I have some issues with derivating D-Q inductance from a five-phase FEA data.
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Hi folks,
I have simulated a five-phase interior PMSM on FEA software to get self and mutual inductance data. The data have been calculated according to mechanical rotor angle (half of the electrical angle in my case). Now, I want to calculate d and q inductances. As you know, there is a five-phase coordinate transformation block in Simulink and one can easily get the equations from the block help page. The equation is given below. However, when I calculate the d-q inductances by electrical rotor angle the results are not constant. The results formed a sinusoidal wave shape. Is there anyone familiar with such calculations to help me figure it out?
Thanks for your help in advance!
The code:
a=length(e_deg);
Ld=zeros(a,1);
Lq=zeros(a,1);
for ind=1:a
Ld(ind,1)=(2/5)*(sind(e_deg(ind,1))*Laa(ind,1)+sind(e_deg(ind,1)-72)*Lbb(ind,1)+sind(e_deg(ind,1)-144)*Lcc(ind,1)+sind(e_deg(ind,1)+144)*Ldd(ind,1)+sind(e_deg(ind,1)+72)*Lee(ind,1));
Lq(ind,1)=(2/5)*(cosd(e_deg(ind,1))*Laa(ind,1)+cosd(e_deg(ind,1)-72)*Lbb(ind,1)+cosd(e_deg(ind,1)-144)*Lcc(ind,1)+cosd(e_deg(ind,1)+144)*Ldd(ind,1)+cosd(e_deg(ind,1)+72)*Lee(ind,1));
end
figure
plot(e_deg,Ld)
hold on
plot(e_deg,Lq)
legend('Ld','Lq')
The transformation matrix I have used: Implement abcde to dqxy0 transform - Simulink (mathworks.com)

The results i got:

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Antworten (1)
Sabin
am 13 Jan. 2025
The code implementing the mathematical transform does not look right as it seems that only mutual inductances are used (Laa, Lbb, Lcc, Ldd, Lee). See below a simple code using symbolic toolbox to check the five-phase transform and inverse.
syms th L1 L2 psim
% Direct transform
T = (2/5)*[sin(th) sin(th-2*pi/5) sin(th-4*pi/5) sin(th+4*pi/5) sin(th+2*pi/5);...
cos(th) cos(th-2*pi/5) cos(th-4*pi/5) cos(th+4*pi/5) cos(th+2*pi/5);...
sin(th) sin(th+4*pi/5) sin(th-2*pi/5) sin(th+2*pi/5) sin(th-4*pi/5);...
cos(th) cos(th+4*pi/5) cos(th-2*pi/5) cos(th+2*pi/5) cos(th-4*pi/5);...
1/sqrt(2) 1/sqrt(2) 1/sqrt(2) 1/sqrt(2) 1/sqrt(2)];
% Inverse transform
Ti = (5/2)*T.';
% Inductance matrix
L = [L1 L2 L2 L2 L2;L2 L1 L2 L2 L2;L2 L2 L1 L2 L2;L2 L2 L2 L1 L2; L2 L2 L2 L2 L1];
% Results
Ldiag = simplify(T*L*Ti);
PsiDQ = simplify(T*psim*[cos(th);cos(th-2*pi/5);cos(th-4*pi/5);cos(th+4*pi/5);cos(th+2*pi/5)]);
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