Implement *abc* to *+-0*
transform

**Library:**Simscape / Power Systems / Simscape Components / Control / Mathematical Transforms

The Symmetrical-Components Transform block implements a symmetrical transform of a set of phasors. The transform splits an unbalanced set of three phasors into three balanced sets of phasors.

In an unbalanced system with balanced impedances, use this block to decouple the system into three independent networks. In a balanced system, use this block to simplify the set of three-phasors to an equivalent one-line network. In this case, the positive set represents the one-line network.

Use the `Power invariant`

property to choose between the Fortescue
transform, and the alternative, power-invariant version.

The symmetrical-components transform separates an unbalanced three-phase signal given in phasor quantities into three balanced sets of phasors:

$$\left[\begin{array}{c}{v}_{a}\\ {v}_{b}\\ {v}_{c}\end{array}\right]=\left[\begin{array}{c}{v}_{a+}\\ {v}_{b+}\\ {v}_{c+}\end{array}\right]+\left[\begin{array}{c}{v}_{a-}\\ {v}_{b-}\\ {v}_{c-}\end{array}\right]+\left[\begin{array}{c}{v}_{a0}\\ {v}_{b0}\\ {v}_{c0}\end{array}\right],$$

*v*,_{a}*v*, and_{b}*v*make up the original, unbalanced set of phasors._{c}*v*,_{a+}*v*, and_{b+}*v*make up the balanced, positive set of phasors._{c+}*v*,_{a-}*v*, and_{b-}*v*make up the balanced, negative set of phasors._{c-}*v*,_{a0}*v*, and_{b0}*v*make up the balanced, zero set of phasors._{c0}

The block calculates the symmetric *a*-phase using the transformation:

$$\left[\begin{array}{c}{V}_{a+}\\ {V}_{a-}\\ {V}_{a0}\end{array}\right]=\frac{K}{3}\left[\begin{array}{ccc}1& a& {a}^{2}\\ 1& {a}^{2}& a\\ 1& 1& 1\end{array}\right]\left[\begin{array}{c}{V}_{a}\\ {V}_{b}\\ {V}_{c}\end{array}\right].$$

$$a={e}^{2\pi i/3},$$

$$\{\begin{array}{cc}K=1& \text{Fortescuetransform}\\ K=\sqrt{3}& \text{Power-invarianttransform}\end{array}$$

```
Power
invariant
```

property. Because the remaining two sets of symmetrical phasors are not often used in calculation, the block does not calculate them. However, they are given in terms of simple rotations of the first set:

$$\left[\begin{array}{c}{V}_{b+}\\ {V}_{b-}\\ {V}_{b0}\end{array}\right]=\left[\begin{array}{ccc}{a}^{2}& 0& 0\\ 0& a& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}{V}_{a+}\\ {V}_{a-}\\ {V}_{a0}\end{array}\right],$$

$$\left[\begin{array}{c}{V}_{c+}\\ {V}_{c-}\\ {V}_{c0}\end{array}\right]=\left[\begin{array}{ccc}a& 0& 0\\ 0& {a}^{2}& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}{V}_{a+}\\ {V}_{a-}\\ {V}_{a0}\end{array}\right].$$

The three sets of balanced phasors generated by the transform have the following properties:

The positive set has the same order as the unbalanced set of phasors

*a-b-c*.The negative set has the opposite order as the unbalanced set of phasors

*a-c-b*.The zero set has no order because all three phasor angles are equal.

This diagram visualizes the separation performed by the transform.

In the diagram, the top axis shows an unbalanced three-phase signal
with components *a*, *b*, and
*c*. The bottom set of axes separates the three-phase signal into
symmetrical positive, negative, and zero phasors.

Observe that in each case, the *a*, *b*, and
*c* components are symmetrical and are separated by:

+120 degrees for the positive set.

*-*120 degrees for the negative set.0 degrees for the zero set.

The symmetrical-components transform is unique and invertible:

$$\left[\begin{array}{c}{V}_{a}\\ {V}_{b}\\ {V}_{c}\end{array}\right]=\frac{1}{K}\left[\begin{array}{ccc}1& 1& 1\\ {a}^{2}& a& 1\\ a& {a}^{2}& 1\end{array}\right]\left[\begin{array}{c}{V}_{a+}\\ {V}_{a-}\\ {V}_{a0}\end{array}\right].$$

[1] Anderson, P. M. *Analysis of Faulted Power Systems.*
Hoboken, NJ: Wiley-IEEE Press, 1995.

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