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Transcription

Figure
3.2: P ID con
3.2: P IDfrom
controller:
points
to plane.
Figure 3.2: PFigure
ID controller:
points tofrom
plane.
nfractional
arecommend
wide
class the
of
oP
For points
a wide
of
controlled
objectsFor
wethe
Dδ
a wide class
of
controlled
objects
we recommend
fractional
P Icontrolled
Figure 3.2: P IDFor
controller:
from
toclass
plane.
CHAPTER 3. FRACTIONAL-ORDER CONTROLLERS
13
δ which is a particular ca
δ
controller,
controller,
which is case
a particular
of P I λ Dwhere
controller,
controller, which
is a particular
of P I λ Dcase
controller,
λ = n, where
n∈ λ=n
n δ N and
∈ R. Integer
order inte
δ ∈ order
R. the
Integer
order
is
for error
steady-state
N andobjects
δ ∈N
R.
Integer
integrator
isIintegrator
important
forδimportant
steady-state
D
For a wide class of controlled
weand
recommend
fractional
P
cancellation
on the
other
hand
but on
thethe
other
hand
fractional
integral
is also
cancellation
but
other
hand
integral
is alsobut
important
for importa
controller, which is a particular
case ofcancellation
P I λon
D δthe
controller,
where
λ fractional
= n,
n ∈the
N and δ ∈ R. Integer order
integrator
is important
fortransfer
steady-state
errorresponse
obtaining
aresponse
Bode’s
loop
trans
obtaining
a Bode’s
ideal
loop
function
with
constantideal
phase
obtaining
a Bode’s
ideal loop
transfer
function
with
constant
cancellation but on the other
hand for
thedesired
fractional
integral
is also
important
margin
range
(e.g.
[1,range
3, for
23,(e.g.
51]).[1, 3,
margin
for23,
desired
marginfrequency
for
desired
frequency
51]). frequency range
obtaining a Bode’s ideal loop transfer function response with constant phase
Figure 3.2: P ID controller: from points to plane.
margin for desired frequency range (e.g. [1, 3,H.W.
23, Bode,
51]). Network Analysis and Feedback Amplifier Design.
Bode’s ideal loop transfer function
Fractional order
systems and controllers
3.2.2
ideal shape
the shape
loop transfer
function
in his
work
Bodeansuggested
an of
ideal
of the
loop
transfer
function
in his of
wo
Bode
suggested
an
idealonshape
Bode’s ideal Bode
loopsuggested
transfer
function
ideal
loopIdeal
transfer
function:
design of feedback
amplifiers
in amplifiers
1945.
transfer
function
hasamplifiers
form:
designBode’s
of feedback
inloop
1945.
Ideal of
loop
transfer
function in
has19fo
design
feedback
Bode suggested an ideal shape of the loop transfer function in his
work
on
!
"α
!
"α
s
s
design of feedback amplifiers
in 1945. Ideal loop transfer function
, =
(3.6)
L(s) = has form:
3.2.2 Bode’s ideal loop transfer function
,
L(s)
L(s)
!
"α
ω
gc
ωgc
Bode suggested an ideal shape
s of the loop transfer function in his work on
design ofL(s)
feedback=amplifiers in 1945.
Ideal
loop
transfer
function
has
form:
,
(3.6)
Hendrik
Wade
Bode
frequency and α is
slope
the
ideal
where ωgc iswhere
"α
ωdesired
gc
and
slopecut-off
of
the ideal
c
ω!gccrossover
is isdesired
crossover
freq
where
ωof
s is desired crossover frequency
(1905-1982)
gc α
,
(3.6)
L(s) =
ωgc
characteristic.
characteristic.
characteristic.
frequency
and
α
is
slope
of
the
ideal
cut-off
where ωgc is desired crossover
where ωgc is desired crossover frequency and α is slope of the ideal cut-off
Phase marginPhase
is Φmmargin
= π(1 +
all+values
The
amplitude
isα/2)
Φm =forπ(1
α/2) of
forthe
all gain.
values
of the
characteristic.
characteristic.
Phase
margin
is Φgain.
π(1 +amp
α/
m = The
o
o
Phase margin
= infinity.
π(1 + α/2) for all
values
of the gain. The
amplitude
The
constant
phase
marginphase
60o , 45margin
and 30
margin
AismΦfor
mis
ocorrespond
o
Phase margin is Φm =margin
π(1
+
α/2)
all
values
of
the
gain.
The
amplitude
is
infinity.
The
constant
60
,
45o and
corre
margin
A
o
o
o
m
is
infinity.
The30
constan
margin
A
Am is infinity. The constant phase margin 60 , 45 and 30 correspond
m
the αslopes
α margin
=and−1.33,
o and −1.66.
to to
the
slopes
= −1.33,
−1.66.60o−1.5
constant
phase
and
30o correspond
margin Am is infinity. The
to−1.5
the
slopes ,α45=
−1.33,
−1.5 and −1.66.
to the slopes α = −1.33, −1.5 and
The Nyquist curve for ideal Bode transfer function is simply a straight line
Nyquist
curveαπ/2
for(seeideal
Bode transfer function is simply a straight line
to the slopes α = −1.33, through
−1.5 The
and
−1.66.
the
origin
with arg(L(jω))
e.g. curve
[1, 39]). for ideal Bode transfer function is simply a straig
The= Nyquist
The Nyquist curve for ideal Bod
Bode’s transfer
function
(3.6) can
be used
as a reference system
in the (see e.g. [1, 39]).
through
the
origin
with
arg(L(jω))
=
απ/2
The Nyquist curve forfollowing
ideal form:
Bode transfer
function
is simply
straight line = απ/2 (see e.g. [1,
through
the origin
witha arg(L(jω))
39]). arg(L(jω))
through the origin4 with
K function (3.6) can be used as a reference system in the
Bode’s
transfer
(0 [1,
< α <39]).
2)
Gc (s) = (see
through the origin with arg(L(jω)) = απ/2
e.g.transfer
Bode’s
function(3.7)
(3.6) can be
used transfer
as a reference
system
sα + K
Bode’s
function
(3.6)
following
form:
K
Bode’s transfer function
(3.6) can
be
used
a< αreference
system(3.8)in the
Go (s)
= α as(0
< 2),
following
form:
s
following
form:
K
following form:
(s) is transfer function in(0 <K
where Gc is transfer function of closed loop
α
<
2)
(3.7)
Gcand
(s)Go=
K
+GKcare:
(0 < α < 2)
(s) = α
K characteristics of Bode’s ideal transfersαfunction
open loop. General
Gc (s) = α
(0 < α < 2)
(3.7)
Gc (s) = α
s +K
s +
K
s +K
Go (s) = α
(0 < αK< 2),
(3.8)
K
K
s Go (s) = α
(0 < α < 2),
Go (s) =
Go (s) = α
(0 < α < 2),
(3.8)
s
s
where sGc is transfer function of closed loop and Go (s) is transfer function in
transfer
function
of closed
loopGand
Go (s) is function
transfer funct
where
GcGiso (s)
of loop.
closedGeneral
loop and
is transfer
function
in where
where Gc is transfer function
is transfer
of cl
open
characteristics
of Bode’s
ideal
transfer
are:
cfunction
open ideal
loop.transfer
General
characteristics
ofopen
Bode’s
ideal
transfer
function are:
open loop. General characteristics of Bode’s
function
are:
loop.
General
characteristics
3.2.2
1
Bode’s
loopideal
transfer
3.2.2 ideal
Bode’s
loopfunction
transfer
3.2.2 function
Bode’s ideal loop
δ
For a wide class of controlled D.
objects
recommend Company,
the fractional Inc.,
P I n DNew
Van we
Nostrand
York, 1945.
controller, which is a particular case of P I λ D δ controller, where λ = n, n ∈
N and δ ∈ R. Integer order integrator is important for steady-state error
cancellation but on the other hand the fractional integral is also important for
margin for desired frequency range (e.g. [1, 3, 23, 51]).
Fractional order systems and
controllers
Fractional-order
systems
Bode’s ideal
loop transfer function
W(s)
W(s)
W(s) +
E(s)
U(s)
Gc(s)
G(s) =
Y(s)
Gs(s)
βn
an s
an D y (t ) + an−1D
• Crossover frequency: a function of K;
π
2
Start
!!
!
β0
y (t ) + ! + a0 D y (t ) = u (t )
14
General characteristics
of the Bode’s
ideal line
transfer
function:
• Phase: horizontal
of −α ;
3.3: Bode’s
ideal loop.
•Figure
Magnitude:
constant
slope of −α20dB/dec.;
1
+ an −1s βn −1 + ! + a1s β1 + a0 s β0
βn −1
K
s!
Figure 3.3: Bode’s ideal loop.
(a) Open loop:
βn
G(s) =
14
Y(s)
Y(s)
K
(a) Open loop:
G(s) =
Figure
s ! 3.3: Bode’s ideal loop.
• Magnitude: constant slope of −α20dB/dec.;
W(s)
–
Gs ( s ) =
Y(s)
K
s!
14
""
"
63 / 90
Back
Full screen
Close
End
• Nyquist: straight line at argument −α π2 .
(a) Open loop:
• Phase: horizontal line of −α π2 ;
(b) Closed loop:
Figure
• Nyquist: straight line at argument
−α π2 .
3.2: P ID controller: from points to plane.
• Gain margin: Am = infinite;
"
!
• Phase margin: constant : Φm = π 1 − α2 ;
• Phase:
line of −α π2 ;
(b)horizontal
Closed loop:
• Step response: y(t) = Ktα Eα,α+1 (−Ktα ) ,
where Ea,b (z) is Mittag-Leffler function of two parameters [43].
a wide class of controlled objects we recommend the fractiona
= infinite;
• Gain margin: AmFor
"
!
λ δ
α
controller,
;a particular case of P I D controller, where λ =
• Phase margin:
constant : Φmwhich
= π 1 −is
2
3.2.3 Illustrative
example
• Gain margin:• AStep
α
α
m = infinite;
response:
y(t)
=
Kt
E
(−Kt
),
α,α+1
N and
δ
∈
R.
Integer
order integrator is important for steady-sta
"
!
α
We can illustrate the fractional order control properties by an example. As;
• Phase margin: where
constant
:
Φ
=
π
1
−
m
function of two parameters [43].
Ea,b (z) is Mittag-Leffler
2
(b) Closed loop:
cancellation but on the other hand the fractional integral is also impo
5
(3.9) with consta
obtaining a Bode’s ideal loop transfer function response
margin
for
desired
frequency
range
(e.g.
[1,
51]).
with J being
the payload inertia.
Our specification
is 3,
the 23,
constant
phase
We can illustrate
the fractional order control properties by an example. AsIllustrative
example
suming that the transfer function of the DC motor is
• Step response: y(t) = Ktα Eα,α+1 (−Ktα ) ,
Km
,
G(s) =
Js(s + 1)
3.2.3 Illustrative example
2
3.2.3
margin independent of the payload changes.
suming that the transfer function of the DC example.
motor that
isAs-we would like to have a closed loop system that is insensitive
We can illustrate the fractional order control properties by an Assume
to gain variations with a constant phase margin of 60o. Bode’s ideal loop
suming that the transfer function of the DC motor is
K
m
(3.9)
G(s) =
transfer, function that gives this phase
margin is
Js(s + 1)
Km
,
(3.9)
G(s) =
1
Js(s + 1)
.
(3.10)
= √
Go (s)
with J being the payload inertia. Our specification is the constant
phase
s3s
with J beingmargin
the payload
inertia. Our
specification
is the constant phase
independent
of the
payload changes.
margin independent of the payload changes.
Assume that we would like to have a closed loop system that is insensitive
Assume that we would like to have a closed loop system that is insensitiveo
to gain variations with a constant phase margin of 60 . Bode’s ideal loop
to gain variations with a constant phase margin of 60o. Bode’s ideal loop
α
transfer
function
that margin
gives this
transfer function
that gives
this phase
is phase margin is
3.2.2
Bode suggested an ideal shape of the loop transfer function in his
design of feedback amplifiers in 1945. Ideal loop transfer function has
!
"
s
,
L(s) =
1
1
ωgc
(3.10)
Go (s) = √ .
(3.10)
G (s) = √ .
o
s3s
s3s
where ωgc is desired crossover frequency and α is slope of the idea
characteristic.
Phase margin is Φm = π(1 + α/2) for all values of the gain. The am
margin Am is infinity. The constant phase margin 60o , 45o and 30o co
to the slopes α = −1.33, −1.5 and −1.66.
The Nyquist curve for ideal Bode transfer function is simply a stra
through the origin with arg(L(jω)) = απ/2 (see e.g. [1, 39]).
transfer
function
can be used as a reference system
Can be used as a Bode’s
reference
system
in (3.6)
the form:
14
following form:
Y(s)
W(s)
K
K
(0 < α < 2)
Gc (s) = α
G(s) = s
s +K
K
Go (s) = α
(0 < α < 2),
Figure 3.3: Bode’s ideal loop.
CHAPTER
15
s
CHAPTER 3.
3. FRACTIONAL-ORDER
FRACTIONAL-ORDER CONTROLLERS
CONTROLLERS
15
(a) Open loop:
where Gc is transfer function of closed loop and Go (s) is transfer fun
open loop. General characteristics of Bode’s ideal transfer function a
PI!Dµ controllers
!
20
20log
log1010 |G(j
|G(j!)|
!)|
--20
20 "" dB/dec
dB/dec
• Phase: horizontal line of −α π2 ;
(b) Closed loop:
log
log !!
arg
arg((G(j
G(j!))
!))
##
• Gain margin:
Am = infinite;
22
"
!
##
""
• Step response:
y(t) = Ktα Eα,α+1 (−Ktα ) ,
22
##
log
log !!
3
3.2.3
Illustrative example
Figure
Figure 3.4:
3.4: Bode
Bode plots
plots of
of transfer
transfer function
function (3.8).
(3.8).
We can illustrate the fractional order control properties by an example. Assuming
the
transfer function
of the
motor transfer
istransfer function
Since
(s)
C(s)G(s),
we
the
controller
Since G
Goothat
(s) =
=
C(s)G(s),
we can
can find
find
theDC
controller
function in
in the
the
following
following form
form
!
"
! G(s) = "
" Km !
!,
"
(3.9)
11
11
JJ
2/3 +
2/3 +
Js(s
+
1) ss2/3
C(s)
ss2/3
=
K
,,
(3.11)
+ 1/3
+
C(s) =
=
=
K
(3.11)
1/3
K
ss1/3
ss1/3
Kmm
with Jis abeing
the payload
inertia. Our
specification
isλ the
constant phase
δ
which
D δ controller
controller (3.1),
(3.1),
which is a particular
particular case
case of
of the
the transfer
transfer function
function of
of the
the PPII λD
marginKindependent
of the payload
changes.
where
where K =
= J/K
J/Kmm isis the
the controller
controller constant.
constant.
Assume that
we would
like to have system
a closed loopa system
that
is insensitive
The
loop
The phase
phase margin
margin of
of the
the controlled
controlled system with
with a forward
forward
loop controller
controller
o
to
gain
isis variations with a constant phase margin of 60 . Bode’s ideal loop
G
(s)
Gcc(s)
transfer function that gives
this
phase
margin
=
arg[C(jω
(3.12)
Φ
)G(jω0 )])]is+
+π,
π,
(3.12)
Φm = arg[C(jω0 )G(jω
m
0
0
where
1
where ωω00 isis the
the crossover
crossover frequency.
frequency.
.
Go (s) = √
The
s3s
The obtained
obtained phase
phase margin
margin isis
##
$$
11
44ππ ππ
=
Φ
+
= ..
= arg[C(jω)G(jω)]
arg[C(jω)G(jω)]+
+ππ =
= arg
arg
Φmm =
+ππ =
= ππ −
−
4/3
(jω)
33 22 33
(jω)4/3
(3.10)
(3.13)
(3.13)
6
• Nyquist: straight line at argument −α .
2
α
• Step response: y(t) = Ktα Eα,α+1 (−Kt
),
(b) Closed loop:
3.2.3
• Gain margin: Am = infinite;
"
!
• Step response:
y(t) = Ktα Eα,α+1 (−Ktα ) ,
Illustrative
example
We can illustrate the fractional order control properties by an example. Assuming that
the Illustrative
transfer function
of the DC motor is
3.2.3
example
PI!Dµ controllers
Bode’s ideal loop transfer function: example
We can illustrate the fractional order control
Kmproperties by an example. As,is
G(s) of=theofDC
suming that
the transfer
transfer function
motormotor
The
function
a DC
is
(3.9)
Js(s + 1)
G(s) =
Km
J is payload inertia
,
(3.9)
Js(s
1)
Our+specification
is the constant
with J being the payload inertia.
phase
margin independent
of
the
payload
changes.
with J being the payload inertia. Our specification is the constant phase
Assume that we would like to have a closed loop
margin
independent
of the
payload
changes.
Assume
that
we would
like
to have
a closed
systemwith
thata is insensitive
that
is
to loop
gainloop
variations
Assumesystem
that we would
likeinsensitive
to have a closed
system
thatois insensitive
.
Bode’s
ideal loop
to gain variations
with
a
constant
phase
margin
of
60
o
constant
margin
60 . Bode’s
Bode’sloop
ideal loop
to gain variations
withphase
a constant
phaseofmargin
of 60o. ideal
transfer function
that
gives
this
phase
margin
is
CHAPTER
3.
FRACTIONAL-ORDER
CONTROLLERS
15
transfer function
that function
gives this phase
is
transfer
that margin
gives this
phase margin is
20 log 10 |G(j !)| 1
Go (s) = √
3
1.
s s - 20 " dB/dec
(3.10)
.
Go (s) =s s√
CHAPTER 3. FRACTIONAL-ORDER
3 CONTROLLERS
Fractional-order PID controllers:
from points to plane
(3.10)
15
15
20 log 10 |G(j !)|
Figure 3.2: P ID controller: from points to plane.
20 log 10 |G(j !)|
CHAPTER 3. FRACTIONAL-ORDER
20 log 10 |G(j !)|
- 20 " dB/dec
CONTROLLERS
15
7
10
For a wide class of controlled objects we recommend the fractional P I n D δ
controller, which is a particular case of P I λ D δ controller, where λ = n, n ∈
N and δ ∈ R. Integer order integrator is important for steady-state error
cancellation but on the other hand the fractional integral is also important for
margin for desired frequency range (e.g. [1, 3, 23, 51]).
- 20 " dB/dec
log !
arg ( G(j !))
#
2
- 20 " dB/dec
log !
arg ( G(j !))
#
" #
22
log !
arg ( G(j !))
## #
" ( G(j
arg
!))
2
#
2
2
log !
3.2.2
log !
#
2
#
"
2
log !
Since Go (s) = Figure
C(s)G(s),
can
findofthe
controller
transfer
3.4:# we
Bode
plots
transfer
function
(3.8). function in the
following form
#
log
!
"
!log !
" !
Since Go (s) = C(s)G(s),
find1 the controller
transfer
1 function in the
J we can
C(s) =
s2/3 + 1/3 = K s2/3 + 1/3 ,
(3.11)
following form
Km
s
s
Figure
3.4:!plots
Bodeof plots
transfer
function
(3.8).
" of function
!
"
Figure
3.4: Bode
transfer
(3.8).
1 function of
Jof the2/3
2/3the P1I λ D δ controller (3.1),
which is a particular
case
transfer
C(s) =
s + 1/3 = K s + 1/3 ,
(3.11)
Kcontroller
s controller transfer
s function in the
m
where
is thewe
constant.
Since
GK
(s)==J/K
C(s)G(s),
can find the
PI!Dµ controllers
Bode’s ideal loop transfer function: example
Sinceo Go (s)
Since
=mC(s)G(s), , we
cancontroller
find the controller
transfer
function
the
transfer
function
is in the
(3.1),
which
a particular
of the
transfer function
of theaPforward
I λ D δ controller
The is
phase
margincase
of the
controlled
system with
loop controller
following
form
following
form
!
"
!
"
K = J/Km is the
constant.
Gwhere
"2/3
"
1!
J controller
c (s) is
2/3 ! 1
+
+
C(s)
=
s
=
K
s
,
(3.11)
1
1 controller
J controlled
The phase margin
with
a forward
2/3
)G(jω
)] K
+s1/3
π,
(3.12) (3.11)
Kmof
ss1/3
m = arg[C(jω
0=
+ 0system
C(s)
=Φthe
s2/3 + loop
,
Km
s1/3
s1/3
Gc (s) is
which
is ω
a 0particular
case of the
transfer function of the P I λ D δ controller (3.1),
where
is the crossover
frequency.
)G(jω0 )] + π,
(3.12)
Φm = this
a 0particular
case
of fractional
PID (3.1),
which
is
a particular
case
of arg[C(jω
the
transfer
function
of the
P I λ D δ controller
where
K=
J/K
the controller
constant.
m is phase
The
obtained
margin
is is
#constant.
$
where
ωK0 is
the
crossover
frequency.
The phase
system
where
=margin
J/K
thecontrolled
controller
o a forward loop
mofisthe
π
1with
4 πcontroller
The
phase margin
is arg
= . loop
(3.13)
arg[C(jω)G(jω)]
π=
+ π with
= π −a forward
Gc (s)Φis
m =obtained
The
phase
margin
of +the
controlled
system
controller
32
3
# (jω)4/3
$
Φm = arg[C(jω0 )G(jω10 )] + π,
π(3.12)
4π
(s)
is
G
=
.
(3.13)
=
arg[C(jω)G(jω)]
+
π
=
arg
Φ
+
π
=
π
−
c
m
The constant
phase margin is not dependent
changes
and the
4/3 of the payload
(jω)
3
2
3
where ω0 is the crossover frequency.
)] +
π,
(3.12)
Φm =isarg[C(jω
0 )G(jω
system
gain K and phase curve
a horizontal
line0at
−2π/3.
The constant
obtained phase margin
The
margin isis not dependent of the payload changes and the
# loop can
$ be expressed as:
Step response
of
closed control
where
theand
crossover
frequency.
0 is K
system ωgain
phase
curve
is
a
horizontal
line
at
−2π/3.
π
1
4π
%
= arg[C(jω)G(jω)]
+1πmargin
= arg& is 4/3 + π = π − ' = . ( (3.13)
ΦmThe
obtained
phase
Step
control loop
canEbe expressed
as:1+1/3
(jω)
3 2−t
3
,
(3.14)
y(t)response
= L−1 %of closed
=
t1+1/3
# 1+1/3, 2+1/3
$
1+1/3 + 1) &
s
(s
'
( 4 πthe π
The constant phase
of the1 payload changes
1 not dependent
−1 margin is
1+1/3
1+1/3 and
,
(3.14)
=
L
=
targ E1+1/3, 2+1/3+ −t
=
.
(3.13)
=Kresponse
arg[C(jω)G(jω)]
+
π
=
π
=
π
−
my(t)
1+1/3
4/3
whereΦgain
step
is
independent
of
the
payload
inertia
and
α
=
4/3.
system
and phase
curve
is
a
horizontal
line
at
−2π/3.
s (s
+ 1)
(jω)
32
3
8
Step response
of closed independent
control loop can
be payload
expressed
as:
where
step response
of the
inertia
α = 4/3.
The
constant
dependent
of the and
payload
changes and the
%phaseis margin
& is not
' 1+1/3 (
1
(3.14)
y(t) =
L−1 K and
= t1+1/3
system
gain
phase curve
is aE1+1/3,
horizontal
line at, −2π/3.
2+1/3 −t
s (s1+1/3 + 1)
The obtained phase margin is 60 :
Step response:
Step response of closed control loop can be expressed as:
'
(
1
y(t) = L−1
= t1+1/3 E1+1/3, 2+1/3 −t1+1/3 ,
s (s1+1/3 + 1)
where step response is %
independent of the
& payload inertia and α = 4/3.
Bode suggested an ideal shape of the loop transfer function in his work on
design of feedback amplifiers in 1945. Ideal loop transfer function has form:
!
"α
s
,
(3.6)
L(s) =
ωgc
Figure 3.4:# Bode plots of transfer function (3.8).
"
13
where ωgc is desired crossover frequency and α is slope of the ideal cut-off
characteristic.
Phase margin is Φm = π(1 + α/2) for all values of the gain. The amplitude
o
margin 60of
, unit
45o and
30o correspond
margin Am is infinity. The constant phaseComparison
step responses
to the slopes α = −1.33, −1.5 and −1.66.of the fractional-order “reality” and
its integer-order “model”
The Nyquist curve for ideal Bode transfer function is simply a straight line
through the origin with arg(L(jω)) = απ/2 (see e.g. [1, 39]).
Bode’s transfer function (3.6) can be used as a reference system in the
following form:
K
(0 < α < 2)
(3.7)
Gc (s) = α
s +K
K
Go (s) = α
(0 < α < 2),
(3.8)
s
where Gc is transfer function of closed loop and Go (s) is transfer function in
open loop. General characteristics of Bode’s ideal transfer function are:
(3.14)
where step response is independent of the payload inertia and α = 4/3.
PI!Dµ controllers
PI!Dµ controllers
Control of the “reality” and the
“model” using a classical PD
controller, which is optimal
for the “model”
9
ss
being developed is the method of dominant roots [37], based on the given
stability measure and the damping ratio of the closed control loop. Assuming
that, the desired dominant roots are a pair of complex conjugate root as follows:
s1,2 = −σ ± jωd ,
PI!Dµ controllers
designed for the damping ratio ζ and natural frequency ωn . The damping
constant (stability measure)
is σ = ζωn and the damped natural frequency of
!
oscillation ωd = ωn 1 − ζ 2 . The design of parameters: Kp , Ti , λ, Td and δ
can be computed numerically from characteristic equation. More specifically,
for simple plant model P (s), this can be done by solving
PI!Dµ controllers: design
min
Kp ,Ti ,λ,Td ,δ
Control of the “reality” and the
“model” using a classical PD
controller, which is optimal
for the “model”,
after detuning (aging) of the
controller (when TD = 1).
||C(s)P (s) + 1||s=−σ±jωd .
The design of PI!D" controllers can be based on gain
Another possible way to obtain the controller parameters is using the tuning
and
phase
margin
specifications:
formula,
based
on gain
and phase
margins specifications [52]:


" [C (jωp )] " [P



" [C (jωp )] # [P
 " [C (jωg )] " [P


 " [C (jω )] # [P
g
(jωp )] − # [C (jωp )] # [P (jωp )] = − A1m ,
(jωp )] + # [C (jωp )] " [P (jωp )] = 0,
(jωg )] − # [C (jωg )] # [P (jωg )] = − cos Φm ,
(jωg )] + # [C (jωg )] " [P (jωg )] = −sinΦm ,
where Φm is a phase margin, Am is a gain margin, ωp and ωg is 0dB (crossover)
frequency.
Last but not least we should mention the optimization algorithm based on
the integral absolute error (IAE) minimization [44]:
& t
& t
|e(t)|dt =
|w(t) − y(t)|dt,
IAE (t) =
0
0
where w(t) is the desired value of closed control loop and y(t) is the real value
of closed control loop.
This method does not insure the desired stability measure of the closed
control loop. Measure of stability has to be checked out additionally. We can
use a frequency method described in [39].
PI!Dµ controllers
Control of the „reality“
using the PD controller,
which is optimal for the
„model“, and using the
PD" controller.
IV.2 A
Design Approach
• The plant model is assumed to be
• and the fractional order PID controller is
• It is expected that the gain and phase margin of
the compensated systems are and
• Question: how to choose
parameters
FOC tutorial III
©Dingyu Xue, NEU, PR China
Design of fractional-order PID controller
parameters is determined by given requirements.
These requirements can be, for example:
• the damping ratio,
• the steady-state error,
• dynamical properties,
• etc.
27
Requiring
it can be found that
• For
• It is found that
PI!Dµ control: MATLAB
I. General Description of Linear
Fractional Order Systems
Fourfour
equations,
withwith
7 variables
We• have
equations
seven variables:
I.1 The normal form
FOC tutorial III
28
The rest of the variables can be determined
by minimizing the ISE criterion
• Thus compared with IO LTI’s, information on
orders are also used
• A FOTF model class/object can be defined in
MATLAB to describe the system model
FOC tutorial III
Sample
plots for
combinations of ! and µ:
µ, various
! combinations
• For different
II. MATLAB Class Design
II.1 Create a @fotf directory
• The fotf class can be defined as
FOC tutorial III
FOC tutorial III
30
4
Courtesy: Dingyü Xue, YangQuan Chen
• A fractional order PID controller can be
!Dµ controller:
The designed
optimal PI
such
that
FOC tutorial III
Courtesy: Dingyü
Xue, YangQuan Chen
3
II.2 Other Overload Functions
• Function call
• A display function
31
FOC tutorial III
5
II.3 Define a fotf Object in MATLAB
• Feedback function
• Example
• MATLAB command
• Display
• These functions are suitable for
interconnections of fractional order systems
FOC tutorial III
6
FOC tutorial III
9 Xue, YangQuan Chen
Courtesy: Dingyü
II.4 Overload functions for block
• A common function unique
• Plus function for parallel connections
FOC tutorial III
7
FOC tutorial III
10
II.5 ofAn
Example for interconnection
Example
interconnection:
• Multiplication for series connection
• Unity negative feedback system with
The closed-loop system can be
obtained as
• Minus
• The closed-loop system can be established
• Uminus
• The closed-loop system
• Inv
FOC tutorial III
FOC tutorial III
8
11
• The closed-loop systems with even more
complicated structures can easily be obtained
with the overloaded functions
FOC tutorial III
12
CHAPTER 4. CONTINUED FRACTION EXPANSION (CFE)
+
+
+
-
-
-
+
+
1
+
+
22
+
+
Y2n* (s)
Z2n -1(s)
Y2n -2 (s)
Z2n-3 (s)
Y2 (s)
Continued fractions (CFE)
CFEs and nested multiple loops
Z 1 (s)
18
Figure 4.6: Nested multiple-loop control system of the first type.
can be expressed in the form:
G(s) ! a0 (s) +
expansion, which is identical with the equation (6.1):
Z(s) = Z1 (s) +
b1 (s)
a1 (s) +
b2 (s)
b3 (s)
a2 (s)+ a (s)+...
3
b1 (s) b2 (s) b3 (s)
= a0 (s) +
...
a1 (s)+
a2 (s)+ a3 (s)+
+
-
+
+
19
+
-
-
(4.1)
Z2n-3 (s)
Nested loop of type I
1
Z(s) =
1
Z1 (s) +
1
CHAPTER 4. CONTINUED
FRACTION EXPANSION
(CFE)
Y
(s)
+
2
Z 1 (s)
1
Z3 (s) +
. . . . . . . . . . . .+. . . . . .+. . . . . . . . +
.......
+
+
+
1
1
+
+
1+
Y2n−2 (s) +
1
expansion, which is identical with the equation (6.1):
Y2n* (s)
Z2n−1 (s) +
Y
(s)
2n
1
Z(s) = Z1 (s) +
1
Z2n -1(s)
Y2 (s) +
Y2 (s)
Z3 (s) +
1
Y2 (s) +
Z3 (s) +
Y2 (s)
Z 1 (s)
1
1
. . . . . . . . . . . . . .expansion,
. . . . . . . . . . . . .which
. . . . . . is identical with the equation (6.1):
1
1
1
Z(s) = Z1 (s) +
Y2n−2 (s) +
1
1
Z2n−1 (s) +
Y2 (s) +
1
Y2n (s)
Z3 (s) +
Similarly to the above considerations, we can obtain a continued fraction
expansion of the transfer function of the other interesting type of a nested
multiple-loop control system, depicted in Fig. 4.7:
Z(s) =
Z1 (s) +
CHAPTER 4. CONTINUED FRACTION EXPANSION (CFE)Y2 (s) +
Figure 4.1: A control loop with From
a negative
+
(4.4)feedback.
it immediately follows that the transfer function
of the circuit
+
+
+
-
-
20
Y * (s)are used for creating
Then this two parts of the characteristic
polynomial
20
Y2n(s)
-
2n
=
P2n−2of(s)
Using the equations (4.4) and (4.5) we obtain the transfer function
the =
1
system shown in Fig. 4.3:
1 + Q2n−1 (s)Y2n−2(s)
system shown in Fig. 4.3:
Y2n−2 (s) +
Q2n−1 (s)
+
1
1
Q2n−1 (s) = Z2n−1 (s) + P2n (s) = Z2n−1 (s) + 1 .
(4.6)
1
Q2n−1 (s) = Z2n−1 (s) + P2n (s) = Z2n−1 (s) + Y2n (s).
(4.6)
Y2n (s)
=
(4.7)
CHAPTER 4. CONTINUED FRACTION EXPANSION
(CFE)
20
1
Combining the equations (4.4) and (4.5) we find the transfer function of
Y2n−2 (s) + Y2n* (s)
Combining the equations (4.4) and (4.5) we find the transfer function of
1
the
nested
multiple-loop
system shown
in Fig.
4.4:
Letthe
usnested
now establish
an interesting
new
relationship
between continued fracZ2n−1 (s) +
multiple-loop system shown in Fig. 4.4:
+
Y2n (s)
1
tions and nested multiple-loopQ2n−1
control
1
(s) systems.
Figure 4.2: Nested multiple-loop
control system – level 1.
1
Q2n−1 (s)
=
2n−2 (s) =
We firstPPrecall
that2n−2
the(s)transfer
function R(s)
of the control
1
=
= known
1 + Qfact
2n−2 (s)the
2n−1 (s)Y
shown in Fig. 4.5 is then given by the
Y2n−2 (s) + The1 transfer function of the system
1 + Q2n−1
(s)Y2n−2
(s)
Y2n* (s)
Y2n−2
+ Qby
(s)
loop with a negative feedback shown in Fig. 4.1
is (s)
given
[8]
+
2n−1
+
relationship
Q2n−1 (s)
1
1
1
+ – level 1.
=
(4.7)
Figure (4.7)
4.2: Nested multiple-loop control system
G(s)
1
=
1
Q2n−3 (s)(4.4)
= Z2n−3 (s) + P
.
Y2n−2R(s)
(s) +=
2n−2 (s)
1
Y2n−2
(s) + 1 + G(s)H(s)
Y2n* (s)
1
Z
(s)
+
+
2n−1
CHAPTER 4. CONTINUED FRACTION
(CFE)
21
+
1
Z2n−1 (s) +EXPANSION
Y2n (s)
1
=
Z
(s)
+
(4.8)
Y2n (s)
+
1
From (4.4) it immediately follows that the
transfer function of the circuit 2n−3
Z2nY-12n−2
(s) (s) +
The transfer function of the system shown in Fig. 4.5 is then given by the
transfer
of the system shown in Fig. 4.5 is then given by the
1
Y2n* (s)
shownThe
in Fig.
4.2 function
is
relationship
Z
(s)
+
2n−1
relationship
Y2n (s)
1
1
Figure 4.3: (4.5)
Nested multiple-loop
Z2n -1(s)
Q2n−3 (s) P
= Z2n−3
+ P2n−2∗(s) =
,
=(s)+
Q2n−3
(s) = 2nZ(s)
(s)(s)
2n−3 (s)
1 +P12n−2
· Y2n
Y12n (s)
Continuing this process, we obtain the transfer function of the nested multiple1
= Z2n−3 (s) +
(4.8)
1 control
loop
system
shown
in Fig.
the
form
of a2. continued
fraction
= Z
(s) +
(4.8)
Figure
4.3: Nested
multiple-loop
control
system
level
∗
equations
(4.4)
and
(4.5)4.6
we in
obtain
the– transfer
function of
the
1 Using the
(s) ++1. 2n−3+
where Y2n (s) = Y2n
Y2n−2 (s) ++
1
1
- Y2n−2 (s) + Z+2n−1system
(s) + 1shown in Fig. 4.3:
Z2n−1 (s) + Y2n (s)
CHAPTER 4. CONTINUED FRACTION EXPANSION
(CFE)
21 (4.4) and (4.5) we obtain the transfer function of the
Using
the equations
Y
(s)
2n
Y * (s)
1
shownmultipleFig.(s)
4.3:
Continuing this process, we obtain the2ntransfer function system
of the nested
Qin2n−1
.
(4.6)
= Z2n−1 (s) + P2n (s) = Z2n−1 (s) +
Continuing this process, we obtain the transfer function of the nested multipleY2n (s)
loop control system shown in Fig. 4.6 in the form of a continued fraction
1
loop control system shown in Fig. Z4.6
in the form of a continued fraction
2n -1(s)
Q2n−1 (s) = Z2n−1 (s) + P2n (s) = Z2n−1 (s) +
.
(4.6)
(s) transfer function of
Combining the equations (4.4) and (4.5) we findY2nthe
Y2n-4 (s)
the nested
multiple-loop
system
shown
in we
Fig.
4.4:
Combining
the equations
(4.4) and
(4.5)
find
the transfer function of
the nested multiple-loop system shown in Fig. 4.4:
1
Q2n−1 (s)
+
+
+
Figure 4.4: Nested
multiple-loop
control
system – level
3. (s) =
=
P2n−2
1
1
(s)
Q2n−1
1
1+Q
(s)
2n−1 (s)Y2n−2=
+
P2n−2 (s) =
Y2n−2 (s) +
1 Q
1 + Q2n−1 (s)Y2n−2(s)
2n−1 (s)
Y2n−2 (s) +
Y2n* (s)
Q2n−1 (s)
1
1
=
1
=
(4.7)
Z2n -1(s)
1
Y2n−2
Y2n−2
(s)(s)
++
1
Z2n−1
(s) +1
Z
(s) +
2n−1
Y2n -2 (s)
Y2n (s)
Y2n (s)
-
-
1
+
+
Y2n* (s)
Z2n -1(s)
Figure 4.7: Nested multiple-loop control system of the second type.
4.5
+
-
Z2n-3 (s)
1
(4.8)
1
Y2n (s)
Continuing this process, we obtain the transfer function of the nested multiple-
+
+
+
control
system – level 4.
*
Y2n(s)
-
CFE and rational and irrational numbers
Every rational and irrational number may be written in CFE form. Basically
the process of finding a continued fraction development consists of two steps:
if the fraction m/n is greater than 1, then divide. Otherwise, write the fraction
m/n as 1/(n/m) and proceed with the first step. Continue until a numerator
of 1 is obtained. Continued fractions are often used to get good rational
CHAPTER 5. APPROXIMATION OF FRACTIONAL OPERATOR
25
approximations for real numbers. Let us consider the numbers written in the
CHAPTER
5. APPROXIMATION OF FRACTIONAL OPERATOR
25
form: a0 + b1 /(a1 + b2 /(a2 + b3 /(a3 + . . . ))). In ”simple” continued fractions,
as [a0 ;5.
a1 , aAPPROXIMATION
all the bi , ∀i, are 1 and the number can be re-written
2 , a3 , . . . ].
CHAPTER
OF FRACTIONAL OPERATOR
APPROXIMATION OF FRACTIONAL OPERATOR
5.2
+
Z2n -1(s)
Y2n -2 (s)
Z2n-3 (s)
Figure 4.5: Nested multiple-loop control system – level 4.
25
General CFE method
5.2 method
General CFE method
General
CFE
CHAPTER
5. APPROXIMATION
OF FRACTIONAL OPERATOR
25
−α
−α
,0<α<1
performing
the
CFE
of the functions:
fractional
integral
the
Laplace
domain)
can
(the
fractional
integral
the
domain)
can be obtained by
where
the operator
following
idea
appeared:
dense
of Laplace
simpleby
poles and
In general [10],(the
a rational
approximation
of thein
function
G(s) a=
s−α , interlacing
0operator
< αbe
<in1obtained
performing
thesome
CFE
the,1functions:
A rational
approximation
In general
a rational
approximation
the
function
G(s)
==obtained
sof−α
0equivalent
<by
α < to1 a branch (5.1)
performing
thezeros
CFE
the
functions:
along
a of
line
in the
sfor
plane is,
in
way,
cut;
(the [10],
fractional
integral
operator
inofthe
Laplace
domain)
can
be
H
h (s)
(1 + sT )α
α
1 the negative
and in
s , 0the
< αLaplace
vieweddomain)
an operator,
branch
"α cut along
(the fractional
can has
be!aobtained
performingintegral
the can
CFEoperator
of
functions:
betheobtained
by< 1,CFE
ofasthe
expressions:
H
(5.1)
1 h (s) = by
1 following
α
(4.7)
(1 + sT
)
Hl (s)
1+
(5.2)
real axis for arguments
but =is otherwise
free of
Hh (s) of
= s on (−π, π)
(5.1)
! poles
"αand zeros.
s
performing the CFE of the functions:
(1 + sT )α
1
1
Hl (s) =
1+
(5.2)
!
"α
(5.1) (ωT >>s 1), and Hl (s)
for high frequencies
where Hh (s) is the approximation
1
α
(1Happroximation
+l (s)
sT )= for
1+
(5.2)
the
low
frequencies
(ω
<<
1).
!
"α where Hsh(s) is the approximation for high frequencies (ωT >> 1), and Hl (s)
h
1 α the approximation for low frequencies (ω << 1).
Example
Hgeneral
1rational
+ 5.2.1.
(5.2)
−α
l (s) =
[10], aPerforming
of the(ωT
function
G(s)
, 0we
< obtain:
α<1
the
CFEfrequencies
of the function
(5.1),
with 1),
T =and
1,=α s=
s forαapproximation
approximation
high
>>
H0.5,
where Hh (s) isInthe
l (s)
Example
5.2.1.
(the
fractional
integral
operator
in
the the
Laplace
can+be
obtained
4 1).
3 CFE ofdomain)
2 function
Performing
the
with
T = 1, by
α = 0.5, we obtain:
the approximation for low frequencies
(ω
<<
+ 1.405s
+ 0.8433s
+ 0.1574s(5.1),
0.008995
0.3513s
= functions:
Hof
l
1 (s)
performing
CFE
the
4
3
2
for the
high
frequencies
(ωT
>>
1),
and
H
(s)
where Hh (s) is the approximation
l
s + 1.333s + 0.478s4 + 0.064s
+ 0.002844
0.3513s + 1.405s3 + 0.8433s2 + 0.1574s + 0.008995
Example
H1 (s) =
the approximation
for low5.2.1.
frequencies (ω << 1).
1 s4 + 1.333s3 + 0.478s2 + 0.064s + 0.002844
Example 5.2.2.
Hh (s) =
(4.8)
5.2
1 CFE method
General
H (s) =
(1 + sT )
!
"
1
H (s) =
1+
Example:
s
(5.1)
(5.2)
Performing the CFE of the function H
(5.1),
= T = 1, αα = 0.5, we obtain: (5.1)
h (s) with
approximation
for highPerforming
frequencies
(ωT
>>
1),
(s)
where H
(1
+ sT
) and
the CFE of
the function
(5.2),
with TH
= l1,
α = 0.5, we obtain:
h (s) is the
Example
5.2.1.
Example 5.2.2.
!
"α
4
3
24
1 2of
3 CFE
+
1.405s
+ 0.8433s
0.1574s
0.008995
0.3513s
Performing
the
the
function
(5.2), with T = 1, α = 0.5, we obtain:
approximation
frequencies
(ω
<<
1).
+=
4s
+obtain:
0.448s
+ 0.0256
s
Performing for
the low
CFE
of
the
function
(5.1),
with
T
=
1,
α
0.5,
we+
Hl (s)
=+
1++2.4s
(5.2)
=
H (s)
H (s) =
loop controlthis
system
shown
Fig. 4.6
the form
of a continued
fraction
Continuing
process,
weinobtain
theintransfer
function
of the nested
multiplethe
loop control system shown in Fig. 4.6 in the form of a continued fraction
Y2n -2 (s)
+
1
11
= Z2n−3 (s) Y+2n−2 (s) +
1
Y2n−2 (s)
+ (s) + 1
Z2n−1
Y2n (s)
Z2n−1 (s)
+
= Z2n−3 (s) +
Nested loop of type II
5.2 approximation
General CFE
method
In general [10], a rational
of the function
G(s) = s , 0 < α < 1
General
CFEmethod
method
−α
5.2 5.2
General
CFE
In general
[10],nominator
a rationalpolynomials.
approximation
of the
function
G(s)
= sdomain)
0<
αbe<
1 in G(s)
(the fractionalProbably,
integral
operator
in the
Laplace
can
obtained
by
was
noted
for ,the
first
time
[17],
In generalthis
[10],fact
a rational
approximation
of
the function
=s
Thetransfer
transfer function
system
shown
in Fig.
thenisgiven
the by the
The
functionofofthethe
system
shown
in 4.5
Fig.is 4.5
thenbygiven
relationship
relationship
Q2n−3 (s) = Z2n−3 (s) + P2n−2 (s)
Q
2n−3 (s) = Z2n−3 (s) + P2n−2 (s)
1
Z3 (s) +
.................................
1
1
Y2n−2 (s) +
1
Z2n−1 (s) +
Y2n (s)
25
nominator
polynomials.
fact for
was the
noted
for the
firstintime
nominator polynomials.
Probably,
this Probably,
fact wasthis
noted
first
time
[17],in [17],
Example
4.5.1.
nominator
polynomials.
Probably,
this fact
was noted
for the
timeofinsimple
[17], OF
5.first
APPROXIMATION
FRACTIONAL
OPERATOR
25
where
theappeared:
following
idea
appeared:
a CHAPTER
dense
interlacing
poles
and
where where
the following
dense
poles
and
The CFE idea
for π, which
gives the ”best”aapproximation
of a given order,
is of simple
nominatorinterlacing
polynomials.
Probably,
this fact was
noted for
the first time in [17],
the
following
idea
appeared:
dense
interlacing
of simple
polesinterlacing
along
the1,a1,swhere
plane
inThesome
way,
equivalent
toand
a branch
cut;poles and
[3, 7, 15, zeros
1, 292, 1,
1, 1, 2,a1,line
3, 1, in
14, 2,
2,
2, 2,the
2,is,
...].
very
following
idealarge
appeared:
a dense
of was
simple
nominator
polynomials.
Probably,
thiscut;
fact
noted for the first time in [17],
zeros along
a
line
in
the
s
plane
is,
in
some
way,
equivalent
to
a
branch
α the
term 292
meansin
a good
approximation
[13].
zeros along
a line
sα plane
is, in iszeros
some
way,
equivalent
tois,ainidea
branch
cut;
along
a line
in the
s aplane
some
way,
equivalent
tointerlacing
a branch cut;
and
sthat
, 0the
<following
< 1,convergent
viewed
as
an
operator,
has
branch
cut
along
the
negative
where
the
following
appeared:
a dense
of simple poles and
sα , 0 <
< 1, viewed
as along
operator,
has
a negative
branch cut along the negative
and sαand
, 0 <sαα
viewed
as
an
operator,
has
branch
cut
along
the
355 and
, 0<
< 1,
α<
1,axis
viewed
as
an
operator,
has
aa. .α.branch
cut
zeros
along
aanline
inthe
the
snegative
plane
is, inand
somezeros.
way, equivalent to a branch cut;
real
for
arguments
on
(−π,
π)
but is
otherwise
free
of
poles
3.14159292
[3, 7, 15,
1] =
[3, 7, 16] = of= s
αof s on (−π, π) but is otherwise free of poles and zeros.
113 real axis for arguments
and s free
, 0 < of
α <poles
1, viewed
an operator, has a branch cut along the negative
realfor
axisarguments
for arguments
s on
(−π,π)π)but
but is
is otherwise
otherwise
andasand
zeros.
real axis
of sof on
(−π,
free
of
poles
zeros.
real axis for arguments of s on (−π, π) but is otherwise free of poles and zeros.
+
1
Both types of nested multiple-loop systems, presented in this section, can
be used for simulations and realizations of arbitrary transcendental transfer
functions. For this, the transfer function should be developed in a continued
fraction, which after truncation can be represented by a nested multiple-loop
system shown in Fig. 4.6 or Fig. 4.7.
CHAPTER 5.
+
(4.9)
1
Z 1 (s)
CFE and nested multiple-loop
control systems
+
1
23
Z2n-3 (s)
.
Y2n (s)
+
+
+
Y2n -2 (s)
Z (s)
Considering the continued fraction
(4.3) as a tool for designing a correZ2n2n-1-1(s)
equations (4.4) and (4.5) we find the transfer function of
sponding LC circuit, we can conclude that stability of a Combining
linear systemthe
is equivFigure 4.3: Nested
multiple-loop
control
system
– level
alent to realizability
its test
function control
R(s) with
the
help2.2.ofmultiple-loop
only passive system shown in Fig. 4.4:
the
Figure 4.3: of
Nested
multiple-loop
system
–nested
level
electric components.
1
Q2n−1 (s)
4. of
CONTINUED
FRACTION
EXPANSION (CFE)
20
Using the equations (4.4) and (4.5) we obtain theCHAPTER
transfer function
the
Y2n -2 (s)
+
+
Z2n -1(s)
1 Using the equations (4.4) and (4.5) we obtain the transfer function of the
+
+
bn−1
1
+
+ s + bsystem
shown in Fig. 4.3:
ns
1
+
is +stable. If some bk is negative,
If bk > 0, k = 1, . . . , n, then theY2nsystem
* (s)
1
Y2n* (s)
Q2n−1 (s) = Z2n−1 (s) + P2n (s) = Z2n−1 (s) +
.
(4.6)
then the system is unstable.
m(s)
1
.................................
1
1
Y2n−2 (s) +
1
Z2n−1 (s) +
Y2n (s)
The rational function R(s) should be written in the form of a continuous
11
1 1
+
form of Pa2nfraction,
the
highest
power of(4.5)
s is
fraction: its test function
,,
(4.5)
(s)P2n
=(s) = in which
+ - in1 the
Z2n -1(s)
∗ ∗ = =
1
1
1 +(4.3)
11 ·+Y12n
· (s)
Y2n (s) Y2n
Y(s)
2n (s)
R(s) =
1
contained
in bthe
denominator:
* (s)
Y
2n
s
+
1
Y2n* (s) where Y
∗ = Y ∗ (s)
1 Y(s)
+"1. 4.3: Nested multiple-loop
where
2n1.Figure
# control system – level 2.
b2 sY
+2n (s) =2n
2n (s) +
. . . . . . .system
. m(s)
. . . . – level 1.
n(s)
Figure 4.2: Nested multiple-loop control
n(s)
Y4 (s) +
20
2n
4.4
Z2n-3 (s)
(4.9)
1
Z1 (s) +
22
Y2n -2 (s)
.................................
1
1
Y2n−2 (s) +
1
Z2n−1 (s) +
Y2n (s)
1
Z(s) =
CHAPTER
4. CONTINUED
FRACTIONbetween
EXPANSION
(CFE)fraccontrol
systems
Let us now establish
an interesting
new relationship
continued
It is also known
continuous
fraction
expansions
tionsthat
and nested
multiple-loop
control
systems. can be used for investiLet us now establish an interesting new +relationship between continued frac1
gating
stabilityFRACTION
of
systems.
Forfact
this,
the
polynomial
Q(s)
Welinear
first
recall
the known
that
thecharacteristic
transfer
function
R(s) of the control
CHAPTER
4. CONTINUED
EXPANSION
(CFE)
19
tions
and
nested
multiple-loop
control
systems.
loop with
aWe
negative
feedback
shown
in Fig.
4.1
is *given
by
[8]
firstofrecall
the
known
fact
that
thebe
transfer
function
the control
of the differential
equation
the
system
should
divided
inR(s)
twoof parts,
the
Y (s)
+
loop
with
a
negative
feedback
shown
in
Fig.
4.1
is
given
by
[8]
G(s)
“even” part (containing
even powers of s) and G(s)
the “odd” part (containing odd
.
(4.4)
R(s)
=
G(s)
+ G(s)H(s)
powers of s):
H(s)
.
(4.4)
R(s)1 =
1 + G(s)H(s)
+
+
Q(s) = m(s)
+ n(s).
1
From (4.4) it immediately
follows
that thetransfer
function of the circuit
(4.9)
1
Y4 (s) +
CFE and nested multiple-loop
or R(s) =
1
+
control
4.4
CFEsystems
and nested
multiple-loop
CFE and
stability
of linear
systems
R(s) =
1
Y2n -2 (s)
sponding LC circuit, we can conclude that stability of a linear system is equivto realizability
of its. .test
function
R(s)i+m
with)).
the help of only passive
the alent
points
(si , G(si)),
. , (s
i+m , G(s
in Fig.
4.2
is (CFE)
CHAPTER 4. CONTINUED shown
FRACTION
EXPANSION
in Fig.(CFE)
4.2 is
CHAPTER 4. CONTINUED FRACTIONshown
EXPANSION
1
.................................
1
1
Y2n−2 (s) +
1
Z2n−1 (s) +
Y2n (s)
Z2n -1(s)
1
bn−1 s +1
b1 s +
b2 s + Pµ (s) bn s
= is. . .stable.
.........
k = 1,!
. . . ,R
n,i(i+1)...(i+m)
then the system
If bk > 0,G(s)
Qν (s) If some bk is negative,
1
then the system is unstable.
µ
p0 + p1 s + . . . + pbn−1
µ ss + b s
Considering the =
continued fraction (4.3) as a tooln for designing a corre(4.2)
ν
q
+
q
s
+
.
.
.
+
q
s
>
0,
k
=
1,
.
.
.
,
n,
then
the
system
is
stable.
If some
bk is negative,
If
b
1
ν
k
sponding LC circuit,
we can0 conclude
that stability
of a linear
system
is equivthen the system is unstable.
alent to realizability
function
R(s) with the help of only passive
m
+ 1 =of its
µ +test
ν+
1
Considering the continued fraction (4.3) as a tool for designing a correelectric components.
4.4
+
+
1
4.3
+
Y4 (s) +
Y2n* (s)
G(s)
where ai s and bi s are rational functions of the variable s, or are constants.
H(s)
! (CFE)which is19an
CHAPTER
4. CONTINUED
FRACTION
EXPANSION
The application of the
method
yields a rational
function,
G(s),
approximation of the irrational
Figure 4.1:function
A controlG(s).
loop
with
a
negative
feedback.
+
G(s)
On the other hand, for interpolation purposes,
rational functions are someThe
rational
function
R(s)
should
be
written
in thedue
formtooftheir
a continuous
times superior to polynomials. This is, roughlyH(s)
speaking,
ability
fraction:
to model functions
with poles. (As it can be seen
1 later, branch points can be
Figure
(4.3)
R(s)4.1:
= A control loop with a negative feedback.
1
considered as accumulations of interlaced
b1 s + poles and zeros). These techniques
1
The
rational
function
R(s)
should
be
written
in
the
form
of
a
continuous
b
s
+
are based on the approximations of an irrational
function, G(s), by a rational
2
............
fraction:
1 in the variable s:
function defined by the
quotient of R(s)
two =polynomials
(4.3)
passing through
+
+
1
-
1
Y2 (s) +
22 Z3 (s) +
1
2 + 0.576s + 0.0256
s0.002844
9s40.064s
+ 12s3 ++4.32s
s4 + 1.333s3 + 0.478s2 +
s4 + 4s3 + 2.4s2 + 0.448s + 0.0256
2
H2 (s) = 4
3
2
Example 5.2.1.
3 + 4.32s2 + 0.576s + 0.0256
1.405s
+ 0.8433s
+ 0.1574s +
0.3513s4 +where
9s + 12s(ωT
is the approximation
for0.008995
high frequencies
>> 1), and Hl (s)
Hh (s)
H1 (s) =
3 + 0.478s
Example
s4 the
+5.2.2.
1.333s
+Carlson’s
0.064s
Performing the
CFE
of
function
(5.1),
with
T+=0.002844
1, α(ω=<<0.5,
5.3 2 for
method
the
approximation
low frequencies
1). we obtain:
Performing the CFE of the function (5.2), with T = 1, α = 0.5, we obtain:
5.3 Carlson’s method
The method
Example
5.2.1.
3
2 proposed by Carlson in [4, 5], derived from a regular Newton
Example 5.2.2.
+ 0.8433s
0.1574s
+ (5.1),
0.008995
0.3513s4 + 1.405s
4the CFE
3 +
2 function
iterative
approximation
of the
can
bewe
considered
Performing
of
the
with
T α-th
= 1,inroot,
α[4,=5],
0.5,
obtain:
+ 4sused
+for
2.4s
0.448s
+ 0.0256
sprocess
The +
method
proposed
by Carlson
derived
from a regular Newton
HPerforming
1 (s) =
as(5.2),
belonging
to this
group.
starting
point
of the method is the statement
(s)
H23function
thesCFE
of the
with
T
=
αThe
=
0.5, we
obtain:
4 + 1.333s
4 2 + 30.064s
2
+=
0.478s
+1,used
0.002844
process
for iterative approximation of the α-th root, can be considered
9sof +
12s
+44.32s
+ 0.576s + 0.0256
the0.3513s
following
relationships:
+
1.405s3 + 0.8433s2 + 0.1574s + 0.008995
as belonging to this group. The starting point of the method is the statement
H1 (s) =
3 + 0.478s2 + 0.064s + 0.002844
2.4s2 + 0.448s
+the
0.0256
s4 + 4s3 +
1/α
s4 + 1.333s
of
following
relationships:
− (G(s))
= 0;
H(s) = (G(s))α
(H(s))
Example 5.2.2. H2 (s) = 9s4 + 12s3 + 4.32s2 + 0.576s
+ 0.0256
1/α
(H(s)) − (G(s)) = 0;
method
Performing the 5.3
CFE ofCarlson’s
the Example
function5.2.2.
(5.2), with T = 1, α = 0.5, we obtain:
(5.3)
H(s) = (G(s))α
Performing the CFE of the function (5.2), with T = 1, α = 0.5, we obtain:
5.3
The method proposed by Carlson in [4, 5], derived from a regular Newton
0.0256
s4 + 4s3 + 2.4s2 + 0.448s
+ 4s
+ 2.4s + 0.448s + 0.0256
s +
Carlson’s
process
for iterative approximation
of the
α-th root, can be considered
H (s) =
= usedmethod
H (s)
4
2
4
3
2
2
3
2
9s4 + 12s3 + 4.32s2 + 0.576s + 0.0256
(5.3)
CHAPTER 6. GENERAL APPROACH TO FRACTANCE DEVICES
General Approach to Fractances
32
CHAPTER 6. GENERAL
APPROACH
TO FRACTANCE DEVICES
6.2 Domino
ladder circuit
32
General
Domino Approach
ladder circuit to Fractances
6.2
Z1
Z3
Z 2n -3
Z 2n -1
6.2
Z1
A devices or a circuit exhibiting fractional-order
behaviour is called a fractance.
Y4
Y
Z(s)
Domino
ladder
circuit
2
Z3
Z1
Z3
Z 2n -3
Y2n -2
Z 2n -3
32
Y2n
Z 2n -1
Z 2n -1
Figure 6.2: Finite ladder circuit.
Y
• domino ladder circuit network,
• a tree structure of electrical elements,
• transmission line circuit
Y
Y
Y2ladder lattice
2n -2
4
Domino
fractional
operator
Y4 can approximate Y
Y2n more 2n
Y2 networks
2n -2
Z(s)
effectively than the lumped networks [10, 30].
Z(s)
Let us consider the circuit depicted in Fig. 6.2, where Z2k−1 (s) and Y2k (s),
k = 1, . . . , n, are given impedances of the circuit elements. The resulting
impedance Z(s) of the entire circuit can be found easily, if we consider it in
Figure 6.2: Finite ladder circuit.
the right-to-left direction:
1 circuit.
Figure 6.2: Finite ladder
Design of fractances can be using a truncated CFE,
which gives a rational approximation.
(6.1)
Z(s) =ladder
Z1 (s)+ lattice networks can approximate fractional operator more
Domino
1
Y2 (s) +
1
effectively than the lumpedZ3networks
[10, 30].
(s) +
Let lattice
us considernetworks
the circuit depicted
in Fig. 6.2, where1fractional
Z2k−1 (s) and Yoperator
2k (s),
Y4 (s)approximate
+
Domino ladder
can
more
. . . .circuit
. . . . . . . . .elements.
. . . . . . . . . . . . The
. . . . . .resulting
..
k = 1, . . . , n, are given impedances of the
1
effectively than
the lumped
networks
[10,
impedance
Z(s) of the
entire circuit
can30].
be found easily, if1we consider it in
Y2n−2 (s) +
1
the right-to-left
direction:
Let us consider
the circuit
depicted in Fig. 6.2,Zwhere
2n−1 (s) + Z2k−1 (s) and Y2k (s),
Y (s)
2n
1
k = 1, . . . , n, Z(s)
are =given
(6.1) resulting
Z1 (s)+ impedances of the circuit elements. The
1
The relationship
Y2 (s) +between the finite domino ladder network, shown in Fig. 6.2,
impedance Z(s) of
the entire
circuit can be found
1 easily, if we consider it in
Z3 (s) +(6.1) provides an easy method for designing a circuit
and the continued fraction
1
Y4 (s)
the right-to-left direction:
with a given impedance Z(s).
For+ this one has to obtain a continued fraction
.................................
Z(s) = Z1 (s)+
expansion for Z(s). Then the obtained particular expressions
for Z2k−1 (s) and
1
1
Y2k (s), k = 1, . . . , n, will give the types 1
of necessary components
of the circuit
Y2n−2 (s) +
and their nominal values.
1 Z2n−1 (s) + Y 1(s)
2n
Y2 (s)
+
Example 6.2.1.
1
Z3 (s)
+ with the impedance
To design
a circuit
(6.1)
The relationship between the finite domino ladder network,1 shown in Fig. 6.2,
Y4(6.1)
(s) provides
+
and the continued fraction
for designing a circuit
+ easy
4s2 +method
1
s4an
Z(s) .=. . . . . . . . . ., . . . . . . . . . . . . . . . . (6.2)
......
with a given impedance Z(s). For this ones3has
+ s to obtain a continued fraction
1
expansion for Z(s). Then the obtained particular expressions for Z2k−1 (s) and
1
Y2k (s), k = 1, . . . , n, will give the types
of
necessary
components
of
the
circuit
Y2n−2 (s) +
1
and their nominal values.
Z2n−1 (s) +
Y2n (s)
Example 6.2.1.
To design a circuit with the impedance
31
6.1.2
Negative impedance converters
It can be shown that the use of CFE for analogue realization of arbitrary
transfer functions may lead to the appearance of negative impedances. This
observation is not unknown. For example, in the paper [10] S. C. Dutta Roy
in [19] and
recalls Khovanskii’s continued fraction expansion for x1/2 found 1/2
makes a remark that
S. C. Dutta Roy on Khovanskii’s CFE for x :
“. . . if x is replaced by the complex frequency variable s, then
the realization would require a negative resistance. Thus, the [Khovanskii’s] CFEs do not seem to be useful for realization of fractional
or capacitor.”
CHAPTER 6. inductor
GENERAL
APPROACH TO FRACTANCE DEVICES 31
Then he describes a method for circumventing this difficulty, which gives
6.1.2
Negative
impedance
converters
a continued fraction
expansion with
positive coefficients.
However,
possibility
of realization
of negative
However,the
the possibility
of realization
of negative impedances
in electric
It can be circuits
shown has
thatbeen
thepointed
use ofoutCFE
for analogue
realization
arbitrary
by Bode
[3, Chapter
Later,pointed
inof1970s,
operaimpedances
in electric
circuits
hasIX].
been
out
transfer functions
may
lead
to
the
appearance
of
negative
impedances.
This
tional amplifiers appeared, which significantly simplified creation of circuits
exby hibiting
H.
W.negative
Boderesistances,
inFor
1945.
observation
is not
unknown.
example,
in
the
paper
[10]
S.
C.
Dutta
Roy
negative capacitances, and negative inductances.
Such circuitscontinued
are called negative-impedance
converters
found in [19] and
recalls Khovanskii’s
fraction expansion
for x1/2[9].
Thethat
simplest scheme of a negative impedance converter (or current inverter)
makes a remark
is shown in Fig. 6.1. The circuit consists of an operational amplifier, two
of replaced
equal resistance
and a component
with
the impedance
“.resistors
. . if x is
by theR,complex
frequency
variable
s, then Z. The
entire circuit,
considered
a single element,
has Thus,
negative
the realization
would
requireasa negative
resistance.
theimpedance
[Kho- −Z.
=
V
/(−Z)).
This
means
that
I
in
in
vanskii’s] CFEs do not seem to be useful for realization of fractional
For example, taking a resistor of resistance RZ instead of the element Z, we
inductor or capacitor.”
obtain a circuit, which behaves like a negative resistance −RZ . The negative
resistance
means
that if such
an element of negative
resistance,which
for instance,
Then he
describes
a method
for circumventing
this difficulty,
gives
kΩ is connected in series with a classical 20 kΩ resistor, then the resistance
a continued−10
fraction
expansion with positive coefficients.
of the resulting connection is 10 kΩ.
However, the possibility of realization of negative impedances in electric
circuits has been pointed out by Bode [3, Chapter IX]. Later, in 1970s, operaR
Iin
tional amplifiers appeared, which significantly
+simplified creation of circuits exhibiting negative resistances, negativeVincapacitances,
and negative inductances.
_
Such circuits are called negative-impedance converters [9].
Z
The simplest scheme of a negative impedanceRconverter (or current inverter)
is shown in Fig. 6.1. The circuit consists of an operational amplifier, two
resistors of equal resistance
R, and
component with converter.
the impedance Z. The
Figure
6.1: aNegative-impedance
entire circuit, considered as a single element, has negative impedance −Z.
This means that Iin = Vin /(−Z)).
For example, taking a resistor of resistance RZ instead of the element Z, we
obtain a circuit, which behaves like a negative resistance −RZ . The negative
resistance means that if such an element of negative resistance, for instance,
−10 kΩ isNegative
connected inimpedance
series with a classical
20 kΩ resistor,
the resistance
converters
are then
available:
of the resulting connection is 10 kΩ.
Iin
Vin
R
+
_
Z
R
Figure 6.1: Negative-impedance converter.
2
The relationship between the finite domino
network, shown in Fig. 6.2,
+1
s4 + 4sladder
Z(s) =
,
(6.2)
s3 +easy
s
and the continued fraction (6.1) provides an
method for designing a circuit
with
a given impedance
Z(s). For
this one has
obtain a continued
fraction 33
CHAPTER
6. GENERAL
APPROACH
TOtoFRACTANCE
DEVICES
expansion for Z(s). Then the obtained particular expressions for Z2k−1 (s) and
Y2k we
(s),have
k = 1,
. . , n, will
giveinthe
types offraction
necessary components of the circuit
to .develop
Z(s)
continued
and their
nominal values.
CHAPTER 6. GENERAL
APPROACH
TO FRACTANCE
DEVICES 33
4s2 + 1
1
s4 +
.
(6.3)
=s+
Example 6.2.1. Z(s) =
1
1
s3 + s
s
+
wedesign
have to
develop with
Z(s) the
in continued
fraction
To
a
circuit
impedance
9
1
3 circuit with
Example
1: design
a domino
ladder
s+
CHAPTER 6.
GENERAL
APPROACH
TO FRACTANCE
DEVICES
2
4 GENERAL
CHAPTER
APPROACH
DEVICES 3333
2 FRACTANCE
2
+ 4s2 +4 1 4s
1TO
s6.
+1
.s
(6.3)
s+
Z(s) = Z(s)3 = s +=
(6.2)
1 3
1,
s
+
s
3
+fraction
s s+
we have to develop
in continued
we have Z(s)
to develop
Z(s) sin continued
fraction
9
1
3
From this expansion it follows that
s+
Find a CFE:
s4 1+ 4s2 + 1
s4 + 4s2 +
2
12 1
s 1. .
(6.3)
=s+
Z(s) =
(6.3)
9
2
s3=+ ss + 1 1 1 s 3+1
s;s
Y2 (s) =s +s; 3
Z3 (s) =s3 +
9Y4 (s)1= s.
2
3
3
s
+
9
1 2
3
s+2
2 3s
2
Therefore, for the analogue 9realization in the
1 form 3ofs the2 first Cauer’s
Z1 (s)
Y2 (s)that
s;
Y4 (s)
= s;From
Z
= s;
= following
= s.of coils and
3 (s)
expansion
follows
canonic LC
circuit
[20] this
we
have
the
values
2 to itchoose
3
3
From this expansion it follows that 9
capacitors:
1
2
Z1 (s) = s;
Z(s) =
Therefore,
Z1 (s) = s; realization
s; theY2form
s.
Z3 (s) = in
(s) = ofs; the Yfirst
Therefore, for the analogue
Cauer’s
4 (s) =
2
3
1
2 3 and
9 =to9choose
2 coils
canonic LC Zcircuit
[20]
the
values
of
s; CY
Y4 (s)
s.].first Cauer’s
s; LweZ=have
(s)1following
=[F ];s;
=
3 (s)the
2=
for
analogue
realization
in
the
form
of
the
[H];=Therefore,
=
L1 = 11 (s)
[H];
C
[F
3
2
4
2
3
capacitors:
For the first
Cauer’s
LC
we
must
2canonic
3circuits
3 3 oftake
canonic LC circuit
[20] we have to choose
the following values
coils and
Therefore,capacitors:
for the analogue
realization 1in the form of 2the first Cauer’s
9
Example
=
L3 =
L16.2.2.
[H]; to choose
C2 = the
[F ];following
C4 =
[F ].of coils and
canonic
LC1 [H];
circuit [20]
we have
values
9
1
2
2
3
3
[H];equation
L3 = (6.2)
L1 = 1by
[H]; canCbe
[F ];
C4 =in [F
]. form
The
function Z(s) given
also
the
2 = written
capacitors:
2
3
3
Example 6.2.2.
4
29
1
2
+
4s +[H];
1
1 C = [F1 ];
s6.2.2.
Example
1Z(s)
[H];=
Lby
L1 = Z(s)
3 =
The function
given
equation
writtenCalso
the]. form
.4 = in [F
(6.4)
= (6.2)
+ 2can 3be
The function
1 be written3 also in the form
1 (6.2) can
s3 Z(s)
+ s2 given bys equation
+
11
s4 + 4s2 + 1 s4 +14s2 + 13s 1 1 9
Example 6.2.2.
(6.4) (6.4)
Z(s) =
.
Z(s) = = 3 + 1 = +2s1+ 2.
1
1
s3 + s
s
ss + s(6.2)+can
The function Z(s) given by equation
be
written
also in the form
+
1
9
9 3s 1 3s
3s
+ 2s + 2
2
2s 1 2
+1
1
s4 + 4sthat
From this expansion
. 3s
(6.4)
= +
Z(s)it=follows
3s
1
1
s
s3 + s
From this expansion
it follows that +
1
9
1
2
1
9
3s
From
expansion
follows
that
+
(s) =
=
(s)
=
(s)
=
Z1this
;
Zit3 (s)
;
Y
;
Y
.
29
4
1
s
3s 2s 1 ;2 Y4 (s)3s= 2 .
Z1 (s) = ; 2s
Z3 (s) = ;
1
1Y2 (s) = 3s3s
2
s 9
2s
= the
=
Z1 (s)for
; analogue
Z3 (s) =
;
Y2 (s)
;
Y
. 3sCauer’s
4 (s) =
Therefore,
realization
in
the
form
of
the
second
s Therefore,
2s
3s
3s second
for the analogue realization in the form of the
Cauer’s
From
this expansion
ithave
follows
that
canonic
LC
circuit
[20]
we
to
choose
the
following
values
of
coils
canonic
LC circuitrealization
[20] we haveintothe
choose
theoffollowing
valuesCauer’s
of coilsand
and
Therefore, for
the analogue
form
the second
capacitors:
capacitors:
1 we have to 9choose the following
1
2 and
canonic LCZcircuit
[20]
= coils
Z3 (s) = ;
Y2 (s) = ; values
Y4 (s)of
.
1 (s) = ;
s
2s 2
3s
3s
3
capacitors:
C1 = 1 [F2];
C3 =
[F ];
9L2 =
L = 3 [H];
L34 =
[H].
Therefore, for the analogue
in the form of the
29 realization
3 2second Cauer’s
General
Fractances
[F ]; to choose
L to
[H].of coils and
= 1 circuit
[F ];Approach
C we
= have
= 3the
[H];
L =values
C LC
canonic
[20]
following
9
2
C1 = 1 [F ];
C3 =
[F ];
2
3 [H];
L4 =
2
[H].
3
2
4
Example 6.2.3.
To design a circuit with the impedance
capacitors:
Example
6.2.3.
CHAPTER
6. GENERAL APPROACH TO FRACTANCE DEVICES 34
Example
To design6.2.3.
a circuit with the impedance
s4 + 3s2 + 8
2
Z(s) L
= = 33 [H]; , L = 3 [H].
[F ]; representation
=circuit
1 [Fa];continuous
3 =impedance
2 2s + of
4
To
with Cthe
onedesign
has C
to1aobtain
fraction
Z(s),
4s the function
1
Example II: design a domino ladder circuit with
(6.5)
CHAPTER 6. GENERAL APPROACH
TO2 +
FRACTANCE
DEVICES
34
9 s4 + 3s
2
8
2
CHAPTER 6. GENERAL
DEVICES 34
1
3sAPPROACH
+=
8 41 TO
s4 +Z(s)
2 FRACTANCE
3
s 2s
+
3s
+
8 ,
(6.6) (6.5)
s+
Z(s) =
=
3
+
4s
1
2
Example
6.2.3.
Z(s)
=
,
(6.5)
2s
+
4s
one has to obtain a continuous fraction representation
of the function Z(s),
3 2s +
2srepresentation
+ 4s
one has to obtain a continuous fraction
1 of the1function Z(s),
To
design a circuits4with
2the impedance
− s1 +
1
4+ 3s 2+ 8
3
12
1
3s + 8 = 1 s +
Z(s)
== s +
− s
3
(6.6)(6.6)
s
+
Z(s)
=
2
2s2s3++4s
1 1 2
22s4 +2s
4s
2s3s
+++ 81,
1
Z(s)
=
(6.5)
1
1
s+
−− s +
2s3 + 4s
3 3
1212
1
1
s(s)s = − 3 s.
Z1 (s) = s;
Z3 (s) = − s;
Y2 (s) = 2s; − 2Y−
42
2
12
2
From
expansionititfollows
follows that
that
From
thisthis
expansion
Therefore,
for the analogue
realization in the form of the first Cauer’s
1
1
3 3
1
1
canonic
LC
Z
Y2 (s)
= s; [20] Zwe
− to s;choose
(s) = −of s.coils and
1 (s)
3 (s)have
Z1 (s)
s;2
Z3 (s)
Y2the
=circuit
==−
(s)=following
=2s;2s; Y4values
Y4 (s) =2 − s.
12 s;
2
12
2
capacitors:
Therefore, for the analogue realization in the form of the first Cauer’s
Therefore,
for the analogue
realization in the form of the 3first Cauer’s
1
1
canonic
circuit [20]
we−have[H];
to chooseC the
following values
L3 =
[F ].and
L1 = LC[H];
C4 = of
− coils
2 = 2 [F ];
canonic
LC 2circuit [20] we have
12 to choose the following values2 of coils and
capacitors:
capacitors:
Here we see
and capacitance. Such elements
cannot
1 negative inductances
1
3
[H];
L3electric
[H];
C2 = However,
[F ]. realized
L1 =
= − components.
2 [F ];
C
4 = −
be realized
using
passive
they
can
1 2
112
3
2 be
[H];
L =inductances
C2 =operating
LHere
− [H]; and
2 [F ]; Such
C
1 =
4 = − [F ].
capacitance.
elements
with the
help
of negative
active3 components,
namely
amplifiers.
We can
2we see
12
2cannot
be realized
passive
However,
beinrealized
realize
this
byusing
negative
inpedance
converter
which
was they
described
previous
Here
we see
negative
inductances
and capacitance.
Suchcanelements
cannot
with theusing
help of
active electric
components,
namely operating
amplifiers.
besubsection.
realized
passive
components.
However,
they can We
be can
realized
realize this by negative inpedance converter which was described in previous
with the help of active components, namely operating amplifiers. We can
subsection.
realize
by negative inpedance
converter
which was described in previous
6.3 thisTransmission
lines
circuit
subsection.
Transmission
lines
circuit
Let6.3
us consider
the circuit depicted
in Fig.
6.3, where Z2k−1(s) and Y2k (s), k =
1, . . . , n, are given impedances of the circuit elements. This structure is known
Let usTransmission
consider the circuit depicted
in Fig.
6.3, where Z2k−1(s) and Y2k (s), k =
6.3
lines
circuit
as a1, transmission
line
or a symetrical
domino
ladderThis
lattice
network
as well.
. . . , n, are given
impedances
of the circuit
elements.
structure
is known
Theasresulting
impedance
Z(s)
of the entire
circuit canlattice
be expressed as
as a CFE
a
transmission
line
or
a
symetrical
domino
Let us consider the circuit depicted in Fig. 6.3,ladder
where Z2k−1network
(s) and Y2kwell.
(s), k =
described by (6.1) with symetrical distribution of elements, Z
(s) = Z (s)
We have:
Notice negative values - can be done using OpAmps
2k−1 as a CFE
2k
The resulting impedance Z(s) of the entire circuit can be expressed
From this expansion it1follows that
L3 = − [H]; 3 C2 = 2 [F ];
C4 = − [F ].
subsection. 1 L1 = [H];
2
3
L1 = [H]; 1 L3 = − [H]; 2 C2 = 2 1[F ]; 312C4 = − [F1].
1
L3 =
[H];
C
2 [F ]; cannot
C4 = − [F ].
2 = − s;
2− capacitance.
1Y=(s) [H];
2 = elements
Here=we
inductances
and
Such
Z1 (s) = s;
Z3 (s)
Y212
(s)
2s;see Lnegative
4 2 = − s.
12
2
2 Here we see negative
12
2 components.
inductances
capacitance.
Such
elements cannot
be realizedand
using
passive electric
However, they can be realized
Hereofwe
negative
inductances and capacitance. Such elements cannot
Therefore, for
the analogue
realization
thehelp
form
thesee
first
Cauer’s
Transmission
lines
circuit
be realized
using6.3
passive
electric
components.
However,
they
cannamely
be realized
within the
of active
components,
operating amplifiers. We can
be realized
using
electric components. However, they can be realized
canonic LC circuit
we have
to choose
the this
following
values
of passive
coils
and
with [20]
the help
of active
components,
namely
operating
amplifiers.
can was described in previous
realize
by negative
inpedance
converterWe
which
with
the
help
of
active
components,
namely
operating
We can
capacitors:
Let us inpedance
consider
the
circuitwhich
depicted
Fig. 6.3,
Z2k−1
(s) and
Y2k (s), amplifiers.
k=
realize this by negative
converter
was in
described
in where
previous
subsection.
realize
this
by
negative
inpedance
converter
which was
described in previous
1 subsection.
11, . . . , n, are given impedances of the
3 circuit elements.
This structure
is known
L3 = − [H];
C2 = 2 [F
L1 = [H];
];
C4 = − [F ].
subsection.
2
12
2
as a transmission
line or a symetrical
dominocircuit
ladder lattice network as well.
6.3 Transmission
lines
Here we see negative inductances
and capacitance.
Such
cannotcircuit can be expressed as a CFE
The resulting
impedance
Z(s)elements
of the entire
6.3
Transmission
lines
circuit
be realized using passive electric
components.
However,
they
can
be
realized
us consider
circuit depicted
in Fig. 6.3,
where circuit
Z2k−1
(s) and
k=
described Let
by (6.1)
with the
symetrical
distribution
oflines
elements,
Z2k−1
(s) Y=2k (s),
Z2k (s)
6.3
Transmission
with the help Let
of active
components,
namely
amplifiers.
We
1, . .=
. ,operating
n,
are
given
impedances
thecan
circuit
elements.
us consider
the
circuit
depicted
Fig.
6.3,
where Z2k−1of
(s)
and
Y2k (s),
k = This structure is known
(s)
Yin
and
Y2k−1
2k (s), k = 1, . . . , n.
as of
awhich
transmission
line or athe
symetrical
domino
ladder
lattice
network
as
well.
inpedance
converter
was
described
in
previous
1, . . . , n, are
given impedances
the
circuit
elements.
This
structure
is
known
Let
us
consider
circuit
depicted
in
Fig.
6.3,
where
Z
(s)
In the case if we assume the same elements in transmision line in2k−1
series and Y2k (s), k =
The resulting
Z(s)
of the
entireof
circuit
can be elements.
expressed as
a CFE
subsection.
as a transmission line or a symetrical
ladder
lattice
network
as
well.
1,domino
. . .impedance
, n, are
given
impedances
the
circuit
This
structure is known
(Z ) and the same in shunt (Z ), we get a structure depicted in Fig. 6.4.
CHAPTER 7.
R
2
3
2
4
6
2n-1
2n
3
5
3
4
5
6
4
2n-1
!5
Magnitude [dB]
!10
2n
1
3
5
!20
!25
!30
!35
!40
2
3
10
4
10
Frequency [rad/sec]
10
!30
!40
!50
!60
!70
!80
!90
2
3
10
4
10
Frequency [rad/sec]
10
Figure 7.5: Bode plots of the I 1/2 controller where half order integral was
approximated by transmission line depicted in Fig. 7.2.
2n-1
as a transmission
line or a symetricallines
dominocircuit
ladder lattice network as well.
6.3 ZaTransmission
Za
The Za
resultinglines
impedance
Z(s) of the entire circuit can be expressed as a CFE
Transmission
circuit
Y6
Y 2n-1
Z6
Y 2n
Z 2n
us consider
circuit depicted
in Fig. 6.3,
where Z2k−1
(s) and
k=
described Let
by (6.1)
with the
symetrical
distribution
of elements,
Z2k−1
(s) Y=2k (s),
Z2k (s)
Figure 6.3: General structure of transmission line.
Zb us Zb
Zb
Zb
Zb
Zb depicted
Zb given
1, . .=
.Zb, n,
are
theand
circuit
elements.
Let
consider
the
circuit
Fig.
6.3,impedances
where Z2k−1of(s)
Y2k (s),
k = This structure is known
(s)
YZain
and
Y2k−1
2k (s), k = 1, . . . , n.
Za
Za
Za
Za line or Za
Za
Za
as
a
transmission
a
symetrical
domino
ladder
lattice
network
as
well.
1, . . . , n, are given impedances
of the
circuit
elements.
is known
In the case
if we
assume
the This
samestructure
elements
in transmision line in series
The resulting
impedance
of the
entire as
circuit
as a transmission line or a symetrical
domino
ladderZ(s)
lattice
well.can be expressed as a CFE
(Zcomposed
same
in
shunt
(Z
), we
getnetwork
a structure
depicted in Fig. 6.4.
a ) and the
Figure 6.4: Transmission line circuit
of two
inpedance
Za and
Zbb.symetrical
described
by (6.1)
with
of elements,
Z2k−1 (s) = Z2k (s)
Zb circuit
Zb Zbcan
Zb expressed
Zb distribution
Zb
Zb
The resulting impedance Z(s)
of theZ(s)
entire
be
as aZbCFE
Impedance
of
such
kind
of
transmision
line,
which
can be used also as a
Za
Za
Za
Za
(s) = Y2k (s),
k = 1, . . . ,Zn.
and Y2k−1
described by (6.1) with symetrical
distribution
of elements,
2k−1 (s) = Z2k (s)
6.4 Tree structure circuit
model
of
real
cable
line
(solved
by
O.
Heaviside
in
1887),
can
be
as
the case if we assume the same elements in transmisionexpressed
line in series
= Y2k (s),TO
k FRACTANCE
= 1, . .In
. , n.
and
Y2k−1 (s)APPROACH
CHAPTER
6. GENERAL
DEVICES 36
!
Figure
6.4:
Transmission
line circuit
of structure
two inpedance
Za and in
Zb .Fig. 6.4.
(Z
same
in 6.5,
shunt
(Z=b ),composed
we
Let us consider the total
impedance
theassume
fractance
circuit
asthe
shown
in Fig.
a ) and
In the
case ifofwe
the
same
elements
in
transmision
line
series depicted
Zaget
.Zba. in
(6.7)
Z(s)
which has a recursive structure with
the combination
Za
such Zkind
of transmision
line, which can be used also as a
a
(Za ) and the sameZa in
shunt
(ZbImpedance
), of
wetwo
getimpedances
a of
structure
depicted
in Fig. 6.4.
6.4
Tree
structure
and Zb . Impedance is derived as geometrical
mean
of Zaof
Zb ,cable
If we substitute
aand
resistance
Za(solved
= circuit
R and
a Heaviside
capacitance
Zb =can
1/sC,
then the
real
be expressed
as
Zb model
Za
Impedance
of such kind
of transmision
line,line
which
canby
beO.used
also asina 1887),
!
!
impedance
is usbyconsider
Za. Let
the total impedance
the fractance
Za .Z
= line
model of realZ(s)
cable
(solved
O. Heaviside
in 1887),ofZ(s)
can
asshown in Fig. 6.5,
Zb
Zacircuit
.Zb . as
(6.7)
= expressed
b
"structure
" be
!
which
has
a
recursive
with
the
combination
of
two
impedances
Z
a
Zb
R −1/2
R mean
.derived
(6.7)
Z(s)
= Za .Z
If
Z
aofcapacitance
Z = 1/sC, then
the
andwe
Zbsubstitute
. Impedance
a and| Zb , . b
s as geometrical
ω −1/2
Z(s) = aisbresistance
=a = R and
eZ−jπ/4
(6.8)
s=jω
Za
= R and a capacitance
Zis
then
If we substitute a resistance Za Za
impedance
!C
b = 1/sC,C
If we substitute a resistance
Z
=
R
and
a
capacitance
Z
=
1/sC,
then
the
a
b
Zb
Z(s) = Za .Z"
"
b.
the impedance shows the
as (see also (6.8))
Zb fractance characteristics
R −1/2
R −1/2 −jπ/4
impedance
Za
" is
s
ω
Z(s) =
=
e
|s=jω .
(6.8)
Zb "
"
"
R −1/2
R −1/2
C
C
Z(s) =
s
ωZb R
=
e−jπ/4
|s=jω
,
R a−1/2
we
substitute
a = R and a capacitance Zb = 1/sC, then
C
sIf−1/2
ω resistance
Z(s)C=
=
e−jπ/4 |Zs=jω
.
(6.8)
Z(s)
the impedance shows
the fractance characteristics as (see also (6.8))
C
C
Z(s)
Take
Then
which is a fractional order integral with absolute value of impedance propor"
"
A phase
self-similar
treeiscircuit
composed
of independent
two inpedance Z
and Z
b.
tional toFigure
ω −1/26.5:
and
angle
constant
−π/4,
ofa the
frequency.
R −1/2
R −1/2 −jπ/4
Z(s) =
s
ω
=
e
|s=jω ,
However, it is impossible to construct such an infinity structure as show in
C
C
circuit, where capacitance C was the same at every stage and resistance R
Fig. 6.5.increased
Therefore,
it seems to be important to study
the effect
of the numby the ratio a at every stage of branching.which
The resulting
impedance
is a fractional order integral with absolute value of impedance proporber of stage
in
the
recursive
self
infinity
(binary)
structure
on
the
impedance
Z(s) had the form of a continued fraction expansion.
tional to ω −1/2 and phase angle is constant −π/4, independent of the frequency.
phase
characteristics
[31].angle is constant !"/4, independent of the frequency
However, it is impossible to construct such an infinity structure as show in
A similar tree structure as a fractal model ofFig.
rough
interface between two
6.5. Therefore, it seems to be important to study the effect of the nummaterials of very different conductivities, e.g. anber
electrode
an electrolyte
of stageand
in the
recursive self infinity (binary) structure on the impedance
was studied in [21]. There was described a slightly
different structure
of tree
characteristics
[31].
CHAPTER 7.
REALIZATION OF FRACTIONAL CONTROLLERS
41
CHAPTER 7.
42
Sample implementations -- transmission lines -- responses:
0.4
0.6
0.4
Output signal: U [V]
Y5
0.2
0
!0.2
!0.4
0.2
0
!0.2
!0.6
!0.8
!0.4
0
0.005
CHAPTER 7.
0.01
0.015
0.02
0.025
0.03
[sec]
REALIZATION OFTime
FRACTIONAL
CONTROLLERS
0.005
0.01
0.005
0.01
0.015
0.02
0.025
0.03
0.015
Time [sec]
0.02
0.025
0.03
[sec]
REALIZATION OFTime
FRACTIONAL
CONTROLLERS
44
1
1
0.5
0
unit step
!0.5
!1
0
CHAPTER 7.
43
Input signal: E [V]
Y4
Z4
0
0.005
0.01
0.015
Time [sec]
0.02
0.025
!0.5
!1
0
0.2
0
!0.2
!0.4
0
0.005
0.01
0.015
Time [sec]
0.02
0.025
1/2
Figure 7.7: Time response of the I
controller to sin input where half order
integral0.4was approximated by transmission line depicted in Fig. 7.2.
0
!0.2
0.03
0
0.005
0.01
0
0.005
0.01
0.015
Time [sec]
0.02
0.025
0.03
0.015
Time [sec]
0.02
0.025
0.03
1
1
0.5
saw
0
!0.5
!1
sine
0
0.6
Figure 7.6: Time response of the I 1/2 controller to unit-step input where half
0.2
order integral
was approximated by transmission line depicted in Fig. 7.2.
!0.4
0.5
0.03
0.4
Y3
Y2
Z2
Input signal: E [V]
Y1
Za
!15
2n
6
Figure 6.3: General structure of transmissionZ(s)
line.
6.3
1k !
Transmission lines circuit
Input signal: E [V]
2
2
1µF
1µF
C
2n-1
6.3
1
1k !
Sample implementations:
Figure 7.4: RC binary tree circuit.
Then
1
1k !
1µF
1µF
Take
2n
6
1k !
40
2n-1
5
4
1µF
1k !
Input signal: E [V]
1
5
1k !
1µF
Z F (s)
Phase [deg]
3
1k !
1µF
a
b
described
by a(6.1)
with can
symetrical
distribution
of elements, Zladder
= Z2k (s)
2k−1 (s) lattice
The resulting impedance Z(s)
of the as
entire
circuit
beline
expressed
as a CFE
transmission
or
aline,
symetrical
Impedance
of such
ofktransmision
which domino
can be used also
as anetwork as well.
(s) = kind
Y2k (s),
= 1, . . . ,Zn.
and Y2k−1
described by (6.1)
with symetrical
distribution
of
elements,
(s)FRACTANCE
=ofZthe
2k−1
2k (s)
The
resulting
impedance
Z(s)
entire
circuit
can be expressed
as a CFE
CHAPTER
6.
GENERAL
APPROACH
TO
DEVICES
34
model
of
real
cable
line
(solved
by
O.
Heaviside
in
1887),
can
be
expressed
as
6.3 Transmission
lines
circuit
In
the
case
if
we
assume
the
same
elements
in
transmision
line
in
series
= Y2k (s),
k = 1, . . . , n. described
and Y2k−1 (s)
CHAPTER 6. GENERAL
APPROACH
35
by (6.1) with
! symetrical distribution of elements, Z2k−1 (s) = Z2k (s)
(Z
)
and
the
same
in
shunt
(Z
),
we
get
a
structure
depicted
in
Fig.
6.4.
a
b
In the case if we assume the same
elements
transmision
.Z . in. ,series
(6.7)
Z(s)
=
a line
(s)in
=fraction
YY2k
(s),
kkZ=
n. which
oneZ has
to 6.3,
obtain
aand
continuous
representation
of the
Z(s),
2k−1
Let us consider the circuit
depicted
in Fig.
where
ZYof
(s)
and
= 1,b . . line,
Z
Z
Z
2k
such
kind
of(s),
transmision
canfunction
be used also
as a
(Za ) and the same in shunt (ZbImpedance
), we get a2k−1
structure
depicted
in Fig. 6.4.
In
case
we
assume
the same
elements
inexpressed
transmision
4thestructure
1, . . . , n, are given impedances of
the
elements.
This
is and
known
Ifsuch
we circuit
substitute
resistance
Z2 a(solved
=if8 can
R
a Heaviside
capacitance
Zb =can
1/sC,
then the
model
ofa real
cable
line
by
O.used
in
1887),
be
as line in series
1
+
3s
+
s
1
Impedance
of
kind
of
transmision
line,
which
be
also
as
a
Z(s) Y
Y
Y
Y
Y
Y
Y
Y
!
(6.6)
s
+
Z(s)
=
=
(Z
)
and
the
same
in
shunt
(Z
),
we
get
a
structure
depicted
in Fig. 6.4.
a
b
as a transmission
line
or
a
symetrical
domino
ladder
lattice
network
as
well.
3
impedance
isZ by O. Heaviside
Z
2
modelZ of realZ cable
line
(solved
in 4s
1887), Z(s)
can
be
expressed
as 1
2s
+
Z
.Z
.
(6.7)
=
a
b
of as
such
kind2sof+ transmision
!Impedance
The resulting impedance Z(s) of the entire circuit can
be "
expressed
a"CFE
1line, which can be used also as a
1
R
R
Za .Z
. real −1/2
(6.7)
Z(s)
=model
If we
substitute
=R
a by
capacitance
Zb = in
1/sC,
then
thebe expressed as
−
Figure
6.3: General
structuredistribution
of transmission
line.
O.s Heaviside
1887),
can
s cable
ω −1/2
= aofbresistance
=aline
e−jπ/4
|+
(6.8)
described by (6.1)
with
symetrical
of Z(s)
elements,
Z2k−1
(s) Z
Z
(s)and
s=jω .3
2k(solved
12
C
!− s
impedance
is a capacitance
Za
Za
we(s),
substitute
ZZa
Zb = C
1/sC,
then
the
a = R and
k = Za1, . .a. ,resistance
n.
and Y2k−1 (s) =IfY2k
CHAPTER
6. GENERAL
FRACTANCE
" APPROACH
" TO
2 b . DEVICES 34
(6.7)
Z(s)
= Za .Z
CHAPTER
6.
GENERAL
APPROACH
TO
FRACTANCE
DEVICES
34
R in−1/2
R −1/2 −jπ/4
impedance
is the same
In the case
if we assume
elements
in transmision
From
expansion
itZ(s)
follows
that
s series
= line
=
.
Z(s) Zb
"this
Zb Zb
Zb Zb
Zb
Zb
Zb
If"we substitute
aC resistance
Zaω = Re and |of
as=jω
capacitance
Z(6.8)
b = 1/sC, then the
C
one
a
continuous
fraction
representation
the
function
Z(s),
(Za ) and the same in
get aR
structure
depicted
in
Fig.
6.4.
Za shuntZa(Zb ), we
Za
Zahas to obtain
R
−1/2 −jπ/41of the function Z(s),
1 =impedance
3
one has to obtain Z(s)
a continuous
fraction
representation
s−1/2
ω(s)
=
eis 4− |s=jω
2 . as Ya2 (s) = (6.8)
Z1 (s)
s;which
=
2s; 1 Y4 (s) = − s.
3
Impedance of such kind of transmision
line,
be
also
C 2=
CZcan
+12
3ss;
+8 "
sused
1
"
2
(6.6)
=
= s+
13
3sinpedance
+28 Za1and ZZ(s)
s4 of+two
Transmission line circuit composed
b.
R −1/2
R1 −1/2 −jπ/4
model ofFigure
real6.4:cable
line (solved
by=
O.
Heaviside
in
1887),
can
be
expressed
as
2
2s
+
4s
(6.6)
s
+
Z(s)
=
Therefore,
first Cauer’s
2s +form
= 1 of ωthe
|s=jω .
(6.8)
1Z(s) = in sthe
2 analogue realization
2s
!3 + 4s for the
1 e
C
C
2s +
6.4 Tree structure circuit
s
+
−
Za .Zcircuit
Z(s)
= LC
1 (6.7) the following
1 to choose
canonic
[20] we have
b.
3 of coils and
12 values
− s+
− s
Let us consider the total impedance of the fractance circuit as shown in Fig. 6.5,
3
12
2
If we substitute
a resistance
Zacapacitors:
= combination
R and aofcapacitance
= 1/sC, then
− s the
which has a recursive
structure with
the
two impedances ZZ
a b
2 that
From
and Zb . Impedance is derived as geometrical mean of Z1
Zb , this expansion1it follows
a and
3
impedance
is
From this
expansion
it
follows
that L
[H];
[H]; 1 C2 = 2 [F ];
L"
C4 = − [F3].
!
1 =
3 = −
"Z(s)
1
= Z .Z .
12
Z
Z
s;
= s;
Y4 (s) 2= − s.
3 (s) = −
R1 −1/2a b
R 2 −1/2
1 1 (s)
3Y2 (s) = 2s;
−jπ/4 2
12
Z(s)
=Z3 (s)
e
|Ys=jω
. = 2s;
(6.8)
Z1=
s;
(s) = ss; Here
− negative
Yand
= − s.
we=ωsee
inductances
capacitance.
Such elements 2cannot
2 (s)
4 (s)
C2
C Therefore,
12
2 in the form of the first Cauer’s
for the analogue realization
Zb = 1/sC, then
If we substitute a resistance Za = R and a capacitance
bethe
realized
using
passive inelectric
components.
However,
they can be realized
Therefore,
for
analogue
realization
the
form
of
the
first
Cauer’s
the impedance shows
the fractance
characteristics
as canonic
(see also
(6.8))
LC circuit [20] we have to choose the following values of coils and
" circuit
" [20]the
with
help
components,
namely
operating
canonic LC
have of
to active
choose the
following values
of coils
and amplifiers. We can
capacitors:
R −1/2
R −1/2we
Z(s) =
s
ω
=
e−jπ/4 |s=jω ,
C
realize
this by negative
which was described in3 previous
capacitors: C
1 inpedance converter
1
[H];
L3 = − [H]; 3 C2 = 2 [F ];
L1 = proporC4 = − [F ].
which is a fractional order integral
absolute value of1impedance
1 with
subsection.
CHAPTER 6.−1/2GENERAL APPROACH
35
2 C = 2 [F
12C = − [F ].
2
L3independent
L1 =
= − [H];
];
tional to ω
and phase
angle is[H];
constant −π/4,
of the frequency.
2
4
2 construct such an infinity12
2 capacitance. Such elements cannot
Here we
see innegative inductances and
However, it is impossible to
structure
as show
Z it seems toZ be important
Z to study the effect
Z of the numFig. 6.5. Therefore,
Here
we see negative
inductances
and
capacitance.
Such
elements
cannot
be realized using passive electric components. However, they can be realized
ber of stage in the recursive self infinity (binary) structure on the impedance
be realized using passive electric
components.
However,
they cannamely
be realized
with the
help of active
components,
operating amplifiers. We can
characteristics [31].
CHAPTER 6. GENERAL APPROACH TO FRACTANCE DEVICES 35
AZ(s)
similarwith
as a fractal
rough
interface
between
Ytree structure
Y
Y
Y active
Ymodel ofYcomponents,
Y
Ynamely
the
help
of
operating
amplifiers.
can was described in previous
realize
this
by two
negative
inpedance
converterWe
which
materials of very Zdifferent conductivities,
e.g.
Z
Z an electrode and Zan electrolyte
Let us inpedance
consider
the
circuit
depicted
in
Fig.
Z which
Z
Z 6.3,
Z Z2k−1 (s) and Y2k (s), k =
converter
was
described
in where
previous
subsection.
was studied in [21]. There was described a slightly different structure of tree
subsection.
1, . . . , n, are given impedances of the circuit elements. This structure is known
1
0
0.005
0.01
0.015
Time [sec]
0.02
0.025
ramp
0
!0.5
!1
0.03
Figure 7.8: Time response of the I 1/2 controller to saw input where half order
integral was approximated by transmission line depicted in Fig. 7.2.
0.5
1/2
Figure 7.9: Time response of the I controller to ramp input where half order
integral was approximated by transmission line depicted in Fig. 7.2.
A similar tree structure as a fractal model of rough interface between two
materials of very different conductivities, e.g. an electrode and an electrolyte
was studied in [21]. There was described a slightly different structure of tree
III.1 Commensurate Order Systems
Sample implementations:
CHAPTER 7.
Stability of fractional order systems
• What is a commensurate order system?
! ofExample
• If one selects the greatest common divisor
An
Commensurate
order
the orders,
suchsystems:
that
fractance
38
• A fractional order system
An Example
An Example
where p and q are the orders of the rational approximation, P and Q are
polynomials of degree p and q, respectively. Block diagram of the analogue
R2
ZF
Ri
E
_
R1
+
+
_
+
+
U
CHAPTER 7.
Figure 7.1: Analogue fractional-order integrator.
CHAPTER 7. REALIZATION OF FRACTIONAL CONTROLLERS
CHAPTER 7. REALIZATION OF FRACTIONAL CONTROLLERS
39
realization of fractional-order operator is shown in Fig. 7.1.
0.47µ F
Z F (s)
0.47µ F
0.47µ F
0.47µ F
0.47µ F
0.47µ F
0.47µ F
7.2
Z (s)
0.47µ F
0.47µ F
0.47µ F
0.47µ F
0.47µ F
R
7.2.1
0.47µ F
0.47µ F
0.47µ F
0.47µ F
0.47µ F
0.47µ F
1k !
1k !
1µF
1k !
1k !
1k !
1µF
1k !
Realization of fractional I λ controller
22k ! 22k ! 22k ! 22k ! 22k ! 22k ! 22k ! 22k ! 22k ! 22k ! 22k !
22k ! 22k ! 22k ! 22k ! 22k ! 22k ! 22k ! 22k ! 22k ! 22k ! 22k !
F
22k !
22k !
0.47µ F
0.47µ F
0.47µ F
0.47µmeasurement
F
0.47µ F
For experimental
we built a fractional-order I λ controller which
0). The conis transmission
a particularline
casecircuit.
of the P I λ D µ controller, (if Kp = 0 and Td =1µF
Figure 7.2: RC
Figure 7.2: RC transmission
troller line
wascircuit.
realized in three forms, namely by: the symetrical domino ladder or
1k !
1k !
1µF
1µF
1k !
1k !
1µF
1µF
1µF
1µFZ
impedance
1µF
1µFF
where
where
Magnitude [dB]
Z F (s)
Transmission line approximation of the s−0.5
Transmission line approximation of the s−0.5
In Fig. 7.5 and Fig. 7.6 - 7.9 the measured characteristics of realized anaIn Fig. 7.5 and Fig. 7.6 - 7.9 the
measured characteristics of realized analogue fractional-order I λ controller, where half order integral was approxilogue fractional-order I λ controller, where half order integral was approximated by transmission line depicted in Fig. 7.2, are presented. It can be seen
mated by transmission line depicted in Fig. 7.2, are presented. It λcan be seen
from Fig. 7.5 that the realized analogue fractional-order I controller provides
from Fig. 7.5 that the realized analogue fractional-order I2λ controller provides
a good approximation in the frequency range [10 rad/sec, 5 · 102rad/sec]. (For
a good approximation in the frequency range [102 rad/sec, 5 · 102rad/sec]. (For
comparison see expected Bode plots shown in Fig. 3.2.2.)
comparison see expected Bode plots shown in Fig. 3.2.2.)
Phase [deg]
2
3
4
10
!80
2
10
3
10
Frequency [rad/sec]
FOC tutorial III
FOC tutorial III
FOC tutorial III
!70
!90
14
!=1/6, thus the original order
• It is obvious that
It
is obvious
!=1/6,
can be
written
as bethat
• system
can
written
as thus the original order
system can be written as
where
1µF
1k !
1k !
1k !line (see
1k !Fig. 7.2)
1k !
Z F (s) ladder (see Fig. 7.3) 1µF
transmission
for n=6, the finite domino
1k !
1k !
1k !
C
for n = 6 and the tree structure (see Fig. 7.4) for n = 4, and Fractances with
1µF
1µFconnected
1µF to feedback
1µF
were
in operational amplifier (Fig. 7.1).
1µF
1µF
It should be noted that the described methods work for arbitrary orders, but
the circuit elements with computed values are not usually available. Because
Figure 7.4: RC binary tree circuit.
of this, in our experiment we proposed and realized the integrator with order
Figure 7.3: RC
domino
ladder circuit.
!5
λ =ladder
0.5. circuit.
Figure 7.3: RC domino
!10
It should be mentioned that this simple case of the
controller order can
!15
we have Ti = 1.4374. The transfer function of the realized analogue fractionalbeofrealized
alsoanalogue
using the
methods described in [31, 33, 38, 54], which do not
The transfer function
the realized
fractionalwe have Ti = 1.4374.
!20
order I λ controller is:
order I λ controller is:
involve explicit
rational approximations.
!25
(7.2)
C(s) = −0.5
1.4374 s−0.5.
(7.2)
C(s) = 1.4374 s In .the case, if we will use
identical resistors (R-series)
and identical ca!30
λ
Adjustment of the integration
constant
T
of
the
fractional-order
I
coni
λ
!35
pacitors
(C-shunt)
in the Ifractances,
then the behaviour
of the circuit will
Adjustment of the integration constant
Ti of the
fractional-order
controller (or half order integrator) depicted in Fig. 7.1 was done by resistor Ri .
!40
half-order
We used the
resistor
troller (or half order integrator) depictedbe
in as
Fig.a 7.1
was doneintegrator/differentiator.
by resistor Ri .
10 values R =
10
If we change the resistor Ri , the integration constant changes the value in the
If we change the resistor Ri , the integration
constant
thethe capacitor values C = 1µF , (Cj = C, j = Frequency [rad/sec]
1kΩ,
(Rj = changes
R, j = the
1, . .value
. , n) inand
required interval.
required interval.
!30
1, . . . , n).
better
measurement
For measuremets we used frequency
100 For
Hz and
amplitude
±10 V. results we used two operational amplifiers
For measuremets we used frequency 100
Hz and amplitude
±10connection.
V.
!40
TL081CN
in inverting
!50
The resistors R1 and R2 are R1 = R2 = 22kΩ.
The integration constant Ti
!
Experimental
!60
7.2.2 7.2.2
Experimental
resultsresults
can be computed from relationship Ti = 1/ R/(Ri2 ∗ C),
and for Ri = 22kΩ
1k !
1µF
Z F (s)
• A fractional order system
If !
can be found,
commensurate order
•• A
fractional
orderthe
system
• It is obvious that !=1/6, thus the origina
For example:
model can easily be written
system can be written as
FOC tutorial III
1µF
Description of circuit
0.47µ F
40
39
4
10
Figure 7.5: Bode plots of the I 1/2 controller where half order integral was
approximated by transmission line depicted in Fig. 7.2.
15
15
Stability of fractional order systems
III.2 Stability of Fractional Order Systems
• The ! curve
– if
, then the stable
condition for the system is
Unstable
region
with four poles in the stable region
FOC tutorial III
17

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