Question C.1: A two-pole synchronous machine is carrying balanced three-ph......
A two-pole synchronous machine is carrying balanced three-phase armature currents
i_a = \sqrt{2}I_a cos ωt \quad i_b = \sqrt{2}I_a cos (ωt – 120°) \quad i_c = \sqrt{2}I_a cos (ωt + 120°)
The rotor is rotating at synchronous speed ω, and the rotor direct axis is aligned with the stator phase-a axis at time t = 0. Find the direct- and quadrature-axis current components.
The angle between the rotor direct axis and the stator phase-a axis can be expressed as
\theta_{\mathrm{me}}=\omega t
From Eq. C. 1
\left[\begin{array}{l}S_{\mathrm{d}} \\S_{\mathrm{q}} \\S_{0}\end{array}\right]=\frac{2}{3}\left[\begin{array}{ccc}\cos \left(\theta_{\mathrm{me}}\right) & \cos \left(\theta_{\mathrm{me}}-120^{\circ}\right) & \cos \left(\theta_{\mathrm{me}}+120^{\circ}\right) \\-\sin \left(\theta_{\mathrm{me}}\right) & -\sin \left(\theta_{\mathrm{me}}-120^{\circ}\right) & -\sin \left(\theta_{\mathrm{me}}+120^{\circ}\right) \\\frac{1}{2} & \frac{1}{2} & \frac{1}{2}\end{array}\right]\left[\begin{array}{c}S_{\mathrm{a}} \\S_{\mathrm{b}} \\S_{\mathrm{c}}\end{array}\right] \quad \quad \quad (C.1)
\\ \begin{aligned}i_{\mathrm{d}} & =\frac{2}{3}\left[i_{\mathrm{a}} \cos \omega t+i_{\mathrm{b}} \cos \left(\omega t-120^{\circ}\right)+i_{\mathrm{c}} \cos \left(\omega t+120^{\circ}\right)\right] \\ \\& =\frac{2}{3} \sqrt{2} I_{\mathrm{a}}\left[\cos ^{2} \omega t+\cos ^{2}\left(\omega t-120^{\circ}\right)+\cos ^{2}\left(\omega t+120^{\circ}\right)\right]\end{aligned}
Using the trigonometric identity \cos ^{2} \alpha=\frac{1}{2}(1+\cos 2 \alpha) gives
i_{\mathrm{d}}=\sqrt{2} I_{\mathrm{a}}
Similarly,
\begin{aligned}i_{\mathrm{q}}= & -\frac{2}{3}\left[i_{\mathrm{a}} \sin \omega t+i_{\mathrm{b}} \sin \left(\omega t-120^{\circ}\right)+i_{\mathrm{c}} \sin \left(\omega t+120^{\circ}\right)\right] \\ \\= & -\frac{2}{3} \sqrt{2} I_{\mathrm{a}}\left[\cos \omega t \sin \omega t+\cos \left(\omega t-120^{\circ}\right) \sin \left(\omega t-120^{\circ}\right)\right.\\ \\& \left.+\cos \left(\omega t+120^{\circ}\right) \sin \left(\omega t+120^{\circ}\right)\right]\end{aligned}
and using the trigonometric identity \cos \alpha \sin \alpha=\frac{1}{2} \sin 2 \alpha gives
i_{\mathrm{q}}=0
This result corresponds directly to our physical picture of the dq0 transformation. From the discussion of Section 4.5 we recognize that the balanced three-phase currents applied to this machine produce a synchronously rotating mmf wave which produces flux along the stator phase-a axis at time t = 0. This flux wave is thus aligned with the rotor direct axis at t = 0 and remains so since the rotor is rotating at the same speed. Hence the stator current produces only direct-axis flux and thus consists only of a direct-axis component.