con1 - DCC

Report
An Overview of Control Theory and
Digital Signal Processing
Luca Matone
Columbia Experimental Gravity group (GECo)
LHO Jul 18-22, 2011
LLO Aug 8-12, 2011
LIGO-G1100863
Syllabus (tentative)
Day Topic
Textbooks
1
Control theory: Physical systems, models, linear systems, block
diagrams, differential equations, feedback loops, cruise control
example, MATLAB implementation.
2
Control theory: Laplace transform and its inverse , transfer
functions, partial fraction expansion, first-order and secondorder systems, dynamic response, bode plots, stability criteria,
MATLAB implementation.
3
Control theory: robustness, typical compensators, noise
suppression, one arm cavity lock example, MATLAB
implementation and time-domain simulations with SIMULINK.
4
DSP: Discrete-time signals and systems, impulse response,
system stability, convolution and correlation, differential to
difference equations, the Z transform, the Discrete-time Fourier
Transform (DTFT), the Discrete Fourier Transform (DFT), MATLAB
implementation.
5
DSP: The Fast-Fourier Transform (FFT), power spectral density,
sampling theorem, aliasing, analog-to-digital transformations,
digital filtering, FIR filters, IIR filters, moving average filter, filter
design, ADC and DACs, MATLAB implementation.
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Chau, Pao C. Process Control:
A First Course with MATLAB®.
Cambridge University Press,
2002. ISBN 0-521-00255-9.
Ingle, Vinay K. and John G.
Proakis. Digital Signal
Processing using Matlab®.
Brooks/Cole 2000. ISBN 0534-37174-4.
Smith, Steven W. The Scientist
and Engineer’s Guide to
Digital Signal Processing.
California Technical
Publishing 1999.
http://www.dspguide.com/
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2
Objective
Control System
• Manages and regulates a set
of variables in a system
– SISO – single-input-singleoutput
– MIMO – multiple-inputmultiple-output
• A quantity is measured then
controlled
• Requirements
–
–
–
–
–
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Bandwidth
Rise time
Overshoot
Steady state error
…
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3
Objective
Digital Signal Processing
• Measure and filter an analog signal
• Digital signal
– Created by sampling an analog signal
– Can be stored
• Analog filters
– Cheap, fast and have a large dynamic
range in both amplitude and frequency
• Digital filters
– Can be designed and implemented “onthe-fly”
– Superior level of performance.
• Example: a low pass digital filter can have a
gain of 1 ± 0.0002, a frequency cutoff at
1000 Hz, and a gain of less than 0.0002
for frequencies above 1001 Hz. A
transition of 1 Hz!
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4
Control Theory 1
• Given a physical system
– Objective: sense and control a variable in the
system
• Examples
– As basic as
• a car’s cruise control (SISO) or
– Not so basic as
• Locking the full LIGO interferometer (MIMO)
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5
Example: Cruise Control
Direction
of motion
Normal
force (n)
Friction
force (ffr)
• Objective
Force from
engine (f)
– Car needs to maintain
a given speed
• Physical system
includes
Weight
(mg)
– car’s inertia
– friction
Force or free body diagram: allows to
analyze the forces at play
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6
Physical model
Direction
of motion
Normal
force (n)
Friction
force (ffr)
Weight
(mg)
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• Physical system described
by the following equation
of motion
Force from
 =  −  = 
engine (f) ∥  
• Simplifying and assuming
friction force  is
proportional to speed
 = 

 = 
=  − 

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7
First-order differential equation
Direction
of motion
• Solving for first-order
differential equation
(assuming  is a constant)
Engine
Friction
force (ffr)
force (f)
Time
constant
 = /
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Speed at
regime
 = /


=  − 

yields the solution
 =  1 −  −/
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8
MATLAB implementation
Direction
of motion
Friction
force (ffr)
The linear differential equation describing the
dynamics of the system


=  − 

Engine
force (f)
Using MATLAB’s Symbolic Math Toolbox
>> dsolve('m*Dy=f-b*y','y(0)=0')
ans =
(f - f/exp((b*t)/m))/b
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9
 = 1000 
 = 50  
 = 500 
Results:  =  ∙ 1 −  −/
 τ = 63% ∙ 
cruise_timedomain.m


 = = 10



τ = = 20 

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10
Block diagram: representing the
physical system
• To illustrate a cause-and-effect relationship
• A single block represents a physical system
• Blocks are connected by lines
– Lines represent how signals flow in the system
• In general, a physical system G has signal x(t)
as input and signal y(t) as output
• G is the transfer function of the system
x(t)
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G
y(t)
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11
Car’s body
Transfer function G represents the car’s body
– G converts the force from the engine  (input
signal, ) to the car’s actual speed  (output
signal, /)
 =∙
with G =
1
(1

– Units:  
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−
−
)
f(t)
G
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v(t)
12
Setting the desired speed
• Second transfer function H (the controller)
– Converts the desired speed (or reference)  to a
required force 
– Sets the throttle
– For simplicity, H is set to a constant
 =  ∙ 
→  =  ∙  ∙ 
 =∙
–  ∙  must be dimensionless
vr
f
H
G
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v
13
Plotting results
Generated force
by controller H
Resulting speed v
cruisefeedback_timedomain.m
• With  =  the actual speed
is the reference:  = 
• Simulate: setting desired
speed to 25 m/s (55 mph)
Desired speed vr
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14
Introducing a disturbance – a hill
• In the presence of a hill the
equation of motion needs
to be re-visited
Force from
engine (f) • Assuming a small angle θ


=  −  −  ∙ θ

Direction
of motion
Friction
force (ffr)
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θ
Weight
(mg)
Added term
Matone: An Overview of Control Theory and Digital Signal Processing (1)
15
Introducing a disturbance – a hill
Assuming  and  are constants


=  −  −  ∙ θ

Direction
of motion
Force from
engine (f)
 =  ∙  −  ∙ θ
Friction
force (ffr)
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ϑ
Weight
(mg)
Added term
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16
Modifying the block diagram
 =∙
θ
 =∙ −∙
K
vr
H
f +
 = 
G
v
Summation junction
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17
Modifying the block diagram
 =∙
θ
 =∙ −∙
K
vr
H
f +
 = 
G
v
Summation junction
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18
Plotting results
Setting desired speed to 25
m/s and slope of  = 5°
Resulting speed v
cruisefeedback_timedomain.m
Generated force
by controller H
Desired speed
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19
Negative Feedback
1. Let’s measure the car’s speed and
2. Correct for it by feeding back into the system
a measure of the actual speed 
θ
K
Error signal
vre
+
-
H
f +
G
v
c
Correction signal
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20
Negative Feedback
1. Let’s measure the car’s speed and
2. Correct for it by feeding back into the system
a measure of the actual speed 
θ
Error signal e: the difference
between the desired speed and
K
the measured speed. If null, then
 = 
f + + e
H
vr
-
G
v
c
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Correction signal c: in this case it is
just a measure of the actual speed
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21
Negative feedback
– Faster response with
feedback (compare blue
against red curves)
– Speed at regime:
23 / (error of
~10%)
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cruisefeedback_timedomain2.m
• Plot of force vs. time
and speed vs. time with
negative feedback
• Setting  = 103 /
• Result:
22
Negative feedback
• Increasing the
controller’s gain (H)
• Setting  = 104 /
• Result:
– Even faster response
– Speed at regime:
24.8 / (error of
~1%)
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cruisefeedback_timedomain3.m
– decreases the rise time
– while decreasing the
steady state error
23
Signal flow and block diagrams
 =  − 
 =  ∙ ( −  ∙ )
 =∙
θ
K
vr +
-
e
H
f +
G
v
c
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 = Matone: An Overview of Control Theory and Digital Signal Processing (1)
24
 =  − 
=  − 
System’s open
loop gain
(dimensionless)
=  −  ∙  −  ∙ 
=  −  ∙  ∙  −  ∙ 
θ
1
∙
=
∙  +
∙
1+∙
1+∙
vr + e
H
-
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K
f +
c
-
Matone: An Overview of Control Theory and Digital Signal Processing (1)
G
v
25
1
∙
=
∙  +
∙
1+∙
1+∙
  ∙  ≫ 1 (high gain,
closed loop, with feedback)
∙

 ≈ 0 ∙  +
∙ ≈ 
  ∙  ≪ 1(low
∙

gain, open loop, or no
θ Error signal in closed loop: close to
feedback)
zero, proportional to angle 
K
 ≈ 1 ∙  +  ∙  ∙ 
vr +
-
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e
H
f +
c



The higher the controller’s gain, the
lower e
=
-
G
v
Matone: An Overview of Control Theory and Digital Signal Processing (1)
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 =∙∙−∙∙
=  ∙  ∙ ( − ) −  ∙ G ∙ 
=  ∙  ∙  −  ∙  ∙  −  ∙  ∙ 
With no feedback
 =  ∙  ∙  −  ∙  ∙ 
θ
∙
∙
=
∙  − (
)∙
1+∙
1+∙
K
vr +
-
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e
H
f +
c
-
Matone: An Overview of Control Theory and Digital Signal Processing (1)
G
v
27
∙
∙
=
 −

1+∙
1+∙
  ∙  ≪ 1 (low gain,   ∙  ≫ 1 (high gain, closed
loop, with feedback)
open loop or no
∙

feedback)
 ≈ 1 ∙  −
∙  ≈  − 
 ≈  ∙  ∙  −  ∙  ∙ 
∙

θ
Actual speed : close to  with an error proportional
to  when in closed loop. The higher the controller’s
gain, the lower the speed error.
vr +
-
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e
H
K
f +
c
-
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G
v
28
 =∙
1
∙
=∙(
 +
)
1+∙
1+∙
θ

∙∙
=
∙  +
∙
1+∙
1+∙
vr +
-
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e
H
K
f +
c
-
Matone: An Overview of Control Theory and Digital Signal Processing (1)
G
v
29

∙∙
=
∙  +
∙
1+∙
1+∙
  ∙  ≫ 1 (high gain,
  ∙  ≪ 1 (low gain,
closed loop, with feedback)
open loop, or no feedback)
1
 ≈  ∙  +  ∙  ∙  ∙ 
 ≈  +  ∙ 

θ
Force : at regime, it does not
depend on the gain in  while
proportional to angle 
vr +
-
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e
K
H
f +
c
-
Matone: An Overview of Control Theory and Digital Signal Processing (1)
G
v
30
Plotting  and  1 + 
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cruisefeedback_timedomain2.m
• Open loop
 =  ∙  ∙ 
• Closed loop
∙
=
∙ 
1+∙
• Setting  = 103 /
• Plotting the open loop
transfer function vs. time
and the closed loop
transfer function vs. time
• Notice the rapid rise time
for the closed loop case
31
The error signal e
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Matone: An Overview of Control Theory and Digital Signal Processing (1)
cruisefeedback_timedomain2.m
• Plot of error signal e vs
time
• Error signal decreases
to 3 m/s.
• Notice a ~10% steady
state error
32
Open and closed loop TF with
 = 104 /
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33
Error signal e with
 = 104 /
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34
Cruise control example
• First-order differential
equation
• Simplest controller:
simply a gain with no
time constants involved
• How to handle more
complicated problems?
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35
Block diagram reduction
x
x
y
P2
P1
P1
+
±
x
P1P2
y
x
P1 ±P2
y
x
1
1 ± 1 2
y
P2
x
+
∓
P1
y
y
P2
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36
Practice
Determine the output C in terms of inputs U and R.
U
R +
-
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1
+
+
2
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C
37
Practice
Determine the output  in terms of inputs
1 , 2 and .
U1
R +
+
1
1
+
+
+
+
2
C
2
U2
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38
More practice
Determine C/R for the
following systems.
(c)
1
1
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R+
+
+C
+
2
1
+
+
C
1
(b)
2
+
R +
(a)
R+
+
2
1
+
+
C
1
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39
How do we MEASURE the OL TF of
a system when the loop is closed?
1. Add an injection point in a closed loop
system
2. Inject signal  and read signal 1 (just before
the injection) and 2 (right after the
injection)
1

3. Solve for the ratio
2
+
+
1
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2
40
So far…
• Control theory builds on differential equations
• Block diagrams help visualize the signal flow in a
physical system
• The cause-and-effect relationship between
variables is referred to as a transfer function (TF)
• The system’s open-loop TF is the product of
transfer functions
– cruise control example:  ∙ 
– Two cases:  ∙  ≪ 1 and  ∙  ≫ 1
• MATLAB implementation
– Functions used: dsolve
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41
Laplace Transforms
• The technique of Laplace transform (and its
inverse) facilitates the solution of ordinary
differential equations (ODE).
• Transformation from the time-domain to the
frequency-domain.
• Functions are complex, often described in
terms of magnitude and phase
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42
Linear systems
• To map a model to frequency space
– System must be linear
– Output proportional to input
• Given system P
x
– Input signals: 1 and 2
– Output signals (response): 1 and 2
P
y
• System P is linear
– If input signal:  1 +  2
– Then output signal:  1 +  2
– Superposition principle
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43
Example
• Is  =


a linear system?
Knowing that 1 =
1

and 2 =
2

If input is 1 1 + 2 2 , output is

1 1 + 2 2 =



1 1 + 2 2 =


1 1 + 2 2
System is linear
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44
Example
• Is  =  2 a linear system?
Knowing that 1 = 1 2 and2 = 2
If input is 1 1 + 2 2 , output is
1 1 + 2 2
2
2
≠ 1 1 + 2 2
System is not linear
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45
In general
Output
 
 −1 
  + −1 −1 + ⋯ + 0 


 
 −1 
=   + −1 −1 + ⋯ + 0 


Input


     =
=0
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=0
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≥
For a stable system
46
Laplace Transform L
• Transforms a linear differential equation into an
algebraic equation
• Tool in solving differential equations
• Laplace transform of function f
  = L ()
• Laplace inverse transform of function F
−1
() = L ()
where  =  is the transform variable
Imaginary unit
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2
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47
Time domain ↔ Laplace domain
=

f()

Input
Output
y()
()
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()
F()
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48
Laplace Transform L
∞
() − 
  = L () =
0
  =L
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−1
1
() =
2
+
()  
−
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49
(Some) Laplace transform pairs
()
()
Unit step
()
1
Unit ramp

1
sin(0 )
Sinusoid
1/ 1 −  −
SHO
0
 −0  ×
1 − 2
× sin( 1 −  2 0 )
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1
 
Exponential

1

2
−
 2 + 0 2
 +
0 2
 2 + 20  + 0 2
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50
Laplace transform properties
• Linearity
L 1 1  + 2 2 
• Derivatives
– First-order: L
()

– Second-order: L
• Integral
= 1 1  + 2 2 
= ()
 2 ()
 2
=  2 ()

1
L
   = ()

0
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51
Solution to ODEs
1. Laplace transform the system’s ODE
2. Solve the algebraic equation in s
3. Inverse transform back to the time domain
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52
Transfer Function  


     =
=0

    
=0

     =
=0
()
  =
=
()
    
Using
derivative
property
=0
Algebraic equation

 in s, the ratio of 2
=0   polynomials in s




=0 
Transfer function   relates input   to output   .53
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Transfer Function  
The roots of the numerator are referred to as zeros.
Transfer function   can
()
be defined by
  =
=
• The coefficients of s or
()
• Its poles and zeros




=0 




=0 
The roots of the denominator are referred to as poles.
54
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Let’s apply it to the cruise control example:
transfer function  (the car’s body)
Direction
of motion
Engine
force (f)
Friction
force (ffr)
f(t)
F(s)
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G(t)
v(t)
()

= () − ()

    =   −  ()
 
V(s)
=     where  
G(s)
1 
= at −  
Pole
 +Signal
/
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Processing (1)
55
Dynamic response: using lookup
tables to inverse transform
Laplace inverse
transform using
lookup tables
1
 +
1
1 −  −

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Input: step function, amplitude 0
  = 0 ()

  = L () = 0 
The response (in frequency space) is
1/
0
  =  ∙  =
∙
 + / 
The time-domain response is
  = L−1 () =
1
−1 0
=L
∙
  ( +  )
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56
Dynamic response: using lookup
tables to inverse transform
Laplace inverse
transform using
lookup tables
1
 +
1
1 −  −

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0
1
  =L
∙
  ( +  )
0 −1
1
= ∙L

 ( +  )
−1

0
− 
  =
1−

1
Pole  =

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57
First-order system step response
63 %

  =
+
firstorder.m
 = 5 /,  = 0.2 
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Pole p
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58
MATLAB implementation
The step response of transfer function
5
  =
+5
>> G=tf(5, [1 5]);
>> step(G);
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59
Partial fraction expansion
1. Reduce a complex function to a collection of
simpler ones
2. Then use lookup table Order m
()

≤
  =
=
()
 + 
Order n

1

−1
  =L
+ ⋯+ L
 + 1
 + 
= 1  −1  + ⋯ +   − 

  =
  −
=0
−1
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60
Comments
  =


 + 

  −
  =
=0
1. Poles of F(s) determine the time evolution of
f(t)
2. Zeros of F(s) affect coefficients
3. Poles closer to origin → larger time constants
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61
Example
Find () of the Laplace transform
6 2 − 12
  = 3
 +  2 − 4 − 4
Sol: Using MATLAB
>> [R,P,K]=residue([6 0 -12],[1 1 -4 -4])
R =
3.0000
3
1
2
1.0000
  =
+
+
+2 −2 +1
2.0000
P =
-2.0000
2.0000
-1.0000
K =
  = 3  −2 +  2 + 2  −
[]
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Example: LRC circuit

 ()


  =   +   +  
 ()


1
=  () +  () +
  


0
11
=  +  +
()

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63
Example: LRC circuit

 ()


 ()
11
 () =     +    +
()


  =    =    

  =   2 +    + 1  ()
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Example: LRC circuit

 ()


 ()
  =   2 +    + 1  ()
L
  =    2 +    + 1  ()
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65
LRC circuit: transfer function

1
 =
∙  
2
 ++1
Setting L = 1 H, C = 1 F and R = 1 Ω

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1
 = 2
∙  
 + +1
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66
LRC circuit: dynamic response to step
Setting the input to a step of amplitude 1 V
1
  =

The unit step response is
1
1
1
  = 2
∙ = 3
 +  + 1   + 2 + 
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67
LRC circuit: dynamic
response to step
1
 () = 3
=
2
 + +


 + 
Using MATLAB for the solution
Coefficient vector for
polynomial at numerator
Coefficient vector for
>> n = [1];
polynomial at denominator
>> d = [1 1 1 0];
>> [α, a, k] = residue(n, d);
  = 1  1 + 2  2 + 3  3
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Plotting results of two methods
>> y=α.'*exp(a*t);
>> plot(t, y, 'bo');
>> y2=step(1,[1 1 1],t);
>> plot(t, y2, ‘r.’);
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Second-order system step
response
secondorder.m
Overshoot
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Period of
oscillation
Settling time: time to
settle to ±5% of final
value (determined by
the  − term)
Note: often the
response of a highorder system is
similar to the
second-order one
Matone: An Overview of Control Theory and Digital Signal Processing (1)
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Verify the following
6 2 − 12
  = 3
 +  2 − 4 − 4
  = 2 − + 3 −2 +  2
6
  = 3
 +  2 − 4 − 4
  = 3 − − 6 −2 + 3 −3
+5
  = 2
 + 4 + 13
  = 2 − − 3 −2 +  2
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Back to cruise control: system’s
step response
∙
() =
∙  ()
1+∙
1 
where   =
and   = 
 + /
θ
K
e
+
vr
-
H
f + -
G
v
c
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Step response
∙
1
  =
∙ =
1+∙ 


(
+

)

2 +




−/
  =
1−
with τ =
 + 
 + 
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Back to cruise control
>>
>>
>>
>>
>>
LIGO-G1100863
n = [k/m];
d = [1 (b+k)/m 0];
[α, a, k] = residue(n, d);
y=α.'*exp(a*t);
plot(t, y, 'bo');
Matone: An Overview of Control Theory and Digital Signal Processing (1)
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Back to cruise control
>>
>>
>>
>>
>>
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G = tf([1/m],[1 b/m]);
H = k;
CL = G * H / (1 + G * H);
[y, t] = step(CL);
plot(t, y, 'b.');
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Transfer function, poles and zeros
• It is convenient to express a transfer function
G(s) in terms of its poles and zeros:
()
  =
()
 − 1 ∙  − 2 …  − 
=∙
 − 1 ∙  − 2 …  − 
• k is the gain of the transfer function
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Summary of pole characteristics
• Real distinct poles (often negative)

↔    
 − 
• Real poles, repeated m times (often negative)
,1
,2
+
 − ,1
 − ,2
2
+ ⋯+
,3
 − ,3
3
+
,
 − ,

↕
1
,
2
,1 + ,2  + ,3  + ⋯ +
 −1 ∙   
2!
−1 !
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Summary of pole characteristics
• Complex-conjugate poles

 ∗
 
∗ 
+
↔


+


 
∗
 −   − 
often re-written as a second-order term
2

↔
~

∙ sin  + 
2
2
 + 2  + 
• Poles on imaginary axis
∗
– Sinusoid
– Pole at zero: step function
• Poles with a positive real part
– Unstable time-domain solution
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Summary
• The Laplace transform is a tool to facilitate solving for ODEs.
• Systems need to be linear
• No need to do the transform (integral)
– Use transform pairs, transform tables
– Laplace transform properties: linearity, derivatives and integrals.
• Once in the Laplace domain, a TF is simply the ratio of two polynomials
in s. Carry out algebra to solve the problem.
• No need to do the inverse transform
– Use transform pairs, transform tables
– For high-order TFs, use the partial-fraction expansion to reduce the
problem to simpler parts
• IMPORTANT:
– Poles of a transfer function determine the time evolution of the system
– Poles with a real positive part correspond to unstable and unphysical
systems
– The system TF needs to have poles with a negative real part
• MATLAB implementation
– Functions used: step, residue, tf
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Solutions
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Practice
Determine the output C in terms of inputs U and R.
Sol:
2
=
∙ 1  + 
1 + 1 2
U
R +
-
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1
+
+
2
Matone: An Overview of Control Theory and Digital Signal Processing (1)
C
81
Practice
Determine the output  in terms of inputs 1 , 2
and .
Sol:
1
=
∙ 1 2  + 2 1 + 1 2 1 2
1 − 1 2 1 2
U
1
R +
+
1
1
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+
+
+
+
U2
2
C
2
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More practice
Determine C/R for the following
systems. Sol:
a)


=
1 +2
1−1 1 −2 2
b)


=
1 +2
1−1 1
c)


=
1 +2 1−1 1
1−1 1
1
(b)
2
1
1
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R+
+
2
+
+
C
1
(c)
+
R +
(a)
+C
+
R+
+
2
1
+
+
C
1
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How do we MEASURE the OL TF of
a system when the loop is closed?
1. Add an injection point in a closed loop
system
2. Inject signal  and read signal 1 (just before
the injection) and 2 (right after the
injection)
1

3. Solve for the ratio
2
Sol:
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1
2
= −
+
+
1
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2
84
Partial-fraction examples
• Denominator: has distinct, real roots
– Example 2.4, 2.5, 2.6
• Denominator: complex roots
– Example 2.7, 2.8
• Denominator: repeated roots
– Example 2.9
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Practice: verify the following
6
  =
( + 1)( + 2)( + 3)
  = 3 − − 6 −2 + 3 −3
+5
  = 2
 + 4 + 13

−2
  = 2
sin(3 + )
4
  =
  =2
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2
+1 3 +2
 2 −
1−+
 −  −2
2
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