Implement *αβ0* to *dq0*
transform

**Library:**Simscape / Electrical / Control / Mathematical Transforms

The Clarke to Park Angle Transform block converts the alpha, beta, and zero components in a stationary reference frame to direct, quadrature, and zero components in a rotating reference frame. For balanced three-phase systems, the zero components are equal to zero.

You can configure the block to align the phase *a*-axis of the
three-phase system to either the *q*- or *d*-axis of
the rotating reference frame at time, *t* = 0. The figures show the
direction of the magnetic axes of the stator windings in the three-phase system, a
stationary *αβ0* reference frame, and a rotating *dq0*
reference frame where:

The

*a*-axis and the*q*-axis are initially aligned.The

*a*-axis and the*d*-axis are initially aligned.

In both cases, the angle *θ* =
*ω**t*, where

*θ*is the angle between the*a*and*q*axes for the*q*-axis alignment or the angle between the*a*and*d*axes for the*d*-axis alignment.*ω*is the rotational speed of the*d*-*q*reference frame.*t*is the time, in s, from the initial alignment.

The figures show the time-response of the individual components of equivalent balanced
*αβ0* and *dq0* for an:

Alignment of the

*a*-phase vector to the*q*-axisAlignment of the

*a*-phase vector to the*d*-axis

The Clarke to Park Angle Transform block implements the transform
for an *a*-phase to *q*-axis alignment as

$\left[\begin{array}{c}d\\ q\\ 0\end{array}\right]=\left[\begin{array}{ccc}\text{sin}(\theta )& -\text{cos}(\theta )& 0\\ \text{cos}(\theta )& \text{sin}(\theta )& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}\alpha \\ \beta \\ 0\end{array}\right]$

where:

*α*and*β*are the alpha-axis and beta-axis components of the two-phase system in the stationary reference frame.*0*is the zero component.*d*and*q*are the direct-axis and quadrature-axis components of the two-axis system in the rotating reference frame.

For an *a*-phase to *d*-axis alignment, the
block implements the transform using this equation:

$\left[\begin{array}{c}d\\ q\\ 0\end{array}\right]=\left[\begin{array}{ccc}\text{cos}(\theta )& \text{sin}(\theta )& 0\\ -\text{sin}(\theta )& \text{cos}(\theta )& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}\alpha \\ \beta \\ 0\end{array}\right]$

[1] Krause, P., O. Wasynczuk, S. D. Sudhoff, and S. Pekarek. *Analysis of
Electric Machinery and Drive Systems.* Piscatawy, NJ: Wiley-IEEE Press,
2013.