## 2 Basic Concepts of DQ0-Based Models
### Example--Modeling a Network Based on DQ0 Quantities  
In this example we consider a linear network containing an ideal power source connected to an infinite bus (voltage source with a constant amplitude and frequency), as described in Figure 8. We will show that the quasi-static model describing this system has no dynamic states, and therefore provides no information regarding the system stability. In addition, with no resistor and no capacitor (\(R=0\), \(C=0\)), an accurate \(dq0\) model shows that the system is unstable. We also provide an explicit value for \(C\) for which the system is stable. Consider the following system data:  
* The network is symmetrically configured. * The sources are balanced (equal amplitudes, \(120^{\circ}\) between phases, \(v_{0}=0\)). * Based on (61), the power source is defined by \[ \begin{split}\frac{1}{2}(v_{d,1}i_{d}+v_{q,1}i_{q})& =P,\\ \frac{1}{2}(v_{q,1}i_{d}-v_{d,1}i_{q})&=0,\end{split} \] (63) meaning that under a quasi-static approximation the active power is \(P\), and the reactive power is zero. * The infinite bus has an RMS voltage of \(V_{g}\), and a frequency of \(\omega_{s}\). * We use a \(dq0\) transformation with a reference angle \(\theta(t)=\omega_{s}t\). The quasi-static model has no dynamic states. The voltages, currents and powers may be found by solving the power flow equations, where bus 1 is a PQ bus (\(P\), \(Q\) known, \(|V|\), \(\delta\) unknown), and bus 2 is the reference bus (\(|V|\), \(\delta=0\) known, \(P\), \(Q\) unknown). Since the quasi-static model does not describe the system dynamics, it provides no information regarding the system stability. Even if the power flow equations has a solution, the system may be either stable or unstable. We will now describe the accurate system dynamics based on \(dq\) signals. The following analysis shows that if \(R=0\) and \(C=0\), then the system is unstable. The infinite bus is described by  
\[ \begin{split} v_{d,2}&=\sqrt{2}V_{g},\\ v_{q,2}&=0.\end{split}  \tag{64}\]  
Using (14) with \(\theta=\omega_{s}t\), the inductor is described by (remember that \(R=0\))  
\[ \begin{split}\frac{\mathrm{d}}{\mathrm{d}t}i_{d}&= \omega_{s}i_{q}+\frac{1}{L}(v_{d,1}-v_{d,2}),\\ \frac{\mathrm{d}}{\mathrm{d}t}i_{q}&=-\omega_{s}i_{d }+\frac{1}{L}(v_{q,1}-v_{q,2}).\end{split}  \tag{65}\]  
The power source is modeled as in (63). We choose to view it as a controlled voltage source. To this end, solving (63) with respect to the voltages yields  
Figure 8: Ideal power source connected to an infinite bus (single-phase diagram). \[ \begin{bmatrix}v_{d,1}\\ v_{q,1}\end{bmatrix} =\frac{2P}{i_{d}^{2}+i_{q}^{2}} \begin{bmatrix}i_{d}\\ i_{q}\end{bmatrix} . \tag{66}\]  
Combination of (64)-(66) provides the (nonlinear) state-space model  
\[ \begin{split}\frac{\mathrm{d}}{\mathrm{d}t}i_{d}& =\omega_{s}i_{q}+\frac{2P}{L}\frac{i_{d}}{i_{d}^{2}+i_{q}^{2}}-\frac{\sqrt{2}V _{g}}{L},\\ \frac{\mathrm{d}}{\mathrm{d}t}i_{q}&=-\omega_{s}i_{d}+\frac{2P}{L} \frac{i_{q}}{i_{d}^{2}+i_{q}^{2}}.\end{split}  \tag{67}\]  
The Jacobian of this model at an equilibrium point \((\vec{i}_{d},\vec{i}_{q})\) is  
\[A= \begin{bmatrix}-\frac{2P}{L}\frac{i_{d}^{2}-i_{q}^{2}}{\left(\vec{i}_{d}^{2} +i_{q}^{2}\right)^{2}}&\omega_{s}-\frac{4P}{L}\frac{i_{d}^{2}i_{q}}{\left( \vec{i}_{d}^{2}+i_{q}^{2}\right)^{2}}\\ -\omega_{s}-\frac{4P}{L}\frac{i_{d}^{2}i_{q}^{2}}{\left(\vec{i}_{d}^{2}+i_{q} ^{2}\right)^{2}}&\frac{2P}{L}\frac{i_{d}^{2}-i_{q}^{2}}{\left(\vec{i}_{d}^{2} +i_{q}^{2}\right)^{2}}\end{bmatrix} . \tag{68}\]  
The poles of the system are found by computing the roots of the characteristic polynomial, which is given for a second-order system by  
\[s^{2}-\mathrm{Tr}(A)s+\mathrm{det}(A)=0, \tag{69}\]  
where \(\mathrm{Tr}(A)\) is the trace of \(A\), and \(\mathrm{det}(A)\) is the determinant of \(A\). It is easy to verify that \(\mathrm{Tr}(A)=0\), and therefore, the poles are \(s_{1,2}=\pm\sqrt{\mathrm{det}(A)}\). * If \(\mathrm{det}(A)>0\), there is a pole in the right half of the complex plane, and the system is unstable.