鎖交磁束関係式のd-q-0変換

発電機巻線と鎖交磁束の関係式」で導出したインダクタンスの式を$d-q-0$変換し、発電機の挙動を$d-q-0$成分量で表すための式をまとめる。

スポンサーリンク

変換行列の書き換え

鎖交磁束関係式の変換の計算をより簡潔に行うための準備として、変換行列の書き換えを行う。

$d-q-0$変換の変換行列$\boldsymbol{D}(t)$および逆変換行列$\boldsymbol{D^{-1}}(t)$は、「パーク変換の導入」の変換行列の定義式より、

$$\begin{align*}
\boldsymbol{D}(t)&=\frac{2}{3}\left(\begin{array}{ccc} \cos\theta_a & \cos\theta_b & \cos\theta_c \\ -\sin\theta_a & -\sin\theta_b & -\sin\theta_c \\ \displaystyle{\frac{1}{2}} & \displaystyle{\frac{1}{2}} & \displaystyle{\frac{1}{2}} \end{array} \right) &・・・(1)\\\\
\boldsymbol{D^{-1}}(t)&=\left(\begin{array}{ccc} \cos\theta_a & -\sin\theta_a & 1 \\ \cos\theta_b & -\sin\theta_b & 1 \\ \cos\theta_c & -\sin\theta_c & 1 \end{array} \right)  &・・・(2)
\end{align*}$$

ここで、$\theta_a=\omega t,\ \theta_b=\omega t-\displaystyle{\frac{2}{3}}\pi,\ \theta_c=\omega t+\displaystyle{\frac{2}{3}}\pi$に注意して、$\cos\theta_a$および$\sin\theta_a$を書き換えると、

$$\begin{align*}
\cos\theta_a=\frac{e^{j\omega t}+e^{-j\omega t}}{2} ・・・(3)\\\\
\sin\theta_a=\frac{e^{j\omega t}-e^{-j\omega t}}{j2} ・・・(4)
\end{align*}$$

$(3),\ (4)$式に従い$(1)$および$(2)$式を書き換えると、$e^{-j\frac{2}{3}\pi}=a^2,\ e^{j\frac{2}{3}\pi}=a$より、

$$\begin{align*}
\boldsymbol{D}(t)&=\small{\frac{2}{3}\left(\begin{array}{ccc} \displaystyle{\frac{e^{j\omega t}+e^{-j\omega t}}{2}} & \displaystyle{\frac{e^{j\left(\omega t-\frac{2}{3}\pi\right)}+e^{-j\left(\omega t-\frac{2}{3}\pi\right)}}{2}} & \displaystyle{\frac{e^{j\left(\omega t+\frac{2}{3}\pi\right)}+e^{-j\left(\omega t+\frac{2}{3}\pi\right)}}{2}} \\ \displaystyle{j\frac{e^{j\omega t}-e^{-j\omega t}}{2}} & \displaystyle{j\frac{e^{j\left(\omega t-\frac{2}{3}\pi\right)}-e^{-j\left(\omega t-\frac{2}{3}\pi\right)}}{2}} & \displaystyle{j\frac{e^{j\left(\omega t+\frac{2}{3}\pi\right)}-e^{-j\left(\omega t+\frac{2}{3}\pi\right)}}{2}} \\ \displaystyle{\frac{1}{2}} & \displaystyle{\frac{1}{2}} & \displaystyle{\frac{1}{2}} \end{array} \right)}\\\\
&=\frac{1}{3}\left(\begin{array}{ccc} \displaystyle{e^{j\omega t}+e^{-j\omega t}} & \displaystyle{a^2e^{j\omega t}+ae^{-j\omega t}} & \displaystyle{ae^{j\omega t}+a^2e^{-j\omega t}} \\ \displaystyle{j\left(e^{j\omega t}-e^{-j\omega t}\right)} & \displaystyle{j\left(a^2e^{j\omega t}-ae^{-j\omega t}\right)} & \displaystyle{j\left(ae^{j\omega t}-a^2e^{-j\omega t}\right)} \\ 1 & 1 & 1 \end{array} \right)\\\\ &・・・(1)’\\\\\\
\boldsymbol{D^{-1}}(t)&=\left(\begin{array}{ccc} \displaystyle{\frac{e^{j\omega t}+e^{-j\omega t}}{2}} & \displaystyle{j\frac{e^{j\omega t}-e^{-j\omega t}}{2}} & 1 \\ \displaystyle{\frac{e^{j\left(\omega t-\frac{2}{3}\pi\right)}+e^{-j\left(\omega t-\frac{2}{3}\pi\right)}}{2}} & \displaystyle{j\frac{e^{j\left(\omega t-\frac{2}{3}\pi\right)}-e^{-j\left(\omega t-\frac{2}{3}\pi\right)}}{2}} & 1 \\ \displaystyle{\frac{e^{j\left(\omega t+\frac{2}{3}\pi\right)}+e^{-j\left(\omega t+\frac{2}{3}\pi\right)}}{2}} & \displaystyle{j\frac{e^{j\left(\omega t+\frac{2}{3}\pi\right)}-e^{-j\left(\omega t+\frac{2}{3}\pi\right)}}{2}} & 1 \end{array} \right)\\\\
&=\frac{1}{2}\left(\begin{array}{ccc} \displaystyle{e^{j\omega t}+e^{-j\omega t}} & \displaystyle{j\left(e^{j\omega t}-e^{-j\omega t}\right)} & 2 \\ \displaystyle{a^2e^{j\omega t}+ae^{-j\omega t}} & \displaystyle{j\left(a^2e^{j\omega t}-ae^{-j\omega t}\right)} & 2 \\ \displaystyle{ae^{j\omega t}+a^2e^{-j\omega t}} & \displaystyle{j\left(ae^{j\omega t}-a^2e^{-j\omega t}\right)} & 2 \end{array} \right)\\\\
 &・・・(2)’
\end{align*}$$

鎖交磁束関係式の変換

電機子巻線鎖交磁束の式の変換

電機子巻線の鎖交磁束と電流との関係式は、「鎖交磁束関係式」の$(12)$式より、

$$\begin{align*}
&\left(\begin{array}{c} \it{\Psi}_a\left(t\right) \\ \it{\Psi}_b\left(t\right) \\ \it{\Psi}_c\left(t\right) \end{array}\right)\\\\
&=\scriptsize{\left(\begin{array}{ccc} -L_{aa0}-L_{aa2}\cos2\omega t & M_{ab0}+M_{ab2}\cos\left(2\omega t-\frac{2}{3}\pi\right) & M_{ab0}+M_{ab2}\cos\left(2\omega t+\frac{2}{3}\right) \\ M_{ab0}+M_{ab2}\cos\left (2\omega t-\frac{2}{3}\pi\right) & -L_{aa0}-L_{aa2}\cos\left(2\omega t+\frac{2}{3}\pi\right) & M_{ab0}+M_{ab2}\cos2\omega t \\ M_{ab0}+M_{ab2}\cos\left(2\omega t+\frac{2}{3}\right) & M_{ab0}+M_{ab2}\cos2\omega t & -L_{aa0}-L_{aa2}\cos2\left(2\omega t-\frac{2}{3}\pi\right) \end{array}\right)}\left(\begin{array}{c} i_a(t) \\ i_b(t) \\ i_c(t)\end{array}\right)\\\\
&\quad+\scriptsize{\left(\begin{array}{ccc} M_{afd}\cos\omega t & M_{aDd}\cos\omega t & -M_{aDd}\sin\omega t \\ M_{afd}\cos\left(\omega t-\frac{2}{3}\pi\right) & M_{aDd}\cos\left(\omega t-\frac{2}{3}\pi\right) (t) & -M_{aDq}\sin\left(\omega t-\frac{2}{3}\pi\right) \\ M_{afd}\cos\left(\omega t+\frac{2}{3}\pi\right) & M_{aDd}\cos\left(\omega t+\frac{2}{3}\pi\right) (t) & -M_{aDq}\sin\left(\omega t+\frac{2}{3}\pi\right) \end{array}\right)\left(\begin{array}{c} i_{fd}(t) \\ i_{Dd}(t) \\ i_{Dq}(t)\end{array}\right)}\\\\
&  ・・・(5)
\end{align*}$$

行列を記号で表記すると、

$$\boldsymbol{\Psi_{abc}\left(t\right)}=\boldsymbol{l_{abc}}(t)\boldsymbol{i_{abc}}(t)+\boldsymbol{m_{abc-fD}}(t)\boldsymbol{i_{fD}}(t) ・・・(5)’$$

$d-q-0$変換後の鎖交磁束$\boldsymbol{\Psi_{dq0}}(t)$は、$(5)’$式の左から変換行列$\boldsymbol{D}(t)$を掛けて、

$$\begin{align*}
\boldsymbol{\Psi_{dq0}}(t)&=\boldsymbol{D}(t)\boldsymbol{\Psi_{abc}}(t)=\boldsymbol{D}(t)\boldsymbol{l_{abc}}(t)\boldsymbol{i_{abc}}(t)+\boldsymbol{D}(t)\boldsymbol{m_{abc-fD}}(t)\boldsymbol{i_{fD}}(t)\\\\
&=\{\boldsymbol{D}(t)\boldsymbol{l_{abc}}(t)\boldsymbol{D^{-1}}(t)\}\boldsymbol{i_{dq0}}(t)+\{\boldsymbol{D}(t)\boldsymbol{m_{abc-fD}}(t)\}\boldsymbol{i_{fD}}(t) ・・・(6)
\end{align*}$$

ここで、$(6)$式の$\boldsymbol{l_{abc}}(t)$および$\boldsymbol{m_{abc-fD}}(t)$を$(3),\ (4)$式を用いて書き換えると、

$$\begin{align*}
&\boldsymbol{l_{abc}}(t)=\scriptsize{\left(\begin{array}{ccc} -L_{aa0}-L_{aa2}\cos2\omega t & M_{ab0}+M_{ab2}\cos\left(2\omega t-\frac{2}{3}\pi\right) & M_{ab0}+M_{ab2}\cos\left(2\omega t+\frac{2}{3}\right) \\ M_{ab0}+M_{ab2}\cos\left(2\omega t-\frac{2}{3}\pi\right) & -L_{aa0}-L_{aa2}\cos\left(2\omega t+\frac{2}{3}\pi\right) & M_{ab0}+M_{ab2}\cos2\omega t \\ M_{ab0}+M_{ab2}\cos\left(2\omega t+\frac{2}{3}\right) & M_{ab0}+M_{ab2}\cos2\omega t & -L_{aa0}-L_{aa2}\cos\left(2\omega t-\frac{2}{3}\pi\right) \end{array}\right)}\\\\
&=\scriptsize{\left(\begin{array}{ccc} -L_{aa0}-\frac{L_{aa2}}{2}\displaystyle{\left(e^{j2\omega t}+e^{-j2\omega t}\right)} & M_{ab0}+\frac{M_{ab2}}{2}\displaystyle{\left(a^2e^{j2\omega t}+ae^{-j2\omega t}\right)} & M_{ab0}+\frac{M_{ab2}}{2}\displaystyle{\left(ae^{j2\omega t}+ae^{-j2\omega t}\right)} \\ M_{ab0}+\frac{M_{ab2}}{2} \displaystyle{\left(a^2e^{j2\omega t}+ae^{-j2\omega t}\right)} & -L_{aa0}-\frac{L_{aa2}}{2} \displaystyle{\left(ae^{j2\omega t}+a^2e^{-j2\omega t}\right)} & M_{ab0}+\frac{M_{ab2}}{2}\displaystyle{\left(e^{j2\omega t}+e^{-j2\omega t}\right)} \\ M_{ab0}+\frac{M_{ab2}}{2}\displaystyle{\left(ae^{j2\omega t}+a^2e^{-j2\omega t}\right)} & M_{ab0}+\frac{M_{ab2}}{2}\displaystyle{\left(e^{j2\omega t}+e^{-j2\omega t}\right)} & -L_{aa0}-\frac{L_{aa2}}{2} \displaystyle{\left(a^2e^{j2\omega t}+ae^{-j2\omega t}\right)} \end{array}\right)}\\\\
&\boldsymbol{m_{abc-fD}}(t)=\small{\left(\begin{array}{ccc} M_{afd}\cos\omega t & M_{aDd}\cos\omega t & -M_{aDq}\sin\omega t \\ M_{afd}\cos\left(\omega t-\frac{2}{3}\pi\right) & M_{aDq}\cos\left(\omega t-\frac{2}{3}\pi\right) & -M_{aDq}\sin\left(\omega t-\frac{2}{3}\pi\right) \\ M_{afd}\cos\left(\omega t+\frac{2}{3}\pi\right) & M_{aDd}\cos\left(\omega t+\frac{2}{3}\pi\right) & -M_{aDq}\sin\left(\omega t+\frac{2}{3}\pi\right) \end{array}\right)}\\\\
&=\left(\begin{array}{ccc} \frac{M_{afd}}{2}\displaystyle{\left(e^{j\omega t}+e^{-j\omega t}\right)} & \frac{M_{aDd}}{2}\displaystyle{\left(e^{j\omega t}+e^{-j\omega t}\right)} & j\frac{M_{aDq}}{2}\displaystyle{\left(e^{j\omega t}-e^{-j\omega t}\right)} \\ \frac{M_{afd}}{2}\displaystyle{\left(a^2e^{j\omega t}+ae^{-j\omega t}\right)} & \frac{M_{aDd}}{2}\displaystyle{\left(a^2e^{j\omega t}+ae^{-j\omega t}\right)} & j\frac{M_{aDq}}{2}\displaystyle{\left(a^2e^{j\omega t}-ae^{-j\omega t}\right)} \\ \frac{M_{afd}}{2}\displaystyle{\left(ae^{j\omega t}+a^2e^{-j\omega t}\right)} & \frac{M_{aDd}}{2}\displaystyle{\left(ae^{j\omega t}+a^2e^{-j\omega t}\right)} & j\frac{M_{aDq}}{2}\displaystyle{\left(ae^{j\omega t}-a^2e^{-j\omega t}\right)} \end{array}\right)
\end{align*}$$

したがって、$(6)$式の$\boldsymbol{D}(t)\boldsymbol{l_{abc}}(t)\boldsymbol{D^{-1}}(t)$および$\boldsymbol{D}(t)\boldsymbol{m_{abc-fD}}(t)$を、$(1)’,\ (2)’$を用いてそれぞれ求めると、「鎖交磁束の関係式」で$L_{aa0}=2M_{ab0},\ L_{aa2}=-M_{ab2}$であったことに注意して、

$$\begin{align*}
&\boldsymbol{D}(t)\boldsymbol{l_{abc}}(t)\boldsymbol{D^{-1}}(t)\\\\
&=\frac{1}{3}\left(\begin{array}{ccc} \displaystyle{e^{j\omega t}+e^{-j\omega t}} & \displaystyle{a^2e^{j\omega t}+ae^{-j\omega t}} & \displaystyle{ae^{j\omega t}+a^2e^{-j\omega t}} \\ \displaystyle{j\left(e^{j\omega t}-e^{-j\omega t}\right)} & \displaystyle{j\left(a^2e^{j\omega t}-ae^{-j\omega t}\right)} & \displaystyle{j\left(ae^{j\omega t}-a^2e^{-j\omega t}\right)} \\ 1 & 1 & 1 \end{array} \right)\\\\
&\quad\times\scriptsize{\left(\begin{array}{ccc} -L_{aa0}-\frac{L_{aa2}}{2}\displaystyle{\left(e^{j2\omega t}+e^{-j2\omega t}\right)} & M_{ab0}+\frac{M_{ab2}}{2}\displaystyle{\left(a^2e^{j2\omega t}+ae^{-j2\omega t}\right)} & M_{ab0}+\frac{M_{ab2}}{2}\displaystyle{\left(ae^{j2\omega t}+ae^{-j2\omega t}\right)} \\ M_{ab0}+\frac{M_{ab2}}{2}\displaystyle{\left(a^2e^{j2\omega t}+ae^{-j2\omega t}\right)} & -L_{aa0}-\frac{L_{aa2}}{2} \displaystyle{\left(ae^{j2\omega t}+a^2e^{-j2\omega t}\right)} & M_{ab0}+\frac{M_{ab2}}{2}\displaystyle{\left(e^{j2\omega t}+e^{-j2\omega t}\right)} \\ M_{ab0}+\frac{M_{ab2}}{2}\displaystyle{\left(ae^{j2\omega t}+a^2e^{-j2\omega t}\right)} & M_{ab0}+\frac{M_{ab2}}{2}\displaystyle{\left(e^{j2\omega t}+e^{-j2\omega t}\right)} & -L_{aa0}-\frac{L_{aa2}}{2} \displaystyle{\left(a^2e^{j2\omega t}+ae^{-j2\omega t}\right)} \end{array}\right)}\\\\
&\quad\times\frac{1}{2}\left(\begin{array}{ccc} \displaystyle{e^{j\omega t}+e^{-j\omega t}} & \displaystyle{j\left(e^{j\omega t}-e^{-j\omega t}\right)} & 2 \\ \displaystyle{a^2e^{j\omega t}+ae^{-j\omega t}} & \displaystyle{j\left(a^2e^{j\omega t}-ae^{-j\omega t}\right)} & 2 \\ \displaystyle{ae^{j\omega t}+a^2e^{-j\omega t}} & \displaystyle{j\left(ae^{j\omega t}-a^2e^{-j\omega t}\right)} & 2 \end{array} \right)\\\\
&=\frac{1}{3}\left(\begin{array}{ccc} A_0\left(\displaystyle{e^{j\omega t}+e^{-j\omega t}}\right) & A_0\left(\displaystyle{a^2e^{j\omega t}+ae^{-j\omega t}}\right) & A_0\left(\displaystyle{ae^{j\omega t}+a^2e^{-j\omega t}}\right) \\ j\left\{A_0\left(\displaystyle{e^{j\omega t}-e^{-j\omega t}}\right)\right\} & j\left\{A_0\left(\displaystyle{a^2e^{j\omega t}-ae^{-j\omega t}}\right)\right\} & j\left\{A_0\left(\displaystyle{ae^{j\omega t}-a^2e^{-j\omega t}}\right)\right\} \\ 0 & 0 & 0 \end{array} \right)\\\\
&\quad\times\frac{1}{2}\left(\begin{array}{ccc} \displaystyle{e^{j\omega t}+e^{-j\omega t}} & \displaystyle{j\left(e^{j\omega t}-e^{-j\omega t}\right)} & 2 \\ \displaystyle{a^2e^{j\omega t}+ae^{-j\omega t}} & \displaystyle{j\left(a^2e^{j\omega t}-ae^{-j\omega t}\right)} & 2 \\ \displaystyle{ae^{j\omega t}+a^2e^{-j\omega t}} & \displaystyle{j\left(ae^{j\omega t}-a^2e^{-j\omega t}\right)} & 2 \end{array} \right)\\\\
&=\frac{1}{6}\left(\begin{array}{ccc} 6A_0 & 0 & 0 \\ 0 & 6A_0 & 0 \\ 0 & 0 & 0 \end{array} \right)\\\\
\end{align*}$$

ただし、

$$\begin{align*}
A_0&=-\frac{3}{2}\left(L_{aa0}+L_{aa2}\right)
\end{align*}$$

したがって、

$$\begin{align*}
\boldsymbol{D}(t)\boldsymbol{l_{abc}}(t)\boldsymbol{D^{-1}}(t)&=-\left(\begin{array}{ccc} \displaystyle{\frac{3}{2}\left(L_{aa0}+L_{aa2}\right)} & 0 & 0 \\ 0 & \displaystyle{\frac{3}{2}\left(L_{aa0}+L_{aa2}\right)} & 0 \\ 0 & 0 & 0 \end{array} \right)\\\\
&\equiv-\left(\begin{array}{ccc} L_d & 0 & 0 \\ 0 & L_q & 0 \\ 0 & 0 & L_0(=0) \end{array} \right) \end{align*}
$$

$$\begin{align*}
&\boldsymbol{D}(t)\boldsymbol{m_{abc-fD}}(t)\\\\
&=\frac{1}{3}\left(\begin{array}{ccc} \displaystyle{e^{j\omega t}+e^{-j\omega t}} & \displaystyle{a^2e^{j\omega t}+ae^{-j\omega t}} & \displaystyle{ae^{j\omega t}+a^2e^{-j\omega t}} \\ \displaystyle{j\left(e^{j\omega t}-e^{-j\omega t}\right)} & \displaystyle{j\left(a^2e^{j\omega t}-ae^{-j\omega t}\right)} & \displaystyle{j\left(ae^{j\omega t}-a^2e^{-j\omega t}\right)} \\ 1 & 1 & 1 \end{array} \right)\\\\
&\quad\times\left(\begin{array}{ccc} \frac{M_{afd}}{2}\displaystyle{\left(e^{j\omega t}+e^{-j\omega t}\right)} & \frac{M_{aDd}}{2}\displaystyle{\left(e^{j\omega t}+e^{-j\omega t}\right)} & j\frac{M_{aDq}}{2}\displaystyle{\left(e^{j\omega t}-e^{-j\omega t}\right)} \\ \frac{M_{afd}}{2}\displaystyle{\left(a^2e^{j\omega t}+ae^{-j\omega t}\right)} & \frac{M_{aDd}}{2}\displaystyle{\left(a^2e^{j\omega t}+ae^{-j\omega t}\right)} & j\frac{M_{aDq}}{2}\displaystyle{\left(a^2e^{j\omega t}-ae^{-j\omega t}\right)} \\ \frac{M_{afd}}{2}\displaystyle{\left(ae^{j\omega t}+a^2e^{-j\omega t}\right)} & \frac{M_{aDd}}{2}\displaystyle{\left(ae^{j\omega t}+a^2e^{-j\omega t}\right)} & j\frac{M_{aDq}}{2}\displaystyle{\left(ae^{j\omega t}-a^2e^{-j\omega t}\right)} \end{array}\right)\\\\
&=\left(\begin{array}{ccc} M_{afd} & M_{aDd} & 0 \\ 0 & 0 & M_{aDq} \\ 0 & 0 & 0 \end{array}\right)
\end{align*}$$

以上より、$(5)$式の$d-q-0$変換後の式は、

$$\begin{align*}
\left(\begin{array}{c} \it{\Psi}_d\left(t\right) \\ \it{\Psi}_q\left(t\right) \\ \it{\Psi}_\rm{0}\left(\it{t}\right) \end{array}\right)&=-\left(\begin{array}{ccc} L_{d} & 0 & 0 \\ 0 & L_{q} & 0 \\ 0 & 0 & L_{0} \end{array}\right)\left(\begin{array}{c} i_d(t) \\ i_q(t) \\ i_0(t)\end{array}\right)\\\\&+\left(\begin{array}{ccc} M_{afd} &M_{aDd}&0 \\ 0 & 0 & M_{aDq} \\ 0&0&0\end{array}\right) \left(\begin{array}{c} i_{fd}(t) \\ i_{Dd}(t) \\ i_{Dq}(t) \end{array}\right) ・・・(7)
\end{align*}$$

となり、$d-q-0$変換後の電流にかかるインダクタンス行列の要素はすべて時間$t$に関係ない定数となる。

スポンサーリンク

界磁・制動巻線鎖交磁束の式の変換

界磁および制動巻線の鎖交磁束と電流との関係式は、「鎖交磁束関係式」の$(13)$式より、

$$\begin{align*}
&\left(\begin{array}{c} \it{\Psi}_{fd}\left(t\right) \\ \it{\Psi}_{Dd}\left(t\right) \\ \it{\Psi}_{Dq}\left(t\right) \end{array}\right)\\\\
&=\small{ -\left(\begin{array}{ccc} M_{afd}\cos\omega t & M_{afd}\cos\left(\omega t-\frac{2}{3}\pi\right) & M_{afd}\cos\left(\omega t+\frac{2}{3}\right) \\ M_{aDd}\cos\omega t & M_{aDd}\cos\left(\omega t-\frac{2}{3}\pi\right) & M_{aDd}\cos\left(\omega t+\frac{2}{3}\right) \\ -M_{aDq}\sin\omega t & -M_{aDq}\sin\left(\omega t-\frac{2}{3}\pi\right) & -M_{aDq}\sin\left(\omega t+\frac{2}{3}\right) \end{array}\right) }\left(\begin{array}{c} i_a(t) \\ i_b(t) \\ i_c(t)\end{array}\right)\\\\
&+\left(\begin{array}{ccc} L_{ffd} & M_{fDd} & 0 \\ M_{fDd} & L_{DDd} & 0 \\ 0 & 0 & L_{DDq} \end{array}\right)\left(\begin{array}{c} i_{fd}(t) \\ i_{Dd}(t) \\ i_{Dq}(t)\end{array}\right) ・・・(8)
\end{align*}$$

行列を記号で表記すると、

$$\begin{align*}
\boldsymbol{\Psi_{dD}}(t)&=-\boldsymbol{m_{fD-abc}}(t)\boldsymbol{i_{abc}}(t)+\boldsymbol{l_{fD}}(t)\boldsymbol{i_{fD}}(t)\\\\
&=-\boldsymbol{m_{fD-abc}}(t)\left\{\boldsymbol{D^{-1}}(t)\boldsymbol{i_{dq0}}(t)\right\}+\boldsymbol{l_{fD}}(t)\boldsymbol{i_{fD}}(t)  ・・・(8)’
\end{align*}$$

ここで、$(8)’$式第1項の$\boldsymbol{m_{fD-abc}}(t)\boldsymbol{D^{-1}}(t)$を計算すると、$(3),\ (4)$式の書き換えを用いて、

$$\begin{align*}
&\boldsymbol{m_{fD-abc}}(t)\boldsymbol{D^{-1}}(t)\\\\
&=\left(\begin{array}{ccc} \frac{M_{afd}}{2}\left(\displaystyle{e^{j\omega t}+e^{-j\omega t}}\right) & \frac{M_{afd}}{2} \left( \displaystyle{a^2e^{j\omega t}+ae^{-j\omega t}} \right) & \frac{M_{afd}}{2} \left(\displaystyle{ae^{j\omega t}+a^2e^{-j\omega t}} \right) \\ \frac{M_{aDd}}{2} \left( \displaystyle{e^{j\omega t}+e^{-j\omega t}} \right) & \frac{M_{aDd}}{2} \left( \displaystyle{a^2e^{j\omega t}+ae^{-j\omega t}} \right) &\quad\frac{M_{aDd}}{2} \left( \displaystyle{ae^{j\omega t}+a^2e^{-j\omega t}} \right) \\ j\frac{M_{aDq}}{2} \left( \displaystyle{e^{j\omega t}-e^{-j\omega t}} \right) & j\frac{M_{aDq}}{2} \left( \displaystyle{a^2e^{j\omega t}-ae^{-j\omega t}} \right) & j\frac{M_{aDq}}{2} \left( \displaystyle{ae^{j\omega t}-a^2e^{-j\omega t}} \right) \end{array}\right)\\\\
&\times\frac{1}{2}\left(\begin{array}{ccc} \displaystyle{e^{j\omega t}+e^{-j\omega t}} & \displaystyle{j\left(e^{j\omega t}-e^{-j\omega t}\right)} & 2 \\ \displaystyle{a^2e^{j\omega t}+ae^{-j\omega t}} & \displaystyle{j\left(a^2e^{j\omega t}-ae^{-j\omega t}\right)} & 2 \\ \displaystyle{ae^{j\omega t}+a^2e^{-j\omega t}} & \displaystyle{j\left(ae^{j\omega t}-a^2e^{-j\omega t}\right)} & 2 \end{array} \right)\\\\
&=\frac{3}{2}\left(\begin{array}{ccc} M_{afd} & 0 & 0 \\ M_{aDd} & 0 & 0 \\ 0 & M_{aDq} & 0 \end{array} \right)
\end{align*}$$

なお、上式から$\boldsymbol{m_{fD-abc}}(t)\boldsymbol{D^{-1}}(t)=\displaystyle{\frac{3}{2}}{}^t\left[\boldsymbol{D}(t)\boldsymbol{m_{abc-fD}}(t)\right]$であることがわかる。

以上より、$(8)$式は、

$$\begin{align*}
\left(\begin{array}{c} \it{\Psi}_{fd}\left(t\right) \\ \it{\Psi}_{Dd}\left(t\right) \\ \it{\Psi}_{Dq}\left(t\right) \end{array}\right)&=-\frac{3}{2}\left(\begin{array}{ccc} M_{afd} & 0 & 0 \\ M_{aDd} & 0 & 0 \\ 0 & M_{aDq} & 0 \end{array} \right) \left(\begin{array}{c} i_d(t) \\ i_q(t) \\ i_0(t)\end{array}\right)\\\\
&+\left(\begin{array}{ccc} L_{ffd} & M_{fDd} & 0 \\ M_{fDd} & L_{DDd} & 0 \\ 0 & 0 & L_{DDq} \end{array}\right)\left(\begin{array}{c} i_{fd}(t) \\ i_{Dd}(t) \\ i_{Dq}(t)\end{array}\right) ・・・(9)
\end{align*}$$

$(7)$および$(9)$式が$d-q-0$成分領域における発電機の鎖交磁束関係式となる。

パークの方程式」と合わせ、これで$d-q-0$領域における発電機の関係式が全て得られた。

スポンサーリンク