Problem in transition from phasor to normal

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Discussion Overview

The discussion revolves around the transition from phasor representation to normal trigonometric functions, specifically focusing on the application of Euler's formula and the treatment of the imaginary part in this context.

Discussion Character

  • Technical explanation, Debate/contested

Main Points Raised

  • Some participants reference Euler's formula, stating it as e^{ix} = cos x + i sin x, while questioning the treatment of the imaginary part in phasor representation.
  • One participant expresses confusion about why the coefficient of sine is represented as the negative of the imaginary part.
  • Another participant corrects the initial reference to Euler's nationality, asserting that he was Swiss, not French, while also addressing the identity involving e^{-ix} and its implications.
  • There is a light-hearted acknowledgment of the correction regarding Euler's nationality and a mention of potential misspelling of his name.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the treatment of the imaginary part in phasor representation, and there are competing views regarding the implications of Euler's formula.

Contextual Notes

Some assumptions about the relationship between phasors and trigonometric functions remain unaddressed, and the discussion does not clarify the mathematical steps involved in the transition.

nhrock3
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2mrf6f6.jpg


oiler formula says
cosx+isinx

but here they multiply by minus the imaginary part of the phasor representation
the coefficient of sine is minus the imaginary part
why??
 
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nhrock3 said:
oiler formula says
cosx+isinx

That's "Euler" formula. A Frenchman, not a oil tanker. And I suppose you mean the identity

[tex]e^{ix} = \cos x + i\sin x[/tex]

but here they multiply by minus the imaginary part of the phasor representation
the coefficient of sine is minus the imaginary part
why??

I have no idea what your image has to do with your question. Do you mean this? :

[tex]e^{-ix} = \cos{(- x)} + i\sin {(-x)} = \cos x - i\sin x[/tex]
 
LCKurtz said:
That's "Euler" formula. A Frenchman, not a oil tanker. And I suppose you mean the identity

A nitpick: Euler was Swiss, not French.
 
CEL said:
A nitpick: Euler was Swiss, not French.

You are right! And I see upon looking him up, I would also have misspelled his first name had I mentioned it. :redface:
 

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