Introduction

Normally, the abc/dq transformation we are talking about is based on the assumptions shown in (1), where $f_d + jf_q = \hat{f}$. However, in “Canadian” convention, the so-called “abc/qd” transformation is defined as in (2), where $f_q - jf_d = \hat{f}$.

We are going to talk more details on how to transform from abc to qd (the Canadian convention) with two different approaches in this note.

Note: variables like $i_{abc}$ and $i_{qd}$ are vectors.

Approach #1

The first approach is to use phaser representations.

Example

Transform the following abc equation to qd form.

\[\begin{equation} \begin{split} L \frac{di_{s,abc}}{dt} + Ri_{s,abc} = V_{t,abc} - V_{s,abc} \end{split} \end{equation}\]

Solution

The vector $i_{s,abc}$ can be represented by $\hat{i_s}e^{j\omega t} = [i_{s,q} - ji_{s,d}]e^{j\omega t}$. By substitution

\[\begin{equation} \begin{split} \boldsymbol{L \frac{d\big[[i_{s,q} - ji_{s,d}]e^{j\omega t}\big]}{dt}} + R[i_{s,q} - ji_{s,d}]e^{j\omega t} = [V_{t,q} - jV_{t,d}]e^{j\omega t} - [V_{s,q} - jV_{s,d}]e^{j\omega t}\\ \Rightarrow \boldsymbol{L \frac{d[i_{s,q} - ji_{s,d}]}{dt}e^{j\omega t} + j\omega L[i_{s,q} - ji_{s,d}]} + R[i_{s,q} - ji_{s,d}]e^{j\omega t} = [V_{t,q} - jV_{t,d}]e^{j\omega t} - [V_{s,q} - jV_{s,d}]e^{j\omega t}\\ \Rightarrow \boldsymbol{L \frac{d[i_{s,q} - ji_{s,d}]}{dt}e^{j\omega t} + \omega L[i_{s,d} + ji_{s,q}]} + R[i_{s,q} - ji_{s,d}]e^{j\omega t} = [V_{t,q} - jV_{t,d}]e^{j\omega t} - [V_{s,q} - jV_{s,d}]e^{j\omega t}\\ \end{split} \end{equation}\]

Cancel the term $e^{j\omega t}$ on both sides and seperate real and imaginary parts, we get the results.

\[\begin{equation} \begin{split} L \frac{di_{s,q}}{dt} + \omega L i_{s,d} + Ri_{s,q} &= V_{t,q} - V_{s,q}\\ L \frac{di_{s,d}}{dt} - \omega L i_{s,q} + Ri_{s,d} &= V_{t,d} - V_{s,d}\\ \end{split} \end{equation}\]

Approach #2

The second approach is to use the transformation matrix $T$ and $T^{-1}$.

\[\begin{equation} i_{qd} = T\cdot i_{abc}\\ i_{abc} = T^{-1}\cdot i_{qd}\\ T = \frac{3}{2} \begin{bmatrix} \cos(\omega t) & \cos(\omega t - \frac{2\pi}{3}) & \cos(\omega t + \frac{2\pi}{3}) \\ \sin(\omega t) & \sin(\omega t - \frac{2\pi}{3}) & \sin(\omega t + \frac{2\pi}{3}) \end{bmatrix} \\ T^{-1} = \begin{bmatrix} \cos(\omega t) & \sin(\omega t) \\ \cos(\omega t - \frac{2\pi}{3}) & \sin(\omega t - \frac{2\pi}{3}) \\ \cos(\omega t + \frac{2\pi}{3}) & \sin(\omega t + \frac{2\pi}{3}) \\ \end{bmatrix} \\ \end{equation}\]

Example

We use the same example.

\[\begin{equation} \begin{split} L \frac{di_{s,abc}}{dt} + Ri_{s,abc} = V_{t,abc} - V_{s,abc} \end{split} \end{equation}\]

Now, by multiplying $T$ on both sides, we get (Note: $[C]$ for $\cos(\omega t)$, $[C-]$ for $\cos(\omega t - \frac{2\pi}{3})$, $[C+]$ for $\cos(\omega t + \frac{2\pi}{3})$)

\[\begin{equation} \begin{split} TL \frac{di_{s,abc}}{dt} + TRi_{s,abc} = TV_{t,abc} - TV_{s,abc} \\ \Rightarrow \frac{2}{3} L \begin{bmatrix} [C] & [C-] & [C+] \\ [S] & [S-] & [S+] \end{bmatrix} \begin{bmatrix} \frac{di_{s,a}}{dt} \\ \frac{di_{s,b}}{dt} \\ \frac{di_{s,c}}{dt} \end{bmatrix} + Ri_{s,qd} &= V_{t,qd} - V_{s,qd}\\ \Rightarrow \frac{2}{3} L \begin{bmatrix} [C]\frac{di_{s,a}}{dt} + [C-]\frac{di_{s,b}}{dt} + [C+]\frac{di_{s,c}}{dt} \\ [S]\frac{di_{s,a}}{dt} + [S-]\frac{di_{s,b}}{dt} + [S+]\frac{di_{s,c}}{dt} \end{bmatrix} + Ri_{s,qd} &= V_{t,qd} - V_{s,qd}\\ \end{split} \end{equation}\]

The matrix here can be written as

\[\begin{equation} \frac{2}{3} \begin{bmatrix} \big[-\omega[S]i_{s,a} + [C]\frac{di_{s,a}}{dt} + \omega[S]i_{s,a}\big] + \big[-\omega[S-]i_{s,a} + [C-]\frac{di_{s,a}}{dt} + \omega[S-]i_{s,a}\big] + \big[-\omega[S+]i_{s,a} + [C+]\frac{di_{s,a}}{dt} + \omega[S+]i_{s,a}\big] \\ \big[\omega[C]i_{s,a} + [S]\frac{di_{s,a}}{dt} - \omega[C]i_{s,a}\big] + \big[\omega[C-]i_{s,a} + [S-]\frac{di_{s,a}}{dt} - \omega[C-]i_{s,a}\big] + \big[\omega[C+]i_{s,a} + [S+]\frac{di_{s,a}}{dt} - \omega[C+]i_{s,a}\big] \end{bmatrix} \\ \Downarrow \\ \frac{2}{3} \begin{bmatrix} \big[\frac{d[C]i_{s,a}}{dt} + \omega[S]i_{s,a}\big] + \big[\frac{d[C-]i_{s,a}}{dt} + \omega[S-]i_{s,a}\big] + \big[\frac{d[C+]i_{s,a}}{dt} + \omega[S+]i_{s,a}\big] \\ \big[\frac{d[S]i_{s,a}}{dt} - \omega[C]i_{s,a}\big] + \big[\frac{d[S-]i_{s,a}}{dt} - \omega[C-]i_{s,a}\big] + \big[\frac{d[S+]i_{s,a}}{dt} - \omega[C+]i_{s,a}\big] \\ \end{bmatrix} \\ \Downarrow \\ \frac{2}{3} \begin{bmatrix} \frac{d\big[[C]i_{s,a} + [C-]i_{s,a} + [C+]i_{s,a}\big]}{dt} \\ \frac{d\big[[S]i_{s,a} + [S-]i_{s,a} + [S+]i_{s,a}\big]}{dt} \end{bmatrix} + \frac{3}{2} \omega \begin{bmatrix} [S]\frac{di_{s,a}}{dt} + [S-]\frac{di_{s,b}}{dt} + [S+]\frac{di_{s,c}}{dt} \\ -[C]\frac{di_{s,a}}{dt} - [C-]\frac{di_{s,b}}{dt} - [C+]\frac{di_{s,c}}{dt} \end{bmatrix} \\ \Downarrow \\ \begin{bmatrix} \frac{di_{s,q}}{dt} + \omega i_{s,d} \\ \frac{di_{s,d}}{dt} - \omega i_{s,q} \\ \end{bmatrix} \end{equation}\]

Substitute it back to the former equation, we get the same results

\[\begin{equation} \begin{split} L \begin{bmatrix} \frac{di_{s,q}}{dt} + \omega i_{s,d} \\ \frac{di_{s,d}}{dt} - \omega i_{s,q} \\ \end{bmatrix} + Ri_{s,qd} &= V_{t,qd} - V_{s,qd}\\ \end{split} \end{equation}\]