Tadeusz Kaczorek
Singular Fractional Continuous-Time and Discrete-Time Linear Systems
SINGULAR FRACTIONAL CONTINUOUS-TIME AND DISCRETE-TIME LINEAR SYSTEMS
Tadeusz KACZOREK*
*Faculty of Electrical Engineering, Bialystok University of Technology, ul. Wiejska 45D, 15-351 Białystok, Poland
[email protected]
Abstract: New classes of singular fractional continuous-time and discrete-time linear systems are introduced. Electrical circuits are example of singular fractional continuous-time systems. Using the Caputo definition of the fractional derivative, the Weierstrass regular pencil
decomposition and Laplace transformation the solution to the state equation of singular fractional linear systems is derived. It is shown
that every electrical circuit is a singular fractional systems if it contains at least one mesh consisting of branches with only ideal supercondensators and voltage sources or at least one node with branches with supercoils. Using the Weierstrass regular pencil decomposition
the solution to the state equation of singular fractional discrete-time linear systems is derived. The considerations are illustrated by numerical examples.
Keywords: Fractional, Singular, Linear Circuit, Regular Pencil, Solution, Supercondensator, Supercoil
1. INTRODUCTION
Singular (descriptor) linear systems have been addressed
in many papers and books (Benvenuti and Farina, 2004; Dodog
and Stosic, 2009; Dail, 1989; Kaczorek, 1992, 2004, 2008, 2010,
2011; Podlubny, 1999). The eigenvalues and invariants assignment by state and output feedbacks have been investigated
in Benvenuti and Farina (2004), Dodog and Stosic (2009), Dail,
(1989), Kaczorek (2004, 2008) and the realization problem
for singular positive continuous-time systems with delays in Kaczorek (2010). The computation of Kronecker’s canonical form
of a singular pencil has been analyzed in Podlubny (1999).
Fractional positive continuous-time linear systems have been
addressed in Kaczorek (2010) and positive linear systems
with different fractional orders in Kaczorek (2007). An analysis
of fractional linear electrical circuits has been presented
in Gantmacher (1960) and some selected problems in theory
of fractional linear systems in the monograph Kaczorek (2007).
In this paper a new class of singular fractional linear systems
and electrical circuits will be introduced and their solution of state
equations will be derived.
The paper is organized as follows. In section 2 the Caputo
definition of the fractional derivative and the solution to the state
equation of the fractional linear system are recalled. The solution
of the state equation of singular fractional linear system is derived
using the Weierstrass pencil decomposition and the Laplace
transform in Section 3. Singular fractional linear electrical circuits
are introduced in Section 4. In section 5 the fractional singular
discrete-time linear systems are introduced and Weierstrass
regular pencil decomposition is recalled. The solution of the state
equation of singular fractional linear discrete-time system is derived using the Weierstrass pencil decomposition in Section 6.
Illustrating numerical examples are given in Section 7. Concluding
remarks are given in Section 8.
To the best of the author’s knowledge singular fractional linear
systems and electrical circuits have not been considered yet.
26
The following notation will be used in the paper.
The set of n m real matrices will be denoted by nm and
n
 : n1. The set of m n real matrices with nonnegative
n
entries will be denoted by  m
and n : n1. The set of

nonnegative integers will be denoted by Z  and the n n
identity matrix by I n .
2. FRACTIONAL CONTINUOUS-TIME LINEAR SYSTEMS
The following Caputo definition of the fractional derivative will
be used (Kaczorek, 2007; Kucera and Zagalak, 1988):
d
1
f ( n)
f (t ) 
d ,

(n   ) (t   ) 1n
dt
0
t

(2.1)
n  1    n  N  {1,2,...}
where
is the order of fractional derivative,
( )
( )
and ( ) ∫
is the gamma function.
Consider the continuous-time fractional linear system described by the state equation:
d
x(t )  Ax(t )  Bu (t ), 0    1
dt 
(2.2)
where ( )
( )
are the state and input vectors
and
,
.
Theorem 2.1. The solution of equation (2.2) is given by:
t

x(t )   0 (t ) x0  (t   ) Bu ( )d , x(0)  x0
0
where:
(2.3)
acta mechanica et automatica, vol.7 no.1 (2013), DOI 10.2478/ama-2013-0005

 0 (t ) 
Ak t k
 (k  1)
(2.4)
k 0
d
dt

(t ) 
Ak t ( k 1) 1
[( k  1) ]
k 0

(2.5)
Remark 2.1. From (2.4) and (2.5) for
 0 (t )  (t ) 

x2 (t )  x2 (t )  B2u (t )
(3.6b)
(3.6c)
Using (2.3) we obtain the solution to Eq. (3.6a) in the form:
t
3. SINGULAR FRACTIONAL CONTINUOUS-TIME
LINEAR SYSTEMS
x1(t )  10 (t ) x10   11(t   ) B1u ( )d
d
E  x(t )  Ax(t )  Bu (t )
dt
where:
10 (t ) 
(3.1a)
y(t )  Cx(t )  Du (t )
(3.1b)
where ( )
( )
, ( )
are the state, input
and output vectors and
,
,
,
.
The initial condition for (3.1a) is given by:
x(0)  x0
(3.1c)
It is assumed that the pencil of the pair (E, A) is regular, i.e.
det[ Es  A]  0
(3.2)
for some
(the field of complex numbers). It is well-known
(Fahmy and O’reill, 1989; Kaczorek, 1992)) that if the pencil
is regular then there exists a pair of nonsingular matrices
such that:
0
I n2 
(3.3)
where:
is equal to degree of the polynomial
,
,
is a nilpotent matrix with the index
(i.e.
and
and
).
Applying to the Eq. (3.1a) with zero initial conditions
the Laplace transform (L) we obtain:
[ Es  A] X (s)  BU (s)
(3.4)
()
( ) . By
where ( )
and ( )
∫ ()
assumption (3.2) the pencil [
] is regular and we may
apply the decomposition (3.3) to Eq. (3.1a).
Premultiplying Eq. (3.1a) by the matrix
and introducing the new state vector:
 x (t ) 
x (t )  Q 1 x(t )   1  , x1(t )  n1 , x2 (t )  n2
 x2 (t )
(3.7a)
0
Consider singular fractional linear system described by the
state equations:
0   A1
s
N   0
dt
B 
PB   1  , B1  n1 m , B2  n2 m
 B2 
we have:
k
( At )
 e At .

(
k

1
)
k 0
I
P[ Es  A]Q   n1
0
d
(3.6a)
where:
Proof is given in Kaczorek (2010a, b).

N
x1(t )  A1x1(t )  B1u (t )
(3.5)
11(t ) 

A k t k
 (k1  1)
(3.7b)
A1k t ( k 1) 1
 [( k  1) ]
k 0
(3.7c)
k 0

and x10  n1 is the initial condition for (3.6a) defined by:
 x10 
1
 x   Q x0 , x0  x(0)
 20 
(3.7d)
To find the solution of Eq. (3.6b) we apply to the equation
the Laplace transform and we obtain:
Ns X 2 (s)  Ns 1x20  X 2 (s)  B2U (s)
(3.8a)
Since Kaczorek (2010a) and (Dodog and Stosic, 2009)
for 0 < α < 1
 d

L   x2 (t )  s X 2 ( s)  s 1x20
 dt

(3.8b)
where X 2 (s)  L[ x2 (t )] . From (3.8) we have:
X 2 (s)  [ Ns  I n2 ]1( B2U (s)  Ns 1x20 )
(3.9)
It is easy to check that:
 1
[ Ns  I n2 ]1    N i si
(3.10)
i 0
since:
  1

[ Ns  I n2 ]   N i s i   I n2


 i 0

(3.11)
and Ni = 0 for i = µ, µ + 1, …
Substitution of (3.10) into (3.9) yields:
we obtain:
27
Tadeusz Kaczorek
Singular Fractional Continuous-Time and Discrete-Time Linear Systems
X 2 ( s)   B2U ( s) 
 1

 N B s
i
i
2
Nx20
s1
U ( s)  N
i 1 (i 1) 1
s
x20

(3.12)
i 1
Example 4.1. Consider electrical circuit shown in Fig. 1 with given
resistance R, capacitances C1, C2, C3 and source voltages e1
and e2.
Using the Kirchhoff’s laws we can write for the electrical circuit
the equations:
Using inverse Laplace transform (
) to (3.12) and the
convolution theorem we obtain for 1 – α > 0:
e1  RC1
t 
x2 (t )  L1[ X 2 ( s)]   B2u (t )  Nx20
(1   )
C1
 1 
t
(3.13)
1
since L   1  
for α +1 >0.
s
 (1   )
Therefore the following theorem has been proved.
Theorem 3.1. The solution to Eq. (3.1a) with the initial condition
(3.1c) has the form
(3.14)
where x1(t) and x2(t) are given by (3.7) and (3.13) respectively.
Knowing the solution (3.14) we can find the output y(t) of the
system using the formula
 x (t ) 
y (t )  CQ 1   Du (t )
 x2 (t )
 C2
 u1  u3
d  u2
dt 
(3.15)
0  
 RC1 0
d
C
C

C3 
2
 1
dt 
 0
0
0 
1 0
e 
 0 0  1 
e
0 1  2 
0
(4.5)
Let the current iC(t) in the supercondensator with the capacity
C be the α order derivative of its charge q(t) (Gantmacher, 1960):
 u1   1 0  1  u1 
u    0 0 0  u 
 2 
 2 
u3   0  1  1 u3 
(4.6)
(4.7)
Note that the matrix E is singular (det E = 0) but the pencil:
RC1s  1
4. SINGULAR FRACTIONAL ELECTRICAL CIRCUITS
dt
dt 
0 
 RC1 0
 1 0  1
1 0
E   C1 C2  C3 , A   0 0 0 , B  0 0
 0
 0  1  1
0 1
0
0 
det[ Es  A] 
d  q(t )
d  u3
In this case we have:

iC (t ) 
 C3
The equations (4.5) can be written in the form:

 x (t ) 
x(t )  Q  1 
 x2 (t )
dt 
dt 
e2  u 2  u 3
 1

d i
d (i 1) 1
   N i B2
u (t )  N i 1
x20 
dt i
dt (i 1) 1

i 1 
d  u1
d  u1
C1s
0

0
C2 s
1

1
 C3s
1
(4.8)
 ( RC1s  1)(C2  C3 ) s  C1s
is regular. Therefore, the electrical circuit is a singular fractional
linear system.
(4.1)
Taking into account that q(t) = CuC(t) we obtain:
iC (t )  C
d  uC (t )
dt
(4.2)
where uC(t) is the voltage on the supercondensator.
Similarly, let the voltage uL(t) on the supercoil (inductor) with
the inductance L be the β order derivative of its magnetic flux
ψ(t):
uL (t ) 
d   (t )
dt 
(4.3)
Taking into account that ψ(t) = LiL(t) we obtain
uL (t )  L
d  iL (t )
dt 
where iL(t) is the current in the supercoil.
28
(4.4)
Fig.1. Electrical circuit
Remark 4.1. If the electrical circuit contains at least one mesh
consisting of branches with only ideal supercondensators and
voltage sources then its matrix E is singular since the row
corresponding to this mesh is zero row. This follows from the fact
that the equation written by the use of the voltage Kirchhoff’s law
is algebraic one.
acta mechanica et automatica, vol.7 no.1 (2013), DOI 10.2478/ama-2013-0005
Example 4.2. Consider electrical circuit shown in Fig. 2 with given
resistances R1, R2, R3 inductances L1, L2, L3 and source voltages
e1 and e2.
Using the Kirchhoff’s laws we can write for the electrical
circuit the equations:
e1  R1i1  L1
d  i1
d  i3

R
i

L
3
3
3
dt 
dt 
e2  R2i2  L2
d  i2
d  i3

R
i

L
33
3
dt 
dt 
(4.9)
i1  i2  i3  0
The equations (4.9) can be written in the form:
 L1
0

 0
0
L2
0
L3  
d
L3 
dt 
0 
0
 i1   R1
i    0
 R2
 2 
i3   1
1
 R3   i1 
 R3  i2 
 1  i3 
1 0
e 
 0 1  1 
e
0 0  2 
(4.10)
In this case we have:
 L1 0 L3 
 R1 0  R3 
1 0




E   0 L2 L3 , A   0  R2  R3 , B  0 1 (4.11)
 0 0 0 
 1
0 0
1
 1 
Note that the matrix E is singular but the pencil:
det[ Es   A] 
L1s   R1
0
0
L2 s   R2
L3s   R3
1
1
1
L3s   R3
 [ L1 ( L2  L3 )  L2 L3 ]s 2 
(4.12)
 [( L2  L3 ) R1  ( L1  L3 ) R2  ( L1  L2 ) R3 ]s 
 R1 ( R2  R3 )  R2 R3
is regular. Therefore, the electrical circuit is a singular fractional
linear system.
at least one zero row. This follows from the fact that the equation
written using the current Kirchhoff’s law for this node is algebraic
one.
In general case we have the following theorem.
Theorem 4.1. Every electrical circuit is a singular fractional system
if it contains at least one mesh consisting of branches with only
ideal supercondensators and voltage source or at least one node
with branches with supercoils.
Proof. By Remark 2.1 the matrix E of the system is singular if the
electrical circuit contains at least one mesh consisting of branches
with only ideal supercondensators and voltage source. Similarly
by Remark 2.2 the matrix E is singular if the electrical circuit
contains at least one node with branches with supercoils.
Using the solution (3.14) of Eq. (3.1a) we may find the
voltages on the supercondensators and currents in the supercoils
in the transient states of the singular fractional linear electrical
circuits. Knowing the voltages and currents and using (3.15)
we may find also any currents and voltages in the singular
fractional linear electrical circuits.
Example 4.3. (an continuation of Example 4.1)
Using one of the well-known methods (Podlubny, 1999; Dodig
and Stosic, 2009; Kaczorek, 1992) we can find for the pencil (4.8)
the matrices:

1

RC

1
1

P  
R(C2  C3 )


0


0

1
C 2  C3
0

C2

RC1 (C2  C3 ) 
C2

,
2 
R(C2  C3 ) 

1




0 
 1 0


C3 
Q   0 1
C 2  C3 


C2 
 0 1

C2  C3 
(4.13)
which transform it to the canonical form (3.3) with:
1
1


  RC

RC1
1
,
A1  
1
1



 R(C2  C3 )
R(C2  C3 ) 
N  [0], n1  2, n2  1
(4.14)
Using the matrix B given by (4.7), (4.13) and (3.6c) we obtain:

1

RC1

 B1 

1
 B   PB  
 2
 R(C2  C3 )

0


Fig.2. Electrical circuit


C2

RC1(C2  C3 ) 

C2
2 
R(C2  C3 ) 

1


(4.15)
from (3.7) we have:
Remark 4.2. If the electrical circuit contains at least one node with
branches with supercoils then its matrix E is singular since it has
29
Tadeusz Kaczorek
Singular Fractional Continuous-Time and Discrete-Time Linear Systems
t
x1(t )  10 (t ) x10   11(t   ) B1u ( )d
(4.16)
0
for any given initial condition x10  n1 and input u(t), where:

A1k t k

 (k  1) , 11(t )  
10 (t ) 
0   1
k 0
k 0
A1k t ( k 1) 1
[( k  1) ]
,
6. SOLUTION OF THE SINGULAR FRACTIONAL
DISCRETE-TIME LINEAR SYSTEMS
Premultiplying the equation (5.1) by the matrix
and introducing the new state vector:
In this case using (3.13) we obtain:
x2 (t )  B2u(t )
Using Lemma 5.1 we shall derive the solution xi to the
equation (5.1) for a given initial conditions x0 and an input vector
ui, i  Z  .
(4.17)
 x (1) 
xi   i(2)   Q 1xi , xi(1)  n1 , xi(2)  n2 , i  Z  (6.1)
 xi 
since N = [0].
In a similar way we may find currents in the supercoils of the
singular fractional electrical circuit shown in Fig. 2.
we obtain:
5. FRACTIONAL DISCRETE-TIME LINEAR SYSTEMS
and after using (5.5) and (6.1):
Consider the singular fractional discrete-time linear system
described by the state equation:
 I n1
0

E xi 1  Axi  Bui , i  Z  {0,1,...}
where:
,
,
of the order α is defined by:
 xi 
 
i
(5.2)
k 0
1
for k  0
  

    (  1)...(  k  1)
for k  1,2,...
k  
k!

(5.3)
(6.3)
where:
 B1 
n m
n m
 B   PB , B1   1 , B2   2
2
 
i 1
 
xi(1)1    (1) k   xi(1)k 1  A1xi(1)  B1ui
k
k 1

det[ Ez  A]  0
 A1 0 
0
, PAQ  


N
 0 I n2 
where:
and
of the polynomial:
and
(5.5)
(5.6)
where:
(6.5)
k
(5.4b)
.
A method for computation of the matrices P and Q has been
given in Van Dooren (1979).
30

k 2
 
(1) k 1  xi(1)k 1  B1ui
i 1


 
N  xi(21)   (1) k   xi(2k) 1   xi( 2)  B2ui


k 
k 1
is a nilpotent matrix with the index (i.e.
),
,
is equal to degree
det[ Es  A]  an1 z n1  ... a1z  a0

i 1
and:
for some
(the field of complex numbers).
Lemma 5.1. (Fahmy and O’Reill, 1989; Kaczorek, 1992) If (5.4)
holds then there exist nonsingular matrices
such
that:
I
PEQ   n1
0
A1 xi(1)
(5.4a)
and:
(6.4)
Taking into account (5.2) from (6.3) we obtain:
It is assumed that:
det E  0
0    xi(1)1   A1 0   xi(1)   B1 
 

   ui , i  Z 

N   xi(21)   0 I n2   xi( 2)   B2 


(5.1)
are the state and input vectors,
and the fractional difference
 (1)k  k  xi k , 0    1
PEQQ 1 xi 1  PEQ Q1xi 1  PAQQ 1xi  PBui (6.2)
(6.6)
.
( )
The solution ̅
to the equation (6.5) is well-known
(Kaczorek, 2007b; 2010) and it is given by the theorem.
( )
Theorem 6.1. The solution ̅
of the equation (6.5) is given
by the formula:
i 1
xi(1)  i x0(1)   i  k 1B1uk , i  Z 
(6.7)
k 0
where the matrices Φi are determined by the equation:
i 1
 
i 1  i A1   (1) k 1 i  k 1 0  I n1
k
k 2
To find the solution ̅
assumed that:
( )
of the equation (6.6) for
(6.8)
it is
acta mechanica et automatica, vol.7 no.1 (2013), DOI 10.2478/ama-2013-0005
0
1

N  0


0
the considerations are similar (dual).
0 ... 0 0
0 ... 0 0
1 ... 0 0  n2

   
0 ... 1 0
(6.9)
For (6.9) the equation (6.6) can be written in the form:
0
1

0


0
( )
Knowing ̅
and
of Eq. (5.1) from (6.1):
0 ... 0 0
 x ( 21)  

 i  j 1  
0 ... 0 0 i 1

 x ( 22)  


1 ... 0 0 (1) j    i  j 1  

 j    
     j  0
 x ( 2, n2 )  

 i  j 1   , i  Z
0 ... 1 0

 x ( 21)   B21 
 i( 22)  

x
  B22 
 i

    ui
   

 xi( 2, n2 )   B2, n2 

(6.10)
 
 
   (1) j   B21ui  j 1  B22ui
 j
j 0
i 1
 
 (1) j  j  xi(22j )1  B23ui
j 0
i 1
 
   (1) j  
 j
j 0

i j2

k 0
(6.11)
 
(1) k   B21ui  j  k  2
k
 
 (1) j  j  B22ui  j 1  B23ui
j 0

i 1
 
 (1) j  j  xi(2,jn11)  B2, n ui
2
 A1
0

2
j 0
If N = 0 then from (6.6) we have:
xi(2)   B2ui , i  Z 
(6.12)
This approach can be easily extended for :
N  blockdiag [ N1
(6.13)
N2 ... Nh ]
where:
has the form (6.9) and ∑
If the matrix N has the form:
0
0

N  

0
0
 1 2 5 
 2 1  1
1 

P   2 4
1 , Q   1 0 0 
11
 4
 0 0 1 
3  2
 I n1
0


( 2, n 2 )
1 0 ... 0
0 1 ... 0
       n2

0 0 ... 1
0 0 ... 0
(6.14)
(7.1)
(7.2)
and:
i 1
xi
 x (1) 
xi  Q  i( 2)  , i  Z 
 xi 
for α = 0.5, ui = u, i  Z  and
(T denotes
the transpose).
It is easy to check that the matrices (7.1) satisfy the
assumptions (5.4). In this case the matrices P and Q have the
forms:
j 0
i 1
xi( 23) 
we can find the desired solution
 1  1  1
0.8 1.7 2.8
1




E   2 4 2 , A  0.4 0.8 1.4 , B   0 
 1 4 1 
2.2 4.6 2.2
 1
 (1) j  j  xi(21j)1  B22ui
xi( 22) 
( )
Example 7.1. Find the solution xi of the singular fractional linear
system (5.1) with the matrices:
  B21ui
i 1
̅
7. EXAMPLES OF SINGULAR FRACTIONAL
DISCRETE-TIME SYSTEMS
From (6.10) we have:
xi( 21)
Note that the matrices (6.9) and (6.9’) are related by:
0 0 ... 0 1
0 0 ... 1 0
.
N  SNS where S  
   ...   


1 0 ... 0 0
.
1 0
0
 PEQ  0 1

N
0 0
0.1
0 
 PAQ   0
I n2 
 0
0
0,
0
1 0
0.2 0,
0 1
(7.3)
  4
 B1  1  
0.6 1 
PB       3, A1  A1  I n1   
,
 0 0.7
 B2  11  6 
 
(n1  2, n2  1)
The equations (6.5) and (6.6) have the forms:
0.6 1  (1) i 1
 0.5 
1  4
xi(1)1  
xi   (1)k 1  xi(1)k 1   ui , i  Z 

11 3
 0 0.7
 k 
k 2
(7.4)
(6.9’)
and:
31
Tadeusz Kaczorek
Singular Fractional Continuous-Time and Discrete-Time Linear Systems
xi(2)   B2ui  
The solution ̅
6
ui , i  Z 
11
( )
(7.5)
of the equation (7.4) has the form:
i 1
xi(1)  i x0(1)   i  k 1B1uk , i  Z 
(7.6)
k 0
where:
1 0
0.6 1 
0  
, 1  A1  

,
0 1 
 0 0.7
 (  1) 0.485 1.300 
 2  A1 2  I n1

,...
0.615
2!
 0
(7.7)
(7.8)
The desired solution of the singular fractional system with
(7.1) is given by:
( )
0 0
1 0,
0 1
(7.12)
 1 0
 B1  
PB      0 1, A1  A1  I n1   [1],
 B2   1 1


(n1  1, n2  2)

 (21)    x (21)   0 1
0 0 i 1
j  0.8   xi  j 1  
i


(

1
)
1 0 
 j  (22)     (22)    1 1ui , i  Z 

 j  0
  xi  j 1    xi  



(7.14)
and:
1 0 0  1 1
0
x0  Q x0  0 1  1 1  0, x0(1)  [1], x0( 2)   
3
2 0 1  1 3
1
(7.9)
(7.15)
( )
( )
where ̅
and ̅
are determined by (6.7) and (7.5),
respectively.
Example 7.2. Find the solution xi of the singular fractional linear
system (5.1) with the matrices:
1 0 0 
 0.2 2  2
1 2




E  0 1  1, A   2
1 0 , B   1 2 
1  1 1 
 1.8 0  1
 2  1
(7.10)
for α = 0.8, arbitrary ui, i  Z  and x0  [1 1 1]T .
It is easy to check that the matrices (7.10) satisfy
the assumptions (5.4). In this case the matrices P and Q have
the forms:
 1 2 2 
 1 0 0


P   1  1  1, Q   2 1 1
 1 2 1 
 2 0 1
0
0,
0
i 1
 0.8 
xi(1)1  xi(1)   (1) k 1  xi(1)k 1  [1 0]ui , i  Z  (7.13)
 k 
k 2
1
 2 1  1 (1)
x 
xi  Qxi   1 0 0   i( 2) 
x
 0 0 1   i 
 A1
0

1 0
0
0 0

PEQ


N 
0 1
0.2
0 
 PAQ   0
I n2 
 0
In this case the equations (6.5) and (6.6) have the forms:
and:
0 1 0   1   2 
x0  Q x0  1 2 1  2    4 ,
0 0 1  1  1
 2
x0(1)   , x0( 2)  [1]
 4
 I n1
0

(7.11)
The solution ̅ of Eq. (7.13) with ̅
found using (6.7) and (6.8).
From (7.14) we have:
( )
can be easily
xi(21)  [0  1]ui , i  Z 
xi(22) 
i 1
 0.8 
[0  1]ui  j 1  [1  1]ui , i  Z 
j 
 (1) j 
j 0
(7.16)
The desired solution of the singular fractional system
with (7.10) is given by:
(1)
 1 0 0  xi 


xi  Qxi   2 1 1  xi( 21) 
 2 0 1  xi( 22) 


( )
where: ̅ , ̅
respectively.
(
)
and ̅
(
)
(7.17)
are determined by (7.13) and (7.16),
and:
8. CONCLUDING REMARKS
The singular fractional linear systems and electrical circuits
have been introduced. Using the Caputo definition of the fractional
derivative, the Weierstrass regular pencil decomposition and the
Laplace transform the solution to the state equation of singular
fractional linear system has been derived (Theorem 3.1). Singular
32
acta mechanica et automatica, vol.7 no.1 (2013), DOI 10.2478/ama-2013-0005
fractional linear electrical circuits have been analyzed. It has been
shown that every electrical circuit is a singular fractional system
if it contains at least one mesh consisting of branches with only
ideal supercondensators and voltage sources or at least one node
with branches with supercoils (Theorem 4.1). The singular fractional linear discrete-time systems have been introduced. Using
the Weierstrass regular pencil decomposition the solution to the
state equation of singular fractional linear discrete-time system
has been derived. The method of finding of the solution to the
singular fractional systems has been illustrated by two examples.
The considerations have been illustrated by singular linear electrical circuits. Those considerations can be extended for singular
fractional linear systems with singular pencils. Open problem are
extension of these considerations for positive singular fractional
linear systems and for singular positive linear systems with different fractional order. The linear systems with different fractional
orders are described by the equation (Kaczorek, 2007a).
 d  x1 
  
 dt    A11 A12   x1    B1 u,
 d  x2   A21 A22   x2   B2 
  
 dt 
p  1    p; q  1    q; p, q  N
where:
(8.1)
,
are the state vectors and
,
,
and
is the input
vector. Initial conditions for (8.1) have the form ( )
and ( )
.
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This work was supported by National Science Centre in Poland under
work no. NN514 1939 33.
33