phasor representation

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elimenohpee

Joined Oct 26, 2008
47
I'm having a little trouble with the derivation. In my electromagnetics book it states:
V(t) = \(Vocos(wt + \phi)\)
and to convert between real time and phasor notation use:
V(t) = Re {Ve\(^{jwt}\)}

But in every example (and by every example I mean the only 2 in the book :( ) the \(e^{jwt}\) is omitted from the answer. Is this because you assume the \(jw\) or \(\theta\) is constant somehow? I was just confused because it seems like you "lose" information about the signal if you don't explicitly tag on the \(e^{jwt}\) on every signal you put into the phasor domain.

Maybe it'll make more sense if someone can explain how to represent the following signals as phasors:

a) \(cos(wt)\) (= \(e^{jwt}\) ? since \(\phi\) is zero? )
b) \(3cos((120\pi t) - \pi /2)\) ( = \(6e^{-j\pi / 2}e^{j120\pi t} \)or just\( 6e^{-j\pi / 2} \)?)
 

steveb

Joined Jul 3, 2008
2,436
But in every example (and by every example I mean the only 2 in the book :( ) the \(e^{jwt}\) is omitted from the answer. Is this because you assume the \(jw\) or \(\theta\) is constant somehow? I was just confused because it seems like you "lose" information about the signal if you don't explicitly tag on the \(e^{jwt}\) on every signal you put into the phasor domain.
Yes, when you don't see the \(e^{jwt}\), you can assume it's there. People leave it out as shorthand to save time. You don't really lose information because everyone knows the meaning of phasors. You will find that linear system responses have frequency dependent amplitude and phase functions. This is the critical information that people care about.
 
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