Page 121 - Mathematical Techniques of Fractional Order Systems

Exact Solution of Linear Fractional Distributed Order Systems Chapter | 4 109

4.3 MAIN RESULTS
The cornerstone of the analytical solution of (4.17) is in fact the impulse
response of the integrator with the same weight function as in the differen-
tiator involved in this equation. This is similar to the case of fractional
order and integer order systems. Recall that the analytical solution of the
c α
fractional order system of linear differential equations D xðtÞ 5 AxðtÞ 1 BuðtÞ
0 t
α
is written by using Mittag Leffler functions E α;β tðÞ 5 P 1N t αk .
k50 Γαk 1 βð Þ
Mittag Leffler functions are constructed on the basis of iterated self-
convolutions of fractional power function, i.e., t α21 =ΓαðÞ which is the
impulse response of fractional integration operator (4.7). A general form of
the analytical solution is obtained in a similar way for distributed order
equations. This idea is in fact analogous to the analytic approach to integral
equations using resolvent kernels. Using this concept here makes sense due
to the fact that it is always possible to turn Eq. (4.17) into the following two
Volterra integral equations of the second kind

x 1 2 Ax 1 i 5 x 0ðÞ
; xtðÞ 5 x 1 tðÞ 1 x 2 tðÞ ð4:19Þ
x 2 2 Ax 2 i 5 Bu i

in which itðÞ is defined in (4.16). In order to show this, let us take the
Laplace transform of (4.19) which yields

X 1 ðsÞ 2 AX 1 ðsÞIðsÞ 5 x 0ðÞ=s
ð4:20Þ
X 2 ðsÞ 2 AX 2 ðsÞIðsÞ 5 BUðsÞIðsÞ
Dividing each relation by IsðÞ and replacing W 2lnsÞ 5 1=IsðÞ afterwards
ð
results

ð
X 1 ðsÞW 2lnsð Þ 2 AX 1 ðsÞ 5 x 0ðÞW 2lnsÞ=s
ð4:21Þ
X 2 ðsÞW 2lnsð Þ 2 AX 2 ðsÞ 5 BUðsÞ
Solving the resultant equations for X 1 ðsÞ and X 2 ðsÞ gives
21
X 1 ðsÞ 5 W 2lnsðð ÞI2AÞ x 0ðÞW 2lnsÞ=s
ð
21 ð4:22Þ
X 2 ðsÞ 5 W 2lnsðð ÞI2AÞ BUðsÞ
In which I is the identity matrix with appropriate dimensions. The solu-
tion of (4.17) is then obtained by adding the homogeneous and particular
solutions XsðÞ 5 X 1 sðÞ 1 X 2 sðÞ as
21
ð
ð
ð
XsðÞ 5 W 2lnsÞI2AÞ x 0ðÞW 2lnsÞ=s 1 BU sðÞ ð4:23Þ
Multiplying each side of (4.23) by W 2lnsÞI 2 A yields
ð
ð
ð W 2lnsÞI 2 AÞXsðÞ 5 x 0ðÞW 2lnsÞ=s 1 BU sðÞ ð4:24Þ
ð