## 2 Basic Concepts of DQ0-Based Models
### Comparison of Time-Varying Phasors and DQ0 Models
Note that the admittance matrix is constant, and is evaluated at a single frequency \(s=j\omega_{s}\). We will now generalize this result, and obtain an equivalent expression for a network that is not quasi-static. Define the \(dq\) signals \(v_{d,n}\), \(v_{q,n}\) to be the voltages on bus \(n\), and \(i_{d,n}\), \(i_{q,n}\) to be the injected currents into bus \(n\). In addition, define the vectors \(V_{d}=[v_{d,1},\ldots,v_{d,N}]^{\mathsf{T}}\) and \(I_{d}=[i_{d,1},\ldots,i_{d,N}]^{\mathsf{T}}\), etc. Using a reference angle \(\theta(t)=\omega_{s}t\), the network is described by  
\[ \begin{bmatrix}I_{d}(s)\\ I_{q}(s)\end{bmatrix} = \begin{bmatrix}\mathcal{N}_{1}(s)&j\mathcal{N}_{2}(s)\\ -j\mathcal{N}_{2}(s)&\mathcal{N}_{1}(s)\end{bmatrix}\begin{bmatrix}V_{d}(s)\\ V_{q}(s)\end{bmatrix} , \tag{54}\]  
where  
\[ \begin{split}\mathcal{N}_{1}(s)&:=\frac{1}{2}\left(\gamma(s+j \omega_{s})+Y(s-j\omega_{s})\right),\\ \mathcal{N}_{2}(s)&:=\frac{1}{2}\left(\gamma(s+j\omega_{s})-Y(s-j \omega_{s})\right),\end{split}  \tag{55}\]  
A proof for these equalities is provided [24]. This model applies to any symmetrically configured linear network. At low frequencies, assuming \(s\to 0\), it may be verified that  
\[ \begin{split}\mathcal{N}_{1}(s)&\approx\frac{1}{2} \left(Y(j\omega_{s})+Y(-j\omega_{s})\right)=\frac{1}{2}\left(Y(j\omega_{s})+Y^ {*}(j\omega_{s})\right)=\operatorname{Re}\{Y(j\omega_{s})\},\\ \mathcal{N}_{2}(s)&\approx\frac{1}{2}\left(Y(j \omega_{s})-Y(-j\omega_{s})\right)=\frac{1}{2}\left(Y(j\omega_{s})-Y^{*}(j \omega_{s})\right)=j\operatorname{Im}\{Y(j\omega_{s})\},\end{split}  \tag{56}\]  
and therefore the \(dq0\) model in (54) and the quasi-static model in (53) are equivalent. This result explains why \(dq0\) models may be viewed as an extension of quasi-static models. In other words, quasi-static models use time-varying phasors, and are therefore accurate only at frequencies much lower than \(\omega_{s}\). In comparison, \(dq\) models are accurate over a wider frequency range. For \(s\to 0\), both models are equivalent [11]. As an example, consider again the symmetrically configured three-phase inductor, which dynamic model is given in (14). Assuming \(\theta=\omega_{s}t\) and \(v_{0}=0\),  
\[ \begin{split} v_{d}&=-\omega_{s}Li_{q}+L\frac{ \operatorname{d}}{\operatorname{d}t}i_{d},\\ v_{q}&=\omega_{s}Li_{d}+L\frac{\operatorname{d}}{ \operatorname{d}t}i_{q}.\end{split}  \tag{57}\]  
This is the accurate model of a symmetrically configured 3-phase inductor. In addition, the admittance matrix describing the inductor is \(Y=1/(j\omega_{s}L)\), and the quasi-static model is \(I=\frac{1}{j\omega_{s}L}V\). By taking the real and imaginary parts of the equation, and using the relations between \(dq\) signals and phasors in (49), equivalent expressions for the quasi-static model are  
\[ \begin{split} v_{d}&=-\omega_{s}Li_{q},\\ v_{q}&=\omega_{s}Li_{d}.\end{split}  \tag{58}\]  
Direct comparison of (57) and (58) reveals that both models are similar, except for the additional time derivative terms in the \(dq\) model, which describes high frequency effects. At low frequencies, when these time-derivatives are negligible, the two models are identical. **Remark 1** (Equilibrium points). : _For balanced systems, since quasi-static models and dq models are equivalent for \(s\to 0\), the equilibrium points of both models are the same, and may be computed by solving the system's power flow equations._