# Why can we drop the Omega term in Euler's expression of AC signal?

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 $$Re(V_p e^{j\phi})$$?

• Probably because $j \omega t$ equals 0 somehow. – Ignacio Vazquez-Abrams Nov 18 '14 at 4:38
• Why do you think that you "can further simplify"? Who told you this? – LvW Nov 18 '14 at 8:39

## 1 Answer

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.