IEEE Circuits and Systems Magazine - Q3 2022 - 12

yn = 32
()
() ()-+ -+ +
11 42
1
2
--=
6
nn1 2
.
Theoretically they are IIR systems, but the IR may go
to zero sufficiently fast to exhibit an FIR behaviour.
Now we are in conditions of defining the frequency
response (FR) of the system (33). We only need to make
the substitution of variable ()/ehs 1
=-
~
jh
in (36). This
involves the transformation of the parameters of the TF
using the binomial coefficients. We conclude that the TF
(36) gives rise to the following FR
M0
Ge
()= N0
j~
/
/
k=0
k=0
where the coefficients Bk
M
Bbh-=1
kl
l
=
/
= f01 2 ,, ,, .
are given by
!
al 1 a k()l
-
k
forkM0 For the Ak
,
tation is similar.
Computing the inverse FT of ()
Gej~
Ae
Be
k
k
- ~
jk
,
- ~
jk
(37)
2) Proper fraction case
Herein, we consider that we have M zeroes and N
poles in (39) with
NM .2 For the sake of simplicity
we assume that the poles have multiplicity one. In
this case we can write ()
Gsd
N
Gs =
where Ak
d () /
k=1
sp
A
k
a
-
k
and poles obtained by substituting w for sa
and p ,k kN are the residues
in
=12 f ,, ,,
(39), respectively. The IR can be obtained by inverting
a combination of partial fractions such:
Fs
(38)
coefficients, the compuwe
obtain a difference
equation equivalent to the differential equation
(33). As observed, for practical purposes, we have to
truncate the sequence.
2) Transient Responses
In general, the systems described by (33) or (36) are IIR systems,
difficult to study due to the problems in computing
the poles and zeros. We only have FIR systems when N 0=
and all the derivative orders are positive integers.
As for the CT we will consider the commensurate
case. The TF assumes the form
M
Gs
d = N
()
/
/
k=0
k=0
The fraction in (39) can be decomposed into a sum
of a polynomial (only zeros) and a proper fraction (polezero).
We are going to study the two cases separately.
1) Polynomial case
Let us consider a transfer function with the form
M
Gs bsk
k
d = () /
=0
To invert this expression, we recall the previous
statement regarding sa
ak
.
(40)
that is the TF of the differentiator.
This is a system with IR given by the
binomial coefficients in agreement with (8). The
IR corresponding to (40) is
M
gnhb nh
=+ k
k=1
0d
12
() () / bh-IEEE
CIRCUITS AND SYSTEMS MAGAZINE
k ()a 1 -ak n
n!
f nh
().
(41)
Fs A
d ()=
as
bs
k
.
ak
k
(39)
ak
fnhA ph=/ k
3
()=k=0
--
ak ()1 ak n
-
1 -n!
f()
. (45)
nh G
However, it will not be an interesting system
in applications, since the values of p must be
large for small values of h.
b) Intersection of the Hilger circle with the disk
a
1
;;2; ;
For this case, we can write
sp
1- ps
s Ap s akE.
-a
-a = ;/ k-1 -
3
(46)
k=1
The corresponding IR, (),fnh is given by
()
fnhA ph=/ k- ak ()-1 ak n
3
1
=
k=1
n!
f() .nh G
d ()= a
sp
A
-
To invert (43) we can insert it into the inversion integral,
(22). The residue theorem tells us that the
inversion is obtained by computing the ()n 1th+
derivative of F(s) and substituting /h1 for s. A
simpler alternative is to use the properties of the
geometric series. We have two possibilities corresponding
to the regions:
a) Intersection of the Hilger circle with the disk
1
;;1;sp; a
In this case we have
()
Fs =d
p
A
1
1p
s
a
=- +1 / -k
k=1
p
A
;
3
ps .
akE
(44)
From the above expression it results that the
IR f(nh), corresponding to a partial fraction of
order one, is given by
.
(43)
,
as
(42)
(47)
This expression is the discrete-time version of
the
a -exponential [29] that we find in CT systems
and that is related to the Mittag-Leffler
function (MLF). Therefore, this case is interesting
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