# Управление качеством переходных процессов в многосвязных системах. Задание на курсовое проектирование, страница 8

В (2.34) исключим Id с помощью (2.26)

U0cosd0     Eq                                                                                             Id = ———– – —–                                                                             (2.36)                     Xd          Xd

EqU0                U0(Xd–Xq)sind0*U0*cosd0                                                    P = ––––– sind0+ ———————————— –                                                                 Xq                                  XqXd

U0(Xd–Xq)sind0*Eq                                                                                         – ————————–                                                                       (2.37)                      XqXd

Таким образом:

EqU0               U02(Xd–Xq)                                                                            P= ––––– sind0+ —————– sin2d0                                                (2.38)                       Xq                    2XqXd

В уравнении (2.38) U0 = const, то есть DU0 = 0. Поэтому линеаризованное уравнение принимает вид:

¶P            ¶P              EqUo                U02(Xd–Xq)                                     DP = –— Dd+ —— DEq=  ——– cosd0 + —————— cos2d0 Dd +

¶d           ¶Eq                Xd                       XqXd

U0sind0

+ ———— DEq                                                                                  (2.39)                Xd

Исключим из (2.28) DId, используя (2.26):

U0cosd0     Eq                                                                                             Id = —–—— – —–                                                                              (2.40)                     Xd         Xd

U0sind0           DEq                                                                                  DId = – ———– Dd – —––                                                                  (2.41)                         Xd                Xd

Xd – X'd                       Xd – X'd                                             (Xd – X'd)DId = – ———— U0sind0Dd – ———— DEq                      (2.42)                                           Xd                                 Xd

Тогда уравнение (2.28) примет вид: