Phasors: How to go from Re{} to Re{}+Im{}

[jaydnul]

I understand that $V=cos(wt+\phi)=Re[e^{j(wt+\phi)}]$

But when doing calculations (like loop voltage analysis or junction current analysis) you're just using $V=e^{j(wt+\phi)}$ (where all of the $e^{jwt}$ will cancel out and you're just left with the phasors)

Example: $A_se^{j\phi _s}{e^{jwt}}=A_1e^{j\phi _1}e^{jwt}+A_2e^{j\phi _2}e^{jwt}$

So how do you get from the initial $V=Re[e^{j(wt+\phi)}]$ to the form $V=e^{j(wt+\phi)}$ to do calcuations? Those two don't equal.

 

[Nick O]

They do not represent the same quantity, even though they represent the same concept. It is an error to equate both to V; my book used v (lower case) for the time-domain value, and V (capital) for the phasor (or "frequency") domain value. The units are the same, but the values are not!

The important thing is that addition and multiplication in the phasor domain are very meaningful. But, to get the final, physically meaningful answer after doing these calculations, you must return to the time domain.

Edit: Transforms (like the phasor transform) are actually very interesting to me. They allow us to solve hard problems by instead working on easy problems that have no real connection to the original ones. It's just an incredibly useful coincidence that changes in the phasor domain can be mapped directly to changes in the time domain.

I'm rambling, but I wish I could convey this idea more clearly. Understanding it is very helpful as you learn more and more transforms.

 

[jaydnul]

Is there a more rigorous mathematical answer? So we define the phasor $P_0$ as $V=Re[P_0e^{jwt}]$.
Now what is the mathematical proof that we can just use $P_0$ as the voltage in the phasor domain?

 

[Nick O]

For a rigorous proof, we need a precise question.

The units of $P_0$ are clearly volts, so that much is self-evident. It is also evident that if $V_0=Re[P_0e^{jwt}]$ and $V_1=Re[P_1e^{jwt}]$, then $V_0+V_1=Re[P_0e^{jwt}]+ Re[P_1e^{jwt}]=Re[(P_0+P_1)e^{jwt}]$.

What do you want to prove?

 

[Nick O]

It's like counting people with your fingers; you are representing each person with a finger, because math works similarly for both fingers and people, but fingers are more convenient.

Or maybe I should sleep.

 

[jaydnul]

Aaaaaaaah. I'm an idiot.

I was confusing myself because $\frac{Re[e^{j\theta}]}{Re[e^{j\phi}]}\neq Re[\frac{e^{j\theta}}{e^{j\phi}}]$

But $Re[e^{j\theta}]+Re[e^{j\phi}] = Re[e^{j\theta}+e^{j\phi}]$

I'm good now. Thanks a bunch Nick O!

 

[Nick O]

No problem! I'm glad that helped, because I haven't used phasors since learning them a few semesters ago, and I wouldn't have been able to make many more useful observations in my sleep-deprived state [emoji14]