## 3 The Synchronous Machine
### Transformation from One Reference Frame to Another
The machine is described by the simplified model presented above, and the infinite bus voltage is \(v_{d,\infty}=\sqrt{2}V_{g}=\text{const}\), \(v_{q,\infty}=0\), \(v_{0,\infty}=0\) with a reference frame \(\omega_{s}t\). We now wish to construct a state-space model of the system, with respect to the reference angle \(\theta_{r}\). A signal flow diagram is shown in Figure 16.  
Figure 16: Signal-flow diagram: synchronous machine connected to an infinite bus. The explicit state-space model is shown next. Using the result above, the induced EMF is described in the reference frame \(\omega_{s}t\) by  
\[ \begin{bmatrix}e_{d}\\ e_{q}\\ e_{0}\end{bmatrix} =\frac{\omega}{\omega_{s}}V_{E} \begin{bmatrix}\cos(\delta)\\ \sin(\delta)\\ 0\end{bmatrix}  \tag{119}\]  
with \(\delta=\theta-\theta_{r}+\frac{\pi}{2}\). In addition, using (105), the machine is modeled by  
\[ \begin{split}\frac{\mathrm{d}}{\mathrm{d}t}\theta& =\omega,\\ \frac{\mathrm{d}}{\mathrm{d}t}\omega&=K\left(3P_{ref}-\frac{1}{ D}(\omega-\omega_{s})-\frac{3}{2}V_{E}\tilde{t}_{q}^{\gamma}\right),\end{split}  \tag{120}\]  
where in this last equation \(\tilde{t}_{q}\) is referenced to \(\theta\). Using (113), this variable may be expressed in terms of the reference angle \(\omega_{s}t\) as  
\[ \begin{bmatrix}\tilde{t}_{d}\\ \tilde{t}_{q}\\ \tilde{t}_{0}\end{bmatrix} = \begin{bmatrix}\cos(\theta-\omega_{s}t)&\sin( \theta-\omega_{s}t)&0\\ -\sin(\theta-\omega_{s}t)&\cos(\theta-\omega_{s}t)&0\\ 0&0&1\end{bmatrix}\begin{bmatrix}i_{d}\\ i_{q}\\ i_{0}\end{bmatrix} , \tag{121}\]  
and using \(\delta=\theta-\omega_{s}t+\frac{\pi}{2}\) yields \(\tilde{t}_{q}=\cos(\delta)i_{d}+\sin(\delta)i_{q}\). Moreover, the machine's series impedance is modeled (with respect to \(\omega_{s}t\)) as  
\[\frac{\mathrm{d}}{\mathrm{d}t} \begin{bmatrix}i_{d}\\ i_{q}\\ i_{0}\end{bmatrix} = \begin{bmatrix}-\frac{R_{a}}{L_{s}}&\omega_{s}&0\\ -\omega_{s}&-\frac{R_{a}}{L_{s}}&0\\ 0&0&-\frac{R_{a}}{L_{s}}\end{bmatrix}\begin{bmatrix}i_{d}\\ i_{q}\\ i_{0}\end{bmatrix} +\frac{1}{L_{s}}\left(\frac{\omega}{\omega_{s}}V_{E} \begin{bmatrix} \cos(\delta)\\ \sin(\delta)\\ 0\end{bmatrix} - \begin{bmatrix}\sqrt{2}V_{g}\\ 0\\ 0\end{bmatrix} \right). \tag{122}\]  
Combining these equations, the resulting state-space model becomes  
\[ \begin{split}\frac{\mathrm{d}}{\mathrm{d}t}\delta& =\omega-\omega_{s},\\ \frac{\mathrm{d}}{\mathrm{d}t}\omega&=K\left(3P_{ref}-\frac{1}{ D}(\omega-\omega_{s})-\frac{3}{2}V_{E}i_{d}\cos(\delta)-\frac{3}{2}V_{E}i_{q}\sin( \delta)\right),\\ \frac{\mathrm{d}}{\mathrm{d}t}i_{d}&=-\frac{R_{a}}{L_{s}}i_{d}+ \omega_{s}i_{q}+\frac{1}{L_{s}}\left(\frac{\omega}{\omega_{s}}V_{E}\cos(\delta )-\sqrt{2}V_{g}\right),\\ \frac{\mathrm{d}}{\mathrm{d}t}i_{q}&=-\omega_{s}i_{d}- \frac{R_{a}}{L_{s}}i_{q}+\frac{1}{L_{s}}\frac{\omega}{\omega_{s}}V_{E}\sin( \delta),\\ \frac{\mathrm{d}}{\mathrm{d}t}i_{0}&=-\frac{R_{a}}{L_{s}}i_{0}. \end{split}  \tag{123}\]  
Note that \(\delta\) is used instead of \(\theta\) as a state variable. As another example, consider two synchronous machines connected to each other, and feeding a resistive load \(R_{L}\). The machines are described using the detailed model (not the simplified one), as given in (92). The reference angle of the first machine is \(\theta_{1}\), and the reference angle of the second machine is \(\theta_{2}\). A signal flow diagram is shown in Figure 17.