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1
Electrical drives
Prof. A.GENON
Universit de Lige (Belgique)
2010
Power electronic systems
Analogic command
Numerical commnd
LAPLACE TRANSFORM
s is a complex variable
Definition
[ ]
==
0
)()()( dtetftfLsFst
)(tf )(sF
Linarit )()( 2211 tfktfk + )()( 2211 sFksFk +
Translation retarde )0()( 00 tttf )(0 sFest
Drivation)(tf
dt
d
)(ssF
Intgration
t
dttf0
)( s
sF )(
Valeur initiale )(lim0
tft
)(lim ssFs
Valeur finale )(lim tft
)(lim0
ssFs
Computational rules
= dsesFj
tfst
)(2
1)(
[ ] == )(
s)(Res)(
2
1)(
i
sFdeples
ststesFdsesF
jtf
[ ]( )
[ ]iss
stk
ik
k
esFssds
d
ksF
=
= )()(
!1
1)(Res
1
1
si
Inverse transformation
NoteNote : is a counterclockwise contour containing the origin ot thecomplex plane and all the poles of the function F(s)(Bromwich
contour .
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TABLES
Inverse transformation
partial fraction expansion
=i
i
ps
rsF )(
=
)()(lim sFpsr i
psi
i
,
= tpi iertf )( .
)()(
)(sH
sE
sS=
Transfer functionContinuous system stability
I m
R e
1
Routh-Hurwitz criterion
1
71
6
13
1
51
4
12
1
31
2
11
=
=
=n
nn
nn
n
nn
nn
n
nn
nn
a
aa
aadtm
Aa
aa
aadtm
Aa
aa
aadtm
A
11
1411
71
23
11
1311
51
22
11
1211
31
21A
AA
aadtm
AA
AA
aadtm
AA
AA
aadtm
A
nnnnnn
=
=
=
21
2421
1411
33
21
2321
1311
32
21
2221
1211
31A
AA
AAdtm
AA
AA
AAdtm
AA
AA
AAdtm
A
=
=
=
01
1
1 ... asasasaDsoitn
n
n
n +++=
...
......
......
......
...
...
...
...
321
232221
131211
531
42
nnn
nnn
nnn
AAA
AAA
AAA
aaa
aaa
The necessary and sufficientcondition to ensure the stability of asystem is that
the coefficients of the denominatorofthe transfer function
and
the elements of the first column of theRouth table
are positive or have the same sign.
Rout-Hurwitz criterion
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Z TRANSFORMATIONNotion of sampled signal
[ ]
=
=0
)()(k
EkTtkftf
Definition
[ ] [ ]
[ ]
[ ] )00(
)(
)()()(
0
00
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TABLES
Inverse Z transform
partial fraction expansion
because
[ ]
=
=
=
1
)()(limavec
)(
k
ii
izz
i
i
i
zrkf
zFzzr
zz
rzF
i
ii
k
i
i
k
i
zzzz
zzzF
zz
zzF
=
=
=
1)(
)(
11
Inverse Z transform
Polynomial division
IfF(z) can be written as the ratio of two
polynomials and if the degree of the
numerator is lower or equal of the degree of
he denominator the polynomial division
gives :
[ ] kfkf
zfzffzDzNzF
=
+++== K
22
110
)()()(
Stability of sampled-data systems in
open loopSampling does not modify the stability of a system inSampling does not modify the stability of a system in
open loop.open loop.
A continuous system is stable in open loop when the real part of
all the poles is negative. As :
( ) ( )EiEiTTjTs
iTjTeeez EiEiiEi
sincos ===
1
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Closed loop stability :
Routh-Hurwitz criterionRH allows to determine if the roots of a polynomial have
negative real parts.
Appying to the characteristic polynomials the following
transformation : z=(1+w)/(1-w) | w=(z-1)/(z+1) one
transforms the inside part of the unit circle onto the negative
real part of the w plane.
Indeed, if :
( )
( )( ) ( )
0sin1cos
sin21
1sincos
1sincos
1
1
10
22
2
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Chapter 3
Mixed systems
Entranements lectriques
Electrical drivesProf. A. GENON, ULg
Is it possible to consider discrete
systems as if they were
continuous ?
If yes, under what conditions?
Pad approximation2/1
2/12/
2/
E
E
sT
sTsT
sT
sT
e
eez
E
E
E
+
==
We observe that the gray areas overlap satisfactorily. We
deduce that the approximation is valid for:
4/4/5,00 EE TetT
Example of a discrete regulator:
the numerical PID
Transfer function of a numerical PID
[ ] [ ] [ ] [ ] [ ]( )
)1(
/)/2()/(
)(
)()(
)(1
)(1
)()(
1
2
0
++++
==
+
+=
++= =
zz
TKzTKKzTKTKK
zE
zYzG
zEz
z
T
KzE
z
zTKzEKzY
kekeT
KieTKkeKky
EDEDPEDEIPRR
E
DEIPR
E
D
k
i
EIPR
1
)(
)(
)()(
+
==z
KzTKK
zE
zYzG PEIPRRPI :
Continuous treatment of a discrete system :
example of the numerical PID
( ) ( )
+++++
+=
+
=
++++
=
4/2/2/1
1)(
2/1
2/1
)1(
/)/2()/()(
2
2
EIEPDI
EIP
E
R
E
E
EDEDPEDEIPR
TKTKKss
KTKK
sTsG
sT
sTz
zz
TKzTKKzTKTKKzG
Hypothesis : Pad conditions are met
If we compare with an analog PID,
we obtain*
***
)(
)()( D
IP
RR Ks
s
KK
sE
sYsG ++==
4/2/ 2*
*
*
EIEPDD
II
EIPP
TKTKKK
KK
TKKK
++=
=
+=
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Q: What must be done if a continuous
system is controlled by a discrete regulator?
R: We must determine the discrete transfer
function of the continuous system ....
Discrete transfer function of the continuous
system preceded by a holding element
( )
( )( )( )
( )( )
==
=
=
=
=
==
=
=
==
i i
i
i
i
i
i
i
i
i
Tp
i
i
tp
i
i
i ii
i
i i
iS
Ts
S
Ts
Ts
EEE
zz
z
p
rzGzzG
zzz
zz
p
rzG
ez
etup
rtg
pssp
r
ps
r
ss
sGsG
sGesGs
esG
s
esGTtututg
Ei
i
E
E
E
1)()1()(
1
1)(
1)()(
111)()(
)()1()(1
)(
1)()()()(
1
1
1
1
1
1
ExampleExample : command
of a DC motor by a
C
Discrete transfer function of the continuous
system preceded by a holding element :
EXAMPLE : 2nd order filter
( )( )
( )( )
21
2211
2211
122211
/
2
/
1
21
21212121
)1()1(
)1()1(
)1()1(
)(
/1
11
/1
11
11
1)(
2
1
TT
zTzTK
zTzT
zzTzzTn
ez
ez
zzzz
nzKzG
TsTTTsTTsTsTsG
TT
TT
S
E
E
=
=
=
=
=
++
+=
++=
Analog to digital conversion
Proper sampling requires that the sampling frequency
is much higher than the highest frequency the signal tobe sampled (3 to 4 times).
If necessary, we must first filter the signal to be
sampled.
Shannon theorem
Analog to digital conversion
Quantization error[ ]
( )
( )
NSBBS
VBS
VV
qeVAR
qdVVpVqdVVpVeeVAR
dVVpVqdVVpVeeE
Vq
V
dVdVVp
dB
N
q
q
N
V
q
q
N
V
N
eff
eff
02,626,4log10/
28
3/
22
12)(
12)()(12)()()(
0)()(12)()()(
12
)dVVV,signal(Prob)(
10
2
2
max
22/3
2/
2
0
2
2/3
2/0
max
max
max,
max,
max
max
+==
=
=
=
==
===
===
=
+==
S/B= signal to noise ratio
Analog to digital conversion
Signal to noise ratio
NSBBSdB 02,626,4log10/ 10 +==
Nombre de bits Rapport signal/bruit
4 21 dB
8 45 dB
16 93 dB
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Parallel analog to digital converter
10 bits 1024 comparators !!!
very fast
Successive approximations ADC
necessits a sample and hold element
works by successive comparisons and accuracy increase with each iteration
Ve > Vref/2, bit1=1
Ve< Vref/2+Vref/4, bit2=0
Ve> Vref/2+Vref/8, bit3=1
etc
slower conversion (10us for 12 bits)
Ramp compare ADC
1. Integration of Ve during a fixed time t1-t02. Negative integration with a known voltage Vrefand measure of elapsed time
t2-t13. Relative slow, but can be very precise
01
12
tt
ttVVrefe
=
Digital to analog converter
Much easier to achieve than the reverse
+++=16842
0123 BBBBVVrefS
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Chapter 4
COMMAND CIRCUITS
Entranements lectriques
Electrical drivesProf. A. GENON, ULg
Conventional circuits
Transistor amplifier : common emitter
Conventional circuits
Transistor amplifier : common collector
Conventional circuits
Push-pull
Conventional circuits
Use of a triac
Transistor chopper command
optocoupler
comM UU #
PWM command
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Transistor chopper command
The LM555 IC
PWM command1
)2(
44.1
CRRf
BA +=
astable monostable
Transistor inverter command
2sin
221,
UU
UU
M
M
=
=
THYRISTOR command THYRISTOR chopper command
THYRISTOR rectifier command
One compares the control signal Ucom to a
sinusoidUrwhose value is maximum at the
instant of natural switching
(ex : -V2 for Th1).
Then
com
r
C
r
comC
Cc
UU
U
U
UU
UU
max
0
max
0
0
arccoscos
cos
=
=
==
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Chapter 5
Analogic controllers
Entranements lectriques
Electrical drivesProf. A. GENON, ULg
Introduction
C o m p a r a t e u r
R g u l a t e u r
C i r c u i t s d e
c o m m a n d e
C o n v e r t i s s e u r
l e c t r o n i q u e
d e p u i s s a n c e
O r g a n e s
d e m e s u r e
S y s t m e
r g l e r
+
-
U
c
The comparator and the analog controller are often combined
in one element.
The purpose of analogic controllers is to compare a reference
and a measurement and to develop a corrective signal applied
to the command circuits.
The analog controllers are mostly designed with operational
amplifiers.
The operational amplifierExample : The uA709
+
R
R
R
1
2
3
Q
Q
Q
Q
Q
Q
Q
Q
Q Q
Q
Q
Q
Q
1
2
3
4
5
6
7
8
9
1 0
1 1
1 2
1 3
1 4
C
C
+
V
E -
V
A -
V
S
V
A +
V
E +
The operational amplifier
+
U
s
U
e
-
In its field of linearity, the ideal OA has the following
characteristics:
infinite gain, therefore no differential input voltage;
infinite input impedance, therefore no input current.
P controller
KR
R
UU
UsG
RRR
emc
sR
e
==
=
+=
1
10
)(
121
PI controller
+=
+=
+=
=
=
ni
n
i
n
emc
sR
e
sTT
T
Ts
Ts
CRs
CRs
UU
UsG
RR
11
11)(
21
1
11
0
Tn=R1C1Ti=ReC1
frequency response step response
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PI controller PI
+=
+=
+=
=
=
ni
n
i
n
emc
sR
e
sTT
T
Ts
Ts
CRs
CRs
UU
UsG
RR
11
11)(
21
1
11
0
If the values ofTn=R1C1 and Ti=ReC1are known :
selectRe between 10 and 100 kohms
C1=Ti/ReR1=Tn/C1
PI controller (adjustable coefficients)
( )ss U
sCRRR
sCRR
UU
sCRRSi
++
+
=
+
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PI controller (adaptation of 1 time constant)
111
1
11
11
11)(
0/1
CRUKTCRT
Ts
Ts
CRUKs
CRssG
sCR
UUK
R
UU
UUKU
evmin
i
n
evm
R
svm
e
cm
svmi
==
+=
+=
=+
+
=
PI controller (adaptation of 2 time constants)
CRUKT
CRUKT
Ts
Ts
CRUKs
CRUKssG
UU
U
evimi
vnmn
i
n
evim
vnmR
mc
s
2
111
2
111 11)(
=
=
+=
+==
Adaptive controller with variable structure
i
n
e
RsT
sT
RR
RsCR
RRRRsC
sG+
=
+
++
=1
1
)(
12
12
122
ie
RsTCsRR
RsG
1)( 3 ==
FET1 ON
FET2 ON
Adaptive controller with variable structure
Note:The sources of the JFETs are connected to the input of the OA (virtual ground)
The diode between drain and ground prevents the drain potential to become too
high
When the JFET leads, the diode does not because UDS is very low
First order passive filter
sTsRCU
U
e
s
+=
+=
1
1
1
1
Active filter of second order
( )( ) ( )( )2121 111
14/1
1
sTsTsRCsRCU
U
e
s
++=
++=
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PID + second order active filter
( )( )( )( ) ( )[ ]
( )( )( )( )
i
vn
eeemc
s
sT
sTsT
sTsT
CsR
CRRsCsRCsRCsRCsR
CsRCsRUU
U
++
++=
++++++
++=
11
11
1
111.
14/1
1
21
1
12123122211
23
PID + reference filter
PID with clipping circuit
If one LED leads, the voltage Us is clamped as shown on
the diagram
The diodes do not
conduct if:
( )( ) 01
01
222
111
>+=
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Chapter 6
Digital controllers
Entranements lectriques
Electrical drivesProf. A. GENON, ULg
Main configurations
Classic method
Consist to use a digital PI or PID
Suitable for systems with 1 or 2 dominant
time constants and who have a well damped
oscillating behaviour
Main configurations.
State control The orders issued by the regulator depends
- from the values of the state variables of the system thatare either measured or estimated (state estimator)
- sometimes from the measurable or estimabledisturbances
The control signal (CS) is a linear combination of
quantities of orders and disturbances The SC is usually an integral component to cancel
the error in steady state
Suitable for control of all systems
Main configurations.
State control
EXAMPLE : the BOOST converter
State variables : Ue et Us
The reference is Usc ; Uem and Usm are measured
The duty cycle to impose is The integral term Ki is added, so
sc
em
U
U=1
=
1
es
UU
)(
1
1 smscii
i
sc
em
UUkKK
KU
U
+=
+=
Main configurations
Sliding mode control
Suitable for systems with a
pulse device (chopper or
inverter)
The output of the controller is
a logical variable depending
from the reference and the
state variables
Main configurations
Cascade control
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Main configurations
Trajectory pursuit control
For fine tuning (robots)
Example: if acceleration and speed are good but position
is too late, R3 acts to increase the speed
RST controlClassic control:
RST control:
More possibilities, numeric, etc
+Um
UsUcN(z)
D(z)
R(z)
S(z)
+Um
UsUcN(z)
D(z)
1T(z)
R(z)
S(z)
)()()()(
)()(
)(
)(
zSzDzRzN
zTzN
zU
zU
c
m
+=
)()()()(
)()(
)(
)(
zSzDzRzN
zRzN
zU
zU
c
m
+=
RST control : effect of disturbances
+ ++++
Um
P2P1
UsUcN(z)
D(z)
1T(z)
R(z)
S(z)
For a null static error in case of disturbance, 1/ S (z) must
contain at least one integrator, and :
)(')1()( zSzzS =
RST control : effect of disturbances
+ ++++
Um
P2P1
UsUcN(z)
D(z)
1T(z)
R(z)
S(z)
Against disturbances, the RST behaviour is the same as
its of a conventional controller
)()()()(
)()()(
)()()()(
)()()(
)()()()(
)()()()( 21
zDzSzNzR
zDzSzP
zDzSzNzR
zNzSzP
zDzSzNzR
zNzTzUzU cm
++
++
+=
RST control : causality+ +
+++Um
P2P1
UsUcN(z)
D(z)
1T(z)
R(z)
S(z)
tztztztzT
rzrzrzrzR
szszszzS
zUzRzUzTzUZS mcs
++++=
++++=
++++=
=
......)(
......)(
......)(
)()()()()()(
2
2
1
10
2
2
1
10
2
2
1
1
( ) ( )( ) )(......
)(......)(......1
2
2
1
10
2
2
1
10
2
2
1
1
zUzrzrzrzr
zUztztztztzUzszszs
m
cs
++++
++++=++++
[ ] [ ] [ ] [ ][ ] [ ] [ ] [ ][ ] [ ] [ ] [ ]
++++++
+++++++++++
+=+
1......211
1......211
1.....11
210
210
21
kUrkUrkUrkUr
kUtkUtkUtkUt
kUskUskUskU
mmmm
cccc
ssss
often ==
RST control : causality+ +
+++Um
P2P1
UsUcN(z)
D(z)
1T(z)
R(z)
S(z)
( )
tztztztzT
rzrzrzrzR
szszszzS
zUzRzUzTzUZS
zUzRzUzTzS
zU
mcs
mcs
++++=
++++=
++++=
=
=
......)(
......)(
......)(
)()()()()()(
)()()()()(
1)(
2
2
1
10
2
2
1
10
2
2
1
1
[ ] [ ] [ ] [ ][ ] [ ] [ ] [ ][ ] [ ] [ ] [ ]
++++++
+++++++++++
+=+
1......211
1......211
1.....11
210
210
21
kUrkUrkUrkUr
kUtkUtkUtkUt
kUskUskUskU
mmmm
cccc
ssss
1
1
+
+
If at position k+1, Uc and/or Um are not available :
often 1==
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RST control : Closed loop transfer fuction
The closed loop transfer function may be selected with some
liberties but :
- avoid to delete zeros of N (z) too close to the unit circle
- meet the following condition :
+Um
UsUcN(z)
D(z)
1T(z)
R(z)
S(z)
)(
)(
)()()()(
)()(
)(
)(
zD
zN
zSzDzRzN
zTzN
zU
zU
m
m
c
m =+
=
[ ] [ ] [ ] [ ])()()()( zNzDzNzD mm
RST control : Closed loop transfer fuction+
Um
UsUc
N(z)
D(z)
1T(z)
R(z)
S(z)
1)1(
)1(=
m
m
D
N
)1/()( = zzzUc
1)1(
)1(
)(
)(
1)1(lim)(
1
==
=
m
m
m
m
zm
D
N
zD
zN
z
zzU
If a null static error against the reference is suited,
the following condition must be met :
Indeed, if UC (z) isa unit step :
RST control : Summary+
UmUsUc
N(z)
D(z)
1T(z)
R(z)
S(z)
)(
)(
)()()()(
)()(
)(
)(
zD
zN
zSzDzRzN
zTzN
zU
zU
m
m
c
m =+
=
== 1== or
)(')1()( zSzzS = if a null static error against disturbances is wished
[ ] [ ] [ ] [ ])()()()( zNzDzNzD mm
1)1(
)1(=
m
m
D
N
ATTENTION : avoid to delete badly damped zeros
if a null static error against reference is wished
RST control : synthesis example
Consider a small motor with a transfer
function between the reference and the
speed is :
( )( )( )45.1264.18
29.1445.20)(
++
+=
ss
ssG
s
It is proposed to control this engine with a digital controller having a sampling period of
100ms.
It is proposed to use a RST controller with a zero static error against the disturbances and
command.
In closed loop, a time constant of 50ms is wished
REMARKS
The transfer function has the following characteristics:
A first time constant of 54ms (time constant electric)
A second time constant of 690ms (mechanical time constant)
A zero corresponds to a delay of 775ms
We note that the sampling period is not negligible against the time constants of the system.
RST control : synthesis example
( )( )( ) ( ) ( )45.1
195.0
264.18
640.20
45.1264.18
29.1445.20)(
+
+=
++
+=
ssss
ssG
s
865.045.1195.0
161.0264.18640.20
1
1
222
111
====
====
E
E
Tp
Tp
ezpr
ezpr
( )( )( ) )(
)(
161.0865.0
878.0930.0
865.0
018.0
161.0
948.01)(
zD
zN
zz
z
zzzz
z
p
rzG
i i
i
i
i =
=
=
=
Transfer function of the engine preceded by a holding element
RST control : synthesis example+
Um
UsUc
N(z)
D(z)
1
T(z)
R(z)
S(z)
A first-order behaviour is wanted (possible because the difference of
degrees between the numerator and denominator in OL is equal to 1).
Moreover, the zero of the numerator OL (0.878) is sufficiently distant
from the unit circle so it can be removed safely:
am
m
c
m
zz
a
zD
zN
U
U
==
)(
)(
( )( )( )161.0865.0878.0930.0
)()(
=zz
zzDzN
Electrical Drives 2010 - A.GENON 17
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4
RST control : synthesis example
135.045.0/1.0/ ===
eezTBFT
aE
135.0)(
)(
==z
a
zD
zN
U
U
m
m
c
m
A time constant of 450 ms is expected
135.01
)1(
)1(
==z
a
D
N
m
m 865.0=a135.0
865.0
)(
)(
==zzD
zN
U
U
m
m
c
m
A null static error against the reference is expected
.
+Um
UsUcN(z)
D(z)
1T(z)
R(z)
S(z)
RST control : synthesis example
+Um
UsUc
N(z)
D(z)
1T(z)
R(z)
S(z)
)()()()(
)()(
)(
)(
zDzSzNzR
zNzT
zD
zN
U
U
m
m
c
m
+==
( )
( )( )161.0865.0
878.0930.0
)(
)(
=
zz
z
zD
zN
( )( ) ( )( ) 135.0
865.0
161.0865.0)(878.0)(930.0
878.0)(930.0
=
+
=
zzzzSzzR
zzT
U
U
c
m
To eliminate the numerators zero and to have a zero static error against
disturbances, S(z) must have the following form :
( )( )
878.0
878.1
1
878.01)(
2
1
0
=
=
=
=
s
s
s
zzzS
21
2
0
21
2
0
)(
)(
rzrzrzR
tztztzT
++=
++=
( )( ) ( )( ) 135.0
865.0
161.0865.0)1(930.0
930.0
21
2
0
21
2
0
=
+++
++=
zzzzrzrzr
tztzt
U
U
c
m
( )( ) ( )( ) ( ) 2
2
21
2
0
21
2
0
135.0
865.0
161.0865.0)1(930.0
930.0
zz
z
zzzrzrzr
tztzt
U
U
c
m
=
+++
++=
RST control : synthesis example
+Um
UsUcN(z)
D(z)
1T(z)
R(z)
S(z)
( )( ) ( )( ) 135.0
865.0
161.0865.0)1(930.0
930.0
21
2
0
21
2
0
=
+++
++=
zzzzrzrzr
tztzt
U
U
c
m
By identifying the coefficients :
( )( ) ( )( ) ( ) 22120
221
20
135.0161.0865.0)1(930.0
865.0930.0
zzzzzrzrzr
ztztzt
=+++
=++
049.10
253.10
033.2930.0
22
11
00
==
==
==
rt
rt
rt
RST control : synthesis example
[ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ]111111 21021021 +++++=+ kUrkUrkUrkUtkUtkUtkUskUskU mmmcccsss
+Um
UsUcN(z)
D(z)
1T(z)
R(z)
S(z)
[ ] [ ] [ ] [ ] [ ] [ ] [ ]1049.1253.11033.21215.01878.0878.11 ++++=+ kUkUkUkUkUkUkU mmmcsss
049.10
253.10
033.2930.0
22
11
00
==
==
==
rt
rt
rt
878.0
878.1
1
2
1
0
=
=
=
s
s
s
Numerical command circuits
Microprocessor :
fas t (3GHz)
32 to 128 bits logic
requires external devices
expensive
Microcontroller
contains CPU + itsminimum environment
autonomy
relatively slow (10MHz, 8 bit)
inexpensive
Numerical command circuits
DSP (Digital Signal Processor)
Evolved microcontroller, rapidity
Contains special functions (FFT, ...)
FPGA (Field Programmable Gate Array)
Circuit with logic functions which may be interconnected byprogramming (VHDL = Very High Speed Description Language)
ASIC (Application Specific Integrated Circuit)
For large series
Electrical Drives 2010 - A.GENON 18
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1
Chapter 7
Electrotechnical applications
of regulators
Entranements lectriques
Electrical drivesProf. A. GENON, ULg
Cascade control
Rectifiers 4 quadrants AC/DC controller with
current circulation
4 quadrants AC/DC controller with
current circulation
4 quadrant AC/DC controller without current
circulation
LOGICLOGIC : the 2 bridges may never
conduct at the same time dead time
Electrical Drives 2010 - A.GENON 19
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2
1 quadrant chopper with 2 position controller1 quadrant chopper with PWM controller
4 quadrant chopper with PWM controller
7
1
6
!
6
"
6
*)
6
8
-
1
J
J
7
J
J
J
J
8
)
J
8
*
J
6
@ 6
,
!
6
"
,
"
6
!
,
!
6
"
,
6
,
6
,
6
VVVU )12()1( ==
(other command modes exist)
Inverter with 2 position controller
Inverter with PWM controller
Electrical Drives 2010 - A.GENON 20
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1
Chapter 8
Command of switched power supplies
Entranements lectriques
Electrical drivesProf. A. GENON, ULg
Boost converter command
v
o
i
o
i
D
i
L
v
i
C
R
i
T
L
D
If the switching frequency is sufficiently high, the system can be studied as a
continuous process although command is discreet
[ ]
+=
+=
))(1)(()(1
))(1)(()(1
tgR
vitg
R
v
Cdt
dv
tgvvtgvLdt
di
oi
oo
oiiL g=1 : switch ON
g=0 : switch OFF
To study this kind of system, we proceed by averaging over one sample period :
=t
TtdttX
TtX )(
1)(
Boost converter command
v
o
i
o
i
D
i
L
v
i
C
R
i
T
L
D
[ ]
+=
+=
))(1)(()(1
))(1)(()(1
tgR
vitg
R
v
Cdt
dv
tgvvtgvLdt
di
oi
oo
oiiL
=t
TtdttX
TtX )(
1)(+
[ ]
+
+=
=
C
i
RC
v
C
tgi
RC
v
dt
vdL
v
L
vtgvv
Ldt
id
LoLoo
oioi
L
)1())(1(
)1())(1(
1
v
i
( 1 - @ )
v
o
i
o
i
L
( 1 - @ )
i
L
v
i
C
R
L
Boost converter command[ ]
+
+=
=
C
i
RC
v
C
tgi
RC
v
dt
vdL
v
L
vtgvv
Ldt
id
LoLoo
oioi
L
)1())(1(
)1())(1(
1
v
i
( 1 - @ )
v
o
i
o
i
L
( 1 - @ )
i
L
v
i
C
R
L
This model is not linear.
Thus, we proceed to the linearization in the vicinity of a static equilibrium point:
+=
+=
+=
+=
~
~
~
~
(
(
(
(
LLL
ooo
iii
iii
vvv
vvv
R
vi
vv
oL
io
(
((
(
(
(
=
=
)1(
1
+=
+
=
C
i
C
i
RC
v
dt
vdL
v
L
v
L
v
dt
id
LLoo
ooiL
~)1(
~~~
~)1(~~
~
((
(
(
By eliminating the static current between these relationships and assuming
constant vi, we obtain the transfer function:
LCs
RCs
LC
v
C
is
voL
o
22 )1(1
)1(
~
~
(
(
(
(
++
=
+
Boost converter command
This transfer function has 1 positive zero and 2 complex poles generally weakly
damped.
This system is very difficult to regulate.
Therefore generally one prefer to use a current control rather than a duty cycle
control.
LCs
RCs
LC
v
C
is
voL
o
22 )1(1
)1(
~
~
(
(
(
(
++
=
v
o
i
o
i
D
i
L
v
i
C
R
i
T
L
D
Current control
principle
Let T be the period of hash and the conduction time. The switch is closed at the
beginning of the period and the current increases. The switch is opened when the
current reaches the value ip-mc .
The slope mc is chosen to ensure stable operation.
v
o
i
o
i
D
i
L
v
i
C
R
i
T
L
D
iL
T=
ip -mc
m1 -m2
0
To ensure stable operation, it is necessary that the
current is amortized if any disturbance happens
=
>
1
12
22112
mmmmc
mc is not necessary if 0,5
Electrical Drives 2010 - A.GENON 21
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2
Current control
Transfer function
v
o
i
o
i
D
i
L
v
i
C
R
i
T
L
DThe transfer function between the outputvoltage and the current iL can be found after
local averaging and linearization of the
equations:iL
T=
ip -mc
m1-m2
0
( )( )csbsas
Ki
v
p
o
++
=~
~
This transfer function has a positive zero and two real poles (one at low frequency
and one at higher frequency).
Under these conditions, the system is much easier to
control
Current control
The IC LM5021
Example : forward converter
with current control
Electrical Drives 2010 - A.GENON 22
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1
Chapter 9
COMMAND OF INDUCTION MOTORS
Entranements lectriques
Electrical drivesProf. A. GENON, ULg
Speed variation history.
1. Ward-Lonard
2. DC motor drives
3. AC motor drives
Dynamics of induction motor :
3 phase model
For each coil :
iiiidt
dIRU +=
+=6
ijjijiii IMIL
Dynamics of induction motor :
3 phase model
;
;
.
;
;.
=
xyz
XYZ
RtSR
SRS
xyz
XYZ
I
I
ZZ
ZZ
U
U
3,3
33;;;;
=
=
=
=
z
y
x
xyz
Z
Y
X
XYZ
z
y
x
xyz
Z
Y
X
XYZ
I
I
I
I
I
I
I
I
U
U
U
U
U
U
U
U
+
+
+
=
SSSS
SSSS
SSSS
S
sLRsMsM
sMsLRsM
sMsMsLR
Z
3333
3333
3333
3
++
+
=RRRR
RRRR
RRRR
R
sLRsMsMsMsLRsM
sMsMsLR
Z
3333
3333
3333
3
+
+
+
=
)cos()3/2cos()3/2cos(
)3/2cos()cos()3/2cos(
)3/2cos()3/2cos()cos(
333
333
333
3
RSRRSRRSR
RSRRSRRSR
RSRRSRRSR
SR
sMsMsM
sMsMsM
sMsMsM
Z
Dynamics of induction motor : 2 phase model
.
xyzt
XYZtAB
ICI
ICI
*
*
=
=
=
2/3
2/3
0
2/1
2/1
1
3
2*C
Dynamics of induction motor : 2 phase model
.
xyzt
XYZtAB
ICI
ICI
*
*
=
=
=
2/3
2/3
0
2/1
2/1
1
32*C
( )( )( )
( )
( )
( )( )( )
( )
( )tgII
tgII
tgII
tgII
tgII
tII
tII
tII
tII
tII
MaxR
MaxR
MaxRz
MaxRy
MaxRx
MaxSB
MaxSA
MaxSZ
MaxSY
MaxSX
cos2
3
sin2
3
240sin
120sin
sin
cos2
3
sin2
3
240sin
120sin
sin
,
,
,
,
,
,
,
,
,
,
=
=
+=
+=
=
=
=
+=
+=
=
Electrical Drives 2010 - A.GENON 23
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2
Induction motor : axes related to magnetic field
=
=
B
A
tS
B
A
aa
aa
qs
ds
I
IP
I
I
I
I,
)cos()sin(
)sin()cos(
=
=
I
IP
I
I
I
ItR
RaRa
RaRa
qr
dr
,)cos()sin(
)sin()cos(
Induction motor : axes related to magnetic field
=
=
B
A
tS
B
A
aa
aa
qs
ds
I
IP
I
I
I
I,
)cos()sin(
)sin()cos(
=
=
I
IP
I
I
I
ItR
RaRa
RaRa
qr
dr
,)cos()sin(
)sin()cos(
MaxRqr
MaxRdr
MaxSqs
MaxSds
II
II
II
II
,
,
,
,
2
3
2
3
2
3
2
3
=
=
=
=
( ) tgt
t
Ra
a
==
=
0
Induction motor : axes related to magnetic field
=
qr
dr
qs
ds
r
s
RtSR
SRS
t
t
tr
ts
qr
dr
qs
ds
I
I
I
I
P
P
C
C
ZZ
ZZ
C
C
P
P
U
U
U
U
0
0
0
0
0
0
0
0
3,3
33
,
,
Induction motor : axes related to magnetic field
=
qr
dr
qs
ds
r
s
RtSR
SRS
t
t
tr
tsqs
ds
I
I
I
I
P
P
C
C
ZZ
ZZ
C
C
P
PU
U
0
0
0
0
0
0
0
0
0
03,3
33
,
,
qrdrR
qsdsS
qsdsS
jIII
jUUU
jIII
+=
+=
+=En posant:
On obtient
Le couple vaut dsqrqsdr IMIIMIC =
dt
dgjIR
dt
djIRU
RRRR
S
SSSS
++=
++=
0
RRSR
RSSS
ILIM
IMIL
+=
+=
Induction motor : axes related to magnetic field
+
+
=
=
=
qs
ds
a
a
a
a
a
a
qs
ds
S
B
A
Z
Y
X
I
I
I
IPC
I
IC
I
I
I
)3/2sin(
)3/2sin(
)sin(
)3/2cos(
)3/2cos(
)cos(**
+
+
=
=
=
qs
ds
a
a
a
a
a
a
qs
ds
S
B
A
Z
Y
X
U
U
U
UPC
U
UC
U
U
U
)3/2sin(
)3/2sin(
)sin(
)3/2cos(
)3/2cos(
)cos(**
Induction motor
In steady state conditions, we find the classical
equations of the equivalent circuit
)( MLX SS =
)( MLXRR =
MX =
Electrical Drives 2010 - A.GENON 24
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3
Command of the induction motor
GENERAL STRATEGY :
We try to maintain constant and maximum the flux in the
machine because the torque is proportional to flux.
To act on the couple, we act on the stator current because :
),sin(.. IIkC =
Command of the induction motor
Four techniques are considered:
1. Scalar control on voltage
2. Vector control on voltage
3. Scalar control on current
4. Vector control on current
Scalar control on voltage
( )
( ) ( )),(
1
11
2
22222
2
22
22
SS
sRs
R
RR
SS
SSS
RR
SS
SSSSSS
gfMLLgLR
RMgpC
LgjR
MgLjR
RU
LgjR
MgLjR
R
j
UIRU
j
R
=+
=
+++
=
+++
==
Scalar control on voltage : simplified version
c
a
a
a
ctt
Z
Y
X
UUCP
U
U
U
+
==
)3/2cos(
)3/2cos(
)cos(
Scalar control on voltage :
action on couple
( ) ( )),(
)3/2cos(
)3/2cos(
)cos(
2
22222
2
SS
sRs
R
c
a
a
a
ctt
Z
Y
X
gfMLLgLR
RMgpC
UUCP
U
U
U
R
=+
=
+
==
Scalar control on voltage :
sophisticated version
( ) ( )),(
)3/2cos(
)3/2cos(
)cos(
2
22222
2
SS
sRs
R
c
a
a
a
ctt
Z
Y
X
gfMLLgLR
RMgpC
UUCP
U
U
U
R
=+
=
+
==
Electrical Drives 2010 - A.GENON 25
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4
Scalar control on voltage :
speed control
( ) ( )),(
)3/2cos(
)3/2cos(
)cos(
2
22222
2
SS
sRs
R
c
a
a
a
ctt
Z
Y
X
gfMLLgLR
RMgpC
UUCP
U
U
U
R
=+
=
+
==
Vector control on voltage
( )
( ) ( )),(
1
11
2
22222
2
1
22
22
SS
sRs
R
RR
SS
SSqsdsS
RR
SS
SSSSSS
gfMLLgLR
RMgpC
LgjR
MgLjR
RjjUUU
LgjR
MgLjR
R
j
UIRU
j
R
=+
=
+++
=+=
+++
==
Vector control on voltage
+
+
==
qs
ds
a
a
a
a
a
a
ctt
Z
Y
X
U
UUCP
U
U
U
)3/2sin(
)3/2sin(
)sin(
)3/2cos(
)3/2cos(
)cos(
Scalar control on current
( ) ( )),(
2
22222
2
12
2
SS
sRs
R
S
RR
SS
S
RR
SRSSS
gfMLLgLR
RMgpC
LgjR
MgjLI
ILgjRMgjLIMIL
R
=+
=
+=
+=+=
Scalar control on current Scalar control on current
Electrical Drives 2010 - A.GENON 26
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5
Scalar control on current
t ( s e c )
C o u p l e
21 , 510 , 5
2
1 , 5
1
0 , 5
0
Vector control on current
dt
dgjIR
dt
djIRU
R
RRR
S
SSSS
++=
++=
0
RRSR
RSSS
ILIM
IMIL
+=
+=
dsqrqsdr IMIIMIC =
In an axes system related to the rotor flux (d in the direction of the rotor flux):
( )dtg
gMR
LIC
gMR
LI
R
Ls
MI
a
R
RRqsR
R
RRqs
R
RRds
+=
==
=
+=
2
)1(
Vector control on current
( )dtg
I
I
I
I
I
a
qs
ds
a
a
a
a
a
a
Z
Y
X
+=
+
+
=
)3/2sin(
)3/2sin(
)sin(
)3/2cos(
)3/2cos(
)cos(
Vector control on current
( )dtg
I
I
I
I
I
a
qs
ds
a
a
a
a
a
a
Z
Y
X
+=
+
+
=
)3/2sin(
)3/2sin(
)sin(
)3/2cos(
)3/2cos(
)cos(
To reach overspeed
conditions, one reduces the
flux
Electrical Drives 2010 - A.GENON 27
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1
Chapter 10
Preparing the lab.
Controlling a DC machine.
Entranements lectriques
Electrical drivesProf. A. GENON, ULg
Transfer function of a DC motor with
independent excitation
( )
IkC
CIksJN
NkIsLRU
r
=
=++=
k
1
s J
1 N
Cr
CIU
R(1 + s Ta)k
+-
ou :
( ) ( )
( )r
a
CCsJ
N
IkC
NkUsTR
NkULsR
I
=
=
+
=+
=
1
)1(
11
Transfer function of a DC motor with
independent excitation
k
1
s J
1 N
Cr
CIU
R(1 + s Ta)k
+-
( ) ( ) ra
aC
k
sTR
k
U
k
RsTsJN
22
)1(1)1(
+
=
+
+
( )
=
2k
RJTm
( )( ) r
aam C
k
sTR
k
UsTsTN
2
)1()1(1
+
=++
We have also :
Principle of cascade control
Niref
+-
+-
Nrefmoteurrgulateur
de vitessergulateurde courant
convertisseur
I
General scheme
Niref
+-
+-
Nrefmoteurrgulateur
de vitessergulateurde courant
convertisseur
I
Technics
To determine the parameters of the regulators,
there exists 2 methods:
analytical method based on model knowledge
experimental method
Electrical Drives 2010 - A.GENON 28
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2
ANALYTICAL METHOD FOR
DETERMINING THE
REGULATORS PARAMETERS
Current regulator (1)
K
m i
1 + s T
m i
K
c o n v
1 + s T
c o n v
1
IU
K
i
1 + s T
i
s T
i
i
r e f
R ( 1 + s T
a )
+
-
Current regulator (2)
mi
mi
aconv
conv
i
ii
ref
mes
sT
K
sTRsT
K
sT
sTK
I
I
+++
+=
1)1(
1
1
1
amiconvcmi TTTT
8/2/2019 Slides Electrical Drives E
30/32
3
Current regulator (6)
22221
1
cmicmiref
mes
TssTI
I
++=
If we apply a step at system input, we can observe
an overshoot of4,3% a rise time of 4,7*Tcmi
K
m i
1 + s T
m i
K
c o n v
1 + s T
c o n v
1
IU
K
i
1 + s T
i
s T
i
i
r e f
R ( 1 + s T
a )
+
-
Current regulator (6b)
Time
0s 2ms 4ms 6ms 8ms 10ms 12ms 14ms 16ms 18ms 20ms
V(LAPLACE4:OUT) V(V13:+)
0V
0.2V
0.4V
0.6V
0.8V
1.0V
1.2V
Behaviour of the closed current loop
overshoot 4,3%
rise time : 4,7*Tcmi= 4,7ms
Current regulator (7)
22221
1
cmicmiref
mes
TssTI
I
++=
Subsequently, we will approximate this transfer function with:
cmimiref sTKI
I
21
11
+=
K
m i
1 + s T
m i
K
c o n v
1 + s T
c o n v
1
IU
K
i
1 + s T
i
s T
i
i
r e f
R ( 1 + s T
a )
+
-
Speed regulator (1)
Speed regulator (2)
mncmic
c
mn
min
nn
mn
mn
cmimin
nn
ref
mes
TTTavec
sT
K
sJ
k
KsT
sTK
sT
K
sJ
k
sTKsT
sTK
N
N
+=
+
+
+
+
+=
2
1
11
121
111
min
min
In OL :
(by grouping small time constants)
Speed regulator (3)
To have a good performance in rejecting disturbances,
one choose ususaly :
min1
11
c
mn
min
nn
ref
mes
sT
K
sJ
k
KsT
sTK
N
N
+
+
mnn
mimn
cn
RKT
KTkK
TT
2
4 min
=
=
Electrical Drives 2010 - A.GENON 30
8/2/2019 Slides Electrical Drives E
31/32
4
Speed regulator (4)
min1
11
c
mn
min
nn
ref
mes
sT
K
sJ
k
KsT
sTK
N
N
+
+
mnn
mimn
cn
RKT
KTkK
TT
2
4 min
=
=
3
min
32
min
2
min
min
8841
41
ccc
c
ref
mes
TsTssT
sT
N
N
+++
+=
In CL
As response to a step at the input, we observe:
an overshoot of 43%
a rise time of 3.2 * Tcmi(valid if the mechanical time constant is muchhigher thanother time
constants)
Speed regulator (4b)
Overshoot : 43%
Rise time : 3,2*Tcmin= 3,2ms
Time
0s 2ms 4ms 6ms 8ms 10ms 12ms 14ms 16ms 18ms 20ms
V(R18:1) V(V13:+)
0V
0.5V
1.0V
1.5V
Closed Speed loop
Speed regulator (5)
3
min
32
min
2
min
min
8841
41
ccc
c
ref
mes
TsTssT
sT
N
N
+++
+=
In CL
As response to a step, we observe :
an overshoot of 8%
a rise time equal to 7,5*Tcmi
To reduce the overshoot (due to the zero in the
numerator), we can add a reference filter :
min41
1)(
csTsF
+=
3
min
32
min
2
min 8841
1
cccref
mes
TsTssTN
N
+++=
Speed regulator (5b)
Closed speed loop with reference filter
Overshoot : 8%
Rise time : 7,5*Tcmi = 7,5 ms
Time
0s 2ms 4ms 6ms 8ms 10ms 12ms 14ms 16ms 18ms 20ms
V(LAPLACE5:OUT) V(V13:+)
0V
0.2V
0.4V
0.6V
0.8V
1.0V
1.2V
EXPERIMENTAL METHOD FOR
THE DETERMININATION OF
REGULATORS PARAMETERS
Reminder: transfer function of a
PID
++= v
n
p sTsT
KsH1
1)(
Electrical Drives 2010 - A.GENON 31
8/2/2019 Slides Electrical Drives E
32/32
Critical oscillations method (1)
Initialy, the system is in CL with :
0
v
n
p
T
T
Klow
Kp is increased gradually until the limit of
oscillation. At this point, we note :
periodnoscillatio
pointat this
=
=
crit
pcrit
T
KK
pv
n
p KsTsT
KsH
++=
11)(
Critical oscillations method (2)
The controller parameters are given in the
following table (Ziegler-Nichols method) :
Kp Tn Tv
PcritK50.0
PIcritK45.0 critT85.0
PIDcrit
K59.0crit
T50.0 critT12.0
Response to a step (1)
The system is initialy in OL : a step K is applied
at the system input and the response is observed.
Tu,Tg and Ks are measured
Response to a step (2)
Coefficients following Ziegler-Nichols :
Kp Tn Tv
P
us
g
TK
T
PI
us
g
TK
T90.0
uT33.3
PID
us
g
TK
T2.1
uT2
uT5.0
Response to a step (3)
Coefficients following Chien, Hrones and Resewick :
Critre de qualit
Rgulation apriodique de trs
courte dure
Rgulation avec dpassement
de 20%
Rejet de
perturbation
Suivi de
consigne
Rejet de
perturbation
Suivi de
consigne
P
us
g
pTK
TK 3.0=
us
g
pTK
TK 3.0=
us
g
pTK
TK 7.0=
us
g
pTK
TK 7.0=
PI
un
us
g
p
TT
TK
TK
4
6.0
=
=
un
us
g
p
TT
TK
TK
2.1
35.0
=
=
un
us
g
p
TT
TK
TK
3.2
7.0
=
=
un
us
g
p
TT
TK
TK
=
= 6.0
PID
uv
un
us
g
p
TT
TT
TK
TK
42.0
4.2
95.0
=
=
=
uv
un
us
g
p
TT
TT
TK
TK
5.0
60.0
=
=
=
uv
un
us
g
p
TT
TT
TK
TK
42.0
2
2.1
=
=
=
uv
un
us
g
p
TT
TT
TK
TK
47.0
35.1
95.0
=
=
=
Electrical Drives 2010 - A.GENON 32