# d-q reference frame

Discussion in 'Math' started by kokkie_d, Nov 10, 2010.

1. ### kokkie_d Thread Starter Active Member

Jan 12, 2009
72
0
I have been learning about d-q reference frame and have figured out the following:

$
V_{s}*e^{j\omega t}=V_{s}cos(\omega t) + j V_{s}sin(\omega t)
$

Where
$
V_{ds} = V_{s}cos(\omega t)
$

$
V_{qs} = V_{s}sin(\omega t)
$

Now if the resulting is j such that
$
-j V_{s}*e^{j\omega t}
$

then apperantly
$
V_{ds} = V_{s}sin(\omega t)
$

$
V_{qs} = V_{s}cos(\omega t)
$

Why does this happen and why is there a sign change?

2. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,297
482
It is surprisingly simple. The key is that (-j)(j)=1. When you expand the exponential into cosine and sine, the result will be -jVcos(wt)+(-j)(j)Vsin(wt). So you see? The Vsin(wt) becomes the real part and -jcos(wt) becomes imaginary part.

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