Space Vector Transform

The space vector transform expresses three-phase quantities in orthogonal \(\alpha\beta0\) coordinates. The \(\alpha\) and \(\beta\) components describe the phase quantities in a two-dimensional plane, while the \(0\) component represents zero sequence.

Author

Key result

\vec v \triangleq v_{\alpha} + \jmath v_{\beta} = \sqrt{\frac{2}{3}}\left( v_a e^{\jmath 0} + v_b e^{\jmath \tfrac{2\pi}{3}} + v_c e^{\jmath \tfrac{4\pi}{3}} \right).

Applications

In power electronics, space-vector modulation represents inverter switching states as discrete vectors in the \(\alpha\beta\) plane and uses them to synthesize a reference voltage vector.

In electric-machine control, balanced sinusoidal phase quantities become a constant-magnitude vector rotating at synchronous speed. This representation is the basis for field-oriented control and related methods.

In power-system analysis, the transform provides a stationary orthogonal representation of instantaneous three-phase voltages and currents.

1) Three independent phase quantities

A set of phase voltages or currents \((v_a,v_b,v_c)\) defines a point in three-dimensional \(abc\) coordinates. If the phase quantities are independent, their possible values occupy the full space.

Random points filling 3D abc space — Independent \(v_a, v_b, v_c\) span a 3D space.

2) The zero-sum constraint

A three-phase set with no zero-sequence component satisfies \[ v_a + v_b + v_c = 0, \] which defines a plane in \(abc\) coordinates. This condition applies to balanced phase quantities and to three-wire systems in which zero-sequence current cannot flow.

Points constrained to plane va+vb+vc=0 — Points satisfying \(v_a+v_b+v_c=0\) lie on a plane.

3) Coordinate rotation

The coordinate system can be rotated so that the \(\alpha\) and \(\beta\) axes lie in the zero-sum plane and the \(0\) axis is normal to it. This rotation produces the Clarke coordinate system.

Reoriented view with va horizontal and vb vertical — Initial orientation of the \(abc\) coordinate system.

Rotate about va by pi/4 — Rotate about \(v_a\) by \(\pi/4\).

Rotate about new beta axis by -atan(1/sqrt(2)) — Rotate about the new \(\beta\) axis by \(-\tan^{-1}(1/\sqrt{2})\).

Plane flattened onto 2D alpha-beta surface — The zero-sum plane expressed in \(\alpha\beta\) coordinates.

4) Rotation matrices in two and three dimensions

Begin with the familiar two-dimensional case. A counterclockwise rotation through an angle \(\theta\) maps \((x,y)\) to \((x',y')\) according to

\begin{bmatrix}x'\\y'\end{bmatrix} = \underbrace{ \begin{bmatrix} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{bmatrix}}_{\mathbf R(\theta)} \begin{bmatrix}x\\y\end{bmatrix}.

A three-dimensional axis rotation applies this same \(2\times2\) rotation within the plane perpendicular to the selected axis and leaves the coordinate along that axis unchanged. For example, rotation about \(x\) rotates the \(yz\) plane while retaining the original \(x\) coordinate.

For a column vector, positive rotations about the \(x\), \(y\), and \(z\) axes are defined by the right-hand rule:

\mathbf R_x(\theta)= \begin{bmatrix} 1 & 0 & 0\\ 0 & \cos\theta & -\sin\theta\\ 0 & \sin\theta & \cos\theta \end{bmatrix}, \qquad \mathbf R_y(\theta)= \begin{bmatrix} \cos\theta & 0 & \sin\theta\\ 0 & 1 & 0\\ -\sin\theta & 0 & \cos\theta \end{bmatrix}, \] \[ \mathbf R_z(\theta)= \begin{bmatrix} \cos\theta & -\sin\theta & 0\\ \sin\theta & \cos\theta & 0\\ 0 & 0 & 1 \end{bmatrix}.

Matrix order is significant. The animations first rotate about the \(v_a\) axis, identified with \(x\), and then about the intermediate \(\beta\) axis, identified with \(y\). The combined rotation is therefore \(\mathbf R_y(\theta_y)\mathbf R_x(\theta_x)\); the rightmost matrix acts first.

5) Derivation of the Clarke transform

The two angles used in the rotation sequence are \(\theta_x=\pi/4\) and \(\theta_y=-\tan^{-1}(1/\sqrt{2})\). Their trigonometric values are

\cos\theta_x=\sin\theta_x=\frac{1}{\sqrt{2}}, \qquad \cos\theta_y=\sqrt{\frac{2}{3}}, \qquad \sin\theta_y=-\frac{1}{\sqrt{3}}. \] Therefore, \[ \mathbf R_x\!\left(\frac{\pi}{4}\right)= \begin{bmatrix} 1 & 0 & 0\\ 0 & \tfrac{1}{\sqrt{2}} & -\tfrac{1}{\sqrt{2}}\\ 0 & \tfrac{1}{\sqrt{2}} & \tfrac{1}{\sqrt{2}} \end{bmatrix} \] and \[ \mathbf R_y\!\left(-\tan^{-1}\!\frac{1}{\sqrt{2}}\right)= \begin{bmatrix} \sqrt{\tfrac{2}{3}} & 0 & -\tfrac{1}{\sqrt{3}}\\ 0 & 1 & 0\\ \tfrac{1}{\sqrt{3}} & 0 & \sqrt{\tfrac{2}{3}} \end{bmatrix}. \] Before substituting the angles, the matrix product has the form \[ \mathbf R_y(\theta_y)\mathbf R_x(\theta_x)= \begin{bmatrix} \cos\theta_y & \sin\theta_y\sin\theta_x & \sin\theta_y\cos\theta_x\\ 0 & \cos\theta_x & -\sin\theta_x\\ -\sin\theta_y & \cos\theta_y\sin\theta_x & \cos\theta_y\cos\theta_x \end{bmatrix}, \] and substitution of the two angles gives \[ \mathbf T_{\mathrm{P}} =\mathbf R_y(\theta_y)\mathbf R_x(\theta_x) = \begin{bmatrix} \sqrt{\tfrac{2}{3}} & -\tfrac{1}{\sqrt{6}} & -\tfrac{1}{\sqrt{6}}\\ 0 & \tfrac{1}{\sqrt{2}} & -\tfrac{1}{\sqrt{2}}\\ \tfrac{1}{\sqrt{3}} & \tfrac{1}{\sqrt{3}} & \tfrac{1}{\sqrt{3}} \end{bmatrix} = \sqrt{\tfrac{2}{3}} \begin{bmatrix} 1 & -\tfrac{1}{2} & -\tfrac{1}{2}\\ 0 & \tfrac{\sqrt{3}}{2} & -\tfrac{\sqrt{3}}{2}\\ \tfrac{1}{\sqrt{2}} & \tfrac{1}{\sqrt{2}} & \tfrac{1}{\sqrt{2}} \end{bmatrix}.

This is the power-invariant Clarke matrix. Its first two rows define axes within the zero-sum plane, and its third row defines the zero-sequence axis normal to that plane:

\begin{bmatrix} v_{\alpha,\mathrm{P}}\\ v_{\beta,\mathrm{P}}\\ v_{0,\mathrm{P}} \end{bmatrix} =\mathbf T_{\mathrm{P}} \begin{bmatrix} v_a\\v_b\\v_c \end{bmatrix}.

6) Compact form using Euler notation

The first two rows of the power-invariant Clarke matrix give

\begin{aligned} v_{\alpha,\mathrm{P}} &=\sqrt{\tfrac{2}{3}} \left(v_a-\tfrac{1}{2}v_b-\tfrac{1}{2}v_c\right),\\ v_{\beta,\mathrm{P}} &=\sqrt{\tfrac{2}{3}} \left(\tfrac{\sqrt{3}}{2}v_b-\tfrac{\sqrt{3}}{2}v_c\right). \end{aligned}

Form the complex \(\alpha\beta\) quantity \(v_{\alpha,\mathrm{P}}+jv_{\beta,\mathrm{P}}\), where \(j^2=-1\). Substitution of the two row equations gives

v_{\alpha,\mathrm{P}}+jv_{\beta,\mathrm{P}} =\sqrt{\tfrac{2}{3}} \left[ v_a +\left(-\tfrac{1}{2}+j\tfrac{\sqrt{3}}{2}\right)v_b +\left(-\tfrac{1}{2}-j\tfrac{\sqrt{3}}{2}\right)v_c \right].

Euler's identity, \(e^{j\theta}=\cos\theta+j\sin\theta\), gives

-\frac{1}{2}+j\frac{\sqrt{3}}{2}=e^{j2\pi/3}, \qquad -\frac{1}{2}-j\frac{\sqrt{3}}{2}=e^{-j2\pi/3}.

Therefore, the compact Euler form is

v_{\alpha,\mathrm{P}}+jv_{\beta,\mathrm{P}} =\sqrt{\tfrac{2}{3}} \left(v_a+v_b e^{j2\pi/3}+v_c e^{-j2\pi/3}\right).

7) Balanced sinusoidal quantities

For an ideal balanced sinusoidal set, \(\vec v(t)\) has constant magnitude and rotates at the electrical angular frequency.

Rotating space vector circle — A balanced three-phase set represented by one rotating vector.

8) Orthonormality and power invariance

Row and column orthonormality

Let the rows of \(\mathbf T_{\mathrm{P}}\) be \(\mathbf r_\alpha\), \(\mathbf r_\beta\), and \(\mathbf r_0\). Their norms are

\begin{aligned} \mathbf r_\alpha\!\cdot\!\mathbf r_\alpha &=\frac{2}{3}+\frac{1}{6}+\frac{1}{6}=1,\\ \mathbf r_\beta\!\cdot\!\mathbf r_\beta &=\frac{1}{2}+\frac{1}{2}=1,\\ \mathbf r_0\!\cdot\!\mathbf r_0 &=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1. \end{aligned} \] Their pairwise dot products are \[ \begin{aligned} \mathbf r_\alpha\!\cdot\!\mathbf r_\beta &=-\frac{1}{\sqrt{12}}+\frac{1}{\sqrt{12}}=0,\\ \mathbf r_\alpha\!\cdot\!\mathbf r_0 &=\frac{\sqrt{2}}{3}-\frac{1}{\sqrt{18}}-\frac{1}{\sqrt{18}}=0,\\ \mathbf r_\beta\!\cdot\!\mathbf r_0 &=\frac{1}{\sqrt{6}}-\frac{1}{\sqrt{6}}=0. \end{aligned}

The columns \(\mathbf c_a\), \(\mathbf c_b\), and \(\mathbf c_c\) are also orthonormal:

\begin{aligned} \mathbf c_a\!\cdot\!\mathbf c_a&=\frac{2}{3}+\frac{1}{3}=1,\\ \mathbf c_b\!\cdot\!\mathbf c_b =\mathbf c_c\!\cdot\!\mathbf c_c&=\frac{1}{6}+\frac{1}{2}+\frac{1}{3}=1,\\ \mathbf c_a\!\cdot\!\mathbf c_b =\mathbf c_a\!\cdot\!\mathbf c_c&=-\frac{1}{3}+\frac{1}{3}=0,\\ \mathbf c_b\!\cdot\!\mathbf c_c&=\frac{1}{6}-\frac{1}{2}+\frac{1}{3}=0. \end{aligned}

Hence,

\mathbf T_{\mathrm{P}}\mathbf T_{\mathrm{P}}^{\mathsf{T}}=\mathbf I, \qquad \mathbf T_{\mathrm{P}}^{\mathsf{T}}\mathbf T_{\mathrm{P}}=\mathbf I.

Inverse transform

Because \(\mathbf T_{\mathrm{P}}\) is orthonormal, its inverse is its transpose:

\mathbf T_{\mathrm{P}}^{-1}=\mathbf T_{\mathrm{P}}^{\mathsf{T}} = \begin{bmatrix} \sqrt{\tfrac{2}{3}} & 0 & \tfrac{1}{\sqrt{3}}\\ -\tfrac{1}{\sqrt{6}} & \tfrac{1}{\sqrt{2}} & \tfrac{1}{\sqrt{3}}\\ -\tfrac{1}{\sqrt{6}} & -\tfrac{1}{\sqrt{2}} & \tfrac{1}{\sqrt{3}} \end{bmatrix}, \] so \[ \begin{bmatrix}v_a\\v_b\\v_c\end{bmatrix} =\mathbf T_{\mathrm{P}}^{\mathsf{T}} \begin{bmatrix}v_{\alpha,\mathrm{P}}\\v_{\beta,\mathrm{P}}\\v_{0,\mathrm{P}}\end{bmatrix}.

Power invariance

Apply the same orthonormal transform to voltage and current. Orthonormality then preserves their inner product:

\begin{aligned} p_{abc} &=\mathbf v_{abc}^{\mathsf{T}}\mathbf i_{abc},\\ p_{\alpha\beta0} &=(\mathbf T_{\mathrm{P}}\mathbf v_{abc})^{\mathsf{T}} (\mathbf T_{\mathrm{P}}\mathbf i_{abc})\\ &=\mathbf v_{abc}^{\mathsf{T}} \underbrace{\mathbf T_{\mathrm{P}}^{\mathsf{T}}\mathbf T_{\mathrm{P}}}_{\mathbf I} \mathbf i_{abc} =p_{abc}. \end{aligned} \] Therefore, \[ v_a i_a+v_b i_b+v_c i_c =v_{\alpha,\mathrm{P}}i_{\alpha,\mathrm{P}} +v_{\beta,\mathrm{P}}i_{\beta,\mathrm{P}} +v_{0,\mathrm{P}}i_{0,\mathrm{P}}.

9) Scaling conventions

The power-invariant Clarke transform uses the factor \(\sqrt{2/3}\):

\mathbf T_{\mathrm{P}} =\sqrt{\tfrac{2}{3}} \begin{bmatrix} 1 & -\tfrac{1}{2} & -\tfrac{1}{2}\\ 0 & \tfrac{\sqrt{3}}{2} & -\tfrac{\sqrt{3}}{2}\\ \tfrac{1}{\sqrt{2}} & \tfrac{1}{\sqrt{2}} & \tfrac{1}{\sqrt{2}} \end{bmatrix}.

This scaling normalizes the three basis vectors, making the transform orthonormal. It therefore preserves inner products and instantaneous power without additional coefficients.

Amplitude-invariant scaling: \(2/3\)

Another common convention uses the factor \(2/3\):

\mathbf T_{\mathrm{A}} =\frac{2}{3} \begin{bmatrix} 1 & -\tfrac{1}{2} & -\tfrac{1}{2}\\ 0 & \tfrac{\sqrt{3}}{2} & -\tfrac{\sqrt{3}}{2}\\ \tfrac{1}{2} & \tfrac{1}{2} & \tfrac{1}{2} \end{bmatrix}.

For a balanced set with phase peak \(V\), this convention gives \(v_\alpha=V\cos\theta\) and \(v_\beta=V\sin\theta\). The magnitude of the \(\alpha\beta\) vector therefore equals the peak magnitude of the phase components. It also gives \(v_0=(v_a+v_b+v_c)/3\). This convention is amplitude invariant, but it is not orthonormal and does not preserve power directly:

p=\frac{3}{2}\left(v_\alpha i_\alpha+v_\beta i_\beta\right)+3v_0i_0.

Unscaled projection: factor \(1\)

An unscaled \(\alpha\beta\) projection is commonly used in electric-machine models, particularly when the transformed variables represent phase-axis flux contributions:

\begin{bmatrix}\psi_\alpha\\\psi_\beta\end{bmatrix} = \begin{bmatrix} 1 & -\tfrac{1}{2} & -\tfrac{1}{2}\\ 0 & \tfrac{\sqrt{3}}{2} & -\tfrac{\sqrt{3}}{2} \end{bmatrix} \begin{bmatrix}\psi_a\\\psi_b\\\psi_c\end{bmatrix}.

This form directly sums the three phase-axis flux contributions to obtain the resultant flux vector. For a balanced set with phase peak \(\Psi\), the resultant magnitude is \(3\Psi/2\). The convention is useful when the net machine flux is the modeled quantity, but it is neither amplitude invariant nor power invariant.