Page 270 - Mathematical Models and Algorithms for Power System Optimization

262 Chapter 7
1
0
When G SðÞ ¼ , a function can be constructed to be similar with the above equation
S n +1
1
and equivalent to , to make:
S n
1 1
¼ lim
S n +1 a!0 ð S + aÞ n +1
and we have:

1 ∂ n 1
0 n T
G ZðÞ ¼ 1ð Þ lim 1 1 Z (7.111)
Z
n! a!0∂a n 1 e aT 1
To sum up, it can be seen that such transformation method must obtain the root of the model
first, yet the root of the equation with higher order cannot be obtained because there is no
available direct algorithm. Therefore, it is necessary to use the computer to make the iteration
solving (such as QR solution) in addition to a great deal of partial methods and combining
the same terms, the following state equation solution is hereby proposed, without needing any
iteration solving.

7.7.3.2 Method of state equation solution

The first method can be used to solve the problem, but it is not fit for the high-order models of
multiple roots or three orders or above, and even difficult to program. According to the detailed
process of the first method, the state equation solution method is conceived in the following
three steps:

Step 1: Transform the differential equation into a difference one.
1. The differential transfer function G(S) is available.
If the transfer function G(S) has been given by the actual engineering system, it must
be transformed into differential state equation.
In case of

n
ySðÞ b 1 S + b 2 S n 1 + ⋯ + b n 1 S + b n
GSðÞ ¼ ¼ (7.112)
n
uSðÞ S + a 1 S n 1 + ⋯ + a n 1 S + a n
it can be transformed into state equation using the following equation:
Let:
X 1 ¼ Y

Then:
_
_
X 2 ¼ Y d 1 u ¼ X 1 d 1 u
⋮
_
X n ¼ Y n 1 d 1 u n 2 d 2 u n 3 ⋯ d n 1 u ¼ X n 1 d n 1 u (7.113)