Page 123 - Mathematical Techniques of Fractional Order Systems
P. 123
Exact Solution of Linear Fractional Distributed Order Systems Chapter | 4 111
It is immediately followed by definition (4.2) that
r 0 2π2 1
r 0 e 2λ 5 e λ ð4:31Þ
which is equivalent to
r 0 π1 1
e λ 5 r 0 e λ ð4:32Þ
Writing (4.32) in a logarithmic form and a little calculation gives
1 π
lnr 0
5 2 ð4:33Þ
r 0 2 1 λ r 0 2 1
|fflffl{zfflffl} |fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl}
ðÞ
ðÞ
f 1 r 0 f 2 r 0
ðÞ) is strictly decreasing (increasing) in
ðÞ
It can be verified that f 1 r 0 (f 2 r 0
the range r 0 A 1; 1 Nð Þ. Therefore, for r . r 0 one could write
lnr 1 π
, 2 ð4:34Þ
r 2 1 λ r 2 1
or equivalently
lnr 1 π 1
p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi , ð4:35Þ
2
r 2 2r 1 1 λ
lnr 1 π lnr 1 π
Note that for any θA 2π; π there holds p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi , ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi.Thus
ð
p
2
r 2 2 2cos θðÞr 1 1 r 2 2r 1 1
by replacing the left side of (4.35) with its lower bound we get
lnr 1 π 1
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi , ð4:36Þ
p 2 λ
r 2 2cos θðÞr 1 1
It is possible to write the left side of (4.36) in the form
q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
ð lnr1πÞ 1
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi , ð4:37Þ
q λ
2 2
ð rcosθ21Þ 1 rsinθð Þ
q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
2
As lnr . 0, it is obvious that ð lnrÞ 1 π , ð lnr1πÞ . On the other
2
hand, θ generally lies in the interval 2π; π. Therefore, as a more conserva-
ð
tive inequality, we obtain the following inequality from (4.37)
q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
lnrÞ 1 θ 2
2
ð 1
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi , ð4:38Þ
q λ
2 2
ð rcosθ21Þ 1 rsinθð Þ
iθ
Finally, writing (4.38) in terms of s 5 re gives
j lnsj 1
, ð4:39Þ
j s 2 1j λ
