Page 288 - Mathematical Models and Algorithms for Power System Optimization

280 Chapter 7

1 A B C D
GZðÞ ¼ ¼ + + +
S 0:53S +1 45:27S +1 0:5S +1 S

1
A ¼ GS ðÞ ð 0:53S +1Þ 1 ¼ 2:68775
S
S¼
0:53

1
B ¼ GSðÞ ð 45:27S +1Þ 1 ¼ 42:1123
S
S¼
45:27

1
C ¼ GSðÞ ð 0:5S +1Þ 1 ¼ 2:5
S
S¼
0:5

1
D ¼ GSðÞ S ¼ 1
S
S¼0
Substitute them into Eq. (7.98) to get:

5:0679 0:93025 4:999999 1 1
GZ ðÞ ¼ + + + 1 Z
Z
Z
Z
1 e 1=0:53 1 1 e 1=45:27 1 1 e 1=0:53 1 1 Z 1
E
¼
1
1
ð
ð 1 0:15156Z Þ 1 0:97815Zð 1 Þ 1 0:1353353Z Þ
where

1 1
E ¼ 5:0697 1 e 0:5Z 1 1 e 0:527Z 1 1 Z 1

1 1
0:93024 1 e 0:53Z 1 1 e 0:5Z 1 1 Z 1

1 1
+4:99997 1 e 0:53Z 1 1 e 45:27Z 1 1 Z 1

1 1 1
+1 e 0:53Z 1 1 e 45:27Z 1 1 e 0:5Z 1
¼ 0:0020285Z 1 +0:0895481Z 2 0:0714933Z 3
Finally

0:0020285Z 1 +0:0895481Z 2 0:714933Z 3
GZðÞ ¼
1 1:265Z 1 +0:3Z 2 0:02Z 3

(2) Solve the transfer function by the State equation method.
First, transform G(S) into a state equation (and in reference to the hydro unit model)