The DQ0-transformation, or direct-quadrature-zero transformation, is a very useful tool for electric power engineers to transform AC waveforms into DC signals. In particular, for three-phase systems, the use of DC signals simplifies the calculations immensely.
The DQ0-transformation is the product of the Clarke and Park transformation. The Clarke transformation converts the abc-phase-components to αβγ-components while the Park transformation converts the αβγ-components to dq0-components.
In this post, I will present a way how to use the DQ0-transformation to calculate the active and reactive power from instantaneous voltage and current signals (follow the link below to download the full Python script). I will not go into more details about the Clarke and Park transformations as this is a well-covered topic in the literature.
The Clarke and Park transformation
Here, I will present the power-invariant Clarke transformation (opposed to the power variant version). The difference between the power-invariant and power-variant transformations is that the power-invariant transformation preserves the active and reactive power. However, the amplitude of the transformed variables, i.e. voltages and currents, is not preserved while it is the other way around when using the power-variant transformation.
The αβγ-components are calculated from the instantaneous phase-voltages (va, vb, vc) as shown below. The same calculation can be applied to the currents.
The pq0-components of the voltages are calculated as shown below. The same calculation can be applied to the currents.
Where f is the frequency of the signal and t is the time.
DQ0-transformation (power-invariant Clarke and Park combined)
Below you can find the power-invariant Clarke and Park transformation combined.
To compute the two transformations efficiently, both are kept separately. The implementation is presented for voltages in the following. However, the same implementation is also valid for currents.
First, solve the Clarke transformation for the αβγ-components as shown below. The below equations are obtained by solving the matrix operation of the Clarke transformation shown above.
Then the Park transformation can be incorporated which leads to the following formulation:
The final step is to calculate the active and reactive power (P and Q). This can be done as follows:
Example of the active and reactive power calculation using the DQ0-transformation
In the following, an example of the calculation of the active/reactive power using the DQ0-transformation is presented.
The first plot shows two periods (40 ms at 50 Hz) of the three instantaneous phase-to-ground voltages and currents. The amplitude of the voltages and currents is chosen to be √2, which means that the RMS-value = 1 V and 1 A, respectively. Within the first period, the current is in phase with the voltage – this means and purely active load. In the second period, the current lags the voltage by 45 degrees – this represents a mixed active and inductive reactive load. Opposed to that, if the current leads the voltage, it would be capacitive.
In the first step, I calculated the DQ0-components according to the computational implementation described above. The results are shown in the plot below for both the voltage and current. It can be seen, that the DQ0-components of the voltage remain unchanged for both periods. This is expected, as the instantaneous voltage does not change. However, the DQ0-components of the currents change from the first to the second period because of the change in the phase of the instantaneous currents.
The final step is to calculate the active, reactive, and apparent power using the DQ0-components as described in the computational implementation above. As expected, in the first period it is purely active power, while in the second period it is a combination of active and reactive power. Since the phase lag of the current is 45 degrees in the second period, the active and reactive power are equal in amplitude. Clearly, the apparent power is constant throughout both periods, because the amplitude of the instantaneous voltages and currents does not change. The apparent power is equal to 3 VA as the voltage and current have both an RMS amplitude of 1 V and 1 A, respectively.
If this was interesting, you might also like the following: Load model identification in electric power systems: a linear least-squares approach
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