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October 28, 2008
Final Work Presentation
1
October 28, 2008
Final Work Presentation
2
Introduction
Application - ESP
• Electro Static Precipitator or Electrostatic air cleaner
- ESP is used to filter and remove particles from a flowing gas
- ESP contains rows of thin wires(electrodes) situated between large metal
plates(collection electrodes)
- High DC voltage between wires and plates
- The particles are ionized by high voltage around the wires(cathode) and
then attracted by the plates(anode)
- The load in the ESP application shows capacitive behaviour
October 28, 2008
Final Work Presentation
3
Introduction
DC/DC Power Converters
• The need to convert a direct current(DC) voltage source from
one voltage level to another
• Linear conversion in compare to Switched Mode conversion
• A DC/DC switching power converter stores the input energy
temporarily and then releases that energy to the output at a
different voltage level (this function motivates the switching
property of the converter)
• The storage may be in magnetic components, like inductors
and transformers, or in capacitors
• Different topologies has been introduced to fulfill our
expectations, like Buck, Boost, Buck-Boost, Resonant power
converter and etc.
October 28, 2008
Final Work Presentation
4
Introduction
ESP application and DC/DC power converters
• The proper high load voltage is
provided by a combination of a
DC/DC power converter and a
transformer
• The resonant converter topology is
chosen for the power converter
• The switching is done by the four
transistors Z1 , Z 2 , Z 3 and Z 4
(Transistors can be used as switches
if they are saturated or cut-off)
• Resonance circuit , switched system
and rectifier
• Switching frequency and resonance
frequency
• The load is connected in series to the
resonance circuit
October 28, 2008
D1
VCC
D2
qw
1 Z1
r
qr
A
Cr
3 Z3
Final Work Presentation
Lr
Z2
Cw 
m
2
B
Lm
Z4
TR
D3
4
D4
qL
CL Load
Rectifier
5
Introduction
• Controller and Observer in a resonant converter
• Resonant phase plane controller
• Control Signal: u AB  - VCC ,
T
• States: x  qr qw qL
D1
1
VCC
0, VCC 
m 
r
D2
qw
Z1
r
qr
A
Cr
3 Z3
Lr
Z2
Cw 
m
Lm
D3
October 28, 2008
Resonant voltage
B
Z4
TR
Resonant circuit
2
Resonant current
Controller and Observer
4
D4
qL
CL Load
Final Work Presentation
Switched System(rectifier)
W-1 - neg. conducting
W0 - not conducting
W1 - pos. conducting
6
Modeling of dynamic systems
• Why modelling is needed? Analysis of system properties, controller design
and observer design is based on the dynamic model of the system
• The traditional differential equation form, ABCD form, A set of equations
derived from physical law governing the system
• The ABCD form is not global for non-linear systems and it eliminates the
algebraic equations (some information of the system are lost)
• Hamiltonian modelling is based on the energy properties of the system
• Hamiltonian modeling DE-form and DAE-form
• Power converter is a non-linear system due to existence of the switching
transistors and rectifier, this motivates using Hamiltonian modelling
October 28, 2008
Final Work Presentation
7
Hamiltonian Modelling
x (t )  J ( x, t ) - R( x)
y (t )  G T ( x, t )

H ( x)  G ( x, t )u (t )
x

H ( x)
x
x(t ) is the Hamiltonian states
H x is the energy storage function (Hamiltonian function)
J ( x, t ) is a skew-symmetric matrix models how energy flows within the system,
R (x ) is a positive semi-definite symmetric dissipation matrix
G ( x, t ) is a skew-symmetric control matrix
*This model structure is used for modelling the converter*
October 28, 2008
Final Work Presentation
8
Hamiltonian Modelling
• Hamiltonian states, x : charge on the capacitors, q i , and the magnetic
flow in the inductors,  i
• The energy function:
q L2  r2  m2  1 T
1  qr2 q w2
  x t M x-1 xt 
H qr , q w , q L ,  r ,  m   




2  C r C w C L
Lr Lm  2
•
where x T  q r q w q L  r  m  and
M x  diag Cr , C w , C L , Lr , Lm 
• The physical interpretation of H x
H  H
 
x  q r
October 28, 2008
H
q w
H
q L
H
 r
H 

 m 
Final Work Presentation
T
 ur 
 
 uw 
 M x-1 x   u L 
 
 ir 
i 
 m
9
Hamiltonian modelling of the converter in DE-form
• Decreasing the number of the states to four. magnetising inductance , Lm, is
high compared to the resonance inductance, making the magnetising
current comparably small and possible to neglect.
• Hamiltonian modelling in DE-form
~
I x x  J A - R A M x-1 x  F - G u
w  F  G  M x-1 x  J D  RD u
T
• Hamiltonian modelling is strongly related to the Graph Theory
• Different block matrices in Hamiltonian DE-form must be defined
according to graph theory
• A Hamiltonian observer can be defined based on 4th order Hamiltonian
model of the precess
October 28, 2008
Final Work Presentation
10
4th order Hamiltonian model of the converter
•
•
•
Power converter is a switching system due
to existence of the transistors and rectifier
The controller produce the proper function
to switch the transistors
The rectifier has 3 possible conducting
states: Positive conducting, Non-
D1
1
D2
qw
Z1
r
qr
A
Cr
3 Z3
conducting and Negative conducting
•
VCC
Rectifier is acting in which subspase!?
Lr
2
Z2
Cw 
m
B
Lm
D3
4
Z4
TR
D4
qL
CL Load
Ω1, positive conduction:
uCL  u w  ir  0 ,
Ω0, no conduction:
- u L  u w  u L  - u L  u w  ir  0  u L  u w  ir  0
Ω-1, negative conduction: u w  - u L  ir  0
October 28, 2008
Final Work Presentation
11
Modelling for subspace Ω0
•
•
•
The graph in the subspase Ω0
Spanning tree, tree branches and links
Component matrices, intermediate block matrices
and the system block matrices are defined according
to the graph theory
 Cr

C  0
0

0
0

Cw 0 
0 CL 
L  ( Lr ) , X  Rr ,Y  0
The intermediate block matrices
T
R22  (Y  QXY
XQXY ) -1  0
T
R11  ( X -1  QXY Y -1QXY
) -1  X  Rr
T
Z  - XQXY Y  QXY
XQXY   0
-1


~
T
I x  diag I C  QCc cQCc
C -1 , I L  QlLT lQlL L-1  diag I C , I L 
October 28, 2008
Final Work Presentation
12
Modelling for subspace Ω0
•
The system block matrices
The system equation:
October 28, 2008
 qr 
0 0
 

q
d  w
0 0
= 
0 0
dt  q L 
 

 
 -1 -1
 r






0 - Rr 
0
0
0
1
1
0
 ur   0 
   
 uw   0 
 u  +  0  u AB
 L   
i 
 r  1
Final Work Presentation
13
Modelling for the subspaces Ω1 and Ω-1
In the subspaces Ω1 and Ω-1 the converter can be modelled by the graphs
Different matrices must be defined for the upper configurations, the final result:
 qr   0 0
  
d  qw   0 0

dt  q L   0 0
  
   -1 -1
 r 
October 28, 2008
0
1   ur   0 
   
0 1 -  2   u w   0 

u
0
   u L   0  AB
   
0 - Rr   ir   - 1
where:
Final Work Presentation

C L
C w  C L
Cw
 1 -  2 ,
2
C w   C L
14
4th order global model
•
A global model is:
•
The 4th order global model of the converter system
xt   J s  - R  M -1 xt   B u AB t 
0
 qr   0
  
0
d  qw   0

0
dt  q L   0
  
    - 1 - 1 - s 2
 r 

•

0
0
0
- s
  ur   0 
   
  uw   0 
  u    0  u AB
 L  
- Rr   ir   - 1
1
1 - s 2
s


where
where s  - 1, 0 ,1

C L
C w  C L
An Hamiltonian observer can be introduced based on the 4th order global model
October 28, 2008
Final Work Presentation
15
Hamiltonian Observer
• The structure of an Hamiltonian observer is given by
dxˆ
  J A - R A M x-1 x  F - G u  K  y - C m xˆ 
dt
where
 qˆ 
xˆ    , where qˆ and ˆ are the estimtes of q and  ,
 ˆ 
y  C m x is the measurment signal
and K is the observer gain
• From the process model and the observer structure, the estimation error
is given as e  x - xˆ , satisfies
.
de
 J A xˆ  - RK M x-1e  J A ( x) - J A ( xˆ ) M x-1 x
dt
where RK  R A  KCm M x
October 28, 2008
The error can be regarded as a
Hamiltonian system!!
Final Work Presentation
16
Hamiltonian Observer
• Error in the observer can be regarded as a Hamiltonian process with the
1
Hamiltonian function(energy function):
H e   e T M -1e
x
it is shown that (Hultgren, lenells 2004) :
2
x
dH e x , t 
0
dt
x
if
x
K0
• Note: The process is modelled by a 4th order Hamiltonian model and the
observer is based on the 4th order Hamiltonian model of the process
October 28, 2008
Final Work Presentation
17
Discretization of the observer
• Controller creates control signal u from the external signal
and ŷ (estimation of the states)
r
(reference)
• Implementation of the controller: Analog or Digital
• In the converter a phase plane feedback controller is used, this controller
uses Resonance voltage and resonance current as feedbacks
• The only measurment in this application is the resonance current
• In discretization of the Hamiltonian observer, the sampling interval is very
important (the error balance must be always negative)
October 28, 2008
Final Work Presentation
18
Implementation in Matlab and Simulink
Process
To Workspace5
U0Nsig
>=
Relational
Operator2
0
s
Demux
Constant4
Scope3
E
I0Nref
Constant3
Dot Product Gain
K*u Out1
In1
s
states
Memory4
kk
Regulator1
To Workspace7
B
Demux -K-
Scope7
Gain4
slrhamilton
|u|
y
1
s
Abs
Integrator
To Workspace3
1
1
s
Scope9
ILNref
Switch
0
Abs1
To Workspace
-KU0N
XY Graph
|u|
Scope6
-KGain1
-K-
Process
ey
To Workspace8
Out1
ex
Out2
Hit
Crossing
Constant
Product
Memory2
Integrator1
In1
In2
Gain2
To Workspace4
In3
Out3 In4
Subsystem1
t
Error
Clock
-K-
To Workspace1
In1
In2
logga i0
Gain3
Observer
Observer
October 28, 2008
Final Work Presentation
19
Implementation in Matlab and Simulink
1
In1
In1
In2
Out1
1
In3
Constant
In4
0
3
A --1
A0* u
x' = Ax+Bu
y = Cx+Du
y2
Demux
y3
State-Space
y4
Multiport
Switch
A0
1
Out2
Out1
V
y1
Subsystem
A2* u
2
In1
Multiport
Switch
In3
G2* u
1
Out1
2
4
In2
In4
D1* u
K*u
Matrix
Gain
A --1
G0* u
In1
slrhamilton
filter
A1
Observer Model
System Model
2
SS
Out2
1
In1
In1
To Workspace2
sigma
In2
Out1
T o Workspace1
In3
1
In4
Subsystem
K*u
0
AND
Constant
y3
y(n)=Cx(n)+Du(n)
Demux
x(n+1)=Ax(n)+Bu(n)
y4
A --1
K*u
K*u
Multiport
Switch
In1
2
y2
A0
To Workspace1
Out2
G1* u
A1
y1
Out1
Matrix
Gain1
A0
A1* u
3
Out3
Dot Product To Workspace6
Discrete State-Space
la
1
Out1
1
uCw>uCL
<=
Out1
AND
y5
-1
uCw<uCL
Gain
2
Constant
In2
>=
lb
3
In3
4
In4
>=
ir>im
A1
<
ir<im
Hamiltonian Observer
October 28, 2008
Logic
Final Work Presentation
20
Working Plane
High Power
vLN
• Normalization of voltages and currents!?
• Where is the operating point practically?
• Does the observer works
Low load current and
the same in all regions?
high load voltage
• Where maximum error in
state estimation happens?
• Does the observer get
unstable anywhere?
• Which state has the most error?
High load current and
low load voltage
• How we can improve state
estimation in the observer?
• Different modelling? Optimal Sampling frequency?
iLN
Result of simulation will answer the questions, hopefully correct!
October 28, 2008
Final Work Presentation
21
High load current and Low load voltage, 10MHz sampling frequency
High load current
Square Wave uCw
Process values for Fourth Order model, K=0,10MHz
600
uCr
uCw
uCL
iLr
400
Process values from Fourth Order observer, K=0,10MHz
500
uCr
uCw
uCL
iLr
400
300
200
200
100
0
0
-100
Different subspaces!
-200
-200
-300
-400
-400
-600
0
1
2
Time s
-500
3
0
1
2
-4
3
-4
x 10
x 10
Spikes! High error!
Esitmation errors for fourth order, K=0 ,10MHz
Error Energy function H(4th order), K=0 ,10MHz
150
14
eCr
eCw
eLr
100
s-sigma
500*H
12
10
50
8
0
6
4
-50
2
-100
0
-150
0
1
2
Time s
October 28, 2008
3
-4
x 10
Final Work Presentation
-2
0
1
2
Time s
3
-4
x 10
22
High load current and Low load voltage, 30MHz sampling frequency
Process values for Fourth Order model, K=0
Process values from Fourth Order observer, K=0
600
500
uCr
uCw
uCL
iLr
400
uCr
uCw
uCL
iLr
400
300
200
200
100
0
0
-100
-200
-200
-400
-300
-400
-600
0
1
2
Time s
3
-500
-4
x 10
0
1
2
3
-4
x 10
Esitmation errors for fourth order, K=0 ,30MHz
Error Energy function H(4th order), K=0 , 30MHz
100
10
eCr
eCw
eLr
80
60
s-sigma
500*H
8
40
6
20
0
4
-20
2
-40
-60
0
-80
-100
0
1
2
Time s
October 28, 2008
3
-4
x 10
Final Work Presentation
-2
0
1
2
Time s
3
-4
x 10
23
High load current and Low load voltage, 3MHz sampling frequency
4th order observer estimations of Process,K=0,3MHz,VL=0.1 IL=0.9
Process values of model with a 4th order observer,K=0,3MHz,VL=0.1 IL=0.9
600
uCr
uCw
uCL
iLr
400
600
400
200
200
0
0
-200
-200
-400
-400
-600
0
1
2
3
Time s
uCr
uCw
uCL
iLr
-600
-4
x 10
Esitmations error for 4th order observer,K=0,3MHz,VL=0.1 IL=0.9
200
0
1
2
3
-4
Error does not
converge to zero!
x 10
Error Energy function H (4th Order),K=0,3MHz,VL=0.1 IL=0.9
45
eCr
eCw
eLr
150
s-sigma
500*H
40
35
100
30
25
50
20
0
15
10
-50
5
-100
0
-150
0
1
2
Time s
October 28, 2008
3
-5
0
-4
x 10
Final Work Presentation
1
2
Time s
3
-4
x 10
24
Hamiltonian observer gain
•
•
How we can improve state estimation? Applying obser gain!? Changing the model!?
Applying an observer gain may improve state estimation but not always!
Esitmation errors for fourth order, K=0.02 ,10MHz
Error Energy function H(4th order), K=0.02 ,10MHz
80
9
eCr
eCw
eLr
60
s-sigma
500*H
8
7
6
40
5
20
4
3
0
2
1
-20
0
-40
0
1
2
Time s
•
-1
3
-4
0
1
2
Time s
x 10
Lets have a look at trajectories:
3
-4
x 10
K 0
uCr - i Lr
K 0
October 28, 2008
Final Work Presentation
25
Low load current and high load voltage, 30MHz sampling frequency
Process values of model with a 4th order observer,K=0,30MHz,VL=0.9 IL=0.1
500
uCr
400
uCw
uCL
300
iLr
200
4th order observer estimations of Process,K=0,30MHz,VL=0.9 IL=0.1
500
Rectifier is operating mostly
In the non-conducting mode
uCr
uCw
uCL
iLr
400
300
200
100
100
0
0
-100
-100
-200
-200
-300
-300
-400
-400
-500
2
1
Time s
0
-500
-4
x 10
0
1
2
-4
x 10
Esitmation errors for fourth order, K=0 ,30MHz
Spikes are much lower
250
eCr
eCw
eLr
200
Error Energy function H(4th order), K=0 , 30MHz
10
s-sigma
500*H
8
150
100
6
50
4
0
-50
2
-100
0
-150
-200
0
1
2
Time s
October 28, 2008
-2
3
0
1
2
Time s
-4
x 10
Final Work Presentation
3
-4
x 10
26
High load current and high load voltage (operating point), 10MHz sampling
frequency
Esitmation errors for fourth order, K=0 ,10MHz
Error Energy function H(4th order), K=0 ,10MHz
250
•
K=0
16
eCr
eCw
eLr
200
150
s-sigma
500*H
14
12
100
10
50
8
0
6
-50
4
-100
2
-150
0
-200
-250
0
1
2
Time s
•
K>0
3
-2
0
1
2
Time s
-4
x 10
3
-4
x 10
Error Energy function H(4th order), K=0.02 ,10MHz
Esitmation errors for fourth order, K=0.02 ,10MHz
9
150
eCr
eCw
eLr
100
s-sigma
500*H
8
7
6
5
50
4
3
0
2
1
-50
0
-100
-1
0
1
2
Time s
October 28, 2008
3
0
-4
x 10
Final Work Presentation
1
2
Time s
3
-4
x 10
27
High load current and high load voltage (operating point),
30MHz sampling frequency
•
Esitmation errors for fourth order, K=0 ,30MHz
K=0
Error Energy function H(4th order), K=0 , 30MHz
250
10
eCr
eCw
eLr
200
150
s-sigma
500*H
8
100
6
50
0
4
-50
2
-100
-150
0
-200
-250
0
1
2
Time s
•
K>0
3
-2
0
1
2
Time s
-4
x 10
3
-4
x 10
Error Energy function H(4th order), K=0.02 , 30MHz
Esitmation errors for fourth order, K=0.02 ,30MHz
10
100
s-sigma
500*H
eCr
eCw
eLr
8
6
50
4
2
0
0
-50
-2
0
1
2
Time s
October 28, 2008
3
0
-4
x 10
Final Work Presentation
1
2
Time s
3
-4
x 10
28
3rd order model of the process
• If we have a lower order model of the system, implementation will be
simpler and we have less calculation in the observer
• By less calculation in the observer, we can discretize the observer with
higher sampling frequency
• Which state is candidate to be eliminated?
• The load capacitance voltage is only slowly varying due to the high value
of CL . During one conduction period of the rectifier the load voltage can
be viewed as a constant voltage source, U 0 , in series with the voltage u AB
• The influence of the parallel capacitance, C w, is only significant when the
rectifier is not conducting. In the case when the rectifier is conducting the
parallel capacitance is connected in parallel with much larger load
capacitance CL .During the periods when the rectifier is not conducting,
the voltage of the parallel capacitance normally is commuting from U 0 to - U 0
or vice versa. The length of the non-conducting time periods is only
significant when the load current is small.
October 28, 2008
Final Work Presentation
29
3rd order model of the process
•
subspace Ω0
 qr   0 0 1   u r   0 
   
d  
q

0
0
0
 L 
  u L    0  u AB
dt   
   
  r   - 1 0 - Rr   ir   - 1
System equation
• subspaces Ω1 and Ω-1
 qr   0 0 1   u r   0 
   
d  
s   u L    0  u AB
 qL    0 0
dt   
   
  r   - 1 - s - Rr   ir   - 1
Global system equation
October 28, 2008
Final Work Presentation
30
High load current and Low load voltage, 10MHz sampling frequency
In this region, the rectifier is in the non-conducting mode for a short time
Esitmation errors for third order, K=0 ,10MHz
Error Energy function H(3rd order), K=0 ,10MHz
150
K=0
16
eCr
eCL
eLr
100
s-sigma
500*H
14
12
10
50
8
0
6
4
-50
2
-100
0
-150
-2
0
1
2
3
Time s
0
1
3
-4
x 10
x 10
Esitmation errors for third order, K=0.02 ,10MHz
Error Energy function H(3rd order), K=0.02 ,10MHz
80
K>0
2
Time s
-4
10
eCr
eCL
eLr
70
s-sigma
500*H
8
60
50
6
40
4
30
2
20
10
0
0
-10
-2
0
1
2
Time s
October 28, 2008
3
-4
x 10
Final Work Presentation
0
1
2
Time s
3
-4
x 10
31
Low load current and High load voltage, 10MHz sampling frequency
In this region, the rectifier is in the non-conducting mode for most of the time
Esitmation errors for third order, K=0 ,30MHz
Error Energy function H(3rd order), K=0 ,30MHz
200
200
eCr
eCL
eLr
150
K=0
s-sigma
500*H
150
100
50
100
0
-50
50
-100
-150
0
-200
-250
0
1
2
Time s
3
-50
0
1
2
Time s
-4
x 10
Error Energy function H(3rd order), K=0.02 ,30MHz
Esitmation errors for third order, K=0.02 ,30MHz
20
80
K>0
s-sigma
500*H
eCr
eCL
eLr
60
3
-4
x 10
15
40
10
20
5
0
0
-20
-5
-40
0
1
2
Time s
October 28, 2008
3
0
1
2
Time s
3
-4
x 10
-4
x 10
Final Work Presentation
32
High load current and high load voltage (operating point),
30MHz sampling frequency
Esitmation errors for third order, K=0 ,30MHz
Error Energy function H(3rd order), K=0 ,30MHz
500
300
eCr
eCL
eLr
400
300
200
K=0
s-sigma
500*H
250
200
100
150
0
100
-100
-200
50
-300
0
-400
-500
-50
0
1
2
3
Time s
0
1
3
-4
x 10
x 10
Error Energy function H(3rd order), K=0.02 ,30MHz
Esitmation errors for third order, K=0.02 ,30MHz
18
80
eCr
eCL
eLr
60
K>0
2
Time s
-4
s-sigma
500*H
16
14
12
40
10
8
20
6
0
4
2
-20
0
-40
-2
0
1
2
Time s
October 28, 2008
3
-4
x 10
Final Work Presentation
0
1
2
Time s
3
-4
x 10
33
Modelling of the system by considering resistance across the load capacitance
The load in this industrial application shows
obvious capacitive behaviour, but due to
existence of corona current around the wires,
the load shows resistivity behaviour too
The modelling takes the same steps as we did for
previous cases. First we define different subspaces
, then we achieve the system equation according to
Hamiltonian modelling and graph theory
October 28, 2008
Final Work Presentation
34
Modelling of the system by considering resistance across the load capacitance
subspace Ω0
 qr 
 
d  qw 
dt  q L 
 
 
 r
=
0
0 0

0
0 0
 0 0 - 1 / R
L

 -1 -1
0






- Rr 
 ur 
 
 uw 
u  +
 L
i 
 r
1
1
0
0
 
0
0
 
1
 
u AB
subspaces Ω1 and Ω-1
0
0

q
 r
0
  0
d  qw  

dt  q L   0 s 1 - s 2
  
RL
 
 r  
2
 -1 - 1- s 



0
-


s
1 - s 2
RL
1
RL
- s
where
October 28, 2008
Final Work Presentation

 u   0 
1 - s 2   r   
  uw    0  u
     AB
s   u L   0 
   
  ir   - 1
- Rr 


1

C L
C w  C L
35
Thank you for your attention!
October 28, 2008
Final Work Presentation
36