EE/ME 574: Intermediate Controls
EE/ME 574: Intermediate Controls
Review-By-Example of some of the Basic
Concepts and Methods of Control System
Analysis and Design
3 Calculating steady-state errors (Franklin et al. sec. 4.2 (6th Ed.))
3.1
Contents
1 Differential Eqns, Transfer Functions & Modeling
4
1.1
Example 1, Golden Nugget Airlines . . . . . . . . . . . . . . . .
4
1.2
Block diagram
. . . . . . . . . . . . . . . . . . . . . . . . . . .
5
1.3
Laplace Transforms and Transfer Functions . . . . . . . . . . . .
6
1.3.1
Laplace transform . . . . . . . . . . . . . . . . . . . . .
6
1.3.2
Some basic Laplace transform Pairs . . . . . . . . . . . .
7
1.3.3
Transfer Functions are Rational Polynomials . . . . . . .
10
1.3.4
A transfer function has poles and zeros . . . . . . . . . .
11
1.3.5
Properties of transfer functions . . . . . . . . . . . . . . .
13
1.3.6
Forms for a transfer function . . . . . . . . . . . . . . . .
14
1.3.7
Proper and strictly proper transfer functions: . . . . . . .
15
Determining steady-state error using Bode standard form .
34
3.2
Table of steady-state errors . . . . . . . . . . . . . . . . . . . . .
35
3.3
Steady-state error example . . . . . . . . . . . . . . . . . . . . .
36
3.4
Summary for steady-state error
36
. . . . . . . . . . . . . . . . . .
4 Characteristics of the Step Response
37
4.1
Rise Time, tr
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
37
4.2
Peak Time, t p . . . . . . . . . . . . . . . . . . . . . . . . . . . .
38
4.3
Settling Time, ts
. . . . . . . . . . . . . . . . . . . . . . . . . .
38
4.4
Percent Overshoot, PO . . . . . . . . . . . . . . . . . . . . . . .
38
5 Working
5.1
with
the
pole-zero
constellation
39
Basics of pole-zero maps, 1st order, p1 = −α . . . . . . . . . . .
39
5.1.1
40
Real Pole location and response characteristics . . . . . .
2nd
order, p1,2 = −α ± j ω
Notation for a complex pole pair . . . . . . . . . . . . . .
42
21
5.2.2
Complex pole location and response characteristics
. . .
43
22
5.2.3
The damping factor: ρ . . . . . . . . . . . . . . . . . . .
47
5.3
Higher order systems: dominant mode . . . . . . . . . . . . . . .
48
5.4
Which mode is the dominant mode ? . . . . . . . . . . . . . . . .
49
5.5
Summary: Pole location and response characteristics . . . . . . .
51
1.3.9
Phasors and the transfer function as complex gain . . . . .
16
1.3.10 The DC gain of a transfer function . . . . . . . . . . . . .
1.3.11 The impulse response of a system . . . . . . . . . . . . .
23
2.1
Analyzing a closed-loop system . . . . . . . . . . . . . . . . . .
27
2.2
Common controller structures: . . . . . . . . . . . . . . . . . . .
28
2.3
Analyzing other loops
29
(Revised: Jan 16, 2013)
3.1.1
5.2.1
16
. . . . . . . . . . . . . . . . . . . . . . .
33
41
System type . . . . . . . . . . . . . . . . . . . . . . . . .
Part 1: Controls Review-By-Example
System type and Bode standard form for the loop gain . . . . . . .
. . . . . . .
1.3.8
2 Closing the loop, feedback control
32
Page 1
5.2
Basis of pole-zero maps,
Part 1: Controls Review-By-Example
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EE/ME 574: Intermediate Controls
EE/ME 574: Intermediate Controls
6 Design
52
1 Differential Eqns, Transfer Functions & Modeling
1.1 Example 1, Golden Nugget Airlines
6.1
Design methods
. . . . . . . . . . . . . . . . . . . . . . . . . .
53
6.2
Root Locus Design . . . . . . . . . . . . . . . . . . . . . . . . .
53
7 Summary
54
8 Glossary of Acronyms
Section 1.1.0
55
• Dynamic systems are governed by differential equations
(or difference equations, if they are discrete-time)
• Example (adapted from Franklin et al. 4thed., problem 5.41, figure 5.79)
GNA
Ω(t), θ(t)
Mp(t) Me(t)
Figure 1: Golden Nugget Airlines Aircraft. Me (t) is the elevator-moment (the
control input), M p (t) is the moment due to passenger movements (a
disturbance input), and θ (t) is the aircraft pitch angle, Ω (t) = θ̇ (t) .
• For example, the aircraft dynamics give:
1
d Ω (t)
d Mt (t)
d 2 Ω (t)
+4
+ 5 Ω (t) = 1
+ 3 Mt (t)
dt 2
dt
dt
Mt (t) = Me (t) + M p (t) ;
1
Ω (t) =
(1)
d θ (t)
dt
(2)
d Me (t)
+ 10 Me (t) = 7 v(t)
dt
(3)
Where Mt (t) is the total applied moment.
Eqn (1) has to do with the velocity of the aircraft response to Mt (t).
Eqn (2) expresses that the pitch-rate is the derivative of the pitch angle.
And Eqn (3) describes the response of the elevator to an input command
from the auto-pilot.
Main Fact: The Differential Equations come from the physics of the system.
Part 1: Controls Review-By-Example
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Part 1: Controls Review-By-Example
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EE/ME 574: Intermediate Controls
Section 1.2.0
1.2 Block diagram
• Eqns (1)-(3) can be laid out graphically, as in figure 2.
Mp(t)
Elevator
servo
7
s + 10
Section 1.3.1
1.3 Laplace Transforms and Transfer Functions
• A block diagram is a graphical representation of modeling equations and
their interconnection.
v(t)
EE/ME 574: Intermediate Controls
Aircraft
Dynamics
+
Me(t)
+
Mt(t)
1.3.1 Laplace transform
Integrator
Ω(t)
s+3
• To introduce the Transfer Function (TF), we need to review the Laplace
transform.
s2 + 4 s + 5
θ(t)
1
s
• The Laplace transform (LT) maps
a signal (a function of time) to a
function of the Laplace variable s .
F(s)
f(t)
L-1{F(s)}
• The Inverse Laplace transform
maps F (s) to f (t).
Figure 2: Block diagram of the aircraft dynamics and signals.
Signal Symbol
Signal Name / Description
Units
v (t)
Voltage applied to the elevator drive
[volts]
Me (t)
Moment resulting from the elevator surface
[N − m]
M p (t)
Moment resulting from movement of the passengers
Ω (t)
[N − m]
Pitch Rate (angular velocity)
[rad/sec]
θ (t)
Aircraft pitch angle
[radians]
Space of all functions
of time
"Time Domain"
Answers:
U(s)
u(t)
• When analyzing a problem from basic principals, we would also have a list
of parameters.
Parameter Symbol
Signal Name / Description
Value
Units
b0
Motion parameter of the elevator
[N-m/volt]
...
...
...
Table 2: List of parameters for the aircraft block diagram.
(Revised: Jan 16, 2013)
Page 5
Space of all
functions of s
"Frequency Domain"
• Why pay attention to Laplace
transforms ?
Table 1: List of signals for the aircraft block diagram.
Part 1: Controls Review-By-Example
L{f(t)}
1. Differential equations in time
domain correspond to algebraic
equations in s domain.
Solve
Algebraic
Eqn.
Solve
Diff Eq.
Y(s)
y(t)
2. The LT makes possible the
transfer function.
3. Time domain: find y (t) for one
u (t).
Space of all functions
of time
Space of all
functions of s
4. Frequency domain: find Guy (s)
for all U (s).
Figure 3: Solve differential equation in time domain or solve an algebraic
equation in s domain.
Part 1: Controls Review-By-Example
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EE/ME 574: Intermediate Controls
Section 1.3.2
EE/ME 574: Intermediate Controls
Section 1.3.2
• Two theorems of the Laplace transform permit us to build transfer functions:
1.3.2 Some basic Laplace transform Pairs
1. The Laplace transform obeys superposition and scaling
Time Domain Signal
Laplace transform
Unit Impulse:
Scaled Impulse:
f (t) = δ (t)
F (s) = 1
f (t) = b δ (t)
F (s) = b
f (t) = 1+ (t) = 1.0, t ≥ 0
F (s) =
1
s
Unit ramp:
f (t) = t, t ≥ 0
F (s) =
Higher Powers of t:
(t) = t n , t
1
s2
n!
Unit step:
f
Decaying Exponential:
≥0
F (s) =
f (t) = b e−αt
F (s) =
f (t) = sin (wt)
Sinusoid:
f (t) = Bc cos (ωt) + Bs sin (ωt)
Sinusoidal oscillation:
f (t) = e−αt (Bc cos (ωt) + Bs sin (ωt))
Oscillatory Exp. Decay:
sn+1
b
s+α
ω
s2 +ω2
c s+Bs ω
F (s) = sB2 +0
s+ω2
Bc s+(Bc α+Bs ω)
F (s) = s2 +2 α s+ω2
n
F (s) =
ωn =
√
α2 + ω2
Given: z (t) = x (t) + y (t)
then
Z (s) = X (s) +Y (s)
z (t) = 2.0 x (t)
then
Z (s) = 2.0 X (s)
Given:
2. The Laplace transform of the derivative of a signal x (t) is s times the LT
of x (t):
Given: X (s) = L {x (t)} ,
• Putting these rules together lets us find the
transfer function of a system from its governing
differential equation.
Area = 1.0
∆
a2
+
1 (t)
1+ (t), unit step
e
L a2
t=0
Decaying expon.
Part 1: Controls Review-By-Example
Oscil. decay exp.
(Revised: Jan 16, 2013)
u(t)
= s X (s)
System
y(t)
Gp(s)
d 2 y (t)
d y (t)
d u (t)
+ a1
+ a0 y (t) = b1
+ b0 u (t)
dt 2
dt
dt
d 2 y (t)
d y (t)
+ a1
+ a0 y (t)
dt 2
dt
(4)
d u (t)
= L b1
+ b0 u (t) (5)
dt
a2 s2 Y (s) + a1 sY (s) + a0 Y (s) = b1 sU (s) + b0 U (s)
a2 s2 + a1 s + a0 Y (s) = (b1 s + b0) U (s)
t=0
A Sinusoid
d x (t)
dt
• Whatever signals y (t) and u (t) are, they have Laplace transforms. Eqn (4)
gives:
Unit ramp
-αt
t=0
(Notice the standard form: output (unknown) on the left, input on the right.)
t=0
t=0
t=0
δ (t) , impulse function
L
• Consider a system that takes in the signal u (t) and gives y (t), governed by
the Diff Eq:
Table 3: Basic Laplace-transform pairs.
1/∆
then
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EE/ME 574: Intermediate Controls
Section 1.3.2
EE/ME 574: Intermediate Controls
• Eqns (5)-(7) tell us something about the ratio of the LT of the output to the
LT of the input:
a2 s2 + a1 s + a0 Y (s) = (b1 s + b0) U (s)
Given
Y (s) = G p (s) U (s)
GNA
(8)
Ω(t), θ(t)
Mp(t) Me(t)
• A Transfer Function (TF) is a ratio of the input and output LT’s
(b1 s + b0)
,
G p (s) =
(a2 s2 + a1 s + a0)
1.3.3 Transfer Functions are Rational Polynomials
(7, repeated)
(b1 s + b0)
Output LT
Y (s)
=
=
= G p (s)
U (s)
(a2 s2 + a1 s + a0)
Input LT
(9)
Where G p (s) is the TF of the plant.
Figure 5: Golden Nugget Airlines Aircraft. Ω (t) [radians/second] is the
pitch-rate of the aircraft, and Mt (t) is the moment (torque) applied by
the elevator surface.
• Consider the differential equation of the Aircraft pitch angle:
• Important fact:
1
– The transfer function is like the gain of the system, it is the ratio of the
output LT to the input LT.
• Basic and Intermediate Control System Theory are present transfer
function-based design.
• By engineering the characteristics of the TF, we engineer the system to
achieve performance goals.
+
e(t)
Controller
u(t)
ys(t)
Plant
y(t)
s2 + 4 s + 5 Ω (s) = (s + 3) Mt (s)
Ω (s)
s+3
= 2
Mt (s)
s +4s +5
(11)
• Note:
– We can write down the TF directly from the coefficients of the differential
equation.
– We can write down the differential equation directly from the coefficients
of the TF.
s+3
numerator polymonial
Ω (s)
=
:
Mt (s) s2 + 4 s + 5
denominator polynomial
Sensor Dynamics
Hy(s)
• We call a TF such as (11) a rational polynomial.
Figure 4: Block diagram of typical closed-loop control.
Part 1: Controls Review-By-Example
(10)
• Transfer functions, such as Eqn (11), are ratios of two polynomials:
Gp(s)
KcGc(s)
d 2 Ω (t)
d Ω (t)
d Mt (t)
+4
+ 5 Ω (t) = 1
+ 3 Mt (t)
dt 2
dt
dt
• From (10) we get the TF, take LT of both sides, and rearrange:
– However, the TF depends only on the parameters of the system
(coefficients a2..a0 and b0..b1 in the example), and not on the actual
values of U (s) or Y (s) (or u (t) and y (t)).
r(t)
Section 1.3.3
(Revised: Jan 16, 2013)
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Part 1: Controls Review-By-Example
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EE/ME 574: Intermediate Controls
Section 1.3.4
• We can use the poles and zeros to write the TF in factored form:
• A TF has a numerator and denominator polynomial, for example
2 s2 + 8 s + 6
N (s)
= 3
D (s) s + s s2 + 4 s + 0
(12)
• The roots of the numerator are called the zeros of the TF, and the roots of
the denominator are called the poles of the TF. For example:
>> num = [2 8 6]
>> den = [1 2 4 0 ]
num =
2
8
6
den =
1
2
4
>> zeros = roots(num)
zeros =
-3
-1
>> poles = roots(den)
poles =
0
-1.0000 + 1.7321i
-1.0000 - 1.7321i
0
• We can also use Matlab’s system tool to find the poles and zeros
>> zero(Gps),
b2 (s − z1) (s − z2)
2 s2 + 8 s + 6
=
s3 + 2 s2 + 4 s (s − p1) (s − p2) (s − p3)
2 (s + 3) (s + 1)
=
(s − 0) (s + 1 − 1.732 j) (s + 1 + 1.732 j)
G (s) =
• With a complex pole pair we can do two things,
1. Use a shorthand
G (s) =
2 (s + 3) (s + 1)
s (s + 1 ± 1.732 j)
(Because poles always come in complex conjugate pairs.)
(a) Write the complex pole pair as a quadratic, rather than 1st order terms
G (s) =
2 (s + 3) (s + 1)
s (s2 + 2 s + 4)
• The zeros are values of s at which the transfer function goes to zero
• The poles are values of s at which the transfer function goes to infinity
%% Build the system object
2 s^2 + 8 s + 6
----------------s^3 + 2 s^2 + 4 s
• We can plot a pole-zero map:
PZ Map
2
>> Gps=tf([2 8 6], [1 2 4 0])
>> pzmap(Gps)
imaginary
>> Gps = tf(num, den)
Transfer function:
Section 1.3.4
Interpreting the poles (p1, p2, ..., pn) and zeros (z1, ..., zm)
1.3.4 A transfer function has poles and zeros
G p (s) =
EE/ME 574: Intermediate Controls
ans = -3
-1
1
0
−1
>> pole(Gps),
ans =
0
-1.0000 + 1.7321i
-1.0000 - 1.7321i
Part 1: Controls Review-By-Example
−2
−4
−2
real
0
Figure 6: Pole-Zero constellation of aircraft transfer function.
(Revised: Jan 16, 2013)
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EE/ME 574: Intermediate Controls
Section 1.3.5
EE/ME 574: Intermediate Controls
Section 1.3.6
1.3.5 Properties of transfer functions
1.3.6 Forms for a transfer function
Just as a differential equation can be scaled by multiplying the left and right sides
by a constant, a TF can be scaled by multiplying the numerator and denominator
by a constant.
1. Rational polynomial form
bm sm + bm−1 sm−1 · · · + b1 s + b0
G (s) =
ansn + an−1 sn−1 + · · · + a1 s + a0
Monic: A TF is said to be monic if an =1. We can always scale a TF to be monic.
If G1 (s) is scaled to be monic, then
b′0
(13)
G1 (s) =
s + a′1
with b′0 = b0/an and a′1 = a1 /an .
Rational Polynomial Form: A TF is in rational polynomial form when the
numerator and denominator are each polynomials. For example
Example:
G (s) =
(16)
3s +8
2 s3 + 16 s2 + 30 s
(17)
2. Factored form
(s − z1) ... (s − zm)
(zi and pi are zeros and poles)
(s − p1) ... (s − pn)
s + 38
3
=
(note: zi and pi are commonly negative values)
2 s (s + 3) (s + 5)
G (s) = Krlg
3. Bode form
1. Bring out any poles or zeros at the origin,
2 s2 + 8 s + 6
G p (s) = 3
s + s s2 + 4 s + 0
(14)
Example:
An example of a TF not in rational polynomial form is:
G3 (s) =
′
2 (s + 3)/ (s)
(s2 + 2 s + 4)/ (s + 1)
(15)
(s) (s + 1)
2 s2 + 8 s + 6
2 (s + 3)(s + 1)
2 (s + 3)/ (s)
=
=
(s2 + 2 s + 4)/ (s + 1) (s) (s + 1) (s2 + 2 s + 4)(s) s3 + 2 s2 + 4 s
(Revised: Jan 16, 2013)
(18)
KBode =
1. Bring out a factor, so that the numerator and denominator polynomials
are monic.
′
G (s) = Krlg
′
′
sm + bm−1 sm−1 + ... + b1 s + b0
′
′
′
sn + an−1 sn−1 + ... + a1 s + a0
s + 83
3
3
K
=
=
rlg
2 s3 + 8 s2 + 15 s
2
Page 13
8
9
4. Root locus form
Example:
Note the middle form above, which can be called factored form.
Part 1: Controls Review-By-Example
′
b sm + ... + b1 s + 1
1
KBode ′ m n−n
′
n
o
s
an−no s o + ... + a1 s + 1
3
1
8
8 s+1
= 1
no = 1 ,
2 2
s
30 30
s + 16
30 s + 1
G (s) =
By clearing the fractions within the fraction, T3 (s) can be expressed in rational
polynomial form
G3 (s) =
2. Bring out a constant, so that the least significant coefficients of the
polynomials are 1.0
Part 1: Controls Review-By-Example
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EE/ME 574: Intermediate Controls
Section 1.3.7
1.3.7 Proper and strictly proper transfer functions:
System Type: A property of transfer functions that comes up often is the system
type. The type of a system is the number of poles at the origin.
– A TF with m ≤ n is said to be proper.
• So, for example, the aircraft transfer function from elevator input to pitch
rate gives a type 0 system:
– When m < n the TF is said to be strictly proper
• Example of a TF that is not proper:
G4 (s) =
s+1
G p1 (s) =
note: m = 2,
n=1
G4 (s) =
s+1
=
Ω (s)
s+3
=
Me (s) s2 + 4 s + 5
poles : s = −2 ± 1 j ,
type : 0
• But the TF from the elevator to the pitch angle gives a type I system:
such a TF can always be factored by long division:
2s2 + 5 s + 4
Section 1.3.9
1.3.8 System type
• m = number of zeros, n = number poles
2s2 + 5 s + 4
EE/ME 574: Intermediate Controls
2 s (s + 1) 3 s + 4
3s +4
1
+
= 2 s+
= (2s + 3)+
(s + 1)
s+1
s+1
s+1
• A non-proper TF such as G4 (s) has a problem: As s = j ω → j ∞, the gain
goes to infinity !
• Since physical systems never have infinite gain at infinite frequency,
physical systems must have proper transfer functions.
G p2 (s) =
s+3
θ (s)
=
Me (s) s (s2 + 4 s + 5)
poles : s = 0, −2 ± 1 j ,
type : I
• If we put a PID controller in the loop, which adds a pole at the origin, the
system will be type II.
1.3.9 Phasors and the transfer function as complex gain
• To consider the complex gain, we need to represent sinusoidal signals
as phasors. A phasor is a complex number that represents a sinusoidal
waveform. A time domain signal, such as
v (t) = Av cos (ωt + θv )
(20)
is represented by the complex number
~V = Av ∠θv = a + j b .
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EE/ME 574: Intermediate Controls
Section 1.3.9
• For example, the time-domain waveform
EE/ME 574: Intermediate Controls
Section 1.3.9
• In the example of figure 7,
v (t) = 3.0 cos (ωt + 40o)
u(t)=Au cos(ωt+θu)
Gp (s)
y(t)=Ay cos(ωt+θy)
is represented by
~V = 3.0 ∠400 = 2.30 + 1.93 j .
Figure 7: An element with sinusoid in and sinusoid out.
• For a phasor to represent a sinusoidal signal, the frequency, ω [rad/sec] must
be given.
• Phasors in polar and rectangular coordinates
Polar
~V = Av ∠θv
p
Av = a2 + b2 , θv = atan2 (b, a)
• The TF evaluated with s = j ω gives the complex gain of the system element
Output Phasor = G p (s = jω) Input Phasor
Ay ∠θy = G p ( j ω)
Rectangular
~V = a + j b
(26)
Au ∠θu
(22)
a = Av cos (θv ) , b = Av sin (θv ) (23)
• Sinusoidal signals in polar and rectangular coordinates
v (t) = Av cos (ωt + θv )
v (t) = a cos (ωt) − b sin (ωt)
(24)
With a and b as given in Eqn (23).
• Notice the difference of ’-’ sign between the phasor in rectangular
coordinates and the time-domain function in rectangular coordinates. This
extra ’-’ comes from the trig identity:
cos (α ± θ) = cos (α) cos (θ) ∓ sin (α) sin (θ)
Part 1: Controls Review-By-Example
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EE/ME 574: Intermediate Controls
Section 1.3.9
Example Question: If v (t) = 1.5 cos (2.0t + 0o) ,
what is the response of the aircraft, Ω (t) ?
EE/ME 574: Intermediate Controls
Section 1.3.9
• The complex gain is a function of frequency. Evaluating
G p (s) =
Solution: Looking back at the transfer functions
s+3
7
Ω (s)
=
V (s)
s + 10 s2 + 4 s + 5
Ω (s)
V (s) s= j ω
as a function of ω gives the Bode plot.
Bode plot of Gp(s)
• Evaluating at s = j ω = j 2 gives the complex gain
(27)
= 0.1512 − 2.2672 j = 0.307∠ − 60.5o
−20
−40
−60
−80
−100
0
• The complex gain determines the output phasor:
~Ω = (0.307∠ − 60.5o) × (1.5∠0o ) = 0.461∠ − 60.5
Magnitude (dB)
Magnitude at w=2.0
(28)
• So the output, in the time domain, is:
Phase (deg)
7
21 + 14 j
Ω (s)
3+2 j
=
=
V (s) s=2 j
10 + 2 j (2 j)2 + 4 (2 j) + 5 −6 + 82 j
0
−45
Phase at w=2.0
−90
−135
−180
−1
10
0
10
1
2
10
10
3
10
Frequency (rad/s)
Ω (t) = 0.461 cos (2.0t − 60.5o)
Figure 8: Bode plot of G p (s).
• Summary, section 1.3.9:
1. The dynamics of the aircraft itself gave us a differential equation
2. For linear differential equations, we can get the transfer function directly
from the Diff Eq.
(a) And for non-linear differential equations, there is no transfer function.
3. With the TF we can determine the complex gain, and find the output for
any sinusoidal input.
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Part 1: Controls Review-By-Example
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EE/ME 574: Intermediate Controls
Section 1.3.10
1.3.10 The DC gain of a transfer function
EE/ME 574: Intermediate Controls
Section 2.0.0
1.3.11 The impulse response of a system
• Following up on the idea that the TF gives the complex gain of a system
block and recalling the Laplace variable
s = jω
• When we have a system such as in figure 9, the Laplace transform of the
output (a signal) is given by
(29)
Y (s) = G p (s) U (s)
(31)
the DC gain is given with s = j0, or simply
DC Gain :
B (s)
=
G (s)
A (s) s=0
s=0
(30)
0
0
• In some cases T (s = j 0) may evaluate to (undefined). In such cases apply
l’Hopital’s rule:
DC Gain :
lim G (s)
• A unit impulse input is a very short pulse with an area under the curve of
the pulse of 1.0.
• Since the Laplace transform of a unit-impulse is 1, then the Laplace
transform of the impulse response is the transfer function
Yimp (s) = G p (s) Uimp (s) = G p (s) 1 = G p (s)
s→0
• The example TF’s give these values of DC gain
s+3
3
Ω (s)
=
V (s)
s + 10 s2 + 4 s + 5
u(t)
y(t)
t
θ (s)
s+3
1
3
=
V (s)
s + 10 s2 + 4 s + 5 s
u(t): impulse
U(s)=1
• Evaluating the DC Gain
t
System
Gp(s)
y(t): impulse response
Y(s)=Gp(s)
Figure 9: The transfer function is the Laplace transform of the impulse response.
Ω (s)
3 3
9
=
=
V (s) s=0
10 5 50
• The connection between the impulse response and TF can be used to
determine the TF
θ (s)
3 31
=∞
=
V (s) s=0
10 5 0
(Many systems have a DC gain of infinity, it is a common property of control
systems.)
– Apply an impulse and measure the output, y (t). Take the LT and use
G p (s) = Y (s).
• The connection between the impulse response and TF helps to understand
the mathematical connection between an LT and a TF
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Part 1: Controls Review-By-Example
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EE/ME 574: Intermediate Controls
Section 2.0.0
2 Closing the loop, feedback control
u(t)
Gp(s)
y(t)
r(t) + u(t)
-
EE/ME 574: Intermediate Controls
Section 2.0.0
• Example, wrapping a simple feedback loop around the aircraft dynamics
r(t)
y(t)
Mt(t)
+
Gp(s)
s+3
1
s +4s+5
s
2
-
θ(t)
sensor
Open Loop
Closed Loop
Figure 11: Simple feedback of aircraft pitch angle.
Figure 10: A plant with TF G p (s) in open and closed loop. Closed loop requires
a sensor.
s+3
G p (s)
Ω (s)
s (s2 +4 s+5)
=
=
R (s) 1 + G p (s) 1 + 2 s+3
s (s +4 s+5)
• Feedback is the basic magic of controls. A feedback controller can
(34)
– Make an unstable system stable . . . . . . . . . . . . . . . . . . Helicopter autopilot
– Make a stable system unstable . . . . . . . . . . . . . . . . Early fly-ball governors
– Make a slow system fast . . . . . . . . . . . Motor drive, industrial automation
– Make a fast system slow . . . . . . . . F16 controls, approach / landing mode
• Eqn (34) is not in rational polynomial form, so
s+3
θ (s)
s (s2 +4 s+5)
=
R (s) 1 + 2 s+3
s (s +4 s+5)
s s2 + 4 s + 5
s+3
=
(35)
s (s2 + 4 s + 5) s (s2 + 4 s + 5) + (s + 3)
– Make an inaccurate system accurate . . . . . . . . . . . . . . . . . . . . machine tools
• The closed-loop TF is still not quite in Rat Poly form, here is the final step:
• The magic comes because closing the loop changes the TF
Open loop:
θ (s)
s+3
=
R (s) s3 + 4 s2 + 6 s + 3
Y (s)
= G p (s)
U (s)
(36)
For the Closed loop use U (s) = R (s) −Y (s), dropping the (s) arguments:
Y = G p (R −Y ) = G p R − G p Y
r(t)
Y (1 + G p) = G p R
Gp
Y (s)
=
R (s)
1 + Gp
Θ(s)
θ(t)
R(s)
(32)
Forward Path Gain
Y (s)
(33)
=
= Try (s)
R (s)
1 + Loop Gain
Part 1: Controls Review-By-Example
(Revised: Jan 16, 2013) Page 23
Figure 12: Block with r (t) as input and θ (t) as output.
Part 1: Controls Review-By-Example
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Page 24
EE/ME 574: Intermediate Controls
Section 2.0.0
EE/ME 574: Intermediate Controls
• Analyzing the response
Gps
= tf([1 3], [1 4 5 0])
%% LTI models
Try = tf([1 3], [1 4 6 3])
figure(1), clf
[Y_open, Top]
= step(Gps, 6);
[Y_closed, Tcl] = step(Try, 6);
plot(Top, Y_open, Tcl, Y_closed)
xlabel(’t [seconds]’);
ylabel(’\Omega, pitch-rate’)
title(’Open- and closed-loop’)
text(3, 1.6, ’Open-loop’, ’rotation’, 45)
text(4, 0.8, ’Closed-loop’)
SetLabels(14)
print(’-deps2c’, ’OpenClosedResponse1’)
Section 2.0.0
• Introduce a proportional controller gain
e(t)
r(t)
Kc
+
Mt(t)
-
s+3
1
s2 + 4 s + 5
s
θ(t)
Figure 14: Feedback for aircraft pitch control, with P-type gain Kc .
• Look at Kc =
1.0, 3.0, 10, 0
Kc = 1; Try1 = tf(Kc*[1
Kc = 3; Try2 = tf(Kc*[1
Kc = 10; Try3 = tf(Kc*[1
figure(1), clf
...
plot(Top, Y_open, Tcl,
...
Open− and closed−loop response, aircraft
3.5
3], [1 4 5 0]+Kc*[0 0 1 3])
3], [1 4 5 0]+Kc*[0 0 1 3])
3], [1 4 5 0]+Kc*[0 0 1 3])
Y_closed1, Tcl, Y_closed2, Tcl, Y_closed3)
Open− and closed−loop response, aircraft
3.5
3
• The response
completely
changes !
• System gets less
stable as Kc
increases
1.5
1
Closed−loop
op
lo
2
1.5
Kc =10
Kc =3
1
Closed−loop, Kc =1
0.5
0
0
2.5
pe
n−
O
pe
n−
lo
2
O
op
• System gets much
faster as Kc
increases
Ω, pitch−rate
• The closed-loop
system is type 0
Ω, pitch−rate
• The open-loop
system is type I
3
2.5
0.5
1
2
3
t [seconds]
4
5
6
0
0
1
2
3
t [seconds]
4
5
Figure 13: Open and Closed loop response of the aircraft, the two responses have
very different characteristics.
Figure 15: Open and Closed loop response of the aircraft, with Kc = 1.0,
Kc = 3.0, and Kc =10.0 .
Part 1: Controls Review-By-Example
Part 1: Controls Review-By-Example
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EE/ME 574: Intermediate Controls
Section 2.1.0
2.1 Analyzing a closed-loop system
EE/ME 574: Intermediate Controls
2.2 Common controller structures:
• A typical, basic loop (such as velocity PI control of a motor drive) has 3
components:
PD
PI
PID
Proportional-Derivative
Prop.-Integral
Prop.-Int.-Deriv.
1. Plant (thing being controlled)
2. Controller or compensator (usually a computer, a often PLC for motor
drives)
3. A sensor
r(t)
+
Section 2.2.0
Gc (s) = kd s + k p
Gc (s) =
k p s +ki
s
Gc (s) =
kd s2 +k p s +ki
s
Lead-Lag
(s+z ) (s+z )
Gc = Kc (s+p11 ) (s+p22 )
• Common Applications
PI: Velocity control of motor drives, temperature control (good speed and
accuracy, acceptable stability)
e(t)
Controller
KcGc(s) =
-
Nc(s)
Kc D (s)
c
u(t)
Plant
Gp(s) =
y(t)
Np(s)
Dp(s)
PID: Many many places, Astrom has estimated that 80% of controllers are
PID (good, speed accuracy, stability)
Sensor Dynamics
Ny(s)
Hy(s) = D (s)
y
ys(t)
PD: Position control where high accuracy is not required (good speed and
stability, so-so accuracy)
Lead-Lag: Used where a pole at the origin is unacceptable, can be as good
as PID (notice, 5 parameters rather than 3)
Figure 16: Basic loop with a plant, compensator and sensor.
• The TF is given as
Try (s) =
Kc Gc G p
Y (s)
Forward Path Gain
=
=
R (s)
1 + Loop Gain
1 + Kc Gc G p Hy
• Often, for the controls engineer the plant, G p (s), is set (e.g., the designer of
a cruise-control does not get to change the engine size).
• As a controls engineer, we get to pick Gc (s) and maybe influence Hy (s)
(e.g., by convincing the project manager to spring $$$ for a better sensor).
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EE/ME 574: Intermediate Controls
Section 2.3.0
EE/ME 574: Intermediate Controls
2.3 Analyzing other loops
2.3 Analyzing other loops (continued)
• As a final example, let’s consider the output arising with sensor noise
d(t)
Disturbance Filter
Nd
Gd(s) = D
d
r(t)
Input Shaping
Hr(s)
+
e(t)
-
Controller
N
KcGc(s) = Kc Dc
c
ys(t)
Section 2.3.0
uc(t)
+
+
up(t)
Tvy (s) =
Plant
Gp(s) =
y(t)
Np
Dp
Vs(t)
+
Figure 17: Basic loop with a disturbance input, d (t) , and sensor noise, Vs (t)
added.
(39)
• The example transfer functions, Eqns (37), (38) and (39) show some
interesting properties. The TFs are repeated here (omitting the many (s)’s)
+
Sensor Dynamics
Ny
Hy(s) = D
y
Y (s) Hy (s) (−1) Kc Gc (s) G p (s)
=
Vs (s)
1 + Kc Gc (s) G p (s) Hy (s)
Try (s) =
Hr Kc Gc G p
1 + Kc Gc G p Hy
,
Tru (s) =
Hr Kc Gc
1 + Kc Gc G p Hy
Ted (s) =
Gd G p Hy (−1)
1 + Kc Gc G p Hy
,
Tvy (s) =
Hy (−1) Kc Gc G p
1 + Kc Gc G p Hy
• In some cases we may want to consider additional inputs and outputs.
– Many systems have a disturbance signal that acts on the plant, think of
wind gusts and a helicopter autopilot.
– The poles are the same for any input/output signal pair
– All systems have sensor noise.
• Any signal in a system can be considered an output. For example, if we want
to consider the controller effort, uc (t), arising due to the reference input
Uc (s) Forward Path Gain
Hr (s) Kc Gc (s)
Tru (s) =
=
=
R (s)
1 + Loop Gain
1 + Kc Gc (s) G p (s) Hy (s)
(Revised: Jan 16, 2013)
– The stability and damping (both determined by the poles) are the same
for any signal pair
• The numerators are different
(37)
• If we wanted to consider the error arising with a disturbance, we would
have
Gd (s) G p (s) Hy (s) (−1)
E (s)
=
(38)
Tde (s) =
D (s) 1 + Kc Gc (s) G p (s) Hy (s)
Part 1: Controls Review-By-Example
• The denominators are all the same
Page 29
– The zeros are in general different for each input/output signal pair
– Since the numerator help determine if the signal is small or large, signals
may have very different amplitudes and phase angles
Part 1: Controls Review-By-Example
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EE/ME 574: Intermediate Controls
Section 2.3.0
• If we consider what happens as Kc → ∞, we can see what happens for very
high gain. For this, assume that Hr (s) = 1.0 and Gd (s) = 1.0, since these
two terms merely pre-filter inputs.
When Kc → ∞ , 1 + Kc Gc G p Hy ≈ Kc Gc G p Hy , so
Try (s) ≈
Kc Gc G p
1
=
Kc Gc G p Hy Hy
Tde (s) ≈
Gd G p Hy (−1) −Gd v
=
Kc Gc G p Hy
Kc Gc
,
Tru (s) ≈
,
Kc Gc
1
=
Kc Gc G p Hy G p Hy
Tvy (s) ≈
Hy (−1) Kc Gc G p
= −1
Kc Gc G p Hy
EE/ME 574: Intermediate Controls
Section 3.0.0
3 Calculating steady-state errors (Franklin et al.
sec. 4.2 (6th Ed.))
• Steady-state errors are computed from the transfer function r (t) → e (t), and
the final value theorem:
Final value theorem: For a stable system with transfer function
Tre (s) = E (s) /R (s),
ess = lim e (t) = lim s E (s)
t→∞
(40)
s→0
• Try (s) ≈ 1/Hy (s) shows that the TF of the plant can be compensated, it
disappears from the closed-loop TF as Kc → ∞.
In Eqn (40) E (s) (a Laplace transform) is given by Tre (s) (a transfer
function) times R (s) (a Laplace transform):
• Try (s) ≈ 1/Hy (s) also shows that the TF of the sensor can not be
compensated. If the characteristics of Hy (s) are bad (e.g., a cheap sensor)
there is nothing feedback control can do about it !
E (s) = Tre (s) R (s)
(41)
ess = lim e (t) = lim s Tre (s) R (s)
(42)
• Tde (s) ≈ −1/Kc Gc shows that disturbances can be compensated, as Kc → ∞,
errors due to disturbances go to zero ;)
• Tru (s) ≈ 1/G p Hy shows that U1 (s) does not go up with Kc , and also, if the
plant has a small gain (G p (s) is small for some s = j ω) then a large control
effort will be required for a given input.
• Tvy (s) ≈ −1 shows that there is no compensation for sensor noise. If there
is sensor noise, it is going to show up in the output !
——– ∼ ——–
• Summary:
– Feedback control can solve problems arising with characteristics of the
plant, G p (s), and disturbances, d (t).
– Feedback control can not solve problems with the sensor, Hy (s), or
sensor noise, vs (t) .
Part 1: Controls Review-By-Example
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Page 31
which gives
t→∞
s→0
– To compute the final value of the error, we have to specify the input r (t).
Signals typically used to compute steady-state error are seen in table 4.
The step, ramp and acceleration input signals and the Laplace Transforms can be
written:
1
1
R (s) = k+1 .
r (t) = t k 1+ (t) ,
(43)
k!
s
For k = 1, 2, 3 the signals and Laplace transforms are given in table 4.
k Test input signal, r (t)
Step input
0
Ramp input
1
Accelerating input 2
r (t) = 1+ (t)
r (t) = t 1+ (t)
r (t) = 12 t 2 1+ (t)
R (s)
R (s) = 1s
R (s) = s12
R (s) = s13
Table 4: Typical inputs to consider for determining steady-state error.
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EE/ME 574: Intermediate Controls
Section 3.1.0
3.1 System type and Bode standard form for the loop gain
r(t)
+
Controller
Nc(s)
Dc(s)
Plant
Gc(s) =
ys(t)
y(t)
Np(s)
Dp(s)
Gp(s) =
EE/ME 574: Intermediate Controls
Section 3.1.1
3.1.1 Determining steady-state error using Bode standard form
• Defining the loop gain as product of gains around the loop. For the standard
loop (figure 18):
GL (s) = Gc (s) G p (s) Hy (s)
(44)
• Writing the loop gain in Bode standard form:
Sensor Dynamics
Ny(s)
Hy(s) = D (s)
y
GL (s) = Gc (s) G p (s) Hy (s) =
Figure 18: Standard feedback loop, with numerator and denominators called out.
• Considering figure 18, the error transfer function is given as:
Ter (s) =
1
1
=
1 + Kc Gc G p Hy 1 + Kc Nc
Dc
N p Ny
D p Dy
=
• Using Eqn (45), a general solution for the steady-state error can be found:
Dc D p Dy
Dc D p Dy + Kc Nc N p Ny
• The poles of the loop gain are the zeros of the error transfer function !
– Therefore, if the loop gain has a pole at the origin, the error transfer
function has a zero at the origin (which means its DC gain is zero)
– If Tre (s) has zeros at the origin, these cancel out
R (s) in the final value theorem.
1
s
terms coming from
• The number of poles at the origin in the loop gain is a sufficiently basic
property of feedback systems that it is given a special name: the System
Type.
• The system type is given as
Type 0:
GL (s) has no pole at the origin, no = 0
Type I:
GL (s) has one pole at the origin, no = 1
Type II:
GL (s) has two poles at the origin, no = 2 etc.
1
bm sm + bm−1 am−1... + b1s + 1
KBode
n
o
s
an sn + an−1 sn−1... a1 s + 1
(45)
ess = lim s Tre (s) R (s)
s→0
= lim s
s→0
1
1 + s1no
m +b
m−1 ...b s+1
m−1 s
1
KBode abmssn +a
n−1 +... a s+1
n
n−1 s
1
R (s)
But in the limit, all of the terms bm sm + bm−1sm−1 + .. + b1s
and am sm + am−1sm−1 + .. + a1s drop out, giving
ess = lim s
s→0
= lim s
(46)
1
R (s)
1 + s1no KBode
sno
s→0 sno + KBode
R (s)
(47)
• Eqn (47) has a particularly simple form, and is valid for step, ramp or
acceleration inputs, and any system type.
where no is the number of open-loop poles at the origin.
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EE/ME 574: Intermediate Controls
Section 3.2.0
3.2 Table of steady-state errors
s→0
• Considering the closed loop example of figure 14, with
sno
R (s)
sno + KBode
Gc (s) =
(48)
• Inserting R (s) from Eqn (43), above, into the equation for steady-state error
gives
ess
s+3
,
s
1
,
s2 + 4 s + 5
r (t)
0
1+ (t)
R (s) =
1
s
1
1+KBode
0
0
Ramp 1
t 1+ (t)
R (s) =
∞
1
KBode
0
Accel 2
1 2 +
2 t 1 (t)
R (s) =
1
s2
1
s3
∞
∞
1
KBode
1
3 1
3s + 1
1
1
5 s 5 s2 + 45 s + 1
which shows a system type of I, and KBode =
find
ess (step) = 0 ,
k
Hy (s) = 1
(50)
we have the loop gain in Bode standard form:
(49)
where no is the system type, and k is the order of the input signal. Using
Eqn (49), steady-state errors for system types 0, I and II are seen in table 5.
Step
G p (s) =
GL (s) = Gc G p Hy =
sno
1
= lim s n
s→0 s o + KBode sk+1
R (s)
Section 3.4.0
3.3 Steady-state error example
• Eqn (47) leads to a general form for the steady-state error. Recall
ess = lim s
EE/ME 574: Intermediate Controls
ess (ramp) =
3
5
(51)
. Considering table 5, we
1
5
= ,
KBode 3
ess (accel) = ∞ .
Type 0 System Type I System Type II System
3.4 Summary for steady-state error
• The standard way to compute steady-state error is through table 5.
– Table 5 shows that steady-state error depends on
1. The system type,
Table 5: Connection between system type and steady-state error.
2. KBode ,
• Types 0 and I are the most common system types in practical systems. Type
II is also seen. Type III and above is quite rare.
3. The order of the input (step, ramp or acceleration).
• We often use the controller to make a Type 0 system into Type I, specifically
to make the steady-state error go to zero for step inputs.
– Controllers that increase the system type by one:
∗ PI
∗ PID
• KBode is the product of the DC gains around the loop. By increasing the
controller gain we increase KBode and reduce steady-state errors.
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Part 1: Controls Review-By-Example
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EE/ME 574: Intermediate Controls
Section 4.1.0
EE/ME 574: Intermediate Controls
4 Characteristics of the Step Response
4.2 Peak Time, t p
Time between application of a step input and the first peak of the
output. Undefined if the output does not have a peak, but taken to be
satisfied in this case if all other speed-related requirements are met.
• We’ve seen that steady-state error is a characteristic of the step (ramp)
response. But SSE we usually calculate from the TF.
• These performance measures are defined from the step response. Generally
speaking, we can’t compute them exactly from the TF, but we’ll see
approximate calculations based on the dominant mode in the next section.
• Rise time
tp ≃
• Settling time
Step response of complex pole pair
2.5
2
2%
e
ss
0.5
0
| |
T : 10−90% →
2
4 |←
6
8
Ts: 98% Settling Time
Time [sec]
(53)
ss
1.5
p
y90
As seen in figure 19, the step response of a linear system lies within an
exponentially decaying envelop. Settling time, ts , is the time required
to assure that the response does not pass again out of a narrow band
around yss .
Typically used: 4 time constants, e−4 = 1.83%. So The 4 time
constant settling time is the time between application of a step input
and the last time that the signal exits a band 1.83% wide around yss .
p
M
y (1)
|
t
p
ts = max{abs(
1
t
0.5
r (0) → y10
ss
r
Rise Time
y
r (1)
ss
Output y(t)
Output y(t)
Overshoot
1
π
ω
4.3 Settling Time, ts
• Peak time
• Overshoot
• When we consider the step response, it is always the closed-loop step
response.
Step response of complex pole pair
1.5
Section 4.4.0
y (0) →
0
ss
r
10
| 1
t
90
2
Time [sec]
(54)
where −αd is the real part dominant pole location.
T : 10−90% Rise Time
0 |
t
y(t) − yss
4
) > e−4 } ≃
yss
αd
3
4.4 Percent Overshoot, PO
Figure 19: Quantity definitions in a CL step response.
Difference between peak response and yss , the steady-state response,
expressed as a fraction of yss :
4.1 Rise Time, tr
The 10% - 90% rise time. For the step response, it is the time between crossing
0.10 yss and 0.90 yss .
1.8
(52)
tr ≃
ωn
Part 1: Controls Review-By-Example
(Revised: Jan 16, 2013) Page 37
PO = 100 ×
max (y(t)) − yss
,
yss
− √πρ
≃e
1−ρ2
(55)
where ρ [·] is the damping factor. If y(t) never goes above yss , PO = 0.
Part 1: Controls Review-By-Example
(Revised: Jan 16, 2013)
Page 38
EE/ME 574: Intermediate Controls
Section 5.1.0
EE/ME 574: Intermediate Controls
Section 5.1.1
5 Working with the pole-zero constellation
5.1.1 Real Pole location and response characteristics
5.1 Basics of pole-zero maps, 1st order, p1 = −α
Example 1: Shifting poles to the left accelerates transient decay
• Two example first-order responses are shown.
• You have seen that:
– Each real pole is associated with a mode of the system response.
– A real pole gives the terms of Eqn (57), as seen in figure 20.
C1
3
=
s+α s+2
y (t) = C1 e−αt = 3 e−2t
⇐⇒
Pole−Zero Map
Amplitude
τ = 1/16 [sec]
(56)
PZ map showing two systems
−α
⇐ Faster
2
h(t) =
e−α t = e−t/τ = e−2 t
C1 / e
1
5
0
⇐ Faster Decay
−5
−20
−15
−10
−5
0
−1
Real Axis [sec ]
5
Figure 21: A change in pole location changes the decay rate and damping.
S−plane
−2
−1
0
Real Axis [sec−1]
1
0
0
|
τ
1
2
Time (secs)
3
Figure 20: First order pole and impulse response.
• A real pole has these characteristics:
–
y (t) ∼ e−t/τ
’∆’ System step response
1.5
τ = 0.06 [sec]
’X’ System step response
1.5
τ = 0.25 [sec]
Amplitude
−2
−3
Amplitude
Imaginary Axis [sec−1]
’∆’ Case:
3
1
−1
τ = 1/4 [sec]
α = 16 [sec−1]
Impulse response
2
0
α = 4 [sec−1]
Imaginary Axis [sec−1]
Y (s) =
’X’ Case:
1
0.5
1
0.5
where τ [sec] = 1/α is the time constant.
– Further to the left indicates a faster response (smaller τ).
0
0
– The pole-zero constellation does not show either the KDC or the Krlg of
the TF or the amplitude of the impulse response.
Part 1: Controls Review-By-Example
(Revised: Jan 16, 2013)
Page 39
0.5
Time (secs)
1
0
0
0.5
Time (secs)
1
Figure 22: A change in α changes the time constant.
Part 1: Controls Review-By-Example
(Revised: Jan 16, 2013)
Page 40
Section 5.2.0
EE/ME 574: Intermediate Controls
5.2 Basis of pole-zero maps, 2nd order, p1,2 = −α ± j ω
Impulse response
Pole−Zero Map
– Each complex pole pair gives a mode of the system response.
– A complex pole pair gives the terms of Eqn (57), as seen in figure 23.
– Using the Laplace transform pair
Bc s + (Bc α + Bs ω)
←→ f (t) = eαt (Bc cos (ωt) + Bs sin (ωt))
s2 + s α s + ω2n
one finds
3
ω
S−plane
−1
Imaginary Axis [sec ]
5
• And we have seen for second order:
F (s) =
Section 5.2.1
2
−αt
0
Amplitude
EE/ME 574: Intermediate Controls
−α
e
←
1
−6 −4 −2
0
Real Axis [sec−1]
←
3.40 cos(4t − 54.0o)
−2
0
2
−1.5t
=e
0
−1
−5
−8
−t/τ
=e
2
4
Time (secs)
Figure 23: Second order pole and impulse response.
5.2.1 Notation for a complex pole pair
⇐⇒
y (t)
=
(57)
• A complex pole pair can be expressed in polar or rectangular coordinates:
*
P−Z Map, p1, p1 =−1.5 ± j 4
A e−αt cos (ωt + φ) = 3.4 e−1.5t cos (4t − 53.97o) (58)
with si = −α ± j ω , ωn =
√
α2 + ω2 ,
ρ: damping factor, ρ =
α
ωn
6
θ1
p
−1
=
Imaginary Axis [sec ]
Y (s)
b1 s + b0
2 s + 14
=
s2 + 2 α s + (α2 + ω2) s2 + 3 s + 18.25
1
4
jω
ωn
2
0
−α
−2
−4
S−plane
p*
1
−6
−3
−2
−1
0
Real Axis [sec−1]
1
2
Figure 24: Complex pole pair with ωn and φ defined.
Part 1: Controls Review-By-Example
(Revised: Jan 16, 2013)
Page 41
Part 1: Controls Review-By-Example
(Revised: Jan 16, 2013)
Page 42
Given by
Units
Decay Rate
Damped Nat. Freq.
Natural Freq.
Pole Angle
Damping Factor
p1 = −α + j ω
p1 = −α + j ω
ω2n = α2 + ω2
θ = atan2 (α, ω)
ρ = α/ωn
[sec−1]
[rad/sec]
[rad/sec]
[deg]
Dim’less, [·]
6
4
2
⇐ Faster Decay
0
−2
−4
−6
−8
Polar
5.2.2 Complex pole location and response characteristics
Example 1: Shifting poles to the left accelerates transient decay
ω
ωn
ρ
PO
’X’ Case: 1 [sec−1] 4 [rad/sec] 4.12 [rad/sec] 0.25 [Dim′ less] 44%
’∆’ Case: 4 [sec−1] 4 [rad/sec] 5.66 [rad/sec] 0.71 [Dim′ less]
Part 1: Controls Review-By-Example
(Revised: Jan 16, 2013)
4%
Page 43
−6
−4 −2
0
2
Real Axis [sec−1]
4
0.7
0.5 0.3
4 0.9
2
0
7
5
3
1
−2
−4
0.9
0.7 0.5 0.3
−6 −4 −2
0
2
−1
Real Axis [sec ]
−6
−8
4
Figure 25: A change in α changes the decay rate and damping.
’∆’ System step response
’X’ System step response
1.5
Amplitude
p1 = −α + j ω
p1 = ωn ∠ (90 + θ)
α = ρ ωn
ρ = sin (θ)
p
ω = 1 − ρ2 ωn
ρ = α / ωn
p
ω
ωn 1 − ρ2
H (s) =
H
(s)
=
s2 + 2 ρ ωn s + ω2n
(s + α)2 + ω2
Table 7: The terms of table 6 relate to rectangular or polar form for the poles.
α
PZ map with sgrid (indicates ρ and ωn)
PZ map showing two systems
6
Table 6: Factors derived from the location of a complex pole.
(Note: Franklin et al. often use σ, ωd and ζ.)
Rectangular
Section 5.2.2
Imaginary Axis [sec−1]
α or σ
ω or ωd
ωn
θ
ρ or ζ
Description
EE/ME 574: Intermediate Controls
ρ = 0.24
46 % Overshoot
1
0.5
0
0
1.5
Amplitude
Term
Section 5.2.2
Imaginary Axis [sec−1]
EE/ME 574: Intermediate Controls
2
4
Time (secs)
ρ = 0.71
4 % Overshoot
1
0.5
0
0
2
4
Time (secs)
Figure 26: Step response: a change in α, ω is unchanged.
Part 1: Controls Review-By-Example
(Revised: Jan 16, 2013)
Page 44
Section 5.2.2
Example 2: Shifting poles out vertically increases oscillation frequency
’X’ Case: 4 [sec−1]
ωn
4 [rad/sec]
ρ
4%
’∆’ Case: 4 [sec−1] 16 [rad/sec] 16.49 [rad/sec] 0.24 [Dim′ less] 44%
0
−10
−20
20
12
−1
4
−10
0.9
−20
−30
0.5 0.3
0.9
28
−20 −10
0
10
Real Axis [sec−1]
−30
1
0.5
0
0
1
Time (secs)
2
4%
16 [rad/sec] 22.63 [rad/sec] 0.71
[Dim′ less]
4%
5.66 [rad/sec]
10
0
−10
−20
0.7 0.50.3
20
0.9
10
0
28 20 12
ρ = 0.24
46 % Overshoot
0.5
0.9
−20
0.7 0.50.3
−30 −20 −10
0
10
−1
Real Axis [sec ]
’∆’ System step response
1.5
ρ = 0.71
4 % Overshoot
1
0.5
0
0
2
1
Time (secs)
ρ = 0.71
4 % Overshoot
1
0.5
τ = 0.25 [sec]
1
Time (secs)
4
−10
’X’ System step response
1
0
0
0.71
4 [rad/sec]
20
1.5
Amplitude
1.5
ρ = 0.71
4 % Overshoot
PO
Figure 29: A radial change in pole location changes the decay rate and oscillation
frequency, but not the damping.
’∆’ System step response
Amplitude
Amplitude
1.5
ρ
[Dim′ less]
−30 −20 −10
0
10
−1
Real Axis [sec ]
0.7 0.5 0.3
−20 −10
0
10
Real Axis [sec−1]
Figure 27: A change in ω changes the oscillation frequency and damping.
’X’ System step response
[sec−1]
ωn
PZ map showing two systems PZ map with sgrid (indicates ρ and ωn)
10
0
4
ω
[sec−1]
’∆’ Case: 16
Imaginary Axis [sec ]
0.7
20
Imaginary Axis [sec−1]
Imaginary Axis [sec−1]
20
⇑
Faster
Oscillation
’X’ Case:
PZ map with sgrid (indicates ρ and ωn)
PZ map showing two systems
10
α
PO
0.71 [Dim′ less]
5.66 [rad/sec]
Example 3: Shifting poles out radially rescales time
−1
ω
Section 5.2.2
Imaginary Axis [sec ]
α
EE/ME 574: Intermediate Controls
Amplitude
EE/ME 574: Intermediate Controls
τ = 0.06 [sec]
2
0
0
0.2
0.4
Time (secs)
Figure 30: Maintaining ρ, time is rescaled.
Figure 28: Step response: a change in ω.
Note: The S-plane has units of [sec−1 ].
Part 1: Controls Review-By-Example
(Revised: Jan 16, 2013)
Page 45
Part 1: Controls Review-By-Example
(Revised: Jan 16, 2013)
Page 46
EE/ME 574: Intermediate Controls
Section 5.2.3
5.2.3 The damping factor: ρ
• The response of a higher order system is the superposition of the responses
of the modes:
b0
b0
=
s2 + a1 s + a0 s2 + 2 ρ ωn s + ω2n
a1
a1
= √
ρ=
2 ωn 2 a0
(59)
(60)
– 1st order modes;
– 2nd order modes.
• Typically (not quite always) one mode will dominate. In the example of
figure 32, the slower second-order mode dominates.
Step response
Pole−Zero Map
q
P.O. = 100 e−π ρ/ 1 − ρ2
(61)
0
−5
−10
1.5
Amplitude
100
1.5
5
Amplitude
• As illustrated in figure 28, above, on the range 0.0 < ρ < 1.0, the damping
factor relates to percent overshoot. For a system with two poles and no
zeros, the percent overshoot is given by Eqn (61) and plotted in figure 31
10
Dom. Mode →
Fast second−order
mode
← Step response
← Dom. mode
1
0.5
0
0
−15 −10 −5
0
−1
Real Axis [sec ]
2
Time (secs)
4
First−order mode
1.5
Amplitude
• ρ is defined by Eqn (60) for either real poles: (ρ ≥ 1.0), or
a complex pole pair: (0.0 < ρ < 1.0).
Imaginary Axis [sec−1]
giving:
Section 5.3.0
5.3 Higher order systems: dominant mode
• Plugging ρ = α/ωn back into the 2nd order form gives:
H (s) =
EE/ME 574: Intermediate Controls
1
0.5
1
0.5
90
0
0
80
Percent Overshoot
70
60
2
Time (secs)
4
0
0
2
Time (secs)
4
Figure 32: Example high-order system showing dominant mode in step response.
50
40
• Recall: the goal of studying the S-plane is to tie controller specifications to
pole locations.
30
20
10
0
0
– The dominant mode dominates the response.
0.1
0.2
0.3
0.4
0.5
0.6
Damping factor [.]
0.7
0.8
0.9
1
Figure 31: Percent overshoot versus damping fact. Exact for a system with two
complex poles and no zeros, and approximate for other systems.
Part 1: Controls Review-By-Example
(Revised: Jan 16, 2013)
Page 47
– The root-locus design method operates by placing the dominant poles in
the desired region of the S-plane
(Other poles can be anywhere in stable region).
Part 1: Controls Review-By-Example
(Revised: Jan 16, 2013)
Page 48
EE/ME 574: Intermediate Controls
Section 5.4.0
Section 5.4.0
5.3 Dominant mode (continued)
• For high-order systems (n ≥ 3), design control by:
• A second example is presented in figure 33.
• Which mode is dominant in figure 33 ?
Imaginary Axis [sec−1]
2. Fine tune controller based on actual (high order) response
⇒ Iterative process of design and analysis.
Step response
10
5
0
−5
−15 −10 −5
0
−1
Real Axis [sec ]
Amplitude
• The dominant mode is:
– Large residue.
• On the S-plane this means that the dominant mode is (almost always):
1
0.5
Slow second−order
mode
5
Time (secs)
10
Step response
Dominant mode
5
0.5
0
0
1
0
0
−10
5.4 Which mode is the dominant mode ?
– A slow mode, with a
Student
Exercise
Pole−Zero Map
1. Use dominant mode concept and approximation for initial design; then
Amplitude
• As the figure 32 “Step response” plot shows, the dominant mode is only an
approximation to the true step response.
Amplitude
Dominant mode (continued)
EE/ME 574: Intermediate Controls
1
5
Time (secs)
10
Fast second−order
mode
0.5
0
0
5
Time (secs)
10
Figure 33: Second example showing dominant mode in step response.
– The slowest mode (other modes die out faster).
• Exceptionally, one must also consider the size of the residue, the dominant
mode should be:
– Near to other poles; for example, s = 0 pole from step input
(Graphical method: being near other poles gives a large residue).
– Far from zeros
(Graphical method: being far from zeros gives a large residue).
Part 1: Controls Review-By-Example
(Revised: Jan 16, 2013)
Page 49
Part 1: Controls Review-By-Example
(Revised: Jan 16, 2013)
Page 50
EE/ME 574: Intermediate Controls
Section 5.5.0
5.5 Summary: Pole location and response characteristics
EE/ME 574: Intermediate Controls
Section 6.0.0
6 Design
• In some sense, Design is the opposite of Analysis
Faster
Oscillation
Better Damping
Faster
Decay
Analysis
p1
X
p1* X
Stable Region
(LHP)
Completed
Controller
Design
Performance
Specifications
S-plane
Unstable Region
(RHP)
Marginally Stable Region
(the jω axis)
Design
Figure 34: Decay rate, oscillation freq. and damping depend upon pole location.
Figure 36: Design = Analysis−1
• In Analysis, we use mathematical methods to determine performance
specifications from a completed controller design (all structure and
parameters specified).
• In Design, we use whatever method works !
– Mathematical
– Gut feeling
– Trial and error
– Calling a colleague with experience
Figure 35: S-plane and responses.
Part 1: Controls Review-By-Example
(Revised: Jan 16, 2013)
to determine a controller structure and parameters to meet performance
goals.
Page 51
Part 1: Controls Review-By-Example
(Revised: Jan 16, 2013)
Page 52
EE/ME 574: Intermediate Controls
Section 6.2.0
The major design methods are:
• Dynamic systems are governed by differential equations
• Root locus
• Every input or output signal of the system has a unique Laplace transform
• For linear systems,
– Speed, Stability: determine by determining pole locations
– Accuracy: increase the system type, check the SSE
– The ratio of Laplace transforms, however, does not depend on the signals.
(∗)
– Speed: Bandwidth and Cross-over frequency directly from bode plot
– The ratio depends only on properties of the system. We call it the transfer
function. For example:
– Stability: Phase margin, Gain margin directly from bode plot
Guy (s) =
– Accuracy: tracking accy., disturbance rejection directly from bode plot
• State Space methods
Section 7.0.0
7 Summary
6.1 Design methods
• Frequency response
EE/ME 574: Intermediate Controls
(62)
• Transfer function gives us many results
(∗)
– Design using state-space design methods, check speed, stability and
accuracy from the step response
6.2 Root Locus Design
– The pole locations tells us the stability and damping ratio
– We can get approximate values for rise time, settling time and other
performance measures
• Control system analysis is the process of determining performance from a
system model and controller design
• Devised by Walter R. Evans, 1948 (1920 1999)
– We have various tools for evaluating performance, including
W.R. Evans, “Control system synthesis by
root locus method,” Trans. AIEE, vol. 69,
pp. 66–69, 1950.
* Pole locations
• Control system
(Amer. Institute of Elec. Engs became
IEEE in 1963)
* Steady-state error
* Step response
* Bode plot
Design =Analysis−1
it is the process of determining a controller design given a system model
and performance goals.
• Evans was teaching a course in controls,
and a student (now anonymous) asked a
question about an approximation.
Part 1: Controls Review-By-Example
b1 s + b0
s 2 + a1 s + a0
– The root locus method is one method for controller design.
(Revised: Jan 16, 2013)
Page 53
Part 1: Controls Review-By-Example
(Revised: Jan 16, 2013)
Page 54
EE/ME 574: Intermediate Controls
Section 8.0.0
8 Glossary of Acronyms
LT: Laplace Transform
TF: Transfer Function
FPG: Forward path gain
LG: Loop Gain, (also Gl (s))
RHP: Right-half of the S plane (unstable region)
LHP: Left-half of the S plane (stable region)
SSE: Steady-state error
PO: Percent Overshoot
LF: Low frequency (as in “LF gain”).
frequency.
Frequencies below the cross-over
HF: High frequency (as in “HF gain”).
frequency.
Frequencies above the cross-over
CL: Closed loop
OL: Open loop
P, PD, PI, PID: Proportional-Integral-Derivative control, basic and very common
controller structures.
Part 1: Controls Review-By-Example
(Revised: Jan 16, 2013)
Page 55
EE/ME 574: Intermediate Controls
EE/ME 574: Intermediate Controls
New notes in section 2.4, not finished !!!
2.3.5
Frequency-Response Based Analysis
. . . . . . . . . . .
36
2.4
Steps for sketching the Bode plot (New for 2014, not yet finished)
38
2.5
Examples: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
39
Contents
1 Introduction
1.1
5
1.1.1
Complex Conjugate: . . . . . . . . . . . . . . . . . . . .
6
1.1.2
Complex Arithmetic:
7
. . . . . . . . . . . . . . . . . . .
2 Frequency Response and the Bode plot
2.1
2.5.1
Example 1: Sketch the bode plot for K G (s)
. . . . . . .
39
2.5.2
Example 2: Sketch the bode plot for a system with a
complex pole pair . . . . . . . . . . . . . . . . . . . . . .
42
3
Review of complex numbers and arithmetic . . . . . . . . . . . .
Sketching a Bode plot by hand
10
. . . . . . . . . . . . . . . . . .
12
2.1.1
Why make a Bode plot by hand . . . . . . . . . . . . . .
12
2.1.2
Sketching individual elements . . . . . . . . . . . . . . .
13
2.1.3
Example 1, One pole no zeros . . . . . . . . . . . . . . .
14
2.1.4
Details of the break point
. . . . . . . . . . . . . . . . .
15
2.1.5
Example 2, Two poles one zero . . . . . . . . . . . . . .
16
2.1.6
Example 3, A pole at the origin . . . . . . . . . . . . . .
17
2.2
The Bode plot of the loop gain . . . . . . . . . . . . . . . . . . .
19
2.3
Bode standard form for the loop gain . . . . . . . . . . . . . . . .
24
−no
2.3.1
Plotting terms of the form KBode ( j ω)
A . . . . . . . .
26
2.3.2
Contrasting KBode with Krlg . . . . . . . . . . . . . . . .
27
2.3.3
Plotting terms of the form ( j ω τ + 1) (a real pole or zero)
28
2.3.4
Plotting a complex pole or zero pair . . . . . . . . . . . .
32
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
Page 1
2.6
Steady state errors
. . . . . . . . . . . . . . . . . . . . . . . . .
45
Determining KBode and SSE using the Matlab bode()
command . . . . . . . . . . . . . . . . . . . . . . . . . .
47
Summary of Frequency Response and the Bode plot . . . . . . . .
51
2.6.1
2.7
Summary of Bode plot plotting rules
3 Stability Considerations, Phase and Gain Margins
52
3.1
Marginal Stability . . . . . . . . . . . . . . . . . . . . . . . . . .
53
3.2
Definitions for phase and gain margin . . . . . . . . . . . . . . .
53
3.3
Examples, stability, marginal stability and instability . . . . . . .
54
3.3.1
A challenge with applying gain and phase margin . . . . .
58
3.3.2
Example with the pattern of figure 38(a) . . . . . . . . . .
60
Interpreting Phase and Gain Margin . . . . . . . . . . . . . . . .
62
3.4
4 Summary: Analysis by Frequency Response
Part 2: Frequency Response Analysis
63
(Revised: Feb 24, 2014)
Page 2
EE/ME 574: Intermediate Controls
Section 1.0.0
1 Introduction
EE/ME 574: Intermediate Controls
Section 1.1.0
• We can determine the steady-state output sinusoid from the input and the
transfer function
• For a linear system a sinusoidal input produces a sinusoidal output,
(after all transients have died out)
• Figure 1 shows the response of the system
G (s) =
– Representing the input as a phasor
Phasor − A Complex Number
Waveform :
u(t)
1
s + 10
y(t)
1
G(s)= s+10
for the input
u (t) = Au cos (ωt + θu)
←→
~U = Au ∠θu
– Looking at the system
u (t) = sin (10t)
Output Phasor ~Y
1
= = Complex Gain = G (s)
=
~U
Input Phasor
j
ω
+ 10
s= j ω
• Once the startup transient dies out, an input with ω = 10 gives and output
with ω = 10.
• For example, with u (t) = sin (10t) and G (s) = 1/ (s + 10), then y (t) is
determined from:
Zero−state Response of G(s) to sin(10 t)
0.2
~Y
1
1
= √ ∠ − 45o
= G (s)
=
~U
j 10 + 10 10 2
s= j 10
0.15
System Output
0.1
• The input as a phasor is give by
0.05
u (t) = sin (10t) = cos (10t − 90o)
0
~U = 1.0∠−90o
(withω = 10[rad/sec])
• The output phasor is given by:
−0.05
−0.1
0
←→
1
2
3
4
5
6
Time [seconds]
7
8
9
~Y = (Input Phasor) (Complex Gain) = (1.0∠ − 90o)
10
Figure 1: A sinusoidal input to a linear system results in a sinusoidal output.
~Y =
1
√ ∠ − 135o
10 2
←→
y (t) =
1
√ ∠ − 45o
10 2
1
√ cos (10t − 135o)
10 2
(This is the waveform in figure 1)
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
Page 3
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
Page 4
EE/ME 574: Intermediate Controls
Section 1.1.0
1.1 Review of complex numbers and arithmetic
(See Franklin et al., Appendix B)
EE/ME 574: Intermediate Controls
Section 1.1.1
• Complex numbers correspond to a point on the plane
Imag
• Certain polynomials have no solutions which are real numbers,
x2 + 1 = 0
or
x2 = −1
A
• Italian mathematicians of the 1500’s began considering polynomials and
√
discussing solutions involving −1
<θ
V
a+b j
Real
V*
• In 1637 René Descartes coins the term ’imaginary number’ to suggest that
work on the subject was an illusion.
Complex Plain
• It turns out that imaginary numbers solve a wide range of problems in signal
processing, medical image reconstruction, impedance analysis, etc.
Figure 2: A complex number = A point on the complex plain.
• Any complex number is made up of a real and imaginary part
• A complex number (point on the complex plain) can be represented in polar
or rectangular coordinates
A = α+ jω
real (A) = α
imag (A) = ω
• j (or i for mathematicians) is the complex number
j2 = −1
(1)
Rectangular
V = Av ∠θv
p
a2 + b2
V = a+ jb
(2)
a = A cos (θv )
(3)
b = A sin (θv )
(4)
Av =
θv = atan2 (b, a)
• Higher powers of j are given from:
etc.
Polar
• What the complex plane reveals is that the real and imaginary parts are
orthogonal.
j3 = j2 j = −1 j
j4 = j2 j2 = −1 − 1 = 1
j5 = j2 j2 j = j
1.1.1 Complex Conjugate:
The complex conjugate of a number V is written V∗ .
With V = Av ∠θv = a + j b , then V∗ = Av ∠ − θv = a − j b
V + V∗ gives a real number, V + V∗ = 2 a + j 0
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
Page 5
V V∗ also gives a real number,
(a + j b) (a − j b) = a2 + b2 + j (a b − a b) = a2 + b2 + j0
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
Page 6
EE/ME 574: Intermediate Controls
Section 1.1.2
1.1.2 Complex Arithmetic:
EE/ME 574: Intermediate Controls
Section 1.1.2
Euler’s Formula:
• The operations of complex arithmetic are given as:
Polar
• Recall the transcendental functions:
Rectangular
ex = 1 + x +
V = A∠θ , W = B∠φ V = a + j b , W = c + j d
Addition :
(none)
V + W = (a + c) + j (b + d)
(5)
Subtraction :
(none)
V − W = (a − c) + j (b − d)
(6)
V ∗ W = A B∠θ + φ
V ∗ W = (a + j b) ∗ (c + j d)
(7)
Multiplication :
= (a c − b d) + j (ad + bc)
Division :
V
W
= A /B∠θ − φ
(ac + bd) + j (bc − ad)
V
(8)
=
W
c2 + d 2
x2 x3 x4
+ + + ···
2! 3! 4!
cos x = 1 −
x2 x4
+ + ···
2! 4!
sin x = x −
x3 x5
+ + ···
3! 5!
• Euler’s formula gives meaning to the complex exponential.
e jθ = 1 + j θ +
j2 θ2 j3 θ3 j4 θ4 j5 θ5
+
+
+
+ ···
2!
3!
4!
5!
(9)
Using the fact that j2 = −1 , j4 = 1, etc.
• Where the equation for division in rectangular coordinates comes from
writing:
V
V W∗
=
W W W∗
and multiplying out the terms.
θ2 j θ3 θ4 j θ5
−
+ +
+ ···
2!
3!
4!
5!
θ2 θ4
θ3 θ5
= 1 − + + ··· + j θ − + + ···
2! 4!
3! 5!
e jθ = 1 + j θ −
e j θ = cos θ + j sin θ
(10)
• From Eqns (2) and (10) we find that
A∠θ = A e j θ = A cos (θ) + j A sin (θ) = a + j b
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
Page 7
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
(11)
Page 8
EE/ME 574: Intermediate Controls
Section 1.1.2
EE/ME 574: Intermediate Controls
Section 2.0.0
2 Frequency Response and the Bode plot
• If we write the phasor
~V = A∠θ
(12)
• Considering the steady state, a sinusoidal input gives a sinusoidal output
then
• The relationship between the input and output is given by the complex gain
v (t) = A cos (ωt + θ) = A real e j(ωt+θ) = A real e jωt e jθ
– Which is given by the transfer function evaluated at s = j ω
• In general, the G (s = j ω) a function of frequency
• We can imagine a complex signal ν (t) = e j(ωt+θ) as a vector on the complex
plane. The real signal is the projection of the vector onto the real axis.
– In polar coordinates, a complex number has a magnitude and phase
– The magnitude and phase are a function of frequency
Imag
Real
v (t) = e j (ωt+θ)
= e jωt e j θ
High−Pass Filter
10
50
0
40
−20dB/decade
−10
−20
G(s)=3/(s+3)
−30
20
0
−50 −1
10
0
1
10
10
2
10
3
10
−10 −1
10
0
1
10
2
10
10
−30
Lagging phase
−60
−90
Leading phase
60
30
0
−1
0
10
10
Figure 3: Phasor rotating on the complex plane.
1
10
Frequency [radians/second]
2
10
Low Pass Filter
• Rotation by an angle φ (either phase shift or by ωt) is just multiplication
by e j φ .
G (s) =
3
s+3
3
10
−1
10
0
10
1
10
Frequency [radians/second]
2
10
High Pass Filter
G (s) =
s + 10
10
Figure 4: Bode plot: First-order response characteristic.
1st order low-pass:
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
Page 9
3
10
90
Phase [deg]
0
Phase [deg]
+20dB/decade
10
−40
Complex Plain
G(s)=(s+10)/10
30
Gain dB
A ej(ωt+θ)
Gain dB
Low−Pass Filter
rotation
ωt
1st order high-pass:
−20dB / decade roll-off
+20dB / decade
−90o
+90o phase at high freq.
phase at high freq.
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
Page 10
3
10
EE/ME 574: Intermediate Controls
• 2nd order
Section 2.0.0
(13)
2.1.1 Why make a Bode plot by hand
• Given a structure for Gc (s) and approximate values, computers can be pretty
good at fine tuning the parameters (adaptive control)
20
0
Gain dB
Gain dB
• Given Gc (s) G p (s) computers are great for making the exact Bode plot
2nd order mode, ρ=0.056
20
−40dB/decade
−20
−40
−60 −1
10
0
10
0
−40dB/decade
−20
−40
G(s)=10/(s2 + 1.0 s + 10)
1
10
10
0
−30
Phase [deg]
0
−30
−60
−90
−120
−150
−180
G(s)=10/(s2 + 0.2 s + 10)
−60 −1
10
2
0
1
10
10
2
10
0
1
10
10
Frequency [radians/second]
2nd order, moderate damping
G (s) =
2
10
−90
−120
−150
−1
0
10
1
10
10
Frequency [radians/second]
2nd order, light damping
10
s2 + 1.0 s + 10
G (s) =
2
10
• The Bode plot is a tool on which Engineering judgement and intuition can
be based.
• Some key ideas of the bode plot are:
10
s2 + 0.2 s + 10
Figure 5: Bode plot: Second-order response characteristic. The smaller the
damping, the higher the resonant peak.
– The axes:
∗ Frequency axis is logarithmic
∗ Magnitude is plotted in decibels (dB)
∗ Phase is plotted in degrees
– Decibels: a logarithmic scale
2nd order response:
|G (s)|dB = 20 log10 (|G (s)|)
−40 dB / decade roll-off
∗ Multiplying the loop gain by a factor corresponds to adding decibels
−180o phase at high frequency
|40|dB = 32 dB , |50|dB = 34 dB ,
Sharpness of the transition depends on the damping:
∗ Multiplicative factors:
2× ≃ +6 dB ,
Moderate to high damping: shallow peak, smooth phase transition
0.5× ≃ −6 dB ,
Low damping: sharp peak, abrupt phase transition
Part 2: Frequency Response Analysis
• Computers are no good at all at determining a good structure for Gc (s), or
finding the correct ball park for parameters
• An approximate sketch of the Frequency Response is a very powerful way
to understand the over-all characteristics of a system.
−60
−180
−1
10
Section 2.1.1
2.1 Sketching a Bode plot by hand
ω2n
G (s) = 2
s + 2ρ ωn s + ω2n
2nd order mode, ρ=0.28
Phase [deg]
EE/ME 574: Intermediate Controls
(Revised: Feb 24, 2014)
Page 11
Part 2: Frequency Response Analysis
|40 × 50|dB = 32 + 34 dB
10× = +20 dB ,
0.1× = −20 dB ,
20× ≃ +26 dB
0.05× ≃ −26 dB
(Revised: Feb 24, 2014)
Page 12
EE/ME 574: Intermediate Controls
Section 2.1.2
• Advantage of working in dB:
u(t)
– Gains of blocks in series multiply,
2.1.3 Example 1, One pole no zeros
G (s) =
One pole no zeros
y(t)
10
G1 G2 = A1 A2 ∠θ1 + θ2 = A12 ∠θ12
−10
Low−Freq Asymptote
• Low Freq Asymp:
−20dB/decade
−20
−30
G ( j 2) ≃
−40
−50 −1
10
– But Bode plot terms in series add
0
10
1
2
10
10
≃ 2.0 dB∠0o
20 log10 (A1 A2 ) = 20 log10 (A1) + 20 log10 (A2 )
Phase [deg]
Phase :
10
0 j+8
3
10
• High Freq Asymp:
(|2 j ω| >> 8)
0
Magnitude :
10
2s +8
• Break Point:
ω = 4 [rad/sec]
← High−Freq Asymptote
0
Gain dB
,
G2
Section 2.1.3
←
G1 = A1 ∠θ1 , G2 = A2 ∠θ2
G1
EE/ME 574: Intermediate Controls
θ12 = θ1 + θ2
−30
−60
G ( j ω) ≃
10/(2 j ω) gives −90o
10
2 jω+0
−90
−1
2.1.2 Sketching individual elements
10
• Low frequency asymptote given as
(14)
lim G ( j ω)
(15)
• Poles and zeros make contributions as a function of ω
– A pole adds a slope of −1 and
2
3
10
10
≃
10
∠ − 90o
2ω
• Break points:
– Freq’s at which poles (or zeros) introduce transitions into the Bode plot.
ω→∞
90o
1
10
Frequency [radians/second]
Figure 6: One pole, no zero.
lim G ( j ω)
ω→0
• High frequency asymptote given as
0
10
– Also, the break-point frequency is the frequency at which the real and
imaginary parts of the pole (or zero) are equal.
– Examples:
s+5
s + 10
Break point frequency of zero: ω = 5 [rad/sec].
G p (s) =
of phase lag
– A pole pair adds a slope of −2 and 180o of phase lag
Break point frequency of pole: ω = 10 [rad/sec].
– A zero adds a slope of +1 and 90o of phase lead
G p (s) =
– A zero pair adds a slope of +2 and 180o of phase lead
3s +1
0.02 ∗ s + 1
Break point frequency of zero: ω = 1/3 [rad/sec].
Break point frequency of pole: ω = 50 [rad/sec].
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
Page 13
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
Page 14
EE/ME 574: Intermediate Controls
Section 2.1.4
2.1.4 Details of the break point
EE/ME 574: Intermediate Controls
2.1.5 Example 2, Two poles one zero
Example:
Break Point
10
−10
s2 + 21 s + 20
=
1
40 (s + 4)
1
4 s+1
=8 0
1
(s + 1)(s + 20)
s (s + 1) 20
s+1
(16)
One pole no zeros
−20
30
−30
−40
−50 −1
10
0
10
1
2
10
10
3
10
• The break point is at
ω = 4.0 [rad/sec]
20
0
Phase [deg]
40 s + 160
G (s) =
Gain dB
Gain dB
10
G (s) =
2s +8
Gain reduce by 1/sqrt(2)
0
Section 2.1.5
10
0
−10
−20
−30 −1
10
• At the break point,
the real and imaginary
parts are equal.
−30
o
Phase shift of 45
−60
0
1
10
2
10
10
3
10
0
−1
10
0
10
1
10
Frequency [radians/second]
2
10
Phase [deg]
−90
3
10
Figure 7: The break point is at ω = 4.0 [rad/sec].
−30
−60
−90
−1
– The break point marks the transition from where the real term dominates
to where the imaginary part dominates
0
10
• The break point is the frequency at which the real and imaginary parts of
the term (pole or zero) are equal
10
1
10
Frequency [radians/second]
2
10
3
10
Figure 8: Two poles, one zero.
• The two asymptotes are very simple
– Low Freq Asymptote:
• Consider
1
G (s) =
s+1
The break point is ω = 1.0 [rad/sec]
lim G ( j ω) = lim
s→ j 0
s→ j 0
40 s + 160
s2 + 21 s + 20
=
160
= 18.1 dB∠0o
20
– High Freq Asymptote:
G (s = 1.0 j) =
1
1
= √ ∠ − 45o
jω+1
2
lim G ( j ω) = lim
s→ j ∞
s→ j ∞
40 s + 160
s2 + 21 s + 20
– Three break points: ω = 1 (down),
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
Page 15
Part 2: Frequency Response Analysis
=
40
40
40 s 40
=
=
= ∠ − 90o
s2
s
jω
ω
ω = 4 (up),
ω = 20 (2x down)
(Revised: Feb 24, 2014)
Page 16
EE/ME 574: Intermediate Controls
Section 2.1.6
2.1.6 Example 3, A pole at the origin
G (s) =
EE/ME 574: Intermediate Controls
– Two break points:
1
4 s+1 1
20 s + 1
1
40 s + 160 40 (s + 4)
=
=8 1
s2 + 20 s
s (s + 20)
s
ω = 4 (up),
Section 2.1.6
ω = 20 (down)
(17)
One pole no zeros
← Mag=40 (32dB)
20
→
Gain dB
40
Mag=8 (16 dB)
0
−20
−1
10
0
1
10
2
10
10
3
10
Phase [deg]
0
−30
−60
−90
−1
10
0
10
1
10
Frequency [radians/second]
2
10
3
10
Figure 9: A Type I system element.
• Asymptotes, think about the points where the asymptotes hit ω = 1.0.
– Low Freq Asymptote:
lim G ( j ω) = lim 8
s→ j 0
s→ j 0
1
s1
1
4 s+1 1
20 s + 1
= lim 8
ω→0
1
8
= ∠ − 90o
jω ω
– High Freq Asymptote (same as example 2):
lim G ( j ω) = lim 8
s→ j ∞
s→ j ∞
1
s1
Part 2: Frequency Response Analysis
1
4 s+1 1
20 s + 1
1
40
40
= 40 =
= ∠ − 90o
s
jω
ω
(Revised: Feb 24, 2014)
Page 17
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
Page 18
EE/ME 574: Intermediate Controls
Section 2.2.0
EE/ME 574: Intermediate Controls
2.2 The Bode plot of the loop gain
Section 2.2.0
H (s)=20/(s+20)
G (s)=1/s
y
p
10
20
0
Gain dB
Gain dB
0
• The Bode plot of elements in series is given by adding the plots of the
individual elements
−20
−40
0
+
2
10
−40 −1
10
3
10
10
Phase [deg]
Phase [deg]
−40
−60
−80
u(t)
Kc Gc(s)
1
2
10
3
10
10
−30
−60
−90
−100 −1
10
0
10
1
10
Frequency [radians/second]
G p (s) =
e(t)
0
10
0
−20
Example 5, A basic loop
r(t)
1
10
0
• It will turn out that we can establish the stability, disturbance rejection, noise
rejection and other properties from the loop gain, without ever solving for
the closed-loop poles.
−20
−30
−60 −1
10
• The loop gain is open loop.
−10
y(t)
2
−1
3
10
10
10
1
s
0
10
1
10
Frequency [radians/second]
Hy (s) =
2
3
10
10
20
s + 20
Figure 11: Components of the loop gain, the plant and sensor.
Gp(s)
ys(t)
• Put together the Bode
plots of G p (s) and
Hy (s)
2
Gc(s) Hy(s)=20/(s +20s)
Hy(s)
20
0
Gain dB
Figure 10: A basic control loop.
With:
−20
G (s) =
−40
−60
−80
−100 −1
10
0
10
1
10
2
10
3
10
G p (s) =
1
s
Hy (s) =
20
s + 20
Phase [deg]
Kc Gc (s) = 2 (s + 5)
K GL = Kc Gc G p Hy =
Part 2: Frequency Response Analysis
• Low Freq Asymp:
G ( j ω) = 20/20 s
= ω1 ∠ − 90o
• High Freq Asymp:
G ( j ω) = 20/s2
= ω202 ∠ − 180o
−60
−90
−120
−150
−180
−1
10
The loop gain is given as:
20
20
=
s2 + 20 s s (s + 20)
0
10
1
10
Frequency [radians/second]
2
10
3
10
• Break Point:
ω = 20 [rad/sec]
Figure 12: Components of the loop gain, combining the plant and sensor.
40 (s + 5)
s (s + 20)
(Revised: Feb 24, 2014)
(18)
Page 19
• Note that at high freq. this system has a roll-off of -40 dB per decade (slope
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
Page 20
EE/ME 574: Intermediate Controls
Section 2.2.0
EE/ME 574: Intermediate Controls
of -2), and has 180o degree of phase lag.
Section 2.2.0
K G (s)=2s+10
G (s) H (s)=20/(s2+20s)
c
c
y
100
Gain dB
0
Gain dB
−20
−40
−60
80
60
40
20
−80
0
1
10
10
2
10
0 −1
10
3
10
Phase [deg]
−100 −1
10
Phase [deg]
c
120
20
−60
−90
−120
−150
−180
0
10
1
10
2
3
10
10
90
60
Break point
30
0
−1
10
0
10
1
10
Frequency [radians/second]
2
10
−1
3
10
10
1 20
s s + 20
G p (s) Hy (s) =
0
10
1
10
Frequency [radians/second]
2
3
10
10
Kc Gc (s) = 2 (s + 5) = 2 s + 10
Figure 13: Components of the loop gain, the plant and sensor.
• Some things to notice:
Loop Gain Kc Gc(s) Gp(s) Hy(s)
• The phase angle corresponds to the slope.
60
Gain dB
40
20
-20 dB per decade
gives -90 [deg]
0
−20
−40 −1
10
0
10
1
10
2
10
3
10
Phase [deg]
0
• The zero comes at
ω = 5 [rad/sec], while
the sensor pole comes at
ω = 20 [rad/sec]
−20
– So the +90o of Gc (s)
wins out for a range
of ω
−40
−60
−80
−100 −1
10
0
10
1
10
Frequency [radians/second]
2
10
3
10
Figure 14: Components of the loop gain, combining the plant and sensor.
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
Page 21
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
Page 22
EE/ME 574: Intermediate Controls
Section 2.2.0
20
120
0
100
−20
80
Gain dB
Gain dB
• What the Bode plot would look like if the sensor pole came first (had lower
frequency) than the PD compensator zero
−40
−60
EE/ME 574: Intermediate Controls
2.3 Bode standard form for the loop gain
• The loop gain of a linear system can be written in a standard form, called
Bode form:
60
K GL (s) = KBode
40
20
−80
−100 −1
10
0
1
10
2
10
0 −1
10
3
10
10
0
10
1
10
2
3
10
10
Phase [deg]
Phase [deg]
= KBode
−60
−90
−120
−150
Section 2.3.0
90
1 ′
G (s)
sno L
(τ1 s + 1) (τ2 s + 1) · · ·
1
(19)
sno (τa s + 1) (τb s + 1)· · · (s2/ω2n + 2 ρ s /ωn + 1)
where
60
30
−180
0
−1
10
0
10
G p (s) Hy (s) =
1
10
Frequency [radians/second]
2
−1
3
10
10
10
1 20
s 2 s + 20
0
10
1
10
Frequency [radians/second]
2
3
10
10
Kc Gc (s) = 2 (0.25 s + 5) = 0.5 s+10
Figure 15: Components of the loop gain, faster compensator, slower sensor.
K GL (s) is the loop gain
KBode is the DC gain of the loop, after poles or zeros at the orgin are
pulled out using the 1/sno term
sno , no is the number of poles at the origin; no is the system type
′
• Some things to notice:
Loop Gain K G (s) G (s) H (s)
c
c
p
y
40
• The Bode plots of the
components look quite
similar,
Gain dB
20
0
−20
−40
−60 −1
10
0
10
1
10
2
10
3
10
Phase [deg]
−40
−60
– The relative speed of
poles and zeros
matters
−80
−100
0
10
1
10
Frequency [radians/second]
2
10
3
10
Figure 16: Components of the loop gain, combining the plant and sensor.
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
τ1 , τ2 , ...[seconds] are time constants of the (real) zeros
τa , τb , ...[seconds] are time constants of the (real) poles
ωn [radians/second] and ρ [·] are the frequency and damping of a
complex pole pair
• Eqn (19) is composed of three classes of terms:
• Conclusion:
−20
−120 −1
10
• The bode plot of the loop
gain is quite different
(especially the phase)
GL (s) is the transfer function of poles and zeros not at the origin,
′
scaled so that the DC gain of GL (s) is 1.0.
Page 23
1. KBode ( j ω)−no
, a constant term and poles or zeros at the origin
±1
2. ( j ω τ + 1)
, first order terms
±1
2
jω
jω
+
1
+
2ρ
, second order terms.
3.
ωn
ωn
• We will see how each of these terms contributes to a Bode plot.
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
Page 24
EE/ME 574: Intermediate Controls
Section 2.3.0
EE/ME 574: Intermediate Controls
Section 2.3.1
2.3.1 Plotting terms of the form KBode ( j ω)−no A
• Example,
• This term combines the DC gain with poles or zeros at the origin.
1 20
1 20 (s + 5)
=2
s s + 20
s s + 20
(20)
• The magnitude curve is a straight line, with a slope of −no × 20 dB / decade
• The phase curve is a flat line, at an angle of −90 no degrees
• Examples of KBode ( j ω)−no :
1. Pull out any singularities, poles or zeros, at the origin
so no = 1
Gain dB
1 20 (s + 5)
K GL (s) = 2
,
s
s + 20
60
40
40
20
Gain dB
K GL (s) = Kc Gc (s) G p (s) Hy (s) = 2 (s + 5)
20
0
−40
−20
no = 1 indicates there is a pole at the origin.
−40 −1
10
0
10
1
2
10
−60 −1
10
3
10
10
1 b (s) 5 ∗ 20 1
=
K GL (s) = KBode n
s o a (s)
20 s
1
5s + 1
1
20 s + 1
1
= 5.0
s
2
10
3
10
Phase [deg]
−40
−60
−100 −1
10
60
40
20
0
0
10
1
10
Frequency [radians/second]
2
−1
3
10
0
10
10
KBode ( j ω)−no = 12 s−1 , no = +1
10
1
10
Frequency [radians/second]
2
10
3
10
KBode ( j ω)−no = 0.05 s+1, no = −1
40
30
Gain dB
3. Identify the time constants
1 (τ1 s + 1)
1
= 5.0
sno (τa s + 1)
s
1
5 s+1
1
20 s + 1
20
10
0
−10 −1
10
(21)
0
1
10
10
2
3
10
10
50
1
[seconds] , KBode = 5.0 and no = 1
which gives τ1 = 15 [seconds], τa = 20
Phase [deg]
K GL (s) = KBode
1
10
80
−20
−80
1
5 s+1
1
20 s + 1
0
10
100
0
Phase [deg]
2. Scale the polynomical terms of GL (s) to have constant coefficients of
1.0, putting the loop gain in Bode standard form:
0
−20
0
−50
• Eqn (21) is the loop gain example (20), expressed in Bode standard form.
−1
10
0
10
1
10
Frequency [radians/second]
KBode ( j ω)
−no
2
3
10
10
= 4 , no = 0
Figure 17: First term of the loop gain
• The most common cases are:
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
Page 25
Part 2: Frequency Response Analysis
no = 0
(Type 0 system)
no = 1
(Type I system)
(Revised: Feb 24, 2014)
Page 26
EE/ME 574: Intermediate Controls
Section 2.3.2
Section 2.3.3
2.3.3 Plotting terms of the form ( j ω τ + 1) (a real pole or zero)
2.3.2 Contrasting KBode with Krlg
• Given an example TF
EE/ME 574: Intermediate Controls
2 s + 10
4s2 + 3 s + 8
we can write the TF in two standard forms:
GL (s) =
(22)
• First-order terms are understood by their asymptotes and the break point
10
Gain dB
0
Bode Form:
Break Point
−10
−20
K GL (s) = KBode
(τ1 s + 1) (τ2 s + 1) · · ·
1
sno (τa s + 1) (τb s + 1)· · · (s2/ω2n + 2 ρ s /ωn + 1)
−30 −1
10
(19, repeated)
0
1
10
10
1 ′
10 1
GL (s) = KBode n GL (s) =
so
8 s0
1
s+1
5 3
1 2
2 s + 8 s+1
(23)
Phase [deg]
0
For example:
−30
Break Point
−60
−90
−1
0
10
10
Frequency [radians/second]
Root Locus Form:
1
10
Figure 18: First-order asymptotes, τa = 0.5 [seconds].
Krlg
(s − z1) (s − z2) · · ·
(s − pa) (s − pb) · · · (s2 + 2 ρ ωn s + ω2n)
The first-order term is
G1 (s) =
where z1 = −1/τ1 , z2 = −1/τ2, pa = −1/τa, pb = −1/τb, etc. For example:
′′
GL (s) = Krlg GL (s) =
2
s+5
4 s2 + 34 s + 2
(24)
• Low-frequency asymptote, | j ω τa | << 1
G1 (s) =
• For making bode plots, we use Bode form and determine KBode
1
1
=
τa s + 1
j ω τa + 1
1
1
1
≈
= = 1∠0o
j ω τa + 1 0 j + 1 1
(25)
• For determining K from a root-locus plot we use root-locus form and
determine Krlg .
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
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Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
Page 28
EE/ME 574: Intermediate Controls
Section 2.3.3
First-order term (continued)
EE/ME 574: Intermediate Controls
Characteristics of the first-order break point
10
10
0
Gain dB
Gain dB
0
Break Point
−10
−20
Break Point
−10
−20
−30 −1
10
0
−30 −1
10
1
10
10
0
0
1
10
10
0
Phase [deg]
Phase [deg]
Section 2.3.3
−30
Break Point
−60
−90
−30
Break Point
−60
−90
−1
0
10
10
Frequency [radians/second]
1
−1
10
Figure 19: First-order term, τa = 0.5 [seconds].
10
Frequency [radians/second]
1
10
Figure 20: First-order term, τa = 0.5 [seconds].
• The break point is at ωbp τa = 1, or ωbp = 1/τa
• High -frequency asymptote, | j ω τa | >> 1
G1 (s) =
0
10
1
1
1
=
∠ − 90o
≈
τa s + 1
j ωτa ω τa
(26)
• At the break point:
G1 (s) =
• The high-frequency asymptote has
1
1
1
=
= √ ∠ − 45o
j ω τa + 1
j1+1
2
(27)
• The slope of the phase curve at ω = ωbp gives the phase transition on the
range
ω = 0.2 ωbp ... 5 ωbp
– A slope of −1 ,
– A phase of −90o
– Intercepts ω = 1 at 20 log10 (1/τa ) (shown at triangle)
(a 25 : 1 range of frequency !)
• On the magnitude curve, the break point is down (or up, for a zero)
√
by 1/ 2 or -3 dB
Part 2: Frequency Response Analysis
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Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
Page 30
EE/ME 574: Intermediate Controls
Section 2.3.3
A zero has the same characteristics as a pole, but with the opposite sense.
60
Section 2.3.4
2.3.4 Plotting a complex pole or zero pair
• Second-order terms are also understood by their asymptotes and break point
40
20
0
Break Point
−40
−60
Intercept
20
−20
Gain dB
Gain dB
EE/ME 574: Intermediate Controls
Intercept
0
1
10
2
10
10
3
10
0
−20
−40
0
1
10
2
10
10
60
0
Break Point
30
0
0
1
10
2
10
10
Frequency [radians/second]
3
10
−60
−90
Break Point
−120
−150
−180o degrees
−180
Figure 21: First-order asymptotes, τ1 = 0.02 [seconds].
0
1
10
The first-order term for a zero is:
2
10
Frequency [radians/second]
10
Figure 22: Second-order asymptotes, ωn = 4.0 [radians/second], ρ = 0.2 [·] .
G1 (s) = (s τ1 + 1)
(28)
• ( j ω τ1 + 1) has a low-frequency asymptote of 1∠0o
The second-order term is
G (s) = • ( j ω τ1 + 1) has a high-frequency asymptote of ω τ1 ∠ + 90o
1
jω
ωn
2
+ 2ρ
jω
ωn
1
=
+1
−
2
ω
ωn
+ 2ρ j
ω
ωn
+1
• Low-frequency asymptote, | j ω/ωn | << 1:
– The high-frequency asymptote has a slope of +1
– The intercept of the high-frequency asymptote is |τ1 |dB at ω = 1.0
• The break point is at
0o degrees
−30
Phase [deg]
Phase [deg]
90
ωbp τ1 = 1 [·]
G (s) ≈
1
1
= = 1∠0o
0 j2 + 0 j + 1 1
(29)
– The phase angle at the break point is +450
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
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Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
Page 32
EE/ME 574: Intermediate Controls
Section 2.3.4
Complex pole pair (continued)
EE/ME 574: Intermediate Controls
Characteristics of the second-order break point
Intercept
20
Section 2.3.4
20
Break Point, 20 log10(1/2ρ)
Gain dB
Gain dB
Break Point
0
−20
0
1
10
0
10
10
0
0o degrees
−30
Phase [deg]
Phase [deg]
1
2
10
−30
−60
−90
Break Point
−120
−150
−180o degrees
−180
0
10
Frequency [radians/second]
1
ω
ωn
+ 2ρ j
ω
ωn
Break Point
1
2
10
Frequency [radians/second]
Figure 24: Second-order break point, detail.
ωn = 4.0 [radians/second], ρ = 0.2 [·] .
• High-frequency asymptote | j ω/ωn | >> 1:
−90
−120
−150
10
Figure 23: Second-order asymptotes, ωn = 4.0 [radians/second], ρ = 0.2 [·] .
2
−60
−180
1
10
−
−20
−40
−40
G (s) =
0
+1
≈
−
• The break point is at ωbp/ωn = 1, or ωbp = ωn
ω 2
1
n
∠ − 180o
2 =
ω
ω
(30)
ωn
• The high-frequency asymptote has
• At the break point:
G (s) = =
– A slope of -2,
1
jω
ωn
2
+ 2ρ
jω
ωn
=
+1
1
−1 + 2ρ j + 1
1
1
=
∠ − 90o
2ρ j 2ρ
– The amplitude at the break point is (this is the resonant peak)
– A phase of −180o
– Intercepts ω = 1 at 20 log10 ω2n (shown at triangle)
Mr =
1
2ρ
– The phase is −90o
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
Page 33
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
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EE/ME 574: Intermediate Controls
Section 2.3.4
Second order break versus damping factor ρ
1. Put the TF in Bode Form
rho=0.02
20
Gain dB
Section 2.3.5
2.3.5 Summary of Bode plot plotting rules
rho: 0.02, 0.10, 0.20, 0.40, 0.60, 0.80, 1.00,
(a) Determine the number of singularities at the origin
0
rho=1.0
(b) Order the poles and zeros in order of ascending frequency
−20
−40 −1
10
0
2. Select a frequency range
1
10
10
• Cover the poles and zeros of the system
0
rho=0.02
−30
Phase [deg]
EE/ME 574: Intermediate Controls
−60
• Consider the frequency range of the inputs and outputs of interest
rho=1.0
−90
3. Determine the KBode ( j ω)−no term
−120
−150
−180
−1
10
0
10
Frequency [radians/second]
4. Plot the magnitude curve
1
10
(a) Plot the low-frequency asymptote, which passes through KBode at ω = 1
.
Figure 25: Second-order break point, detail.
ωn = 1.0 [radians/second], ρ = {0.2 } [·] .
• Low-frequency asymptote slope: −no (×20 dB / decade)
– Often no = 0 or no = 1
• As damping decreases
– The transition region gets narrower
(b) Plot the high-frequency asymptote, which passes through Krlg at ω = 1 .
– The peak gets higher
(c) Complete the Magnitude Asymptotes
• Go from low frequency to high frequency, putting in break points
• What value of ρ corresponds to Mr = 1.0 ?\SetScrunchOne
• Extend the low-frequency asymptote to the next break point
• Change the slope by ±20dB / decade or ±40 dB / decade
• Continue through all break points
(d) Sketch the approximate magnitude curve
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
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Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
Page 36
EE/ME 574: Intermediate Controls
Section 2.3.5
EE/ME 574: Intermediate Controls
Section 2.4.0
2.4 Steps for sketching the Bode plot (New for 2014, not yet
5. Plot the phase curve
finished)
(a) Plot the low-frequency phase, φ = −no 90o for ω τ << 1
(b) Plot the high-frequency asymptote, which occurs above the highest
frequency singularity
(c) As a guide, put in ±90o or ±180o steps in phase at the break points
(starting with the lowest frequency break point)
Steps for one method of sketching the Bode plot.
1. Put the loop gain in Bode standard form
2. Determine the Low-Frequency Asymptote (LFA), and break points.
(a) Choose the range of the frequency axis, to cover the break points
φ (ω) ≃ −90 (n − m)
(31)
3. Sketch the Bode plot,
where n is the number of poles to the left of ω, and m is the number of
zeros to the left of ω .
(a) Sketch the low-frequency asymptote,
(b) Add construction lines for each break point:
(d) Setup the width of the phase transitions
• First order, transition width: 15 ωbp ... 5 ωbp
• Second Order: transition width depends on damping for 2nd order
(wide for heavy damping, narrow for light damping)
i. Slope decrements by one, and phase decrements by 90 degrees, at
each pole break point. Slope increments by one, and phase increments
by 90 degrees, at each zero break point.
(c) Sketch in the smooth Bode-plot curve.
(e) Draw the phase curve
4. Use the High-Frequency Asymptote to validate the Bode-plot sketch.
——– ∼ ——–
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
Page 37
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
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EE/ME 574: Intermediate Controls
Section 2.5.1
2.5 Examples:
EE/ME 574: Intermediate Controls
4. Plot the magnitude curve,
2.5.1 Example 1: Sketch the bode plot for K G (s)
K G (s) =
(a) Put in the low-frequency asymptote
2000 (s + 0.5)
s (s + 10) (s + 50)
K G ( j ω) = KBode
lim 2.0
(32)
s→ j 0
1 ′
( j ω 2 + 1)
1
G (s) = 2.0
sno L
j ω ( j ω 0.1 + 1) ( j ω 0.02 + 1)
(b) Put in the high-frequency asymptote
lim 2.0
s→ j ∞
(a) Poles at origin: no = 1
(b) Poles/zeros in order of ascending ωbp :
zeros: τ1 = 2.0 [seconds], (ωbp = 0.5 [rad/sec])
poles: τa = 0.1 [seconds], τb = 0.02 [seconds], ωbp = 10, 50 [rad/sec]
2. Select the frequency range
( j ω 2 + 1)
( j ω 2)
1
1
1
≃ 2.0
= 2000
j ω ( j ω 0.1 + 1) ( j ω 0.02 + 1)
j ω ( j ω 0.1) ( j ω 0.02)
( j ω)2
• Gives slope of -2, phase: −180o, and an intercept
(ω = 1) of 2000 = 66 dB.
(c) Put in the break points
(d) Proceed from low frequency to high frequency
(a) Poles and zeros:
• slowest: τ1 →
1
( j ω 2 + 1)
1
= 2.0
j ω ( j ω 0.1 + 1) ( j ω 0.02 + 1)
jω
• Gives a slope of -1, phase of −90o, and an intercept (ω = 1) of |2.0|dB
1. Put the TF in Bode form:
• fastest:
Section 2.5.1
(e) Sketch the magnitude curve
(1)
ωbp
(b)
τb → ωbp
= 0.5 [radians/second]
5. Plot the phase curve
= 50 [radians/second]
(1)
(a) Put in φ = −90o at low frequency (below 1/5 ωbp )
(b) Choose ω = [0.01 ... 1000][radians/second]
3. Determine the KBode ( j ω)
−no
(b)
(b) Put in φ = −180o at high frequency (above 5 ωbp )
term
KBode ( j ω)−no = 2.0
1
jω
(c) Put in ±90o or ±180o steps
,
no = 1
(d) Setup the width of the phase transitions
(e) Draw the phase curve
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
Page 39
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
Page 40
EE/ME 574: Intermediate Controls
Section 2.5.2
EE/ME 574: Intermediate Controls
Section 2.5.2
2.5.2 Example 2: Sketch the bode plot for a system with a complex pole pair
Gain dB
50
0
K G (s) =
−50
1. Put the TF in Bode Form:
−100 −2
10
−1
0
10
10
1
10
2
3
10
10
0
Phase [deg]
10
s (s2 + 0.4 s + 4)
K G ( j ω) = KBode
10 1
1
1 ′
G (s) =
sno L
4 j ω ( j ω)2 /4 + 0.1 j ω + 1
(a) Poles at origin: no = 1
−30
(a)
(b) Poles/zeros in ascending order: poles: ωn = 2 [radians/second]
−60
−90
Note: standard forms for a quadratic form:
−120
−150
Monic :
−180
−2
−1
10
10
0
1
10
10
Frequency [radians/second]
2
3
10
10
Bode :
Figure 26: Bode plotting example 1.
( j ω 2 + 1)
1
2000 (s + 0.5)
= 2.0
K G (s) =
s (s + 10) (s + 50)
j ω ( j ω 0.1 + 1) ( j ω 0.02 + 1)
• Break points:
(1)
ωbp = 1/τ1 = 0.5 ,
(a)
(b)
ωbp = 1/τa = 10 , ωbp = 1/τb = 50 , [rad/sec],
0 -1
Poles at :
1
s2/ω2n + 2 ρ s/ωn + 1
b1 + b0
s2 + 2 α s + (α2 + ω2 )
s1,2 = −α ± j ω
2. Select the frequency range
• The only singularity is at ωn = 2 [radians/second],
choose ω = [0.1..100] [radians/second]
• Slopes:
-1
Rectangular :
b1 + b0
s2 + 2 ρ ωn + ω2n
-2
• Phase angles:
−90o
0o
−90o
−180o \SetScrunchOne
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
Page 41
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
Page 42
EE/ME 574: Intermediate Controls
Section 2.5.2
EE/ME 574: Intermediate Controls
1
3. Determine the KBode ( j ω)
no term
1
1
= 2.5
( j ω)no
jω
20
0
Gain dB
KBode
Section 2.5.2
no = 1
,
4. Plot the magnitude curve,
−20
−40
−60
−80 −1
10
(a) Low-frequency asymptote:
• Slope of −1, Phase of −90o, intercept (ω = 1) of |2.5|dB
0
10
1
10
2
10
0
lim =
s→ j∞
Phase [deg]
(b) High-frequency asymptote:
1
10 1
1
= 10
4 j ω ( j ω)2 /4
( j ω)3
• Slope of −3, Phase of −270o, intercept (ω = 1) of |10.0|dB
−90
−180
−270
(c) Put in the break points (ωn = 2.0)
−1
0
10
1
10
10
Frequency [radians/second]
(d) Sketch the magnitude curve
2
10
Figure 27: Bode plotting example 2.
5. Plot the phase curve
• Break points:
(a) Put in φ = −90o at low frequency (below ωn )
(a)
ωn = 2.0 [rad/sec],
(b) Put in φ = −270o at high frequency (above ωn )
• Slopes:
(c) Put in −180o steps at the break point
-1
(d) Setup the width of the phase transitions
• Phase angles:
(e) Draw the phase curve
K G (s) =
-3
−90o
−270o
2000 (s + 0.5)
( j ω 2 + 1)
1
= 2.0
s (s + 10) (s + 50)
j ω ( j ω 0.1 + 1) ( j ω 0.02 + 1)
Part 2: Frequency Response Analysis
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Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
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EE/ME 574: Intermediate Controls
Section 2.6.0
2.6 Steady state errors
EE/ME 574: Intermediate Controls
• If no = 0, (Type 0 system):
• When the Loop Gain is in Bode Form, there is a straight-forward way
to determine the steady-state error for step, ramp or constant acceleration
signal.
ess (step) = lim s Tre (s) R (s) = lim s
s→0
ess (step) =
• Starting with the basic block diagram of figure 28
d(t)
r(t)
+
-
Section 2.6.0
e(t)
Kc Gc(s)
u(t)
+
1
1 + KBode
Gp(s)
ess (step) = lim s
s→0
v(t)
+
Hy(s)
ess (ramp) = lim s
s→0
Figure 28: Block Diagram of a Basic Loop.
ess (ramp) =
• The input-error TF is given as:
1
=
Tre (s) =
1 + Loop Gain 1 + KBode
(τ1 s+1) (τ2 s+1) ···
1
sno (τa s+1) (τb s+1)···(s2 /ω2n +2 ρ s /ωn +1)
(33)
• When we consider the SSE, we take the limit as s → 0
ess (step) = lim s
s→0
s→0
ess (accel) = lim s
s→0
s→0
• And in the limit, all the terms except KBode s1no go to 1/1
Part 2: Frequency Response Analysis
s→0
1
1 + KBode s1no
1
s
1
1
1
1
= lim s
= lim
=
1 + KBode s−1 s2 s→0 s + KBode s2 s→0 s + KBode KBode
1
KBode
ess (ramp) = lim s
ess = lim s Tre (s) R (s)
s→0
1
s
1
1
s
=0
= lim s
= lim
1 + KBode s−1 s s→0 s + KBode s s→0 s + KBode
• If no = 2, (Type II system):
1
ess = lim s Tre (s) R (s) = lim s
1
1
1
= lim
1 + KBode s0 s s→0 1 + KBode 1
• If no = 1, (Type I system):
y(t)
ys(t)
s→0
ess (accel) =
1
1
1
1 + KBode
s−2
s2
1
1
s2
=0
= lim s 2
= lim 2
s s→0 s + KBode s s→0 s + KBode
1
1
1
s
s2
= lim s
= lim
=0
1 + KBode s−2 s2 s→0 s2 + KBode s2 s→0 s2 + KBode
1
1 + KBode
1
s−2 s3
= lim s
s→0
s2
1
s2 + KBode s3
= lim
1
s→0 s2 + KBode
1
KBode
R (s)
(Revised: Feb 24, 2014)
Page 45
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
Page 46
2
EE/ME 574: Intermediate Controls
Section 2.6.1
• We come back to the basic SSE table
Ramp
>> KBode = bode(Kc*Gc*Gp*Hy, 0)
KBode = 3.0000
r (t) = 1, t > 0; R (s) =
1
s
1
1+KBode
0
0
– The steady-state error is given by:
r (t) = t, t > 0; R (s) =
1
s2
∞
1
KBode
0
∞
∞
>> SSEstep = 1/(1+KBode)
SSEstep = 0.25
Accel r (t) =
1 2
2t ,t
> 0; R (s) =
1
s3
1
KBode
Table 1: Connection between system type and steady state error.
• If the system is type I, the situation is a little more complex, the loop gain
goes to infinity as s → 0 . Considering
• In each case, the SSE is determined by KBode .
• In all cases, KBode determines where the low-frequency asymptote hits
ω = 1.0 .
2.6.1 Determining KBode and SSE using the Matlab bode() command
• If we setup the transfer function in Matlab, we can directly determine KBode
• Type 0 system: KBode is given by bode(Kc*GL, 0).
Kc Gc = 800
>> Gc = tf([0.1 1], [1 0]);
0.1s + 1
s
(34)
Kc = 800
– We see that the loop gain is infinity for s = 0.
>> KDC = bode(Kc*Gc*Gp*Hy, 0)
KDC = Inf
– Looking to the Bode form
– For example, considering the block diagram of figure 28 with
Kc Gc (s) = 3000
Section 2.6.1
– Now KBode is given by:
Type 0 System Type I System Type II System
Step
EE/ME 574: Intermediate Controls
′
K GL (s) = Kc Gc (s) G p (s) Hy (s) = KBode s−no GL (s)
0.1s + 1
s + 0.5
1
, Gp = 2
, Hy (s) =
0.01 s + 1
s + 60 s + 500
0.01 s + 1
with no = 1.
The Matlab code is set up with:
>>
>>
>>
>>
Gp
Hy
Gc
Kc
=
=
=
=
tf([1 0.5], [1 60 500])
tf(1, [0.01 1])
tf([0.1 1], [.01 1])
3000
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
Page 47
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
Page 48
EE/ME 574: Intermediate Controls
Section 2.6.1
– If we cancel the pole at the origin, the bode() command will give KBode .
EE/ME 574: Intermediate Controls
• If the system is type II, then the Bode form is:
>> KBode = bode( tf([1 0], 1), Kc*Gc*Gp*Hy, eps)
KBode = 0.8
K GL (s) = KBode
Note that the bode() command is evaluating
= s 800
s Kc Gc
s=ε
0.1s + 1
s + 0.5
1
s
s2 + 60 s + 500 0.01 s + 1
1 ′
G (s)
s2 L
– We can determine KBode by canceling two poles at the origin. For
example, considering
(35)
Kc Gc (s) = 800
where the s in the numerator cancels the s in the denominator. The
bode() command does not make the cancelation, so we evaluate with
ω = ε.
– For the Type I system
0.1s + 3
s + 0.5
1
, Hy (s) =
, Gp =
s
0.01 s + 1
s (s2 + 60 s + 500)
– The system is now type II, so we cancel two poles at the origin in the
bode() command
>> KBode = bode(tf([1 0 0 ], 1) * Kc*Gc*Gp*Hy, eps)
KBode = 2.4000
>> SSEaccel = 1/KBode
SSEaccel = 0.4167
>> SSEramp = 1/KBode
SSEramp = 1.2500
Part 2: Frequency Response Analysis
Section 2.6.1
(Revised: Feb 24, 2014)
Page 49
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
Page 50
EE/ME 574: Intermediate Controls
Section 2.7.0
• The loop gain is the product of transfer functions of blocks around the loop
• The Bode form is a standard way to algebraically arrange the loop gain into
the form
1 ′
(36)
K GL (s) = KBode n GL (s)
so
where no determines the system type.
• We have seen how to sketch the bode plot based on the break points of the
zeros and poles
– First order terms introduce a +1 or -1 slope and ±90o of phase shift
– Second order terms introduce a +2 or -2 slope, and
of phase shift
• Two measures of stability that are useful for analysis and design are:
– Phase Margin
– Gain Margin
• These stability measures can be read directly from the Bode plot, and also
engineered to desired values by shaping the frequency response curve.
• Engineering the phase and gain margins to desired values is a topic of
controller design, the subject of the next Part.
• Here we look at the determination of phase and gain margin from a Bode
plot, and determination of stability.
• Consider a general closed-loop TF for a block diagram like figure 29(a).
• Computers can accurately draw a Bode plot graph
– The importance of sketching lies in the understanding of how changes to
the controller design effect system response.
• In the next section we will see how system stability and transient
characteristics can be determined from the Bode plot.
• Hendrik Wade Bode, born 1905 in Madison, Wisconsin, died 1982. H. Bode
made several including
– Bode plots, 1938
Try ( j ω) =
Forward Path Gain
1 + K GL ( j ω)
(37)
• Figure 29(b) presents a simplified loop, which will have the same stability
as that of figure 29(a).
r(t)
+
e(t)
u(t)
Kc Gc(s)
y(t)
r(t)
+
Gp(s)
-
e(t)
ys(t)
K GL(s)
ys(t)
Hy(s)
– 1944, became director of the Mathematical Research Group at Bell
Laboratories
– In 1948, President Harry S. Truman awarded Bode the President’s
Certificate of Merit, in recognition of his remarkable scientific
contributions
Part 2: Frequency Response Analysis
Section 3.0.0
3 Stability Considerations, Phase and Gain Margins
2.7 Summary of Frequency Response and the Bode plot
±180o
EE/ME 574: Intermediate Controls
(Revised: Feb 24, 2014)
Page 51
(a) Control loop
(b) Simplified control loop with loop gain
Figure 29: Control loop with compensator, plant and sensor.
• Using the phase and gain margins (and one additional test), we can
determine the stability of control loop such as figure 29 (a) or (b).
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
Page 52
EE/ME 574: Intermediate Controls
Section 3.2.0
3.1 Marginal Stability
EE/ME 574: Intermediate Controls
Section 3.3.0
3.3 Examples, stability, marginal stability and instability
• A system is marginally stable when poles lie on the s = j ω axis.
• Consider the example system of figure 30, with GL (s) as given.
• Considering Eqn (37), marginal stability occurs when one or more values of
ω give:
(38)
1 + K GL ( j ω) = 0
r(t)
+
-
e(t)
u(t)
K
ys(t)
GL (s) =
GL(s)
s+4
(42)
s (s2 + 3 s + 2)
• The marginal stability condition, Eqn (38) can be re-written:
K GL ( j ω) = 1.0∠ − 180o
Figure 30: Control loop with controller gain Kc and loop gain GL .
(39)
Step Response
Step Response
Amplitude
|K GL ( j ω)| = 1.0 (0 dB)
∠K GL ( j ω) = −180o (±n 360o)
Amplitude
1.5
1
0
0
5
Time (sec)
10
(a) K = 0.25
3.2 Definitions for phase and gain margin
1.5
50
1
Cases:
(a) K = 0.25 : stable
(b) K = 6.00 :
marginally stable
(c) K = 144 : unstable
0
0.5
−50
0
−100
0
5
Time (sec)
10
(b) K = 6.00
0
5
Time (sec)
10
(c) K = 144
Figure 31: Step response of system for three values of K.
• Cross-over frequency, ωc : Frequency (-ies) at which the magnitude curve
crosses the 0 dB line.
• Phase cross-over frequency, ωcp : Frequency (-ies) at which the phase curve
crosses over the −180o line
• Phase Margin [degrees]: Degree by which the phase curve misses −180o at
ω = ωc
PM = φ (ωc ) − (−180o) = φ (ωc ) + 180o
(40)
• Gain Margin [dB]: Degree by which the magnitude curve misses 0 dB at
ω = ωcp
GM = −20 log10 (K GL ( j ωcp )) dB
(41)
(Revised: Feb 24, 2014)
100
0.5
• If a value of ω exists satisfying Eqn (39), the system is marginally stable.
Part 2: Frequency Response Analysis
Step Response
2
Amplitude
Which gives:
Page 53
• Depending on the value of K, the system can be stable, marginally stable or
unstable.
• Clearly it is very desirable to know the critical value of K , K = Kcrit , at
which the system transitions from stability to instability.
• Knowing that
K GL (s) = 1∠ − 180o
when K = Kcrit is the basis for determining stability from the Bode plot.
• The gain margin and phase margin are simply the distance (in gain and
phase) by which the K GL (s) curve misses the critical point.
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
Page 54
EE/ME 574: Intermediate Controls
Section 3.3.0
• Looking at the Bode plot for the example, with K = 0.25 (stable).
Critical Point, |G|=0dB
ω =2.83 [r/s]
c
Gm = 27.6 dB (at 2.83 rad/sec) , Pm = 59.6 deg (at 0.448 rad/sec)
%% Build the model, GL (s)
>> GL = tf([1 4], [1 3 2 0])
−50
−100 −1
10
0
1
10
10
%% Plot the margins
>> margin(0.25*GL)
2
10
50
Magnitude (dB)
Mag K G_{L1}
Section 3.3.0
• Matlab’s margin() command draws the bode plot and marks the gain and
phase margins.
Bode Diagram
50
0
EE/ME 574: Intermediate Controls
0
−50
0
−90
−120
Critical Point, <G=−180,
ω =0.45 [rad/sec]
−150
cp
−180
−1
10
0
1
10
10
Frequency [radians/second]
2
10
−100
−90
−135
−180
−225
−2
10
Figure 32: Bode plot of GL (s), corresponding to K = 1.0 .
−1
10
0
1
10
Frequency (rad/sec)
2
10
10
Figure 33: Output of the margin command, showing gain and phase margins.
• Figure 32 shows two special values of ω,
ωc : Cross-over frequency, where the magnitude |GL ( j ωc )| = 1.0 .
ωcp : Phase cross-over frequency, where ∠GL ( j ωcp ) = −180o
• With K = 0.25, ωc < ωcp , and K GL (s) does not hit the critical value of
K GL (s) = 1∠ − 180o.
• Evaluating K GL (s) at the cross-over and phase cross-over frequencies:
• Now consider all three levels of gain.
magnitude curve, as seen in figure 34.
Increasing K moves only the
100
Changed: ωc and gain and phase
margins
Kc= 0.25
Kc= 6.00
50
Mag K GL1
Phase [deg]
−60
Phase (deg)
Results:
ωc = 0.448 [rad/sec]
ωcp = 2.83 [rad/sec]
PM = 59.6 deg
GM = 27.6 dB
−30
Kc=144.00
0
wc
−50
Unchanged: ωcp (where the
gain margin is measured)
−100 −1
10
0
1
10
2
10
10
0
K GL (s = j ωc ) = 1.0∠ − 120.4o
Phase [deg]
−30
K GL (s = j ωcp ) = 0.0417∠ − 180o
−60
−90
−120
−150
– At ωc , the phase lag is not sufficient for marginal stability.
−1
10
– At ωcp , the gain is not sufficient for marginal stability.
(Revised: Feb 24, 2014)
0
1
2
10
10
Frequency [radians/second]
10
Figure 34: Bode plot for three levels of gain.
−→ This system is not marginally stable.
Part 2: Frequency Response Analysis
wcp
−180
Page 55
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
Page 56
EE/ME 574: Intermediate Controls
Section 3.3.0
EE/ME 574: Intermediate Controls
Section 3.3.1
3.3.1 A challenge with applying gain and phase margin
Bode Diagram
Gm = 0.000207 dB (at 2.83 rad/sec) , Pm = 0.000681 deg (at 2.83 rad/sec)
100
Results:
ωc = 2.83 [rad/sec]
ωcp = 2.83 [rad/sec]
PM = 0 deg
GM = 0 dB
Magnitude (dB)
System is marginally stable.
• The GM and PM critera give us an exact definition of marginal stability,
but do not indicate whether the system is stable for increasing or decreasing
gain.
50
0
−50
• Define the level of gain that sets cross-over where φ = −180o (or,
equivalently choose K to give ωc = ωcp , and GM=0 dB, PM=0o )
−100
−90
Phase (deg)
%% Plot the margins
>> margin(6.00*GL)
– This critical value of K can be called Kcrit
−135
• Most control loops are low pass filters, with Bode plots that look similar to
figures 32-36, above.
−180
−225
−2
10
−1
10
0
10
Frequency (rad/sec)
1
10
2
10
– But not all.
Figure 35: Output of the margin command, K = 6.00
• For typical systems, the system is stable when K < Kcrit , giving the pattern
of figure 37.
Bode Diagram
Gm = −27.6 dB (at 2.83 rad/sec) , Pm = −4.15 deg (at 12.2 rad/sec)
100
Results:
ωc = 12.2 [rad/sec]
ωcp = 2.83 [rad/sec]
PM = -4.15 deg
GM = -27.6 dB
Magnitude (dB)
System is unstable.
stable
unstable
50
K=0
0
−135
K increasing
• In cases with the pattern of figure 37, the system is stable when
∠K GL ( j ωc ) > −180o and K GL ( j ωcp ) < 0 dB .
−180
−225
−2
10
Kcrit
Figure 37: Stability and instability versus K. Marginal stability when K = Kcrit
−50
−90
Phase (deg)
%% Plot the margins
>> margin(144*GL)
−1
10
0
10
Frequency (rad/sec)
1
10
2
10
– Giving the standard definitions of phase and gain margin:
GM = − K GL ( j ωcp )dB
Figure 36: Output of the margin command, K = 144.
PM = 180 + φ (ωc )
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
Page 57
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
Page 58
EE/ME 574: Intermediate Controls
Section 3.3.1
• However, other patterns are possible, as seen in figure 38.
unstable
EE/ME 574: Intermediate Controls
Section 3.3.2
3.3.2 Example with the pattern of figure 38(a)
• Consider a system described by:
stable
GL (s) =
K=0
Kcrit
(a)
K increasing
Step Response
Step Response
50
Kcrit1
(43)
Step Response
1.5
1.5
Kcrit2 K increasing
0
Amplitude
unstable
Amplitude
K=0
(b)
stable
Amplitude
unstable
s2 + 8 s + 16
s3
1
0.5
Figure 38: Stability and instability versus K, alternative possible patterns.
−50
0
5
Time (sec)
10
(a) K = 1.00
• When using the Bode stability criterion, we need to check that the resulting
closed-loop system is stable. We can check stability by examining:
0
Cases:
(a) K = 1.00 : unstable
(b) K = 10.0 : stable
(c) K = 100 : stable
1
0.5
0
1
Time (sec)
2
(b) K = 10.0
0
0
0.2
0.4
Time (sec)
(c) K = 100
Figure 39: Step response of system for three values of K.
Bode Diagram
Gm = 6.02 dB (at 4 rad/sec) , Pm = −18.1 deg (at 2.9 rad/sec)
2. The step response
%% Plot the margins
>> margin(1.00*GL)
3. A root-locus plot
• When the regions of stability follow an atypical pattern, the Nyquist stability
criterion can be used.
Results:
ωc = 2.90 [rad/sec]
ωcp = 4.00 [rad/sec]
PM = -18.1 deg
GM = 6.02 dB
100
Magnitude (dB)
%% Build the model, GL (s)
>> GL = tf([1 8 16], [1 0 0 0])
50
0
−50
−90
Phase (deg)
1. The closed-loop poles
−180
−270
−1
10
0
1
10
2
10
10
Frequency (rad/sec)
Figure 40: Output of the margin command, showing gain and phase margins,
system of Eqn (43).
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
Page 59
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
Page 60
Section 3.3.2
EE/ME 574: Intermediate Controls
Section 3.4.0
3.4 Interpreting Phase and Gain Margin
3.3.2 Example with the pattern of figure 38(a) (continued)
• The example above shows a case where the phase and gain margins are
difficult to interpret.
• For typical systems, a positive PM and GM indicate a stable system.
• For example, for the TF given we find the bode plot and margins of fig. 41.
Bode Diagram
Gm = −4.27 dB (at 1.43 rad/sec) , Pm = −8.28 deg (at 1.64 rad/sec)
• Cases where the open-loop system is unstable (right-half-plane poles) or
non-minimum phase (right-half-plane zeros) often give rise to complex
frequency response patterns.
• The Nyquist stability criterion (Franklin et al., section 6.3) is a frequency
domain tool capable of handling with complex Bode plots.
20
10
GL (s) =
s+3
s3 + 1.4 s2 + 1.44 s + 1.04
Magnitude (dB)
EE/ME 574: Intermediate Controls
0
−10
−20
−30
−40
0
• We will not be studying the Nyquist stability criterion in detail.
Phase (deg)
−45
– It is complex
– With modern computer tools such as Matlab, stability questions can be
answered other ways, the Nyquist stability criterion is seldom used.
– The Nyquist stability criterion is important for robust control, such as
H∞ control. Students who take an advanced course in robust control will
encounter the Nyquist stability criterion.
• Bode phase and gain margin remain valuable tools
−90
−135
−180
−225
−1
10
0
1
10
Frequency (rad/sec)
10
Figure 41: Bode plot and margins for GL (s).
• Interpretations that can be made from the Bode plot and margins are:
– The overall shape of the magnitude and phase curves indicate the
qualitative properties of the system.
– They apply in a straight forward way to most systems
– The gain margin indicates change in gain required before the system
becomes stable/unstable (an aspect of robustness).
– Phase and gain margin give insight into how to engineer controllers to
achieve performance goals
– The phase margin indicates
– Bode stability results should be validated by an alternative method:
1. Determine the poles
2. Examine the step response of the final system, or as a function of K ,
as illustrated in figures 31 and 39
3. Examine the root locus
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
Page 61
∗ The damping, we’ll see how to determine the approximate overshoot
from from the PM
∗ The amount of additional phase lag the control loop can tolerate
before becoming unstable (an aspect of robustness).
– In the next part, we will learn a number of additional control system
performance measures that can be determined from the Bode plot.
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
Page 62
EE/ME 574: Intermediate Controls
Section 4.0.0
4 Summary: Analysis by Frequency Response
• For linear systems, a sinusoid input gives an sinusoid output.
– The transfer function can be interpreted as a complex-valued gain:
~Y = T (s = j ω) U
~
~ is the input phasor and T (s = j ω).
where ~Y is the output phasor, U
• By evaluating the transfer function as a function of ω, we can make the Bode
plot.
• The Bode plot can also be sketched by hand, by putting together the break
points for each pole and zero.
• The bode plot indicates overall characteristics of the system, as well as
specific features that indicate the stability of the system.
– Cross-over frequency, ωc
– Phase cross-over frequency, ωcp
– Phase margin, PM
– Gain margin, GM
• Since the standard interpretation of PM and GM does not apply to all
systems, we need to validate results by checking the poles, step response
or root locus of the system.
• Bode analysis is a valuable tool, used in conjunction with analysis of the
poles, root locus and other tools.
Part 2: Frequency Response Analysis
(Revised: Feb 24, 2014)
Page 63
EE/ME 574: Intermediate Controls
EE/ME 574: Intermediate Controls
Designing Control by Frequency-Response
2.2.2
2.2.3
Contents
2.3
1 Performance measures
1.1
1.2
1.3
1.4
1.5
3
Phase-lead design example . . . . . . . . . . . . . . . . .
37
PI compensator
. . . . . . . . . . . . . . . . . . . . . . . . . .
42
Basics of PI design . . . . . . . . . . . . . . . . . . . . .
42
Phase-Lag compensator . . . . . . . . . . . . . . . . . . . . . . .
43
2.3.1
4
1.1.1
Bandwidth (ωb) . . . . . . . . . . . . . . . . . . . . . .
5
2.4.1
Basics of phase-lag design . . . . . . . . . . . . . . . . .
45
1.1.2
Cross-over frequency (ωc ) . . . . . . . . . . . . . . . . .
7
2.4.2
Stability related performance measures . . . . . . . . . . . . . . .
8
Phase-lag compensation design procedure (see Franklin,
section 6.7.4) . . . . . . . . . . . . . . . . . . . . . . . .
46
1.2.1
Gain Margin, GM and Phase Margin, PM . . . . . . . . .
8
Phase-lag design example #1
. . . . . . . . . . . . . . .
47
Accuracy related performance measures . . . . . . . . . . . . . .
11
PID compensation . . . . . . . . . . . . . . . . . . . . . . . . . .
59
1.3.1
Steady State Error . . . . . . . . . . . . . . . . . . . . .
11
2.5.1
Basics of PID design . . . . . . . . . . . . . . . . . . . .
60
1.3.2
Disturbance rejection, magnitude Edist at frequency ωd . .
12
2.5.2
Design procedure . . . . . . . . . . . . . . . . . . . . . .
60
1.3.3
Tracking Error, Etrack at frequency ωt . . . . . . . . . . .
17
2.5.3
1.3.4
Noise Rejection, Ngain at frequency ωs . . . . . . . . . . .
20
Ziegler-Nichols Method 1: Quarter-Cycle Damping
Method . . . . . . . . . . . . . . . . . . . . . . . . . . .
61
Connection between ωc , ωb and tr . . . . . . . . . . . . . . . . .
22
2.5.4
Ziegler-Nichols Method 2: Ultimate Sensitivity Method .
64
1.4.1
23
2.5.5
Tyreus-Luyben Tuning Chart (“TLC” tuning rules) . . . .
67
2.5.6
PID design by frequency response . . . . . . . . . . . . .
68
2.5.7
Realizable form for PD and PID controllers . . . . . . . .
71
Lead-Lag compensation . . . . . . . . . . . . . . . . . . . . . .
73
2.6.1
74
Table of approximate dominant-mode relationships . . . .
Summary of performance measures and interpretation on the Bode
plot . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4
2.4.3
2.5
24
2.6
PD compensator
30
. . . . . . . . . . . . . . . . . . . . . . . . . .
31
Basics of PD design . . . . . . . . . . . . . . . . . . . .
31
Phase-lead compensator . . . . . . . . . . . . . . . . . . . . . .
32
2.2.1
33
2.1.1
2.2
35
Speed related performance measures . . . . . . . . . . . . . . . .
2 Design of frequency compensation
2.1
Phase-lead compensation design procedure (Franklin,
section 6.7.2) . . . . . . . . . . . . . . . . . . . . . . . .
Basics of phase-lead design . . . . . . . . . . . . . . . .
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 1
Basics of Lead-Lag compensation . . . . . . . . . . . . .
3 Summary of frequency-response based
compensator design
Part 3: Frequency Response Design
75
(Revised: Mar 26, 2013)
Page 2
EE/ME 574: Intermediate Controls
Section 1.0.0
1 Performance measures
• Several performance measures are best understood based on the frequency
response.
• The complete list of performance measures of interest is repeated in table 1.
• The items based on frequency analysis are highlighted, and will be defined
with the aid of examples.
From table 1, there are seven performance measures typically evaluated from the
bode plot. We will look at the seven using an example.
• The example system is our basic control loop, including a reference input,
disturbance and sensor noise.
disturbance
r(t)
Symbol
Domain
Evaluated From
Rise Time
tr
Time
Step Response or Dominant Mode
Peak Time
tp
Time
Step Response or Dominant Mode
Settling Time
ts
Time
Step Response or Dominant Mode
Pole Locations
pi
Freq
Transfer Function, S-plane
Bandwidth
ωb
Freq
Closed-loop Freq. Resp.
Cross-over Frequency
ωc
Freq
Frequency Response
TF
Transfer Function
Overshoot
Mp
Time
Step Response or Dominant Mode
Damping Factor
ρ
Freq
Dominant Mode
Pole Locations
pi
Freq
Transfer Function, S-plane
Phase Margin
PM
Freq
Frequency Response
Gain Margin
GM
Freq
Frequency Response
+
e(t)
Controller
u1(t)
KcGc(s)
-
Speed:
+
d(t)
+
Plant
u2(t)
y(t)
Gp(s)
+
Sensor Dynamics
ys(t)
Hy(s)
sensor
noise
vs(t)
+
Figure 1: Basic control loop.
• The example transfer function is:
Degree of Stability:
Stable / Unstable
Section 1.1.0
1.1 Speed related performance measures
• Frequency response gives us a distinct way to understand system behavior
Performance Measure
EE/ME 574: Intermediate Controls
G p1 (s)
in Bode form:
Accuracy:
Steady State Error
SSE
Time
Step/Ramp Response or Trans. Fun.
Integrated Squared Error
ISE
Time
Step Response
Integrated Absolute Error
IAE
Time
Step Response
Disturbance Rejection
Edist , ωd
Freq
Freq Resp, or closed-loop Freq Resp
Tracking Error
Etrack , ωt
Freq
Freq Resp, or closed-loop Freq Resp
Noise Rejection
Ngain , ωs
Freq
Freq Resp, or closed-loop Freq Resp
s + 10
s + 10
=
s (s2 + 5 s + 4) s (s + 1) (s + 4)
1
10 −1
10 s + 1
s
G p1 (s) =
4
(s + 1) 14 s + 1
• For our examples will use
Hy (s) = 1.0
(1)
(2)
(3)
which is included for completeness.
Table 1: A list of standard performance measures and how they are evaluated.
Highlighted performance measures are based on frequency analysis.
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 3
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 4
EE/ME 574: Intermediate Controls
Section 1.1.1
EE/ME 574: Intermediate Controls
• The Bode plot of G p1 (s) is seen in figure 2.
Section 1.1.2
– Example of bandwidth, taking Kc Gc (s) = 1.0
20
40
0
Gain dB
Gain dB
20
0
−20
−40
−40
−60
−60
−80 −1
10
−80 −1
10
0
1
10
Try (s) =
Determines Bandwidth
−20
0
10
ωb
1
10
2
10
2
10
10
−30
−30
−60
Phase [deg]
Phase [deg]
0
0
−60
−90
−150
−150
−180
−1
−180
10
−1
0
10
1
2
10
10
Frequency [radians/second]
10
>> Gp = tf([1 10], [1 5 4]);
>> Gc = tf(1, 1); Hy = tf(1,1)
>> Try = ...
feedback(Gc*Gp*Hy, 1);
>> bode(Try);
>> Wb = bandwidth(Try);
ωb = 2.12 [rad/sec]
−90
−120
−120
Kc Gc G p
1 + Kc Gc G p Hy
(5)
0
1
10
10
Frequency [radians/second]
2
10
Figure 3: Frequency-response plot of Try (s) (closed loop).
Figure 2: Bode plot of G p1 (s) (open loop).
– Notice the difference between the frequency-response plot of figures 2
and 3.
1.1.1 Bandwidth (ωb )
• The first performance measure is actually measured from the closed-loop
frequency-response plot.
1. At low frequency, the loop gain, G p1 (s), is large with ∠ − 90o , but
the closed-loop TF, Try (s), is about 1.0 with ∠0o
2. At high frequencies the two curves are very comparable
ωb is the lowest frequency at which the magnitude of the closed-loop
transfer function equals ( √1 ) Try (s = j 0), where Try (s = j 0) is the
2
DC gain of the closed-loop system .
Try ( j ωb ) =
1
√
2
Try (0)
(4)
Equivalently: the bandwidth is the frequency at which the magnitude
of the closed-loop frequency-response plot as dropped of by 3 dB.
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 5
– These two statements reveal basic facts about feedback control:
∗ Within the bandwidth of the closed-loop system, feedback gives the
system very nice behavior (Try (s) ≃ 1.0∠0o)
∗ Outside the bandwidth, feedback does not significantly change the
response
• Bandwidth, ωb , is commonly used to specify the speed of a system.
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 6
EE/ME 574: Intermediate Controls
Section 1.1.2
EE/ME 574: Intermediate Controls
Section 1.2.1
1.1.2 Cross-over frequency (ωc )
1.2 Stability related performance measures
We have already seen cross-over frequency when considering the phase margin.
1.2.1 Gain Margin, GM and Phase Margin, PM
• ωc is the frequency at which the Bode magnitude curve first crosses 0 dB.
We have already seen phase and gain margin. These are determined
on the Bode plot, which is to say the frequency-response plot of the
loop gain (open loop).
– If the magnitude curve crosses 0 dB several times,
∗ For phase margin, they must all be considered
Gain dB
20
ωc
0
−40
20
−60
−80 −1
10
0
Cross−over
ωc = 1.39 [rad/sec]
Gain Margin 12.0 dB
−20
40
0
1
10
10
2
10
−20
−40
0
ωc = 1.39 [rad/sec]
−60
−80 −1
10
0
10
1
10
2
10
0
−30
−60
ωcp = 2.83 [rad/sec]
−90
−120
Phase Margin 24.4 deg
−150
−30
Phase [deg]
40
Phase [deg]
Gain dB
∗ For performance specification, the lowest frequency of crossing is
taken as ωc
ω
−180
−60
−1
10
−90
cp
0
1
10
10
Frequency [radians/second]
2
10
−120
Figure 5: Bode plot of G p1 (s) (open loop).
−150
−180
−1
10
0
1
10
10
Frequency [radians/second]
2
10
Note: Generally (but not always) ωc < ωcp for a stable system.
Figure 4: Bode plot of G p1 (s) (open loop).
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 7
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 8
EE/ME 574: Intermediate Controls
Section 1.2.1
• The phase margin is also connected to closed-loop damping. Consider the
2nd order system (see Franklin 5th ed. Eqn (6.31) and figures 6.34-6.36)
ω2n
s (s + 2 ρ ωn)
ω2n
2
s + 2 ρ ωn s + ω2n
2ρ
1 + 4 ρ4 − 2ρ
(8)


• By evaluating the derivative of Eqn (8) an the approximate relation between
PM and damping is seen to be:
PM
(9)
ρ≈
100
1
0.9
(10)
where 100 × M p is the percent overshoot.
(7)
For this particular case, the phase margin can be solved explicitly and comes
out to be:



PM = tan−1  qp
Mp = e
100
90
p
Try (s) =
− √ πρ
(1−ρ2 )
(6)
• When we set up G p2 (s) with the feedback Hy (s) = 1.0, Kc Gc (s) = 1.0 , we
get the closed-loop transfer function
Section 1.2.1
• Another view that shows the relationship between phase margin and
damping is given by evaluating
Percent Overshoot, 100*M
G p2 (s) =
EE/ME 574: Intermediate Controls
80
70
60
50
40
30
20
10
0
0
10
20
30
40
50
60
Phase Margin [deg]
70
80
90
Figure 7: Phase margin vs percent overshoot, for a system with 2 poles and no
zeros.
ρ vs PM
Approximation
0.8
• Note that Eqns (8) - (10) and figures 6 and 7 are all based on the 2-pole
no-zero system.
Damping ratio, ρ
0.7
0.6
0.5
• These approximations show that damping and overshoot are strongly linked
to phase margin
0.4
0.3
0.2
– PM and damping factor are both approximate indicators of overshoot,
0.1
0
0
10
20
30
40
50
Phase Margin [deg]
60
70
– We must check the actual step response to assure that an overshoot
specification is met.
80
Figure 6: Damping ratio vs. phase margin, for a system with 2 poles and no zeros.
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 9
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 10
EE/ME 574: Intermediate Controls
Section 1.3.1
EE/ME 574: Intermediate Controls
Section 1.3.2
1.3 Accuracy related performance measures
1.3.2 Disturbance rejection, magnitude Edist at frequency ωd
1.3.1 Steady State Error
Many closed-loop systems operate with disturbances. For these systems, it is
often necessary to specify the maximum influence the disturbance can have on
the output. Defining the disturbance transfer function as
The steady state error is determined by the system type and KBode .
We can read both directly off the Bode plot, as illustrated in fig 8.
• We see a slope of -1 in low-frequency range, so the system is type I
• The DC gain is 8.0 dB (2.5)
Therefore, the steady state error is:
(11)
The magnitude of the disturbance transfer function Tdy (s) will
not be greater than Edist at any frequency at or below ωd . Or,
mathematically
SSE (step) = 0
1
= 0.4
2.5
SSE(ramp) =
G p (s)
Y (s)
=
D (s) 1 + Kc Gc (s) GP (s) Hy (s)
A disturbance rejection specification of Edist at frequency ωd specifies:
– KBode is where the low freq. asymptote hits ω = 1.0 [rad/sec]
Tdy ( j ω) ≤ Edist
∀ ω ≤ ωd
(12)
For example, if we combine PID control with G p1 (s),
40
20
Gain dB
Tdy (s) =
DC Gain
Low−fre
q asym
0
Kc Gc (s) =
ptote
kd s2 + k p s + kI
s2 + 2.4 s + 2.44
= 1.5
s
s
(13)
−20
−40 −1
10
0
The transfer function is given as
1
10
10
0
Try (s) =
Phase [deg]
−30
−60
Kc Gc G p
1.5 s3 + 18.6 s2 + 39.66 s + 36.6
=
1 + Kc Gc GP Hy s4 + 6.5 s3 + 22.6 s2 + 39.66 s + 36.6
−90
And the disturbance transfer function is given as:
−120
−150
−180
−1
10
0
10
Frequency [radians/second]
1
Tdy (s) =
10
Gp
s2 + 10 s
= 4
3
1 + Kc Gc GP Hy s + 6.5 s + 22.6 s2 + 39.66 s + 36.6
(14)
Figure 8: Bode plot of G p1 (s) , showing low-frequency asymptote and DC gain.
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 11
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 12
EE/ME 574: Intermediate Controls
Section 1.3.2
1.3.2 Disturbance rejection, magnitude Edist at frequency ωd (continued)
EE/ME 574: Intermediate Controls
• Figure 10 shows that the system also meets a specification of
• For example, Tdy (s) is plotted in figure 9. The frequency-response plot
shows that the closed-loop system meets a requirement of:
Edist = −20 dB ,
Section 1.3.2
Edist = −25 dB ,
ωd = 0.2 [rad/sec]
– Note that only the magnitude curve is needed to evaluate disturbance
response.
ωd = 0.3 [rad/sec]
>> Tdy = feedback(Gp, -Hy*Kc*Gc)
>> bode(Tdy)
0
Gain dB
−20
0
Dist. Rej.
Gain dB
−20
−40
−60
−80 −1
10
−40
0
1
10
10
Frequency [radians/second]
2
10
−60
−80 −1
10
Figure 10: Disturbance transfer function, Tdy (s)
0
10
1
10
2
10
• Figure 11 shows that the system does not meet a specification of
Phase [deg]
90
0
Edist = −25 dB ,
−90
ωd = 0.3 [rad/sec]
−180
0
1
10
10
Frequency [radians/second]
0
2
10
−20
Gain dB
−1
10
Figure 9: Disturbance transfer function, Tdy (s)
−40
−60
−80 −1
10
0
1
10
10
Frequency [radians/second]
2
10
Figure 11: Disturbance transfer function, Tdy (s)
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 13
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 14
EE/ME 574: Intermediate Controls
Section 1.3.2
• Considering the two transfer functions:
Disturbance :
Gp
=
1 + Kc Gc GP Hy
Tdy (s) =
1
Try (s)
Kc Gc (s)
and
Try (s) ≃ 1.0∠0o ω << ωb
– Additionally, ωd << ωb , since, very reasonably, the closed loop can
only compensate for disturbances at frequencies well below the closedloop bandwidth.
we see that the disturbance transfer function is given as
1
Try (s)
Kc Gc (s)
(15)
– And finally, in most cases, the controller has relatively high gain at low
frequencies (this improves accuracy). The Bode plot of Kc Gc (s) is seen
in figure 13.
40
• Eqn (15) can be understood by considering the general characteristics of
closed-loop systems.
30
Gain dB
Input − Output :
Tdy (s) =
Section 1.3.2
– Since
Kc Gc G p
Try (s) =
1 + Kc Gc G p Hy
Tdy(s)
EE/ME 574: Intermediate Controls
20
10
– In most cases Try (s) ≃ 1.0 ∠0o for ω << ωb . For example, Try (s) for the
plant of figure 3 and controller of Eqn (13) is seen in figure 12
0 −1
10
20
0
1
10
10
90
Phase [deg]
Gain dB
60
0
−20
30
0
−30
−60
−90
−40 −1
10
0
10
1
10
−1
10
2
10
1
10
Figure 13: Gain of the controller Kc Gc (s) , Eqn (13), for a PID controller.
0
Phase [deg]
0
10
Frequency [radians/second]
−30
• Putting these facts together, we find:
−60
−90
−120
−150 −1
10
0
1
10
10
Frequency [radians/second]
Tdy (s) ≃
2
10
Figure 12: Closed-loop transfer function, Try (s) , plant with controller (13).
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 15
1
Kc Gc (s) s= j ω , ω<ωb
(16)
• For good disturbance rejection, we want a large controller gain !
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 16
EE/ME 574: Intermediate Controls
Section 1.3.3
1.3.3 Tracking Error, Etrack at frequency ωt
• A tracking error specification refers to accurately tracking input signals of
any frequency up to a specified frequency.
0 ≤ ω ≤ ωt ,
for
Section 1.3.3
• The example system (figure 14) meets the requirement
(note: 1% = -40 dB)
Etrack = 1%, ωt = 0.3 [radians/second]
This means that for any input signal
r (t) = A cos (ωt + φ)
100 × |Tre ( j ω)| ≤ Etrack %
the output will track the input with an error of less than 1%, for ω < ωt .
• For the basic system of figure 1, the error response transfer function is:
Tre (s) =
EE/ME 574: Intermediate Controls
1
E ( j ω)
=
R ( j ω) 1 + Kc Gc G p Hy
• Accurate tracking requires
(17)
Tre (s) =
• We can plot the frequency-response of Tre (s). For example,
1
<< 1
1 + Kc Gc G p Hy
so
– The example system (Eqns (1) and (13)) gives the tracking error
frequency response of figure 14.
Kc Gc G p Hy = K GL (s) >> 1 for ω ≤ ωt
• When |K GL (s)| >> 1 , tracking error can be approximated by:
Gain dB
0
Meets:
Etrack ≤ 1% ,
ωt = 0.3 [r/s]
−20
−40
0
10
1
10
2
10
1
K GL (s)
(19)
• For example,
0
Phase [deg]
Tre (s) ≈
so even though tracking is a closed-loop property, we can read the
approximate tracking error directly off the Bode plot.
−60
−80 −1
10
(18)
−90
– For a tracking error ≤ 2%,
−180
– We require |K GL ( j ωt )|≥ 50. And 50x = 34 dB
−270
−360
−1
10
0
1
10
10
Frequency [radians/second]
– To track a signal with less than 2% error, the loop gain must be greater
than 34 dB at ω ≤ ωt
2
10
Figure 14: Error response transfer function for the example system (closed-loop).
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 17
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 18
EE/ME 574: Intermediate Controls
Section 1.3.3
• The bode plot for the example system with PID controller (Eqns (1) and
(13)) is shown in figure 16.
– The system can meet tracking requirement up to ωt = 0.4 [rad/sec]
EE/ME 574: Intermediate Controls
Section 1.3.4
1.3.4 Noise Rejection, Ngain at frequency ωs
• Consider the noise response transfer function:
Tvy (s) =
Bode plot, Gp1, with PID control
−Kc Gc G p Hy
−K GL (s)
=
1 + Kc Gc G p Hy 1 + K GL (s)
(20)
60
20
0
−20
−40 −1
10
• We would like the noise contribution to the output to be small
ωc
Tracking
Error
Gain dB
40
ωt
0
10
• At low frequency, ω < ωb , where |K GL (s)| >> 1
1
10
Tvy ≃ 1.0∠ − 180o
2
10
(21)
0
there is nothing we can do (except buy a better sensor)
Phase [deg]
−30
−60
• At high frequency, ω >> ωb , we can have low sensor-noise sensitivity by
having
|K GL (s)| << 1 for
ω >> ωb
−90
−120
−150
−180
−1
10
0
1
10
10
Frequency [radians/second]
2
10
which gives
Tvy (s) ≃ −K GL (s)
Figure 15: Loop gain (Bode plot) for the example system, Eqn (1) with example
PID control, Eqn (13).
• Some sensors (e.g., a tachometer) may have considerable noise at
frequencies higher than the closed-loop bandwidth of the system
– So noise rejection in the control loop is important
• A noise-gain specification requires that the closed-loop gain of the system
be below some level Ngain for all frequencies above a specified frequency
ωs .
for
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 19
ω ≥ ωs ,
Part 3: Frequency Response Design
Tvy ( j ω) ≤ Ngain
(22)
(Revised: Mar 26, 2013)
Page 20
EE/ME 574: Intermediate Controls
Section 1.3.4
• We can form the closed-loop TF Tvy (s) and plot the frequency response.
• For design, it is more useful to consider the Bode plot.
2. Bandwidth, ωb
3. Rise time, tr
– For ω >> ωb , the noise gain is approximately the loop gain.
• For design, it will be useful to consider the connection, so we can see how
to adjust one in order to achieve performance goals in another
60
• The cross-over frequency is defined by a property of the loop gain
Noise Rejection
Gain dB
40
ωc
0
1.4 Connection between ωc , ωb and tr
1. Cross-over frequency, ωc
−K GL (s)
−K GL (s)
≃
= −K GL (s)
1 + K GL (s)
1
20
Section 1.4.0
• These speed characteristics are closely linked:
– Since |K GL (s)| << 1 in the region of interest
Tvy (s) =
EE/ME 574: Intermediate Controls
−20
ωc : |K GL ( j ωc )| = 1.0
(24)
Tnoise
−40 −1
10
0
ω
1
10
s
10
2
10
• The bandwidth is defined by a property of the closed-loop transfer function
0
Kc Gc ( j ω) G p ( j ω)
1 + K GL ( j ω)
(25)
√
• Eqn (25) holds at a frequency where K GL ( j ω) ≈ 1/ 2 (not exactly,
because of the complex values in the denominator)
Phase [deg]
−30
1
ωb : Try ( j ωb ) = √ Try ( j 0) , with
2
−60
−90
−120
−150
−180
−1
10
0
1
2
10
10
Frequency [radians/second]
10
Figure 16: Loop gain for the example system, Eqn (1) with example PID control,
Eqn (13). Noise rejection plotted.
• We meet the sensor noise specification if:
∀ ω ≥ ωs
Part 3: Frequency Response Design
Try ( j ω) =
• Since the magnitude curve typically trends downward in the region of crossover, with a slope between -1 and -1.5 , we find that
– The bandwidth will always be somewhat greater than the cross-over
frequency
Tvy (s) ≤ Ngain
(Revised: Mar 26, 2013)
(23)
Page 21
– We can often use the approximation
√
ωb ≈ 2 ωc
Part 3: Frequency Response Design
(26)
(Revised: Mar 26, 2013)
Page 22
EE/ME 574: Intermediate Controls
Section 1.4.1
EE/ME 574: Intermediate Controls
Section 1.5.0
1.5 Summary of performance measures and interpretation on
• Similarly, we can approximate the 10%-90% rise time as:
the Bode plot
√
1.8
2
2
≃
≃
tr ≃
ωn ωb
ωc
(27)
• The Bode plot of the loop gain shows us 3 regions:
– Region I: low-frequency, the magnitude curve shape in this region will
determine the accuracy:
• The main point is that if we increase the speed in one measure (such as ωc ,
for frequency-domain design), then the speed in all measures (such as ωb or
tr ) will increase.
* SSE ,
* Tracking accuracy ,
* Disturbance rejection
– Region II: cross-over,
∗ the cross-over frequency, ωc determines the speed of the system,
1.4.1 Table of approximate dominant-mode relationships
PM←→ ρ
− √ πρ
1−ρ2
P.O.=100 × e
ρ ≃ PM
100
ωc ←→ ωb
ωb ≃
ωc ←→ ωn
√
PM 2
100
ωn ≃ ωb
ωn ←→ tr
≃ ω2b ≃ ωc2
p
ω = ωn 1 − ρ2 ≃ ωc
PM, ωc
←→ ω
ωc , PM →
ts
ts =
4
α
=
√
4
ρ ωn
π
ωc
≃
From dominant mode,
b
, numerical study
GL (s) = s(s+a)
Close with ρ = 0.7, else approx.
(Franklin et al. sect. 6.1)
1.8
ωn
t p = ωπ ≃
ω ←→ t p
– Region III: high frequency, there is little useful performance in this
region,
∗ The roll-off of the gain limits noise sensitivity
From dominant mode, and
b
GL (s) = s(s+a)
ωb ←→ ωn
tr ≃
From dominant mode
From dominant mode, and
b
GL (s) = s(s+a)
2 ωc
ωn ≃ ωc 1 +
∗ phase curve in this region determines the phase margin and the
stability of the system
Notes
4
PM
ωc ( 100
)
50
Gain dB
P.O ↔ ρ
Expression
0
Region III
(Roll−off)
Region I
(Accuracy)
Region II
(Speed/Damping)
−50
−2
10
−1
10
0
10
1
10
G p (s) =
1
s2 + s + 0.1
Hy (s) =
1
s + 50
2
10
From dominant mode
0
Connecting approximations
From dominant mode
Phase [deg]
Approx.
−90
−180
Connecting approximations
−2
10
Table 2: Table of approximations linking PM and ωc to ρ, P.O., ωb , ωn , tr and ts .
−1
10
0
10
Frequency [radians/second]
1
10
2
10
Figure 17: Three regions for consideration of design specifications
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 23
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 24
EE/ME 574: Intermediate Controls
Section 1.5.0
• Often in controller design we have three big goals and one minor goal
EE/ME 574: Intermediate Controls
Section 1.5.0
• Example 1: Boost the gain in Region I, to improve accuracy
– Put in a compensator with high gain at low frequencies
– Big goals:
– Extra slope on the the gain curve increases phase lag at low frequencies
1. Achieve sufficient accuracy.
∗ That is, get good low-frequency gain.
– If the frequency range of added gain is well below ωc , the damage to
phase margin is not too bad.
(Region I)
2. Achieve sufficient speed.
20
∗ That is, get sufficient ωc .
(Region II)
Gain dB
3. Achieve sufficient damping.
∗ That is, get good phase margin.
Phase lag:
10
0
(Region II)
−20 −2
10
– Minor goal:
Kc Gc (s) = Kc
Region I
(Accuracy)
−10
−1
0
10
1
10
10
2
10
1. Achieve sufficient roll-off at high frequency. Don’t amplify noise, if
present.
∗ That is, get good roll-off for ω > ωc .
(Region III)
Phase [deg]
0
τs + 1
ατs + 1
α > 1
Example: α = 10
−20
−40
Kc Gc (s) = 10
−60
10 s + 1
100 s + 1
Added Phase Lag
−80
−2
−1
10
0
10
10
Frequency [radians/second]
1
10
2
10
Figure 18: Bode plot of compensator with high gain in Region I (phase-lag)
Gain dB
50
- -: Original system
0
Region I
(Accuracy)
−50
−2
10
−1
10
0
10
1
10
2
—: With phase-lag
compensation
10
Phase [deg]
0
−90
Added Phase Lag
−180
−2
10
−1
10
0
10
Frequency [radians/second]
1
10
2
10
Figure 19: Improve performance by boosting low-frequency gain.
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 25
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 26
EE/ME 574: Intermediate Controls
Section 1.5.0
• Example 2: Add positive phase (phase lead) in the region of cross-over
EE/ME 574: Intermediate Controls
• Example 3: Add phase lead to push out the cross-over frequency
– Put in a compensator with phase lead around ωc
– Put in a compensator to support the phase angle at the new ωc
– Adjust Kc to move up ωc
– This example set up to leave speed unchanged
60
Gain dB
– Big boost in phase lead (recall ρ ≃ PM/100 )
Boosts HF gain
30
Phase lead:
Gain dB
20
40
Kc Gc (s) = 30
20
0.3 s + 1
0.03 s + 1
Boosts HF gain
10
0
Kc Gc (s) = Kc
−20 −2
10
−1
0
10
10
1
10
2
10
100
τs + 1
ατs + 1
α < 1
Example: α = 0.1
80
−2
10
−1
0
10
1
10
2
10
10
100
Phase [deg]
0
−10
Phase [deg]
Section 1.5.0
80
60
Added Phase Lead
At new crossover
40
20
60
4s +1
Kc Gc (s) = 0.316
0.4 s + 1
Added Phase Lead
At crossover
40
20
0 −2
10
−1
10
0
10
Frequency [radians/second]
1
10
0 −2
10
−1
10
0
10
Frequency [radians/second]
1
2
10
10
Figure 22: Bode plot of compensator with phase lead in Region II (phase-lead)
2
10
Figure 20: Bode plot of compensator with phase lead in Region II (phase-lead)
50
Gain dB
- -: Original system
0
Region II
(Speed/Damping)
−50
−2
−1
0
10
10
1
10
2
Region II
(Speed/Damping)
−2
10
—: With phase-lead
compensation
−90
−2
−1
10
0
10
Frequency [radians/second]
1
10
1
10
2
10
Added Phase Lead
−90
−180
−2
10
0
10
0
10
Added Phase Lead
−180
−1
10
10
0
Phase [deg]
0
−50
Phase [deg]
10
Gain dB
Increase speed
50
2
−1
10
0
10
Frequency [radians/second]
1
10
2
10
Figure 23: Improve performance by boosting low-frequency gain.
10
Figure 21: Improve performance by boosting low-frequency gain.
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 27
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 28
EE/ME 574: Intermediate Controls
Section 1.5.0
EE/ME 574: Intermediate Controls
Section 2.0.0
2 Design of frequency compensation
• Basis of controller design:
– Compensators re-shape the loop gain
To approach compensator design we will
• Some compensators boost gain (improve accuracy)
1. Review standard compensators and their properties
– These compensators will add phase lag
2. Consider examples of compensator design
– Generally the frequency range has to be low enough that the phase lag
does not damage stability
• Some compensators boost phase angle (improve damping)
Standard compensators
• There are four basic standard compensators
– These compensators can be used for
1. Proportional-Derivative (PD)
1. Improve phase margin at the existing cross-over
2. Phase-lead
∗ No change to gain
3. Proportional-Integral (PI)
2. Increase the frequency of cross-over
4. Phase-lag
∗ Increased gain
• Trade-offs can be made by adjusting the frequency or gain of compensation
• These can be combined into larger units. Common examples are
1. Proportional-Integral-Derivative (PID)
The most common type of controller, K. Aström estimates that 80% of
series compensators are PID
2. Lag-lead compensation
3. Double phase lead
4. Double phase lag
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 29
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 30
EE/ME 574: Intermediate Controls
Section 2.1.1
EE/ME 574: Intermediate Controls
2.1 PD compensator
Section 2.2.0
2.2 Phase-lead compensator
u (t) = kd ė (t) + k p e (t)
(28)
Kc Gc (s) = kd s + k p = Kc (τD s + 1)
Kc Gc (s) = Kc
(29)
Kc = k p , τD = kd / k p
so:
(τ s + 1)
(α τ s + 1)
,
α < 1,
τ : [seconds] , α : [·]
(30)
30
Root Locus
0.3
40
30
Gain dB
0
−10 −2
10
−1/τD
−1
0
10
1
10
10
2
10
0 −2
10
0.1
−1/(ατ)
−1/τ
0
90
Phase [deg]
−0.05
−0.1
−0.15
−0.2
60
Phase Lead
30
−0.3
0
−0.4
−6
−2
−3.5
−3
−2.5
−2
−1.5
−1
−0.5
0
0.5
−1
10
10
Real Axis
0
10
Frequency [radians/second]
1
10
1
10
30
Zero
−5
−4
−3
−2
−1
0
1
0 −2
10
−1
10
Pole
0
10
Frequency [radians/second]
1
10
2
10
Figure 25: Phase Lead Compensator
Figure 24: PD Compensator
• The phase lead compensator puts a zero and a pole in the transfer function.
• The PD compensator puts a zero in the transfer function.
• At the frequency of the zero, the PD compensator introduces phase lead
2.1.1 Basics of PD design
• Use the phase lead to improve stability and/or increase the speed of the
closed loop
• Cross-over should occur where the compensator gives a reasonable amount
of phase lead
(at or above 3 [rad/sec] in the figure)
• The increasing gain with increasing frequency is not helpful
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
zero: s = −1/τ ,
pole: s = −1/α τ ,
α<1
• The phase is concentrated at the center frequency between the zero and pole.
• The phase-lead compensator is basically a PD compensator with a pole
above the cross-over frequency to limit the high-frequency gain
– Since high-frequency gain is rarely desirable, phase-lead is nearly always
more desirable than PD
– Phase lead has 1 more parameter, and the designer has to take care to
place ωc in the center of the phase-lead region
∗ This makes phase-lead a bit trickier than PD, and perhaps less robust
than PD.
(increased gain in Region III does no good at all)
Page 31
2
10
60
Real Axis
−0.2
−4
0
10
90
Phase [deg]
real
−1
10
real
−0.1
0
10
0.2
10
0.1
0.05
20
20
Imaginary Axis
imag
0.15
Gain dB
50
Root Locus
0.2
Imaginary Axis
imag
0.4
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 32
2
10
EE/ME 574: Intermediate Controls
Section 2.2.1
EE/ME 574: Intermediate Controls
Section 2.2.1
• Eqn (32) is plotted in figure 27.
2.2.1 Basics of phase-lead design
• First we have to understand the role of the parameters α and τ .
Gain dB
Maximum phase lead
30
α=0.05
Kc Gc (s) = Kc
−1
10
0
10
10
For these
√ plots,
τ = 1/ α
Phase [deg]
α=0.05
60
α=0.2
10
−1
10
0
10
Frequency [radians/second]
1
2
10
0.1
0.01
• α = 0.1 is a typical value, giving 55o of phase lead.
• One thing that must be carefully considered in phase-lead design is to be
sure that cross-over occurs at or near the frequency of maximum phase lead,
that is
(33)
ωc ≃ ωmax
2
10
10
Figure 26: Phase Lead Compensator, α determines the amount of phase lead.
• We need to determine the relationship between α and the phase lead
• Otherwise, the benefit of the phase lead is lost.
• And we need to know the frequency of the peak phase lead (and set Kc so
that this is the cross-over frequency)
• Looking at figure 26, and considering symmetry, ωmax is 1/2 way between
the zero and pole frequencies, on a logarithmic scale. Giving
• These are derived in Franklin et al. ,
1−α
sin φmax =
1+α
which gives
1
10
1/α
ωmax , the frequency of maximum phase lead
α=0.5
0 −2
10
30
Figure 27: Maximum phase increase for lead compensation (adapted from
Franklin et al., 5th Ed.)
2
90
30
40
α: 1
α<1
1
10
50
0 0
10
α=0.5
0 −2
10
60
20
(τ s + 1)
(α τ s + 1)
α=0.2
10
80
70
• α determines the separation between the zero and pole.
20
90
1 − sin φmax
α=
1 + sin φmax
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
1
ωmax = √
τ α
(34)
(31)
Or if we write:
(32)
Page 33
Kc Gc (s) = Kc
s+z
s+ p
Part 3: Frequency Response Design
then
ωmax =
p
|z| |p|
(Revised: Mar 26, 2013)
Page 34
EE/ME 574: Intermediate Controls
Section 2.2.2
2.2.2 Phase-lead compensation design procedure (Franklin, section 6.7.2)
• Need to find Kc (step 1) , α (step 4) and τ (step 5)
Kc Gc (s) = Kc
τs + 1
ατs + 1
Step 2: Evaluate the phase margin (PM) of G p (s = j ωc ) using ωc and Kc
2. Verify that ωmax ≃ ωc , since any mismatch will result in phase lead that
is not utilized
3. Plot the step response of the system, verify overshoot, rise and settling
time, etc., depending on requirements
Step 7: Iterate on the design, make adjustments as needed.
• If more speed is required, increase ωc (and ωmax )
• If more damping is required, reduce α and to increase the phase margin
Step 3: Allow an extra margin (about 10o ), and determine the needed phase lead
to meet the stability requirement.
Step 4: Determine α from Eqn (32) or figure 27
– Add a second phase-lead filter (double-lead compensation) if
necessary
• If more accuracy is required, consider phase-lag compensation to
increase low-frequency gain.
Step 5: Set ωmax = ωc , find τ
√
ωmax α
2.2.2 Phase-lead compensation design procedure (continued)
1. Draw the bode plot for the compensated system, verify the phase margin,
tracking accuracy, etc., as needed.
1. Determine Kc to meet an accuracy requirement. Shifting the magnitude
curve according to Kc determines ωc
√
2. A speed requirement determines ωc (such as ωc ≃1/ 2 ωb). Shifting the
magnitude curve to set ωc determines Kc
1
Section 2.2.2
Step 6: Verify the design
Step 1: Determine where the cross-over frequency, ωc , will fall, and the gain Kc .
Typically
τ=
EE/ME 574: Intermediate Controls
Analysis (exact)
[seconds]
(35)
• This completes the design
Specifications
Complete
Design Details
Design (approximate, iterative, intuitive)
Figure 28: The design cycle: an asymmetric loop.
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 35
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 36
EE/ME 574: Intermediate Controls
Section 2.2.3
Kc will be driven by the SSE requirement. To achieve SSE (step) ≤ 0.01 we
need a LF gain of 100. The current LF gain is about 10, so we need Kc = 10.
• Consider the plant of figure 17 (repeated here)
Step 2: Kc = 10 gives the dashed magnitude of figure 31. And we find
Gm = 34.1 dB (at 7.08 rad/sec), Pm = 54.2 deg (at 0.82 rad/sec)
50
Gain dB
Section 2.2.3
Step 1: Determine Kc and ωc
2.2.3 Phase-lead design example
1
G p (s) = 2
s + s + 0.1
(1)
ωc = 3.1 [rad/sec] ,
PM (1) = 14.5o
0
(1)
1
Hy (s) =
0.02 s + 1
−50
−2
10
−1
0
10
10
1
where ωc and PM (1) are the first measurements of ωc and PM.
50
2
10
10
Gain dB
0
−90
0
−50
−2
−180
10
−2
10
−1
10
0
10
Frequency [radians/second]
1
10
Figure 29: A third order plant
• The design specifications are:
0
1
10
10
2
10
−90
−180
−2
10
1. SSE (step) ≤ 0.01
−1
10
0
10
Frequency [radians/second]
1
10
2
10
Figure 30: G p (s) with adequate gain
2. PO ≤ 20%
• The steady-state error requirement tells us that we need a low-frequency
gain of at least 99 (100, rounding off)
• For a PO of 20% we should have a damping ratio of at least ρ = 0.46 , which
suggests PM ≥ 46o
Part 3: Frequency Response Design
−1
10
0
2
10
Phase [deg]
Phase [deg]
EE/ME 574: Intermediate Controls
(Revised: Mar 26, 2013)
Page 37
Step 3: The additional phase lead needed is
δPM = 46 − 14.5 + 10 ≃ 42o
where +10o have been added to have a little extra margin.
Step 4: Reading off of figure 27, for a phase-boost of 42o, we find α ≃ 0.16 .
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 38
EE/ME 574: Intermediate Controls
Section 2.2.3
(1)
Step 5: ωc was found from the Bode plot to be ωc = 3.1(1) . To get the greatest
benefit from the phase-lead compensation, we need ωmax ≃ ωc . With
√
ωmax = 1/τ α, solving for τ:
1
1
√ =
√
= 0.806 [seconds]
ωc α
3.1 0.16
τ=
Section 2.2.3
Step 7: Iterating on the design, making adjustments as needed
• In this case we want to adjust τ to center the phase lead on the cross-over
frequency (cf. figure 25)
(36)
>> wmax = 6.27
>> tau = 1/(wmax*sqrt(alpha))
tau = 0.3987
>> Gc = tf( [tau, 1], [alpha*tau, 1])
Step 6: Verify the design
>>
>>
>>
>>
>>
>>
EE/ME 574: Intermediate Controls
Gps = tf(1, [1 1 0.1]); Hys = tf(1, [0.02, 1]);
Gp1a = series(Gps, Hys); Kc = 10;
alpha = 0.16; tau = 0.806;
Gc = tf( [tau, 1], [alpha*tau, 1])
figure(1), clf
margin(Kc*Gc*Gp1a)
Bode Diagram
Gm = 23.1 dB (at 26.2 rad/sec) , Pm = 52.6 deg (at 4.33 rad/sec)
50
Magnitude (dB)
0
Bode Diagram
Gm = 16.8 dB (at 19.3 rad/sec) , Pm = 41.8 deg (at 6.27 rad/sec)
50
−50
−100
Magnitude (dB)
0
−150
0
−50
Phase (deg)
−100
Phase (deg)
−150
0
−90
−180
−270
−3
10
−180
−270
−3
10
−90
−2
10
−1
10
0
10
Frequency (rad/sec)
1
10
2
10
3
10
Figure 32: Kc Gc (s) G p (s) for verification II.
−2
10
−1
10
0
10
Frequency (rad/sec)
1
10
2
10
3
10
• The phase margin comes out right at the target of 52o
Figure 31: Kc Gc (s) G p (s) for verification.
• The phase margin came out a little short. On examination, it is because the
cross-over frequency is not 3.1 [rad/sec] as we planned, but 6.27 [rad/sec].
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 39
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 40
Section 2.2.3
EE/ME 574: Intermediate Controls
2.3 PI compensator
Verify that requirements are met
• The original specification had a steady state error and PO specifications,
Z t
e (τ) dτ
(37)
0
(τI s + 1) k p s + ki
=
s
s
Kc = ki ; τI = k p / kI
so:
(38)
50
Root Locus
High LF Gain
40
0.06
imag
0.04
30
20
10
0
0.02
−10 −2
10
0
−1
0
10
1
10
10
2
10
real
0
Phase [deg]
Imaginary Axis
– Tre (s) has a DC gain of 0.1/10.1, so the steady state error
u (t) = k p e (t) + kI
Kc Gc (s) = Kc
– Checking the error transfer function Tre (s) = E (s) /R (s)
>> Tre = feedback(1, Kc*Gc*Gp1a)
Transfer function:
0.001276 s^4 + 0.08507 s^3 + 1.084 s^2 + 1.008 s + 0.1
------------------------------------------------------0.001276 s^4 + 0.08507 s^3 + 1.084 s^2 + 4.996 s + 10.1
Section 2.3.1
Gain dB
EE/ME 574: Intermediate Controls
−0.02
SSE (step) = 0.0099
−0.04
−30
−60
−90
– Checking the step response, we just meet the 20% overshoot requirement
−0.06
−1.2
−2
−1
−0.8
−0.6
−0.4
−0.2
0
0.2
−1
10
10
Real Axis
(It is a good thing we put the extra
ρ = 0.46 is for a 2nd order system)
10o
of phase margin, recall that
1.4
Figure 34: PD Compensator
• Below the frequency of the zero, the PI compensator increases gain, for
improved accuracy
1.2
• The PI compensator introduces phase lag.
1
Amplitude
1
10
• The PI compensator puts a pole at the origin, and zero determined by τI
Step Response
– We avoid damaging the phase margin by putting the PI controller well
below ω = ωc
0.8
0.6
– Contrast this with the Phase-Lead compensator, were we deliberately
(and carefully !) setup the dynamics to have effect right at ω = ωc .
0.4
0.2
0
0
0
10
Frequency [radians/second]
2.3.1 Basics of PI design
0.5
1
1.5
2
• Use the increased gain to improve accuracy at low frequencies
Time (sec)
• Cross-over should occur above the range of significant phase lag.
Figure 33: Step response, for verification.
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 41
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 42
2
10
EE/ME 574: Intermediate Controls
Section 2.4.0
EE/ME 574: Intermediate Controls
2.4 Phase-Lag compensator
Kc Gc (s) = Kc α
• The phase-lag compensator is basically a PI compensator with a finite pole,
and thus finite DC gain
(τ s + 1)
(α τ s + 1)
,
α > 1,
Root Locus
(39)
40
∗ Phase-lag compensation can be used
Gain dB
imag
30
0.3
0.2
Imaginary Axis
τ : [seconds] , α : [·]
– Certain systems (e.g, with static friction) don’t work well with an
integrator in the control loop (that is, a pole at the origin)
0.4
20
10
– In many cases, a finite gain boost is sufficient
0
−10 −2
10
0.1
0
real
−1
0
10
1
10
10
2
10
0
Phase [deg]
−0.1
−0.2
−0.3
Pole
−5
−4
−3
−60
−2
−1
0
1
−2
−1
10
10
Real Axis
– The phase lag has one more parameter that PI control (the speed of the
pole)
Zero
−30
−90
−0.4
−6
Section 2.4.0
0
10
Frequency [radians/second]
1
10
Figure 35: Phase Lag Compensator
2
10
∗ Phase-lag design is basically no trickier than PI design, in both cases
we must assure that the zero is well below ωc , to avoid damaging the
phase margin.
∗ The desired gain boost determines the location of the pole.
• The phase lag compensator puts a pole and a zero in the transfer function.
pole: s = −1/α τ ,
zero: s = −1/τ ,
α>1
• The phase-lag compensator boosts the low-frequency gain.
– The gain boost occurs over the region between the pole and zero
– The magnitude of the gain boost is α (α > 1 , for phase lag)
• In the region between the pole and zero, the phase lag compensator
introduces phase lag, we’ve got to place the phase-lag away from ωc
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 43
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 44
EE/ME 574: Intermediate Controls
Section 2.4.1
2.4.1 Basics of phase-lag design
Step 1: Determine ωc and the open-loop gain Kc that will meet the phase-margin
requirement without compensation
• α determines the separation between the zero and pole.
Step 2: Draw the Bode plot of the uncompensated system with ωc and Kc from
step 1
Gain dB
30
α=20
Kc Gc (s) = Kc α
α=2
α>1
0
−2
10
−1
0
10
1
10
10
2
10
For these
√ plots,
τ = 1/ α
0
Phase [deg]
(τ s + 1)
(α τ s + 1)
α=5
10
α=2
−30
Step 3: Choose α to meet the low-frequency gain requirement (determined from
an accuracy requirement)
Step 4: Choose the corner frequency ω = 1/τ, (the zero frequency of the lag
compensator) to be one octave (2×) to one decade (10×) below the crossover frequency, ωc . That is:
1 1
1
1
< ωc ... < ωc
τ 2
τ 10
α=5
−60
α=20
10
−1
10
0
10
Frequency [radians/second]
1
10
2
10
Step 6: Verify the design
Figure 36: Phase Lag Compensator, α determines the amount of gain boost and
phase lag.
1. Draw the bode plot for the compensated system, verify the phase margin,
tracking accuracy, etc., as needed.
2. Plot the step response of the system, verify overshoot, rise and settling
time, etc., depending on requirements
• Aspects to consider:
– The relationship between α and the LF gain boost:
Gain boost = α
– Phase-lag
1 1
< ωc
τ 5
(40)
(Recall: the zero frequency is α× above the pole frequency.)
(Revised: Mar 26, 2013)
Step 7: Iterate on the design, make adjustments as needed.
• If more speed is required, increase ωc (may require phase-lead
compensation)
∗ The phase lag is substantial up to about 5×the zero frequency, so
Part 3: Frequency Response Design
(41)
Step 5: The second corner frequency (the pole location) is then at ω = 1/ατ.
−90
−2
Section 2.4.2
2.4.2 Phase-lag compensation design procedure (see Franklin, section 6.7.4)
• First we have to understand the role of the parameters α and τ .
20
EE/ME 574: Intermediate Controls
• If more damping is required, increase the phase margin, either by
reducing the zero frequency (ωz = 1/τ) or by added a phase-lead
compensator.
• If more accuracy is required, increase α
Page 45
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 46
EE/ME 574: Intermediate Controls
Section 2.4.3
2.4.3 Phase-lag design example #1
• Consider the plant of figure 37
Motor and load time
constants:
0
τm = 1.0, τe = 0.05 [sec]
−20
−40
• The PO suggests a damping factor of ρ ≥ 0.6, which suggests a phase
margin of at least PM ≥ 60o.
• A blow-up of the Bode plot of G p is seen in figure 38
• The essential concepts for PI or phase-lag compensation are:
1. The phase margin of the plant should be good enough
2. The compensator boosts the LF gain
−60
−80 −1
10
0
1
10
10
2
10
3
G p (s) =
10
Km /b
(τm s + 1) (τe s + 1)
0
−30
G p (s) =
−60
−90
2
(s + 1)(0.05 s + 1)
10
Gain dB
Gain dB
Section 2.4.3
Phase-lag design example #1 (continued), Design:
Velocity control of a motor is a common application of PI or phase-lag
compensation.
−120
−150
−180
−1
10
0
10
1
10
Frequency [radians/second]
2
10
3
10
Figure 37: A motor with velocity control.
• With the design goals
– ωb ≥ 10 [radians/sec]
0
−10
−20 0
10
1
Hy (s) =
0.02 s + 1
Phase [deg]
Phase [deg]
EE/ME 574: Intermediate Controls
1
10
−60
−70
−80
−90
−100
−110
−120
−130
−140
−150 0
10
ωc=7.2 [rad/sec]
1
10
Frequency [radians/second]
Figure 38: Blow-up of Bode plot of motor velocity response.
– PO ≤ 10%
– Tracking accuracy, Etrack ≤ 2%, ωt = 0.1 [rad/sec]
• Figure 38 shows we can achieve a phase margin of 70o at ωc = 7.2 [rad/sec].
– Upeak ≤ 5.0 [Amps]
– A phase margin of 70o gives us a little extra PM
– ωc = 7.2 [rad/sec] should give a closed-loop bandwidth of
ωb ≈ 10 [rad/sec], which gives us a little extra speed
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 47
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 48
EE/ME 574: Intermediate Controls
Section 2.4.3
Step 1: Determine ωc and Kc to meet PM requirement
For ωc = 7.2 [rad/sec],
Section 2.4.3
Step 4: Choose the zero location
Kc = +12 dB = 4.0
Step 2: Draw the Bode plot of the uncompensated system with ωc and Kc from
step 1.
• We don’t want to add very much phase lag at ω = ωc , set the compensator
zero a full decade below ωc
– Choose wz = 0.72 [rad/sec] , which gives: τ = 1/0.72 = 1.39 [sec]
Step 5: The pole is located at ω p = 1/ατ = 1/ (8 ∗ 1.39) = 0.090 [rad/sec]
PM = 69.16 at wcp= 7.36 [rad/sec]
40
EE/ME 574: Intermediate Controls
Step 6: Verify the design, recall the requirements
Accuracy Requirement
Gain dB
20
• ωb ≥ 10 [radians/sec]
0
• PO ≤ 10%
−20
−40 −2
10
−1
0
10
1
10
10
• Tracking accuracy, Etrack ≤ 2%, ωt = 0.1 [rad/sec]
2
10
• Upeak ≤ 5.0 [Amps]
Phase [deg]
0
−90
−180
−270
−2
10
−1
10
0
10
Frequency [radians/second]
1
10
2
10
Figure 39: Motor velocity control, K = 4.0 .
Step 3: Choose α to meet the LF gain requirement.
• For a tracking error less than 2%, we need a gain at ωt = 0.1 [rad/sec] of at
least 50.
– The current gain is 18 dB or 8.0,
%% Plant Model
>> Gp1a = tf(2, poly([-1 -20 -50])/1000)
%% Compensator
>> Tau = 1.39; Alpha = 8; Kc = 4;
>> Gc1a = tf(Alpha*[Tau 1], [Alpha*Tau, 1])
%% Closed Loop TF
>> Try = feedback( Kc*Gc1a*Gp1a, 1)
Transfer function:
88.9 s + 64
--------------------------------------------------0.0111 s^4 + 0.790 s^3 + 12.0 s^2 + 101.1 s + 65
>> Bandwidth = bandwidth(Try)
Bandwidth =
12.7847 (CHECK)
– So we need to increase the gain by 50/8=6.25
• Choose α = 8 (a little extra)
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 49
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 50
EE/ME 574: Intermediate Controls
Section 2.4.3
EE/ME 574: Intermediate Controls
Section 2.4.3
• Examine the tracking-error Bode plot,
Phase-lag design example #1 (continued)
• Examining the step response
>>
>>
>>
>>
>> figure(2), clf
>>
step(Try)
>> y = step(Try);
>> PO = 100* (max(y)-y(end))/y(end)
PO = 4.2410 %
Tre = feedback( 1, Kc*Gc1a*Gp1a);
figure(1), clf
bode(Tre)
Wt = 0.1; MagE1 = bode(Tre, Wt)
MagE1 = 0.0229
%% R->E TF
10
Step Response
0
Gain dB
1.4
1.2
Amplitude
−20
−30
1
−40 −2
10
0.8
0.6
−1
0
10
1
10
10
2
10
Phase [deg]
90
0.4
0.2
0
0
−10
1
2
3
4
60
30
0
−30
5
Time (sec)
−2
10
Figure 40: Motor velocity control, closed-loop step response
• The step response shows about 4.2% overshoot
(CHECK)
−1
10
0
10
Frequency [radians/second]
1
10
2
10
Figure 41: Motor velocity control, frequency response plot of Tre (s), the error
response.
• Requirement:
Etrack ≤ 2% at ωt = 0.1 [rad/sec]
• Actual performance:
Tre (s = j ωt ) = 0.0229
• The requirement is not met !
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 51
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 52
EE/ME 574: Intermediate Controls
Section 2.4.3
EE/ME 574: Intermediate Controls
Section 2.4.3
Phase-lag design example #1 (continued), 2nd iteration
• Check the Upeak
>> Tru = feedback( Kc*Gc1a, Gp1a);
>> figure(2),
>> step(Tru)
4.5
>> Upeak = ...
4
max(step(Tru))
• We can increase the LF gain by increasing α
%% R->U TF
• But the frequency of the pole is already below ωt = 0.1 [rad/sec]
Step Response
ω pole =
1
= 0.090 [rad/sec]
ατ
3.5
• Additional gain below ω = 0.1 [rad/sec] does not help us.
Amplitude
3
2.5
• Figure 43 shows the bode plot of the compensator,
2
– We are introducing a little additional phase lag at ωc
1.5
1
– Strategy: increase α by a small amount, holding α τ constant
0.5
0
0
0.1
0.2
0.3
0.4
Time (sec)
0.5
0.6
0.7
0.8
(hold the pole location constant, move up the zero).
∗ We will loose some phase margin, but pick up some low-freq gain
Figure 42: Motor velocity control, closed-loop step response
PM = 64.01 at wcp= 7.39 [rad/sec]
• Considering all of the performance requirements
ωb ≥ 10 [rad/sec]
ωb = 12.78 [rad/s]
P.O. ≤ 10%
P.O. = 4.2%
√
Etrack ≤ 2.29%
Upeak ≤ 5.0 [Amps]
Upeak = 4.03
20
10 −2
10
√
zero
−1
0
10
1
10
10
2
10
0
Step 7: Iterate on the design, make adjustments as needed.
−20
−40
−60 −2
10
• We have plenty of headroom on ωb , P.O. and Upeak . We need to improve
Etrack by a small amount.
(Revised: Mar 26, 2013)
pole
25
15
Table 3: Performance requirements of the motor velocity control example.
Part 3: Frequency Response Design
Gain dB
2nd iteration 3rd iteration
√
Etrack ≤ 2%, ωt = 0.1 [rad/s]
30
Phase [deg]
1st iteration
Requirement
35
Page 53
−1
10
0
10
Frequency [radians/second]
1
10
2
10
Figure 43: Motor velocity control, Bode plot of compensator.
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 54
EE/ME 574: Intermediate Controls
Section 2.4.3
Phase-lag design example #1 (continued), 2nd iteration
>>
>>
>>
>>
>>
>>
%% Compensator 2, move up zero by 10%
Tau = 1.38/1.1; Alpha = 8*1.1
Gc1b = tf(Alpha*[Tau 1], [Alpha*Tau, 1]);
Try = feedback(Kc*Gc1b*Gp1a, 1);
Tre = feedback(1, Kc*Gc1b*Gp1a);
Tru = feedback(Kc*Gc1b, Gp1a);
Bandwidth2 = bandwidth(Try)
Bandwidth2 =
12.8159
>> y = step(Try);
PO2 = 5.4665
EE/ME 574: Intermediate Controls
Section 2.4.3
Phase-lag design example #1 (continued), 3rd iteration
%% R->Y TF
%% R->E TF
%% R->U TF
PO2 = 100*(max(y)-y(end)) / y(end)
>>
>>
>>
>>
>>
>>
%% Compensator 3, move up zero by 20%
Tau = 1.35/1.2; Alpha = 8*1.2
Gc1b = tf(Alpha*[Tau 1], [Alpha*Tau, 1]);
Try = feedback(Kc*Gc1b*Gp1a, 1);
Tre = feedback(1, Kc*Gc1b*Gp1a);
Tru = feedback(Kc*Gc1b, Gp1a);
Bandwidth3 = bandwidth(Try)
Bandwidth3 =
12.8495
>> y = step(Try);
PO3 = 5.6338
PO3 = 100*(max(y)-y(end)) / y(end)
>> u = step(Tru); Upeak2 = max(u)
Upeak2 = 4.0272
>> u = step(Tru); Upeak3 = max(u)
Upeak3 = 4.0311
>> [MagE2, PhaseE] = bode(Tre, Wt)
MagE2 = 0.0209
>> [MagE3, PhaseE] = bode(Tre, Wt)
MagE3 = 0.0192
Requirement
1st iteration
2nd iteration
3rd iteration
ωb ≥ 10 [rad/sec]
ωb = 12.78 [rad/s]
√
ωb = 12.82 [rad/s]
√
P.O. ≤ 10%
P.O. = 4.2%
√
P.O. = 5.5%
√
Etrack ≤ 2%, ωt = 0.1 [rad/s]
Etrack ≤ 2.29%
-
Etrack ≤ 2.09%
-
Upeak ≤ 5.0 [Amps]
Upeak = 4.03
√
Upeak = 4.03
√
1st iteration
Requirement
2nd iteration
3rd iteration
ωb ≥ 10 [rad/sec]
ωb = 12.78 [rad/s]
√
ωb = 12.82
√
ωb = 12.85
√
P.O. ≤ 10%
P.O. = 4.2%
√
P.O. = 5.5%
√
P.O. = 5.6%
√
Etrack ≤ 2%, ωt = 0.1 [rad/s]
Etrack ≤ 2.29%
-
Etrack ≤ 2.09%
-
Etrack ≤ 1.92%
Upeak ≤ 5.0 [Amps]
Upeak = 4.03
√
Upeak = 4.03
√
Upeak = 4.03
Table 4: Performance requirements of the motor velocity control example.
Table 5: Performance requirements of the motor velocity control example.
• Close, but not quite there. We lost a little on the PO, but we still have room
to make additional adjustment.
• Success !
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 55
√
√
>> Alpha,
Alpha = 9.6000
>> Tau,
Tau =
1.1583
>> Gc1b/Alpha
Transfer function:
(1.158 s + 1) / (11.12 s + 1)
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 56
EE/ME 574: Intermediate Controls
Section 2.4.3
EE/ME 574: Intermediate Controls
Section 2.4.3
• Design process tips:
Summary, phase-lag design example #1
• What suggested phase lag, as opposed to phase lead, or some other
compensator ?
– Carrying the design through step 2, and
– Notice the code to evaluate each of the performance criteria
– Setup the compensator
%% Compensator 3
>> Tau = 1.39/1.2; Alpha = 8*1.2
>> Gc1c = tf(Alpha*[Tau 1], [Alpha*Tau, 1])
∗ Seeing that the speed and damping were adequate,and that
∗ The challenging requirement is tracking accuracy
– Setup the transfer functions that will be needed
– Trying a phase-lead compensation, and seeing problems
(Phase-lead would not be good for this application, try it and see !)
– Knowing that motor velocity servos are often PI controllers
• Why did it take 3 iterations
>> Try = feedback(Kc*Gc1c*Gp1a, 1);
>> Tre = feedback(1, Kc*Gc1c*Gp1a);
>> Tru = feedback(Kc*Gc1c, Gp1a);
– bandwidth() is a routine I wrote (write needed routines)
– There is a 73x frequency spread between ωc and ωt , but it was still a little
difficult to get the needed low-frequency gain boost.
>> Bandwidth3 = bandwidth(Try)
– Directly evaluate the percent overshoot and Upeak
∗ Boost needed at ωt
>>
>>
>>
>>
∗ Can’t tolerate very much added phase lag at ωc
∗ Examine the phase lag put in by the compensator
>> [temp, PhaseGc1c] = bode(Kc*Gc1c, wc2)
PhaseGc1b = -5.9657
Gc (s) has −6.0o of phase at ωc (a bit of damage to the phase margin)
• If needed, we could have gotten more low-frequency gain by putting in a
double-lag compensator,
Double lag compensation: 2 lag compensators, with the pole repeated and
the zero repeated.
y = step(Try);
PO3 = 100*(max(y)-y(end)) / y(end)
u = step(Tru);
Upeak3 = max(u)
– Directly evaluate the error magnitude
>> [MagE3, PhaseE] = bode(Tre, Wt)
– Pay careful attention to validation, it is the most important step !
– If you’re not sure how an adjustment will impact the design, try it !
– 2x as much low-frequency boost
– 2x as much damage to the phase margin, may need to increase τ
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 57
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 58
EE/ME 574: Intermediate Controls
Section 2.5.0
EE/ME 574: Intermediate Controls
2.5 PID compensation
Section 2.5.2
2.5.1 Basics of PID design
• Determine desired cross-over frequency, ωc
• Proportion-Integral-Derivative compensation combines the elements of PI
and PD control
– For example, from a speed requirement
• It is very widely used for motion controllers, in the process industry and in
other applications
u (t) = kd ė (t) + k p e (t) + kI
Z t
– Often, the zeros will be set to the same speed:
e (τ) dτ
(42)
0
τD =≃ τI
kd s2 + k p s + ki
1
Kc Gc (s) = Kc (τD s + 1) 1 +
=
s τI
s
– Define ωz = 1/τD = 1/τI as the speed of the PID zeros, then the phase
lead is given for several ratios ωc /ωz
40
Gain dB
imag
0.15
0.1
ωc /ωz ∠ PID
High LF Gain
30
20
0.5
10
0
1.0
−10 −2
10
0.05
0
real
−0.1
−0.15
−1
10
0
1
10
10
Phase Lead
0
−90
−0.2
−3.5
−2
−3
−2.5
−2
−1.5
−1
−0.5
0
0.5
−1
10
10
Real Axis
0
10
Frequency [radians/second]
1
10
−37o
00
2
10
90
Phase [deg]
−0.05
(44)
(43)
50
Root Locus
0.2
Imaginary Axis
• Set the zeros to give good phase lead at ωc
2
10
2.5
45o
3.8
∞
60o
90o
2.5.2 Design procedure
1. Determine the desired value of cross-over, ωc
Figure 44: PID Compensator
2. Set ωz , determine τD , τI
• The PID compensator puts
3. Determine Kc to achieve cross-over at ωc
– A pole at the origin, for high LF gain (and to increase the system type)
4. Validate
– Two zeros, at 1/τI and 1/τD , for phase lead (and improved damping)
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 59
5. Iterate
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 60
EE/ME 574: Intermediate Controls
Section 2.5.3
EE/ME 574: Intermediate Controls
Section 2.5.3
• The form for Gc (s) for Ziegler and Nichols tuning is:
2.5.3 Ziegler-Nichols Method 1: Quarter-Cycle Damping Method
• Proposed in 1942 by J.G. Ziegler and N. B. Nichols
1
τD s2 + s + 1/τI
Gc (s) = k p 1 +
+ τD s = k p
τI s
s
• Provides a systematic approach when no transfer function is available
• Widely used in the process industry (cf. Franklin et al. (5th Ed.) sec. 4.4.2).
• Also called the “Process Reaction Curve Method”
• With a series of trials on an analog computer, Ziegler and Nichols
determined the following recommended gain values of table 6.
• The first Ziegler-Nichols method is based on the open-loop step response of
the system.
Recommended Gains
τI
kp
– This method is suitable for over-damped systems, especially with time
delay (examples: mayonnaise processing, heat or flow in an oil refinery)
P-Control
PI-Control
– A typical curve is seen in figure 46. The curve defines three values:
Amplitude of response
L:
Time Delay [seconds]
τ:
Time constant [seconds]
R = A/τ
0.9/RL L/0.3
Process
Reaction
Curve
2L
0.5 L
Example, Quarter-Cycle Damping method
Step Response
G p (s) =
A e−s L
τs + 1
Reaction Rate:
G p2 (s) =
2
A = 2
1
L = 0.3 [seconds]
R = A/τ
τ = 2.2 [seconds]
0.5
L=td
τ
t
0
Figure 45: Typical process reaction curve.
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
2
(s + 1) (0.5 s + 1) (0.2 s + 1)
1.5
Amplitude
A
Recall,
Reaction Rate:
Table 6: Ziegler-Nichols tuning for PID controller, for quarter-cycle damping.
Approximate transfer
function:
y(t)
τD
1/RL
PID-Control 1.2/RL
A:
(45)
0
1
2
3
4
5
Time (sec)
6
7
8
9
Figure 46: Process reaction curve for G p2 (s)
Page 61
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 62
EE/ME 574: Intermediate Controls
Section 2.5.4
• The calculation gives:
>>
>>
>>
>>
>>
EE/ME 574: Intermediate Controls
Section 2.5.4
2.5.4 Ziegler-Nichols Method 2: Ultimate Sensitivity Method
A=2; L=0.3; tau = 2.2;
R = A/tau„
kp
= 1.2/(R*L)„
tauI = 2*L„
tauD = 0.5*L„
R
kp
tauI
tauD
=
=
=
=
• The ultimate sensitivity method is based on performing a test in which the
closed-loop system is driven to marginal stability.
0.9091
4.4000
0.6000
0.1500
– Wrap a loop around the process, with proportional gain, as shown in
figure 48.
>> Gc2 = tf(kp*[tauD, 1, 1/tauI], [1 0])
Transfer function:
(0.66 s^2 + 4.4 s + 7.333) / s
– Increase the gain until marginal stability is reached (figure 49), this
determines two values
Ku :
Ultimate Gain
Pu :
Ultimate Period
Step Response
1.8
P
1
1.6
r(t)
1.4
2
1.2
Amplitude
y(t)
e(t)
K
For Quarter-Cycle Damping:
P
(P2 − 1) =
1
1
(P1 − 1)
4
Process
(46)
0.8
Figure 48: Feedback for the ultimate sensitivity method. Kc is the proprotional
feedback gain.
0.6
0.4
0.2
1
0
0
2
4
6
8
10
12
Pu
0.8
14
Time (sec)
0.6
Figure 47: Closed-loop step response for system tuned by the Quarter-Cycle
Damping method.
0.4
y(t)
0.2
• The objective is “Quarter-Cycle Damping,” which means that each
successive peak is reduced to 1/4th the amplitude of the previous peak.
0
−0.2
−0.4
−0.6
−0.8
• In this example, we don’t quite get quarter-cycle damping,
−1
0
5
– But the parameters give a starting point for final tuning.
10
time [seconds]
15
20
Figure 49: Output of the system at marginal stability.
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 63
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 64
EE/ME 574: Intermediate Controls
Section 2.5.4
τD
>> Ku = 6;
1
8 Pu
Table 7: Ziegler-Nichols tuning for PID controller, for ultimate sensitivity
method.
Example, Ultimate Sensitivity Method
• Example plant TF:
G p (s) =
Section 2.5.4
• Applying the Ultimate Sensitivity method
Recommended Gains
kp
τI
P-Control
0.5 Ku
PI-Control 0.45 Ku Pu /1.2
1
PID-Control 0.6 Ku
2 Pu
EE/ME 574: Intermediate Controls
2
s2 + 2 s + 5
>>
>>
>>
>>
Pu = 4.6/3;
%% Raw measurements
kp
= 0.6*Ku
%% Determine compensator
tauI = Pu/2
tauD = Pu/8
Gc = kp * tf( [tauD, 1, 1/tauI], [1 0] )
Transfer function:
0.69 s^2 + 4.5 s + 4.696
kd s^2 + kp s + ki
-----------------------=
------------------s
s
• Test values of gain until marginal sensitivity is reached
>> figure(2), clf
%% Plot CL step response
>> step( Gc*Gp2a) ; grid on
Kc = 1.0000
Kc = 2.0000
0.8
1.4
0.7
1.2
0.5
0.4
0.4
0.2
0.2
2
4
6
8
0
0
10
4
6
Time (sec)
Kc = 6.0000
1.6
1.4
1.4
1.2
1.2
Amplitude
1.6
1
0.8
10
Ku = 6
0.6
0.4
SSE(step):
0.00
SSE(ramp):
0.5
0.11
Pu = 1.533 [s]
0.8
0.4
0.2
8
1
0.6
0
0
1
2
Time (sec)
Kc = 5.0000
Amplitude
0.1
Amplitude
0.8
0.6
0.3
0
0
Step response of system, by Ziegler−Nichols UC tuning
1.5
1
Amplitude
Amplitude
0.6
0.2
2
4
6
8
10
0
0
Time (sec)
2
4
6
8
10
0
0
Time (sec)
2
4
6
8
10
12
Time (sec)
Figure 50: Step response for K = 1, 2, 5, 6
Figure 51: Controller tuned by the Ultimate Sensitivity method.
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 65
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 66
EE/ME 574: Intermediate Controls
Section 2.5.5
2.5.5 Tyreus-Luyben Tuning Chart (“TLC” tuning rules)
EE/ME 574: Intermediate Controls
Section 2.5.6
2.5.6 PID design by frequency response
• An alternate set of rules due to B.D. Tyreus and W.L. Luyben gives:
• The Bode plot for a PID controller with τI = τD is seen in figure 53.
Recommended Gains
60
kp
τI
τD
PI-Control Ku/3.2 2.2 Pu
PID-Control Ku/2.2 2.2 Pu Pu /6.3
Gain dB
50
40
30
High gain for accuracy
in this region
20
Table 8: Tyreus-Luyben alternate tuning for PID controller. Ku and Pu
determined as for the ultimate sensitivity method.
10 −2
10
• Relative to the ultimate sensitivity method, TLC tuning gives:
−1
0
10
1
10
10
2
10
Phase [deg]
90
– Slightly lower gain
– Slightly longer τ′ s
60
Crossover
in this region
30
0
−30
−60
– Will give a slightly slower, more damped response
−90
−2
10
−1
10
0
10
Frequency [radians/second]
1
10
2
10
Step response of system, by frequency response design
Figure 53: PID controller tuned by frequency response.
1
0.9
• Frequency response design of PID control comprises two main ideas:
0.8
Amplitude
0.7
1. Determine cross-over, to get adequate speed and stability
0.6
SSE(step):
0.00
0.5
SSE(ramp):
0.47
0.4
Bandwidth:
6.81
– Boosting the phase margin with phase lead, if necessary
2. Boost the low frequency gain, as needed, to meet accuracy requirements
0.3
0.2
0.1
0
0
2
4
6
8
10
12
Time (sec)
Figure 52: Controller tuned by TLC tuning rules.
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 67
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 68
EE/ME 574: Intermediate Controls
Section 2.5.6
2.5.6 PID design by frequency response (continued)
• Consider the uncompensated plant, G p2 (s)
EE/ME 574: Intermediate Controls
2.5.6 PID design by frequency response (continued)
2
G p2 (s) =
(s + 1) (0.5 s + 1) (0.2 s + 1)
• Let’s set ωc = 4.1 [rad/sec] (this determines k p)
• To get good phase margin
Gain dB
50
– Both compensator zeros at ωz = ωc /5 [rad/sec] = 0.8 [rad/sec]
0
τD = τI = 1/ωz
−50
−100 −2
10
−1
0
10
1
10
10
(47)
2
10
• Including Gc (s) in the bode plot, k p for ωc = 4.1[rad/sec] comes out to 1.3
0
Phase [deg]
Section 2.5.6
• Writing the compensator transfer function as:
−90
Gc (s) = k p
−180
−270
−2
10
−1
10
0
10
Frequency [radians/second]
1
10
= kp
2
10
Figure 54: The uncompensated plant, G p2 (s).
• We find
– ZN tuning gives about 25o of phase lead around 4 [rad/sec],
Gm = NaN dB (at
Gain dB
0.9
0.8
(Revised: Mar 26, 2013)
SSE(step):
0.00
SSE(ramp):
0.6
SSE(step):
0.5
SSE(ramp):
−50 −2
10
0.00
0.47
−1
10
0
10
1
2
10
10
0.47
0
Bandwidth:
6.81
Phase [deg]
−30
0.1
0
0
−60
−90
−120
−150
−180
2
4
6
8
10
Time (sec)
12
−2
10
−1
10
0
10
Frequency [radians/second]
1
2
10
10
Figure 55: Step response (closed-loop) and Bode plot (open-loop), Gc3 G p2 .
Steady state error for a ramp input is 47%.
– Z-N is a coarse tuning, a starting point for fine tuning !
Part 3: Frequency Response Design
NaN rad/sec), Pm = 65.0 deg (at 4.38 rad/sec)
0
0.7
0.2
∗ This is why the Z-N tuned controller looks lightly damped
(50)
50
0.3
∗ 90o of phase lead would be the maximum possible
(49)
1.22 s2 + 2 s + 0.82
s
Step response of system, by frequency response design
0.4
∗ Certainly not as much as possible
1
Amplitude
• For comparison, the compensation given by Ziegler-Nichols Quarter-Cycle
Damping type tuning is seen figure 47.
(48)
τD s2 + (1 + τD/τI ) s + 1/τI
s
Gc3 (s) = 1.3
• Phase cross-over (φ = −180o) occurs at 4.1 [rad/sec]
• With effective phase boost (the phase-lead portion of the PID) we can get
60o of phase margin at ωc = 4.0 [rad/sec]
(τD s + 1) (s + 1/τI )
s
Page 69
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 70
EE/ME 574: Intermediate Controls
Section 2.5.7
2.5.6 PID design by frequency response (continued)
• Study questions, for the previous example
EE/ME 574: Intermediate Controls
Section 2.5.7
Solution:
• To make Gc3 (s) proper, we can add one or several poles at frequencies well
above ωc
Gc3 (s) ≃ 1.3
1.22 s2 + 2 s + 0.82
1.22 s2 + 2 s + 0.82
≃ 1.3
(53)
s (0.01s + 1)
s (0.01s + 1) (0.01s + 1) ...
– What would we change to increase the bandwidth of the system ?
Gc (s) = 3 s + 7 ≃
∗ What would be the side effects ?
3s +7
3s +7
≃
...
(0.01 s + 1) (0.01 s + 1) (0.01 s + 1)
(54)
– What would we change to increase the accuracy of the system ?
50
∗ What would be the other consequences ?
Gain dB
40
– What would we change to increase the damping of the system ?
∗ What would be the other consequences ?
30
ωc
20
10
0 −2
10
−1
0
10
1
10
10
2
10
2.5.7 Realizable form for PD and PID controllers
90
Gc3 (s) = 1.3
1.22 s2 + 2 s + 0.82
s
Phase [deg]
• The transfer function for the PID controller of the last example is
(51)
60
30
0
−30
−60
−90
−2
10
• A PD controller might have a TF of the form
Gc (s) = 3 s + 7
(52)
−1
10
0
10
Frequency [radians/second]
1
10
2
10
Figure 56: An additional pole at s = −100 has little influence near ωc .
• Additional poles should be at frequencies well above cross-over
• Neither of these transfer functions is implementable (infinite gain at infinite
frequency)
• They should be put in so that their low-frequency gain is 1, that is:
Not
>> step(Gc3)
??? Error using ==> ltipack.ltidata.utCheckComputability at 13
Not supported for models with more zeros than poles.
1/ (0.01 s + 1)
1/ (s + 100)
• These poles are physical, actual controller devices have finite bandwidth !
• That is, Gc3 (s) is not proper.
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 71
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 72
EE/ME 574: Intermediate Controls
Section 2.6.0
EE/ME 574: Intermediate Controls
2.6 Lead-Lag compensation
Section 2.6.1
2.6.1 Basics of Lead-Lag compensation
• The structure of the lead-lag compensator is illustrated in figure 58.
• Lead-lag compensation is like PID compensation, except that:
– The low-frequency pole is at finite frequency, not the origin
ωc somewhere in here
– There is an additional high-frequency pole.
frequency, ω
x
oo
Gc (s) = Kc α1
(τ1 s + 1) (τ2 s + 1)
(α1 τ1 s + 1) (α2 τ2 s + 1)
with
lag section: α1 > 1,
lead section: α2 < 1.
x
S-plane
Lead Part
Lag Part
Figure 58: Structure of the poles and zeros in a Lead-Lag compensator
• Design ideas:
Root Locus
40
– Use the lag part to boost the low-frequency gain
1.5
Gain dB
imag
30
1
0
−0.5
High gain for
10 accuracy in
this region
0 −2
−1
10
10
– The lag zero should be well below the cross-over frequency
0
1
10
10
2
10
– The lead pole should be well above the cross-over frequency
real
90
Phase Lead Portion
Phase [deg]
Imaginary Axis
0.5
– Use the lead part to boost the phase margin
20
Phase Lag Portion
−1
60
– Choose Kc so that cross-over occurs in the desired region.
30
0
Crossover
in this region
−30
−60
−90
−1.5
−25
−20
−15
−10
−5
0
5
−2
−1
10
10
Real Axis
0
10
Frequency [radians/second]
1
10
2
Figure 57: Phase Lag Compensator
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
– The DC gain is finite
∗ The infinite DC gain of the PID compensator may not be
advantageous, if, for example, we have a disturbance rejection
requirement, at ωd .
%% Example Lead-Lag Compensation
>> Alpha1 = 20„
Tau1 = 2
%% Lag portion
>> Alpha2 = 0.1„ Tau2 = 0.5
%% Lead portion
>> Kc = 2
>> GcLeadLag = Kc* series( ...
tf( Alpha1 * [Tau1 1], [(Alpha1*Tau1) 1]), ...
tf( [Tau2 1], [(Alpha2*Tau2) 1]) )
• Comparing with the PID compensator
10
– The Lead-Lag compensator has a 2nd pole, above the cut off frequency
of the controller.
∗ We’ve seen that PID controllers (and everything really) actually have
additional high-frequency poles.
%% Lag Part
%% Lead Part
Page 73
∗ The explicit HF pole of Lead-Lag compensation may be good for
controlling noise sensitivity.
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 74
EE/ME 574: Intermediate Controls
Section 3.0.0
EE/ME 574: Intermediate Controls
Section 3.0.0
• Look at the phase-margin at ωc
3 Summary of frequency-response based
compensator design
– If the speed and phase margin are adequate,
∗ PI or Phase-Lag control may be suitable
• Broadly speaking, controller design comprises 4 steps:
– If the speed or phase margin needs boosting
1. Determining specifications
∗ PD, Phase-Lead control may be suitable
2. Modeling the system to be controlled
3. Choosing a controller structure
• Look at the accuracy requirement
4. Tuning the controller parameters
– If the accuracy is adequate
• We’ve seen a number of different controller structures, these are listed in
table 9, with indications of their virtues.
∗ PD or Phase-lead control may be suitable
– If the accuracy needs improving
Compensator
Type:
Speed Damping
Accuracy
Number of
Parameters:
P-type
+
-
+
1
PD
++
++
0
2
Phase-Lead
++
++
0
3
PI
0
-
++
2
Phase-Lag
0
-
++
3
lag-lead
++
++
++
5
PID
++
++
++
3
State Space
++
++
++
n
Notation
∗ PI or Phase-Lag control may be suitable
• If both (speed or phase margin) and accuracy need a boost
– PID or Lead-Lag control may be suitable
’+’: improves, ’-’: makes worse, ’0’: unchanged.
Table 9: Controller Structures
• The most important step of any frequency-domain design is to determine
the cross-over frequency.
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 75
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 76
EE/ME 574: Intermediate Controls
Section 3.0.0
• It is permitted to use multiple compensators.
– For example, to get a phase boost of more than about 60o we can use a
double-lead compensator
Gc (s) = Kc
τs + 1 τs + 1
,
ατs + 1 ατs + 1
α<1
(55)
– A double-lag compensator would give greater boost of the low-frequency
gain
– A Lag-Lead-Lead compensator would give some accuracy improvement
with considerable phase support
– Et cetera for other combinations
• Poles at ω p >> ωc (frequency significantly above ωc ) have little influence
on the closed-loop response of the system.
– They can be added to the compensator to make it realizable
– They can be added as needed to reduce sensitivity to high-frequency
noise.
——– ∼ ——–
• The Bode plot reveals many features of the requirements and performance
of feedback control systems, including
– The concepts of cross-over frequency, phase and gain margin,
– Tools for assessing the speed, stability and accuracy of feedback control,
– Aspects to examine, to help select a controller structure, and
– Design techniques to realize controllers of many different types.
Part 3: Frequency Response Design
(Revised: Mar 26, 2013)
Page 77
EE/ME 574: Intermediate Controls
EE/ME 574: Intermediate Controls
State-Variable Modeling for
Controller Analysis and Design
Contents
1 Introduction to state-variable models
1.1 Voice-coil motor: an example contrasting classical and statevariable modeling . . . . . . . . . . . . . . . . . . . . . . . . . .
1.1.1 Developing a classical model . . . . . . . . . . . . . . . .
1.1.2 Some observations on developing the classical model . . .
1.1.3 Writing the model in state-variable form . . . . . . . . .
1.1.4 Form for the state-variable model . . . . . . . . . . . . .
1.1.5 Steps of building a state-variable model, with the voicecoil motor example . . . . . . . . . . . . . . . . . . . . .
1.1.6 Matlab code to set up the voice-coil motor example . . . .
1.1.7 Some observations on developing the state-variable model
1.2 State-variable modeling, conclusions . . . . . . . . . . . . . . . .
12
21
24
26
2 Review of basics of linear algebra
2.1 Scalars, matrices and vectors . . . . . . . . . . . . . . . . .
2.2 Transpose . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3 Basic arithmetic: +, - and * are well defined . . . . . . . . .
2.3.1 Multiplication . . . . . . . . . . . . . . . . . . . . .
2.4 Commutative, associative, distributive and identity properties
2.4.1 Commutative property . . . . . . . . . . . . . . . .
2.4.2 Associative property . . . . . . . . . . . . . . . . .
2.4.3 Distributive property . . . . . . . . . . . . . . . . .
2.4.4 Identity Matrix . . . . . . . . . . . . . . . . . . . .
2.4.5 Doing algebra with vectors and matrices . . . . . .
2.5 Linear independence and the determinant . . . . . . . . . .
2.5.1 Computing the determinant . . . . . . . . . . . . .
27
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28
29
31
31
31
32
32
33
34
35
Part 4: State-Space Control
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37
38
39
41
44
46
47
3 Working with state-variable models
3.1 Deriving the transfer function from the state-variable model . . . .
3.1.1 Interpreting J p . . . . . . . . . . . . . . . . . . . . . . .
3.2 Getting a state-space model from a transfer function . . . . . . . .
3.2.1 Example, state-space model from a transfer function . . .
3.2.2 One more complication . . . . . . . . . . . . . . . . . .
3.3 Coordinate transformation of a state-variable model . . . . . . . .
3.3.1 Derivation of the state model transformation . . . . . . .
3.3.2 Example: transforming a system model into modal
coordinates . . . . . . . . . . . . . . . . . . . . . . . . .
3.3.3 Summary of coordinate transformation . . . . . . . . . .
49
49
51
52
56
57
58
59
4 Introducing state-feedback control
4.1 Determination of a new model with state feedback
4.2 Determining N f . . . . . . . . . . . . . . . . . . .
4.2.1 Note on N f and SSE(step) = 0 . . . . . . .
4.3 Example system with state feedback . . . . . . . .
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70
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75
2.6
2.7
2.8
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(Revised: Jan 27, 2016)
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2.5.2 Some relations involving determinants . . . . . . . .
Rank . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The norm of a vector . . . . . . . . . . . . . . . . . . . . . .
A quick review of eigenvalue and eigenvectors . . . . . . . .
2.8.1 Some properties of Eigenvectors and Eigenvalues . .
2.8.2 A few more notes on Eigenvectors and Eigenvalues .
2.8.3 Positive definite and positive semi-definite matrices .
2.8.4 One final fact about the eigensystem: V diagonalizes A
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5 State-space controller design, determine K and N f
5.1 State-Space Controller Design Method #1: Pole Placement
5.1.1 Pole placement example . . . . . . . . . . . . . . .
5.1.2 Try pole-placement control design, twice as fast . . .
5.1.3 Taking a look at the control effort and tracking error
Part 4: State-Space Control
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(Revised: Jan 27, 2016)
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EE/ME 574: Intermediate Controls
5.2
EE/ME 574: Intermediate Controls
State-Space Controller Design Method #2: The Linear Quadratic
Regulator, an optimal control method . . . . . . . . . . . . . . .
5.2.1 The LQR cost function . . . . . . . . . . . . . . . . . . .
5.2.2 Designing an LQR controller . . . . . . . . . . . . . . .
5.2.3 Tuning an LQR controller . . . . . . . . . . . . . . . . .
5.2.4 Procedure for tuning Q and R . . . . . . . . . . . . . . .
5.2.5 Further notes on choosing Q and R . . . . . . . . . . . . .
5.2.6 Connection between LQR and the Symmetric Root Locus
6 State-variable models with various inputs and outputs
6.1 Considering various inputs . . . . . . . . . . . . . . . . . .
6.1.1 Considering a disturbance input . . . . . . . . . . .
6.2 Considering various outputs . . . . . . . . . . . . . . . . .
6.3 Determining the Steady-State error for a state-variable model
6.3.1 Example . . . . . . . . . . . . . . . . . . . . . . .
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9 Summary, state-variable modeling and control
121
78
79
82
84
86
87
88
89
. 89
. 93
. 95
. 97
. 100
7 Integral control: augmenting the model with an integral control state 103
7.1 Augment the state vector with controller states . . . . . . . . . . . 103
7.2 Block diagram of state-variable controller with
controller states . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
7.3 Example, integral control for the voice-coil motor . . . . . . . . . 108
7.3.1 Now, set N f . . . . . . . . . . . . . . . . . . . . . . . . 111
7.4 Summary of integral control . . . . . . . . . . . . . . . . . . . . 113
8 Controllability and observability
8.1 Main facts about uncontrollable or unobservable modes: . .
8.2 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.3 Mathematical tests: . . . . . . . . . . . . . . . . . . . . . .
8.3.1 Test for controllability . . . . . . . . . . . . . . . .
8.3.2 Test for observability . . . . . . . . . . . . . . . . .
8.4 Summary Controllability and Observability . . . . . . . . .
8.4.1 Uncontrollable and unobservable modes and rocketry
Part 4: State-Space Control
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Part 4: State-Space Control
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Page 4
EE/ME 574: Intermediate Controls
Section 1.1.0
EE/ME 574: Intermediate Controls
Section 1.1.0
1 Introduction to state-variable models
• To introduce the the subject and advantages of state-variable modeling, we
will consider an example.
• We will compare and contrast state-variable modeling with “classical
modeling”
Figure 1: Voice-coil motor schematic (Example 2.10, Figure 2.19 in Franklin et
al. 5th Ed.).
– Classical modeling:
∗ Algebraic simplification of equations to 1 nth order differential
equation in 1 variable.
• Figure 1 shows a schematic diagram. The a voice-coil motor combines
– An electrical side
– State-variable modeling:
– A mechanical side
∗ Little algebraic simplification of differential equations required
• The model equations are:
∗ Represent the model as n first order differential equations
1.1 Voice-coil motor: an example contrasting classical and
cone :
coil current :
state-variable modeling
back emf
M z̈ (t) = −bż (t) + ks ia (t)
d
La ia (t) = −Ra ia (t) − ecoil (t) + va (t)
dt
ecoil (t) = ks ż (t)
(1)
(2)
(3)
• The model for voice-coil motor will be used as an example to develop the
ideas of state variable modeling and control.
• Eqn (1) comes from the mechanics of the voice coil
• Voice-coil motors are linear motors used in high speed applications, such as
the hummingbird robot developed by IBM research laboratories (cf. Zai et
al., 1992).
• Eqn (2) comes from the circuit and the inductor
• Eqn (3) reflects the back EMF of the coil
• A model for a voice-coil motor is developed in chapter 2 of Franklin et al.
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 5
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 6
EE/ME 574: Intermediate Controls
Section 1.1.1
EE/ME 574: Intermediate Controls
• Signals and parameters are listed in table 1
Section 1.1.2
– Now the ia (t) term on the right has to be eliminated, use Eqn (1) to write:
Signal
Description
Units
Parameter
Description
z (t)
Cone position
[m]
M
Cone mass
ia (t)
Coil current
[A]
b
Cone damping
va (t)
Applied voltage
[V]
ks
Coil motor constant
Ra
Coil resistance
La
Coil inductance
Table 1: Signals and parameters of the voice coil model.
Units
i
h
N
[kg] = m/s
2
h i
N
A
N
m/s
=
h
V
m/s
ia (t) =
(6)
– Plugging (6) back into 5 gives
i
V
[Ω] = A
[H]= V A−s
1
(M z̈ (t) + b ż (t))
ks
M ˙z˙˙(t) = −b z̈ (t) + ks
1
1
(M z̈ (t) + b ż (t)) − ks ż (t) + va (t)
−Ra
La
ks
– After a certain amount of algebra, one finds:
1.1.1 Developing a classical model
L M ˙z˙˙(t) + La b z̈ (t) + Ra M z̈ (t) + Ra b ż (t) + ks2ż (t) = ks va (t)
• A classical model is developed first, to show the contrast with state-variable
modeling.
• We can see from Eqns (1)-(3) that the system will have a 3rd order model.
• If the system input is va (t) and the output is z (t), we should eliminate
ecoil (t) (easy) and i (t) (harder) from the equations
(7)
– Which gives the transfer function:
Z (s)
ks
=
Va (s) La M s3 + (La b + Ra M) s2 + (Ra b + ks2) s
(8)
1.1.2 Some observations on developing the classical model
• It requires many algebra steps
– Taking the derivative of Eqn (1) will introduce d ia (t) /dt :
Actually, the algebra steps required goes up rapidly in the system order
M ˙z˙˙(t) = −b z̈ (t) + ks i˙a (t)
(4)
– 3rd is 4 − 5×more algebra than 2nd order
∗ 4th order is 10 − 20×more algebra than 2nd order
– The d ia (t) /dt term can be eliminated with Eqn (2) :
M ˙z˙˙(t) = −b z̈ (t) + ks
Part 4: State-Space Control
1
(−Ra ia (t) − ks ż (t) + va (t))
La
(Revised: Jan 27, 2016)
∗ Imagine deriving a 6th order model !
(5)
Page 7
∗ Imagine deriving a 200th order model !
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 8
EE/ME 574: Intermediate Controls
Section 1.1.2
• By eliminating the intermediate variables we throw away information
– Suppose we have an ammeter in the circuit and measure ia (t), how do
we relate that to Eqn (5) ?
EE/ME 574: Intermediate Controls
Section 1.1.3
1.1.3 Writing the model in state-variable form
• A State Variable (SV) model addresses all of the above concerns about
classical models
• To construct a n SV model, we need n, 1st order differential equations
• To solve Eqn (5) we need the initial conditions z (t0 ), ż (t0 ), z̈ (t0)
• The form for a state-variable model is
– What is the physical significance of z̈ (t0 )?
ẋ (t) = Fp x (t) + G p u (t)
(9)
y (t) = H p x (t) + J p u (t)
(10)
• What are the mathematical properties of Eqn (5) ?
u(t)
– Is it the simplest form ?
Fp, Gp, Hp, Jp
y(t)
x(t)
– Can we determine the states from measurements of z (t)?
∗ Or from ia (t) ?
(ia (t) is even in the model !)
Figure 2: Basic block diagram of a state-variable model. Fp , G p , H p , J p are
model matrices. x (t) is the state vector.
• Paying attention to details, there are five steps in writing a state variable
model:
– By choosing va (t) can we control z (t) ?
• What other forms for the model are possible,
1. Write the relevant relations for the system
2. Identify the differential equations
– For example, we could have developed the model Ia (s) /Va (s)
3. Write the differential equations in state-variable form
– Can we describe all possible alternative model forms ?
4. Write the equation of the output signal (or signals) using the states and
inputs
5. Verify correctness, by verifying the individual rows of the state variable
model, and check units in the state and output equation.
——– ∼ ——–
• Let’s look at each step and break it out into sub-steps, and see an example.
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 9
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 10
EE/ME 574: Intermediate Controls
Section 1.1.4
1.1.4 Form for the state-variable model
EE/ME 574: Intermediate Controls
Section 1.1.5
1.1.5 Steps of building a state-variable model, with the voice-coil motor
example
• The form for a state-variable model is
1. Write the relevant relations for the system
ẋ (t) = Fp x (t) + G p u (t)
(11)
y (t) = H p x (t) + J p u (t)
(12)
(a) Define symbols for the signals and parameters
(b) Write the equations
u(t)
Fp, Gp, Hp, Jp
i. Constituent relations (for elements)
y(t)
ii. Continuity constraints (for how elements are linked into a system)
x(t)
(c) Record the units, verify that units balance in the equations
Figure 2 (repeated). Basic block diagram of a state-variable model.
Signal
x (t)
u (t)
y (t)
name
description
Parameter
State vector n-element vector
Input vector m-element vector
Output vector p-element vector
name
• The equations express laws of physics, the units must balance
System matrix n × n matrix
Input matrix n × m matrix
Output matrix p × n matrix
DT term
p × m matrix
Fp
Gp
Hp
Jp
——– ∼ ——–
description
• For the VCM:
– Going back to the basic modeling equations, and dividing through by
any leading constant
Table 2: Signals and parameters of the state-variable model.
cone :
• For a linear model with constant coefficients , Fp , G p , H p , J p are constant
coil current :
– Matrices Fp , G p , H p , J p represent the system
back emf
ks
b
ż (t) + ia (t)
(15)
M
M
Ra
1
1
d
ia (t) = − ia (t) − ecoil (t) + va (t) (16)
dt
La
La
La
ecoil (t) = ks ż (t)
(17)
z̈ (t) = −
• Signals x (t), u (t) and y (t) are signals
– Re-write Eqn (15) as two first order equations, define v (t) = ż (t) ,
so v̇ (t) = z̈ (t):
• 2nd order example (to see what the model looks like):

x (t) = 
z (t)
ż (t)

,
Part 4: State-Space Control

ẋ (t) = 
−1
3


 x (t) + 
2
−2
0
h
i
y (t) =
3 4 x (t) + [0] u (t)
1

 u (t)
(Revised: Jan 27, 2016)
cone :
(13)
b
ks
v (t) + ia (t)
M
M
ż (t) = v (t)
v̇ (t) = −
(18)
(19)
(14)
– Table of parameters with units is given in table 1.
Page 11
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 12
EE/ME 574: Intermediate Controls
Section 1.1.5
2. Identify the differential equations
(a) Put derivatives in the left-hand side, these must be the state derivatives
• The number of independent “storage” elements
• The system order will almost always be sum of the orders of the
contributing differential equations
• Rarely, differential equations may be inter-dependent in a way that
reduces the order
(b) Select the state variables
(b) All signals on the right-hand side must be expressed in terms of the states
and inputs
(c) Higher order differential equations are written as a chain of first-order
equations
(d) Put the model in state-variable form
nth
order system
——– ∼ ——–
• State variables must be independent, no state variable can be a linear
combination of other state variables
(a) From Eqns (15)-(19) we can write:
• The highest derivatives in the differential equations are not state
variables, they go on the left-hand side of ẋ (t) = A x (t) + Bu (t)
ks
b
v (t) + ia (t)
M
M
ż (t) = v (t)
Ra
1
1
i˙a (t) = − ecoil (t) − ia (t) + va (t)
La
La
La
v̇ (t) = z̈ (t) = −
• The choice of state variables is not unique
• Often the “storage” variables are a good choice
• For the VCM:
——– ∼ ——–
(a) Eqns (16) (18) and (19) are 3 first-order differential equations,
system order: n = 3.
(b) Select the variables of the state vector


for Eqn (18)
v (t)



x (t) = 
for
Eqn (19)
z
(t)


for Eqn (16)
ia (t)
(Revised: Jan 27, 2016)
v̇ (t) = −
Page 13
(21)
(22)
(23)
(b) Looking at the right-hand side, all signals must be states or inputs.
Signals v (t) , ia (t) are states, va (t) is the input, however ecoil (t) is not a
state or input. Use the algebraic modeling equations to rewrite ecoil (t) in
terms of states+inputs:
ecoil (t) = ks ż (t)
Giving
– Now, for writing ẋ (t) = Fp x (t) + G p u (t) we need ẋ (t), which


is
v̇ (t)
∗ These are precisely the derivatives



ẋ (t) = 
of Eqns (16) (18), and (19)
 ż (t)  (20)
i˙a (t)
Part 4: State-Space Control
Section 1.1.5
3. Write the differential equations in state-variable form
(a) Determine the system order
• n state variables for an
EE/ME 574: Intermediate Controls
ks
b
v (t) + 0 z (t) + ia (t) + 0 va (t)
M
M
ż (t) = 1 v (t) + 0 z (t) + 0 ia (t) + 0 va (t)
Ra
1
ks
i˙a (t) = − v (t) + 0 z (t) − ia (t) + va (t)
La
La
La
Part 4: State-Space Control
(Revised: Jan 27, 2016)
(24)
(25)
(26)
Page 14
EE/ME 574: Intermediate Controls
Section 1.1.5
(c) Starting with
EE/ME 574: Intermediate Controls
Section 1.1.5
With matrices
b
ks
z̈ (t) = − ż (t) + ia (t)
M
M
(27)
Introduce
v (t) = ż (t)
v̇ (t) = z̈ (t)
(28)
Using Eqn (28) in (27) gives a chain of two first-order differential
equations, capturing the second order relationship of Eqn (27) :
ks
b
v (t) + ia (t)
M
M
ż (t) = v (t)
v̇ (t) = −

− Mb
ks
M
0



Fp =  1 0 0


− Lksa 0 − RLaa







and


 0 




Gp =  0 




1
La
(29)
(30)
(d) Eqns (24)-(26) is written in state-variable form with:
•
 
ks
b
 v (t)   − M 0 M
 

 

ẋ (t) =  z (t)  =  1 0 0
 

 

− Lksa 0 − RLaa
ia (t)

 

  v (t)   0 
 


 


  z (t)  +  0  [va (t)] (31)
 


 


1
ia (t)
La

• Which is the state equation:
ẋ (t) = Fp x (t) + G p u (t)
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 15
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 16
EE/ME 574: Intermediate Controls
Section 1.1.5
4. Write the equation of the output signal (or signals) using the states and
inputs
——– ∼ ——–
• For the VCM, the output is the position, z (t)
y (t) = [z (t)] =

v (t)



0 1 0 
 z (t)

ia (t)
EE/ME 574: Intermediate Controls
Section 1.1.5
• Keeping in mind the units of the state vector:
 m
 s

x (t) =  [m]

[A]





• Checking units in the state equation, ẋ (t) = Fp x (t) + G p u (t)




 + [0] [va (t)]


 m
 s2  m
 sec
h
i
Amp
sec
• Which is the output equation:

 "h
N
2
i#
[·]
 hm/sN i

m/s
 
 
=
1
[·]
  "h i#

V
m/s

[·]
[ VA−s ]
"h
kg−m/s2
A
kg
[·]
[ VA ]
[
V −s
A
]
i#


 m

 s

  [m]


 [A]
 
[·]
 
 
 +  [·]
 
A
V −s



 [volts]

• Simplifying the units expression:
y (t) = H p x (t) + J p u (t)
With
Hp =
0 1 0
and
 m
 s2  m
 sec
h
i
J p = [0]
Amp
sec
5. Verify correctness, by verifying the individual rows of the state variable
model, and check units in the state and output equation.
——– ∼ ——–
• For the VCM, we can examine the individual rows, and assure that the
correspond to the original modeling equations (from page 6).
 h 2i
m/s
[·]
s−1
A
 
−1
 
= 1
s
[·]
  −1
A
[·]
s
m

 m
 s

  [m]

[A]


[·]
 
 
 +  [·]
 
A
V −s



 [volts]

• Looking at the output equation
 m
h
i s

[m] = [·] 1 [·]  [m]

[A]



 + [·] [volts]

– The output signal is one of the states, so H p and J p are pretty simple
in this example.
– In some cases there is more going on in the output equation.
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 17
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 18
EE/ME 574: Intermediate Controls
Section 1.1.5
• Some things to notice about the example state variable model
EE/ME 574: Intermediate Controls
Section 1.1.5
• Further remarks on state-variable models
– All of the information about the system is in the Fp , G p , H p , J p matrices.
(any parameter that doesn’t show up in the matrices, doesn’t influence the
system).
– The middle row of the state equation, Eqn (31), is
– The state variable model is a very general form.
– In fact, any linear, time-invariant constant coefficient system model can
be written as a state variable model.
ż (t) = 1 v (t) + 0 z (t) + 0 ia (t) + 0 va (t)
ẋ (t) = Fp x (t) + G p u (t)
y (t) = H p x (t) + J p u (t)
∗ Or simply
ż (t) = 1 v (t)
∗ This derivative equation might seem trivial, since we defined v (t) =
ż (t). However, it is needed.
∗ The derivative equation is needed for each state variable, no matter
how simple.
– The state variable model can directly handle multi-input, multi-output
systems (like a helicopter)
∗ For a system with m inputs and p outputs:
• Some essential things to keep in mind for state-variable modeling:
– Always distinguish between signals and parameters
∗ Signals are functions of time and change when the input (signals)
change.
x (t) ∈ Rn×1
Fp ∈ Rn×n
u (t) ∈ Rm×1
G p ∈ Rn×m
y (t) ∈ R p×1
H p ∈ R p×n
J p ∈ R p×m
∗ Parameters are generally constant (or slowly varying) and are
properties of the system.
∗ Both have physical units
– Pay special attention to the states
∗ Systems have many signals, only n signals are states
∗ States correspond to the needed initial condition needed to determine
the output of the system
∗ With experience, it is often quite straight-forward to determine the
states for a state-variable model.
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 19
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 20
EE/ME 574: Intermediate Controls
Section 1.1.6
1.1.6 Matlab code to set up the voice-coil motor example
%% Setup the parameters as constants
ks = 0.63;
b = 0.03;
M = 0.2;
R = 0.3;
L = 0.02;
>> Fp = [ -(b/M) 0
(ks/M);
1
0
0
;
-(ks/L) 0 -(R/L) ]
>> Gp = [0; 0; (1/L)]
>> Hp = [ 1 0 0 ];
Jp = 0;
Fp =
-0.1500
1.0000
-31.5000
0
0
0
%% Define the state equations
%% using the constants
%% (not numbers)
EE/ME 574: Intermediate Controls
Section 1.1.6
>> OLPoles = pole(Ga_ss)
OLPoles =
0
-7.5750 + 6.6404i
-7.5750 - 6.6404i
>> OLZeros = zero(Ga_ss)
OLZeros = Empty matrix
>> figure(1), clf
%% Examine the system response
>> step(Ga_ss)
>> title(’Open Loop Step Response’); grid
>> print(’-deps2c’, ’FEStepOL’)
3.1500
0
-15.0000
Gp =
0
0
50
Open Loop Step Response, position
3
2.5
0
1
Jp =
0
0
2
Amplitude
Hp =
%% Build Matlab Sys model, examine poles and zeros
>> Ga_ss = ss(Fp, Gp, Hp, Jp);
1.5
1
0.5
Matlab prints
a =
b =
0
0
0.5
1
1.5
2
Time (sec)
x1
x2
x3
x1
-0.15
1
-31.5
x2
0
0
0
c =
x3
3.15
0
-15
x1
x2
x3
u1
0
0
50
Figure 3: Open-Loop system step response, note: system is type I.
d =
y1
x1
0
x2
1
x3
0
Part 4: State-Space Control
y1
u1
0
(Revised: Jan 27, 2016)
Page 21
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 22
EE/ME 574: Intermediate Controls
Section 1.1.6
EE/ME 574: Intermediate Controls
• By changing the H p and J p matrices, we can consider the velocity or the
current as the output of the system
>>
>>
>>
>>
Section 1.1.7
1.1.7 Some observations on developing the state-variable model
(Compare with section 1.1.2)
• Classical modelling: requires many steps of algebra
Hp_v = [1 0 0]; Jp_v = [0];
%% Cone Velocity as output
Hp_i = [0 0 1]; Jp_i = [0];
%% Coil current as output
figure(1), clf, step(ss(Fp, Gp, Hp_v, Jp_v), 2)
figure(2), clf, step(ss(Fp, Gp, Hp_i, Jp_i), 2)
– Writing the state variable model required almost no algebra
– For a large model, this would be a significant advantage
Open Loop Step Response, velocity
Open Loop Step Response, current
1.6
2.5
• Classical modeling: eliminates intermediate variables
1.4
2
1.2
– State variables are not thrown out !
Amplitude
Amplitude
1
0.8
1.5
– Information about the evolution of the states is explicitly available in the
model
1
0.6
0.4
0.5
0.2
• Classical modeling: What is the significance of the initial conditions ?
0
0
0.5
1
1.5
2
0
0
0.5
Time (sec)
1
1.5
2
Time (sec)
Figure 4: Velocity response and current response.
– The initial conditions, x (t = 0) are the states at t = 0 !
– The physical significance of the initial conditions, as well as how to
accurately determine them, is readily apparent.
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 23
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 24
EE/ME 574: Intermediate Controls
Section 1.1.7
• Classical modeling: transfer function provides limited mathematical insight
into the system
– State-variable modeling brings a wealth of insight into the details of the
physical system
EE/ME 574: Intermediate Controls
Section 1.2.0
1.2 State-variable modeling, conclusions
• An overview of basic system modeling (TF and SS) is found chapter 2 of
Franklin et al.
• State variable modeling in 5 steps:
∗ The modes of response
∗ Geometric interpretation of the model and system response (EE/ME
701)
1. Write the relevant relations for the system
2. Identify the differential equations
∗ Rigorous definitions of controllability and observability
· Very difficult with transfer functions, and a key original motivation
for state-space development
– Optimal control design methods
3. Write the differential equations in state-variable form
4. Write the equation of the output signal (or signals) using the states and
inputs
5. Verify correctness, by verifying the individual rows of the state variable
model, and check units in the state and output equation.
• Classical modeling: what other forms for the model are possible ?
– Model transformation shows the set of all possible state-variable models
for a system.
• Matlab manipulates state variable-models as a type of LTI model
• State variable models carry information (the states) that TF models throw
away.
• And we can get the TF from the SS model !
• New control design methods for state-variable models (coming up!)
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 25
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 26
EE/ME 574: Intermediate Controls
Section 2.2.0
2 Review of basics of linear algebra
EE/ME 574: Intermediate Controls
Section 2.3.0
2.3 Basic arithmetic: +, - and * are well defined
Define:
2.1 Scalars, matrices and vectors

1 2
A=
;
3 4
• Matrices are rectangular sets of numbers or functions, examples:
• Matrices have zero or more rows and columns:



v (t)
1 0 7






x (t) =  w (t)  , x (t) ∈ R3×1
A= 4 5 0



z (t)
2 3 6



,


3×3
A∈R
• Vectors are special cases of Matrices, with only one row or column,
– x is a column vector,
h
i
– w = 3 4 ∈ R1×2 is a row vector
• Scalar values (numbers or functions with one output variable) can also be
treated as matrices or vectors:
3 = [3] ∈ R1×1
• ’Array’ is a synonym for Matrix


 −1 1 
B=
;
2 3


2
 
x =  ;
3
w=


3 1 




C =  4 −1 




5 2
3 4
• Operations +, - and * are well defined. The dimensions of the operands
must be compatible.
• For addition and subtraction, the operation is element-wise, and the
operands must be the same size:

0
3


A+B = 

5 7


2
1


A−B = 

1 1

– If not the same size, there is no defined result (the operation is
impossible)
A + C = undefined
2.2 Transpose
Transposing an array rearranges each column to a row:

3
1





C =  4 −1 


5 2

CT = 
3
4
5
1 −1 2
We say that A and C are incommensurate for addition.


• C ∈ R3×2 , CT ∈ R2×3
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 27
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 28
EE/ME 574: Intermediate Controls
Section 2.3.1
EE/ME 574: Intermediate Controls
Section 2.3.1
• Examples:
2.3.1 Multiplication
• For multiplication, the operation is by row on the left and column on the
right.
– To produce one element of the result, go across each row on the left and
multiply with elements of the column on the right. With




1
2
−1
1




A=
, B=

3 4
2 3
 (1 ∗ (−1) + 2 ∗ 2)
A ∗ B = AB = 
(3 ∗ (−1) + 4 ∗ 2)


3 1 


 1 2 

AC = 
  4 −1  = undefined



3 4 
5 2


With n = 3 , m = 2 , j = 2 , k = 2 : then m = j :
We get

With n = 2 , m = 2 , j = 3 , k = 2 , then m 6= j :

 
(1 ∗ 1 + 2 ∗ 3)   3 7 

=
5 15
(3 ∗ 1 + 4 ∗ 3)






3 1 
 6 10 

 1 2




 

C A =  4 −1  
= 1 4 





 3 4


5 2
11 18
– A and B don’t have to be the same size to multiply

  
 

1
2
2
(1
∗
2
+
2
∗
3)
8

  
 

Ax = 
 =
=

3 4
3
(3 ∗ 2 + 4 ∗ 3)
18
– For A B = C , given A ∈ Rn×m and B ∈ R j×k
∗ The number of columns in A must match the number of rows in B
that is, m = j
∗ The size of the result matrix is determined by the number of rows in
A and columns in B
∗ To multiply A and B, we must have m = j , then C ∈ Rn×k
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 29
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 30
EE/ME 574: Intermediate Controls
2.4 Commutative,
associative,
Section 2.4.2
distributive and identity
properties
EE/ME 574: Intermediate Controls
Section 2.4.4
2.4.3 Distributive property
• Like scalar algebra, * distributes over + and - :
2.4.1 Commutative property
(A + B) C = A C + B C
• Like scalar algebra, addition commutes: A + B = B + A
• Like scalar algebra, subtraction commutes with a ’-’ sign:
A − B = − (B − A)
C (A − B) = C ∗ A − C ∗ B
• Unlike scalar algebra, multiplication does not generally commute:
A B 6= B A
Note: left-to-right order of multiplications in second example.
2.4.4 Identity Matrix
– Example:


3
7


AB = 
;
5 15


2
2


BA = 

11 16
• Like scalar algebra, linear algebra has a multiplicative identity:
IL C = C IR = C
– ’Generally’: There are many properties of linear algebra that are
’generally’ true, but may not be true for some special cases.
– Examples:
2.4.2 Associative property
• Like scalar algebra, +, - and * have the associative property:
(A + B) + C = A + (B + C)



1  3 1 
 

 

=  4 −1 
−1 
 

 

2
5 2



3 1 

0 


 =  4 −1 




1
5 2
– If A is square, then IL = IR , and we just call it I, the identity matrix:
(A + B) − C = A + (B − C)

 

 

1
0
1
2
1
2
1
0
1
2

 


 


=


=

0 1
3 4
3 4
0 1
3 4

(A ∗ B) ∗ C = A ∗ (B ∗ C)
(Revised: Jan 27, 2016)

 1 0 0  3




 0 1 0  4




0 0 1
5



3 1 

 1


 4 −1  



 0
5 2
– ’Does not generally commute’ means that most matrices do not commute,
but that perhaps special matrices can be found that do commute.
Part 4: State-Space Control

Page 31
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 32
EE/ME 574: Intermediate Controls
Section 2.4.5
EE/ME 574: Intermediate Controls
Section 2.5.0
2.5 Linear independence and the determinant
2.4.5 Doing algebra with vectors and matrices
• Starting with an equation,
• The columns (or rows) of a matrix are linearly independent if there are no
coefficients a1, a2, ... such that
– We can add, subtract or multiply on the left or right by any allowed term
and get a new equation.
– Examples, given:
a1 c1 + a2 c2 + ... + an−1 cn−1 = cn
(32)
Or, equivalently, there is no vector x such that C x = 0, where C = [c1 , c2 , ...]
is the array of vectors ci , and 0 is the null vector (vector of zeros).
A+B = C
• The determinant is a scalar measure of the ’size’ of a square matrix.
Then
det (A) = |A| ∈ R1
(33)
(A + B) + D = C + D
– The determinant is not defined for a non-square matrix.
• The determinant of a matrix will be non-zero if and only if the rows (or,
equivalently, columns) of a matrix are linearly independent. Examples:
E (A + B) = E A + E B = E C
(A + B) F = C F
(a)
– Where A, B, C, D, E and F are compatible sizes.
• Matrices that are the appropriate size for an operation are called
commensurate .
1 2 3 4 5 6 =0
10 14 18 (b)
1 2 3
4 5 6
7 8 −9
= 54
(34)
– In case (a), the third column is given by 2xCol2 - 1xCol1.
In case (b), the three columns are independent.
– Notice that in case (a), the third row is given by 2xRow1 + 2xRow2.
∗ Always for a square matrix, if the columns are dependent, the rows
will be dependent.
Part 4: State-Space Control
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Page 33
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 34
EE/ME 574: Intermediate Controls
Section 2.5.1
2.5.1 Computing the determinant
Section 2.5.2
2.5.2 Some relations involving determinants
The determinant of a square matrix (the only kind !) is defined by Laplace’s
expansion (following Franklin et al.):
1. det (I) = 1, where I is the identity matrix
2. det (A B) = det (A) det (B)
n
det A =
EE/ME 574: Intermediate Controls
∑ ai j γi j
for any i = 1, 2, ..., n
(35)
j=1
3. Given M, an invertible matrix,
where ai j is the element from the ith row and jth column of A; and γi j is called the
cofactor, given by:
(36)
γi j = (−1)i+ j det Mi j
where Mi j is called a minor. Mi j is the same matrix as A, with the ith row and jth
column removed.
(a) det M −1 = 1/ det (M)
(b) det M A M −1 = det (M) det (A) 1/ det (M) = det (A)
• For example, with i = 1,


j=1
j=2
j=3






a b c 


e
f
d
f
d
e








det  d e f  = a det 
 − b det 
 + c det 





h i
g i
g h
g h i
• Closed-form expressions for 2x2 and 3x3 matrices are sometimes handy.
They are:


a b
 = ad − bc
 det 

c d
a b c 




det  d e f  = a e i − a h f − b d i + b g f + c d g − c g e




g h i
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 35
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 36
EE/ME 574: Intermediate Controls
Section 2.6.0
2.6 Rank
EE/ME 574: Intermediate Controls
2.7 The norm of a vector
• The rank of a matrix is the number of independent rows and columns
r = rank (A)
• The norm of a vector, written ||x|| is a measure of the size of the vector.
(37)
• The number of independent rows is always equal to the number of
independent columns
• In mathematics many “measures of size” are defined for special purposes.
• The most common measure by far is the "2-norm", (a.k.a. the Euclidean
norm)
s
n
• Example:
||x||2 =

1 2 3 4


A = 5 6 7 8

6 8 10 12
rank (A) = 2
3rd
∑ x2i
(38)
i=1

example:





– The third row is the sum of the first two.
– Notice that the
Section 2.7.0
||x||2 =
column is 2 x Col2 - Col1
– And the 4th column is 3 x Col2 - 2 x Col1
r

2






x =  −3 




−4
(39)
22 + (−3)2 + (−4)2 = 5.39
(40)
• The rank of a matrix can not be greater than the number of rows or columns
rank (A) ≤ min (n , m)
• A matrix is said to be full rank if rank (A) takes its maximum value, that is
rank (A) = min (n , m)
• Otherwise the matrix is said to be rank deficient.
• A square matrix is invertible if and only if it is full rank.
• The determinant of a square matrix is zero if the matrix is rank deficient.
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 37
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 38
EE/ME 574: Intermediate Controls
Section 2.8.0
EE/ME 574: Intermediate Controls
2.8 A quick review of eigenvalue and eigenvectors
A quick review of eigenvalue and eigenvectors (continued)
These are the main properties of eigenvectors and eigenvalues, which are typically
covered more thoroughly and rigorously in a first course in linear algebra.
• For any square matrix A, certain vectors have the property that the matrix
changes magnitude only, not direction.
– For A, such vectors are:
• A vector has both a direction and a magnitude. Consider





Section 2.8.0



1
 −1.5 
 1.58 
v1 =   , v2 = 
 , v3 = 

2
−3.0
1.58
1.5
1.5


=
2.0 −0.5
1.0
0.5


1
1


;


1
2


=
2.0 −0.5
1.0
0.5


1
2


• These special vectors are called the eigenvectors
– v1 has the same direction as v2 , but has a different magnitude.
• An eigenvector is scaled by the matrix

– v3 has the same magnitude as v1 , but has a different direction.

x2
v1
• In general, multiplying a vector v by a matrix
v3
x1
y4 = A v4
1.5


2.0 −0.5

1


1


1

 = 
   = 1.5   , so v1 =   , λ1 = 1.5
1
1.5
1.0 0.5
1
1

 
 
 
 
2.0 −0.5
1
1
1
1
  = 
   = 1.0   , so v2 =   , λ2 = 1.0
1.0 0.5
2
2
2
2
• We can write:
introduces both a change of magnitude and a change of
direction
v2
 


1
2.5
2.0
−0.5
 



y4 = A v4 = 
=


−1
0.5
1.0 0.5
λ i vi = A vi ,
x2


for example 1.5 
1
1


 = A
1
1


(41)
where the λi are the eigenvalues and the vi are the eigenvectors.
y4
x1
v4
• Notice that in general a x 6= A x, because multiplication by matrix A will
rotate a general vector x
(Choosing vectors x at random, what is the probability of selecting an
eigenvector ?)
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 39
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 40
Student
Exercise
EE/ME 574: Intermediate Controls
Section 2.8.1
EE/ME 574: Intermediate Controls
Section 2.8.1
4. A complete set of eigenvectors exists when an n × n matrix has n distinct
eigenvectors (so V, the matrix of eigenvectors, is invertible).
2.8.1 Some properties of Eigenvectors and Eigenvalues
1. Only square matrices have an eigensystem
• For the example matrix
2. The eigenvalues are the solutions to the equation
det (A − λ I) = 0 .
• Example, using
(42)


2.0
−0.5


A=

1.0 0.5
with >> [V, U] = eig(A);


• The matrix of eigenvectors is
a b
det 
 = ad − bc
c d
This leads to a polynomial equation in λ





2.0
−
λ
−0.5
2.0
−0.5
1
0






det 

−λ 
 = det 
1.0 0.5 − λ
1.0 0.5
0 1

(a) A 3x3 matrix gives an equation in λ3, a 4x4 gives an equation in λ4 , etc.
(b) Abel’s theorem states: there is no closed-form solution for the roots of
a polynomial of 5th order and above, therefore there is no closed-form
solution for the eigenvalues a 5x5 matrix or larger.

1
1


V=

1 2
and the matrix of eigenvalues is


1.5
0


U=

0 1
= (2.0 − λ)(0.5 − λ) − 1.0 (−0.5) = λ2 − 2.5λ + 1.5 = 0
3. The polynomial equation given by Eqn (42) is called the characteristic
equation of the matrix A

5. A matrix that lacks a complete set of eigenvectors is said to be defective,
but this name is meaningful only from a mathematical perspective. For
many control systems, some combinations of parameter values will give a
’defective’ system matrix.
• This case corresponds to the case of repeated roots in the study of
ordinary differential equations, where solutions of the form y (t) = t e−λt
arise.
• A defective matrix can only arise when there are repeated eigenvalues.
• In this case, a special tool called the Jordan Form is required to solve
the equation ẋ (t) = A x (x).
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 41
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 42
EE/ME 574: Intermediate Controls
Section 2.8.1
6. The eigensystem is computed in Matlab with the eig() command,
• V is the matrix of eigenvectors, U is a matrix with the corresponding
eigenvalues on the main diagonal.
[
1
2
-2
4
5
6
>> [v,u] = eig(A)
v =
0.5276
0.2873
-0.7994
u =
-2.4563
0
0
Section 2.8.2
2.8.2 A few more notes on Eigenvectors and Eigenvalues
>> [v,u] = eig(A)
>> A =
EE/ME 574: Intermediate Controls
3
1
3 ]
1. Don’t use the determinant equation to determine the characteristic equation
to find eigenvalues. Don’t solve:
det (A − λI) = 0
(44)
λn + an−1λn−1 + · · · + a1λ + a0 = 0
(45)
to get
and then solve Eqn (45) to find the eigenvalues λi
0.5267 - 0.0250i
-0.1294 + 0.1030i
0.8335
0.5267 + 0.0250i
-0.1294 - 0.1030i
0.8335
0
5.2282 + 0.5919i
0
0
0
5.2282 - 0.5919i
• Some reasons to avoid the determinant are:
(a) Going from Eqn (44) to Eqn (45) by solving the determinant involves
symbolic manipulation of matrix A. It is a lot of work.
(b) The determinant does not give the eigenvectors
(c) Once you get Eqn (45), how are you going to find the roots of the
polynomial ? (Answer on the next page)
7. Notice that eigenvalues can be real or complex.
>> Poly = poly(diag(u))
Poly =
1.0000
-8.0000
(d) Working with the determinant does not lead to a numerically stable
algorithm.
2.0000
68.0000
The characteristic equation corresponding to 3 × 3 matrix A is
λ3 − 8 λ2 + 2 λ + 68 = 0
(43)
• As with many things, wikipedia has a very nice article on eigenvectors and
eigenvalues:
“Everything you wanted to know about the eigensystem of A, in 3 pages. “
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 43
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 44
EE/ME 574: Intermediate Controls
Section 2.8.2
EE/ME 574: Intermediate Controls
Section 2.8.3
2. Matlab does not create the characteristic equation using det (A − λ I) = 0,
and then solving for the polynomial roots.
• Relationship:
• Matlab actually goes the other way, and solves for the roots of a polynomial
by forming a matrix and finding the eigenvalues of that ! The polynomial
• For a state-variable model, the poles of the system are the eigenvalues of the
F matrix
λ3 − 8 λ2 + 2 λ + 68 = 0
Roots of Characteristic Equation ⇔ Eigenvalues of the Companion Matrix
– If the F matrix has poles in the right half plane, the system is unstable.
is represented in Matlab as the vector
• Note: I haven’t told you an algorithm for eig(F).
Poly = [ 1, -8, 2, 68]
The algorithm is: Use Matlab !
and the command roots() finds the roots:
2.8.3 Positive definite and positive semi-definite matrices
>> roots(Poly)
ans = [ 5.2282 + 0.5919i
5.2282 - 0.5919i
-2.4563
• A matrix is positive definite if
– Actually ‘under the hood’ Matlab forms a matrix (called the companion
matrix) and applies the eigenvalue routine to that:
>> compan(Poly)
ans =
8.0000
1.0000
0
-2.0000
0
1.0000
-68.0000
0
0
xT A x > 0 ∀ x ∈ Rn
– A symmetric matrix A is positive definite (PD) if all eigenvalues λi > 0
– For diagonal matrices, A is positive definite if each diagonal element is
greater than zero.
• A matrix is positive semi-definite if
>> eig(compan(Poly))
ans = [ 5.2282 + 0.5919i
5.2282 - 0.5919i
-2.4563 ]
Part 4: State-Space Control
λ 3 + a2 λ 2 + a1 λ + a0 = 0


 −a2 −a1 −a0 




C= 1

0
0




0
1
0
(Revised: Jan 27, 2016)
Page 45
xT A x ≥ 0 ∀ x ∈ Rn
– A symmetric matrix A is positive semi-definite if all eigenvalues λi ≥ 0
– For diagonal matrices, A is positive definite if each diagonal element is
greater than or equal to zero.
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 46
EE/ME 574: Intermediate Controls
Section 2.8.4
2.8.4 One final fact about the eigensystem: V diagonalizes A
EE/ME 574: Intermediate Controls
V diagonalizes A (continued)
• When a complete set of eigenvectors exists, the matrix of the eigenvectors
• A numerical example:

– which is an n × n matrix
can be used to transform A into a diagonal matrix, by similarity transform
• General form for the similarity transform:
b = T A T −1
A

1
2
3






A= 4 5 6 




7 8 9
[V, Λ] = eig (A)


−0.2320
−0.7858
0.4082






V =  −0.5253 −0.0868 −0.8165 




−0.8187 0.6123 0.4082
(46)
• Properties of a similarity transform:

0
0
 16.1168


U = 
0
−1.1168
0


0
0
0.0000
– T must be invertible
– The similarity transform preserves eigenvalues
b
eig (A) = eig(A)
Diagonalizing A :
U = V−1 A V
(47)
Recovering A from [V, U] :
A = V U V−1
(48)

0
0
 16.1168


−1
V AV = 
0
−1.1168
0


0
0
0.0000
−1
• Eqns (47) and (48) are both similarity transforms
(Revised: Jan 27, 2016)







– We have:
• The equations for V diagonalizing A are, with V the array of eigenvectors,
and U with the eigenvalues on the diagonal:
Part 4: State-Space Control
Section 2.8.4
VUV
Page 47
Part 4: State-Space Control






=U


1 2 3




=  4 5 6 =A




7 8 9
(Revised: Jan 27, 2016)
Page 48
EE/ME 574: Intermediate Controls
Section 3.1.0
3 Working with state-variable models
EE/ME 574: Intermediate Controls
Section 3.1.0
• Or we can use Matlab !
3.1 Deriving the transfer function from the state-variable
model
>> [Np, Dp] = ss2tf(Fp, Gp, Hp, Jp); %% One method (old fashioned)
• Or, using LTI models
• It is straight forward to get the transfer function from a state variable model.
Starting with the state equation
>> SSp
>> Gps
= ss(Fp, Gp, Hp, Jp);
= tf(SSp);
%% Create a state variable model object
%% Second method (uses tf object)
ẋ (t) = Fp x (t) + G p u (t)
Transfer function:
157.5 s - 1.553e-29
-------------------------------s^3 + 15.15 s^2 + 101.5 s
• Considering the Laplace transform of the state equation:
s X (s) = Fp X (s) + G p U (s)
(49)
Y (s) = H p X (s) + J p U (s)
(50)
• Eqn (49) leads to:
• Returning to Eqn (53)
Y (s)
= H p (s I − Fp)−1 G p + J p
U (s)
(s I − Fp) X (s) = G p U (s)
(Note round-off error)
(53, repeated)
or
X (s) = (s I − Fp)−1 G p U (s)
(51)
• With (51), Eqn (50) leads to:
• From Eqn (53) we can determine the DC Gain of a state-variable model,
KDC = lim
s→0
Y (s) = H p (s I − Fp)−1 G p U (s) + J p U (s)
Y (s)
= lim H p (s I − Fp)−1 G p + J p = H p (−Fp)−1 G p + J p
U (s) s→0
(54)
(52)
– If the system is a Type 0 system, Eqn (54) will give a finite result.
• If m = 1 (one input) and p = 1 (one output), then Eqn (52) gives the transfer
function:
Y (s)
(53)
= H p (s I − Fp)−1 G p + J p
U (s)
• Eqn (53) can be solved symbolically by Cramer’s rule, to give the symbolic
transfer function.
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 49
– If the system is a Type I system, (−Fp )−1 will blow up, giving ∞ gain.
∗ limω→0 G (s = jω) does go to ∞ for a Type 1 system
∗ More on this later .
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 50
EE/ME 574: Intermediate Controls
Section 3.1.1
• If (−Fp)−1 is undefined, Matlab can still determine the transfer function (by
carrying out the calculation of Eqn (53) symbolically, and canceling terms
that go to zero).
>> Fp
Fp = -0.1500
0
3.1500
1.0000
0
0
-31.5000
0 -15.0000
>> inv(-Fp)
Warning: Matrix is singular to working precision.
ans =
Inf
Inf
Inf
Inf
Inf
Inf
%% ss() gives an SS object
Inf
Inf
Inf
%% tf() converts to tf object
>> Gtf = tf( ss(Fp, Gp, Hp, Jp) )
Transfer function:
157.5
------------------------s^3 + 15.15 s^2 + 101.5 s
EE/ME 574: Intermediate Controls
3.2 Getting a state-space model from a transfer function
• The usual way to derive a state-variable model is from the basic modeling
equations of the system
• But it can be desirable to obtain a state variable model directly from the
transfer function. One approach (used by the Matlab tf2ss() command)
is to express the model in controllable canonical form.
• Given a system with a transfer function
u(t)
y(t)
• It can be considered a poles filter, A (s), and a zeros filter B (s). The two
components can be arranged
u(t)
• If J p ∼=0, then
B(s)
A(s)
3.1.1 Interpreting J p
• A direct transmission term corresponds to a transfer function that is not
strictly proper.
Section 3.2.0
1
A(s)
xn(t)
y(t)
B(s)
Figure 5: Transfer function with zeros in the output path.
• If we build
the number of zeros equals the number of poles.
– A system matrix that produces 1/A (s), and
– An output matrix that produces B (s)
• In many controls situations,
Jp = 0
=⇒ Then we will have a state-variable model corresponding to the
transfer function B (s) /A (s).
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 51
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 52
EE/ME 574: Intermediate Controls
Section 3.2.0
• Consider the strictly proper transfer function
T (s) =
• For example,
T (s) =
bm−1sm−1 + · · · + b1s + b0
an sn + an−1sn−1 + · · · + a1s + a0
(55)
(56)
• Eqn (56) comes from 1/A (s), and can be implemented on an analog
computer as shown in figure 6
+
.
x1
..
.
x3(t)
x3(t)
x. 2
x1
∫
+
-a2
∫
x3(t)
x. 3
x2
-a1
∫
x3(t)
x3
.
x1
x. 2
x1
∫
x. 3
x2
∫
x3(t)
x3(t)
x3
∫
-a1
-a0
(Notation follows Bay)
Figure 6 (repeated). 1/A (s) constructed in phase-variable form. A TF
denominator of any order can be constructed by chaining n integrators,
introducing n states.

ẋ1 (t)


ẋ (t) =  ẋ2 (t)

ẋ3 (t)


−a2 −a1 −a0
 
 
= 1
 
0
0
1



0   x2 (t)

0
x3 (t)
——– ∼ ——–
u(t)
1
A(s)
x1 (t)
xn(t)


1

  
  
 +  0  u (t) (57)
  
0
y(t)
B(s)
Figure 5 (repeated). xn (t) is generated by Eqn (57).
-a0
• Looking back to figure 5, the numerator term corresponds to
(Notation follows Bay)
Figure 6: Realization of controllable canonical form, incorporates Phase
Variable Feedback.
iT
h
• The states in figure 6, x (t) = x1 x2 x3 , are called the “phase
variables” because they are successive derivatives of xn (t).
B (s) =
Y (s)
= b2 s 2 + b1 s + b0
Xn (s)
(58)
for the m = 2 case. Converting the TF, Eqn (58), back to a diff. eq.,
y (t) = b2
x2 (t) = ẋ3 (t) ,
.
x3(t)
• The state equation for the realization of figure 6 may be written:
˙x˙˙3 (t) = −a2 ẍ3 (t) − a1ẋ3 (t) − a0 x3 (t) + u (t)
u(t)
..
x3(t)
x3(t)
note : a3 = 1
Which can be written:
x3(t)
u(t)
Section 3.2.0
-a2
b2 s 2 + b1 s + b0
,
s 3 + a2 s 2 + a1 s + a0
˙x˙˙3 (t) + a2 ẍ3 (t) + a1 ẋ3 (t) + a0 x3 (t) = u (t)
Signal:
...
Signal:
• The transfer function denominator corresponds to the differential equation
...
EE/ME 574: Intermediate Controls
d 2 x3 (t)
d 1 x3 (t)
+ b1
+ b0 x3 (t)
dt 2
dt
(59)
x1 (t) = ẋ2 (t) = ẍ3 (t) , and ẋ1 (t) = ẍ2 (t) = ẋ˙˙3 (t)
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 53
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 54
EE/ME 574: Intermediate Controls
Section 3.2.0
EE/ME 574: Intermediate Controls
• The output equation can be built from Eqn (59)
3.2.1 Example, state-space model from a transfer function
• If the system is strictly proper, then b3 =0, and
• Example
y (t) = b2 ẍ3 (t) + b1 ẋ3 (t) + b0 x3 (t)
(59, repeated)
>> num = [2 3 4]
>> den = [1 6 7 8]
– But recall the state vector in phase-variable form:
 


x1 (t)
ẍ3 (t)


 


 
x (t) =  x2 (t)  =  ẋ3 (t) 


 
x3 (t)
x3 (t)
%%
>> TF = tf(num, den)
Transfer function:

x1 (t)
 h

 = b2 b1 b0

h
i
h
i

y (t) = b2 b1 b0 x (t) = b2 b1 b0  x2 (t)

x3 (t)
– And the state model can be written:
 


1
−a2 −a1 −a0
 


 


x
(t)
+
ẋ (t) =  1
 0  u (t)
0
0 
 


0
1
0
0
h
i
y (t) =
b2 b1 b0 x (t) + [0] u (t)
T(s) =
i

 ẋ3 (t)

x3 (t)
b1
u(t)
+
.
x1
∫
.
x2
x1
-a2
∫
(61)
-a1
∫
%%
a0 = 1
Obtain the tf model
%% Generate the state-variable model
%% in Controllable Canonical Form
>> [Fp, Gp, Hp, Jp] = tf2ss(num, den)
Hp
=
-6
1
0
-7
0
1
-8
0
0
Gp =
1
0
0
2
3
4
Jp =
0
• Compare with:

x3
y (t) =
-a0
(Revised: Jan 27, 2016)
−a2 −a1 −a0


ẋ (t) =  1

0
h
0
1


1

 

 

0  x (t) +  0  u (t)
 

0
0
i
b2 b1 b0 x (t) + [0] u (t)
(62)
• The coefficients of A (s) and B (s) are simply plugged into Fp and H p
Figure 7: Strictly proper system in controllable canonical form.
Part 4: State-Space Control




y(t)
b0
.
x3
x2

(60)
s3 + a2 s2 + a1 s + a0
b2
ẍ3 (t)
Fp =
+
b2 s2 + b1 s + b0

Note
2 s^2 + 3 s + 4
--------------------s^3 + 6 s^2 + 7 s + 8
– So Eqn (59) can be written:

Section 3.2.1
Page 55
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 56
EE/ME 574: Intermediate Controls
Section 3.2.2
EE/ME 574: Intermediate Controls
Section 3.3.0
3.3 Coordinate transformation of a state-variable model
3.2.2 One more complication
• To reduce sensitivity in certain numerical computations, the ss() routine
‘balances’ the state-space realization when converting from a TF
• Question: given a state variable model of a system
u(t)
Fp, Gp, Hp, Jp
• This results in modified numerical values
>> TF1 = tf( num, den)
>> SS1 = ss( TF1 )
a =
x1
x2
x1
-6 -1.75
x2
4
0
x3
0
1
x(t)
x3
-2
0
0
b =
x1
x2
x3
u1
2
0
0
d =
c =
y1
x1
1
x2
0.375
y(t)
x3
0.5
– Is it possible to find a different realization of the model for the same
system ?
u(t)
Fp, Gp, Hp, Jp
y(t)
z(t)
– Is it possible to find all possible alternative realizations ?
y1
u1
0
• Example of two model realizations of a given system:
>> SSuy1=ss(Fp, Gp, Hp, Jp)
>> SSuy2 = ss(Fp2, Gp2, Hp2, Jp2)
Continuous-time model.
• Compare with controllable canonical form (given by tf2ss()):
– Structurally the same, but with ’balanced’ values
• For EE/ME 574, the take home points are:
Fp=
x1
x2
x1
-4
1
x2
0
-3
Gp=
u1
5
0
Fp2= x1
x1 -4
x2
0
x2
0
-3
Gp2= u1
5
5
Hp=
y1
x1
0
x2
2
Jp=
u1
0
Hp2= x1
y1
2
x2
-2
Jp2= u1
0
1. The state variable model is not unique.
>>
2. tf2ss() gives the model in standard controllable canonical form.
3. ss() gives the model in ’balanced’ controllable canonical form.
4. tf2ss() is used in the EE/ME 574 project.
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 57
tf(SSuy1)
Transfer function:
10
-------------s^2 + 7 s + 12
Part 4: State-Space Control
>>
tf(SSuy2)
Transfer function:
10
-------------s^2 + 7 s + 12
(Revised: Jan 27, 2016)
Page 58
EE/ME 574: Intermediate Controls
Section 3.3.1
3.3.1 Derivation of the state model transformation
EE/ME 574: Intermediate Controls
3.3.2 Example: transforming a system model into modal coordinates
• Starting with the voice-coil motor system model (Ra increased to have real
poles in the example, method works for real or complex poles).
• Starting with the state model based on state vector x (t)
ẋ (t) = Fp x (t) + G p u (t)
(63)
y (t) = H p x (t) + J p u (t)
(64)
• The new state vector z (t) is related to the original by a transformation
matrix T :
z (t) = T x (t)
(65)
• We can derive a new state model based on z (t). From Eqn (65) we can solve
for x (t):
x (t) = T −1 z (t)
(66)
ẋ (t) = T −1 ż (t)
(67)
T −1 ż (t) = Fp T −1z (t) + G p u (t)
Fp =
-0.1500
1.0000
-31.5000
0
0
0
3.1500
0
-25.0000
Gp =
0
0
50
Hp =
0
1
0
Jp =
0
>> SSuy1 = ss( Fp, Gp, Hp, Jp)
>> Guy1 = tf( SSuy1 )
Transfer function:
157.5
----------------------s^3 + 25.15 s^2 + 103 s
• Find the eigensystem of Fp
Plugging (65)-(67) into Eqns (63) and (64) gives the state model
(68)
y (t) = H p T −1 z (t) + J p u (t)
>> [V, U] = eig(Fp)
V =
0
1.0000
0
U =
0
0
0
• Left-multiplying Eqn (68) by T gives:
ż (t) = T Fp T −1 z (t) + T G p u (t)
y (t) = H p T
which can be rewritten as:
−1
Section 3.3.2
(69)
z (t) + J p u (t)
0.5303
-0.1030
-0.8415
0
-5.1483
0
-0.1567
0.0078
0.9876
0
0
-20.0017
With:
(Recall: Fbp = T Fp T −1 and U = V−1 A V)
• Choose
cp u (t)
ż (t) = Fbp z (t) + G
cp z (t) + J p u (t)
y (t) = H
Part 4: State-Space Control
(70)
(71)
−1
Fbp = T Fp T
c
Gp = T Gp
cp = H p T −1
H
J p : unchanged
(Revised: Jan 27, 2016)
(72)
(73)
(74)
(75)
Page 59
>> T = inv(V)
T = 0.2428
2.5201
2.1473
Part 4: State-Space Control
1.0000
0
0
0.0306
0.3999
1.3533
(Revised: Jan 27, 2016)
Page 60
EE/ME 574: Intermediate Controls
Section 3.3.2
• Form the new model matrices:
>> Gphat = T*Gp
Gphat = 1.5295
19.9945
67.6637
And any invertible matrix T
1.2
1.2
1
1
0.6
0.4
0.4
0.2
0.2
1
cp u (t)
ż (t) = Fbp z (t) + G
cp z (t) + J p u (t)
y (t) = H
1.5
Time (sec)
2
2.5
Fbp = T Fp T −1
cp = T G p
G
cp = H p T −1
H
J p : unchanged
0.8
0.6
0.5
• We can construct a new realization of the system model with
Step Response
1.4
Amplitude
Amplitude
Step Response
1.4
x (t) , Fp , G p , H p , J p
ẋ (t) = Fp x (t) + G p u (t)
0.0078
>> SSuy2 = ss( Fphat, Gphat, Hphat, Jp)
>> Guy2 = tf( SSuy2 )
Transfer function:
157.5
----------------------s^3 + 25.15 s^2 + 103 s
0.8
• Given a model
y (t) = H p x (t) + J p u (t)
>> Hphat = Hp*inv(T)
Hphat = 1.0000
-0.1030
0
Section 3.3.3
3.3.3 Summary of coordinate transformation
>> Fphat = T*Fp*inv(T)
Fphat = 0 -0.0000
0.0000
0 -5.1483
0.0000
0 -0.0000 -20.0017
0
EE/ME 574: Intermediate Controls
0
• The set of all possible forms for the system model is given by the set of all
possible invertible matrices T .
0
0.5
1
1.5
o
n
cp , H
cp , J p
z (t) , Fbp , G
2
2.5
Time (sec)
• Some forms have particular theoretical or practical interest, such as modal
coordinates.
Figure 8: Step response of two realizations of the voice-coil motor.
• The model above and on page 60 are two model realizations for one system.
• Look at the Fbp matrix. The states are decoupled.
• The elements of the state vector z (t) are called the modes of the system.
• The state model in the form using Fbp is called the model in modal
coordinates.
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 61
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 62
EE/ME 574: Intermediate Controls
Section 4.0.0
4 Introducing state-feedback control
EE/ME 574: Intermediate Controls
4.1 Determination of a new model with state feedback
• Put in State Feedback control
• To determine the state-variable model of the system with feedback, start
with the open-loop model (figure 9)
u (t) = −K x (t) + N f r (t)
(76)
– The control signal depends on the state vector and a reference input.
• Putting in feedback control is illustrated in figures 9, 10 and 11.
u(t)
Section 4.1.0
Fp, Gp, Hp, Jp
y(t)
ẋ (t) = Fp x (t) + G p u (t)
(11, repeated)
y (t) = H p x (t) + J p u (t)
(12, repeated)
• Plugging the control law
x(t)
Figure 9: State-variable model of the open-loop system (the plant). This is the
plant before feedback control is applied, u (t) as input, and y (t) as
output.
r’(t)
r(t)
Nf
+
u(t)
x(t)
+
(76, repeated)
into the state equation, we find
ẋ (t) = Fp x (t) + G p −K x (t) + N f r (t) = Fp x (t) − G p K x (t) + G p N f r (t)
y(t)
Fp, Gp, Hp, Jp
u (t) = −K x (t) + N f r (t)
x(t)
= (Fp − G p K) x (t) + G p N f r (t)
-K
• So we can write
New System
Figure 10: State-variable model of the closed-loop system with feed-forward gain
in input. The closed-loop system has r (t) as input and y (t) as output.
r(t)
Fcl, Gcl, Hcl, Jcl
New System
ẋ (t) = Fcl x (t) + Gcl u (t)
(77)
Fcl = Fp − G p K
(78)
Gcl = G p N f
(79)
with
y(t)
x(t)
Figure 11: State-space model of the closed-loop system, with r (t)as input and
y (t) as output.
• Feedback control fundamentally transforms the system. Changing the statevariable model from figure 9 to 11.
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 63
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 64
EE/ME 574: Intermediate Controls
Section 4.1.0
EE/ME 574: Intermediate Controls
4.2 Determining N f
• Plugging the control law into the output equation, we find
y (t) = H p x (t) + J p −K x (t) + N f r (t) = H p x (t) − J p K x (t) + J p N f r (t)
• With state-feedback control, the model is given as:
= (H p − J p K) x (t) + J p N f r (t)
– So we can write
y (t) = Hcl x (t) + Jcl r (t)
(80)
Hcl = H p − J p K
(81)
Jcl = J p N f
(82)
with
• Eqns (77)-(82) describe how we determine the state-variable model of the
system with feedback control.
with
(85)
y (t) = Hcl x (t) + Jcl r (t)
(86)
Gcl = G p N f
Hcl = H p − J p K ,
Jcl = J p N f
• The transfer function for a state variable model was derived in Eqn (53).
Applying Eqn (53) to the closed-loop system gives:
Try (s) = Hcl , (s I − Fcl )−1 Gcl + Jcl
= Hcl (s I − Fcl )−1 G p N f + J p N f
• Evaluating Try (s = 0), the DC gain is given as:
→ Feedback is based on the state vector
Try (s = 0) = Hcl (−Fcl )−1 G p + J p N f
→ There is no compensator transfer function Gc (s)
∈ Rm×n
(87)
• We can make the DC gain of Try (s) = 1.0 by choosing
m is the number of inputs
1
Nf = Hcl (−Fcl )−1 G p + J p
n is the number of states
• For the voice-coil motor introduced in section 1.1:
u (t) = −K x (t) + N f r (t)
K =
k1 k2 k3
Part 4: State-Space Control
ẋ (t) = Fcl x (t) + Gcl u (t)
Fcl = Fp − G p K ,
• State feedback control is fundamentally different from single-loop,
compensated feedback:
→ The control gains form a vector, K
Section 4.2.1
(Revised: Jan 27, 2016)
(83)
• Eqn (88) looks a bit complicated. But as we will see, it is straight forward
to calculate. In the common case that J p = 0, Eqn (88) simplifies to
(84)
Nf =
Page 65
(88)
Part 4: State-Space Control
1
H p (−Fcl )−1 G p
(Revised: Jan 27, 2016)
(89)
Page 66
EE/ME 574: Intermediate Controls
Section 4.3.0
EE/ME 574: Intermediate Controls
Section 4.3.0
4.2.1 Note on N f and SSE(step) = 0
Closed−Loop Step Response, current
1
>> figure(1), clf
>> step(ss(Fcl1, Gcl1,...
Hcl, Jcl), 2)
• When the model knowledge is ideal, the choice for N f will give zero steady
state error for step input.
0.9
0.8
Amplitude
0.7
• Practically: if the DC gain is an approximation, N f will be an
approximation, and the steady state error will be greater than zero, and
depend on the loop gain in the usual way.
0.6
0.5
0.4
0.3
0.2
4.3 Example system with state feedback
• For a first experiment, try K =
0.1
0
0
2 5 1 ,
0.5
1
1.5
2
Time (sec)
Figure 12: Closed-Loop system step response.
Which corresponds to u (t) = −2.0 v (t) − 5.0 z (t) − 1.0 i (t) + N f r (t)
• Looking at the closed-loop poles
>> K1 = [ 2 5 1]
%% Determine Fcl and Hcl from K1
>> Fcl1 = Fp - Gp * K1
Fcl1 = -0.1500
1.0000
-131.5000
– First, the eigenvalues of the system matrix
>> Hcl = Hp - Jp*K1
0
0
3.1500
0
-250.0000
-65.0000
%% Determine Nf1
>> Nf1 = 1 / (Hcl * inv(-Fcl1) * Gp + Jp),
Hcl =
0
1
>> CLPoles = eig(Fcl1)
CLPoles = -58.0841
-3.5330 + 1.0374i
-3.5330 - 1.0374i
Nf1 =
%% Poles are eigenvalues
%% of the system matrix
5
– Now using the pole command, applied to the CL system model
%% Determine the rest of the CL model
>> Gcl = Gp * Nf1
>> Jcl = Jp * Nf1
Gcl =
0
Jcl =
0
0
250
Part 4: State-Space Control
0
(Revised: Jan 27, 2016)
>> SSry1 = ss(Fcl1, Gcl1, Hcl, Jcl);
>> CLPoles = pole(SSry1)
%% Poles from state model
CLPoles = -58.0841
-3.5330 + 1.0374i
-3.5330 - 1.0374i
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Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 68
EE/ME 574: Intermediate Controls
Section 5.0.0
5 State-space controller design, determine K and N f
• State feedback design is fundamentally different from frequency-response
or root-locus design
• We will see the two most common methods for state-space controller design:
EE/ME 574: Intermediate Controls
Section 5.1.0
5.1 State-Space Controller Design Method #1: Pole Placement
(See Franklin et al. 5th ed. section 7.5)
• Under very general conditions, the poles of the closed-loop system can be
arbitrarily placed by proper choice of K
– System has n poles (n degrees of freedom)
1. Pole placement
– K has n elements (n degrees of freedom)
– Design procedure:
(a) Choose the desired pole locations
– Original system matrix, Fp
(b) Calculate K to give the desired closed-loop poles
– System matrix with feedback, Fcl = Fp − G p K
%% Pd is a vector of desired pole locations
>> Pd = [p1, p2, p3]
>> K = place(Fp, Gp, Pd);
• Design procedure:
1. Choose the desired pole locations, S =
(c) Calculate N f for y (t) = r (t) in steady state
s1 s2 · · · sn
2. Calculate K to give the desired closed-loop poles
2. Linear Quadratic Regulator (LQR) design, a form of optimal control
– Design procedure:
>> K = place(Fp, Gp, S)
>> K = acker(Fp, Gp, S)
%% Kautsky-Nichols algorithm
%% Ackerman’s formula
– Mathworks recommends the Kautsky-Nichols algorithm, place(),
for most cases.
(a) Specify a ’cost’ function
(b) Calculate the K to minimize the cost function
– Ackerman’s formula may succeed for repeated roots where the
Kautsky-Nichols algorithm fails.
(c) Calculate N f for y (t) = r (t) in steady state
• Pole placement and LQR design determine K. N f is given by Eqn (89).
3. Calculate N f for y (t) = r (t) in steady state
>> Fcl = Fp - Gp * K
%% Form closed-loop Syst. Mat.
>> Hcl = Hp - Jp * K
%% Form closed-loop Output
>> Nf = 1/(Hcl*inv(-Fcl)*Gp+Jp) %% Nf for CL DC Gain = 1.0
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EE/ME 574: Intermediate Controls
Section 5.1.1
EE/ME 574: Intermediate Controls
• Examine the closed-loop poles to be sure we got the desired poles.
5.1.1 Pole placement example
%% Check the CL Poles
>> CLPoles2 = eig(Fcl2)
CLPoles2 =
-2.0000 + 0.2000i
-2.0000 - 0.2000i
-2.0000
• Steps:
1. Select the desired pole locations
Suggestion: do not choose repeated poles.
2. Compute the needed K using pole placement
>>
• Example:
SSry2 = ss(Fcl2, Gp*Nf2, Hp-Jp*K2, Jp*Nf2)
%% The ss command pretty prints the model
a =
>> DesiredPoles0 = [ -2, -2+0.2*j, -2-0.2*j]
>> K2 = place(Fp, Gp, DesiredPoles0)
K2 =
-0.5591
0.0513
-0.1830
%% Desired poles
%% K by K-N algorithm
x1
x2
x3
c =
>> K3 = acker(Fp, Gp, DesiredPoles0)
K3 =
-0.5591
0.0513
-0.1830
%% Determine CL Poles
• Form a state-variable model of the closed-loop system:
3. Compute the needed N f using Eqn (89)
%% K by Ack. formula
y1
x1
-0.15
1
-3.544
x2
0
0
-2.565
x1
x2
x3
0
1
0
b =
x3
3.15
0
-5.85
u1
0
0
2.565
x1
x2
x3
d =
u1
y1
>> figure(4), clf ;
>> Fcl2 = Fp - Gp * K2
Fcl2 = -0.1500
0
1.0000
0
-3.5437
-2.5651
Section 5.1.1
0
step(SSry2, 4)
%% Form the CL Syst. Mat.
Step Response
3.1500
0
-5.8500
1
0.9
0.8
0.7
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Amplitude
%% Determine Nf2
>> KDC0 = Hcl2 * inv(-Fcl2) * Gp + Jp
KDC0 =
19.4926
%% Determine Gcl
>> Nf2 = 1/KDC0
>> Gcl2 = Gp*Nf2
Nf2 =
0.0513
Gcl2 =
0
0
2.5651
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.5
1
1.5
2
2.5
3
3.5
4
Time (sec)
Figure 13: Closed-Loop system with K = K2 , step response.
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Page 72
EE/ME 574: Intermediate Controls
Section 5.1.2
5.1.2 Try pole-placement control design, twice as fast
Section 5.1.2
Notes:
>> DesiredPoles2b = [ -4, -4+0.2*j, -4-0.2*j]
>> K2b = place(Fp, Gp, DesiredPoles2b)
K2b = -0.3363
0.4074
-0.0630
>> Fcl2b = Fp - Gp * K2b
Fcl2b =
-0.1500
0
1.0000
0
-14.6865 -20.3683
EE/ME 574: Intermediate Controls
• Pole placement will be successful, except for uncontrollable or
unobservable systems
– Uncontrollable or unobservable systems often require modification of the
system hardware, see section 8.
3.1500
0
-11.8500
>> CLPoles2 = eig(Fcl2b)
%% CL poles are eigenvalues of CL SS model
CLPoles2 = -4.0000 + 0.2000i
-4.0000 - 0.2000i
-4.0000
>> KDC0 = Hcl * inv( -Fcl2b ) * Gp
%% Determine DC gain (with Nf=1)
KDC0 = 2.4548
• Pole placement can be sensitive to small variations in parameters
• The pole locations relate directly to speed and damping, but not so directly
to accuracy. The desired pole locations may not be obvious.
>> step( ss(Fcl2b, Gp, Hcl, Jp) )
Step Response
2.5
KDC0 = 2.4548
2
Amplitude
1.5
1
0.5
0
0
0.5
1
1.5
2
Time (sec)
2.5
3
3.5
4
Figure 14: Step response of closed-loop system with K2b , real part of poles
increased 2X. Note change in DCgain, N f must be re-computed
whenever K is changed.
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EE/ME 574: Intermediate Controls
Section 5.1.3
EE/ME 574: Intermediate Controls
5.1.3 Taking a look at the control effort and tracking error
Section 5.1.3
>> step(SSry)
>> step(SSru)
>> step(SSre)
• With Matlab and transfer-function modelling, we could easily set up various
transfer function models
Step Response, y(t)
1
0.9
%%
%%
%%
TF: Input -> Output
TF: Input -> Error
TF: Input -> Control
0.8
0.7
Amplitude
>> Try = feedback( Gc*Gp, Hy)
>> Tre = feedback( 1, Gc*Gp*Hy)
>> Tru = feedback(Gc, Gp*Hy)
• Similarly, we can build state models with various outputs.
For each of these models,
0.6
0.5
0.4
0.3
0.2
0.1
– The state equation remains the same
0
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
Time (sec)
ẋ (t) = Fcl x (t) + Gcl r (t)
Figure 15: Plot of control effort during a step response.
– Only the output equation changes.
Step Response, e(t)
Step Response, u(t)
0.8
∗ For output y (t)
0.7
1
0.9
−U
peak
0.8
0.6
>> SSry = ss(Fcl2b, Gcl2b, Hp-Jp*K2b, Jp*Nf2b)
0.7
Amplitude
0.5
Amplitude
y (t) = H p x (t) + J p u (t) = (H p − J p K) x (t) + J p N f r (t)
0.4
0.6
0.5
0.4
0.3
0.3
∗ For control signal u (t)
0.2
0.2
0.1
u (t) = −K x (t) + N f r (t)
0
0
>> SSru = ss(Fcl2b, Gcl2b, -K2b, Nf2b)
0.1
0.5
1
1.5
2
Time (sec)
2.5
3
3.5
0
0
0.5
1
1.5
2
2.5
3
3.5
Time (sec)
Figure 16: Plot of control effort during a step response.
∗ For error signal e (t)
e (t) = r (t) − y (t) = r (t) − (Hcl x (t) + Jcl u (t))
= −Hcl x (t) + (1 − Jcl) r (t)
>> SSre = ss(Fcl2b, Gcl2b, -Hcl, 1-Jcl)
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4
4.5
EE/ME 574: Intermediate Controls
Section 5.1.3
• We can also use the step() command to generate the response data.
EE/ME 574: Intermediate Controls
Section 5.2.0
5.2 State-Space Controller Design Method #2: The Linear
Quadratic Regulator, an optimal control method
– Determine Upeak from a state-variable model
>> uu = step(SSru);
>> Upeak = max(uu)
Upeak = 0.7016
%% Determine the u(t) signal
%% Determine the max of u(t)
– The pole locations or other characteristics determine the control gains
and thus the final, closed-loop performance.
– Determine yss and P.O. from a state-variable model
>> yy = step(SSry);
>> Yss = yy(end)
Yss = 1.0000
%% Determine the y(t) signal
%% Determine the steady-state value
>> PO = 100 * (max(yy)-Yss)/Yss
PO = 0
• With other control design methods, the controls engineer selects closedloop pole locations (root locus, or pole placement) or other response
characteristics (e.g, ωc and PM in frequency domain design).
• Designing an LQR controller is completely different, for an LQR design,
the controls engineer selects a cost function.
%% Determine the PO
– Linear Quadratic Regulator (LQR) controller design determines the
controller gains which will minimize the cost function.
– Determine the steady-state error to a step input from a state-variable
model
– The cost function determines the control gains and the thus final, closedloop performance.
>>
>>
ee = step(SSre);
Ess = ee(end)
Ess =
1.1576e-05
%% Determine the e(t) signal
%% Determine e(t) at the end
%% of the transient
– The controls engineer selects the cost function to achieve the design goals
(more on this later).
• And we can use the bode() command to address questions of frequency
response
>> bode(SSre)
>> Etrack = 100 * bode(SSre, Wt)
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EE/ME 574: Intermediate Controls
Section 5.2.1
EE/ME 574: Intermediate Controls
• An example is seen in figure 17.
5.2.1 The LQR cost function
%% Zero-input response of LQR design
>> Q = eye(3); R = eye(1)
>> K4 = lqr(Fp, Gp, Q, R)
%% LQR Design
K4 = [ 0.7711 1.0000 0.7896 ]
>> Fcl4 = (Fp - Gp*K4);
%% Build CL model
>> Sys = ss(Fcl4, [0;0;0], Hp, Jp)
• A general quadratic cost function has the form:
J (x, u) =
Z ∞
0
shapes :
x (t)T
xT (t) Q x(t) + uT (t) R u(t) + 2 ∗ xT (t) N u (t) dt








Q



  x (t)




(90)




 h


i



T
 + u (t) [ R][ u (t)] + 2 x (t)T  N  [ u (t)]






• lsim() command simulates response of the system with initial cond. x (0).
>> tt = 0:0.01:10;
X0 = [0;1;0];
>> [y, t, x] = lsim(Sys, zeros(size(tt)), tt, X0);
>> u = - K4 * x’;
%% Calculate u (t)
where: J (x, u) is a ’cost’ function, to be minimized.
and
u(t) ∈ Rm
Rn×n
1
0.8
are the state and input signals.
Rm×m
Q∈
and R ∈
are square matrices that put weights the state
and input signals, determining their contributions to the total cost.
y(t)
x(t)
∈ Rn
Section 5.2.1
0.6
0.4
0.2
0
0
N ∈ Rn×m determines a contribution to the cost from a product of x and
u. This term is often zero.
2
4
2
4
6
8
10
6
8
10
0
• First, the “regulator” case is considered, that is
u(t)
−0.5
−1
u (t) = −K x (t)
−1.5
0
with no reference input.
t [sec]
– In this case, for any stable system the state and output go to zero,
– The control signal goes to zero
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 79
Figure 17: Response of the voice-coil motor to initial condition
h
iT
x (0) = 0 1 0
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 80
EE/ME 574: Intermediate Controls
Section 5.2.1
• It turns out that for the quadratic cost function
J (x, u) =
Section 5.2.2
5.2.2 Designing an LQR controller
xT (t) Q x(t) + uT (t) R u(t) + 2 ∗ xT (t) N u (t) dt
Z ∞
0
EE/ME 574: Intermediate Controls
There is a unique value of K that minimizes the cost function for any initial
condition.
• This is the K given by the LQR method !
– The details of how K is computed is a subject in optimal control
• Using Matlab’s lqr() function
Example, starting with the voice-coil motor model:



0






Gp =  0 




50

−0.15
0
3.15






Fp =  1

0
0




−31.5 0 −15
Hp =
– The LQR method gives K, N f can be computed in the usual way
0 1 0
(91)
J p = [0]
• Choose the state-error cost matrix (the identity matrix is a good place to
start):


1 0 0





Q = diag
= 0 1 0 
1 1 1




0 0 1
• Choose the control-effort cost matrix:
R = [1]
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EE/ME 574: Intermediate Controls
Section 5.2.2
• In Matlab:
>> K4a = lqr(Fp, Gp, Q, R);
K4a = [ 0.7711 1.0000 0.7896 ]
>> Fcl4a = Fp - Gp*K4a;
Hcl4a = Hp - Jp * K4a
>> KDC = Hcl4a*inv(-Fcl4a)*Gp + Jp;
>> Nf4a = 1/KDC;
%% Determine Nf
Nf4a = 1.0000
>> Gcl4a = Gp*Nf4a;
Jcl4a = Jp*Nf4a;
%% Define Gcl and Jcl
>> SSry = ss(Fcl4a, Gcl4a, Hcl4a, Jcl4a) %% Define SS object
EE/ME 574: Intermediate Controls
5.2.3 Tuning an LQR controller
• For the moment, let’s just try a different Q matrix. Lets increase the cost on
position error by 100×
• So choose:
• Recall that the state vector is:

1 0 0




Qb =  0 100 0  ,




0 0 1


 
 v (t)   Cone velocity 


 

 

x (t) =  z (t)   Cone position 

 


 

Winding current
ia (t)

• Verify the response:
>>
>>
>>
>>
Section 5.2.3
subplot(2,1,1)
%% Plot step response of output
step(Fcl4a, Gcl4a, Hcl4a, Jcl4a)
subplot(2,1,2)
%% Plot step response of control effort
step(Fcl4a, Gcl4a, -K4a, Nf4a)
>> K4b = lqr(Fp, Gp, Qb, R)
K4b = [ 2.2962 10.0000
Nf4b = 10.0
R=1
0.8744 ]
Step Response
1.5
• Characteristics:
Amplitude
Step Response
1
(A)
• Characteristics:
0.4
0.2
0
(A)
0
0
1
2
– Settling time: ts ≈ 5.0 [sec]
0
1
2
3
4
5
6
7
Time (sec)
Step Response
1
3
Time (sec)
4
5
6
7
Step Response
– Damping: over damped
– ISE =0.85
– Upeak = 1.0
(B)
0.8
Amplitude
– Settling time: ts ≈ 0.46 [sec]
(much faster)
0.5
0.6
– Damping:
slightly under damped
(PO ~ 2%, damping reduced)
– ISE =0.0022
(much lower ISE)
10
Amplitude
Amplitude
0.8
1
5
(B)
0
0.6
0.4
−5
0.2
0
0
1
2
3
4
5
6
7
Time (sec)
Figure 18: Step response of the closed-loop system with first LQR controller.
(A) Output, (B) Control effort.
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 83
0
1
2
3
Time (sec)
4
5
6
7
Figure 19: Step response of the closed-loop system with second LQR controller.
(A) Output, (B) Control effort.
Part 4: State-Space Control
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EE/ME 574: Intermediate Controls
Section 5.2.3
• By increasing the cost on the position variable, we made
EE/ME 574: Intermediate Controls
Section 5.2.4
5.2.4 Procedure for tuning Q and R
– The position error go to zero more quickly
1. Set up the system in state space.
– At the expense of a larger velocity transient and greater use of control.
2. Choose initial values for Q and R:
• Identity matrices are often a good starting point
• LQR design is really pretty simple:
(Note: R will be a 1x1 matrix for a single input system)
1. Work with diagonal matrices
3. Adjust cost on elements of Q and R matrices to achieve design goals
2. All values must be non-negative
– R matrix must be positive definite
• If error is too large, increase the cost on error
– Q matrix must be positive semi-definite
• If overshoot is too large, increase the cost on velocity
(Often some diagonal elements of Q can be zero)
• If control use is too large, increase R
• General idea of LQR design:
——– ∼ ——–
– Increase cost on states or input that should be reduced or converge to
steady state quickly
• LQR design often gives a desirable controller:
– Reduce cost on states or input that can be increased
– Well damped (we have seen)
– Make large adjustments of Q and R terms, of 10x or 100x (not 20%)
– Low to moderate control usage
– Examples:
– Low to moderate controller gains
∗ If the system is too slow, increase cost on the position element of x.
∗ If the system has too much overshoot, increase cost on the velocity
element of x
∗ If Upeak is too high, increase R
– Use trial and error (a.k.a. Iterative Design)
Part 4: State-Space Control
– LQR-designed controllers tend to be quite robust to variations in plant
parameters
• By switching from tuning compensator poles and zeros and loop gain to
tuning Q and R elements, one is trading one set of adjustments for another.
– Which is more suitable will depend on the application and on the
experience of the controls designer.
(Revised: Jan 27, 2016)
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Part 4: State-Space Control
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EE/ME 574: Intermediate Controls
Section 5.2.5
5.2.5 Further notes on choosing Q and R
EE/ME 574: Intermediate Controls
Section 5.2.6
5.2.6 Connection between LQR and the Symmetric Root Locus
1. R ∈ Rm×m will be a scalar if the system is single-input
(m is the number of inputs).
• In section 7.6 (5thed.) Franklin et al. give a method of LQR based on the
symmetric root locus
2. Often choose Q and R to be diagonal matrices with positive values on the
main diagonal
• The method uses the cost function
• Zero is OK for some diagonal elements of Q.
JF (x, u) =
• It is possible for Q and R to have non-zero off-diagonal elements. The
requirements are:
Z ∞
0
ρ y2 (t) + u2(t) dt
(92)
– Q and R must be symmetric, that is Q = QT , R = RT
which is equivalent to the general LQR cost function, Eqn (90), with the
special choice:
Q = ρ H pT H p ,
R=1
(93)
– All eigenvalues of R must be positive
(Student activity: prove this, use J p = [0].)
– All eigenvalues of Q must be non-negative
• The Symmetric Root Locus (SRL) theoretical result is a neat mathematical
result which gives the CL poles when Q is chosen by Eqn (93).
3. The relative values of Q and R matter, not the absolute magnitude
• If Q0 and R0 give K0
– LQR K can be found using the SRL by
→ then Q = 10 Q0 and R = 10 R0 also give K = K0 (no change).
1. Determine the CL poles by the SRL method,
2. Determine K by pole placement.
• However, there is no real need to use the SRL, since we can find K directly
with the lqr() command
1. Choose Q and R (possibly set Q and R by Eqn (93))
2. Call K = lqr (Fp , G p, Q, R)
Part 4: State-Space Control
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Part 4: State-Space Control
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EE/ME 574: Intermediate Controls
Section 6.1.0
EE/ME 574: Intermediate Controls
6 State-variable models with various inputs and
Section 6.1.0
• The model matrices are:

outputs
ks
M
− Mb 0


Fp =  1 0 0

− Lksa 0 − RLaa
• State space models give the controls engineer the freedom to make many
changes or examine the model in many different ways.
6.1 Considering various inputs
Hp =
• We can add an input just be adding either a new column on the G matrix, or
adding more G matrices.
h
0 1 0
i
,







Gp =  0 




,

J p = [0]
0
1
La
(position output)
• Let’s suppose there are some additional inputs. As an example, we’ll
consider fc (t), an external disturbance force acting on the speaker cone;
and vn (t), voltage noise pickup in the wiring.
Consider the voice-coil motor, the original modeling equations are:
M z̈ (t) + b ż (t) = ks ia (t)
1. Determine how the new input signals enter into the original model (as
always, be sure units balance). In this case:
d i (t)
+ Ra ia (t) = va (t) − ks ż (t)
La
dt
• fc (t) enters the equation for z̈ (t) ,
We wrote the SS model as:
M z̈ (t) + b ż (t) = ks i (t) + fc (t)
(94)
ẋ (t) = Fp x (t) + G p u (t)
y (t) = H p x (t) + J p u (t)
with
State vector :
Input vector :
Output vector :

 
• And vn (t) enters the equation for d ia (t) /dt,

 v (t)   Cone velocity 
 


 


x (t) =  z (t)   Cone position 
 


 


i (t)
Winding current
u (t) = [va (t)] [Applied Voltage]
(Revised: Jan 27, 2016)
Page 89
d i (t)
+ R i (t) = va (t) − ks ẋ (t) + vn (t)
dt
(95)
2. Introduce the signals into the state-variable model. Considering Eqns (94)
and (95), we can write the model as:

z̈ (t)


ẋ (t) =  ż (t)

i˙a (t)
y (t) = [z (t)] [Cone position]
• In the original model there is a single input and a single output.
Part 4: State-Space Control
L


− Mb 0
ks
M
 
 
= 1 0 0
 
− Lksa 0 − RLaa
Part 4: State-Space Control

ż (t)


  z (t)

ia (t)
 
0


1
M


0






 





 
 +  0  va (t)+  0  fc (t)+  0  vn (t)





 
1
1
0
La
La
(96)
(Revised: Jan 27, 2016)
Page 90
EE/ME 574: Intermediate Controls
Section 6.1.0
EE/ME 574: Intermediate Controls
• Eqn (96) has the form
Now:
ẋ (t) = Fp x (t) + Gva va (t) + G fc fc (t) + Gvn vn (t)
(97)
• The system matrix for feedback control is constructed:
Fcl = (Fp − Gva K)
– Gva reflects how control input va (t) enters the dynamic model,
– G fc reflects how disturbance fc (t) enters the dynamic model,

 0 




Gva =  0 





1

 M 




G fc =  0 




0
,
1
La
,
– Notice the physics of the system determining the modeling.


• If we want to consider the response of the closed-loop system to a unit step
of the disturbance force fc (t), then y (t) = Hcl x (t) + Jd d (t), so use
 0 




Gvn =  0 




>> SSdy = ss(Fcl, Gfc, Hcl, Jd);
1
La
– For the model in standard form, we put all input signals on u (t) and the
three Gx vectors in one, 3 × 3 G p matrix. ẋ = Fpx + G pu (t) becomes:



ks
M
− Mb
0
 z̈ (t)  
 

 

 ż (t)  =  1 0 0
 

 

− Lksa 0 − RLaa
i˙a (t)
where

 0


Gp =  0


1
La
1
M
0
0

0 


,
0 


1
La

 
1
M
  ż (t)   0

 

 
  z (t)  +  0 0

 

 
1
ia (t)
La 0
and


0   va (t) 




 fc (t) 
0 




1
vn (t)
La


v
(t)
 a





u (t) =  fc (t) 




vn (t)
(Revised: Jan 27, 2016)
step( SSdy )
• If we want to consider the frequency response to voltage noise vn (t), then
y (t) = Hcl x (t) + Jvn vn (t), so use
>> SSvy = ss(Fcl, Gvn, Hcl, Jvn) ;
bode( SSvy
)
• If we want the transfer function of the open-loop system for a disturbance
force input, setup
Gdy = tf( ss(Fp, Gfc, Hp, Jd) )
• If we want the transfer function of the closed-loop system for a voltage noise
input and control effort output, then u (t) = −K x (t) + 0 vn (t), so use
>> SSvu
– The system has n states and m inputs, and G p ∈ Rn×m .
Part 4: State-Space Control
(98)
because the control signal is applied on the va (t) input.
– Gvn reflects how noise vn (t) enters the dynamic model,

Section 6.1.0
= ss(Fcl, Gvn, -K, 0) ;
Tvu
= tf( SSvu )
• And many other input/output combinations can be considered.
Page 91
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 92
EE/ME 574: Intermediate Controls
Section 6.1.1
6.1.1 Considering a disturbance input
The most general model for a disturbance input is seen in figure 20, with
ẋ (t) = (Fp − G p K) x (t) + G p N f r (t) + Gd d (t)
(99)
y (t) = Hcl x (t) + Jcl r (t) + Jd d (t)
EE/ME 574: Intermediate Controls
In the special case that the disturbance is a signal added to the plant input
signal, u (t), then the block diagram looks as in figure 21. And for equations (99)
to (102)
Gd = G p , Jd = J p .
(100)
where Gd reflects the impact of disturbance d (t) on the state derivative, and will
come from considering the basic physical modeling equations of the system, and
Jd is the direct-transfer term between the disturbance input and the output.
Section 6.1.1
d(t)
r(t)
r’(t)
Nf
+
+
+
y(t)
Fp, Gp, Hp, Jp
u(t)
x(t)
x(t)
d(t)
-K
r’(t)
r(t)
Nf
+
u(t)
New System
y(t)
Fp, Ga, Hp, Jp
+
x(t)
x(t)
Figure 21: Control additive disturbance input.
-K
New System
Figure 20: General disturbance input.
Considering r (t) and d (t) together, one can write Eqns (99) and (100) in a
single, standard state-variable model with:
ẋ (t) = Fcl x (t) + Ga u (t)
(101)
y (t) = Hcl x (t) + Ja u (t)
(102)
where

u (t) = 
r (t)
d (t)

,
Part 4: State-Space Control

|
|





Ga =  G p N f Gd  ,


|
|
Ja =
h
Jcl Jd
i
(Revised: Jan 27, 2016)
Page 93
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 94
EE/ME 574: Intermediate Controls
Section 6.2.0
6.2 Considering various outputs
EE/ME 574: Intermediate Controls
Section 6.2.0
6.2 Considering various outputs (continued)
To consider various signals the as output, adjust the H and J matrices accordingly.
For example:
• If we want to consider the cone velocity, then use:
Hcv =
• u (t) = −K x (t) + N f r (t) so to consider the control effort as output use:
Hu = −K;
h
1 0 0
i
Jcv = [0]
• For example, to plot the step response of the cone velocity of the closed-loop
voice-coil motor to an input step of an external disturbance force applied to
the speaker cone, use:
Ju = N f
then
>> SSru = ss( Fcl, Gcl, -K, Nf )
>> bode( SSru )
%% Frequency response of u(t) to input r(t)
>> step( ss(Fcl, Gfc, Hcv, Jcv) )
• Important fact of state-variable modeling:
• Given the state vector, if we want to consider the winding current, then use:
Hi =
0 0 1
>> SSri = ss( Fcl, Gcl, Hi, Ji )
– Any signal in a linear system can be expressed as a linear combination
of states and inputs.
Ji = [0]
%% Closed-loop system, r(t)->i(t)
• For example, to plot the step response of the winding current of the
open-loop voice-coil motor to an input step of applied voltage, use:
%% Note open loop system matrix.
>> SSvi_ol = ss(Fp, Gva, Hi, Ji) ;
step( SSvi_ol) ;
• To plot the step response of the winding current of the closed-loop voice-coil
motor to a unit step disturbance force, use:
%% Note disturbance input, Gfc, and current output
>> SSfi = ss(Fcl, Gfc, Hi, Ji) ;
step( SSfi )
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 95
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 96
EE/ME 574: Intermediate Controls
Section 6.3.0
6.3 Determining the Steady-State error for a state-variable
model
EE/ME 574: Intermediate Controls
Section 6.3.0
System Type using Eqn (104)
• For the state variable model, there is not a direct analog to the loop gain.
• We can determine the system type by one of two methods:
• Steady-state error is given by the final value theorem as:
ess = lim e (t) = lim s E (s) = lim s
t→∞
s→0
s→0
E (s)
R (s)
R (s)
(103)
2. Count the number of zeros of the closed-loop error transfer function at
the origin.
For ess , what is needed is the Laplace transform of the error signal.
• With a state variable model, Fcl , Gcl , Hcl , Jcl reflecting the closed-loop
system, we can determine E (s) /R (s).
since :
e (t) = r (t) − y (t) = r (t) − (Hcl x (t) + Jcl r (t))
>> SSre = ss( Fcl, Gcl, -Hcl, (1-Jcl) )
>> Tre = tf( SSre )
sn + a
• For example:
– Type 0 system might give the TF
Tre (s) =
%% Model r(t) -> e(t)
%% E(s) / R(s)
bm s m + · · · + b2 s 2 + b1 s + b0
n−1 + a
n−2 + · · · + a s + a
n−1 s
n−2 s
1
0
(104)
bm s m + · · · + b2 s 2 + b1 s + b0
n−1 + a
n−2 + · · · + a s + a
n−1 s
n−2 s
1
0
(105)
sn + a
bm s m + · · · + b2 s 2 + b1 s + 0
n−1 + a
n−2 + · · · + a s + a
n−1 s
n−2 s
1
0
(106)
– Type II system might give the TF
Tre (s) =
Then we can plug Eqn (104) into (103) and find ess .
sn + a
– Type I system might give the TF
Tre (s) =
• Writing the transfer function E (s) /R (s) as
Tre (s) =
1. Count the open loop poles at the origin, that is the number of eigenvalues
of Fp that are equal to zero.
sn + a
bm sm + · · · + b2 s2 + 0s + 0
n−1 + a
n−2 + · · · + a s + a
n−1 s
n−2 s
1
0
(107)
• Just by looking at Tre (s), we can determine the system type.
• Notice Eqns (105)-(107) have a0 6= 0
– a0 = 0 corresponds to a pole of the closed-loop system at the origin.
– Such a system would be marginally stable, and is rarely (if ever)
encountered in practice
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 97
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 98
EE/ME 574: Intermediate Controls
Section 6.3.0
Determining the steady-state error from Tre (s)
• Considering the voice-coil motor and controller K4a , developed in section
5.2.2
lim s Tre (s) R (s) = lim s
s→0
= lim s
s→0
bm s m + · · · + b2 s 2 + b1 s + b0
R (s)
sn + an−1 sn−1 + · · · + a1s + a0
ess =
b0
a0
(108)
This is logical, b0/a0 is the DC gain of Tre (s)
ess = lim s
s→0
b0 1
=∞
a0 s 2
(109)
= lim s
s→0
0
0
0
3.1500
0
-15.0000
Gp =
0
0
50
Hp =
0
1
0
Jp =
0
0.7896 ]
• The model r (t) − > e (t) is obtained
>> Fcl4a = Fp - Gp*K4a;
Gcl4a = Gp*Nf4a ;
>> Hcl4a = Hp - Jp*K4a;
Jcl4a = Jp*Nf4a;
%% Consider SSre for K4a
>> SSre = ss(Fcl4a, Gcl4a, -Hcl4a, (1-Jcl4a))
• For a Type I system, b0 = 0 , so
s→0
-0.1500
1.0000
-31.5000
K4a = [ 0.7711 1.0000
Nf4a = 1.0000
– For a ramp input, R (s) = 1/s2 , so
lim s Tre (s) R (s) = lim s
Fp =
b0
R (s)
a0
– For a step input, R (s) = 1/s, so
s→0
Section 6.3.1
6.3.1 Example
• For a Type 0 system, b0 6= 0 , so
s→0
EE/ME 574: Intermediate Controls
bm s m + · · · + b2 s 2 + b1 s + 0
R (s)
n
s + an−1 sn−1 + · · · + a1s + a0
b1 s + 0
b1 s
R (s) = lim s
R (s)
s→0 a0
a0
• Determine Tre (s)
– For a step input, R (s) = 1/s, so
b1 s 1
=0
ess = lim s
s→0 a0 s
(110)
– For a ramp input, R (s) = 1/s2 , so
ess = lim s
s→0
Part 4: State-Space Control
s b1 1
b1
=
a0 s 2 a0
(Revised: Jan 27, 2016)
(111)
Page 99
>> Tre=tf(SSre)
Transfer function:
s^3 + 54.63 s^2 + 228.8 s - 1.192e-13
------------------------------------s^3 + 54.63 s^2 + 228.8 s + 157.5
– 1.192 × 10−13 ≃ 0 , system is type I
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 100
EE/ME 574: Intermediate Controls
Section 6.3.1
Note the round-off error:
Section 6.3.1
Conclusions
• The TF is given as
• For series compensation, we have a short cut for steady-state error based on
the System Type and DC gain.
Tre (s) = (−Hcl ) (sI − Fcl )−1 Gcl
(112)
• But SSE is defined in terms of the final value theorem, based on Tre (s),
which is easily determined from the state-variable model.
which evaluates to give
Tre(s) =
EE/ME 574: Intermediate Controls
s^3 + 54.63 s^2 + 228.8 s - 1.192e-13
------------------------------------s^3 + 54.63 s^2 + 228.8 s + 157.5
• For state-space systems, go back to the basics, and use Tre (s) to determine
SSE.
• With infinite precision arithmetic, the calculation would give
Tre(s) =
s^3 + 54.63 s^2 + 228.8 s - 0
------------------------------------s^3 + 54.63 s^2 + 228.8 s + 157.5
• In the calculation of Eqn (112), round-off error produces the
b0 = −1.192 × 10−13 term.
• b0 = −1.192 × 10−13 is effectively zero.
• The system is Type I
• SSE (ramp) = 228.8/157.5=145%
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 101
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 102
EE/ME 574: Intermediate Controls
Section 7.1.0
7 Integral control: augmenting the model with an
EE/ME 574: Intermediate Controls
• Augmenting the state vector increases the size of the F, G, H and J matrices
integral control state
• The PID control law is:
u(t) = k pe(t) + kd ė(t) + ki
Z t
e(τ) dτ
(113)
0
=⇒ To introduce integral control, we must add a state to the state vector
• Define
ie (t) = −
Z t
d
ie (t) = y (t) − r (t)
dt
e(τ) dτ , or
0
(114)
If ie (t) were in the state vector, we could put ki in the control vector, and we
would have integral control !
• Adding a controller state, such as ie (t), results in an augmented state vector.
•

x p (t)
h
.
H p .. 0


ẋa (t) =  · · ·

xc (t)
y (t) =
• Integral control has an internal state (the integral of error)
Section 7.1.0


Fp
..
.
0
 
 
 =  ··· ··· ···
 
.
Fcp .. Fcc
i

x p (t)


 ···

xc (t)


Gp
 
 
 +  ···
 
Gcu






 u (t) +  · · ·


Gcr
xa (t) + [J p] u (t)
• The control signal can be written
u (t) = −K p x p (t) − Kc xc (t) + N f r (t) = −Ka xa (t) + N f r (t)
(117)
h
i
where Ka = K p ... Kc is the controller gains designed for the augmented
system.
• Writing out the individual equations, we find

where

..
 . 


 x p(t) 




xa (t) =  ... 




 ··· 


xc (t)
ẋ p (t) = Fp x p (t) + G p u (t)
(unchanged)
y (t) = H p x p (t) + J p u (t)
(unchanged)
• Controller state equation is:
(115)
ẋc (t) = Fcp x p (t) + Fcc xc (t) + Gcu u (t) + Gcr r (t)
(118)
Fcp determines how the controller state(s) depend on x p (t)
xa (t) is the augmented state vector (the state vector of the complete
system)
x p (t) are the original plant states, and
Fcc determines in internal dynamics of the controller state(s)
Gcu is a coupling of u (t) to ẋc (t), internal to the controller
Gcr is a coupling of r (t) to ẋc (t), internal to the controller
xc (t) is the controller states, such as xc (t) = [ie (t)]
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 103
Part 4: State-Space Control



 r (t)

(116)
7.1 Augment the state vector with controller states
• We can augment the state vector with controller states,
0
(Revised: Jan 27, 2016)
Page 104
EE/ME 574: Intermediate Controls
Section 7.1.0
EE/ME 574: Intermediate Controls
• Returning to the augmented state model, and defining


.
0
Fp ..




Fa =  · · · · · · · · ·  is the augmented state transition matrix ;


.
Fcp .. Fcc




0
Gp








Gau =  · · ·  , Grr =  · · ·  are the augmented input matrices;




Gcr
Gcu
Section 7.1.0
• Writing:



Gar = Gau N f + Grr = 

N f Gp
···
Gcu N f + Gcr





(122)
• Putting together Eqns (121), (120) and (122) gives the closed-loop model:
ẋa (t) = Fcl xa (t) + Gar r (t)
Ha = [H p , 0] is the augmented output vector
(123)
y (t) = Hcl xa (t) + Jcl r (t)
J p is unchanged
where
• The augmented state model becomes:
ẋa (t) = Fa xa (t) + Gau u (t) + Grr r (t)
Fcl = Fa − Gau K
(119)
(124)
Gar = Gau N f + Grr
y (t) = Ha xa (t) + J p u (t)
Hcl = Ha − J p K
Jcl = J p N f
• Using the control law:
u (t) = −Ka xa (t) + N f r (t)
(120)
gives the closed-loop model is given as
ẋa (t) = (Fa − Gau Ka ) xa (t) + Gau N f + Grr r (t)
(121)
y (t) = (Ha − J p Ka) xa (t) + J p N f r (t)
Notes
• All matrices and matrix components are the appropriate size,
e.g., Fcp ∈ R1×n p if there is 1 controller state, and n p plant states.
• Student exercise: determine the sizes of each element in Fa , Ga , Ha , Ja .
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 105
Student
Exercise
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 106
EE/ME 574: Intermediate Controls
Section 7.2.0
7.2 Block diagram of state-variable controller with
xc xa
(123, repeated)
y (t) = Hcl xa (t) + Jcl r (t)
Fp , Gp , Hp , Jp
+
xp
ẋa (t) = Fcl xa (t) + Gar r (t)
Plant
Nf
Section 7.3.0
• The complete system of figure 22 is encapsulated as a single state-variable
model with input r (t) and output y (t) and governed by:
controller states
r(t)
EE/ME 574: Intermediate Controls
u(t)
y(t)
-Ka
xp (t)
xp(t)
r(t)
Plant + Controller
Fcl , Gar , Ha , Jp
y(t)
Controller
xa(t)
xp
xa(t)
Fcc , Gx , 0, 0
Figure 23: Block diagram of the complete Plant + Controller system.
Gcu
xc (t)
Gcr
xc(t)
Figure 22: Block diagram of a state-space model with controller states, such as
integral control.
• A block diagram for integral control is seen in figure 22.
• Everything outside the plant is implemented as equations in the controller
(typically a program in an embedded micro processor)
x p (t)


u (t) = −Ka  · · ·

xc (t)



 + N f r (t)

(Revised: Jan 27, 2016)
xc (t) = [ie (t)] =
Z t
−e (t) dτ =
0
Z t
(y (τ) − r (τ)) dτ ,
0
nc = 1
• Assuming J p = 0, so that y (t) = H px p (t) , determine the derivative equation
for the new controller state
x˙c (t) = i˙e (t) = −e (t)
• Comparing with Eqn (118)
x˙c (t) = Fcp x p (t) + Fcc xc (t) + Gcu u (t) + Gcr r (t)
• The controller has internal state xc (t) (values inside the control CPU). These
are updated according to Eqn (118)
Part 4: State-Space Control
• Add the error integrator state (note sign, could be done either way):
ẋc (t) = −e (t) = y (t) − r (t) = (H p x p (t) + J p u (t)) − r (t)
= (H p − J p K p) x p (t) − J p Kc xc (t) + J p N f r (t) − r (t)
= (H p − J p K p ) x p (t) − J p Kc xc (t) + −1 + J p N f r (t) (125)
• The control vector Ka processes both x p (t) and xc (t) terms:

7.3 Example, integral control for the voice-coil motor
Page 107
we find:
Fcp = H p − J p K p ,
Part 4: State-Space Control
Fcc = −J p Kc ,
Gcu = 0 ,
(118, repeated)
Gcr = (−1 + Jcl )
(Revised: Jan 27, 2016)
Page 108
Section 7.3.0
EE/ME 574: Intermediate Controls
• Voice-coil motor example
.
Section 7.3.0
%% Now design Ka (needed to determine Nf)
>> Q = diag([ 1 100 1 2]);
R = 1;
>> K5a = lqr(Fa, Gau, Q, R)
K5a =
2.3484
10.4203
0.8772
1.4142
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% Voice Coil with Integral Control
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% For initial analysis, set Nf to 0
>> Nf0 = 0
>> Gar0 = [Nf0 * Gp; Nf0*Gcu + Gcr]
>> SSry0 = ss( (Fa-Gau*K5a), Gar0, Ha,
%% Setup Model open-loop model (Ka = [Kp, Kc] = Null)
Fcp = Hp;
Fcc = 0;
%% Controller state terms
Gcu = 0;
Gcr = (-1 + Jp);
%% Augmented model terms
>> Fa = [Fp , [0;0;0] ;
Fcp, Fcc
]
>>
>>
>>
>>
Jp)
• Build the error model from e (t) = r (t) − y (t) = r (t) − (Ha xa (t) + Ja r (t))
>> SSre0 = ss( (Fa-Gau*K5a), Gar0, -Ha, (1-Jp))
Step SSry0
Bode SSry0
1
Magnitude (dB)
0
>> Gau = [Gp; Gcu]
>> Ha = [Hp, 0]
0.9
−50
0.8
−100
0.7
−150
Amplitude
EE/ME 574: Intermediate Controls
−200
0
0.6
0.5
Phase (deg)
0.4
• The augmented model is
−90
0.3
−180
0.2
−270
−360
−3
10
0.1
−2
10
−1
0
10
1
10
2
10
10
0
0
3
10
10
20
Frequency (rad/sec)
Ha =
0
0
0
0
1.0000
1
0
3.1500
0
-15.0000
0
0
0
0
0
0
Gau =
Jp =
0
0
50
0
30
40
50
40
50
Time (sec)
Figure 24: r (t) → y (t) Bode plot and step response, SSry0
Bode SSry0
0
Step SSre0
20
1
0
0.9
0.8
−20
0.7
−40
Amplitude
-0.1500
1.0000
-31.5000
0
Magnitude (dB)
Fa =
−60
90
0.6
0.5
Phase (deg)
0.4
45
0.3
0.2
0
0.1
−45
−3
10
−2
10
−1
10
0
10
Frequency (rad/sec)
1
10
2
10
0
0
10
20
30
Time (sec)
Figure 25: r (t) → e (t) Bode plot and step response of error, SSre0
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 109
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 110
Section 7.3.1
EE/ME 574: Intermediate Controls
7.3.1 Now, set N f
%% Final Closed loop system
>> Gar = [Nf5a * Gp; Nf5a*Gcu+Gcr]
>> SSry = ss( (Fa-Gau*K5a), Gar, Ha, Jp)
>> SSre = ss( (Fa-Gau*K5a), Gar, -Ha, 1-Jp)
• With integral control, the closed-loop DC gain of SSry is 1.0, no matter what
the value of N f (including N f = 0!)
– One method to choose N f : choose N f so that ie (t) goes to zero
Step SSry
Magnitude (dB)
Bode SSry
• Examining the response of ie (t) for N f = 0
>> SSri0 = ss(
Section 7.3.1
(Fa-Gau*K5a), Gar0, [0 0 0 1], 0)
50
1.4
0
1.2
−50
1
−100
Amplitude
EE/ME 574: Intermediate Controls
−150
0
Bode SSri0
0
−1
0
−2
0.4
−180
0.2
0
1
10
2
10
0
0
3
10
10
1
2
3
Frequency (rad/sec)
−3
−40
180
4
5
6
7
8
6
7
8
Time (sec)
Figure 27: r (t) → y (t) Bode plot and step response, SSry
−4
−5
135
Bode SSry
−6
Step SSre
50
1.2
0
1
−2
10
−1
10
0
10
Frequency (rad/sec)
1
10
2
10
−8
0
5
10
15
20
25
30
35
40
45
Time (sec)
Figure 26: r (t) → ie (t) Bode plot and step response of error, SSre0
• Choose N f to make
ie (t) → 0 by canceling the contributions due G p and
Gcr , with Hie = 0 0 0 1
%%
>>
%%
>>
DC gain from r(t) input to ie(t), through Gcr
iss1 = [0 0 0 1] * inv( -Fcl ) * [0*Gp; 0*Gcu+1*Gcr]
DC gain from r(t) input to ie(t), through Gp
iss2 = [0 0 0 1] * inv( -Fcl ) * [1*Gp; 1*Gcu+0*Gcr]
%% Choose Nf so that
Nf iss2 + iss1 = 0;
>> Nf5a = - iss1/ iss2
Nf5a = 10.2975
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 111
0.8
−50
Amplitude
−7
45
−3
10
Magnitude (dB)
90
−100
135
Phase (deg)
Phase (deg)
0.6
−90
−270
−1
10
−20
Amplitude
Magnitude (dB)
20
Phase (deg)
Step SSri0
0.8
0.6
0.4
90
0.2
45
0
0
−45
−3
10
−2
10
−1
10
0
10
1
10
2
10
−0.2
0
Frequency (rad/sec)
1
2
3
4
5
Time (sec)
Figure 28: r (t) → e (t) Bode plot and step response of error, SSre
• Comparing figures 24 and 27, changing N f from 0 to 10.2975
1. Enormously accelerates the system
2. Introduces some overshoot.
• Like any controller parameter, N f can be tuned to meet system requirements.
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 112
EE/ME 574: Intermediate Controls
Section 7.4.0
7.4 Summary of integral control
EE/ME 574: Intermediate Controls
Section 7.4.0
2. Select N f for good transient response (as given in example 7.3.1, or by
trial and error, or other method)
• To use a controller design with internal state(s), we augment the state vector
with the controller state(s), using Eqn (119).
• Integral control can significantly improve performance, especially accuracy.
• Specifically, to put in integral control introduce
ie (t) = −
Z t
e (τ) dτ ,
which gives
0
d
d
ie (t) = xc (t) = y (t) − r (t)
dt
dt
by choosing
Fcp = H p − J p K p ,
Fcc = −J p Kc ,
Gcu = 0 ,
Gcr = −1 + Jcl
which leads to:
Fa




= 


Fp
..
.
0
···
···
···
Hp − Jp Kp
..
.
−J p Kc




,



G
 p




Gau =  · · ·  ,




0

where the controller gain vector is partitioned according to Ka =

N
G
f p 





Gar =  · · · 




−1 + Jcl

K p : Kc
.
• Control can then be designed by:
1. Determine the Ka vector by
(a) LQR
(b) Pole placement
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 113
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 114
EE/ME 574: Intermediate Controls
Section 8.0.0
8 Controllability and observability
EE/ME 574: Intermediate Controls
Section 8.2.0
8.1 Main facts about uncontrollable or unobservable modes:
• No discussion of State-Variable models would be complete without a
discussion of observability and controllability (R. E. Kalman [1960, 1962]).
• Even if a pole and zero cancel, the system still has that mode in its response.
Challenge: if a pole and zero cancel, it means that feedback control can do
nothing about the mode:
• Modes correspond to energy storage elements in the system
– When a pole and a zero cancel, the mode remains in the system:
• Example: a pole-zero cancellation (fig 29(A)).
– If the mode is unstable, feedback control cannot stabilize the system.
– If the mode is slow or lightly damped, it will remain slow or lightly
damped.
– The mode disappears from the transfer function.
– The mode - and energy storage element - remains in the system:
∗ Either the signal of the mode does not appear in the output
(Unobservable, fig 29(B));
∗ Or the input signal can’t reach the mode (Uncontrollable, fig 29(C)).
– If the mode is fast and well damped:
unobservability is often not a problem.
uncontrollability or
• How do uncontrollable or unobservable modes get excited:
– Disturbances (don’t go through zeros of control input path),
u(t)
B(s)
y(t)
A(s)
y(t)
s+3
u(t)
(s+3) (s+4)
– Initial conditions,
(A) System with pole-zero cancellation.
– Non-linearities, etc.
1
u(t)
8.2 Definitions
y(t)
s+3
(s+3) (s+4)
Controllable System: The system can be driven from any state, x0, to any state,
x1 , by suitable choice of input signal. Equivalently, the input signal reaches
each mode.
(B) Unobservable system (state hidden from output)
u(t)
s+3
y(t)
1
Figure 29(C) shows an uncontrollable system.
(s+3) (s+4)
Observable System: The state x (t = 0) can be determined by observing y (t).
Equivalently, the response of each mode reaches the output signal.
(C) Uncontrollable system (state untouched by input)
Figure 29: A pole-zero cancellation can physically occur in either the input or
output path.
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 115
Figure 29(B) shows an unobservable system.
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 116
EE/ME 574: Intermediate Controls
Section 8.3.1
EE/ME 574: Intermediate Controls
8.3 Mathematical tests:
Section 8.3.2
8.3.2 Test for observability
8.3.1 Test for controllability
Test 1: If the observability matrix is full rank, then the system is observable:

Test 1: If the controllability matrix is full rank, then the system is controllable:
C = G p Fp G p Fp2 G p · · ·
(n−1)
Fp
Gp






O =






(126)
• System is controllable if C is full rank.
• Example:


 −1 0 
Fp = 
,
1 1
C =
G p , Fp G p


 −2 
Gp = 

1




Hp













H p Fp
H p Fp2
..
.
(n−2)
H p Fp
(127)
• System is observable if O is full rank.
• Example:




  −2   −1 0   −2    −2 2 
= 
, 

 =

1
1 1
1
1 −1
since the rank of C is 1 (not 2), the system is uncontrollable.

−1
0


F =

1 1

H=
0 1



0
1





H
0
1
 
 

O = 
=

=


 −1 0  
 0 1 
HF
1 1

1 1



since the rank of O is 2, the system is observable.
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 117
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 118
EE/ME 574: Intermediate Controls
Section 8.4.0
8.4 Summary Controllability and Observability
EE/ME 574: Intermediate Controls
Section 8.4.1
8.4.1 Uncontrollable and unobservable modes and rocketry
• The order of the system is based on the number of storage elements (or 1st
order differential equations).
– If a pole and zero cancel, it does not remove an energy storage element
from the system: the mode is still in there.
• Space launch vehicles are one type of system where controllability and
observability problems arise:
– Flexible structures
– Restricted freedom to place sensors or actuators
• Controllability and observability are properties of the system, not the
controller.
– There is no way to fix a controllability or observability problem by
choosing a different controller.
• Most often controllability or observability problems are fixed by changing
(or adding) an actuator; or changing (or adding) a sensor.
– The poles and zeros move around vigorously during launch, as the
aerodynamic condition changes and mass changes.
• Uncontrollable or unobservable modes must be very carefully considered
for rocketry
– Open-loop unstable mode (without control, the rocket will fall over)
– Lightly damped structural modes:
∗ Can be excited by vibrations during launch
∗ Moderate oscillations can (and did!) exceed structural limits.
• Controllability and observability make no sense with transfer functions.
State variable models were needed to see what was going on (R.E. Kalman
[1960, 1962])
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 119
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 120
EE/ME 574: Intermediate Controls
Section 9.0.0
9 Summary, state-variable modeling and control
• State space models can be written down directly from system equations.
– Often easier than transfer function modeling, since internal variables do
not need to be eliminated.
– When numerical values are available, the plant TF can be obtained from
from the SS model
>> SSuy = ss(Fp, Gp, Hp, Jp)
>> Guy = tf(SSuy)
EE/ME 574: Intermediate Controls
Section 9.0.0
• Integral control
– Any linear system can be modeled in state-variable form, including a
linear plant with PID control.
– Control states are appended to the state vector, giving the augmented
state vector.
– The Fa , Ga and Ha matrices are produced as expected.
– Integral control can be designed with standard tools, e.g., the lqr()
command.
• State-variable modeling
• Observability and Controllability
– A state-variable model is nothing other than a collection of modeling
equations stacked up in matrix-vector form.
– State-variable models have units.
– We can easily represent a variety of different input or output
configurations, by writing the corresponding Gx (for inputs) and Hx and
Jx (for outputs).
– Unobservable and uncontrollable modes arise when poles and zeros
cancel.
– The cancellation is mathematical, the energy storage elements remain in
the system.
– Observability and controllability are intrinsic properties of the system,
independent of the way it is modelled.
• Controller Design
– We have seen two state feedback control design methods:
– Observability and controllability are easily determined with a statevariable model.
1. Pole placement
2. The linear quadratic regulator (LQR design)
– Pole placement design may result in very high controller gains.
– LQR design gives fast, well damped control in many situations.
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 121
Part 4: State-Space Control
(Revised: Jan 27, 2016)
Page 122
EE/ME 574: Intermediate Controls
EE/ME 574: Intermediate Controls
Discrete-Time Control Systems,
the basis of Computer Control
3.2.1
3.2.2
3.2.3
Contents
1 Introduction, big picture of discrete control
1.1 Doing signal processing with sampled signals . . . . . . . . .
1.2 Connecting a controller computer to the continuous-time world
1.2.1 Converting from an analog signal to a sampled signal .
1.2.2 Converting a sampled signal back to an analog signal .
1.3 Building a control system with a discrete-time controller . . .
1.4 Terminology and synonyms . . . . . . . . . . . . . . . . . . .
. .
.
. .
. .
. .
. .
3
6
10
11
13
14
15
2 Introduction to the Z-transform
2.1 Recall, for continuous-time systems: . . . . . . . . . . . . . . . .
2.2 The Z-transform . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.1 Definition of the Z-transform and table of Z-transform pairs
2.2.2 Z-transform gives transfer function of a difference equation
2.2.3 Essential properties of a Z-domain transfer function . . . .
2.2.4 Discrete final value theorem . . . . . . . . . . . . . . . .
2.2.5 Time shifting, one extra bit of nuisance: . . . . . . . . . .
2.3 Discrete system / Z-transform example . . . . . . . . . . . . . . .
2.3.1 Nyquist frequency . . . . . . . . . . . . . . . . . . . . .
2.3.2 Tips . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4 Conclusions on z-transform . . . . . . . . . . . . . . . . . . . . .
16
16
18
19
20
21
22
23
25
27
27
29
3 Interpretation of pole locations on the Z-plane
3.1 Examining Z-domain poles by example . . . . . . . . .
3.1.1 Real poles . . . . . . . . . . . . . . . . . . . .
3.1.2 Complex pole pairs . . . . . . . . . . . . . . .
3.2 Determining the frequency and damping of Z-plane poles
30
30
30
32
34
Part 5: Z-Transform and discrete control
.
.
.
.
.
.
.
.
.
.
.
.
(Revised: Feb 13, 2013)
.
.
.
.
.
.
.
.
Page 1
Matched s-plane and Z-plane poles . . . . . . . . . . . .
Pictoral examples of matched pole pairs . . . . . . . . .
Example, determining ωn and ρ for Z-plane poles . . . . .
34
35
39
4 Converting between continuous and discrete systems
4.1 Determining G pz (z) , the z-transform of G p (s) . . . . . . . . . .
4.1.1 Discrete control example . . . . . . . . . . . . . . . . .
40
40
42
5 Compensator design by discrete equivalent TF
5.1 Matlab’s c2d() command . . . . . . . . . . . . . . . . . . . . .
5.1.1 Why 6 different design methods . . . . . . . . . . . . . .
5.2 Tustin’s rule to find Gcz (z) that approximates Gcs (s) . . . . . . .
5.2.1 Deriving Tustin’s rule . . . . . . . . . . . . . . . . . . .
5.2.2 Example with Tustin’s rule . . . . . . . . . . . . . . . .
5.2.3 Example controller design by discrete equivalent, done by
hand . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2.4 Example controller design by discrete equivalent, using
Matlab . . . . . . . . . . . . . . . . . . . . . . . . . . .
48
50
52
53
53
55
6 DC Gain, Steady State Error and Complex Gain
6.1 DC Gain . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.2 Steady-State error . . . . . . . . . . . . . . . . . . . . . . .
6.2.1 Determining steady-state error from Trez (z) (or Tre (s))
6.3 Complex gain from a discrete TF . . . . . . . . . . . . . . .
7 Conclusion
Part 5: Z-Transform and discrete control
.
.
.
.
.
.
.
.
57
60
65
65
67
67
71
72
(Revised: Feb 13, 2013)
Page 2
EE/ME 574: Intermediate Controls
Section 1.0.0
1 Introduction, big picture of discrete control
EE/ME 574: Intermediate Controls
Section 1.0.0
• Computer control brings a fundamental change to the nature of signals
− Analog control (Op Amps, capacitors, etc). operate on analog signals
• A basic control loop is illustrated in figure 1.
1
The controller is represented by Gc (s). So far we haven’t talked about how
controllers are actually built.
0.5
Trajectory
Generation
r(t)
+
Process under control
f(t)
Controller
0
e(t)
Controller
Plant
u(t)
Gc(s)
y(t)
Gp(s)
-
−0.5
0
2
4
6
8
10
t [sec]
Sensor
Figure 2: An analog signal, f (t) = e−αt sin (ωt).
Hy(s)
Figure 1: Basic control loop. The Controller comprises the compensator, Gc (s),
and may include trajectory generation, generating r (t), and other
elements.
• How controllers are actually built
There are two broad approaches
1. Analog (e.g., Op Amps, capacitors and inductors, or pneumatics)

0


 
 
1

k=
 
2
 
..
.
0.0000






 0.6113 

f (k) = 


 0.7258 


..
.
0.0



 

 
 0.8  
=
tk = 

 
 1.6  

 
..
.
0 τs



1 τs 


2 τs 

..
.
• When a sampled system is connected to a continuous system, two special
pieces of hardware are needed,
2. Discrete or sampled (by computer)
• Today the vast majority of controllers are sampled
− An A/D converter, and a D/A converter
− Process control, oil refineries, computer controlled in the 1960’s with
’mini’ computers
− Machine tools / robotics, computer controlled in the 1970’s
Continuous
Signal
f(t)
− Aviation, computer controlled in the 1980’s
Sampled
Signal
f(k)
Sampled
Signal
Continuous
Signal
g(t)
g(k)
ZOH
tk = k Ts
− Today, virtually all control (car engines, building environment, ...)
is implemented by computer
Part 5: Z-Transform and discrete control
− Computers operate on discrete sequences of numbers, or samples, of
sampled signals
(Revised: Feb 13, 2013)
Page 3
(A/D Converter)
(D/A Converter)
Figure 3: Block diagrams for A/D and D/A converts.
Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
Page 4
EE/ME 574: Intermediate Controls
Section 1.0.0
• A sampled signal can be represented graphically by a stem plot, as seen in
figure 4.
EE/ME 574: Intermediate Controls
1.1 Doing signal processing with sampled signals
Controller
Gc(z)
e(k)
>>
>>
>>
>>
Ts = 0.8
td = 0:Ts:10;
fd = exp(-alpha*td) .* sin(omega*td);
stem(td, fd)
1
0.5
0.5
• A continuous signal is a wave-form, such as u (t). A sampled signal is a
sequence of numbers, such as u (k). Examples:

0
−0.5
0
−0.5
2
4
6
8
k [sample]
10
12
14
(a) f (k)
0
0





 1 Ts 

t (k) = 


 2 Ts 


..
.
s
f(k)
0
u(k)
Figure 5: Block implementing Gc (z), the compensator in a computer.
f(k T )
1
Section 1.1.0
2
4
6
8
10
t [sec]
(b) f (k Ts )

,
0.0000





 0.6113 

e (k) = 


 0.7258 


..
.

,

 
 
?

u (k) = 
 
 
 
..
.
how can we do signal processing to create a new sampled signal ?
• Sampled signals are processed using difference equations.
Figure 4: A stem plot showing a sampled signal, (a) versus k (which is what the
computer knows), (b) versus tk (which relates to the analog world).
• Figure 4 shows something basic about sampled signals:
− In a difference equation, the output is given by sums and differences of
samples. For example:
u (k) = −a1 u (k − 1)− a2 u (k − 2)+b0 e (k) +b1 e (k − 1)+· · ·
(1)
A sampled signal is only defined at the sample instants.
• At each sample k, the output is calculated as a combination of
− The input, e (k)
− Past inputs, e (k − 1) , e (k − 2)...
− Past output, u (k − 1) , u (k − 2) ...
Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
Page 5
Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
Page 6
EE/ME 574: Intermediate Controls
Section 1.1.0
EE/ME 574: Intermediate Controls
• Difference equation example:
• A difference equation is linear.
For example, if e (k) is doubled, u (k) is doubled.
u (k) = 0.8 u (k − 1) + 0.1 e (k) + 0.1 e (k − 1)
(2)
u (k) = 0.8 u (k − 1) + 0.1 e (k) + 0.1 e (k − 1)
.
• Impulse response, an digital impulsive signal has e (0) = 1 , e (1, 2 ...) = 0
.
k
1
2
3
4
e(k-1)
0
1
0
0
Section 1.1.0
e(k)
1
0
0
0
u(k-1)
0
0.1
0.180
0.144
u(k)
0.100
0.180
0.144
0.115
k
1
2
e(k-1)
0
2
e(k)
2
0
(2, repeated)
u(k-1)
0
0.2
u(k)
0.200
0.360
...
• Since difference equations are linear, a sampled sinusoidal input gives a
sampled sinusoidal output. For example,
t (k) = k Ts = 0.8 k ,
e (k) = 1.0 sin (0.75 t (k))
(Ts = 0.8 seconds)
u (k) = 0.338 sin (0.75t (k) − 70.2o)
0.2
0.16
0.14
u(k)
e(k)
0.12
0.1
0.08
0.06
1
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0
−0.2
−0.4
−0.4
−0.6
−0.6
−0.8
−1
0
0.02
5
10
t [seconds]
15
20
(a) Signal e (k) = 1.0 sin (0.75t)
5
10
0
−0.2
−0.8
0.04
0
0
1
0.8
u(k)
0.18
−1
0
5
10
t [seconds]
15
20
(b) u (k): output of Gc1 (z)
15
k
Figure 6: Stem plot of the impulse response of difference Eqn (2).
Notice the exponential decay.
Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
• The new magnitude and phase are determined by:
>>
>>
>>
>>
Page 7
Ts = 0.8;
Num = [0.1 0.1]; Den = [1 -0.8];
Gc1z = tf(Num, Den, Ts)
[Mag, Phase] = bode(Gc1z, 0.75)
Mag = 0.3380
Phase = -70.2421
Part 5: Z-Transform and discrete control
%%
Setup model
%% Calc Mag and Phase
%% With bode command
(Revised: Feb 13, 2013)
Page 8
EE/ME 574: Intermediate Controls
Section 1.1.0
− Specifically, since a sinusoidal input gives a sinusoidal output
world
• We can make a bode plot of Gc1 (z), which plots the magnitude and phase
of the output for sinusoidal inputs.
10
• To use digital control, the controller computer has to receive inputs from
sensors, and supply outputs to actuators.
• A typical configuration is seen in figure 8.
0
Gain dB
Section 1.2.0
1.2 Connecting a controller computer to the continuous-time
• Since the difference equation is linear
−10
− The sensor itself is analog (such as a pressure sensor). The analog sensor
signal y (t) is converted to a digital signal y (k) by an A/D converter. (An
Analog To Digital converter)
−20
−30
−40
−50 −2
10
−1
10
0
10
1
10
0
Phase [deg]
EE/ME 574: Intermediate Controls
− The computer command, u (k), is converted from a digital signal by a
D/A converter. (A Digital To Analog converter)
∗ The D/A converter implements a “Zero-Order Hold” (ZOH), which
must be modelled in the control loop (more later).
−30
−60
Discrete part, Computer Controller
−90
−2
10
−1
0
10
10
Frequency [radians/second]
1
10
r(k)
Trajectory
Generator
+
e(k)
Figure 7: Bode plot for difference equation (2). Eqn (2) gives a low-pass filter.
ys(t)
ys(k)
Kc Gc(z)
u(k)
-
ZOH
(D/A
Conv.)
u(t)
Plant
y(t)
Gp(s)
t=kTs
(A/D Conv.)
>> Gc1z = tf(Num, Den, Ts)
>> bode(Gc1z)
Sensor
%%
%%
%%
Hy(s)
Ts = 0.8 [seconds]
Fs = 1.25 [Hz]
Fnyquist = Fs/2 = 0.625 [Hz]
Analog part
Figure 8: Configuration of basic computer control. Sensor signals are prepared
for the computer by
Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
Page 9
Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
Page 10
EE/ME 574: Intermediate Controls
Section 1.2.1
EE/ME 574: Intermediate Controls
Section 1.2.1
− Voice is sampled at 8 bits, 8,000 samples per second
1.2.1 Converting from an analog signal to a sampled signal
• This is the job of the A/D converter. The sampled signal is defined only at
the sampling instants.
− One voice channel on a telephone trunk line is 64,000 bits per second
(uncompressed)
1
0.5
y(k)
tk = k Ts
f(k Ts)
y(t)
(A/D Converter)
0
−0.5
0
2
4
6
8
10
t [sec]
Figure 9: Sampling, converting from y (t) to y (k).
• A/D converters are defined by
− A range, such as e (t) = −10..10 [volts]
− A resolution, normally in bits, such as 8 bit, 10 bit, 16 bit.
− A maximum speed,
GSamples/second
such as 100k samples/second,
up to 1
• Example: telephone
− Today essentially all telephone exchanges are digital
Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
Page 11
Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
Page 12
EE/ME 574: Intermediate Controls
Section 1.2.2
EE/ME 574: Intermediate Controls
Section 1.3.0
1.3 Building a control system with a discrete-time controller
1.2.2 Converting a sampled signal back to an analog signal
• If we are going to control continuous-time equipment, we need to convert
u (k) back into u (t)
• This is the job of the D/A converter. The output signal is u (t), defined at all
instants.
1
• Most control systems comprise a discrete-time part (the controller) and a
continuous-time part (the thing being controlled), figure 11.
− The computer controller implements Gc (z), a difference equation
− The plant is governed by differential equations
− The two pieces are joined together by an A/D and a D/A
u(k)
u(t)
Discrete part, Computer Controller
0.5
ZOH
Trajectory
Generator
f(t)
(D/A Converter)
r(k)
+
e(k)
0
ys(t)
ys(k)
Kc Gc(z)
u(k)
-
ZOH
(D/A
Conv.)
u(t)
Plant
y(t)
Gp(s)
t=kTs
(A/D Conv.)
−0.5
0
2
4
6
8
10
t [sec]
Sensor
Hy(s)
Figure 10: Sampling, converting a signal from u (k) to u (t) with zero-order hold.
Analog part
• The Zero-Order-Hold (ZOH)
Figure 11: System with sampled control.
− A level u (k) is given at t = tk = k Ts
Broad approach to design and analysis:
− The ZOH has a hold circuit (Op Amp + capacitor) that holds the signal
until u (k + 1) arrives.
Z-transform. (Analogous to the Laplace transform)
• D/A converters are defined by
2. Convert Ys (s) /U (s) to discrete-time (including the zero-order hold), to
analyze the complete discrete-time system.
− A range, such as u (k) = −10..10 [volts]
− A resolution, normally in bits, such as 8 bit, 10 bit, 16 bit.
− A maximum speed, such as 100 kS/sec, up to 1 GS/second
3. Obtain the digital compensator, Gc (z) by:
• Example: telephone
(a) Discrete Equivalent Design: design Gc (s), convert to discrete-time,
− At the phone switch, a D/A converter converts 8,000 8-bit samples per
second to an analog signal, which goes out on the local subscriber line.
Part 5: Z-Transform and discrete control
1. Develop transform method to analyze discrete-time system, the
(Revised: Feb 13, 2013)
Page 13
or
(b) Direct Digital Design: directly design Gc (z) based on discrete-time
analysis.
Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
Page 14
EE/ME 574: Intermediate Controls
Section 1.4.0
EE/ME 574: Intermediate Controls
Section 2.1.0
2 Introduction to the Z-transform
1.4 Terminology and synonyms
• Discrete-time systems
• The Z-transform can be understood by analogy with the Laplace transform
for continuous-time systems.
− Digital control
− Discrete system
2.1 Recall, for continuous-time systems:
− Sampled system
• Consider a controller with an input e (t) and an output u (t)
− z-domain system
input e(t)
• Sampled signals
− Discrete-time signal
Controller
output u(t)
Figure 12: System with input and output.
− Discrete signal
− Digital signal
• Continuous signals are functions of time, such as e (t) and u (t).
− z-domain signal
• Continuous time systems are governed by differential equations, such as
• Continuous-time systems
ü (t) + a1 u̇ (t) + a2 u (t) = bo ë (t) + b1 ė (t) + b2 e (t)
− Analog system
− s-domain system
(3)
• For continuous equations, the highest derivative of the output is given by:
− The lower derivatives of the output, and
− The input and derivatives of the input
ü (t) = −a1 u̇ (t) − a2u (t) + bo ë (t) + b1 ė (t) + b2 e (t)
Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
Page 15
Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
(4)
Page 16
EE/ME 574: Intermediate Controls
Section 2.1.0
2.1 Recall, for continuous-time systems (continued)
L
=⇒
L {u (t)} =
Z ∞
0−
e−st u (t) dt = U (s)
(5)
if v (t) = 3 y (t) → V (s) = 3Y (s)
2. Superposition
• For the controls engineer, the transfer function is the value
proposition
− As we’ve seen, the control’s engineer rarely forms the Laplace transform
of a signal, or inverse Laplace Transform of U (s)
− The transfer function enables us to study the properties of systems
without solving differential equations.
• The essential properties of the Laplace Transform are:
1. Scaling
Section 2.2.0
2.1 Recall, for continuous-time systems (continued)
• The Laplace transform is a transform between a time-domain signal and a
function of s
u (t)
EE/ME 574: Intermediate Controls
if v (t) = y (t) + u (t) → V (s) = Y (s) +U (s)
− The transfer function is the reason we use Laplace transforms in the
analysis of control systems.
3. It handles derivatives nicely:
2.2 The Z-transform
L
Given:
L
y (t) =⇒ Y (s)
then ẏ (t) =⇒ sY (s)
• The Z-transform is a transformation between a sampled time-domain signal
and a function of z :
• Properties 1, 2 and 3 enable us to define the transfer function, given
y (k)
Z
→
∞
Z {y (k)} =
∑ u (k) z−k = Y (z)
(8)
k=0
ü (t) + a1 u̇ (t) + a2 u (t) = bo ë (t) + b1 ė (t) + b2 e (t)
• As one would expect, the important properties of the Z-transform are:
U (s) s2 + a1 s + a2 = E (s) b0 s2 + b1 s + b2
2. Superposition
(6)
Forward :
s2 + b
U (s)
b0
1 s + b2
= 2
E (s)
s + a1 s + a2
(Revised: Feb 13, 2013)
if v (k) = y (k) + u (k) → V (z) = Y (z) +U (z)
3. It handles time shifts:
• And obtain the transfer function
Part 5: Z-Transform and discrete control
if v (k) = 3 y (k) → V (z) = 3Y (z)
1. Scaling
• We can form the Laplace transform of the left and right sides
(7)
and
Backward :
Page 17
Z {y (k + 1)} = z Z {y (k)} = zY (z)
Z {y (k − 1)} = z−1 Z {y (k)} = z−1 Y (z)
Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
Page 18
EE/ME 574: Intermediate Controls
Section 2.2.1
EE/ME 574: Intermediate Controls
2.2.1 Definition of the Z-transform and table of Z-transform pairs
2.2.2 Z-transform gives transfer function of a difference equation
1. Starting with difference equation (10)
• The one-sided Z-transform is defined by:
∞
Z { f (k) } = F (z) =
Section 2.2.2
∑
f (k) z−k
u (k) + a1 u (k − 1) + a2 u (k − 2) =
(9)
(10)
k=0
bo e (k) + b1 e (k − 1) + b2 e (k − 2)
− Like the Laplace transform integrals, Eqn (9) is not often used directly
by the controls engineer.
2. Taking the Z-transform of the each term gives:
− What is important is properties (1) - (3) above, because they allow one to
define the transfer function of a difference equation.
Z {u (k)} + Z {a1 u (k − 1)} + Z {a2 u (k − 2)} =
(11)
Z {bo e (k)} + Z {b1 e (k − 1)} + Z {b2 e (k − 2)}
• Some Z-transform pairs are given in table 1.
3. Now applying the scaling and time-shift properties of the Z-transform:
Signal
Impulse
Z-transform, t (k) = k Ts
Laplace transform
f (t) = δ (t)
F (s) = 1
f (k) = 1
Z {u (k)} + a1 z−1 Z {u (k)} + a2 z−2 Z {u (k)} =
F (z) = 1
b0 Z {e (k)} + b1 z−1 Z {e (k)} + b2 z−2 Z {e (k)}
0 , otherwise
Step
f (t) = 1+ (t)
Ramp
f (t) = t
Exp. Decay
f (t) = e−at
F (s) =
1
s
F (s) =
1
s2
f (k) = k Ts , k ≥ 0
F (z) =
Ts z
(z−1)2
1
s+a
f (k) = e−a k Ts
F (z) =
z
z−e−a Ts
F (s) =
Sine
f (t) = sin (ωt) F (s) =
a
s2 +a2
Cosine
f (t) = cos (ωt) F (s) =
s
s2 +a2
f (k) = 1 , k ≥ 0
F (z) =
f (k) = sin (ω k Ts ) F (z) =
z sin(ω Ts )
z2 −2 cos(ω Ts ) z+1
f (k) = cos (ω k Ts ) F (z) =
z(z−cos(ω Ts ))
z2 −2 cos(ω Ts ) z+1
Table 1: Laplace and Z-transform transform pairs
Part 5: Z-Transform and discrete control
z
z−1
(Revised: Feb 13, 2013)
Page 19
(12)
which gives:
U (z) 1 + a1 z−1 + a2 z−2 = E (z) b0 + b1 z−1 + b2 z−2
(13)
and finally:
U (z) b0 + b1 z−1 + b2 z−2 bo z2 + b1 z + b2
=
= 2
E (z)
1 + a1 z−1 + a2 z−2
z + a1 z + a2
(14)
• Eqn (14) is the Transfer Function of Eqn (10)
• The steps of forming a discrete transfer function, Eqns (10) to (14), are
entirely analogous to the steps of forming a continuous transfer function.
Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
Page 20
EE/ME 574: Intermediate Controls
Section 2.2.3
2.2.3 Essential properties of a Z-domain transfer function
EE/ME 574: Intermediate Controls
Section 2.2.4
2.2.4 Discrete final value theorem
The essential properties of a discrete transfer function are also analogous to the
properties of the continuous transfer function. Given:
• The final value theorem for the z-transform is:
lim x (k) = xss = lim 1 − z−1 X (z)
(15)
(16)
z→1
k→∞
U (z) bo z2 + b1 z + b2
= 2
Gc (z) =
E (z)
z + a1 z + a2
− Steady-state error applies !
• All information about the system is captured in the transfer function, Gc (z)
− Gc (z) is a rational polynomial of constants,
− Example, final value of u (k) with Gc1 (z) and e (k) given by a unit step
U (z)
E (z)
lim u (k) = lim 1 − z−1 U (z) = lim 1 − z−1
z→1
z→1
E (z)
k→∞
− It is independent of the particular input and output signals.
(17)
− Since e (k) is a unit step, E (z) = z/ (z − 1) , and so
• The Discrete Transfer Function has poles and zeros
0.1 z + 0.1 z
lim u (k) = lim 1 − z−1
z→1
k→∞
z − 0.8 z − 1
− They are the roots of the denominator and numerator polynomials
− The poles correspond to the modes.
− Notice that
− We can test the poles for stability
− Root locus applies !
(But special interpretation is needed for pole locations on the Z-plane)
z−1
1
=
1 − z−1 = 1 −
z
z
(18)
(19)
Giving
lim u (k) = lim
k→∞
z→1
z−1
z
0.1 z + 0.1
0.1 z + 0.1 z
= lim
z − 0.8 z − 1 z→1 z − 0.8
• Example, considering the difference equation (2),
− Finally
u (k) = 0.8 u (k − 1) + 0.1 e (k) + 0.1 e (k − 1)
(2, repeated)
lim u (k) = lim
k→∞
0.1 z + 0.1
= 1.0
z − 0.8
The final value of u (k) with e (k) = 1+ (k) is u (k) = 1.0
Then
Gc1 (z) =
Pole: z = 0.8 ,
z→1
U (z) 0.1 + 0.1 z−1 0.1 z + 0.1
=
=
E (z)
1 − 0.8 z−1
z − 0.8
− Notice the analogy between the cancellation in Eqn (19) and the
cancellation in:
U (s) 1
lim u (t) = lim s
t→∞
s→0 E (s) s
Zero: z = −1.
Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
Page 21
Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
Page 22
EE/ME 574: Intermediate Controls
Section 2.2.5
EE/ME 574: Intermediate Controls
2.2.5 Time shifting, one extra bit of nuisance:
Section 2.2.5
Conclusion on time shifting
• Consider the difference equation
u (k) = +1.6 u (k − 1) − 0.68 u (k − 2) + 0.4 e (k) − 0.28 e (k − 1) + 0.048 e (k − 2)
u (k) = +1.6 u (k − 1) − 0.68 u (k − 2) + 0.4 e (k) − 0.28 e (k − 1) + 0.048 e (k − 2)
(20)
(Eqn 20, repeated)
• We can also write the difference equation directly from Eqn (22)
• Gives the Z-transform:
− Remembering that z2U (z)
−1
−2
−1
→
u (k + 2) = +1.6 u (k + 1) − 0.68 u (k) + 0.4 e (k + 2) − 0.28 e (k + 1) + 0.048 e (k)
(24)
• If difference Eqn (20) is true for all k, then difference Eqn (24) is likewise
true for all k.
U (z) 0.4 − 0.28 z−1 + 0.048 z−2
=
= Geuz (z)
E (z)
1 − 1.6 z−1 + 0.68 z−2
• This is analogous to saying,
• But we work with poles and zeros of the TF written with z, so clear the z−n:
Gc (z) =
U (z) b0 + b1 z−1 + b2 z−2
=
E (z)
1 + a1 z−1 + a2 z−2
bo z 2 + b1 z + b2
z2
=
z2
z 2 + a1 z + a2
= u (k + 2), Eqn (22) becomes:
−2
U (z) − 1.6 z U (z) + 0.68 z U (z) = 0.4 E (z) − 0.28 z E (z) + 0.048 z E (z)
U (z) 1 − 1.6 z−1 + 0.68 z−2 = E (z) 0.4 − 0.28 z−1 + 0.048 z−2
• And so:
Z −1
z2
Gc (z)
z2
(21)
• In a sense, the distinction between Eqns (20) and (24) lies is the definition
of k.
• So we have to shift back and forth between two forms of the TF:
− We normally set up the difference equation to compute u (k),
0.4 z2 − 0.28 z + 0.048
Gc (z) =
z2 − 1.6 z + 0.68
Gc (z) =
0.4 − 0.28 z−1 + 0.048 z−2
1 − 1.6 z−1 + 0.68 z−2
For analysis
(22)
For implementation
(23)
− In some cases we might compute u (k + 1). For example, if output u (k)
depends only on values up to e (k − 1), in this case we might set up the
the difference equation
∗ To compute u (k + 1) based on e (k) (pre-computing)
∗ Rather than computing u (k) based on e (k − 1).
• Eqns (22) and (23) are the two forms of the same TF.
• In general, we can time-shift the difference equation as needed
− The TF in the form of Eqn (23) gives the Diff Eqn (20)
Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
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Part 5: Z-Transform and discrete control
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EE/ME 574: Intermediate Controls
Section 2.3.0
EE/ME 574: Intermediate Controls
2.3 Discrete system / Z-transform example
Section 2.3.0
Step response from Eqn ()
2.4
2.2
• A TF model of the difference equation is setup with the tf() command,
including a value for ts .
>> figure(1), clf
>>
step(Guez);
>>
grid on
2
1.8
1.6
Amplitude
• Starting with the difference equation, and Fs = 20 [samples/second]
1.4
1.2
u (k) = +1.6 u (k − 1)−0.68 u (k − 2)+0.4 e (k)−0.28 e (k − 1)+0.048 e (k − 2)
1
0.8
• Setup the TF with the Matlab code:
0.6
0.4
0
0.5
1
1.5
Time (sec)
>> Fs = 20; Ts = 1/Fs;
>> Num = [0.4 -0.28 0.048]
>> Den = [1 -1.6 +0.68]
>> Geuz = tf(Num, Den, Ts)
%%
%%
%%
%%
Figure 13: Discrete step response of first example system.
Set the sample rate,
often convenient to set Fs
Set the Num and Den polynomials
Setup the TF
• We can also determine the frequency response
Bode Diagram
0.4 z^2 - 0.28 z + 0.048
-----------------------z^2 - 1.6 z + 0.68
20
15
Magnitude (dB)
Transfer function:
>> figure(2), clf
>>
bode(Geuz)
10
5
Sampling time: 0.05
0
90
45
Phase (deg)
• Printing “Sample time” indicates that the TF is discrete-time.
0
−45
−90
−135
−180
−2
10
+
−1
10
0
10
Frequency (rad/sec)
1
10
Figure 14: Discrete step response. Note the Nyquist frequency limit at 10 Hz
(62.8 [rad/sec]), from Fs = 20 [samples/sec].
Part 5: Z-Transform and discrete control
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Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
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2
10
EE/ME 574: Intermediate Controls
Section 2.3.2
EE/ME 574: Intermediate Controls
Section 2.3.2
• Pay attention to get the numerator powers correct, when multiplying by
zn/zn to convert between analysis form and implementation form.
2.3.1 Nyquist frequency
• In a sampled system, the Nyquist frequency highest frequency sinusoidal
signal that can be uniquely determined. It is given by:
FNyquist =
1
Fs
2
(25)
• Example: CD music format:
Fs = 44, 000 [samples/sec]
FNyq = 22, 000 [Hz]
• Matlab draws the discrete Bode plot up to the Nyquist frequency.
2.3.2 Tips
• Notice the ’-’ signs in the denominator
− The difference equation
u (k) = +1.6 u (k − 1) − 0.68 u (k − 2)
+0.4 e (k) − 0.28 e (k − 1) + 0.048 e (k − 2)
− Gives the TF
Geuz (z) =
0.4 z2 − 0.28 z + 0.048 U (z)
=
E (z)
z2 − 1.6 z + 0.68
• Pay attention to the minus signs in the denominator; the +ai terms in the TF
are −ai terms in the implementation equation.
Part 5: Z-Transform and discrete control
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Part 5: Z-Transform and discrete control
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EE/ME 574: Intermediate Controls
Section 2.4.0
EE/ME 574: Intermediate Controls
Section 3.1.1
3 Interpretation of pole locations on the Z-plane
2.4 Conclusions on z-transform
• Starting with the difference equation
• Discrete time systems are implemented as difference equations
u (k) = −a1 u (k − 1) − a2 u (k − 2)
• The z-transform gives us the z-domain transfer function, such as Gc (z)
(26)
+bo e (k) + b1 e (k − 1) + b2 e (k − 2)
We get the TF
− Gc (z) can be obtained directly from the difference equation
Geuz (z) =
− Gc (z) is a rational polynomial, and has poles and zeros
U (z) bo z2 + b1 z + b2
= 2
E (z)
z + a1 z + a2
(27)
• The poles are the roots of
• Commands such as tf(), step(), and bode() operate with discrete as well
as continuous transfer functions.
z 2 + a1 z + a2 = 0
• It is the feedback of past outputs to the current output that determines
stability, that is, the poles of G (z).
3.1 Examining Z-domain poles by example
3.1.1 Real poles
u (k) = 0.8 u (k − 1) + e (k) ,
• Consider
1
Gc (z) =
z
z − 0.8
1.5
0.9
Unit Circle
1
0.8
0.7
0.5
Imaginary
u(k)
0.6
0.5
0.4
0
−0.5
0.3
0.2
−1
0.1
0
0
2
4
6
k [sample]
8
10
12
−1.5
−1.5
−1
−0.5
0
Real
0.5
1
1.5
Figure 15: Impulse response and pole-zero map.
− Pole at z1 = +0.8 , yet the system is stable.
Part 5: Z-Transform and discrete control
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Part 5: Z-Transform and discrete control
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EE/ME 574: Intermediate Controls
Section 3.1.1
• Now consider
EE/ME 574: Intermediate Controls
Section 3.1.2
3.1.2 Complex pole pairs
Gc (z) =
u (k) = 1.1 u (k − 1) + e (k) ,
3.5
z
z − 1.1
• Consider a system with a complex pole pair
3
• This system has pole pair at:
0.5
Imaginary
u(k)
1.5
z2
z2 − 1.6 z + 0.8
Unit Circle
1
2.5
2
Gc (z) =
u (k) = 1.6 u (k − 1) − 0.8 u (k − 2) + e (k) ,
1.5
0
z1,2 = 0.8 ± j 0.4 = 0.89∠ ± 26.6o
−0.5
1
−1
0
0
2
4
6
k [sample]
8
10
−1.5
−1.5
12
−1
−0.5
0
Real
0.5
1
1.5
u(k)
Figure 16: Impulse response and pole-zero map.
Pole at z1 = +1.1 , System unstable.
• Intuitively, with
u (k) = a1 u (k − 1) + b0e (k)
− If |a1| < 1.0 , the successive values of u (k) will get smaller.
− If a1 = 1.0 , the successive values of u (k) will remain a constant size.
− If |a1| > 1.0, the successive values of u (k) will grow larger
f (k) = e
− Pole at z1
− If α < 0,
= eα Ts
,
z
F (z) =
z − eα Ts
1
1
0.5
0.5
0
0
−0.5
−0.5
−1
−1
0
2
4
6
k [sample]
8
10
12
Unit Circle
−1.5
−1.5
−1
−0.5
0
Real
0.5
1
1.5
• The difference equation with a complex pole pair:
(28)
.
|z1| < 1.0 ,
1.5
Figure 17: Impulse response and pole-zero map. The system is stable.
• Mathematically, from table 1, we have the Z-transform pair:
α k Ts
2
1.5
Imaginary
0.5
− Gives a damped sinusoidal impulse response,
− If we write:
z1,2 = A ∠ ± θ
(stable)
− If α = 0, |z1| = 1.0 ,
(marginally stable)
∗ Then the envelop evolves by a factor of A with each sample,
− If α > 0, |z1| > 1.0 ,
(unstable)
∗ And the sinusoid rotates by an angle θ with each sample.
Part 5: Z-Transform and discrete control
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Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
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Page 32
EE/ME 574: Intermediate Controls
Section 3.1.2
Section 3.2.1
3.2 Determining the frequency and damping of Z-plane poles
• Finally, consider another complex pole pair
u (k) = 2.0 u (k − 1) − 1.8 u (k − 2) + e (k) ,
Gc (z) =
z2
z2 − 2.0 z + 1.8
z1,2 = 1.0 ± j 0.89 = 1.34∠ ± 41.8o
• This system has pole pair at:
EE/ME 574: Intermediate Controls
− Rotation per sample: 41.8o
• There are two ways to think about frequency and damping on the Z-plane
1. Matching Z-plane and s-plane poles, and interpreting the s-plane pole
locations
2. Directly considering pole locations on the z-plane.
− Magnitude change per sample: 1.34 .
3.2.1 Matched s-plane and Z-plane poles
30
1.5
• The relationship between an s-plane si and a Z-plane pole zi is:
25
Unit Circle
1
20
0.5
Imaginary
u(k)
15
10
zi = esi Ts ,
si =
0
ln (zi )
Ts
(30)
5
−0.5
0
• Eqn (30) leads to a couple of useful forms:
−1
−5
−10
0
2
4
6
k [sample]
8
10
12
−1.5
−1.5
−1
−0.5
0
Real
0.5
1
1.5
s − plane
Figure 18: Impulse response and pole-zero map. The system is unstable.
s1,2 = −α ± j ω
s1,2 =
• The difference equation with a complex pole pair with |zi | > 1.0
− Gives an unstable oscillatory response, as with real (si ) > 0 .
θ
ln (A)
±j
Ts
Ts
z − plane
z1,2 = e−α Ts ∠ ω Ts
(31)
z1,2 = A ∠ ± θ
(32)
− In Eqns (31) and (32), the s-plane poles are in rectuangar coordinates,
while the z-plane poles are in polar coordinates.
Conclusions
• Poles that lie within the unit disk give stable modes
• A stable complex pole pair gives a damped oscillatory response
Part 5: Z-Transform and discrete control
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Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
Page 34
EE/ME 574: Intermediate Controls
Section 3.2.2
3.2.2 Pictoral examples of matched pole pairs
EE/ME 574: Intermediate Controls
Section 3.2.2
1. Consider contours of constant decay rate on the s and Z-planes
• Lines of constant decay rate are circles in the Z-plane.
• The smaller the radius, the faster the decay.
5
4
4
3
3
Faster Decay
1
Faster Decay
2
1
1
0
0.6
0.4
imaginary
2
imaginary
imaginary
0.8
0
−1
−1
−2
−2
−3
−3
−4
−4
0.2
0
−0.2
−0.4
−0.6
s−plane
−0.8
Z−plane
−1
s−plane
−5
−5
0
real
−4
−3
−2
−1
0
1
2
3
−1
−0.5
real
5
Figure 19: s-plane pole locations, ’+’ and ’x’ are stable, ’♦’ is marginally stable,
’∇’ and ’△’ are unstable.
0
real
0.5
1
(a) s-plane
(b) Z-plane
Figure 21: s and Z-plane, contours of constant
decay rate,
α = −1 , −2 , −3 , −4 sec−1 , Ts = 0.2 [sec].
2. Consider contours of constant oscillatory frequency, ω
• Angle of pole gives oscillation frequency.
• Negative real axis corresponds to the Nyquist frequency.
2
1.5
1
4
1.5
Faster ω
3
1
2
0
0.5
1
imaginary
−0.5
−1
Faster ω
imaginary
imaginary
0.5
0
0
Nyquist
Freq
−1
−0.5
Z−plane
−1.5
−2
s−plane
−2
−2
0
real
1
2
−4
−4
−3
−2
−1
0
1
2
3
real
Figure 20: Corresponding Z-plane pole locations, ’+’ and ’x’ are stable, ’♦’ is
marginally stable, ’∇’ and ’△’ are unstable.
Part 5: Z-Transform and discrete control
Z−plane
−1
−3
−1
(Revised: Feb 13, 2013)
Page 35
−1.5
−1.5
−1
−0.5
0
real
0.5
1
1.5
(a) s-plane
(b) Z-plane
Figure 22: s and Z-plane, contours of constant ω, ω = 1 , 2 , 3 [r/s], Ts = 0.2 [s].
Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
Page 36
EE/ME 574: Intermediate Controls
Section 3.2.2
3. Consider contours of constant damping, ρ
EE/ME 574: Intermediate Controls
Section 3.2.2
4. Consider contours of constant natural frequency, ωn
• On the s-plane, damping corresponds to angle
• On the s-plane, ωn corresponds to radius.
• On the Z-plane, contours of constant damping correspond to complex
curves
• On the Z-plane, contours of constant ωn correspond to complex curves
• Generally, the nearer the pole lies to the real axis, the greater the damping
• The character of the contours near z = 1.0 match the character of the
contours near s = 0
• The character of the contours near z = 1.0 match the character of the
contours near s = 0
1.5
4
3
Increasing
Wn
Increasing
Wn
1
2
1.5
4
imaginary
imaginary
1
Increasing
Damping
2
0
−1
0.5
imaginary
0
0
−0.5
−2
1
imaginary
0.5
1
Increasing
Damping
3
s−plane
−3
0
Z−plane
−1
−4
−1
−0.5
−4
−2
0
2
−1.5
−1.5
−1
−0.5
real
−2
s−plane
(a) s-plane
Z−plane
−1
−3
−4
−4
−3
−2
−1
0
1
2
3
−1.5
−1.5
−1
−0.5
real
(a) s-plane
0
real
0.5
1
1.5
0
real
0.5
1
1.5
(b) Z-plane
Figure 24: s and Z-plane, contours of constant ωn , ωn = 1 , 2 , 3 , 4 [rad/sec],
Ts = 0.2 [sec].
(b) Z-plane
Figure 23: s and Z-plane, contours of constant ρ,
ρ = 0.24 , 0.44 , 0.71 , 0.89 , 0.97 [·], Ts = 0.2 [sec].
• Units of the s-plane:
Decay or rotation per second, seconds−1
• Units of the Z-plane:
Decay or rotation per sample (dimensionless)
Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
Page 37
Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
Page 38
EE/ME 574: Intermediate Controls
Section 3.2.3
3.2.3 Example, determining ωn and ρ for Z-plane poles
Answer: The decay rate, α, and oscillatory frequency, ω, can be found
directly from zi . Writing
systems
• We need to convert between discrete and continuous systems for two
reasons:
1. Analysis of the complete system using z-transforms
o
− The decay rate α is given by the pole radius, we have A = e−α Ts , so
α=−
Section 4.1.0
4 Converting between continuous and discrete
• Given Ts = 0.15 [seconds], and z1,2 = 0.8 ± 0.2 j, find the equivalent s-plane
poles, and find α, ω, ωn and ρ for the poles.
z1,2 = A∠ ± θ = 0.8246∠ ± 14.03
EE/ME 574: Intermediate Controls
ln (A)
ln (0.8246)
=−
= 1.2855 [sec−1]
Ts
−0.15
2. To implement compensators, Gc (s) design in continuous-time and
implemented in discrete-time.
4.1 Determining G pz (z) , the z-transform of G p (s)
− The oscillatory frequency is given by the angle of the z-plane pole
θ degrees/sample
θ degrees
ω =
=
Ts seconds/sample
Ts second
Discrete part, Computer Controller
r(k)
Trajectory
Generator
+
e(k)
ys(t)
ys(k)
u(k)
Gc(z)
ZOH
(D/A
Conv.)
-
u(t)
Plant
y(t)
Gp(s)
t=kTs
(A/D Conv.)
ω = 14.03/0.15 = 93.57 [degrees/second]
= 1.633 [radians/second]
Sensor
Hy(s)
− We can also directly find the equivalent s-plane pole
ln z1,2
s1,2 = −α ± j ω =
Ts
−0.1928 + j 0.2450
= −1.2855 + j 1.6332 [sec−1]
=
0.15
− ωn and ρ can be found from α and ω
ωn =
p
Part 5: Z-Transform and discrete control
Analog part, u(t)->y(t)
Figure 25: Build a discrete model of the system.
• The plant is itself is modelled as a continuous-time system, G p (s)
• But if we combined the plant with the D/A and A/D, we have a true discrete
system.
G pz (z) : u (k) − > ys (k)
(33)
α2 + ω2 = 2.0784 [radians/second]
ρ = α/ωn = 1.2855/2.0784 = 0.6185
Discrete signal path, u(k)->ys(k)
[·]
(Revised: Feb 13, 2013)
Page 39
Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
Page 40
EE/ME 574: Intermediate Controls
Section 4.1.0
• By converting G p (s) to G pz (z), including the Zero-Order-Hold of the D/A
converter, we can form G pz (z) and put together an all-discrete model.
EE/ME 574: Intermediate Controls
Section 4.1.1
4.1.1 Discrete control example
• Consider a system governed by
d(k)
r(k)
+
e(k)
+ u(k)
u(k)
Kc Gcz(z)
-
+
G p (s) =
ys(k)
Gpz(z)
0.2
s+1
• Implement a discrete controller according to (more on discrete design in the
next section):
ys(k)
Figure 26: Block of all-discrete model.
Ts = 1/20 [sec]
• Standard transfer-function analysis applies, using the z-transform.
u (k) = 0.9 u (k − 1) + 1.0 e (k) − 0.5 e (k − 1)
Kc Gcz G pz
1 + Kc Gcz G pz
(34)
1
Trez (z) =
1 + Kc Gcz G pz
(35)
Tryz (z) =
Tdyz (z) = =
G pz
1 + Kc Gcz G pz
(36)
Etc.
(38)
• Find G pz (z)
>> Gps = tf(0.2, [1 1])
>> Fs = 20; Ts = 1/Fs
>> Gpz = c2d(Gps, Ts, ’zoh’)
Transfer function:
0.009754
---------z - 0.9512
Sampling time: 0.05
%% Gps from physics of system
%% Setup Ts
%% Find Gpz
• The discrete TF G pz (z) is given by
G (s) p
G pz (z) = 1 − z−1 Z
s
• Find Gcz (z) from the controller implementation, Eqn (38),
(37)
Where Z {F (s)} is the z-transform of the sampled time series whose
Laplace transform is F (s) (see Franklin et al., section 6.8).
• Practically, G pz (z) is determined using Matlab’s c2d() command.
Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
Page 41
>> Kc = 10
>> Gcz = tf([1 -0.5], [1 -.9], Ts)
Transfer function:
z - 0.5
------z - 0.9
Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
Page 42
EE/ME 574: Intermediate Controls
Section 4.1.1
• Find the closed-loop transfer function
EE/ME 574: Intermediate Controls
Section 4.1.1
• Find the equivalent s-plane pole locations
1. By converting the z-plane poles directly
>> Tryz = feedback(Kc*Gcz*Gpz, 1)
Transfer function:
>> CLPolesz = pole(Tryz)
CLPolesz =
0.8768 + 0.1962i
0.8768 - 0.1962i
>> CLPoless = log(CLPolesz)/Ts
CLPoless =
-2.1402 + 4.4018i
-2.1402 - 4.4018i
0.09754 z - 0.04877
---------------------z^2 - 1.754 z + 0.8073
Sampling time: 0.05
• Plot step response. The step() command recognizes that Tryz is a
z-transform.
>> figure(1), clf
>>
step(Tryz)
%% Find z-plane poles
%% Convert to s-plane
2. By converting Tryz (z) to Trys (s)
>> Trys = d2c(Tryz)
%%
Convert Tryz to Trys
Step Response
Transfer function:
1.622 s + 21.78
-------------------s^2 + 4.28 s + 23.96
1.4
1.2
Amplitude
1
0.8
>> pole(Trys)
ans =
-2.1402 + 4.4018i
-2.1402 - 4.4018i
0.6
0.4
0.2
0
0
0.5
1
1.5
Time (sec)
2
2.5
3
3.5
• Find ρ , ω , ωn from s-plane poles
Figure 27: Step response of Tryz (z).
• Find step-response properties (overshoot, settling time, rise time) from the
discrete step response.
Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
Page 43
>>
>>
>>
>>
alpha
omegan
rho
omega
=
=
=
=
-real(CLPoless(1))
abs(CLPoless(1))
alpha / omegan
omegan * sqrt(1-rho^2)
Part 5: Z-Transform and discrete control
alpha =
2.1402
omegan =
4.8945
rho =
0.4373
omega =
4.4018
(Revised: Feb 13, 2013)
Page 44
EE/ME 574: Intermediate Controls
Section 4.1.1
• Take a look at the Bode plot for the loop, with margins
>> margin(Kc*Gcz*Gpz);
EE/ME 574: Intermediate Controls
Section 4.1.1
• Take a look at the frequency response of the compensator
grid on
>> bode(Kc*Gcz);
grid on
Bode Diagram
Gm = 28.1 dB (at 62.8 rad/sec) , Pm = 50.2 deg (at 4.42 rad/sec)
Bode Diagram
20
35
30
Magnitude (dB)
Magnitude (dB)
10
0
−10
25
−20
20
−30
0
15
0
Phase (deg)
Phase (deg)
−45
−90
−135
−30
−180
−225
−2
10
−1
10
0
10
Frequency (rad/sec)
1
10
−60
−1
10
2
10
0
1
10
10
2
10
Frequency (rad/sec)
Figure 28: Bode response of Trez (z).
Figure 29: Frequency response of Gcz (z).
• Questions:
− What type of controller is implemented by Gcz (z)
− What are the approximate frequencies of the pole and zero ?
Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
Page 45
Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
Page 46
EE/ME 574: Intermediate Controls
Section 4.1.1
Section 5.0.0
5 Compensator design by discrete equivalent TF
• Take a look at the error response
>> Trez =
EE/ME 574: Intermediate Controls
• One way to design a compensator for computer-based control is by
emulating a continuous-time controller.
feedback(1, Kc*Gcz*Gpz)
Transfer function:
z^2 - 1.851 z + 0.8561
---------------------z^2 - 1.754 z + 0.8073
1. Design
− Design a continuous-time, Gcs (s), by any suitable method
Sampling time: 0.05
>> bode(Trez); grid on
2. Implementation
• System will meet a tracking error requirement of Etrack ≤ 10% with ωt
below about 0.3 [rad/sec].
Bode Diagram
(a) Select the sample rate Fs [samples/second]. Ts = 1/Fs .
(b) Convert Gcs (s) to a discrete-time compensator, Gcz (z)
3. Analysis
10
5
(a) Done in the z-domain
Magnitude (dB)
0
− To correctly consider sampling effects.
−5
−10
(b) Convert the continuous-time model for the plant, G ps (s) to discrete
time, G pz (z)
−15
−20
−25
Phase (deg)
−30
90
(c) Analyze the complete discrete time system
• Many discrete controllers (compute-based controllers) are designed in this
fashion, because the tools and understanding of continuous-time can be
brought to bear.
45
0
−2
10
−1
10
0
10
Frequency (rad/sec)
1
10
• Synonyms
2
10
− Design by discrete equivalent
Figure 30: Bode response of Trez (z).
− Design by emulation
Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
Page 47
Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
Page 48
EE/ME 574: Intermediate Controls
Section 5.0.0
• Block diagrams for 3 steps, designing a discrete compensator by emulation
1. Design
Controller
Trajectory
Generation
r(t)
+
Controller
e(t)
Plant
u(t)
Section 5.1.0
• Design by discrete equivalent:
1. Design Gcs (s) by standard methods
Process under control
Gc(s)
EE/ME 574: Intermediate Controls
2. Select Ts and convert Gcs (s) to Gcz (z)
y(t)
Gp(s)
-
3. Analyze Kc Gcz (z) G pz (z)
Sensor
− For step 2, we must use a method to find Gcz (z) that approximates Gcs (s)
Hy(s)
——– ∼ ——–
Figure 31: Block diagram for design.
5.1 Matlab’s c2d() command
2. Implementation
Discrete part, Computer Controller
r(k)
Trajectory
Generator
+
e(k)
ys(t)
u(k)
Kc Gc(z)
ZOH
(D/A
Conv.)
-
ys(k)
u(t)
Plant
y(t)
Gp(s)
t=kTs
(A/D Conv.)
Sensor
Hy(s)
Analog part
Figure 32: Block diagram for implementation.
3. Analysis
Discrete Controller
Trajectory
Generator
r(k)
+
e(k)
-
Computer
Controller
Gc(k)
u(k)
ZOH + Plant
y(k)
Gp(z)
ys(k)
ys(k)
Sensor
Hy(z)
Discrete Model of Plant
Figure 33: Block diagram for analysis.
Part 5: Z-Transform and discrete control
• Matlab’s c2d() command as six possible methods converting a continuoustime TF into a discrete-time TF
(Revised: Feb 13, 2013)
Page 49
>> help c2d
C2D Conversion of continuous-time models to discrete time.
SYSD = C2D(SYSC,Ts,METHOD) converts the continuous-time LTI
model SYSC to a discrete-time model SYSD with sample time Ts.
The string METHOD selects the discretization method among the
following:
’zoh’
Zero-order hold on the inputs
’foh’
Linear interpolation of inputs (triangle appx.)
’imp’
Impulse-invariant discretization
’tustin’
Bilinear (Tustin) approximation
’prewarp’
Tustin approximation with frequency prewarping.
The critical frequency Wc (in rad/sec) is specified
as fourth input by
SYSD = C2D(SYSC,Ts,’prewarp’,Wc)
’matched’
Matched pole-zero method (for SISO systems only).
The default is ’zoh’ when METHOD is omitted.
Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
Page 50
EE/ME 574: Intermediate Controls
Section 5.1.1
• ZOH
EE/ME 574: Intermediate Controls
Section 5.1.1
5.1.1 Why 6 different design methods
The “Zero-Order Hold” conversion models a zero-order-hold in the input
path of the continuous time system. This option exactly models G p (s) in
figure 32.
• When the physical system incorporates a zero-order hold, we can find a
G (z) that exactly matches G (s).
− This is the case for converting G ps (s) to G pz (z)
• FOH
Models a first-order-hold, correct for signal processing applications where
a FOH is used.
• imp
• When we matching transfer functions (no ZOH in the physical system) there
is not an exact match
− This is the case for converting Gcs (s) to Gcz (z)
Matches the impulse response of the transfer function. Note that a
continuous-time input is not the same signal as a discrete-time input
Impulse type:
Height
Width
− Tustin’s rule gives a good approximation for controller design.
Continuous-time Infinite Infinitesimal
Discrete-time
− The reason the conversion is approximate lies with the difference
between the continuous and discrete input signal. When this is modelled
exactly (e.g., ZOH) the correspondence is exact.
1
Ts
——– ∼ ——–
Ts
• Tustin’s method
See the next section.
• prewarp
Used for digital filters design, adjusts the conversion to give a more accurate
match in a ’critical’ frequency range.
• matched
Transforms poles and zeros according to z = es Ts .
Part 5: Z-Transform and discrete control
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Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
Page 52
EE/ME 574: Intermediate Controls
Section 5.2.1
5.2 Tustin’s rule to find Gcz (z) that approximates Gcs (s)
EE/ME 574: Intermediate Controls
Section 5.2.1
• Rearranging to derive the transfer function
5.2.1 Deriving Tustin’s rule
Ts
Ts
e (k) + e (k − 1)
2
2
u (k) − u (k − 1) =
• Consider
u (t) =
Z t
e (τ) dτ
or
Gc (s) =
Z k Ts
e (τ) dτ
o
− Then
u (k) = u (k − 1) +
k Ts −Ts
1
s
Ts U (z) 1 − z−1 = E (z)
1 + z−1
2
(39)
(40)
• From which we get the TF
Ts
U (z)
=
Gc (z) =
E (z)
2
• Eqns (39) and (40) are exact, but with knowledge of e (t) only at sample
instants, there is no exact way to integrate e (t) in Eqn (40).
1 + z−1
1 − z−1
(42)
• One approximate method to integrate e (t) is the trapezoidal-rule.
• Comparing Gc (z) with Gc (s), one finds an approximation for s:
e(t)
e(k-1)
e(k)
1 Ts
≃
s
2
t
R
Figure 34: The trapezoidal rule of Tustin’s method approximates e (t).
Z k Ts
k Ts −Ts
e (τ) dτ ≃
←→
1 + z−1
1 − z−1
so
Gc (z) =
s≃
2
Ts
Ts
2
1 + z−1
1 − z−1
1 − z−1
1 + z−1
=
2
Ts
z−1
z+1
(43)
1. Replace s in Gs (s) by
[e (k − 1) + e (k)] Ts
2
2
Ts
Plugging this into Eqn (40) gives:
[e (k − 1) + e (k)] Ts
2
Ts
= u (k − 1) + [e (k − 1) + e (k)]
2
u (k) = u (k − 1) +
Part 5: Z-Transform and discrete control
1
s
• Tustin’s rule, to form Gz (z) from Gs (s):
• The trapezoidal approximation gives:
•
Gc (s) =
(Revised: Feb 13, 2013)
z−1
z+1
2. Simplify Gz (z), and express it in rational polynomial form .
(41)
Page 53
Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
Page 54
EE/ME 574: Intermediate Controls
Section 5.2.2
• Example
4s +5
s+2
Gc (z) =
(phase − lead compensator)
2 z−1
,
Ts z + 1
3.333 − 1.111 z−1
1 − 0.1111 z−1
(44)
• Next, form the difference equation
Using Ts = 0.8 and
s∼
Section 5.2.2
• For implementation, first rearrange into past-sample form:
5.2.2 Example with Tustin’s rule
Gc (s) =
EE/ME 574: Intermediate Controls
Gc (z) =
Gives
4
2
0.8
2
0.8
1.0 u (k) − 0.1111 u (k − 1) = 3.333 e (k) − 1.111 e (k − 1)
z−1
z+1 + 5
z−1
z+1 + 2
(45)
• Which is implemented in the controller as
• With some re-arrangement we get:
1.0 u (k) = +0.1111 u (k − 1) + 3.333 e (k) − 1.111 e (k − 1)
(46)
15 z − 5
10 (z − 1) + 5 (z + 1)
=
Gc (z) =
2.5 (z − 1) + 2 (z + 1) 4.5 z − 0.5
• Putting in monic form
Gc (z) =
3.333 z − 1.111
z − 0.1111
• With Matlab c2d() command
>> Gcs = tf( [4 5], [ 1 2])
>> Ts = 0.8
>> Gcz = c2d(Gcs, Ts, ’tustin’)
Transfer function:
3.333 z - 1.111
--------------z - 0.1111
Sampling time: 0.8
Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
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Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
Page 56
EE/ME 574: Intermediate Controls
Section 5.2.3
EE/ME 574: Intermediate Controls
Section 5.2.3
− Thus, if Ts changes, the discrete transfer functions change.
5.2.3 Example controller design by discrete equivalent, done by hand
• Consider example 6.17 in Franklin, et al., where
G ps (s) =
1
0.5 s + 1
9
(s + 1)
1
2s + 1
,
Gcs (s) = 3
5s +1
15 s + 1
(47)
• Find Gcz (z), plug in the Tustin’s rule approximation for s:
5
2 z−1
,
s∼
Ts z + 1
Gc (z) = 3
h
15
2
Ts
h
2
Ts
1−z−1
1+z−1
i
1−z−1
1+z−1
+1
i
+1
(48)
• Put expression in rational polynomial form:
Gc (z) = 3
5 T2s 1 − z−1 + 1 1 + z−1
(49)
15 T2s (1 − z−1) + 1 (1 + z−1)
−1
+ 1 + 1 − 10
Ts z
= 3
30
30
−1
Ts + 1 + 1 − Ts z
=
c0 = 3
10
Ts
c0 + c1 z−1
c0 /d0 + c1/d0 z−1
=
−1
d0 + d1 z
1 + d1/d0 z−1
(50)
10
10
30
30
+ 1 , c1 = 3 1 −
+ 1 , d1 = 1 −
, d0 =
Ts
Ts
Ts
Ts
where the last express is to put coefficient a0 = 1 (the standard form).
• Notice: that the expressions for terms c0.. d1 depend on Ts
Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
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Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
Page 58
EE/ME 574: Intermediate Controls
Section 5.2.3
5.2.3 Example controller design using emulation (continued)
10
Ts
+1
>> Kc = 3;
>> Gc_617_s = tf([5 1], [15 1])
Transfer function:
5 s + 1
-------15 s + 1
= 63
c1 = 3 1 − 10
= −57
Ts
30
Ts + 1
d0
=
d1
= 1 − 30
= −59
Ts
%% Form the discrete-time TF by Tustin’s rule
>> Fs = 2; Ts = 1/Fs;
>> Gc_617_z = c2d(Gc_617_s, Ts, ’tustin’)
= 61
Transfer function:
• And finally:
Gc (z) =
63/61 − 57/61 z−1 1.033 − 0.9344 z−1
=
1 − 59/61 z−1
1 − 0.9672 z−1
(51)
0.3443 z - 0.3115
---------------z - 0.9672
Sampling time: 0.5
• Examine the step response of the system
• Which gives the implementation equation
u (k) = 0.9672 u (k − 1) + 1.033 e (k) − 0.9344 e (k − 1)
(52)
• Equations (47) - (51) can be a bit of algebra, especially for higher-order
Gc (s), but fortunately there is an easier way to make this calculation.
Part 5: Z-Transform and discrete control
Section 5.2.4
5.2.4 Example controller design by discrete equivalent, using Matlab
• Using Ts = 0.5[seconds], gives
c0 = 3
EE/ME 574: Intermediate Controls
(Revised: Feb 13, 2013)
Page 59
− Write down the continuous-time model of the plant (still from example
Franklin et al. 6.17)
>> Nps = K = 3
>> Dps = [
1.0000
3.5000
>> Guy_617_s = tf(Nps, Dps)
Transfer function:
Gp(s) =
Part 5: Z-Transform and discrete control
3.5000
1.0000 ]
3
------------------------s^3 + 3.5 s^2 + 3.5 s + 1
(Revised: Feb 13, 2013)
Page 60
EE/ME 574: Intermediate Controls
Section 5.2.4
• Convert the plant to discrete
EE/ME 574: Intermediate Controls
Section 5.2.4
• Let’s look at the discrete and continuous system responses
%% Convert plant to discrete, using zero order hold method
>> Fs = 2; Ts = 1/Fs;
>> Guy_617_z = c2d(Guy_617_s, Ts, ’zoh’)
Transfer function:
0.0409 z^2 + 0.1071 z + 0.01706
---------------------------------z^3 - 1.753 z^2 + 0.982 z - 0.1738
>>
>>
[Ycont, Tcont] =
step(Try_617_s, TStepfinal);
hold on; plot(Tcont, Ycont, ’--’); hold off
Step Response
1.4
1.2
• Form the closed-loop TF and plot the step response
>> TStepfinal = 20;
>> Try_617_z = feedback(Kc*Gc_617_z*Guy_617_z, 1)
Transfer function:
0.04224 z^3 + 0.07239 z^2 - 0.08246 z - 0.01594
----------------------------------------------z^4 - 2.678 z^3 + 2.75 z^2 - 1.206 z + 0.1521
>> figure(3)
>>
step(Try_617_z, TStepfinal)
Amplitude
1
0.8
0.6
0.4
0.2
0
0
2
4
6
8
10
Time (sec)
12
14
16
18
20
Figure 36: Step response of discrete model for example 6.17; with continuous
model and control in - - , Ts = 0.5.
Step Response
1.4
1.2
• It is seen that the discrete system is a bit slower and with more overshoot.
Amplitude
1
• The loss of speed and damping is due delay introduced by the ZOH.
0.8
0.6
− The added delay introduces phase lag, it subtracts from the phase margin
0.4
− The added delay is approximately τd = Ts /2,
0.2
− The lost phase margin depends on the delay and the period of one cycle
at the cross-over frequency
0
0
2
4
6
8
10
12
14
16
18
20
Time (sec)
Figure 35: Step response of discrete model for example 6.17, Ts = 0.5 [sec].
Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
Page 61
φlost ≃ 360
Ts /2
0.25
τd
= 360
= 360
= 13o
τcycle
2 π/ωc
6.9
Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
(53)
Page 62
EE/ME 574: Intermediate Controls
Section 5.2.4
EE/ME 574: Intermediate Controls
− In this case, about 13o of phase margin are lost through discrete
implementation.
Section 5.2.4
• The lost phase margin can be reduced by increasing the sample rate.
• Try accelerated sampling the sample with Fs = 8 [samples/second]
Step Response
1.4
• Changed Fs
1.2
Was: Fs = 2 [samples/sec]
1
Amplitude
Now: Fs = 8 [samples/sec]
0.8
0.6
0.4
0.2
0
0
2
4
6
8
10
12
14
16
18
20
Time (sec)
Figure 37: Step response of discrete model for example 6.17; with continuous
model and control in - - , Ts = 0.125 [sec].
• Response is quite a lot better. Check the closed-loop bandwidth, to compare
with the sampling rate:
%% CL bandwidth, in Hz
>> CLWb = (bandwidth(Try_617_z) / (2*pi))
CLWb = 0.2491 [Hz]
• The first try, Fs = 1/Ts = 2 [Hertz] was about 8x ωb, pretty slow.
• Second try, Fs = 1/Ts = 8 [Hertz] was about 32x ωb, and shows much better
performance.
• Performance can also be improved by adding phase lead to Gc (s)
Notes: Pay attention to distinguish [cycles/sec] and [rad/sec].
The limit for the project is Fs ≤ 50 ωb/2π [samples/sec].
Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
Page 63
Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
Page 64
EE/ME 574: Intermediate Controls
Section 6.1.0
6 DC Gain, Steady State Error and Complex Gain
EE/ME 574: Intermediate Controls
Section 6.1.0
• If we have numerator and denominator polynomials, with can use
− Polynomial evaluation, evaluating with z = 1, or
6.1 DC Gain
− The sum of the numerator and denominator
• The DC gain of a discrete system is given as
KDC = T (z)
(54)
z→1
>> num = Gcz.num{1}
num =
1.0328
-0.9344
• For example, the compensator above had the transfer function
Gc (z) =
1.0333z − 0.9344
z − 0.9672
>> den = Gcz.den{1}
den =
1.0000
-0.9672
(55)
>> KDC = polyval(num, 1) / polyval(den, 1)
KDC =
3
So the DC gain of the compensator is:
KDC = Gc (z = +1) =
1.0333 − 0.9344 0.0984
=
=3
1 − 0.9672
0.0328
(56)
>> KDC = sum(num) / sum(den)
KDC =
3
• Recall that for the example, the original Gc (s) was
• The bode() command also works
5s + 1
Gc (s) = 3
15 s + 1
• Notice in Eqn (56) that it is not easy to examine Gc (z) and determine the
DC gain.
>> [Mag, Phase] = bode(Gcz, 0)
Mag
=
3
Phase =
0
%%
bode(TF, omega)
• Notice that bode() is evaluated with omega, not z
Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
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Part 5: Z-Transform and discrete control
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EE/ME 574: Intermediate Controls
Section 6.2.1
6.2 Steady-State error
EE/ME 574: Intermediate Controls
Section 6.2.1
• Build two systems
• The analysis of discrete-time steady-state error is analogous to the
continuous-time case.
1. Type 0 system: with phase-lag control
Gc1 (s) = 3
5s +1
15 s + 1
• This can be done from either the open-loop loop gain, or the closed-loop
Trez (z)
>> Gc_617_PhaseLag_z = c2d(Gc_617_PhaseLag_s, Ts, ’tustin’)
Transfer function:
0.3443 z - 0.3115
----------------z - 0.9672
Sampling time: 0.5
• Methods
1. Apply the final value theorem from theory (see section 2.2.4)
2. Determine the Bode form for the discrete loop-gain transfer function
(open-loop), and determine the system type and SSE from the Bode form.
− Bode form for discrete TF’s:
Kc Gc (z) G p (z) = KBode
z−1
Ts
no
Glz (z)
(57)
where all poles and zeros at z = +1 are placed in the term (z − 1)no ,
and Glz (z) has a DC gain of 1.0.
>> Trez1 = feedback(1, Kc*Gc_617_PhaseLag_z*Guy_617_z)
Transfer function: z^4 - 2.72 z^3 + 2.678 z^2 - 1.124 z + 0.1681
--------------------------------------------z^4 - 2.678 z^3 + 2.75 z^2 - 1.206 z + 0.1521
Sampling time: 0.5
2. Type I system: with PI control
Gc2 (s) = 0.1
5s +1
s
6.2.1 Determining steady-state error from Trez (z) (or Tre (s))
• We can determine steady state error for step or ramp inputs from the bode
plot of Trez (z) (or Tre (s))
• Consider the example plant of section 5.2.4
>> Guy_617_z= c2d(Guy_617_s, Ts, ’zoh’)„
Transfer function:
0.0409 z^2 + 0.1071 z + 0.01706
---------------------------------z^3 - 1.753 z^2 + 0.982 z - 0.1738
Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
Page 67
>> Gc_617_PI_z = c2d(Gc_617_PI_s, Ts, ’tustin’)
Transfer function:
5.25 z - 4.75
------------z - 1
Sampling time: 0.5
>> Trez2 = feedback(1, Kc*Gc_617_PI_z*Guy_617_z)
Transfer function:
z^4 - 2.753 z^3 + 2.735 z^2 - 1.156 z + 0.1738
---------------------------------------------z^4 - 2.732 z^3 + 2.772 z^2 - 1.198 z + 0.1657
Sampling time: 0.5
Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
Page 68
EE/ME 574: Intermediate Controls
Section 6.2.1
EE/ME 574: Intermediate Controls
• Looking at Bode plots of the two transfer functions:
• Type I system (continued)
Bode: Trez1
0
Magnitude (dB)
Magnitude (dB)
20
−10
−20
−30
90
Trez z = e j ω Ts ≃
0
−20
−40
1
jω
KBode
(58)
Where KBode is KBode of the loop gain in Bode form.
−60
− Evaluating Trez (z) at a low frequency ωa, KBode is given by:
90
Phase (deg)
Phase (deg)
− At low frequencies, the Bode plot of figure 38.b is approximated as:
Bode: Trez2
10
Section 6.2.1
45
0
−45
−3
10
−2
−1
10
10
0
10
1
10
45
−45 −3
10
Frequency (rad/sec)
(a) Trez1 (z) (Type 0 system)
KBode ≃
0
−2
−1
10
0
10
10
ωa
|Trez (z = e j ωa Ts )|
(59)
1
10
Frequency (rad/sec)
(b) Trez2 (z) (Type I system)
− The the steady-state error works out to be (for Trez2 (z)):
Figure 38: Bode plots of Trez1 (z) and Trez2 (z).
SSE (ramp) =
1
|Trez (z)|
, with kv = KBode so SSE (ramp) ≃
(60)
kv
ωa
• Type 0 system:
− The limω→0 Terz z = e j ω Ts is evident in figure 1. For the Type 0 system,
the steady-state error for a step input is given as:
>> SSEstep1 = bode(Trez1, 0)
SSEstep1 = 0.1000
− Simply evaluate the DC gain of Trez (z) (or Tre (s)) to determine SSE(step)
• Type I system
− We can recognize from the Bode plot of Trez2 (z) that the system is Type I.
∗ The Trez2 (z) increases 20 dB per decade in the low-frequency region,
and
∗ The angle is +90o
Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
Page 69
− Example:
>> wa = 0.001
>> KBode = wa / bode(Trez2, wa),
>> SSEramp = bode(Trez2, wa)/wa,
KBode = 0.3000
SSEramp = 3.3333
− KBode is not sensitive to changes in ωa , so long as ωa is sufficiently small
>> wa = 0.0001
>> SSE = bode(Trez2, wa)/wa
SSE = 3.3333
>> wa = 0.01
>> SSE = bode(Trez2, wa)/wa
SSE = 3.3333
• The method of Eqns (59)-(60) works equally well with continuous time
systems,
|Tre (s = j ωa )|
(61)
SSE (ramp) ≃
ωa
Part 5: Z-Transform and discrete control
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EE/ME 574: Intermediate Controls
Section 6.3.0
EE/ME 574: Intermediate Controls
Section 7.0.0
7 Conclusion
6.3 Complex gain from a discrete TF
• An important idea that underlies frequency-domain analysis and the Bode
plot is complex gain.
− A sinusoidal input gives a sinusoidal output, with a changed magnitude
and phase
− The complex gain is given as:
• The vast majority of controllers are actually implemented in
microprocessors, as discrete-time systems.
• Discrete-time systems are implemented as difference equations, such as
u (k) = −a1 u (k − 1) − a2 u (k − 2) − · · · + bo e (k) + b1 e (k − 1) + · · ·
~Y
= G (z)
~
U
z=1∠(ω T )
− Outputs are governed by a sum of:
s
∗ Current input and past inputs, and
• Example:
Sys167_cz =
∗ Past outputs
0.3443 z - 0.3115
----------------z - 0.9672
• We have seen the z-transform, which
Sampling time: 0.5
>> Nz = Sys617_cz.num{1}
Nz =
0.3443
-0.3115
− Gives us many of the methods we use with Laplace transforms
>> Dz = Sys617_cz.den{1}
Dz =
1.0000
-0.9672
>> w = 2;
>> z = 1.0*exp(j*w*Ts),
z =
0.5403 + 0.8415i
>> Gain0 = polyval(Nz, z) / polyval(Dz, z)
Gain0 = 0.3340 - 0.0203i
− Has certain differences with the Laplace transform, such as the region on
the z-plane for stable poles.
• Discrete controller design
− Can be done by designing Gcs (s) and converting it to Gcz (z)
%% Determine Gcz at w=2 with bode command, and convert to complex val.
>> [Mag1, Th1] = bode( Sys617_cz, 2 );
>> Gain2 = Mag1*exp(j*(Th1*pi/180))
Gain2 =
0.3340 - 0.0203i
− Can be done directly in the discrete domain
∗ Analyze the Bode plot and root locus of z-transforms
• Controller performance is analyzed entirely in the z-domain.
• The calculations above underlie the bode plot for discrete systems.
Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
Page 71
Part 5: Z-Transform and discrete control
(Revised: Feb 13, 2013)
Page 72
EE/ME 574: Intermediate Controls
EE/ME 574: Intermediate Controls
Section 1.0.0
1 Building a discrete state-space model
Discrete State Space Methods
• The discrete state-space model takes the form:
Contents
x (k + 1) = A p x (k) + B p u (k)
1 Building a discrete state-space model
1.1 Finding B p . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Converting continuous state space models to discrete . . . .
1.2.1 The output equation . . . . . . . . . . . . . . . . .
1.3 Discrete Transfer Function from a discrete state-space model
2 Control of discrete state-space systems
2.1 Pole placement . . . . . . . . . .
2.2 Linear Quadratic Regulator . . . .
2.2.1 Try with new Q . . . . .
2.2.2 Try another new Q . . . .
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3 Integral control for discrete state-space systems (draft)
18
4 Controllability and Observability
19
5 Conclusions, Discrete State-variable model formulation and control
22
(1)
y (k) = C p x (k) + D p u (k)
Notes: we’ll use Fp , G p, H p , J p for the continuous time model, and
A p , B p, C p, D p for the discrete (sampled) model.
• Starting with the continuous model, the discrete model is computed by
solving the differential equation for one sample interval:
− Starting with:
ẋ (t) = Fp x (t) + G p u (t)
(2)
y (t) = H p x (t) + J p u (t)
− And integrating the differential equation, gives:
x (t) = eFp (t−t0 ) x (t0 ) +
Z t
t0
eFp (t−τ) G p u (τ) dτ
(3)
− By letting t0 = k Ts and t = t0 + Ts , Eqn (3) becomes:
x ((k + 1) Ts ) = eFp Ts x (k Ts ) +
Part 6: Discrete State Space Methods
(Revised: May 06, 2014)
Page 1
Part 6: Discrete State Space Methods
Z (k+1)Ts
k Ts
eFp ((k+1)Ts−τ) G p u (τ) dτ
(Revised: May 06, 2014)
(4)
Page 2
EE/ME 574: Intermediate Controls
Section 1.1.0
− Which can be put in the discrete form:
EE/ME 574: Intermediate Controls
1.2 Converting continuous state space models to discrete
• Referring to Franklin et al. example 2.14, the voice-coil motor, with
Fp, G p , H p , J p as given previously, then the discrete model is obtaned by
starting with the continuous time state-space model, and converting it to
discrete:
x (k + 1) = A p x (k) + B p u (k)
where
A p = eFp Ts
(5)
1.1 Finding B p
• To find B p , the integral of Eqn (4) must be solved.
• An assumption must be made about the input signal u (t) over the sample
interval.
>> SSuys = ss(Fp, Gp, Hp, Jp)
a =
x1
x2
x3
x1
-0.15
0
3.15
x2
1
0
0
x3
-31.5
0
-15
b =
x1
x2
x3
• Assuming that the input to the continuous system is a Zero Order Hold
(ZOH),
− Term u (τ) is constant over the integration interval in Eqn (3), and G p and
u (τ) come out of the integral.
− Assuming ZOH, the integral of the input becomes:
Z (k+1)Ts
k Ts
eFp ((k+1)Ts −τ) G p u (τ) dτ =
so
Bp =
Z
0
Ts
Section 1.2.0
Z
0
Ts
eFp τ dτ G p u (k Ts )
eFp τ dτ G p
(6)
(7)
c =
y1
x1
0
x2
1
x3
0
d =
y1
u1
0
Continuous-time model.
%% Build the discrete SS model, set the sample rate to 6/second
>> Fs = 6; Ts = 1/Fs„
Ts = 0.1667
>> SSuyz = c2d(SSuys, Ts, ’zoh’)
a =
x1
x2
x3
x1
0.4096
0.1245
-1.2
x2
0
1
-0
c =
x3
0.12
0.01815
-0.1563
b =
x1
x2
x3
u1
0.9076
0.06405
1.948
y1
u1
0
d =
y1
• The integral of Eqn (7) can be solved numerically, giving B p.
u1
0
0
50
%% First define the
%% continuous-time model
x1
0
x2
1
x3
0
Sampling time: 0.16667
Part 6: Discrete State Space Methods
(Revised: May 06, 2014)
Page 3
Part 6: Discrete State Space Methods
(Revised: May 06, 2014)
Page 4
EE/ME 574: Intermediate Controls
Section 1.2.0
EE/ME 574: Intermediate Controls
Section 1.3.0
1.2.1 The output equation
Discrete-time model.
• Since
y (t) = H p x (t) + J p u (t)
the sampled y (k) is given (simply) by:
y (k) = H p x (k) + J p u (k)
− In other words
· The output equation does not change, C p = H p , D p = J p .
1.3 Discrete Transfer Function from a discrete state-space
model
• Starting with
x (k + 1) = A p x (k) + B p u (k)
(1, repeated)
y (k) = C p x (k) + D p u (k)
We get:
z Z {x (k)} = A p Z {x (k)} + B p Z {u (k)}
(8)
This gives:
(z I − A p) x (z) = B p U (z)
(9)
x (z) = (z I − A p)−1 B p U (z)
(10)
Y (z) = C p x (z) + D p U (z)
(11)
• Which gives the discrete transfer function
Y (z)
= C p (z I − A p)−1 B p + D p
U (z)
(12)
• Eqn (12) has precisely the form of Y (s) /U (s) for continuous systems.
Part 6: Discrete State Space Methods
(Revised: May 06, 2014)
Page 5
Part 6: Discrete State Space Methods
(Revised: May 06, 2014)
Page 6
EE/ME 574: Intermediate Controls
Section 1.3.0
EE/ME 574: Intermediate Controls
• Obtaining the discrete transfer function of the plant
Section 1.3.0
• We can work directly with the discrete state-variable model to investigate
the characteristics of the open-loop system
1. Approach one, discretize the State-variable model, then obtain the TF
>> OLPolesz = pole(SSuyz)
OLPolesz =
1.0000
0.1266 + 0.2530i
0.1266 - 0.2530i
. %% Use SSuyz from above
>> Guyz = tf(SSuyz)
Transfer function:
0.06405 z^2 + 0.1322 z + 0.01766
-----------------------------------z^3 - 1.253 z^2 + 0.3333 z - 0.08006
Sampling time: 0.16667
• Note pole at z = +1.0, the system is Type1
− Note: The Discrete SS model and Discrete TF model have the sample
period, Ts recorded, so the tf(·) command knows Ts .
• Examining the pole-zero map and bode plot of the OL system
2. Approach two, Convert the continuous-time state-variable model to a
continuous-time TF, the discretize that
>> Guys = tf(SSuys)
Transfer function:
>> figure(1), clf
>> bode(SSuyz)
>> figure(2), clf
>> pzmap(SSuyz)
157.5
------------------------s^3 + 15.15 s^2 + 101.5 s
Pole−Zero Map
Bode Diagram
40
Magnitude (dB)
1
0.8
0.6
0.4
Imaginary Axis
20
0
−20
0.2
0
−40
−90
−0.2
−0.4
Phase (deg)
>> Guyz2 = c2d(Guys, Ts, ’zoh’)
Transfer function:
0.06405 z^2 + 0.1322 z + 0.01766
-----------------------------------z^3 - 1.253 z^2 + 0.3333 z - 0.08006
Sampling time: 0.16667
−0.6
−0.8
−1
−2
−1.5
−1
−0.5
0
0.5
1
Real Axis
−180
−270
−360
−450
−1
10
0
1
10
10
2
10
Frequency (rad/sec)
Figure 1: PZ Map and Bode plot for SSuyz.
Part 6: Discrete State Space Methods
(Revised: May 06, 2014)
Page 7
Part 6: Discrete State Space Methods
(Revised: May 06, 2014)
Page 8
EE/ME 574: Intermediate Controls
Section 2.1.0
2 Control of discrete state-space systems
EE/ME 574: Intermediate Controls
Section 2.1.0
3. Design the controller by pole-placement
2.1 Pole placement
• Pole-placement design is carried out exactly as with continuous-time, statevariable model,
− Except that the poles must make sense on the Z-plane
>>
>>
>>
>>
Ap = SSuyz.a;
%% Extract the Ap and Bp matrices
Bp = SSuyz.b;
Cp = SSuyz.c; Dp = SSuyz.d;
Kz = place(Ap, Bp, DesiredPolesz)
Kz = [ -0.0211
0.7572
0.0047 ]
1. Select S-plane closed-loop poles, si
>> DesiredPoless = [-2, -8+j*2,
>> Acl = (Ap - Bp*Kz)
Acl =
0.4287
-0.6872
0.1259
0.9515
-1.1591
-1.4754
-8-j*2]
2. Convert the poles to the z-plane
zi = esi Ts
0.1157
0.0178
-0.1655
(13)
>> DesiredPolesz = exp(DesiredPoless * Ts)
DesiredPolesz = [0.7165, 0.2491 + 0.0862i, 0.2491 - 0.0862i]
%% Determine the DCGain, and Nf. Notice the discrete-time form
>> Nfz = 1 / (Ccl*inv(eye(3)-Acl)*Bp + Dp)
Nfz = 0.7572
• Examining the PZ map
>> Bcl = Nfz * Bp
Bcl =
0.6872
0.0485
1.4754
1.5
1
imaginary
0.5
0
>> Dcl = Dp*Nfz
Dcl =
0
%% Form the closed-loop system
%% Notice Ts on each line, this is what tells
%% the ss() command to build a discrete-time system model
>> SSryz = ss(Acl, Bcl, Ccl, Dcl, Ts);
>> SSrez = ss(Acl, Bcl, -Ccl, (1-Dcl), Ts);
>> SSruz = ss(Acl, Bcl, -Kz, Nfz, Ts )
0
−0.5
−1
−1.5
−1.5
>> Ccl = Cp - Dp*Kz
Ccl =
0
1
−1
−0.5
0
real
0.5
1
1.5
Figure 2: Pole-Zero map showing ’x’ open-loop and ’△’ desired CL poles.
Part 6: Discrete State Space Methods
(Revised: May 06, 2014)
Page 9
Part 6: Discrete State Space Methods
(Revised: May 06, 2014)
Page 10
EE/ME 574: Intermediate Controls
Section 2.1.0
EE/ME 574: Intermediate Controls
• Examine the step response, output and error
>> figure(1), clf
>>
step(SSryz)
>> figure(2), clf
>>
step(SSrez)
Step Response
1
0.9
0.9
0.8
0.8
0.7
0.7
0.6
0.6
Amplitude
Amplitude
Step Response
0.5
0.4
0.3
0.3
0.2
0.2
0.1
0.1
0.5
1
1.5
2
2.5
3
3.5
>> Gcl = Nfs * Gp
Gcl =
0
0
43.1746
0.5
0.4
0
0
• Continuous controller given for comparison
>> Ks = place(Fp, Gp, DesiredPoless)
Ks = [ -0.0121 0.8635
0.0570
>> Fcl = (Fp - Gp*Ks)
Fcl =
-0.1500
0
3.1500
1.0000
0
0
-30.8960 -43.1746 -17.8500
>> Nfs = 1 / (Hcl*inv(-Fcl)*Gp + Jp)
Nfs =
0.8635
>> figure(3), clf
>>
step(SSruz)
1
0.5
1
1.5
Time (sec)
2
2.5
3
3.5
>> Hcl = Hp - Jp*Ks
Hcl =
0
1
0
>> Jcl = Jp*Nfs
jcl =
0
4
Time (sec)
(a) Step response
]
>> SSrys = ss(Fcl, Gcl, Hp, Jp);
0
0
4
Section 2.1.0
%% Plot both continuous and discrete system response
>> [Yz, Tz] = step(SSryz); step(SSrys);
>> hold on, stem(Tz, Yz); hold off
(b) Error response to a step input
Figure 3: Step and error response
Step Response
Step Response
0.8
1
0.7
• The
continuous
and discrete systems agree
quite well at the sample
instants.
0.6
0.8
Amplitude
Amplitude
0.5
0.4
0.3
0.6
0.4
− They do not exactly
agree
0.2
0.2
0.1
0
0
0.5
1
1.5
2
2.5
3
3.5
0
0
4
Time (sec)
1
1.5
2
Time (sec)
(Revised: May 06, 2014)
2.5
3
− The control signals are
not quite the same.
Figure 5: Continuous and discrete system response.
Figure 4: Control effort response to a step input.
Part 6: Discrete State Space Methods
0.5
Page 11
Part 6: Discrete State Space Methods
(Revised: May 06, 2014)
Page 12
EE/ME 574: Intermediate Controls
Section 2.1.0
• Notice that Ks is comparable to Kz
>> Ks = place(Fp, Gp, DesiredPoless)
Ks = [ -0.0121 0.8635
0.0570
• To determine the LQR controller of a discrete system,
− Choose Q and R (identical to continuous-time Q and R)
− Matlab’s lqr() command automatically determines whether the system
is discrete or continuous from the LTI model.
]
>>
>>
>>
>>
>>
>>
Kz → Ks
• Looking at N f
>> Nfz
Nfz =
>> Nfs
Nfs =
Section 2.2.0
2.2 Linear Quadratic Regulator
>> Kz = place(Ap, Bp, DesiredPolesz)
Kz = [ -0.0211
0.7572
0.0047 ]
• In the limit as Ts → 0 ,
EE/ME 574: Intermediate Controls
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% LQR Example
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Q = diag([1 1 1]);
Ks = lqr(SSuys, Q,
Ks = [ 0.7711
>> Fcl = Fp-Gp*Ks;
0.7572
R = 1;
R)
%% Continuous lqr
1.0000
0.7896 ]
Hcl = Hp - Jp*Ks
0.8635
>> Kz =lqr(SSuyz, Q, R)
%% Discrete lqr
Kz = [ -0.1314
0.3369
0.0083 ]
>> Acl = Ap-Bp*Kz;
Ccl = Cp - Dp*Kz
− Again comparable, but not identical.
• Notice the difference in calculation to determine N f :
− Continuous
N f = 1/ Hcl (sI − Fcl )−1 G p + J p
s=0
= 1/ Hcl (−Fcl )−1 G p + J p
%% Set up correct feed-forward gain, continuous
%% Note (0*I - Fcl)
>> Nfs = 1/(Hcl*inv(-Fcl)*Gp+Jp)„
%% Continuous: DC = lim s -> 0
Nfs = 1.0000
>> Gcl = Nfs * Gp;
= 1/ Dcl (I − Fcl )−1 B p + D p
%%Set up correct feed-forward gain, discrete
>> Nfz = 1/(Ccl*inv(eye(3)-Acl)*Bp+Dp)„
%% Note (Eye - Acl)
Nfz = 0.3369
%% Discrete: DC = lim z -> 1
>> Bcl = Nfz * Bp;
− Discrete
N f = 1/ Dcl (zI − Acl )−1 B p + D p
Part 6: Discrete State Space Methods
z=+1
(Revised: May 06, 2014)
Page 13
Part 6: Discrete State Space Methods
(Revised: May 06, 2014)
Page 14
EE/ME 574: Intermediate Controls
Section 2.2.0
• Determine closed-loop poles and build system models
EE/ME 574: Intermediate Controls
Section 2.2.1
2.2.1 Try with new Q
>> CLPoless = eig(Fcl)
CLPoless = -50.1257
-0.8634
-3.6392
%% Continuous
>> CLPolesz = eig(Acl)
CLPolesz =
0.8673
0.4440
0.0236
%% Discrete

0 0 0


Q2 = 100CTp C p =  0 100 0


0 0 0




,


R = [1]
>> K2z = lqr(SSuyz, Q2, R)
>>
...
>> figure(4), clf
>>
step(SSryz, 7)
>> SSrys = ss(Fcl, Gcl, Hp-Jp*Ks, Jp*Nfs);
>> SSryz = ss(Acl, Bcl, Cp-Dp*Kz, Dp*Nfz, Ts)
%% Note Ts
Discrete LQR, with continuous ’−−’
1.4
Discrete LQR, with continuous ’−−’
1
1.2
0.9
1
0.8
Amplitude
0.6
Q = [ 1.00
[ 0.00
[ 0.00
0.5
0.00
1.00
0.00
Amplitude
0.7
PO: 0%,
0.00]
0.00]
1.00]
T : 2.50,
0.8
Q=[
[
[
PO: 5.54%,
0.00 0.00 0.00]
0.00 100.00 0.00]
0.00 0.00 0.00]
T : 0.17,
r
0.6
r
T : 0.67,
s
T : 5.00,
0.4
s
0.4
ISE: 0.1677
ISE: 0.8383
0.3
0.2
0.2
0.1
0
0
1
2
3
4
5
6
7
Time (sec)
0
0
1
2
3
4
5
6
7
Figure 7: LQR design, response with a large cost on output error.
Time (sec)
Figure 6: LQR response for the sampled system,Ts = 1/6 [seconds.
Dashed line: continuous-time system response.
Part 6: Discrete State Space Methods
(Revised: May 06, 2014)
Page 15
Part 6: Discrete State Space Methods
(Revised: May 06, 2014)
Page 16
EE/ME 574: Intermediate Controls
Section 2.2.2
EE/ME 574: Intermediate Controls
Section 3.0.0
3 Integral control for discrete state-space systems
2.2.2 Try another new Q
(draft)


1 0 0




Q =  0 100 0  ,




0 0 0
• To use a controller design with internal state(s), we augment the state vector
with the controller state(s), using Eqn (??).
R = [1]
• Specifically, to put in integral control introduce
>> K3z = lqr(SSuyz, Q3, R)
>>
...
ie (k) = − ∑ e (k) = ie (k − 1)−e (k − 1)
>> figure(4), clf
>>
step(SSryz, 7)
by choosing
Fcp = C p − D p K p ,
Discrete LQR, with continuous ’−−’
which gives ie (k + 1) = ie (k)+y (k)−r (k)
Fcc = −D p Kc ,
Gcu = 0 ,


Gcr = −1 + Jcl
1
which leads to:
0.9
0.8
Q=[
[
[
Amplitude
0.7
PO: −0.00%,
1.00 0.00
0.00]
0.00 100.00 0.00]
0.00 0.00 0.00]
Tr: 0.33,

0.6
T : 0.50,
0.5
ISE: 0.1785
s



Fa = 


0.4
0.3
Ap
..
.
···
···
···
Cp − D p Kp
..
.
1 − D p Kc
0



,



B
p






Gau =  · · ·  ,




0
0.2
where the controller gain vector is partitioned according to Ka =
0.1
0
.
0
1
2
3
4
5
6




Gar = 


N f Bp
···
−1 + Dcl
K p : Kc
7
Time (sec)
Figure 8: LQR design, response with a modest additional cost on rate.
• Control can then be designed by:
1. Determine the Ka vector by
Part 6: Discrete State Space Methods
(Revised: May 06, 2014)
Page 17
Part 6: Discrete State Space Methods
(Revised: May 06, 2014)
Page 18







EE/ME 574: Intermediate Controls
Section 4.0.0
EE/ME 574: Intermediate Controls
Section 4.0.0
− Controllability
(a) DLQR
(b) Pole placement
2. Select N f for good transient response (as given in example ??, or by trial
and error, or other method)
• Integral control can significantly improve performance, especially accuracy.
>> CCntl = [Bp, Ap*Bp, Ap*Ap*Bp]
CCntl =
0.9076
0.6056
0.0807
0.0640
0.2124
0.2626
1.9484
-1.3938
-0.5090
>> rank(CCntl)
ans =
3
4 Controllability and Observability
− The system is controllable
• The concepts of controllability and observability carry over to discrete
systems.
− The mathematical foundations are somewhat different
− The equations for controllability and observability are the same.
C = B A B A2 B · · · A(n−1) B







O =






C
CA
C A2
..
.
C A(n−2)
(14)














(15)
• Considering the voice coil motor:
Part 6: Discrete State Space Methods
(Revised: May 06, 2014)
Page 19
Part 6: Discrete State Space Methods
(Revised: May 06, 2014)
Page 20
EE/ME 574: Intermediate Controls
Section 4.0.0
− Observability:
EE/ME 574: Intermediate Controls
5 Conclusions,
Discrete
Section 5.0.0
State-variable
model
formulation and control
>> OObsvr = [Cp; Cp*Ap; Cp*Ap*Ap]
OObsvr =
0
1.0000
0
0.1245
1.0000
0.0182
0.1538
1.0000
0.0303
>> rank(OObsvr)
ans =
3
• Discrete state-variable models can be obtained by
− Converting a continuous-time state-variable model
− Converting a discrete transfer-function model, with tf2ss()
− Some systems are naturally discrete (e.g., an inventory control system,
or message queues in the controls of a communication network)
− The system is observable
• Poles are the eigenvalues of the system matrix, A p in the notation used here
− Poles must be interpreted on the z-plane, or
− Converted to equivalent s-plane poles and interpreted.
• Controller design can be done in several ways
− Pole placement, by determining the desired z-plane poles
− By the Linear-Quadratic-Regulator algorithm
• Matlab’s tools make it straight forward to generate various models, to
address
− Output response
− Error response
− Control response
− Disturbance response
Part 6: Discrete State Space Methods
(Revised: May 06, 2014)
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Part 6: Discrete State Space Methods
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EE/ME 574: Intermediate Controls
EE/ME 574: Intermediate Controls
Section 1.0.0
1 Introduction
Direct Digital Design
or
The Discrete Root Locus
• As we’ve seen, the digital equivalent of an S-plane controller design can
break down if the sample rate is not 20 - 50 times faster than the closedloop bandwidth.
– Sampling with 20 FBW ≤ Fs ≤ 50 FBW is sometimes called ’rapid
sampling’
Contents
1 Introduction
1.1 Illustration of the effects of rapid sampling . . . . . . . . . . . . .
1.1.1 High noise sensitivity . . . . . . . . . . . . . . . . . . .
1.1.2 High sensitivity to parameter variation . . . . . . . . . .
2 Direct Digital Control Design (section 8.6 in
edition))
2.1 Direct digital design, example 1 . . . . . .
2.1.1 Example: apply PI compensation .
2.2 A more complex example . . . . . . . . . .
Franklin et al.
• Advantages of rapid sampling:
– Minimal phase lag introduced by the Zero-Order-Hold
– More accurately implement the discrete equivalent of Gc (s)
• Rapid sampling, however, has some disadvantages. Disadvantages of rapid
sampling:
(5th
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
3 Conclusions
2
3
3
4
10
12
16
21
27
– Sensitivity to sensor and round-off noise can increase with the sample
rate (especially for PD, PID control)
∗ Rapid sampling gives high sensitivity to small variations in gain.
∗ Variations in gain arise through component variations, and also
through round-off error
• Direct digital design can permit slow sampling, 5 FBW ≤ Fs < 20 FBW which
avoids the disadvantages of rapid sampling
– In direct digital design all design work is done in the Z-domain
– Phase lag introduced by the Zero-Order-Hold is explicitly considered,
and compensated by phase lead
– There is no Gc (s)
– Sensitivity to sensor noise and round-off error can be lower with slower
sampling
– Sensitivity to coefficient variations is lower with slower sampling
Part 7: Direct Digital Design
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Part 7: Direct Digital Design
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EE/ME 574: Intermediate Controls
Section 1.1.1
1.1 Illustration of the effects of rapid sampling
EE/ME 574: Intermediate Controls
1.1.2 High sensitivity to parameter variation
1.1.1 High noise sensitivity
• Consider the plant given by:
• Consider the PD compensator:
Gc (s) = kd s + k p
with kd = k p = 1
Section 1.1.2
G p (s) =
(1)
1
,
s2 + s + 0.1
Hy (s) =
1
,
0.02 s + 1
Gc (s) = 6
3s +1 3s +1
0.48 s + 1 30 s + 1
where the compensator is a Lead-Lag compensator chosen for a phase
margin of PM ≥ 600, with steady-state error, SSE (step) ≤ 10%.
– Gc (z), with Fs = 10 [samples/second]
>> Ts=1/Fs
Ts =
0.1000
>> GPDz = c2d(GPDs, Ts, ’tustin’)
Transfer function:
21 z - 19
Gc(z) = --------z + 1,
• The Bode plot and step response are seen in figure 1.
Fbw =0.35 [Hz]
Bode Diagram
Gm = 30.7 dB (at 11.1 rad/sec) , Pm = 67.6 deg (at 1.34 rad/sec)
Step Response
50
1.4
Magnitude (dB)
0
The implementation equation is
1.2
−50
1
u (k) = −u (k − 1) + 21 e (k) − 19 e (k − 1)
Amplitude
−100
−150
0
0.8
0.6
>> Ts=1/Fs
Ts =
0.0100
>> GPDz = c2d(GPDs, Ts, ’tustin’)
Transfer function:
Gc(z) =
Phase (deg)
– Gc (z), with Fs = 100 [samples/second]
−90
0.4
−180
0.2
−270
−3
10
−2
10
−1
10
0
10
Frequency (rad/sec)
1
10
2
10
3
10
0
0
2
4
6
8
10
Time (sec)
12
14
Figure 1: Bode plot and step response for example system.
201 z - 199
-----------
The implementation equation is
• The closed-loop bandwidth is 2.2 [rad/second], so we can choose the
sample rate to be as high as 17 samples/second, and stay within the range
Fs ≤ 50 × FBW
u (k) = −u (k − 1) + 201 e (k) − 199 e (k − 1)
• Changing Fs from 10 to 100, the numerator terms (coefficients multiplying
e (k)and e (k − 1)) have become 10x larger.
• Choosing Fs = 17 [samples/second] gives
• If there is noise on the e (k) signal, it will have a 10x larger impact on u (k) !
>> Gcz
Part 7: Direct Digital Design
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16
18
EE/ME 574: Intermediate Controls
Section 1.1.2
• Looking at the closed-loop pole constellation, figure 2(a), the closed-loop
poles seem reasonable.
3.6 z^2 - 7.059 z + 3.461
------------------------z^2 - 1.883 z + 0.8828
• Figure 2(b) is more worrisome. It shows the constellation of poles given by
a random variation with σ =0.01 in the compensator parameters.
imaginary
Transfer function:
Section 1.1.2
1.5
1.5
1
1
0.5
0.5
imaginary
EE/ME 574: Intermediate Controls
0
0
−0.5
−0.5
−1
−1
−1.5
−1.5
−1
−0.5
0
real
0.5
1
1.5
−1.5
−1.5
−1
−0.5
0
real
0.5
1
1.5
Cluster of poles given with
(a) Pole constellation, ideal parameters
(b)
σ=0.01 parameter variation
Figure 2: Closed-loop pole constellation for the digitized system, design by
discrete equivalent. Fs = 17, or 50x Fbw .
• Code to draw figure 2(b)
sigma = 0.01;
clear Poles;
for jj = 1:100,
Num = Gcz.num{1};
Den = Gcz.den{1};
Num = Num + sigma*randn(size(Num));
%% Add 0.01 variation
Den = Den + sigma*randn(size(Den)); Den(1) = 1;
Try_z = feedback(tf(Num, Den, Ts)*Gpz, 1);
Poles(:,jj) = pole(Try_z);
end
Part 7: Direct Digital Design
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EE/ME 574: Intermediate Controls
Section 1.1.2
EE/ME 574: Intermediate Controls
1.1.2 High sensitivity to parameter variation (continued)
Section 1.1.2
(Although controller parameters have to be at least 16 bit, and preferably
32 or 64 bit).
• Figure 3 shows the range pole locations corresponding to Fs = 1.4 [Samples
per second], (Fs ≃ 4 × Fbw ) , and coefficient variations of 0.01 standard
deviation.
1.5
1.5
1
1
0.5
0.5
imaginary
imaginary
• While this system has variation in the closed-loop poles, it does not go
unstable for any value with this range of variation.
0
0
−0.5
−0.5
−1
−1
−1.5
−1.5
−1
−0.5
0
real
0.5
1
1.5
−1.5
−1.5
−1
−0.5
0
real
0.5
1
1.5
Cluster of poles given with
σ=0.01 parameter variation
Figure 3: Closed-loop pole constellation for the digitized system, Fs = 1.4
[samples/second], or 4x Fbw , with variations with σ =0.01.
(a) Pole constellation, ideal parameters
(b)
• Variation in the effective value of controller parameters can come about
through round-off error in the digital control calculations.
– This is especially true because controllers are often implemented with 16
or even 8-bit microprocessors
Part 7: Direct Digital Design
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Part 7: Direct Digital Design
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EE/ME 574: Intermediate Controls
Section 1.1.2
EE/ME 574: Intermediate Controls
Section 2.0.0
2 Direct Digital Control Design (section 8.6 in
1.1.2 High sensitivity to parameter variation (concluded)
• Random variations with σ = 0.001 and Fs = 100 [samples/second] are seen
in figure 4.
Franklin et al. (5th edition))
• The design process bears a strong relationship to continuous-time design:
– Many of the cases give poles outside the unit circle
– Analysis tools:
• Round-off errors of ±0.001 are hard to avoid
(especially with integer arithmetic)
* Root locus
• Fs = 100 will work very nicely in theory, but practically speaking, it is
sampling too fast.
0.2
* Bode plot
* Step response
– Compensators:
* Phase - lead
* PD
* Phase - lag
* PI
* Lead - Lag
* PID
0.15
• For continuous design, learning to interpret pole-locations on the S-plane is
essential.
imaginary
0.1
0.05
• For discrete design it is essential to learn to interpret pole locations on the
Z-plane.
0
−0.05
−0.1
−0.15
0.7
0.75
0.8
0.85
0.9
real
0.95
1
1.05
Figure 4: Closed-loop pole constellation for the digitized system, Fs = 100
[samples/second], or 290x Fbw , with variations with σ =0.001.
Part 7: Direct Digital Design
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EE/ME 574: Intermediate Controls
Section 2.0.0
EE/ME 574: Intermediate Controls
Section 2.1.0
2.1 Direct digital design, example 1
• Direct digital design proceeds by these three steps
1. Select sample rate, convert the models of continuous-time system
elements to discrete-time (Z-transform) models.
• A typical control loop is seen in figure 5.
Discrete part, Computer Controller
r(k)
Trajectory
Generator
2. Design the control, using, for example
+
e(k)
ys(t)
– Root locus
ys(k)
Kc Gc(z)
u(k)
ZOH
(D/A
Conv.)
-
u(t)
Plant
y(t)
Gp(s)
t=kTs
(A/D Conv.)
– Frequency response / Bode plot
Sensor
Hy(s)
– Discrete Final Value Theorem
Analog part
3. Analyze the resulting discrete time system in discrete-time
Figure 5: Basic sampled control loop.
4. Iteratively adjust the design parameters to achieve performance goals.
• Consider the example system
G p1s (s) =
• Direct digital design has the advantages:
– It is exact, the sampled model obtained in step 1 is exact.
1
s+1
,
Hy1 (s) = 1
(2)
• Performance Requirements:
– Generally speaking, lower sample rates are possible
– ωb ≥ 1 [rad/sec]
• Something that some people may consider a disadvantage is that pole-zero
maps in the Z-plane must be considered; requiring a developed sense of how
to interpret the Z-plane.
– PO ≤ 20%
– SSE(step) ≤ 10%
• The performance requirements translate to:
– ωc ≥ 0.5 [rad/sec]
– PM ≥ 50o
Part 7: Direct Digital Design
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Part 7: Direct Digital Design
(Revised: May 09, 2013)
Page 12
Section 2.1.0
• As a first selection, choose Ts = 0.25 [seconds] (we can revise Ts later)
EE/ME 574: Intermediate Controls
Section 2.1.0
• Looking at CL poles and response with proportional feedback Kc
Root Locus
>> jj=0;
>> for Kc = [0.5, 2, 5, 9],
>> jj=jj+1;
>> Try =feedback(Kc*Gp1z, 1)
>> Poles(:,jj) =pole(Try);
>> end
>> Gp1s = tf(1, [1 1]) ;
Ts = 0.25
>> Gp1z = c2d(Gp1s, Ts, ’zoh’)
Transfer function:
0.2212
---------z - 0.7788
1
0.8
0.6
0.4
Imaginary Axis
EE/ME 574: Intermediate Controls
0.2
9.0
5.0
2.0
0.5
0
−0.2
−0.4
−0.6
−0.8
• The root locus indicates pole-location for proportional control
−1
−1.5
−1
−0.5
0
0.5
1
Real Axis
Root Locus
>> rlocus(Gp1z)
Figure 7: Root locus for plant Eqn (2).
1
1
1
0.8
Kc = 2.00
Kc = 0.50
0.9
0.9
0.6
0.8
0.8
0.7
0.7
0.2
0.6
0.6
0
y(k)
Imaginary Axis
0.4
−0.2
0.5
0.5
0.4
−0.4
0.4
−0.6
0.3
0.3
−0.8
0.2
0.2
−1
0.1
0
−1.5
−1
−0.5
0
0.5
1
0.1
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
0
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
4
4.5
5
Real Axis
40
Figure 6: Root locus for plant Eqn (2).
Kc = 9.00
Kc = 5.00
30
1.2
20
1
10
0
y(k)
0.8
−10
0.6
−20
0.4
−30
0.2
0
−40
0
0.5
1
1.5
2
2.5
k [sample]
3
3.5
4
4.5
5
−50
0
0.5
1
1.5
2
2.5
k [sample]
3
3.5
Figure 8: Step responses for four values of K.
Part 7: Direct Digital Design
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Part 7: Direct Digital Design
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EE/ME 574: Intermediate Controls
Section 2.1.0
• Kc = 2 is the largest tested value for smooth proportional control.
EE/ME 574: Intermediate Controls
Section 2.1.1
2.1.1 Example: apply PI compensation
• Consider the steady state error with Kc = 2.0. The discrete FVT:
z−1
Trez (z) R (z)
lim e (k) = lim 1 − z−1 E (z) = lim
z→1
z→1
k→∞
z
(3)
• Since the speed and damping are adequate, but SSE is too large, consider PI
control. Discrete PI control has the structure:
Kc Gc1z (z) = Kc
Note that
1 − z−1
=
z−1
,
z
and
E (z) = Trez (z) R (z)
z + b1
z−1
(4)
• Initial choice for b1, b1 = −0.2 (placing the zero on the real axis)
with
R (z) =
z
z−1
Ts z
>> Gc1z= tf([1 -0.2], [1 -1], Ts)
1
step input
0.8
Transfer function:
z - 0.2
Gc1z(a) =
------z - 1
0.6
ramp input
2
(z − 1)
0.4
Imaginary
R (z) =
• Forming the TF Trez (z)
0.2
0
−0.2
−0.4
−0.6
−0.8
>> Trez = feedback(1, K*Gp1z)
Transfer function:
z 0.7788
------------z - 0.3364
−1
−1.5
−1
−0.5
0
0.5
1
Real
Figure 9: PZ map with PI compensator. ⋄: compensator pole, o: comp. zero.
• Forming the right-hand side of Eqn (3)
SSE(step) = lim
z→1
z−1
z
z − 0.7788
z − 0.3364
z
z−1
• Cancel the zero and pole at z = +1
SSE(step) = lim
z→1
z − 0.7788 0.2212
=
= 33%
z − 0.3364 0.6636
• SSE(step) is much too large.
Part 7: Direct Digital Design
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EE/ME 574: Intermediate Controls
Section 2.1.1
• Consider the Bode plot and root-locus for Gc1z (z) G p1z (z) ,
(initially set Kc = 1.0)
EE/ME 574: Intermediate Controls
Section 2.1.1
• Consider the step response:
>> figure(3), clf
>> step(Tryz)
>> Kc = 1.0;
>> figure(1), clf
>>
margin(Kc*Gc1z*Gp1z)
Step Response
>> Tryz = feedback(Kc*Gc1z*Gp1z, 1)
>> Axis1 = [ -1.5 1.2 1.2*[-1 1]]; tt = 0:0.01:(2*pi);
>> figure(3), clf
>>
rlocus(Kc*Gc1z*Gp1z)
>>
axis(Axis1); set(gca, ’DataAspectRatio’, [1 1 1])
>>
hold on
>>
plot(cos(tt), sin(tt), ’k:’, ’linewidth’, 2);
%% plot unit circle
>>
plot(real(pole(Tryz)), imag(pole(Tryz)), ’rd’) %% plot closed-loop poles
>>
hold off
1.4
1.2
Amplitude
1
0.8
0.6
0.4
0.2
0
0
Bode Diagram
Gm = 22.5 dB (at 12.6 rad/sec) , Pm = 35.8 deg (at 1.7 rad/sec)
2
4
6
8
10
12
Time (sec)
Root Locus
Figure 11: Step response with K = 1 and PZ map of figure 10.
40
1
20
0.8
0
0.6
0.4
−20
Imaginary Axis
Magnitude (dB)
60
Phase (deg)
−40
−90
• Consider the bandwidth
0.2
0
−0.2
>> Wb = bandwidth(Tryz)
Wb = 2.7175
−0.4
−135
−0.6
−0.8
−1
−180
−2
10
−1
10
0
10
1
10
2
10
−1.5
Frequency (rad/sec)
−1
−0.5
0
0.5
1
Real Axis
• Consider the steady-state error
Figure 10: Margins plot and root locus for Gc1 (z) and CL poles with K = 1.0 .
>>
• The Bode plot shows more than adequate ωc
• The phase margin is insufficient, but can be increased by reducing Kc .
SSEStep = bode(Trez, 0),
SSEStep =
8.8818e-16
– SSEstep is effectively 0, corresponding to the Type I system.
• Since the control is PI, SSE(step) will be 0.
Part 7: Direct Digital Design
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Section 2.1.1
EE/ME 574: Intermediate Controls
• Considering SSE for the ramp input (since the system is Type I)
• Returning to the root locus, we know that if we move the zero closer to the
poles, it will bring the root-locus branches closer to the real axis.
• Method 1: Final value theorem
lim e (k) = ess = lim 1 − z−1 Tre (z) R (z)
– Moving the compensator zero to z = 0.5
(5)
z→1
(z − 1) z2 − 1.779 z − 0.7788 Ts z
z→1
1 z2 − 1.558 z + 0.7346 (z − 1)2
(z − 1) (z − 1) (z − 0.7788) Ts z
= lim
z→1
1 z2 − 1.558 z + 0.7346 (z − 1)2
(1 − 0.7788) 1
= Ts
= 0.3125
1 − 1.558 + 0.7346
= lim
(6)
Bode Diagram
Gm = 11.1 dB (at 12.6 rad/sec) , Pm = 46.1 deg (at 2.9 rad/sec)
60
(7)
(8)
Magnitude (dB)
k→∞
>> SSE_ramp = bode( tf([1 -1], 1, Ts) ...
* Trez
...
* tf(Ts*[1 0], [1 -2 1], Ts), 0)
SSE_ramp =
0.3125
Phase (deg)
– Setup the Z-transform of Eqn (6) directly for Matlab’s bode command:
0
−0.2
−0.6
−0.8
−1
0
10
1
10
2
10
10
−1.5
−1
−0.5
0
0.5
1
Real Axis
Figure 12: Margins plot and root locus for Gc2 (z) and CL Poles for Kc = 1.0 .
>> Gc2z= tf([1 -0.5], [1 -1], Ts)
Transfer function:
z - 0.5
------z - 1
SSEStep =
0
Wb = 2.04 [rad/sec]
PO = 15.6%
Step Response
1.4
1
Amplitude
0.8
0.6
• 1st PI controller:
0.2
−0.4
−135
1.2
term
0.4
0.2
– ωb OK,
• Design Complete !
0
0
– SSE(step) OK,
1
2
3
4
5
6
7
8
9
Time (sec)
Figure 13: Step response with Kc = 1 and Gc2z (z).
– PO too large.
Part 7: Direct Digital Design
0.4
−1
term
Ts z
(z−1)2
0.6
0
Frequency (rad/sec)
2. The Tre (z) term
3. The
1
0.8
20
−180
−2
10
Note that the transfer function given the bode() command has three
parts, corresponding to the three terms in Eqn (6)
(z−1)
z
Root Locus
40
−20
−90
• Method 2: use matlab bode command:
1. The
Section 2.1.1
Imaginary Axis
EE/ME 574: Intermediate Controls
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Part 7: Direct Digital Design
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EE/ME 574: Intermediate Controls
Section 2.2.0
EE/ME 574: Intermediate Controls
2.2 A more complex example
• Phase-lead compensation, first engineered in continuous-time
>> Kcs = 2.0; tau = 0.5; alpha = 0.2;
>> Gcs = tf([tau 1], [tau*alpha 1])
Transfer function:
0.5 s + 1
Gcs(s) =
--------0.1 s + 1
• Consider the Boeing 777 pitch control example from CTM:
1.151 s + 0.1774
------------------------s^3 + 0.739 s^2 + 0.921 s
Bode Diagram
20
15
50
10
0
−50
0
−100
0
−50
−100
0
−45
0
−45
−90
−10
−90
−15
−135
−180
−3
10
−2
10
−1
0
10
10
1
10
2
10
S−plane
−20
−20
3
10
−10
0
10
Real Axis
Frequency (rad/sec)
−135
−180
−2
10
5
−5
Phase (deg)
Magnitude (dB)
50
Root Locus
Bode Diagram
Gm = Inf dB (at Inf rad/sec) , Pm = 55.4 deg (at 1.93 rad/sec)
100
Magnitude (dB)
Guy(s) =
Imaginary Axis
>> num=[1.151 0.1774];
>> den=[1 0.739 0.921 0];
>> Guy = tf(num,den)
Transfer function:
Phase (deg)
Section 2.2.0
Figure 15: Margins plot and root locus for Gc (s) G p (s).
−1
10
0
10
1
10
2
10
Frequency (rad/sec)
Step Response
Figure 14: Bode plot for Boeing 777 pitch control.
1
0.9
0.8
• Design goals:
– Put cross-over above the resonance, increase accuracy and avoid a very
low ωb
Amplitude
0.7
0.6
0.5
0.4
0.3
• Plan:
0.2
0.1
– Apply phase-lead compensation to boost phase margin around ωc = 5.0
[rad/sec]
Part 7: Direct Digital Design
(Revised: May 09, 2013)
Page 21
0
0
10
20
30
40
50
Time (sec)
Figure 16: Continuous system step response with Kc = 10 and PZ map of Fig. 15.
Part 7: Direct Digital Design
(Revised: May 09, 2013)
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EE/ME 574: Intermediate Controls
Section 2.2.0
EE/ME 574: Intermediate Controls
• First iteration: design by discrete equivalent
Section 2.2.0
• Look at the root locus.
• Notice many poles and zeros in a small region near z = +1.0 .
– Start with rapid sampling, FBW = 0.43 , choose Fs = 20 or 47x FBW
>> Fs = 20; Ts = 1/Fs;
>> Gcz = c2d(Gcs, Ts, ’tustin’)
Transfer function:
Gcz(z) =
• This will lead to the condition where small variations in parameters make
large variations in closed-loop pole locations.
4.2 z - 3.8
----------z - 0.6
• Several poles and zeros crowded into the region near z = +1 indicates that
the sample rate is too high.
Sampling time: 0.05
Root Locus
• Taking a look at the margins and step response of the discrete system
0.8
Root Locus
0.6
2
• The response of the discrete implementation looks very
0.4
Imaginary Axis
1.5
1
Imaginary Axis
comparable to the continuous system response.
0.5
0
Step Response
−0.4
−1
−0.6
1.4
100
0
−0.2
−0.5
Bode Diagram
Gm = 29.4 dB (at 17.1 rad/sec) , Pm = 52.7 deg (at 1.93 rad/sec)
0.2
−3
−2.5
−2
−1.5
−1
−0.5
0
0.5
1
0
0.5
1
Real Axis
1.5
Real Axis
1
Figure 18: Root locus corresponding to figure 17.
−100
0
Phase (deg)
−2
−3.5
−50
0.8
0.6
−90
0.4
−180
−270
−3
10
−0.8
Z−plane
1.2
0
Amplitude
Magnitude (dB)
−1.5
50
0.2
−2
10
−1
10
0
10
Frequency (rad/sec)
1
10
2
10
0
0
5
10
15
20
25
30
35
40
45
Time (sec)
Figure 17: Margins plot and step response for Gcz (z) G pz (z).
Part 7: Direct Digital Design
(Revised: May 09, 2013)
Page 23
Part 7: Direct Digital Design
(Revised: May 09, 2013)
Page 24
EE/ME 574: Intermediate Controls
Section 2.2.0
• Second design iteration. Reduce the sample rate, Fs = 2.4 [Hz], or 5.6xFBW .
>> Gcz = c2d(Gcs, Ts, ’tustin’)
Transfer function:
2.297 z - 0.9459
---------------z + 0.3514
Sampling time: 0.41667
EE/ME 574: Intermediate Controls
Section 2.2.0
• Examine loop-gain poles, to determine system Type
>> OLPoles = pole( Kcz*Gcz*Guyz)
OLPoles =
1.0000
0.7996 + 0.3093i
0.7996 - 0.3093i
-0.3514
• From Bode plot, choose Kc = 1.5 (ωc above resonance, reasonable PM)
Bode Diagram
Gm = 13 dB (at 4.19 rad/sec) , Pm = 39.6 deg (at 1.66 rad/sec)
1.2
20
>> Trez = feedback(1, Kz*Gcz*Guyz)
Transfer function:
z^4 - 2.248 z^3 + 1.421 z^2 + 0.08513 z - 0.2582
-----------------------------------------------z^4 - 1.934 z^3 + 1.286 z^2 - 0.1843 z - 0.1464
1
0
−20
Amplitude
Magnitude (dB)
40
−40
0
Phase (deg)
• Steady-state error
Step Response
1.4
60
0.8
0.6
−90
0.4
−180
0.2
−270
−3
10
−2
−1
10
0
10
0
0
1
10
10
10
20
30
Frequency (rad/sec)
40
50
60
70
Time (sec)
• One OL Pole at z = +1, system is Type I
Figure 19: Margins plot and step response for Gc (z) G p (z).
Root Locus
• Check SSE(step)
0.8
Root Locus
2
0.6
1.5
>> SSEStep = bode(Trez, 0)
SSEStep = 3.8733e-14
(0 to within round-off error)
0.4
Imaginary Axis
Imaginary Axis
1
0.5
0
0.2
0
−0.2
• Check SSE(ramp)
−0.5
−0.4
−1
−0.6
−1.5
−0.8
Z−plane
−2
−3.5
−3
−2.5
−2
−1.5
−1
−0.5
0
0.5
1
1.5
0
0.5
1
Real Axis
Real Axis
Figure 20: Root locus corresponding to figure 19.
• Result of reduced Fs : 3 poles have been moved away from the z = +1 region.
Part 7: Direct Digital Design
(Revised: May 09, 2013)
Page 25
>> Wa = 0.001;
>> SSE_ramp = 100/bode( Wa* Kcz*Gcz*Guyz , Wa)
SSE_ramp = 346.10
– 346% SSE(ramp) appears large, but does not violate the specification.
Part 7: Direct Digital Design
(Revised: May 09, 2013)
Page 26
EE/ME 574: Intermediate Controls
Section 3.0.0
3 Conclusions
• Design can combine a variety of processes
– Direct Digital Design
– Design by Discrete Emulation + Iteration based on Direct Digital Design
• Direct Digital Design
– Direct analysis of the discrete domain Bode plot and root locus needed
to understand behavior of the discrete system.
– Especially when Fs is only a little greater than the bandwidth.
• Analysis of system response is done in the discrete domain
– Step response
– Margins
– Bode plot
– Steady-state error
– Root locus
Part 7: Direct Digital Design
(Revised: May 09, 2013)
Page 27