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附录A
推导 De (p) |p= -jωm 。在工作点线性化正文式（1）~式（3）得：
ì Δud + jΔuq = jωc (Δψd + jΔψ q )+ j(Δψd0 + jΔψ q0 )Δωk - ra (Δid + jΔiq )
ï
ï ΔuCd + jΔuCq = Δud + jΔuq - j(Δid0 + jΔiq0 )LΔωk - (R + jωc L)(Δid + jΔiq ) í
U ¥ (cosδk - jsinδk )Δδk
ï
ï jω C(Δu + jΔu )+ jC(Δu + jΔu )Δω = Δi + jΔi
Cd
Cq
Cd0
Cq0
k
d
q
î c
(A1)
其中， Δδk = Δωk /p = Δωk /(-jωm ) 。将 Δud + jΔuq ， ΔuCd + jΔuCq 作为中间变量消去，再实部/虚部分
T
别展开解得 éë Δid , Δiq ùû 与 Δωk 的关系，如下式所示：
é Δid ù
é1 - ωc 2 LC
1
ê Δi ú =
2
2 ê
ë q û ( ωc 2 LC - 1 ) + ( ωc CR0 ) ë -ωc CR0
ωc CR0 ù é A1 ù
ú ê ú Δωk
1 - ωc 2 LC û ë B1 û
(A2)
其中，A0 = LCωc + 1/ωk - ωc C/ωm [ ωk L - 1/(ωk C)] ，B0 = ωc C/(ωm R0 ) , A1 = ωc Cψd0 ωc /ωm + A0 id0 -
B0 iq0 , B1 = ωc Cψ q0 ωc /ωm + B0 id0 + A0 iq0 。
进一步，
De (p) |p= -jωm Δωk » ψ d0 Δiq - ψ q0 Δid = -
其中， Dep =
ωc (ψ q0 2 + ψ d0 2 )
2
é 1 æ
ù
1 ö
ωm ê ç ωc L ÷ + R0 ú
ωc C ø
êë R0 è
úû
Δωk + Dep Δωk
(A3)
(ωc L - 1/ωc C)(B1ψ d0 - A1ψ q0 )/(ωc CR0 ) - (A0 id0ψ d0 - B0 iq0 ψd0 + B0 id0ψq0 + A0 iq0ψ q0 )
é 1
ωc ê
êë R0
2
ù
æ
1 ö
ω
L
+
R
ç c
÷
0ú
ωc C ø
úû
è
在式（A3）中，通常 Dep Δωk 相对第一项较小，在发电机出力比较低即 id0 、 iq0 较小时该项约为 0。从而
得到 De (p) |p= -jωm ，又 De (p) |p= jωm » 0 ，由式正文（9）得到 De 的近似表达式如正文式（11）
。