Digital Control Design - University of...
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Digital Control Design
© 2017 School of Information Technology and Electrical Engineering at The University of Queensland
TexPoint fonts used in EMF.
Read the TexPoint manual before you delete this box.: AAAAA
http://elec3004.com
Lecture Schedule: Week Date Lecture Title
1 28-Feb Introduction
2-Mar Systems Overview
2 7-Mar Systems as Maps & Signals as Vectors
9-Mar Systems: Linear Differential Systems
3 14-Mar Sampling Theory & Data Acquisition
16-Mar Aliasing & Antialiasing
4 21-Mar Discrete Time Analysis & Z-Transform
23-Mar Second Order LTID (& Convolution Review)
5 28-Mar Frequency Response
30-Mar Filter Analysis
6 4-Apr Digital Filters (IIR) & Filter Analysis
6-Apr Digital Filter (FIR)
7 11-Apr Digital Windows
13-Apr FFT
18-Apr
Holiday 20-Apr
25-Apr
8 27-Apr Active Filters & Estimation
9 2-May Introduction to Feedback Control
4-May Servoregulation/PID
10 9-May PID & State-Space
11-May State-Space Control
11 16-May Digital Control Design 18-May Stability
12 23-May Digital Control Systems: Shaping the Dynamic Response
25-May Applications in Industry
13 30-May System Identification & Information Theory
1-Jun Summary and Course Review
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G. Franklin,
J. Powell,
M. Workman
Digital Control
of Dynamic Systems
1990
TJ216.F72 1990
[Available as
UQ Ebook]
Follow Along Reading:
B. P. Lathi
Signal processing
and linear systems
1998
TK5102.9.L38 1998
State-space [A stately idea! ]
• FPW
– Chapter 4: Discrete Equivalents to
Continuous
– Transfer Functions: The Digital
Filter
• Lathi Ch. 13 – § 13.2 Systematic Procedure for
Determining State Equations
– § 13.3 Solution of State Equations
Today
16 May 2017 - ELEC 3004: Systems 3
Final Exam Information announcement • Date:
Saturday, June/10 (remember buses on Saturday Schedule)
• Time:
4:30-7:45 (+/-)
• Location:
TBA
• UQ Exams are now “ID Verified”
Please remember your ID!
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Final Exam Information • Section 1:
– Digital Linear Dynamical Systems
– 5 Questions
– 60 Points (33 %)
• Section 2:
– Digital Processing / Filtering of
Signals
– 5 Questions
– 60 Points (33 %)
• Section 3:
– Digital & State-Space Control
– 5 Questions
– 60 Points (33 %)
16 May 2017 - ELEC 3004: Systems 5
NEXT WEEK: Lab 4 – LeviLab II:
• AKA “Revenge of the !”
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Lab 4 News:
Digital PID Controls
(AKA: Magic “PID Made Easy”
Equations)
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Implementation of Digital PID Controllers
Source: Dorf & Bishop, Modern Control Systems, §13.9, pp. 1030-1
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Implementation of Digital PID Controllers (2)
Source: Dorf & Bishop, Modern Control Systems, §13.9, pp. 1030-1
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Back to State-Space …
Solving State Space
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Great, so how about control? • Given 𝒙 = 𝐅𝒙 + 𝐆𝑢, if we know 𝐅 and 𝐆, we can design a
controller 𝑢 = −𝐊𝒙 such that
eig 𝐅 − 𝐆𝐊 < 0
• In fact, if we have full measurement and control of the states of 𝒙,
we can position the poles of the system in arbitrary locations!
(Of course, that never happens in reality.)
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• Recall:
• For Linear Systems:
• For LTI:
Solving State Space…
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𝑥 = 𝐴𝑥 + 𝐵𝑢
𝑠𝑋 𝑠 − 𝑥 0 = 𝐴𝑋 𝑠 + 𝐵𝑈 𝑠
𝑋 𝑠 = 𝑠𝐼 − 𝐴 −1𝑥 0 + 𝑠𝐼 − 𝐴 −1𝐵𝑈 𝑠
𝑋 𝑠 = ℒ 𝑒𝐴𝑡 𝑥 0 + ℒ 𝑒𝐴𝑡 𝐵𝑈 𝑠
𝑥 𝑡 = 𝑒𝐴𝑡𝑡
0
𝐵𝑢 𝜏 𝑑𝜏
⇒ 𝑒𝐴𝑡
Solutions to State Equations
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• Φ 𝑡 = 𝑒𝐴𝑡 = ℒ−1[ 𝑠𝐼 − 𝐴 −1]
• It contains all the information about the free motions of the
system described by 𝑥 = 𝐴𝑥
LTI Properties:
• Φ 0 = 𝑒0𝑡 = 𝐼
• Φ−1 𝑡 = Φ −𝑡
• Φ 𝑡1 + 𝑡2 = Φ 𝑡1 Φ 𝑡2 = Φ 𝑡2 Φ 𝑡1
• Φ 𝑡 𝑛 = Φ 𝑛𝑡
The closed-loop poles are the eignvalues of the system matrix
State-Transition Matrix Φ
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• Difference equations in state-space form:
• Where: – u[n], y[n]: input & output (scalars)
– x[n]: state vector
Digital State Space:
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Digital Control Law Design
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Discretisation (FPW!) • We can use the time-domain representation to produce
difference equations!
𝒙 𝑘𝑇 + 𝑇 = 𝑒𝐅𝑇 𝒙 𝑘𝑇 + 𝑒𝐅 𝑘𝑇+𝑇−𝜏 𝐆𝑢 𝜏 𝑑𝜏𝑘𝑇+𝑇
𝑘𝑇
Notice 𝒖 𝜏 is not based on a discrete ZOH input, but rather
an integrated time-series.
We can structure this by using the form:
𝑢 𝜏 = 𝑢 𝑘𝑇 , 𝑘𝑇 ≤ 𝜏 ≤ 𝑘𝑇 + 𝑇
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State-space z-transform
We can apply the z-transform to our system:
𝑧𝐈 − 𝚽 𝑿 𝑧 = 𝚪𝑈 𝑘 𝑌 𝑧 = 𝐇𝑿 𝑧
which yields the transfer function: 𝑌 𝑧
𝑿(𝑧)= 𝐺 𝑧 = 𝐇 𝑧𝐈 − 𝚽 −𝟏𝚪
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• Design for discrete state-space systems is just like the
continuous case. – Apply linear state-variable feedback:
𝑢 = −𝐊𝒙
such that det 𝑧𝐈 − 𝚽 + 𝚪𝐊 = 𝛼𝑐 𝑧
where 𝛼𝑐(𝑧) is the desired control characteristic equation
Predictably, this requires the system controllability matrix
𝓒 = 𝚪 𝚽𝚪 𝚽2𝚪 ⋯ 𝚽𝑛−1𝚪 to be full-rank.
State-space control design ¿¿¿Que pasa????
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Solving State Space
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1. Choose all independent capacitor voltages and inductor
currents to be the state variables.
2. Choose a set of loop currents; express the state variables and
their first derivatives in terms of these loop currents.
3. Write the loop equations and eliminate all variables other
than state variables (and their first derivatives) from the
equations derived in Steps 2 and 3.
A Systematic Procedure for Determining State Eqs.
See also: Lathi § 13.2-1 (p. 788)
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1. The inductor current q1 and the capacitor voltage q2 as the state variables.
2.
3.
A Quick Example
See also: Fig. 13.2, Lathi p. 789
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•
•
•
•
Another Example
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Another Example
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• N1(t)=N1(0)exp(-λ1t)
•
•
•
Another Example
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Example: PID control • Consider a system parameterised by three states:
– 𝑥1, 𝑥2, 𝑥3
– where 𝑥2 = 𝑥 1 and 𝑥3 = 𝑥 2
𝒙 =1
1−2
𝒙 − 𝐊𝑢
𝑦 = 0 1 0 𝒙 + 0𝑢
𝑥2is the output state of the system;
𝑥1is the value of the integral; 𝑥3 is the velocity.
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Example: PID control [2] • We can choose 𝐊 to move the eigenvalues of the system
as desired:
det
1 − 𝐾11 −𝐾2
−2 − 𝐾3
= 𝟎
All of these eigenvalues must be positive.
It’s straightforward to see how adding derivative gain 𝐾3 can stabilise the system.
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Can you use this for
more than Control?
YES! 16 May 2017 - ELEC 3004: Systems 28
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The Approach:
• Formulate the goal of control as an optimization (e.g. minimal impulse response,
minimal effort, ...).
• You’ve already seen some examples of optimization-based design: – Used least-squares to obtain an FIR system which matched (in the least-squares sense)
the desired frequency response.
– Poles/zeros lecture: Butterworth filter
Frequency Response in State Space
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Discrete Time Butterworth Filters
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• Constrained Least-Squares …
How?
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Break
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Example 1:
16 May 2017 - ELEC 3004: Systems 39
TF 2 SS – Control Canonical Form)
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Control Canonical Form as a Block Diagram
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• CCF is not the only way to tf2ss
• Partial-fraction expansion of the system
System poles appear as diagonals of Am
• Two issues: – The elements of matrix maybe complex if the poles are complex
– It is non-diagonal with repeated poles
Modal Form
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Modal Form
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Modal Form Block Diagram
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• Given: 𝑌 𝑠
𝑈 𝑠=
25.04𝑠+5.008
𝑠3+5.03247𝑠2+25.1026𝑠+5.008
Get a state space representation of this system
• Matlab:
• Answer:
Matlab’s tf2ss
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Example 2:
Obtaining a Time Response
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• Given: 𝑥 = 𝐴𝑥 + 𝐵𝑢
• Solution:
– Substituting 𝑡0 = 0 into this:
– Write the impulse as:
– where w is a vector whose components are the magnitudes of r
impulse functions applied at t=0
From SS to Time Response — Impulse Functions
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• Given: 𝑥 = 𝐴𝑥 + 𝐵𝑢
• Start with 𝑢 𝑡 = 𝒌
Where k is a vector whose components are the magnitudes
of r step functions applied at t=0.
– Assume A is non-singular
From SS to Time Response — Step Response
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• Given: 𝑥 = 𝐴𝑥 + 𝐵𝑢
• Start with 𝑢 𝑡 = 𝑡𝒗
Where v is a vector whose components are magnitudes of ramp
functions applied at t = 0
– Assume A is non-singular
From SS to Time Response — Ramp Response
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Example: Obtain the Step Response • Given:
• Solution:
– Set k=1, x(0)=0:
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Example II: Obtain the Step Response • Given:
• Solution:
– Assume x(0)=0:
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Example 3:
Command Shaping
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Command Shaping
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Command Shaping
• Zero Vibration (ZV)
• Zero Vibration and Derivative (ZVD)
20
11
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di
i
TK
K
Kt
A
21
eK
ddi
i
TT
K
K
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)1()1(
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12
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22
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• Digital Feedback Control
• Review: – Chapter 2 of FPW
• More Pondering??
Next Time…
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