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I know that for an AC signal like $$V=V_p\cos(\omega t+ \phi)$$ we can simplify it to $$Re(V_p e^{j(\omega t+\phi)})$$ But why can we further simplify it by dropping the Omega term $$Re(V_p e^{j\phi})$$?

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    \$\begingroup\$ Probably because \$j \omega t \$ equals 0 somehow. \$\endgroup\$ – Ignacio Vazquez-Abrams Nov 18 '14 at 4:38
  • \$\begingroup\$ Why do you think that you "can further simplify"? Who told you this? \$\endgroup\$ – LvW Nov 18 '14 at 8:39
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I think you should give the book or docs where the equations come. Maybe it's just for convenient. Because the results of signals addition/subtraction with same frequency also have the same frequency with themselves.

Below is another version, and i think this version is more clear.

In Fundamentals of Electric Circuits, 5th by Sdiku, Matthew, Page. 378

$$ v(t)=V_{m}cos(\omega t+\phi)=Re(V_{m}e^{j(\omega t + \phi)}) $$

Thus

$$ v(t)= Re(\dot{V}e^{j \omega t}) $$

where

$$ \dot{V}=V_{m}e^{j\phi}=V_{m} \angle \phi $$

\$\dot{V}\$ is thus the phasor representation of the sinusoid \$v(t)\$.

Note: for clarity, i put one dot on top of \$V\$.


In your equations, apparently, the second and the third equation can't be equal, in your second equation, \$V_{p}\$ should be a scalar, and in your third equation \$V_{p}\$ should be a vector.

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