The direct-quadrature-zero (DQZ or DQ0[1] or DQO,[2] sometimes lowercase) transformation or zero-direct-quadrature[3] (0DQ or ODQ, sometimes lowercase) transformation is a tensor that rotates the reference frame of a three-element vector or a three-by-three element matrix in an effort to simplify analysis. The DQZ transform is the product of the Clarke transform and the Park transform, first proposed in 1929 by Robert H. Park.[4]
The DQZ transform is often used in the context of electrical engineering with three-phase circuits. The transform can be used to rotate the reference frames of AC waveforms such that they become DC signals. Simplified calculations can then be carried out on these DC quantities before performing the inverse transform to recover the actual three-phase AC results. As an example, the DQZ transform is often used in order to simplify the analysis of three-phase synchronous machines or to simplify calculations for the control of three-phase inverters. In analysis of three-phase synchronous machines, the transformation transfers three-phase stator and rotor quantities into a single rotating reference frame to eliminate the effect of time-varying inductances and transform the system into a linear time-invariant system
The DQZ transform is made of the Park and Clarke transformation matrices. The Clarke transform (named after Edith Clarke) converts vectors in the ABC reference frame to the XYZ (also called αβγ) reference frame. The primary value of the Clarke transform is isolating that part of the ABC-referenced vector, which is common to all three components of the vector; it isolates the common-mode component (i.e., the Z component). The power-invariant, right-handed, uniformly-scaled Clarke transformation matrix is
KC=\sqrt{
| 2 | |
| 3 |
To convert an ABC-referenced column vector to the XYZ reference frame, the vector must be pre-multiplied by the Clarke transformation matrix:
\vec{u}XYZ=KC ⋅ \vec{u}ABC
\vec{u}ABC=
| -1 | |
| K | |
| C |
⋅ \vec{u}XYZ
The Park transform (named after Robert H. Park) converts vectors in the XYZ reference frame to the DQZ reference frame. The Park transform's primary value is to rotate a vector's reference frame at an arbitrary frequency. The Park transform shifts the signal's frequency spectrum such that the arbitrary frequency now appears as "dc," and the old dc appears as the negative of the arbitrary frequency. The Park transformation matrix is
KP=\begin{bmatrix} \cos{\left(\theta\right)}&\sin{\left(\theta\right)}&0\\ -\sin{\left(\theta\right)}&\cos{\left(\theta\right)}&0\\ 0&0&1 \end{bmatrix}
uDQZ=KP ⋅ uXYZ
uXYZ=
| -1 | |
| K | |
| P |
⋅ uDQZ
The Clarke and Park transforms together form the DQZ transform:
KCP=KP ⋅ KC
\to\begin{bmatrix} \cos{\left(\theta\right)}&\sin{\left(\theta\right)}&0\\ -\sin{\left(\theta\right)}&\cos{\left(\theta\right)}&0\\ 0&0&1 \end{bmatrix} ⋅ \sqrt{
| 2 | |
| 3 |
\to\sqrt{
| 2 | |
| 3 |
| -1 | |
| K | |
| CP |
=\sqrt{
| 2 | |
| 3 |
To convert an ABC-referenced vector to the DQZ reference frame, the column vector signal must be pre-multiplied by the DQZ transformation matrix:
uDQZ=KCP ⋅ uABC
uABC=
| -1 | |
| K | |
| CP |
⋅ uDQZ
To understand this transform better, a derivation of the transform is included.
The Park transform is based on the concept of the dot product and projections of vectors onto other vectors. First, let us imagine two unit vectors,
\hat{u}D
\hat{u}Q
\vec{v}XY
\hat{u}D= \cos{\left(\theta\right)}\hat{u}X+\sin{\left(\theta\right)}\hat{u}Y
\hat{u}Q= -\sin{\left(\theta\right)}\hat{u}X+\cos{\left(\theta\right)}\hat{u}Y
\vec{v}XY= vX\hat{u}X+vY\hat{u}Y
\hat{u}X
\hat{u}Y
\theta
\hat{u}X
\hat{u}D
vD= \hat{u}D ⋅ \vec{v}XY
\to\cos{\left(\theta\right)}vX+\sin{\left(\theta\right)}vY
vQ= \hat{u}Q ⋅ \vec{v}XY
\to-\sin{\left(\theta\right)}vX+\cos{\left(\theta\right)}vY
vD
\vec{v}XY
\hat{u}D
vQ
\vec{v}XY
\hat{u}Q
vD
vQ
\vec{v}DQ
\vec{v}XY
Notice that the positive angle
\theta
A single matrix equation can summarize the operation above:
\vec{v}DQ=\begin{bmatrix} \cos{\left(\theta\right)} &\sin{\left(\theta\right)}\\ -\sin{\left(\theta\right)} &\cos{\left(\theta\right)} \end{bmatrix} ⋅ \vec{v}XY
KP=\begin{bmatrix} \cos{\left(\theta\right)} &\sin{\left(\theta\right)} &0\\ -\sin{\left(\theta\right)} &\cos{\left(\theta\right)} &0\\ 0&0&1 \end{bmatrix}
From a linear algebra perspective, this is simply a clockwise rotation about the z-axis and is mathematically equivalent to the trigonometric difference angle formulae.
Consider a three-dimensional space with unit basis vectors A, B, and C. The sphere in the figure below is used to show the scale of the reference frame for context and the box is used to provide a rotational context.
Typically, in electrical engineering (or any other context that uses three-phase systems), the three-phase components are shown in a two-dimensional perspective. However, given the three phases can change independently, they are by definition orthogonal to each other. This implies a three-dimensional perspective, as shown in the figure above. So, the two-dimensional perspective is really showing the projection of the three-dimensional reality onto a plane.Three-phase problems are typically described as operating within this plane. In reality, the problem is likely a balanced-phase problem (i.e., vA + vB + vC = 0) and the net vector
\vec{v}=vA\hat{u}A+vB\hat{u}B+vC\hat{u}C
To build the Clarke transform, we actually use the Park transform in two steps. Our goal is to rotate the C axis into the corner of the box. This way the rotated C axis will be orthogonal to the plane of the two-dimensional perspective mentioned above. The first step towards building the Clarke transform requires rotating the ABC reference frame about the A axis. So, this time, the 1 will be in the first element of the Park transform:
K1=\begin{bmatrix} 1&0&0\\ 0&\cos{\left(-
| \pi | |
| 4 |
\right)}&\sin{\left(-
| \pi | |
| 4 |
\right)}\\ 0&-\sin{\left(-
| \pi | |
| 4 |
\right)}&\cos{\left(-
| \pi | |
| 4 |
\right)} \end{bmatrix}
\to\begin{bmatrix} 1&0&0\\ 0&
| 1 | |
| \sqrt{2 |
Next, the following tensor rotates the vector about the new Y axis in a counter-clockwise direction with respect to the Y axis (The angle was chosen so that the C' axis would be pointed towards the corner of the box.):
K2=\begin{bmatrix} \cos{\left(\theta\right)}&0&-\sin{\left(\theta\right)}\\ 0&1&0\\ \sin{\left(\theta\right)}&0&\cos{\left(\theta\right)} \end{bmatrix}
\theta=\cos-1\left(\sqrt{
| 2 | |
| 3 |
K2=\begin{bmatrix} \sqrt{
| 2 | |
| 3 |
| \vec{m}=\left(0, | \sqrt{2 |
\vec{n}=\left(
| 1 | |
| \sqrt{3 |
\vec{n}=\left(1,1,1\right)
\vec{m} ⋅ \vec{n}=|\vec{m}||\vec{n}|\cos\theta,
\theta
\vec{m}
\vec{n},
| \left(0, | \sqrt{2 |
\cos\theta=0+
| \sqrt{2 | |
\theta=\cos-1\left(\sqrt{
| 2 | |
| 3 |
\theta=35.26\circ.
The norm of the K2 matrix is also 1, so it too does not change the magnitude of any vector pre-multiplied by the K2 matrix.
At this point, the Z axis is now orthogonal to the plane in which any ABC vector without a common-mode component can be found. Any balanced ABC vector waveform (a vector without a common mode) will travel about this plane. This plane will be called the zero plane and is shown below by the hexagonal outline.
The X and Y basis vectors are on the zero plane. Notice that the X axis is parallel to the projection of the A axis onto the zero plane. The X axis is slightly larger than the projection of the A axis onto the zero plane. It is larger by a factor of . The arbitrary vector did not change magnitude through this conversion from the ABC reference frame to the XYZ reference frame (i.e., the sphere did not change size). This is true for the power-invariant form of the Clarke transform. The following figure shows the common two-dimensional perspective of the ABC and XYZ reference frames.
It might seem odd that though the magnitude of the vector did not change, the magnitude of its components did (i.e., the X and Y components are longer than the A, B, and C components). Perhaps this can be intuitively understood by considering that for a vector without common mode, what took three values (A, B, and C components) to express, now only takes 2 (X and Y components) since the Z component is zero. Therefore, the X and Y component values must be larger to compensate.
The power-invariant Clarke transformation matrix is a combination of the K1 and K2 tensors:
KC=\underbrace{\begin{bmatrix} \sqrt{
| 2 | |
| 3 |
KC=\sqrt{
| 2 | |
| 3 |
\to\begin{bmatrix}
| 2 | |
| \sqrt{6 |
Notice that when multiplied through, the bottom row of the KC matrix is 1/, not 1/3. (Edith Clarke did use 1/3 for the power-variant case.) The Z component is not exactly the average of the A, B, and C components. If only the bottom row elements were changed to be 1/3, then the sphere would be squashed along the Z axis. This means that the Z component would not have the same scaling as the X and Y components.As things are written above, the norm of the Clarke transformation matrix is still 1, which means that it only rotates an ABC vector but does not scale it. The same cannot be said for Clarke's original transform.
It is easy to verify (by matrix multiplication) that the inverse of KC is
| -1 | |
| K | |
| C |
=\begin{bmatrix}
| 2 | |
| \sqrt{6 |
It is sometimes desirable to scale the Clarke transformation matrix so that the X axis is the projection of the A axis onto the zero plane. To do this, we uniformly apply a scaling factor of and a to the zero component to get the power-variant Clarke transformation matrix:
K\hat{C
\to
| 2 | |
| 3 |
\begin{bmatrix} 1&-
| 1 | |
| 2 |
&-
| 1 | |
| 2 |
\\ 0&
| \sqrt{3 | |
K\hat{C
This will necessarily shrink the sphere by a factor of as shown below. Notice that this new X axis is exactly the projection of the A axis onto the zero plane.
With the power-variant Clarke transform, the magnitude of the arbitrary vector is smaller in the XYZ reference frame than in the ABC reference frame (the norm of the transform is), but the magnitudes of the individual vector components are the same (when there is no common mode). So, as an example, a signal defined by
\begin{bmatrix} A\\ B\\ C \end{bmatrix}=\begin{bmatrix} \cos{\left(\omegat\right)}\\ \cos{\left(\omegat-
| 2\pi | |
| 3 |
\right)}\\ \cos{\left(\omegat+
| 2\pi | |
| 3 |
\right)} \end{bmatrix}
\begin{bmatrix} X\\ Y\\ Z \end{bmatrix}=\begin{bmatrix} \cos{\left(\omegat\right)}\\ \cos{\left(\omegat-
| \pi | |
| 2 |
\right)}\\ 0 \end{bmatrix}
The DQZ transformation uses the Clarke transform to convert ABC-referenced vectors into two differential-mode components (i.e., X and Y) and one common-mode component (i.e., Z) and then applies the Park transform to rotate the reference frame about the Z axis at some given angle. The X component becomes the D component, which is in direct alignment with the vector of rotation, and the Y component becomes the Q component, which is at a quadrature angle to the direct component. The DQZ transform is
KCP=KP ⋅ KC
\to\begin{bmatrix} \cos{\left(\theta\right)}&\sin{\left(\theta\right)}&0\\ -\sin{\left(\theta\right)}&\cos{\left(\theta\right)}&0\\ 0&0&1 \end{bmatrix} ⋅ \sqrt{
| 2 | |
| 3 |
In electric systems, very often the A, B, and C values are oscillating in such a way that the net vector is spinning. In a balanced system, the vector is spinning about the Z axis. Very often, it is helpful to rotate the reference frame such that the majority of the changes in the abc values, due to this spinning, are canceled out and any finer variations become more obvious. This is incredibly useful as it now transforms the system into a linear time-invariant system
The DQZ transformation can be thought of in geometric terms as the projection of the three separate sinusoidal phase quantities onto two axes rotating with the same angular velocity as the sinusoidal phase quantities.Shown above is the DQZ transform as applied to the stator of a synchronous machine. There are three windings separated by 120 physical degrees. The three phase currents are equal in magnitude and are separated from one another by 120 electrical degrees. The three phase currents lag their corresponding phase voltages by
\delta
\omega
\theta=\omegat
ID
IQ
The transformation originally proposed by Park differs slightly from the one given above. In Park's transformation q-axis is ahead of d-axis, qd0, and the
\theta
P=
| 2 | |
| 3 |
\begin{bmatrix}\cos(\theta)&\cos(\theta-
| 2\pi | |
| 3 |
)&\cos(\theta+
| 2\pi | |
| 3 |
)\\ \sin(\theta)&\sin(\theta-
| 2\pi | |
| 3 |
)&\sin(\theta+
| 2\pi | |
| 3 |
)\\
| 1 | & | |
| 2 |
| 1 | & | |
| 2 |
| 1 | |
| 2 |
\end{bmatrix}
and
P-1=\begin{bmatrix}\cos(\theta)&\sin(\theta)&1\\ \cos(\theta-
| 2\pi | |
| 3 |
)&\sin(\theta-
| 2\pi | |
| 3 |
)&1\\ \cos(\theta+
| 2\pi | |
| 3 |
)&\sin(\theta+
| 2\pi | |
| 3 |
)&1\end{bmatrix}
D. Holmes and T. Lipo, Pulse Width Modulation for Power Converters: Principles and Practice, Wiley-IEEE Press, 2003, and
P. Krause, O. Wasynczuk and S. Sudhoff, Analysis of Electric Machinery and Drive Systems, 2nd ed., Piscataway, NJ: IEEE Press, 2002.
See main article: article and αβγ transform.
The dqo transform is conceptually similar to the αβγ transform. Whereas the dqo transform is the projection of the phase quantities onto a rotating two-axis reference frame, the αβγ transform can be thought of as the projection of the phase quantities onto a stationary two-axis reference frame.