Takashi Ohira
The sinusoidal voltage and current of the system shown in Figure 1 are orthogonally decomposed into \begin{align*}\left[{\begin{array}{c}{{v}\left({t}\right)}\\{{i}\left({t}\right)}\end{array}}\right] = \left[{\begin{array}{c}{\begin{array}{cc}{{V}_{P}}&{{V}_{Q}}\end{array}}\\{\begin{array}{cc}{{I}_{P}}&{{I}_{Q}}\end{array}}\end{array}}\right]\left[{\begin{array}{c}{\sin{\omega}{t}}\\{\cos{\omega}{t}}\end{array}}\right]{.} \tag{1} \end{align*}
Figure 1. A simple system comprising a sinusoidal source and linear passive load.
The matrix product can be expanded and rewritten in the traditional ${e}^{{j}{\omega}{t}}$expression as \begin{align*} & = \left[{\begin{array}{c}{{V}_{P}\sin{\omega}{t} + {V}_{Q}\cos{\omega}{t}}\\{{I}_{P}\sin{\omega}{t} + {I}_{Q}\cos{\omega}{t}}\end{array}}\right] \\ & = {\mathfrak{R}}\left[{\begin{array}{c}{\left({{V}_{Q}{-}{jV}_{P}}\right){e}^{{j}{\omega}{t}}}\\{\left({{I}_{Q}{-}{jI}_{P}}\right){e}^{{j}{\omega}{t}}}\end{array}}\right] \tag{2} \end{align*} where the operator $\mathfrak{R}$ denotes the real part of a complex. Extracting the parenthesized factors from (2), the complex voltage-to-current ratio defines the impedance \begin{align*}{Z} & = \frac{{V}_{Q}{-}{jV}_{P}}{{I}_{Q}{-}{jI}_{P}} \\ & = \frac{{V}_{P} + {jV}_{Q}}{{I}_{P} + {jI}_{Q}}{.} \tag{3} \end{align*}
On the other hand, the complex impedance of a circuit is commonly expressed as \[{Z} = {R} + {jX}{.} \tag{4} \]
To make (3) consistent with (4), the voltage and current vectors should relate as \begin{align*}\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]{.} \tag{5} \end{align*}
Therefore, the correct answer to last month’s quiz is (c). Conversely, (5) can be solved for the impedance as \begin{align*}\left[{\begin{array}{c}{R}\\{X}\end{array}}\right] = {\left[{\begin{array}{cc}{{I}_{P}}&{{-}{I}_{Q}}\\{{I}_{Q}}&{{I}_{P}}\end{array}}\right]}^{{-}{1}}\left[{\begin{array}{c}{{V}_{P}}\\{{V}_{Q}}\end{array}}\right]{.} \tag{6} \end{align*}
The matrix formulas (5) and (6) serve as a mutual interpreter between the R-X and P-Q domains. An instructive exercise applying them to the RF diode rectifier will be demonstrated in next month’s issue.
Digital Object Identifier 10.1109/MMM.2023.3314353