Question 6.3: A generator is operating with balanced positive sequence vol......

This is a simple case of transformation using (6.65):

\left|\begin{matrix} i_{0} \\ i_{d} \\ i_{q} \end{matrix} \right| =\sqrt{\frac{2}{3} } \left|\begin{matrix} \frac{1}{\sqrt{2} } & \frac{1}{\sqrt{2} } & \frac{1}{\sqrt{2} } \\ \cos \theta & \cos \left(\theta -2 \frac{\pi }{3} \right) & \cos \left(\theta +2\frac{\pi }{3} \right)) \\ \sin \theta & \sin \left(\theta -2\frac{\pi }{3} \right) & \sin \left(\theta +2\frac{\pi }{3} \right) \end{matrix} \right| \left|\begin{matrix} i_{a} \\ i_{b} \\ i_{c} \end{matrix} \right|       (6.65)

A solution of this equation gives
v_{d} = \sqrt{3} |V| \sin [(ω_{0} – ω_{1})t + < V -δ]
v_{q} = \sqrt{3} |V| \cos [(ω_{0} – ω_{1})t + < V -δ]      (6.97)
These relations apply equally well to the derivation of i_{d}, i_{q}, λ_{d}, and λ_{q}.
For synchronous operation ω_{1} = ω_{0} and the equations reduce to
v_{q} = \sqrt{3} |V| \cos ( \angle V -δ)
v_{d} = \sqrt{3} |V| \sin ( \angle V -δ)     (6.98)
Note that v_{q} and v_{d} are now constant and do not have the slip frequency term ω_{1} – ω_{0}. We can write
v_{q} + jv_{d} = \sqrt{3} |V|e^{j(\angle V -δ) }= \sqrt{3} V_{a} e^{-jδ}            (6.99)
Therefore, V_{a} can be written as
V_{a}= (\frac{v_{q}}{\sqrt{3}}+ j \frac{v_{d}}{\sqrt{3}}) e^{jδ}= (V_{q} + jV_{d}) e^{jδ}       (6.100)
where
V_{q}= \frac{v_{q}}{\sqrt{3}}   and V_{d}= \frac{v_{d}}{\sqrt{3}}           (6.101)
This is shown in the phasor diagram of Figure 6.11.

We can write these equations in the following form:

\left|\begin{matrix} V_{q} \\ V_{d} \end{matrix} \right|=\left|\begin{matrix} \cos\delta & \sin \delta \\ -\sin \delta & \cos \delta \end{matrix} \right|\left|\begin{matrix} Re & V_{a} \\ Im & V_{a} \end{matrix} \right|        (6.102)