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:

    <br />
V_{s}*e^{j\omega t}=V_{s}cos(\omega t) + j V_{s}sin(\omega t)<br />

    Where
    <br />
V_{ds} = V_{s}cos(\omega t)<br />
    <br />
V_{qs} = V_{s}sin(\omega t)<br />

    Now if the resulting is j such that
    <br />
-j V_{s}*e^{j\omega t}<br />
    then apperantly
    <br />
V_{ds} = V_{s}sin(\omega t)<br />
    <br />
V_{qs} = V_{s}cos(\omega t)<br />

    Why does this happen and why is there a sign change?
     
  2. shteii01

    AAC Fanatic!

    Feb 19, 2010
    3,397
    497
    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.
     
    kokkie_d likes this.
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