Problem in "abc - dq0" Transformation using "Phase-Locked Loop" in Simulink Model.

Hello all,
I am using a very simple block to transfer from 3 pahse abc voltage to 2 phase dq using the attached model. I am not using the built-in model as I am told by my instructor to implement the 3-phase PLL (phase-locked Loop) to implement the transformation.
I am expected to see the output DQ as 1 and 0. 0 for Q voltage and 1 for D, if the input magnitude is 1 ofcourse.
Now, in the attached file, I have a simple function and the output of the DQ block is fed into a PLL loop made by me to feedback the (wt = omega*t) component which is used in the function block.
Can anyone tell me why am I getting a -1 instead of a 1 for D voltage.
Function block codes...
First block -
function [Va,Vb,Vc] = ThreePhaseSystem(a,f,t)
x=2*pi*f*t;
Va = a*sin(x);
Vb = a*sin(x-(2*pi/3));
Vc = a*sin(x+(2*pi/3));
end
2nd Block -
function [Vd,Vq] = fcn(Va,Vb,Vc,w)
Vd = (2/3)*((Va*sin(w))+(Vb*sin(w-(2*pi/3)))+(Vc*sin(w+(2*pi/3))));
Vq = (2/3)*((Va*cos(w))+(Vb*cos(w-(2*pi/3)))+(Vc*cos(w+(2*pi/3))));
end
The "Q" output of this block is fed into a PI controller with Kp = -100, and Ki = -10000. Output is integrated again and fed into the 2nd block.
Your help will be greatly appreciated.
Thanks, Faisal Khan

3 commentaires

Azzi Abdelmalek
Azzi Abdelmalek le 9 Déc 2012
Modifié(e) : Azzi Abdelmalek le 9 Déc 2012
What is w?
w is the phase. theta from the equation.
Its not easy to guess without your model

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Réponses (1)

Sabin
Sabin le 23 Déc 2025
Without a model is difficult to tell. My guess is that the angle is not correct, may have an offset that is not accounted for.

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le 8 Déc 2012

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