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If the $$e^{j\omega} = \cos(\omega)+j \sin(\omega)$$ and I have the following equation $$\phi_s = {2\over3}\left(\phi_u+\phi_ve^{{2\pi\over3}j}+\phi_we^{{4\pi\over3}j}\right)e^{-j\theta}$$

The imaginary and real part is the following according the lecture: $$\mathop{\rm Re}(\phi_s)={2\over3}\left(\cos(\theta)\phi_u+\cos\left(\theta-{2\pi\over3}\right)\phi_v+\cos\left(\theta+{2\pi\over3}\right)\phi_w\right)$$ $$\mathop{\rm Im}(\phi_s)={2\over3}\left(\sin(\theta)\phi_u+\sin\left(\theta-{2\pi\over3}\right)\phi_v+\sin\left(\theta+{2\pi\over3}\right)\phi_w\right)$$

According my calculation all the sin is with a minus. Where do I make the mistake(s)?

  • 2
    First, in mathematics it is way more customary to use $\,i\,$ and not $\,j\,$ to denote the imaginary unit. Second, what is $\,\phi_u\,$ etc. and how come none of them appear in neither the real nor the imaginary parts??2012-09-09
  • 1
    sorry, I fixed the equation, In my case the i is the current, so I don't want to mix that.2012-09-09
  • 0
    I would trust your calculation...2012-09-09
  • 0
    As the son of an electrical engineer, I learned to call the imaginary unit $j$ before I found out that some people call it $i$.2012-09-09

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Note that you're subtracting $\theta$ from the angle in each term, getting "$-\theta$", "$2\pi/3-\theta$", and "$4\pi/3-\theta$".

Now, cosine is an even function ---that is, $\cos(-x)=\cos(x)$--- so that you can simply reverse the order of the subtraction if desired (for instance, because it just looks better to have $\theta$ first): $$\begin{align} \cos(-\theta)&= \cos(\theta) \\ \cos\left(\frac{2\pi}{3}-\theta\right)&=\cos\left(\theta-\frac{2\pi}{3}\right)\\ \cos\left(\frac{4\pi}{3}-\theta\right)&=\cos\left(\theta-\frac{4\pi}{3}\right) = \cos\left(\theta +\frac{2\pi}{3}\right) \end{align}$$

On the other hand, sine is an odd function ---$\sin(-x)=-\sin(x)$--- so that

$$\begin{align} \sin(-\theta) &= -\sin(\theta) \\ \sin\left(\frac{2\pi}{3}-\theta\right) &=-\sin\left(\theta-\frac{2\pi}{3}\right) \\ \sin\left(\frac{4\pi}{3}-\theta\right)&=-\sin\left(\theta-\frac{4\pi}{3}\right) = -\sin\left(\theta+\frac{2\pi}{3}\right) \end{align}$$

So, if you had negatives, then you are correct. Perhaps there was a typo (or write-o) in the lecture notes. (Maybe the "2/3" on the imaginary part was supposed to have a minus sign?)