IEEE Circuits and Systems Magazine - Q2 2022 - 43

that generalizes the expression corresponding to a 1 .=
Knowing relations (28) and (30) and attending to (24) we
are able to compute the impulse and step responses of
any LTIS defined by the TF (22).
2) Using the Inverse LT
The inversion of a TF based on the MLF has an important
drawback: the solution relies on one, or several,
series, that create severe computational problems. Furthermore,
such solution masks the underlying structure
of the system, in the sense that it does not highlight the
presence of two different terms
■ One component of integer order that inherits
the classical behavior, mainly oscillations and
(un)stability;
■ Another component of
fractional order responsible
for the long range behavior of the fractional
linear systems, that is intrinsically stable as we
will demonstrate in the sequel.
We start from the Bromwich inversion integral (2)
and decompose it in two parcels according to each path
section [80]. We get [81]
K0
gt = / k
k= 1
()
2r
1 # jj t
+3
rr v
Ae t
j
ptf()
1/a
6Ge uG eu eu t-- d·f(),
() ()@
where the constants A ,k
of G(s) at
p .
1/a
kK0
= f are the residues
12
=+
,, ,,
The LT of both sides in (31) leads to
()
Gs Gs Gs() (),
if
where the integer order part is
()
K0
Gs = /
i
k= 1 -
sp
A
a Re
and the fractional part is
()
(32)
(31)
for () 2 max Re
ht=- o
n=0
f ()
2 j
1
r / 6Ge uG eu e
-jrr -ut
() ()@
j
n
n
n==o
01 f
n
L
(35)
withun ,, ,, .nL The sampling interval o and
the positive integer L are chosen to guarantee that the
fraction in (35) is small for uLL
= o.
Remark
III.3. Consider the TF in (22) with
nk11 2== f
k ,, ,
N
Gs
where Rk
Qp arg pk N12kk
hti
=-ra 1
"
# ra
=
ht =
ik
kp Q
()
= 1;
k ,, ,, ,
K = a k
11/- a
/ RB et·
k !
k
pt (),
k
1
/a
e
k
() = /
k= 1
sp
R
k
a - k
pk = f are the residues and pseu12
N
f ,.
(37)
where Rk = f12 N represent the coefficients obtained
from the partial fraction decomposition and
Bp /1
are the residues of /(
corresponding LT becomes [82]
()
Hs = /
i
(( )),
k
kp Q
N
= ;1 k !
sp
RB·
Resp pQ/1 a
k !
kk
1/a
- k
, as defined above.
With generality, consider the situation where we have
K pseudo-poles. The result in (35) allows us to state that:
■ For
,, ,, ,kK
a 11 2 f== we have no fractional
component. The TF is a sum of partial fractions
and each one has an exponential for inverse LT.
The corresponding TF is the quotient of two polynomials
in s.
k ,()( ())
1/ sp1/a
2 max Re
(33)
■ For a 111 2 f=
k
nents depending on the location of pk
Gs=f
11djj
r 0
2 j # 6 () ()
3 rrGe
uG eu @
su
+
plex plane:
* If () ,, ,, ,
arg pk K12
f
;;2ra =
u
(34)
which is valid for Re(s) > 0. The integer order part of
the impulse response (33) is the classical sum of exponentials
(or sinusoids), eventually multiplied by integer
powers. The possible sinusoidal behavior comes from
this term. The numerical computation of this part does
not put any significant problem. The fractional part is
expressed as an integral of the product of an exponential,
with negative real exponent, by a bounded function,
that is zero at the origin and at infinite. This means that
the integral is easily computed by means of a simple numerical
procedure. Using a uniform sampling interval
o ,
we can write
SECOND QUARTER 2022
* If ()arg pk
arg pk
,, ,, ,kK we may have two compoin
the comthen
we do not
have the integer order component: it is a pure
fractional system;
;;# ra, for some kK
,, ,
= 12 f , then
the system is mixed, in the sense that we have
both components;
;; ra=
= 12 f ,, , ,
then the integer order component is sinusoidal,
but the fractional component exists as well.
Remark III.4. From the above considerations we conclude
that we can have purely fractional systems. We only
have to choose the pseudo-poles with arguments in the
region defined by () .
;;2ra
arg pk
Example III.1. Consider the fractional RC circuit with
TF expressed by:
IEEE CIRCUITS AND SYSTEMS MAGAZINE
43
* If () (/2) , for some kK
1 sp )k
a -
at
p .
/
k
1 a
The
(38)
and ,, ,, ,
do-poles. Define the set
:( ), ,, ,
The integer part () of the impulse response is given by:
N
,
(36)
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