Takashi Ohira
As depicted in Figure 1, a sinusoidal voltage source is loaded with the complex impedance R + jX. For algebraic elegance in this “Enigmas, etc.” series, we regularly characterize the voltage v(t) and current i(t) using their orthogonal decomposition: \begin{align*}\left[{\begin{array}{c}{v(t)}\\{i(t)}\end{array}}\right] = \left[{\begin{array}{cc}{{V}_{P}}&{{V}_{Q}}\\{{I}_{P}}&{{I}_{Q}}\end{array}}\right]\left[{\begin{array}{c}{{\text{sin}}\,{\omega}{t}}\\{{\text{cos}}\,{\omega}{t}}\end{array}}\right]\end{align*}
Figure 1. A simple system including a sinusoidal source and linear passive load. How does the voltage v(t) relate to the current i(t) based on the complex load impedance R + jX?
where the subscripts P and Q stand for in-phase and quadrature components, respectively. Find a conversion matrix to bridge the gap between the P–Q domain and usual R–X domain. Which of the following is correct? \begin{align*} & \left({\text{a}}\right)\left[{\begin{array}{c}{{V}_{P}}\\{{V}_{Q}}\end{array}}\right] = \left[{\begin{array}{cc}{R}&{X}\\{X}&{R}\end{array}}\right]\left[{\begin{array}{c}{{I}_{P}}\\{{I}_{Q}}\end{array}}\right] \\ & \left({\text{b}}\right)\left[{\begin{array}{c}{{V}_{P}}\\{{V}_{Q}}\end{array}}\right] = \left[{\begin{array}{cc}{R}&{X}\\{{-}{X}}&{R}\end{array}}\right]\left[{\begin{array}{c}{{I}_{P}}\\{{I}_{Q}}\end{array}}\right] \\ & \left({\text{c}}\right)\left[{\begin{array}{c}{{V}_{P}}\\{{V}_{Q}}\end{array}}\right] = \left[{\begin{array}{cc}{R}&{{-}{X}}\\{X}&{R}\end{array}}\right]\left[{\begin{array}{c}{{I}_{P}}\\{{I}_{Q}}\end{array}}\right] \\ & \left({\text{d}}\right)\left[{\begin{array}{c}{{V}_{P}}\\{{V}_{Q}}\end{array}}\right] = \left[{\begin{array}{cc}{R}&{X}\\{X}&{{-}{R}}\end{array}}\right]\left[{\begin{array}{c}{{I}_{P}}\\{{I}_{Q}}\end{array}}\right]{.}\end{align*}
Note that we use the time-varying factor ${e}^{{j}{\omega}{t}}$ to solve the problem. The correct answer will be revealed in next month’s issue.
Digital Object Identifier 10.1109/MMM.2023.3303688