Inverse Clarke Transform

Implementαβtoabctransformation

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Motor Control Blockset / Controls / Math Transforms

Description

TheInverse Clarke Transformblock computes the Inverse Clarke transformation of balanced, two-phase orthogonal components in the stationaryαβreference frame. It outputs the balanced, three-phase components in the stationaryabcreference frame.

The block accepts theα-βaxis components as inputs and outputs the corresponding three-phase signals, where the phase-aaxis aligns with theα-axis.

Theαandβinput components in theαβreference frame.
The direction of the equivalenta,b, andcoutput components in theabcreference frame and theαβreference frame.
The time-response of the individual components of equivalent balancedαβandabcsystems.

Equations

The following equation describes the Inverse Clarke transform computation:

$[\begin{matrix} f_{a} \\ f_{b} \\ f_{c} \end{matrix}] = [\begin{matrix} 1 & 0 & 1 \\ - \frac{1}{2} & \frac{\sqrt{3}}{2} & 1 \\ - \frac{1}{2} & - \frac{\sqrt{3}}{2} & 1 \end{matrix}] [\begin{matrix} f_{α} \\ f_{β} \\ f_{0} \end{matrix}]$

For balanced systems like motors, the zero sequence component calculation is always zero:

$i_{a} + i_{b} + i_{c} = 0$

Therefore, you can use only two current sensors in three-phase motor drives, where you can calculate the third phase as,

$i_{c} = - (i_{a} + i_{b})$

By using these equations, the block implements the Inverse Clarke transform as,

$[\begin{matrix} \begin{matrix} f_{a} \\ f_{b} \end{matrix} \\ f_{c} \end{matrix}] = [\begin{matrix} \begin{matrix} 1 & 0 \\ - \frac{1}{2} & \frac{\sqrt{3}}{2} \\ - \frac{1}{2} & - \frac{\sqrt{3}}{2} \end{matrix} \end{matrix}] [\begin{matrix} f_{α} \\ f_{β} \end{matrix}]$

where:

$f_{α}$ and $f_{β}$ are the balanced two-phase orthogonal components in the stationaryαβreference frame.
$f_{0}$ is the zero component in the stationaryαβreference frame.
$f_{a}$ , $f_{b}$ , and $f_{c}$ are the balanced three-phase components in theabcreference frame.

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