Documentation

### This is machine translation

Translated by
Mouseover text to see original. Click the button below to return to the English version of the page.

# abc to dq0, dq0 to abc

Perform transformation from three-phase (abc) signal to dq0 rotating reference frame or the inverse

## Library

Simscape / Electrical / Specialized Power Systems / Control & Measurements / Transformations

## Description

The abc to dq0 block performs a Park transformation in a rotating reference frame.

The dq0 to abc block performs an inverse Park transformation.

The block supports the two conventions used in the literature for Park transformation:

• Rotating frame aligned with A axis at t = 0, that is, at t = 0, the d-axis is aligned with the a-axis. This type of Park transformation is also known as the cosine-based Park transformation.

• Rotating frame aligned 90 degrees behind A axis, that is, at t = 0, the q-axis is aligned with the a-axis. This type of Park transformation is also known as the sine-based Park transformation. Use it in Simscape™ Electrical™ Specialized Power Systems models of three-phase synchronous and asynchronous machines.

Deduce the dq0 components from abc signals by performing an abc to αβ0 Clarke transformation in a fixed reference frame. Then perform an αβ0 to dq0 transformation in a rotating reference frame, that is, −(ω.t) rotation on the space vector Us = uα + j· uβ.

The abc-to-dq0 transformation depends on the dq frame alignment at t = 0. The position of the rotating frame is given by ω.t (where ω represents the dq frame rotation speed).

When the rotating frame is aligned with A axis, the following relations are obtained:

$\begin{array}{l}{U}_{s}={u}_{d}+j\cdot {u}_{q}=\left({u}_{a}+j\cdot {u}_{\beta }\right)\cdot {e}^{-j\omega t}=\frac{2}{3}\cdot \left({u}_{a}+{u}_{b}\cdot {e}^{\frac{-j2\pi }{3}}+{u}_{c}\cdot {e}^{\frac{j2\pi }{3}}\right)\cdot {e}^{-j\omega t}\\ {u}_{0}=\frac{1}{3}\left({u}_{a}+{u}_{b}+{u}_{c}\right)\\ \left[\begin{array}{c}{u}_{d}\\ {u}_{q}\\ {u}_{0}\end{array}\right]=\frac{2}{3}\left[\begin{array}{ccc}\mathrm{cos}\left(\omega t\right)& \mathrm{cos}\left(\omega t-\frac{2\pi }{3}\right)& \mathrm{cos}\left(\omega t+\frac{2\pi }{3}\right)\\ -\mathrm{sin}\left(\omega t\right)& -\mathrm{sin}\left(\omega t-\frac{2\pi }{3}\right)& -\mathrm{sin}\left(\omega t+\frac{2\pi }{3}\right)\\ \frac{1}{2}& \frac{1}{2}& \frac{1}{2}\end{array}\right]\left[\begin{array}{c}{u}_{a}\\ {u}_{b}\\ {u}_{c}\end{array}\right]\end{array}$



Inverse transformation is given by

$\left[\begin{array}{c}{u}_{a}\\ {u}_{b}\\ {u}_{c}\end{array}\right]=\left[\begin{array}{ccc}\mathrm{cos}\left(\omega t\right)& -\mathrm{sin}\left(\omega t\right)& 1\\ \mathrm{cos}\left(\omega t-\frac{2\pi }{3}\right)& -\mathrm{sin}\left(\omega t-\frac{2\pi }{3}\right)& 1\\ \mathrm{cos}\left(\omega t+\frac{2\pi }{3}\right)& -\mathrm{sin}\left(\omega t+\frac{2\pi }{3}\right)& 1\end{array}\right]\left[\begin{array}{c}{u}_{d}\\ {u}_{q}\\ {u}_{0}\end{array}\right]$

When the rotating frame is aligned 90 degrees behind A axis, the following relations are obtained:

$\begin{array}{l}{U}_{s}={u}_{d}+j\cdot {u}_{q}=\left({u}_{\alpha }+j\cdot {u}_{\beta }\right)\cdot {e}^{-j\left(\omega t-\frac{\pi }{2}\right)}\\ \left[\begin{array}{c}{u}_{d}\\ {u}_{q}\\ {u}_{0}\end{array}\right]=\frac{2}{3}\left[\begin{array}{ccc}\mathrm{sin}\left(\omega t\right)& \mathrm{sin}\left(\omega t-\frac{2\pi }{3}\right)& \mathrm{sin}\left(\omega t+\frac{2\pi }{3}\right)\\ \mathrm{cos}\left(\omega t\right)& \mathrm{cos}\left(\omega t-\frac{2\pi }{3}\right)& \mathrm{cos}\left(\omega t+\frac{2\pi }{3}\right)\\ \frac{1}{2}& \frac{1}{2}& \frac{1}{2}\end{array}\right]\left[\begin{array}{c}{u}_{a}\\ {u}_{b}\\ {u}_{c}\end{array}\right]\end{array}$

Inverse transformation is given by

$\left[\begin{array}{c}{u}_{a}\\ {u}_{b}\\ {u}_{c}\end{array}\right]=\left[\begin{array}{ccc}\mathrm{sin}\left(\omega t\right)& \mathrm{cos}\left(\omega t\right)& 1\\ \mathrm{sin}\left(\omega t-\frac{2\pi }{3}\right)& \mathrm{cos}\left(\omega t-\frac{2\pi }{3}\right)& 1\\ \mathrm{sin}\left(\omega t+\frac{2\pi }{3}\right)& \mathrm{cos}\left(\omega t+\frac{2\pi }{3}\right)& 1\end{array}\right]\left[\begin{array}{c}{u}_{d}\\ {u}_{q}\\ {u}_{0}\end{array}\right]$

## Parameters

Rotating frame alignment (at wt=0)

Select the alignment of rotating frame a t = 0 of the d-q-0 components of a three-phase balanced signal:

(positive-sequence magnitude = 1.0 pu; phase angle = 0 degree)

When you select Aligned with phase A axis, the d-q-0 components are d = 0, q = −1, and zero = 0.

When you select 90 degrees behind phase A axis, the default option, the d-q-0 components are d = 1, q = 0, and zero = 0.

## Inputs and Outputs

abc

The vectorized abc signal.

dq0

The vectorized dq0 signal.

wt

The angular position of the dq rotating frame, in radians.

## Examples

The power_Transformations example shows various uses of blocks performing Clarke and Park transformations.