## 3 The Synchronous Machine
### Transformation from One Reference Frame to Another  
Recall that the dynamic models discussed so far are defined in the rotor reference frame, with respect to the angle \(\theta\). A typical question is how to construct a dynamic model of a system that has more than one natural reference frame. Such a need arises when the system includes more than one machine, or when a machine is connected to an infinite bus. The typical solution is to transform the \(dq0\) variables from one reference frame to another. Assume a system with two sets of signals: \(x_{dq0}\) are defined with respect to a reference angle \(\theta_{r}\), and \(\tilde{x}_{dq0}\) are defined with respect to a reference angle \(\theta\). The relations between these signals can be found by transforming the signals to the \(abc\) reference frame and back, which may be written as \(x_{dq0}=T_{\theta_{r}}T_{\theta}^{-1}\tilde{x}_{dq0}\) or \(\tilde{x}_{dq0}=T_{\theta}T_{\theta_{r}}^{-1}x_{dq0}\). Using the \(dq0\) identities in Appendix A,  
\[\theta\rightarrow\theta_{r}:\qquad x_{dq0}= \begin{bmatrix}\cos(\theta- \theta_{r})&-\sin(\theta-\theta_{r})&0\\ \sin(\theta-\theta_{r})&\cos(\theta-\theta_{r})&0\\ 0&0&1\end{bmatrix} \tilde{x}_{dq0}, \tag{112}\]  
\[\theta\leftarrow\theta_{r}:\qquad\tilde{x}_{dq0}= \begin{bmatrix}\cos(\theta- \theta_{r})&\sin(\theta-\theta_{r})&0\\ -\sin(\theta-\theta_{r})&\cos(\theta-\theta_{r})&0\\ 0&0&1\end{bmatrix} x_{dq0}. \tag{113}\]  
Consider now a general unit, which model is given with respect to the angle \(\theta\). This model can be linked to a system with a reference angle \(\theta_{r}\), as described in Figure 15.  
Figure 14: Energy conversion in the machine, based on the simplified model. Figure 15: Transformation from one reference frame to another. As a special important case, consider the EMF source \(\tilde{e}_{d}=0\), \(\tilde{e}_{q}=\frac{\omega}{\omega_{s}}V_{E}\), \(\tilde{e}_{0}=0\), with a reference angle \(\theta\). What is the model of this source with respect to \(\theta_{r}\)? Direct transformation yields  
\[ \begin{bmatrix}e_{d}\\ e_{q}\\ e_{0}\end{bmatrix} = \begin{bmatrix}\cos(\theta-\theta_{r})&-\sin(\theta-\theta_{ r})&0\\ \sin(\theta-\theta_{r})&\cos(\theta-\theta_{r})&0\\ 0&0&1\end{bmatrix}\begin{bmatrix}0\\ \frac{\omega}{\omega_{s}}V_{E}\\ 0\end{bmatrix} =\frac{\omega}{\omega_{s}}V_{E} \begin{bmatrix}-\sin(\theta- \theta_{r})\\ \cos(\theta-\theta_{r})\\ 0\end{bmatrix}  \tag{114}\]  
or,  
\[ \begin{bmatrix}e_{d}\\ e_{q}\\ e_{0}\end{bmatrix} =\frac{\omega}{\omega_{s}}V_{E} \begin{bmatrix}\cos(\delta) \\ \sin(\delta)\\ 0\end{bmatrix} ,\qquad\delta=\theta-\theta_{r}+\frac{\pi}{2}, \tag{115}\]  
where \(e_{d}\), \(e_{q}\), \(e_{0}\) are defined with respect to \(\theta_{r}\). The reader may recognize \(\delta\) as the power angle of the machine. Therefore, at low frequencies, the simplified machine model is identical to the quasi-static (time-varying phasors) model. To see this, the phasor describing the induced EMF is  
\[E=\frac{1}{\sqrt{2}}\left(e_{d}+je_{q}\right)=\frac{\omega V_{E}}{\sqrt{2} \omega_{s}}\left(\cos(\delta)+j\sin(\delta)\right). \tag{116}\]  
Further define  
\[|E|=\frac{\omega V_{E}}{\sqrt{2}\omega_{s}}=\frac{1}{\sqrt{2}}\omega L_{af}I_{ f}. \tag{117}\]  
This is the induced EMF (RMS value) at steady state, and at the rotor frequency \(\omega\). Substitute (117) in (116) to get \(E=|E|\angle\delta\). This phasor describes the induced EMF of the quasi-static model. Note that \(|E|\) depends on \(\omega\); however, in a quasi-static model we typically assume that \(\omega\approx\omega_{s}\) and \(|E|\) is constant. Also recall that in a time-varying phasor model \(p_{e}=3P\). This is true since  
\[3P=3\operatorname{Re}\{EI^{*}\}=3\operatorname{Re}\left\{\frac{1}{\sqrt{2}} \left(e_{d}+je_{q}\right)\frac{1}{\sqrt{2}}\left(i_{d}+ji_{q}\right)^{*}\right\} =\frac{3}{2}\operatorname{Re}\{e_{d}i_{d}+e_{q}i_{q}\}=p_{e}. \tag{118}\]  
For example, consider a synchronous machine connected to an infinite bus.