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In this question, he converted the sinusoidal voltage to exponential voltage.

what's the formula used here ? This is not the same one used in this video

enter image description here

Here is the formula derived in the video

enter image description here

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    \$\begingroup\$ well you don't seriously expect me to watch the video, do you? \$\endgroup\$ – user_1818839 Feb 22 at 13:12
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    \$\begingroup\$ I’m voting to close this question because this is the most basic mathematical expression in harmonic analysis: Euler's relation. This is basic math, not EE. \$\endgroup\$ – Marcus Müller Feb 22 at 13:18
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    \$\begingroup\$ I agree with @ahmedosama , they are not mathematically equivalent. Mathematically, sin x = (e^jx - e^-jx)/2j. What is going on, is that electrical engineers tend to ignore the fact that one needs to add or subtract the complex conjugate to get a real value (or take the Re part). \$\endgroup\$ – Math Keeps Me Busy Feb 22 at 14:40
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    \$\begingroup\$ I think they are phase shifting the Euler formula 90 degrees with the J at the front since the real part of Euler is given in terms of cosine but your source function is given in sin. A sin is a 90 degree phase shifted cosine. More obvious if you change Euler to complex cartesian form and multiply by J. And they omitted taking the real part. It's sloppy. \$\endgroup\$ – DKNguyen Feb 22 at 15:12
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    \$\begingroup\$ By multiplying by J and not -J they also phase shifted it 90 degrees in the wrong direction to get the real parts strictly equal, but since it is the only source it doesn't matter since phases only have meaning relative to each other, so long as you shift all phases by the same amount. As written the real parts are actually negatives of each other and not even equal. Real sloppy. In light of all this sloppiness and omissions I vote to reopen. \$\endgroup\$ – DKNguyen Feb 22 at 15:44
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I don't know about the video, but the Euler's relation

\$e^{jx} = \cos(x) + j\sin(x)\$

where j = \$\sqrt{-1}\$, x = \$ \omega t\$

...means that the real portion Re(j\$e^{j\omega t}\$) = -\$\sin(\omega t) \$ or Re(j\$e^{-j\omega t}\$) = \$\sin(\omega t) \$

So I think there are two things:

  1. Take the real part
  2. There is a missing minus sign in the book.
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  • \$\begingroup\$ But they are not mathematically equal, right ? \$\endgroup\$ – ahmed osama Feb 22 at 14:57
  • \$\begingroup\$ No they are definitely not mathematically equal. Equivalent representations (modulo that minus sign). \$\endgroup\$ – Spehro Pefhany Feb 22 at 14:58
  • \$\begingroup\$ I don't know, but he did this exact same thing in many examples in the book, so It's probably not a mistake, he meant to write it this way. I guess I will ask my professor for it. \$\endgroup\$ – ahmed osama Feb 22 at 15:02
  • \$\begingroup\$ Usually examples use cos() rather than sin(). \$\endgroup\$ – Spehro Pefhany Feb 22 at 15:03
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    \$\begingroup\$ @BrianDrummond Using cos gets rid of the j multiplier and the minus sign, so it's simpler, I assume. \$\endgroup\$ – Spehro Pefhany Feb 22 at 15:28

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