I have a problem with adding harmonics to park-clark transformed 3 phase value. Here is transformations I'm using:

Vsd= 1, Vsq= 1, x=angle;
sinA = sin(x);
cosA = cos(x);
a= Vsd * cosA - Vsq * sinA;
b= Vsd * cosA + Vsq * sinA;
Va= a;
Vb= -1/2*a + sqrt(3)/2*b;
Vb= -1/2*a - sqrt(3)/2*b;

and on the graphs I get following result: enter image description here

When I'm trying to add harmonics to sine and cosine, for example 3rd harmonic:

sinA = sin(x) + sin(3*x);
cosA = cos(x) + cos(3*x);

I'm getting mess instead of M shaped waves on all 3 phases:

enter image description here

what I'm doing wrong ?

  • \$\begingroup\$ What do you expect to get? I suggest making the harmonic components lower in amplitude - multiply them by say 0.1 or 0.2 and then look at the result. \$\endgroup\$ Commented Jan 29, 2017 at 1:32
  • \$\begingroup\$ I'm trying to get M shaped wave form on 3 phases, so it should look like MW (0-2PI) on all 3 phases \$\endgroup\$
    – Harry
    Commented Jan 29, 2017 at 3:32

1 Answer 1


cosB = cos(x-π) + cos(3*x-π)
enter image description here cosA = cos(x) + cos(3*x)

ref http://www.falstad.com/fourier/

  • \$\begingroup\$ What is π ? is it PI or Period ? In both case it doesn't solve the problem... \$\endgroup\$
    – Harry
    Commented Jan 29, 2017 at 3:30
  • \$\begingroup\$ Pi solves it, see the "M" \$\endgroup\$ Commented Jan 29, 2017 at 3:31
  • \$\begingroup\$ not in my case, try to apply it to sinA = sin(x) + sin(3*x); cosA = cos(x) + cos(3*x); and then use inverse Park and Clark transform... you will get the same mess \$\endgroup\$
    – Harry
    Commented Jan 29, 2017 at 11:20

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