s)
= G( s)
R( s)
∂T ( s ) G ( s )
⋅
=1
∂G ( s ) T ( s )
• Closed-loop
Transfer function: T ( s ) =
∫
T (s)
G(s)
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=
C ( s)
G( s)
=
R( s ) 1 + G ( s ) H ( s )
1
∂T ( s ) G ( s )
⋅
=
∂G ( s ) T ( s ) 1 + G ( s ) H ( s )
6
Rf
Vin
Ri
en
Ka
K a = 3000
Vo
Homework
Answer the questions:
1. The necessary assumptions, such that the transfer function of the inverse amplifier can
be obtained as follows: V (s) = − R
f
o
Vin ( s )
2. The largest ratio of the
3.
S ATo
4.
S RT f
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Rf
Rin
and R , why?
in
7
Control of then Transient Response of Control System
• Open-loop
Transfer function:
G( s) =
T (s) =
Y (s)
b
=
R(s) s + a
b
A
, R( s) =
s+a
s
Output:
b A
⋅
s+a s
Ab
y (t ) = L−1{Y ( s )} =
(1 − e − at ), ∀t ≥ 0
a
Y ( s) = G ( s) ⋅ R( s) =
Time constant: τ =
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1
a
Settling time:
5τ
8
Error signal:
As + (a − Ab)
s( s + a)
a − Ab Aa − (a − Ab) − at
e(t ) = L−1{E ( s )} =
+
e
a
a
E ( s) = R( s) − Y ( s) =
Rise time: tr = t2 − t1
y (t ) |t =t2 = 0.9 ⋅ Cy (∞)
1
⇒ t 2 = − ln(0.1)
a
c(t ) |t =t1 = 0.1⋅ y (∞)
1
⇒ t1 = − ln(0.9)
a
1 .2
1
0 .8
0 .6
0 .4
0 .2
t1
t2
0
0
1
2
3
4
5
1
1
tr = − ln(0.1) + ln(0.9)
a
a
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• Closed-loop
( H ( s ) = 1)
Transfer function: T ( s ) =
Output:
C ( s)
b
=
R ( s ) s + ( a + b)
b
A
⋅
s + ( a + b) s
Ab
y (t ) = L−1{Y ( s )} =
(1 − e −( a +b )t ), ∀t ≥ 0
a+b
Y ( s) = T ( s) ⋅ R( s) =
Time constant: τ =
Error signal:
1
a+b
As + Aa
s ( s + a + b)
Aa
Ab − at
e(t ) = L−1{E ( s )} =
e
+
a+b a+b
E ( s) = R( s) − Y ( s) =
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Disturbance Signal in a Feedback Control System
• Open-loop
d
+
R(s)
+
Y(s)
G(s)
Output due to R(s): YR ( s ) ( s ) |d =0 = G ( s ) ⋅ R ( s )
Output due to d: Yd ( s ) |R =0 = 1 ⋅ d ( s )
Total output: Y ( s ) = YR ( s ) ( s ) + Yd ( s ) ( s )
= G ( s) ⋅ R( s) + d ( s)
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• Closed-loop
Td (s )
R (s )
+
−
G (s )
+
+
Y(s)
H (s )
Output due to R(s): YR ( s ) ( s ) |Td =0 =
Output due to Td(s): YTd ( s ) |R =0 =
G(s)
⋅ R( s)
1 + G (s) H (s)
1
⋅ Td ( s )
1 + G (s) H (s)
Total output: Y ( s ) = YR ( s ) + YTd ( s )
=
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G (s)
1
⋅ R( s) +
⋅ Td ( s )
1 + G (s) H (s)
1 + G (s) H (s)
12
Td (s )
Homework #5
Y (s )
R (s )
8
s+5
Fig. H5a
3
r (t )
t
0
td (t )
1
t
5
y (t )
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t
13
Y (s )
R (s )
8
s+5
Fig. H5b
N d (s )
3
r (t )
t
0
nd (t )
1
t
5
y (t )
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t
14
Steady-state Error
ess = e(t ) |t →∞ = e( ∞ )
= lim e(t ) = lim s ⋅ E ( s )
t →∞
s →0
• Open-loop
Error signal: E ( s ) = R ( s ) − Y ( s ) = (1 − G ( s )) R ( s )
A
s
ess = lim(1 − G ( s )) R ( s ) = A(1 − G (0))
Steady-state error: if R ( s ) =
• Closed-loop
Error signal:
s →0
E ( s) =
Steady-state error:
1
R( s)
1 + G( s) H ( s)
if R ( s ) =
A
s
ess = lim( sE ( s )) =
s →0
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when H ( s ) = 1
A
1 + G ( 0) H ( 0)
1
s →0 1 + G ( s )
ess = lim
15
The Cost of Feedback System
1.
Loss of gain
2.
Increased number of components and complexity
3.
Possibility of instability
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16
Lab #1
° Introduction to Matlab and Simulink
± Simulation (Lab #3,4,5,6)
² How can the Simulink be applied to measure the angular
velocity and angular displacement in which an incremental
encoder are used in control systems?
Exercises:
E4.1 E4.3 E4.6 E4.7 P4.4 P4.14 AP4.6 AP4.7
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CHAPTER 5
The Performance of Feedback
Control Systems
•
•
•
•
•
•
Introduction
Test Input Signal
Performance of a Second-order System
The S-Plane Root Location and The Transient Response
The Steady-state Error of Feedback Control System
Performance Indices
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1
Introduction
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2
Test Input Signal
• Step-function
⎧A t ≥ 0
u (t ) = ⎨
⎩0 t < 0
U ( s ) = L{u(t )} =
or
u (t ) = A ⋅ us (t )
u (t )
A
s
t
• Ramp-function
⎧ At t ≥ 0
r (t ) = ⎨
⎩0 t < 0
R ( s ) = L{r (t )} =
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or
A
s2
r (t ) = At ⋅ us (t )
r (t )
t
3
• Parabolic-function
⎧⎪ 1 2
At
p(t ) = ⎨ 2
⎪⎩0
t≥0
t<0
P ( s ) = L{ p (t )} =
or p (t ) =
1 2
At ⋅ us (t )
2
A
s3
p (t )
t
General Form
tn
A
r (t ) = A ⋅ ⋅ us (t ) ⇒ R ( s ) = n +1
n!
s
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4
Performance of a Second-order System
ωn2
G(s)
K
Y (s) =
R( s ) = 2
R( s) = 2
1 + G(s)
s + ps + K
s + 2ξωn s + ωn2
With a unit step input, we obtain
ωn2
Y (s) =
s ( s 2 + 2ξωn s + ωn2 )
ωn is held constant while the damping ratio ζ is varied
ζ >1 : Overdamped ⇒ s1, 2 = −ζω n ± ω n ζ 2 − 1
ζ =1 : Critically damped ⇒ s1, 2 = −ω n
0<ζ <1 : Underdamped
ζ =0: Undamped
⇒ s1, 2 = −ζω n ± jω n 1 − ζ 2
⇒ s1, 2 = ± jω n
ζ <0 : Negatively damped ⇒ s1, 2 = −ζω n ± jω n 1 − ζ 2
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5
Performance of a Second-order System
1
ωn2
Y ( s) =
s ( s 2 + 2ξωn s + ωn2 )
when 0 < ξ < 1
y (t ) = 1 −
1
β
e
−ξωn t
sin(ωn βt + θ ) β = 1 − ξ
1− ξ 2
θ
ξ
2
y (t )
y max
Overshoot
1
0 .9
ε = 2 % or 5 %
0 .5
0 .1
0 td
tp
ts
t (sec)
tr=t2t1
Typical unit-step response of a control system
t1 t2
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6
ωn
dy (t )
=
e −ξωnt sin ω n 1 − ξ 2 t = 0
dt
1−ξ 2
e −ξωnt = 0
⇒t→∞
sin ω n 1 − ξ 2 t = 0 ⇒ ω n 1 − ξ 2 t = nπ
t=
nπ
ωn 1 − ξ
π
tp =
n =1→ t = tp
2
ωn 1 − ξ 2
M p = 1 + e −ξπ /
P.O =
1−ξ 2
y max − y fin
y fin
× 100% = e −ξπ /
1− ξ
ts = −
2
× 100%
Damping Factor : α = ξωn
Settling Time:
1
1−ξ 2
e −ξωnt s < 0.02
1
ξωn
ln(0.02 1 − ξ 2 ) ≈
2007年1月31日
4
ξωn
, for ess < 0.02
7
2007年1月31日
8
The S-plane Root Location and the Transient Response
θ↑
α↑
↓
↓
↑
⇔
⇔
ωd = ωn 1 − ξ 2 ↑
2007年1月31日
↑
⇔
9
B
S-plane
Second-order step response
A
C
System A
Draw the step responses of the system B and
C,. respectively
2007年1月31日
10
ω n 3 > ω n 2 > ω n 1 jω
ω n1
ζ 3 > ζ 2 > ζ1
ωn2
σ
ζ2
ζ3
ζ1
jω
ζ =0
σ
ωn3
constant ζ
constant ωn
jω
α 2 < α1 < 0
α3 > 0
α1
α2
α3
constant α
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ωd 2 > ωd1
jω
ωd 2
ωd1
σ
σ
constant ωd
11
0<ζ<1
ζ⌫ ∞
ζ=1
0<ζ<1
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ζ=0
0>ζ>-1
ζ=-1
ζ>1 ζ<-1
ζ=0
ζ⌫ -∞
0>ζ>-1
12
jω
jω
c(t )
ω
ζ >1
c(t )
S - 平面
S - 平面
ζ =0
1
0
ω
1
0
0
jω
t
t
0
c(t )
jω
S - 平面
c(t )
S - 平面
ω
1
ω
0
0
1
ζ =1
0
jω
c(t )
S - 平面
0 <ζ <1
t
ω
−1 < ζ < 0
jω
S - 平面
c(t )
ω
1
0
0
0
t
t
0
1
ζ < −1
0
t
Step response
2007年1月31日
13
2007年1月31日
14
The Steady-state Error of Feedback Control System
• Type of Control Systems
R (s )
Loop transfer function:
Q( s)
G( s) H ( s) =
P( s )
+
−
E (s )
G (s )
C(s)
H (s )
bm s m + bm −1s m −1 + L + b1s + b0
=
an s n + an −1s n −1 + an −2 s n −2 + L + a1s + a0
m
=
K ∏( s + z j )
j =1
n− N
s N ∏ ( s + pi )
i =1
The loop transfer function as s approaches zero depends on the
number of integrations N.
The number of integrations is often indicated by labeling a
system with a type number that simply is equal to N.
2007年1月31日
15
Error function: E ( s ) =
Steady-state error:
1
⋅ R( s)
1 + G( s) H ( s)
ess = lim s ⋅ E ( s )
s →0
if H ( s) = 1
Step-function input
1
A
⋅
s →0
1 + G(s) s
A
=
1 + lim G ( s )
ess = lim s ⋅
s →0
Define the position error constant as:
K p = lim G ( s )
s →0
2007年1月31日
ess =
A
1+ Kp
16
Ramp-function input
ess = lim s ⋅
s →0
1
A
A
⋅ 2 =
1 + G ( s) s
lim sG ( s )
s →0
K v = lim sG ( s )
Define the velocity error constant as:
s →0
A
Kv
ess =
Parabolic-function input
ess = lim s ⋅
s →0
A
A
1
⋅ 3=
1 + G ( s) s
lim s 2G ( s )
s →0
Define the acceleration error constant as:
K a = lim s 2G ( s )
s →0
ess =
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A
Ka
17
Summary of steady-state errors (unit feedback)
Type of Error Constant
system
N
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K p Kv Ka
Steady-state error
A
1+ Kp
A
Kv
ess
A
Ka
0
K
0
0
Constant
∞
∞
1
∞
K
0
0
Constant
∞
2
∞
∞
K
0
0
Constant
3
∞
∞
∞
0
0
0
18
Performance indices
A performance index is a
quantitative measure of the
performance of a system and is
chosen so that emphasis is given to
the important system specifications.
A system is consider an optimum
control system when the system
parameters are adjusted so that the
index reaches an extreme value,
commonly a minimum value.
The Integral of the Square of the Error:
T
ISE = ∫ e 2 dt
0
2007年1月31日
19
2007年1月31日
20
CHAPTER 6
The Stability of Linear Feedback Systems
• The Concept of Stability
• The Routh-Hurwitz Stability Criterion
•The Relative Stability of Feedback Control Systems
2007年1月31日
1
The Concept of Stability
• Stability
ƒ Absolute stability
ƒ Relative stability
• The methods for the determination of stability
of linear continuous-time systems:
ƒ Routh-Hurwitz criterion
stable
ƒ Nyquist criterion
ƒ Bode diagram
stable
jω
S-plane
unstable
σ
unstable
A stable system is a dynamic system with a
bounded response to a bounded input
2007年1月31日
2
S-plane
2007年1月31日
3
2007年1月31日
4
2007年1月31日
5
• Bounded-input bounded-output (BIBO) stability:
u (t )
System
δ (t )
y (t )
h (t )
∞
∫
y (t ) = ∫ u (t − τ ) ⋅ h (τ )dτ ⇒ y (t ) ≤ ∫ u (t − τ ) ⋅ h(τ ) dτ
y (t ) = u (t − τ ) ⋅ h (τ )dτ
0
∞
∞
0
0
If u(t) is bounded, u (t ) ≤ M
∫
∞
⇒ y (t ) ≤ M h (τ ) dτ
0
Thus, if y(t) is to be bounded, or
∫
y (t ) ≤ N < ∞
∞
⇒ M h(τ ) dτ ≤ N < ∞
0
∫
∞
⇔ h (τ ) dτ ≤ Q < ∞
0
M , N and Q are finite positive number
2007年1月31日
6
; The Routh-Hurwitz Stability Criterion
Consider the characteristic equation of a linear SISO system
q ( s ) = an ( s − r1 )( s − r2 ).....( s − rn ) = 0
=a n s n − an (r1 + r2 + ... + rn ) s n −1 + an (r1r2 + r2 r3 + ...) s n − 2
+ an (r1r2 r3 + r1r2 r4 + ...) s n −3 + .... + an (−1) n r1r2 ... = 0
q( s) = s n +
an −1
an
an −2
an
an −3
an
a
an −1 n −1
a
s +L+ 1 s + 0 = 0
an
an
an
∑ all roots
= ∑ products of the roots taken two at a time
= − ∑ products of the roots taken three at a time
=−
M
a0
an
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= ( −1) n products of the roots
7
The necessary condition to guarantee that all roots of q(s)=0 with
negative real part are:
Š All the coefficients of the equation have the the same sign.
Š None of the coefficients vanishes.
Necessary but not sufficient!
q( s) = an s n + an −1s n −1 + ..... + a1s + a0 = 0
sn
an
a n−2
a n−4
s n −1
a n −1
a n −3
a n −5
s n−2
b1
b2
b3
s n −3
.
.
s
c1
.
.
.
.
s0
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a n −6 ......
a n −7 ....
b1 =
a n −1a n − 2 − a n a n −3
an
c1 =
b1a n −3 − a n −1b2
b1
b2 =
a n −1a n − 4 − a n a n −5
an
The Routh-Hurwitz criterion states that the number of roots of q(s)
with positive real parts is equal to the number of changes in sign
of the first column of the Routh array
8
Case1:No element in the first column is zero
Second-order system
Third-order system
q ( s ) = a2 s 2 + a1s + a0 = 0
q ( s ) = a3 s 3 + a2 s 2 + a1s + a0 = 0
s2
a2
a0
s3
a3
a1
s1
a1
0
s2
a2
a0
s0
b1 =
a1a0 − a2 0
= a0
a1
The system is stable if all the coefficients
have the same sign.
1
bs1 =
s0
a2 a1 − a3a0
a2
a0
set
0
a3 ~ a0 > 0
The system is stable if a a
2 1
2007年1月31日
> a3a0
9
Case 2: Zeros in the first column while some other elements of the row containing a
zero in the first column are nonzero.
q( s ) = s 5 + 2 s 4 + 2 s 3 + 4s 2 + 11s + 10 = 0
ε >0
s5
s3
1
2
0
s2
c1
s1
d1
s0
10
s4
11
10
0
0
0
0
ε
d1
4ε − 12
ε
1
x5
+ 2 x1 + 2 x1 + 4 x1 + 11 x1 + 10
=
1
x5
⋅ (1 + 2 x1 + 2 x 2 + 4 x 3 + 11x 4 + 10 x 5 )
≡
1
x5
⋅ F ( x)
4
3
2
≈−
12
ε
≈ −∞
6c1 − 10ε
≈6
c1
F ( s) s= =
1
x
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2
4
6
10
0
0
c1 =
1
10
Case 3: Zeros in the first column, and the other element of the row containing the
zero are also zero.
This condition occurs when the polynomial contains singularities that are symmetrically
located about the origin of the s-plane.
S-plane
2007年1月31日
S-plane
S-plane
11
q( s) = s 3 + 2s 2 + 4s + 8 = 0
s3 1
s
2
4
2
8
s1 0
0
A( s ) = 2 s 2 + 8 = 0
= s2 + 4 = 0
s = ± j2
s0
(1)
s+2
s 2 + 4 s 3 + 2s 2 + 4s + 8
+ 4s
s3
( 2)
dA( s )
= 2s
ds
s3 1
4
2
+8
s2
2
8
2s 2
+8
s1 2
0
2s
q ( s ) = ( s + 2)( s + j 2)( s − j 2) = 0
s0
8
The system has not any pole on the RHP,
but system is marginally stable.
2007年1月31日
12
Case 4: Repeated roots of the characteristic equation
on the imaginary axis.
(2)
(1)
A( s ) = s 4 + 2 s 2 + 1 = ( s 2 + 1) 2 = 0
s = j , j ,− j ,− j
q( s) = s 5 + s 4 + 2s 3 + 2s 2 + s + 1 = 0
s
5
s
4
s
3
s
2
A( s ) = s 4 + 2 s 2 + 1 = ( s 2 + 1) 2 = 0
s = j , j ,− j ,− j
1
2
1
1
2
1
0
0
0
.
s
5
s
4
1
2
1
s3
4
4
0
s2 1
1
0
s1 0
0
0
.
1
2
1
dA( s )
= 4s 3 + 4s
ds
s +1
s 4 + 2s 2 + 1
s5 + s 4 + 2s 3 + 2s 2 + s + 1
+ 2s 3
s5
+s
s4
+ 2s 2
+1
s4
+ 2s 2
+1
The system has not any root on the RHP,
but the system is unstable.
Repeated roots on the imaginary axis
The resulted of Routh array is Falsely
indicated.
2007年1月31日
13
q( s) = s 3 + 2s 2 + 4s + k = 0
s3
s
2
s1
s0
1
4
2
8−k
2
k
k
q ( s ) = s 3 + 3ks 2 + (k + 2) s + 4 = 0
Ans: k>0.528, system is stable.
0
(1) k > 8
(2) k > 0
8
0
∞
∞
Ans: k>8, system is stable.
2007年1月31日
14
2007年1月31日
15
q ( s ) = s 4 + 8s 3 + 17 s 2 + (k + 10) s + ka = 0
s4 1
17
s3 8
(k + 10)
s 2 b1
ka
ka
0
s1 c1
s 0 ka
b1 =
126 − k
b (k + 10) − 8ka
, and c1 = 1
8
b1
k < 126
ka > 0
(k + 10)(126 − k ) − 64ka > 0
2007年1月31日
16
2007年1月31日
17
Exercises and Problems
E6.1,E6.2,E6.6,E6.17
P6.6,P6.12,P6.15,P6.18
AP6.4,DP6.1
2007年1月31日
18
CHAPTER 7
The Root Locus Method
•The Root Locus Concept
•The Root Locus Procedure
•The Root contour
•The Root Locus Using Matlab
2007年1月31日
1
T ( s) =
Y ( s ) p( s)
=
R ( s ) q( s )
Roots
to find out the solution of the characteristic
equation.
q( s ) = 1 + G ( s ) H ( s ) = 0
R (s )
+
−
E (s )
G (s )
C(s)
H (s )
q( s ) = 1 + KG1 ( s ) H1 ( s ) = 0
G1 ( s ) H1 ( s ) = −1 •
| G1 ( s ) H1 ( s ) |=
1
K
1
|K |
⎧180 ± 2kπ , for K > 0 ⎫
⎪
⎪
∠G1 ( s ) H1 ( s ) = ⎨ 00 ± 2kπ , for K < 0 ⎬
⎪ where k = 0,1,2...
⎪
⎩
⎭
2007年1月31日
q( s ) = s 3 + 2 s 2 + Ks + 4 K = 0
= s 3 + 2 s 2 + K ( s + 4) = 0
s+4
= 1+ K 2
=0
s ( s + 2)
s+4
G1 ( s ) H1 ( s ) = 2
s ( s + 2)
2
2007年1月31日
3
q( s ) = 1 + G ( s ) H ( s ) = 1 + K
| G1 ( s ) H1 ( s ) |s = s1 =
1
=
s( s + 2) s = s1
1
=0
s ( s + 2)
s1
∠G1 ( s ) H1 ( s ) |s = s1
2007年1月31日
4
2007年1月31日
5
2007年1月31日
6
; The Root Locus Procedure
Step
Related equation or Rule
1. Write the characteristic equation so that F ( s ) = 1 + KG1 ( s ) H 1 ( s ) = 0
the parameter of interest K appears as a
multiplier.
m
2. Factor G1 ( s ) H 1 ( s ) in terms of n poles and
∏(s + z j )
j =1
m zeros.
G1 ( s ) H 1 ( s ) = n
∏( s + pi )
i =1
3. Locate the open-loop poles and zeros of : poles, {: zeros,
F (s ) in the s-plane with selected symbols. Δ or †: roots of characteristic equation
4. Locate the segments of the real axis that a). Locus begins at a pole and ends at zero.
b). Locus lies to left of an odd number of
are root locus.
poles and zeros ( K ≥ 0 ).
5. The number of branch on the root loci, ρ. ρ = n, when n ≥ m;
n: number of finite poles, m: number of finite zeros
6. The root loci are symmetrical with respect
to the horizontal real axis.
7. Intersect of the asymptotes (Centroid)
∑ pi − ∑ z j or σ = ∑ Re( pi ) − ∑ Re( z j )
σ=
8. Angles of asymptotes of the root loci.
2007年1月31日
n−m
n−m
⎧ ( 2k + 1)π
, ∀K ≥ 0
⎪
θk = ⎨ n − m
; k = 0,1,2,L, n − m − 1
2kπ
⎪
,
∀K < 0
⎩ n−m
7
9. Breakaway points (saddle points) on the Roots of d G ( s ) H ( s ) = 0 or d K = 0
1
1
ds
ds
root loci.
10. Intersection of root loci with imaginary Routh-Hurwitz criterion.
axis.
11. Angles of departure and angles of arrival
⎧( 2k + 1)π ∀K ≥ 0
∠G1 ( s ) H 1 ( s ) = ⎨
, k = 0,1,2,L
of the root loci.
∀K < 0
⎩ 2 kπ
at s = pi or s = z j .
n
12. Calculation of K at a specific root si .
∏ (s + p )
K=
i =1
m
i
∏ (s + z j )
j =1
2007年1月31日
s = si
8
2007年1月31日
9
Step 4: The root locus on the real axis always lies in a section of the
real axis to the left of an odd number of poles and zeros.
Odd segments
Step 5: Determine the number of separate loci,SL.The number
of separate loci is equal to the number of poles
2007年1月31日
10
Step 6: The root loci must be symmetrical with respect to the horizontal real
axis.
Step 7: The linear asymptotes are centered at a point on the real axis given
by σ = ∑ poles − ∑ zeros The angle of the asymptotes with respect to the
A
real axis is
2007年1月31日
n−m
φA =
( 2q + 1)
× 1800
n−m
n = 4, m = 1, φ1 =
( 2q + 1)
× 1800 , q = 0,1,2, n − m − 1
4 −1
11
Step 8:The actual point at which the root locus crosses the imaginary axis is
readily evaluated by utilizing the Routh-Hurwitz criterion.
Step 9: Determine the breakaway point
dK
=0
ds
the tangents to the loci at the breakaway point are equally over 3600
1 + KG1 ( s ) = 1 + K
2007年1月31日
1
=0
( s + 2)( s + 4)
12
K = −( s + 2)( s + 4)
dK
= −( 2 s + 6) = 0, s = −3
ds
dK
ds
2
s
-2
2007年1月31日
13
Step 10:Determine the angle of departure of the locus from a pole and the
angle of arrival of the locus at a zero,using the phase angle criterion.
2007年1月31日
14
; Examples
F ( s) = 1 + K
1
F ( s) = 1 + K
s ( s + 2)
2
1
s( s + 2)( s + 3)
6
1.5
4
1
2
Imag Axis
Imag A xis
0.5
0
0
-0.5
-2
-1
-4
-1.5
-2
-3
-2
-1
0
Real A xis
1
2
-6
-4
-3
-2
-1
Real Axis
0
1
2
Effects adding poles to G1(s)H1(s)
2007年1月31日
15
F ( s) = 1 + K
1
s( s + 2)( s + 3)( s + 4)
F ( s) = 1 + K
4
5
3
4
s+4
s( s + 2)( s + 3)
3
2
2
1
Imag A xis
Imag Axis
1
0
-1
0
-1
-2
-2
-3
-3
-4
-6
-4
-5
-4
-3
-2
Real Axis
-1
0
1
Effects adding poles to G1(s)H1(s)
2007年1月31日
2
-5
-5
-4
-3
-2
-1
Real A xis
0
1
2
Effects adding zeros to G1(s)H1(s)
16
F ( s) = 1 + K
1
s 4 + 12 s 3 + 64 s 2 + 32 s
F ( s) = 1 + K
1
s( s + 3)( s 2 + 2 s + 2)
4
6
3
4
2
1
Imag A xis
Imag A xis
2
0
0
-1
-2
-2
-4
-6
-6
-3
-5
-4
-3
-2
Real A xis
-1
0
1
Effects adding poles to G1(s)H1(s)
2
-4
-6
-5
-4
-3
-2
-1
Real A xis
0
1
2
3
-∞<K<∞
Lab #3
Written a M-file to plot the root loci step by step. (rlocfind)
2007年1月31日
17
q( s ) = s 3 + s 2 + β s + α = 0
1+
βs
s + s +α
3
2
=0
s3 + s2 + α = 0
1+
α
s ( s + 1)
2
=0
2007年1月31日
18
R (s )
Y (s )
k1
s(s + 2)
k2s
Specifications:
1.
Steady-state error for a ramp input ≤ 35%
2.
Damping ratio of dominant roots
3.
Settling time to within 2 % of the final value
≥ 0.707
sec.
≤3
sec.
1 + GH ( s ) = s 2 + 2 s + βs + α = 0
β = k2k1 , α = k1
2007年1月31日
19
2007年1月31日
20
2007年1月31日
21
1 + G1 ( s ) H1 ( s ) = 1 + K
s +1
=0
s (s + a)
2
Root Locus
Root Locus
10
Root Locus
8
8
6
6
4
4
8
6
4
-2
2
Imaginary Axis
Imaginary Axis
2
0
0
-2
-6
0
-2
-4
-4
-4
-8
-6
-10
-10
-9
-8
-7
-6
-5
-4
-3
-2
-1
-6
0
Real Axis
-8
-9
a = 10
-8
-7
-6
-5
-4
-3
-2
-1
-8
-8
0
-7
-6
-5
-4
Real Axis
-3
-2
-1
0
0
0.2
Real Axis
a=9
a =8
Root Locus
8
Root Locus
Root Locus
5
6
0.4
4
0.3
3
4
0.2
2
2
0.1
0
-2
Imaginary Axis
1
Imaginary Axis
Imaginary Axis
Imaginary Axis
2
0
-1
0
-0.1
-2
-4
-0.2
-3
-6
-0.3
-4
-8
-7
-6
-5
-4
-3
Real Axis
a=7
2007年1月31日
-2
-1
0
-5
-2
-1.8
-1.6
-1.4
-1.2
-1
Real Axis
a=2
-0.8
-0.6
-0.4
-0.2
0
-0.4
-1.2
-1
-0.8
-0.6
-0.4
-0.2
Real Axis
a =1
22
2007年1月31日
23
2007年1月31日
24
2007年1月31日
25
2007年1月31日
26
CHAPTER 8
Frequency Response Methods
Introduction
Frequency Response Plots
Š Polar Plot
Š Bode Plot
Š Log Magnitude and Phase Plot
Frequency Response Measurements
Performance Specifications in The Frequency Domain
2007年1月31日
1
;Introduction
r(t)
T(s)
c(t)
C(s)
R(s)
Output: C ( s ) = T ( s ) R ( s )
If
r (t ) = A sin(ω ⋅ t ) ⇒ c(t ) = C sin(ω ⋅ t + φ )
For sinusoidal steady-state analysis, we replace s by jω, and
output c(t) becomes
C ( jω ) = T ( jω ) ⋅ R ( jω ) = C ( jω ) ∠C ( jω ) = C ( jω ) e j∠C ( jω )
For a closed-loop system
T ( jω ) = T ( s ) s = jω =
G ( jω )
1 + G ( jω ) H ( jω )
= T ( jω ) ∠T ( jω )
or R( w) + jX ( w)
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2
where
T ( jω ) =
G ( jω )
1 + G ( jω ) H ( jω )
∠T ( jω ) = ∠G ( jω ) − ∠1 + G ( jω ) H ( jω )
and
R (ω ) = Re[T ( jω )]
X (ω ) = Im[T ( jω )]
T ( jω ) = Re[T ( jω )] + j Im[C ( jω )]
ΔR (ω ) + jX (ω )
T ( jω ) = R 2 (ω ) + X 2 (ω )
∠T ( jω ) = tan −1
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X (ω )
R (ω )
3
The Laplace transform pair
C ( s ) = L{c(t )} = ∫0 c(t )e − st dt
∞
c(t ) = L−1{C ( s )} =
1 σ + j∞
st
C
(
s
)
e
ds
∫
2πj σ − j∞
The Fourier transform pair
C ( jω ) = F {c(t )} = ∫−∞ c(t )e − jωt dt
∞
and
c(t ) = F −1{C ( jω )} =
1
2π
j∞
jωt
C
(
j
ω
)
e
dω
∫−∞
The Fourier transform exists for f(t) when
∞
∫−∞ c(t ) dt < ∞
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;Frequency Response Plots
Š Polar plot
A plot of the magnitude versus phase in the polar coordinates as ω is
varied from 0 → ∞.
ƒ Find T(jω); | T(jω)|, ∠ T(jω), R (jω) and X(jω).
ƒ To Determine the behavior of the magnitude and phase of T(jω) at
ω→0 and ω→∞.
lim T ( jω ) ; lim ∠T ( jω )
ω →0
ω →0
lim T ( jω ) ; lim ∠T ( jω )
ω →∞
ω →∞
ƒ To Determine the the intersection
R (ω ) ω =ω = 0 ⇒ X (ω ) ω =ω = β
i
i
X (ω ) ω =ω = 0 ⇒ R (ω ) ω =ω = α
r
r
ƒ To Determine the the asymptotic line
lim T ( jω ) = lim[R (ω ) + jX (ω )] = a + jb
ω →0
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ω →0
5
‘Case 1: Integral and derivative factor s ±1
ω →∞
ω →∞
ω →0
ω →0
A). T ( jω ) =
1
jω
B ). T ( jω ) = jω
‘Case 2: First-order factor T ( s ) = (1 + τs ) ±1
ω →∞
ω →∞
ω →0
1
ω →0
1
ω = ωi
A). T ( jω ) =
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1
1 + jωτ
B ). T ( jω ) = 1 + jωτ
6
‘Case 3: Quadratic factor
ω n2
A). T ( s ) = 2
s + 2ζω n s + ω n2
1
=
1
1
ωn
ωn
T ( jω ) =
2
s
+ 2ζ
2
1 + 2ζ (ωjω ) + (ωjω )
ζ : small
2
n
1
2
1 − (ωω ) + j 2ζ (ωω )
n
B ). T ( s ) =
ωn
s +1
1
n
=
ω →0
ω →∞
ω →∞
n
s 2 + 2ζω n s + ω n2
ω →0
ω n2
1
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ŠBode plot
The Bode plot of the function T(jω) is composed of two plots:
° The amplitude of T(jω) in decibels (dB) versus ω or log10 ω.
± The phase of T(jω) in degree versus ω or log10 ω.
T ( jω ) = T ( jω ) e j∠T ( jω )
Logarithmic gain = 20 ⋅ log10 T ( jω )
100
Phase (deg); Magnitude (dB)
50
0
-50
-100
-50
-100
-150
-200
-250
-300
-2
10
-1
10
0
10
1
10
2
10
Frequency (rad/sec)
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ƒ Constant factor: K
K dB = 20 ⋅ log K
and
⎧0°
∠K = ⎨
⎩± 180°
for K > 0
for K < 0
15
Phase (deg); Magnitude (dB)
14.5
14
13.5
13
12.5
200
100
0
-100
-1
10
0
10
1
10
2
10
Frequency (rad/sec)
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T ( jω ) dB = 20 ⋅ log ( jω ) ± p = ±20 p ⋅ log(ω )
d
[ ±20 p ⋅ log(ω )] = ±20 p dB / decade
d (log ω )
and
200
∠( jω ) ± p = ± p × 90°
Phase (deg); Magnitude (dB)
100
0
-100
-200
200
100
0
-100
-200
-300
-1
10
0
1
10
10
2
10
Frequency (rad/sec)
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ƒ Simple zero: 1 + sτ
T ( jω ) dB = 20 ⋅ log T ( jω ) = 20 ⋅ log 1 + (ωτ ) 2
ωτ << 1 ⇒ T ( jω ) dB ≅ 20 ⋅ log(1) = 0
ωτ >> 1 ⇒ T ( jω ) dB ≅ 20 ⋅ log (ωτ ) 2 = 20 ⋅ log(ωτ )
and
25
∠T ( jω ) = tan (ωτ )
−1
15
Phase (deg); Magnitude (dB)
1 ⎧ T ( jω ) dB = 20 ⋅ log 2 ≅ 3
ω = ⇒⎨
−1
τ
⎩∠T ( jω ) = tan (1) = 45°
20
10
5
0
-5
100
80
60
40
20
0
0
10
1
10
2
10
Frequency (rad/sec)
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ƒ Simple pole:
1
1 + sτ
T ( jω ) dB = 20 ⋅ log T ( jω ) = 20 ⋅ log
1
1+ ( ωτ ) 2
ωτ << 1 ⇒ T ( jω ) dB ≅ 20 ⋅ log(1) = 0
ωτ >> 1 ⇒ T ( jω ) dB ≅ −20 ⋅ log(ωτ )
and
5
∠T ( jω ) = − tan (ωτ )
0
⎧ T ( jω ) dB = −20 ⋅ log 2 ≅ −3
ω = ⇒⎨
−1
τ
⎩∠T ( jω ) = − tan (1) = −45°
1
Phase (deg); Magnitude (dB)
−1
-5
-10
-15
-20
0
-20
-40
-60
-80
-100
0
10
1
2
10
10
Frequency (rad/sec)
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2
ω
n
ƒ Quadratic poles:
s 2 + 2ζω n s + ω n2
T ( jω ) dB = 20 ⋅ log
1
[1−(
] [2ζ ( ) ]
ω 2 2+
ωn )
ω
ωn
2
ω
<< 1 ⇒ T ( jω ) dB ≅ 20 ⋅ log(1) = 0
ωn
ω
2
>> 1 ⇒ T ( jω ) dB ≅ −20 ⋅ log(ωω ) = −40 ⋅ log(ωω )
ωn
n
n
and
⎡ 2ζ ( ωω ) ⎤
∠T ( jω ) = − tan ⎢
ω 2⎥
1
(
−
ω ) ⎦
⎣
−1
n
n
ω
<< 1 ⇒ ∠T ( jω ) = 0°
ωn
ω
>> 1 ⇒ ∠T ( jω ) = −180°
ωn
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ω = ω n ⇒ T ( jω ) dB = −20 ⋅ log(2ζ )
ω = ω n ⇒ ∠T ( jω ) = −90°
20
ζ = 0.1, 0.2
Phase (deg); Magnitude (dB)
0
-20
-40
0
-50
-100
-150
-200
0
10
1
10
2
10
Frequency (rad/sec)
Š Log Magnitude and Phase Plot
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14
;Performance Specifications in The Frequency Domain
Š Resonant peak: Mpω
The resonant peak is the maximum value of |T(jω)|.
Š Resonant frequency: ωr
The resonant frequency is the frequency at which the peak resonant occur.
Š Bandwidth: BW (ωB)
The bandwidth is the frequency at which T(jω) drops to 70.7% of, or 3 dB
down from, its zero-frequency value.
Š Cutoff rate
The cutoff rate is the slop of log-magnitude curve near the cutoff frequency.
M pω
1
0.707
ωr
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ωB
15
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CHAPTER 9
Stability in the Frequency Domain
Introduction
Mapping Contours in the s-Plane
The Nyquist Criterion
Relative Stability and the Nyquist Criterion
Time-Domain performance Criteria Specified in
the Frequency Domain
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1
Introduction
Time-domain
Routh-Hurwitz criterion
Root-Locus
by locating the roots of characteristic equation in the s-plane.
Frequency domain
Nyquist stability criterion
based on the theorem of complex variables due to Cauchy,
commonly known as principle of argument.
H. Nyquist 1932
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2
F ( s) = 2s + 1
Enclose
A : F ( s ) |s =1+ j = 2 s + 1 = 2 + 2 j + 1 = 3 + 2 j
Clockwise and eyes right
B : F ( s ) |s =1− j = 2 s + 1 = 3 − 2 j
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3
F (s) =
Γs
s
s+2
ΓF
Cauchy’s Theorem:If a contour Γ in the s-plane encircles Z zeros and P poles of
s
F(s) and does not pass through any poles or zeros of F(s) and the traversal is in the
clockwise direction along the contour, the corresponding contour ΓF in the F(s)-plane
encircles the origin of the F(s)-plane N=Z-P time in the clockwise direction
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F ( s) =
Γs
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s
s + 0.5
ΓF
5
F ( s ) =| F ( s ) | ∠F ( s ) =
| s + z1 || s + z 2 |
(∠s + z1 + ∠s + z 2 − ∠s + p1 − ∠s + p2 )
| s + p1 || s + p2 |
=| F ( s ) | ∠(φ z1 + φ z2 − φ p1 − φ p2 ) =| F ( s ) | ∠φF
φF = φ z − φ p
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6
P = 1, Z = 3,
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N = Z − P = 3 −1 = 2
7
Nyquist contour (Nyquist path)
Since the Nyquist contour must not pass through any poles and zeros
of F(s), the small semicircles shown along the jω-axis are used to
indicate that the path should go around these poles and zeros if they
fall on the jω-axis.
jω
j∞
d
c
jω 1 ×
b
a
0×
j
e
σ
i
h
− jω1×
g
f
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− j∞
8
jω
j∞
d
c
jω1 ×
b
a
0×
j
i
h
− jω1 ×
g
f
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− j∞
e
σ
Path ab:
s=jω ; 0<ω<ω1
Path bc:
s=jω1+ε⋅ejθ; ε→0 and -90 º <θ<90º
Path cd:
s=jω ; ω1<ω<∞
Path def:
s=R⋅ejθ; R→∞ and 90 º <θ<-90º
Path fg:
s=jω ; -∞<ω<-ω1
Path gh:
s=-jω1+ε⋅ejθ; ε→0 and -90 º <θ<90º
Path hi:
s=jω ; -ω1< ω<0
Path ija:
s=ε⋅ejθ; ε→0 and -90 º <θ<90º
9
GH ( s ) =
1.ω : 0 → 0 , φ : −90 ~ 90, s = εe
−
+
K
s (τs + 1)
jφ
K
= ∞e − jφ ,900 ~ −900
j
φ
j
φ
ε →0 εe (ττe
+ 1)
G ( s ) = lim
2. ω : 0 + → ∞, s = jω
K
K
GH ( s ) =
=
, ω → ∞, | GH ( jω ) |≈ 0, φ ≈ −180 0
2
jω ( jωτ + 1) − ω τ + jϖ
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3. ω : ∞ ~ -∞ , s = Re jφ , R → ∞, φ : 90 ~ −90
K
K
GH ( s ) |s = Re jφ = lim
= 2 e − j 2φ ≈ 0e − j 2φ ,−180 ~ +180
j
φ
j
φ
R → ∞ Re (τ Re + 1)
τR
4. ω : −∞ → 0 −
The portion of the polar plot from ω = −∞ to ω = 0 −
is symmetrical to the polar plot from ω = 0 + to ω = +∞
10
G ( s) =
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100
( s + 1)( s + 1)
10
11
Nyquist criterion and the GH(s) plot
Since the open-loop transfer function GH(s) is generally known, it
would be simpler to construct the GH(s) plot that corresponds to the
Nyquist path, and the same stability conditions of the closed-loop
system can be obtained by observing the number of encirclements of
the (-1,j0) point in the GH(s)-plane.
F ( s ) = 1 + GH ( s ) ⇔ GH ( s ) = 1 − F ( s )
jω
+ j∞
s-plane
jv
GH(s)
Γs
σ
( −1, j 0)
u
− j∞
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12
Stability requirements
For the closed-loop system to be stable, there should be no zeros
of F(s) in right half s-plane; i.e.,
Z=0
this condition is met if
N=Z–P=0 – P= – P
The special case of P=0, means open-loop stable system, the
closed-loop system is stable if
N=Z–P=0
N 0 = Z 0 − P0
Open − loop : GH ( s) =
N ( s)
D( s)
closed − loop : F ( s ) = 1 + GH ( s ) = 1 +
P−1 = P0 ⇐ poles of L( s ) = F ( s )
N ( s) D( s) + N ( s)
=
D( s)
D( s)
N −1 = Z −1 − P−1
Z −1 = 0 ⇐ stable for closed - loop
N −1 = 0 − P−1 = − P−1
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13
GH ( s ) =
K
s (τ 1s + 1)(τ 2 s + 1)
1.ω : 0 − → 0 + , s = εe jφ , ε → 0, φ : −90 ~ 90
K
= ∞e − jφ , | GH ( S ) |= ∞, φ : 90 ~ −90
j
φ
ε → 0 εe (τ εe
+ 1)(τ 2εe + 1)
1
GH ( s ) = lim
jφ
jφ
2. ω : 0 + → ∞, s = jω ,
GH ( s ) =
=
K
jω ( jτ 1ω + 1)( jτ 2ω + 1)
− K (τ 1 + τ 2 ) − jK ( 1 )(1 − ω 2τ 1τ 2 )
ω
1 + ω 2 (τ 12 + τ 22 ) + ω 4τ 12τ 22
Let Im=0
ω=
Re =
= Re + J Im
1
τ 1τ 2
The magnitude of the real part,Re of the GH(jw) at this frequency is
Re =
1
− K (τ 1 + τ 2 )
,ω =
2
2
4 2 2
1 + ω (τ 1 + τ 2 ) + ω τ 1 τ 2
τ 1τ 2
2
- Kτ 1τ 2
=
τ1 + τ 2
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3. ω : +∞ → -∞, s = re jφ , r → ∞, φ : 90 ~ −90
K
GH ( s ) = lim jφ
≈ 0e − j 3φ
j
φ
jφ
r →∞ re ( rτ e
+ 1)(rτ 2 e + 1)
1
14
GH ( s ) =
K
s ( s + 1)( s + 1)
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K =1
stabl
K =2
K =3
unstab
15
System with a pole in the right-hand s-plane
GH ( s ) =
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K1
,
s ( s − 1)
K2 = 0
1.ω = 0 − → 0 +
16
1.w : 0 − → 0 + , s = εe jφ , φ : −90 ~ 90
K
= −∞e − jφ = ∞∠ − 180 − φ
j
j
φ
φ
ε →0 εe (εe
− 1)
GH ( s ) = lim
2.w : 0 + → ∞, s = jw
GH ( jw) =
=
K
=
jw( jw − 1)
K
1
2 2
K
1
2 2
∠ − 900 − tan −1 (− w)
( w4 + w )
∠90 + tan −1 ( w)
( w4 + w )
3.w : ∞ → −∞, s = re jφ , φ : 90 ~ −90
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closed − loop
Y ( s)
K1
= 2
, F ( s ) = s 2 + K1 K 2 s − s + K 1 = 0
R( s ) s + ( K1 K 2 − 1) s + K1
F (s) = 1 +
K1 (1 + K 2 s)
=0
s ( s − 1)
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System with a zero in the right-hand s-plane
GH ( s ) =
N −1 = 1
N0 = 1
K ( s − 2)
( s + 1) 2
Z 0 = 1, N 0 = 1 ⇒ P0 = 0
P−1 = P0 = 1, N −1 = 1, ⇒ Z −1 = 2
∴ Closed-loop system is unstable
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Relative Stability and the Nyquist Criterion
The Nyquist stability criterion is defined in term of (-1,j0) points on the
polar plot or the 0 dB, -180 º point on the Bode diagram or Log-magnitudephase diagram. Clearly, the proximity of L(jω) locus to this stability point is
a measure of the relative stability of the system.
Definition
Phase Crossover
A phase crossover on the L(jω) plot
is a point at which the plot intersects
the negative real axis.
Phase Crossover Frequency: ωp
The phase crossover frequency is the
frequency at the phase crossover, or
where
∠L(jωp)=± 180 º
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ωp
ω1
ω2
20
Gain Crossover
The gain crossover is a point on L(jω) plot at which the
magnitude of L(jω) is equal to 1.
Gain Crossover Frequency: ωg
The gain crossover frequency is the frequency of L(jω) at the
gain crossover, or where
|L(jωg)|=1
( −1, j 0)
ωg
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Gain Margin: (GM)
The gain margin is the increase in the system gain when phase = - 180 º
that will result a marginally stable system with intersection of (-1,j0)
point on the Nyquist plot.
Gain margin is the amount of gain in decibels (dB) that can be added to
the loop before the closed-loop system become unstable.
GM = 20 log{1 − L( jω p ) }
= −20 log{ L( jω p ) }
1
or 20 log{
}
L( jω p )
ω = ωp
(−1, j0)
L ( jω p )
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22
Bode plot
Gm=15.563 dB (at 2.2361 rad/sec), Pm=43.21 deg. (at 0.77934 rad/sec)
50
Phase (deg); Magnitude (dB)
0
GM
-50
-100
-150
-50
-100
PM
-150
-200
-250
-300
-2
10
-1
10
ω = ω g 10
0
ω = ωp
1
10
2
10
Frequency (rad/sec)
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23
K
GH ( s ) =
s (τ 1s + 1)(τ 2 s + 1)
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24
60
40
20
PM
0
dB
ω = ωg
GM
-20
ω = ωp
-40
-60
-80
-100
-120
-280 -260 -240 -220 -200 -180 -160 -140 -120 -100
-80
degree
Log-magnitude-phase curve
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25
ζ = 0.01φPM
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26
Y ( jw)
G ( jw)
= T ( jw) =
= M ( w)e jφ ( w)
R ( jw)
1 + GH ( jw)
G ( jw) = u + jv
G ( jw)
u + jv
u 2 + v2
M=
=
=
1 + G ( jw) 1 + u + jv (1 + u ) 2 + v 2
1
2
(1 − M 2 )u 2 + (1 − M 2 )v 2 − 2 M 2u = M 2
2 M 2u ⎛ M 2 ⎞ ⎛ M 2 ⎞ ⎛ M 2 ⎞
⎟=⎜
⎟+⎜
⎟
u +v −
+⎜
1 − M 2 ⎜⎝ 1 − M 2 ⎟⎠ ⎜⎝ 1 − M 2 ⎟⎠ ⎜⎝ 1 − M 2 ⎟⎠
2
2
2
2
⎛
M2 ⎞
⎛ M ⎞
⎜⎜ u −
⎟ + v2 = ⎜
2 ⎟
2 ⎟
⎝1− M ⎠
⎝ 1− M ⎠
2
⎛v⎞
⎝u⎠
⎛ v ⎞
⎟
⎝1+ u ⎠
φ = ∠T ( jw) = ∠(u + jv) /(1 + u + jv) = tan −1 ⎜ ⎟ − tan −1 ⎜
let N = tan φ
v
=0
N
1 2 1⎛
1 ⎞
(u + 0.5) 2 + (v −
) = ⎜1 + 2 ⎟
2N
4⎝ N ⎠
u2 + v2 + u −
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29
Time-Domain performance Criteria Specified in
the Frequency Domain
Nichols Charts
40
30
Mpω=8dB
Open-Loop Gain (dB)
20
10
ω = ωr
0
ω = ωB
-10
-3dB
-20
-30
-40
-280
k = k2
-260
-240
k = k1
-220
-200
-180
-160
-140
-120
-100
-80
Open-Loop Phase (deg)
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Table 9.6 Transfer Function Plots for Typical Transfer Functions
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Table 9.6 Transfer Function Plots for Typical Transfer Functions
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Table 9.6 Transfer Function Plots for Typical Transfer Functions
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Table 9.6 Transfer Function Plots for Typical Transfer Functions
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Table 9.6 Transfer Function Plots for Typical Transfer Functions
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Table 9.6 Transfer Function Plots for Typical Transfer Functions
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Table 9.6 Transfer Function Plots for Typical Transfer Functions
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Table 9.6 Transfer Function Plots for Typical Transfer Functions
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38
Table 9.6 Transfer Function Plots for Typical Transfer Functions
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Table 9.6 Transfer Function Plots for Typical Transfer Functions
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Table 9.6 Transfer Function Plots for Typical Transfer Functions
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41
Table 9.6 Transfer Function Plots for Typical Transfer Functions
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42
Table 9.6 Transfer Function Plots for Typical Transfer Functions
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Table 9.6 Transfer Function Plots for Typical Transfer Functions
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44
Table 9.6 Transfer Function Plots for Typical Transfer Functions
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45
Exrcises
E9.1,E9.7,E9.8,E9.15,E9.21,P9.2,P9.4,P9.17,P9.21,AP9.4,
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CHAPTER 10
The Design of Feedback Control Systems
The Design of Feedback Control System
§ Introduction
§ Time Domain Design
P-Control PI-Control PD-Control PID-Control
Phase-Lead Compensation Design
Phase-Lag Compensation Design
§ Frequency Domain Design
Phase-Lead Design Using Bode Diagram
Phase-Lag Design Using Bode Diagram
§Design for Deadbeat Response
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1
;Introduction
Š Design procedure
ƒ Determine what the system should do and how to it (design specifications).
ƒ Determine the controller or compenstaor configuration relative to how it is
connected to the controlled plant.
ƒ Determine the parameter values of the controller or compensator to achieve
the design goals.
ƒ Simulation verifications and recheck the specifications.
ƒ Experimental results.
Š Design specification
ƒ Time domain: maximum overshoot, rise time, settling time and steady-state
accuracy
ƒ Frequency domain: gain margin, phase margin, resonant peak, bandwidth
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Š Controller (Compensator) configuration
ƒ
ƒ
ƒ
ƒ
Cascade (series) compensation
Feedback compensation
Series-feedback compensation
Forward compensation with series compensation (two degree of
freedom)
ƒ Feedforward compensation (two degree of freedom)
ƒ State-feedback compensation
r (t )
e(t )
Gc (s )
u (t )
G p (s )
y (t )
r (t )
e(t )
u (t )
G p (s )
y (t )
Gc (s )
Figure 1 Cascade
Figure 2 Feedback
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r (t )
Gc (s )
e(t )
u (t )
G p (s )
y (t )
G H (s )
Figure 3 Series-feedback compensation
r (t )
GcF (s )
e(t )
u (t )
Gc (s )
G p (s )
y (t )
Figure 4Forward compensation with series compensation (two degree of freedom)
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GcF (s )
r (t )
u (t )
Gc (s )
y (t )
G p (s )
Figure 5 Feedforward compensation (two degree of freedom)
r (t )
u (t )
G p (s )
x (t )
C
y (t )
K
Figure 6 State-feedback compensation
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§ Time Domain Design
r (t )
e(t )
Gc ( s )
u (t )
G p ( s)
y (t )
PID Controller Design
d
e(t )
dt
U ( s ) = ( K p + K i 1s + K d s ) E ( s )
u (t ) = K p e(t ) + K I ∫ e(t ) + K d
Kp
U (s)
U (s)
Gc ( s ) =
= K p + K i 1s + K d s
E ( s)
E (s)
db
1
KI
s
Kds
db
20 log K p
− 20db / decade
ω
p − Controller
− 900
db
ω
I − Controller
+ 20db / decade
+ 900
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D − Controller
ω
6
§ Closed-loop control with P-controller
PM → ∞, GM → ∞
P-control for first order system
R (s )
a
s+a
Kp
Y (s)
a>0
r (t )
Kp = 3
Kp = 2
Kp =1
t
aK p
Y (s)
=
R ( s ) s + (aK p + a )
y (t ) =
S − plane
GH ( jω ) − plane
∞ ← Kp
−a
KP
(1 − e − a ( K P +1)t )
KP +1
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P-control for second order system
R (s )
aK p
Y ( s)
= 2
R( s ) s + as + aK p
ωn = aK p ,
ξ=
K P1
Kp
a
2 aK p
Y (s)
a>0
K p1 > K p 2 > K p 3
GH ( jω ) − plane
K P2
K P3
a
s( s + a)
−1
S − plane
ω→∞
GM →∞
ω → 0+
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PI-control for first order system
1
s
a( K p s + K I )
Gc ( s ) = K p + K I
R (s )
Kp
Y (s)
=
R( s ) s 2 + (a + bK p ) s + bK I
ωn = bK I , ξ =
a + bK p
bK I
,
KI
A = 1, B = ( a + bK p ), C = bK I
1
s
Y (s)
b
s+a
a, b > 0
B 2 − 4 AC > 0, ⇒ ξ > 1
B 2 − 4 AC = 0, ⇒ ξ = 1
B 2 − 4 AC < 0, ⇒ ξ < 1
GH ( jω ) − plane
−1
S − plane
ω→∞
GM →∞
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ω → 0+
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PI-control for second order system
K I R (s )
bK
(
s
+
)
b
p
Y ( s)
K
s
P
=
, F (s) = 1 +
2
b( K s + K I )
R(s)
s ( s + a)
1+ 2 p
s (s + a)
K ps + KI
Kp
KI
Y (s)
b
s(s + a)
a, b > 0
1
s
KI
<< a
KP
S − plane
GH ( jω ) − plane
ω→∞
0+ ← ω
−a
−
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KI
KP
10
PD-control for second order system
R (s )
Gc ( s ) = K p + K d s
Kp
a( K d s + K p )
Y (s)
= 2
R( s ) s + (a + bK d ) s + bK p
ωn = bK p , ξ =
a + bK d
,
bK p
KDs
b
s(s + a)
a, b > 0
A = 1, B = (a + bK d ), C = bK I
B 2 − 4 AC > 0, ⇒ ξ > 1
B 2 − 4 AC = 0, ⇒ ξ = 1
B 2 − 4 AC < 0, ⇒ ξ < 1
GH ( jω ) − plane
S − plane
−1 ω → ∞
GM →∞
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ω → 0+
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PID-control for second order system
1
s
b( K d s 2 + K p s + K I )
R (s )
Gc ( s ) = K p + K d s + K I
Y (s)
=
R( s ) s 3 + (a + bK d ) s 2 + bK p s + bK I
Kp
KI
1
s
Y (s)
b
s(s + a)
a, b > 0
(a + bK d ) • bK p > bK I
The Optimum Coefficients of T(s) Based on the ITAE
Criterion for a step Input (P.252 Table 5.6)
Kds
s 3 + 1.75ωn s 2 + 2.15ωn s + ωn3
GH ( jω ) − plane
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S − plane
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Phase-lead Compensation Network
G(s) =
τ=
V2 ( s )
R2
( R1Cs + 1)
=
V1 ( s) R1 + R2 {[ R1 R2 /( R1 + R2 )]Cs + 1}
R1 R2
R + R2
C,α = 1
R1 + R2
R2
G(s) =
(1 + ταs)
,α > 1
α (1 + τs )
ωm = zp =
1
z=
1
1
, p = ,z < p
ατ
τ
τ α
αωτ − ωτ
φ = tan −1
, when ω = ωm ,
1 + (ωτ )) 2 α
(α / α − (1 / α ) α − 1
=
1+1
2 α
1 + sin φm
α −1
sin φm =
,α =
α +1
1 − sin φm
tan φm =
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Phase-Lag Compensation Network
R2 + ( 1 )
Vo
R2Cs + 1
Cs
=
G(s) =
=
Vin R1 + R2 + ( 1 ) ( R1 + R2 )Cs + 1
Cs
1 + τs
1 s+z
( R + R2 )
=
=
,τ = R2C , α = 1
R2
1 + ατs α s + p
1
1
z= ,p=
, α > 1, ωm = zp
τ
ατ
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Root Locus Method
S − plane
s+z
,z < p
Gc ( s ) =
s
+
p
Phase-Lead Compensation Design
The s-plane root locus method is as follows:
−p
1.
List the system specifications and translate them into a desired root
location for the dominant roots.
2.
Sketch the uncompensated root locus, and determine whether the
desired root locations can be realized with an uncompensated system.
3.
If a compensator is necessary, place the zero of the phase-lead network
directly below the desired root location (or to the left of the first two
real poles).
4.
Determine the pole location so that the total angle at the desired root
location is 1800
5.
Evaluate the total system gain at the desired root location and then
calculate the error constant.
6.
Repeat the steps if the error constant is not satisfactory.
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−z
15
Gc ( s ) = K
s+z
s+ p
R (s )
Gc (s)
The specification for the system are
Y (s)
2
s2
a>0
Settling time(2% criterion) Ts ≤ 4 sec onds
Percent overshoot for a step input, ≤ 35%
Sol:
S − plane
−πξ
r1
P.O = e 1−ξ ≤ 0.35 ⇒ ξ ≥ 0.32, θ = cos −1 ξ = 710
4
Ts ≤
= 4 ⇒ ξωn ≥ 1
2
−1
ξωn
r2
Select the desired roots location are
r1r2 = −1 ± j 2 θ = 63.40 ⇒ ξ = 0.45
r1
φ = −2(1800 − tan −1 2) + 900 = −2(1160 ) + 900 = −142
− 180 = φ − θ p ⇒ θ p = 380
θp
−p
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−1
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R + jQ
Q
Q
= tan θ p , M =
, p = z+m
M
tan θ p
M=
2
= 2.6 ⇒ p = 1 + 2.6 = 3.6
0.7812
A
B
C
θp
GcG p ( s ) =
K=
2 K ( s + 1)
, F ( s ) = 1 + KG1 H1 ( s ) = 0
s 2 ( s + 3.6)
−p
M
−z
1
A * A * B 2.232 * 3.25
=
=
= 4.1
G1 H1 ( s ) s = r
2C
2*2
1
Gc ( s ) = 4.1
s +1
s + 3.6
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Phase-Lag Compensation Design
The steps necessary for the design of a phase-lag network on the s-plane are as
follows:
1.
Obtain the root locus of the uncompensated system.
2.
Determine the transient performance specifications for the system and locate
suitable dominant root location on the uncompensated root locus that will
satisfy the specifications.
3.
Calculate the loop gain at the desired root location and thus the system error
constant,
4.
Compare the uncompensated error constant with the desired error constant,
and calculate the necessary increase that must result from the pole-zero ratio
of the compensator,α.
5.
With the known ratio of the pole-zero combination of the compensator,
determine a suitable location of the pole and zero of the compensator so that
the compensated root locus will still pass through the desired root location.
Locate the pole and zero near the origin of the s-plane in comparison to ωn .
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R (s )
Gc (s)
Sol:
Y (s)
K
s ( s + 10) 2
Spec. : ξ = 0.707, K v = 20
K v = 20 = lim G ( s ) =
s →0
K
, K = 2000
10 2
If K=2000, the system will be unstable
ξ = 0.707 ⇒ s1, 2 = −2.9 ± j 2.9
⇒
K
α
α=
= 236
z 2000
=
= 8.5
p
236
select
z = 0.1, p = 0.1 / 9 = 0.0111
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§ Frequency Domain Design
Phase-Lead Design Using Bode Diagram
1. Determined the system DC-gain such that the error constant can be satisfied.
2. Evaluate the uncompensated system phase margin when the error constant are satisfied
3. Allowing for a small of safety(5~10 degree), determine the necessary additional phase
lead, φm
φm = φd − φn + 30 ~ 100 α =
4. Evaluate α from Eq.(10.11).
sin φm + 1
1 − sin φm
5. Evaluate 10 logα and determined the frequency where the uncompensated magnitude
curve is equal to -10 logα
. Because the compensation network provides a gain of
10 logα atωm ,this frequency is the new 0-dB crossover frequency and
simultaneously.
p
6. Calculate the pole p = ωm α and z = α
7. Draw the compensated frequency response, check the resulting phase margin, and repeat
the steps if necessary. Finally, for an acceptable design, raise the gain of the amplifier in
order to account for the attenuation( 1α )
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dB
ωm
log ω
− 10 log α
φ
− 1800
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R (s )
Example 10.1
Gc ( s ) = K
p s+z
,
z s+ p
z=
1
ατ
,
p=
1
Y (s)
2
s ( s + 2)
⎛ p⎞ s+z
K⎜ ⎟
⎝ z ⎠s+ p
τ
Spec. : K v < 5%, ξ ≥ 0.4
The system dc-gain>=20
0
Phase margin
20:45
GH ( jw) =
jw(0.5 jw + 1)
0 − dB = 20 log 20 − 20 log w − 20 log[(0.5w) 2 + 1]
= 20log20 - 20logw - 10log[(0.5w) 2 + 1]
wc ≈ 6.2
φn = 180 − 90 − tan −1 (0.5wc ) = 180 − 162 = 180
φm = φd − φn + 30 ~ 100 = 450 − 180 + 30 = 300
1 + sin φm 1 + 0.5
α=
=
=3
1 − sin φm 1 − 0.5
− 10 log α = −4.77 dB
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2
find the new crossover frequency from the bode diagram
p
ωm = 8.4 ⇒ p = ωm α = 14.4, z = = 4.8
α
p
K * ( ) = 20 * 3 = 60
z
s + 4.8
Gc ( s ) = 60
s + 14.4
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Bode Diagram
40
Magnitude (dB)
20
0
-20
-40
-60
Phase (deg)
-80
-90
-135
-180
-2
10
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-1
10
0
10
Frequency (rad/sec)
1
10
2
10
23
Bode Diagram
80
Magnitude (dB)
60
40
20
0
-20
Phase (deg)
-40
-90
-135
-180
-2
10
-1
10
0
10
1
10
2
10
Frequency (rad/sec)
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Phase-lag Design Using the Bode Diagram
1.
Obtain the Bode diagram of the uncompensated system with the gain adjusted for
the desired error constant.
2.
Determined the phase margin of the uncompensated system and , if it is
insufficient, proceed with the following steps.
3.
Determined the frequency where the phase margin requirement would be
0
satisfied if the magnitude curve crossed the 0-dB line at this frequency. ( allow 5
phase lag from the phase-lag network when determining the new crossover
frequency.
4.
Place the zero of the compensator one decade below the new crossover frequency
and thus ensure only 50 of additional phase lag at ωc' due to the lag network.
5.
Measure the necessary attenuation at ωc' to ensure that the magnitude curve
crosses at this frequency.
6.
Calculate by noting that the attenuation introduced by the phase-lag network is 20logα at .
7.
Calculate the pole as
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ω p = 1ατ = ω z / α
,and the design is completed.
25
dB
20 log α
p
z
ω
p=
z
α
z=
'
c
log ω
ωc'
10
φ
− 1800
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R (s )
Example:
⎛ p⎞ s+z
⎜ ⎟
⎝ z ⎠s+ p
1
Y (s)
20
s ( s + 2)
τ ( + s)
1 + τs
p s+z
τ
=
=
Gc ( s ) =
1 + ατs ατ ( 1 + s ) z s + p
ατ
p=
1
ατ
GH ( jw) =
,z =
1
τ
Kv
20
=
, K v = 20 / 2
jw( jw + 2) jw( j 0.5w + 1)
Spec. :phase margin=450 the uncompensated system
has a phase margin of 200
Allowing for the phase-lag compensator, we locate
the frequency ω where φ (ωc' ) = −1300 which is to be
our new crossover frequency ωc = 1.5
20dB = 20 log α ⇒ α = 10
ωc' 1.5
ω
z = ωz =
=
= 0.15, p = ω p = z = 0.015
10
α 10
Gc ( s ) =
1 s + 0.15
10 s + 0.015
The attenuation necessary to cause
to be the new
crossover frequency is equal to 20 dB.
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Bode Diagram
60
40
M
a
g
n
itu
d
e(d
B
)
20
0
-20
-40
P
h
a
se(d
e
g
)
-60
-90
-135
-180
-2
-1
10
10
0
1
10
Frequency
2
10
10
(rad/sec)
Bode Diagram
100
M
a
g
n
itu
d
e(d
B
)
50
0
-50
P
h
a
s
e(d
e
g
)
-100
-90
-120
-150
-180
-2
10
-1
10
0
10
Frequency
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1
10
2
10
(rad/sec)
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§Design for Deadbeat Response
Deadbeat response: proceeds rapidly to the desired level and
holds at that level with minimal overshoot.
F ( s ) = s 3 + αω n s 2 + βω n2 s + ωn3
A deadbeat response has the following characteristics:
1.
steady-state error=0
2.
Fast response: minimum rise time and settling time
3.
0.1%<= percent overshoot<2%
4.
Percent undershoot<2%
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F ( s ) = s 4 + αω n s 3 + βω n2 s 2 + γωn3 s + ωn4
29
Spec. : settling time 1.2 sec
R (s )
z
s+z
K (s + z)
2
s + p s ( s + 2)
Pre-filter
Gc G p ( s )
z
2 Kz
T ( s) =
= 3
s + z 1 + Gc G p ( s ) s + (2 + p ) s 2 + (2 p + 2k ) s + 2kz
Gc G p ( s ) =
s+z
K
s+ p
Y (s)
2
s(s + 2)
4.04
= 3.37
1.2
T (s) = s3 + αωn s2 + βωn2s + ωn3
sol : ωnTs = 4.04,ωn =
α = 1.9, β = 2.2
T (s) = s3 + 6.4s2 + 24.98s + 38.27 = s3 + (2 + p)s2 + (2 p + 2k)s + 2kz
p = 4.4, K = 8.09, z = 2.36
s + 2.36
Gc ( s ) = 8.09
s + 4.4
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1.4
1.2
1
0.8
Uncompensated system step response
0.6
0.4
system step response with deadbeat compensated
0.2
0
0
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10
15
32
PD controller
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PI controller
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PID controller
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Exercises:
E10.1, E10.2, E10.4,
E10.8, P10.9 P10.11,
P10.14, P10.21, P10.39 ,AP10.9 AP10.10, DP10.4
DP10.5
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