Application of the Park transform, matrix derivative without an input vector

Question

MathJax wont produce a matrix for me so I have to post pictures:

I am trying to perform a Park transform on a three phase balanced inverter. Every paper I come across does not explain the transformation from $abc$ to $dq0$. I have found an explanation but it also says "All identities may be proved by straightforward algebraic operations." I cant see how this works:

if

and its inverse is

Then how is it possible that:

I tried to take the time derivative of $T^{-1}_{\theta}$ with $\theta = \omega*t$ element wise but that didn't work. Every resource I came come across on Matrix differentiation relies on an input vector.

Does anyone have a link to a resource that could help me come up with the bottom result.

Answer 1

The element-wise differentiation was the good path. Indeed, one has : $$ \frac{\mathrm{d}}{\mathrm{d}t} T_\theta^{-1} = -\dot{\theta} \begin{pmatrix} \sin(\theta) & \cos(\theta) & 0 \\ \sin(\theta-2\pi/3) & \cos(\theta-2\pi/3) & 0 \\ \sin(\theta+2\pi/3) & \cos(\theta+2\pi/3) & 0 \end{pmatrix} $$ where $\dot{\theta} = \frac{\mathrm{d}\theta}{\mathrm{d}t}$. You can verify by a mere matrix multplication that the above expression is indeed equal to $-T_\theta W$. And it goes the same way for $\frac{\mathrm{d}T_\theta}{\mathrm{d}t}$.

Answer 2

From $$ \frac{\partial \mathbf{T}}{\partial \theta} = -\frac23 \begin{bmatrix} \sin \theta & \sin (\theta-2\pi/3) & \sin (\theta+2\pi/3) \\ \cos \theta & \cos (\theta-2\pi/3) & \cos (\theta+2\pi/3) \\ 0 & 0 & 0 \end{bmatrix} $$ we immediately observe that $$ \frac{\partial \mathbf{T}}{\partial \theta} = \begin{bmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \mathbf{T} = \mathbf{UT} $$

Taking differentials give \begin{eqnarray*} d \mathbf{T} &=& \frac{\partial \mathbf{T}}{\partial \theta} d\theta = \frac{\partial \mathbf{T}}{\partial \theta} \dot{\theta} dt = \dot{\theta}\mathbf{UT} dt= \mathbf{WT} dt \end{eqnarray*} with $$ \mathbf{W} =\dot{\theta}\mathbf{U} =\begin{bmatrix} 0 & \dot{\theta} & 0 \\ -\dot{\theta} & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} $$ This indicates that $ \frac{\partial \mathbf{T}}{\partial t} = \mathbf{WT} $.

\begin{eqnarray*} d \mathbf{T}^{-1} &=&- \mathbf{T}^{-1} (d\mathbf{T}) \mathbf{T}^{-1} \\ &=&- \mathbf{T}^{-1} \left(\frac{\partial \mathbf{T}}{\partial \theta}\right) \mathbf{T}^{-1} d\theta = - \mathbf{T}^{-1} \mathbf{U}d\theta \\ &=& - \mathbf{T}^{-1} \mathbf{U} \dot{\theta} dt = -\mathbf{T}^{-1} \mathbf{W} dt \end{eqnarray*} This indicates that $ \frac{\partial \mathbf{T}^{-1}}{\partial t} = -\mathbf{T}^{-1} \mathbf{W} $.