ELECTROMECHANICAL MOTION DEVICES
Rotating Magnetic Field Based Analysis
3^rd Edition

In Chapter 5, we derived the q and d axis variables for the Arbitrary Reference Frame. Traditionally, these transformed variables have been derived mathematically and given little intuitive physical meaning. We have animated Tesla’s rotating magnetic field and viewed it from various reference frames, an intuitive understanding of q axis and d axis variables becomes evident. Choose a reference frame speed, $\omega$, and position yourself on its q axis. The q and d axis variables in your frame are those variables necessary to produce Tesla’s rotating magnetic field as you view it in that reference frame. In other words, the q and d axis circuit variables (i.e., voltages and currents) in the selected reference frame are simply those that would produce the rotating magnetic field, or the stator mmf as viewed from the reference frame. For rotating symmetrical circuits replace $\omega$ with $\beta$ where $$\beta = (\omega-\omega_r)~~~~~~~~~~\text{(1)}$$ and we will find that with the rotor windings short circuited the mmf$_r$ rotates with the mmf$_s$. Thus mmf$_s$ and mmf$_r$ are always stationary relative to each other. A necessary condition to produce a constant torque.

The air-gap mmf may be expressed $$\text{mmf}_s=\frac{N_s}{2}\left[i_{as}\cos\phi_s+i_{bs}\sin\phi_s\right]~~~~~~~~~~\text{(2)}$$ where from the above figure $$\phi_s=\theta+\phi~~~~~~~~~~\text{(3)}$$ Substituting (3) into (2) yields $$\text{mmf}_s=\frac{N_s}{2}\left[i_{as}\cos(\theta+\phi)+i_{bs}\sin(\theta+\phi)\right]~~~~~~~~~~\text{(4)}$$ which may be written $$\text{mmf}_s=\frac{N_s}{2}\cos\phi(i_{as}\cos\theta+i_{bs}\sin\theta)+\frac{N_s}{2}\sin\phi~~~~~~~~~~\text{(5)}$$ Let $\phi = 0$, we get $i_{qs}$ $$i_{qs} = i_{as}\cos\theta+i_{bs}\sin\theta~~~~~~~~~~\text{(6)}$$ Let $\phi=-\pi/2$ we get $i_{ds}$ $$i_{ds}=i_{as}\sin\theta-i_{bs}\cos\theta~~~~~~~~~~\text{(7)}$$ Now $$\theta = \omega t+\theta(0)~~~~~~~~~~\text{(8)}$$ where $\theta(0)$ is generally set equal to zero and $\omega$ can a function of time. From (6) and (7), the transformation to the arbitrary reference frame becomes $$\left[\begin{array}{c} f_{qs}\\f_{ds}\end{array}\right]=\left[\begin{array}{cc} \cos\theta & \sin\theta\\ \sin\theta & -cos\theta\end{array}\right]\left[\begin{array}{c} f_{as}\\f_{bs}\end{array}\right]~~~~~~~~~~\text{(9)}$$ which may be written $$\mathbf{f}_{qds} = \mathbf{K}_s\mathbf{f}_{abs}~~~~~~~~~~\text{(10)}$$ We will deal primarily with (2) the stationary reference frame where $\omega = 0$ and the transformation is $\mathbf{K}_s^s$, where the $s$ superscript denotes the stationary reference frame variables and (3) the synchronously rotating reference frame where $\omega = \omega_e$ and the transformation is $\mathbf{K}_s^e$ and $e$ denotes synchronously rotating reference frame variables.
The voltage equations in the arbitrary reference frame for the symmetrical stator are $$v_{qs}=r_si_{qs}+\omega\lambda_{ds}+p\lambda_{qs}~~~~~~~~~~\text{(11)}$$ $$v_{ds}=r_si_{ds}-\omega\lambda_{qs}+p\lambda_{ds}~~~~~~~~~~\text{(12)}$$ Note that with $\omega=0$ for the stationary reference frame the $\omega\lambda_{qs}^s$ and $\omega\lambda_{ds}^s$ terms are zero. Also, with $\omega=\omega_e$ the $p\lambda_{qs}^e$ and $p\lambda_{ds}^e$ both become zero for balance steady-state operation.