Getting from exponentials to phasors

Thread Starter

jaydnul

Joined Apr 2, 2015
152
I actually have two questions about phasors.

1. I still don't understand why you can just add an imaginary part \(jsin(wt)\) to a sinusoidal voltage source to make it exponential \(e^{j(wt+\theta)}\). How does this not mess up the calculations? (I understand how it becomes exponential, Euler's formula. Just how does adding that imaginary part not screw it up in calculations?)

2. How do you get from \(V=V_0e^{j(wt+\theta)}\) to the phasor \(V_0e^{j\theta}\)? Where does the \(e^{jwt}\) go? Are we setting t=0?
 

Papabravo

Joined Feb 24, 2006
13,759
I don't exactly understand your notion of "Getting from". Adding an imaginary part to a sinusoidal voltage to make it exponential is not what is happening. What is happening is that a sinusoidal voltage can be expressed as a combination of complex exponential functions, or as the real part of a complex exponential function.

The answer to the second question is that we are performing a transformation of the exponential to eliminate time from consideration because we are only interested in magnitude and phase. Details of how the transformation is accomplished can be found here:
http://en.wikibooks.org/wiki/Circuit_Theory/Phasors
and here:
http://en.wikipedia.org/wiki/Phasor
 
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Thread Starter

jaydnul

Joined Apr 2, 2015
152
The answer to the second question is that we are performing a transformation of the exponential to eliminate time from consideration because we are only interested in magnitude and phase. Details of how the transformation is accomplished can be found here:
http://en.wikibooks.org/wiki/Circuit_Theory/Phasors
and here:
http://en.wikipedia.org/wiki/Phasor
I don't see how the transformation is derived in these links, all I see is how to do it.

If you start with a sinusoidal voltage, it has amplitude, frequency, and phase shift. You can represent that as \(Ae^{j(wt+\phi)}\)

But to transform that original sinusoidal function into phasor, you just use this \(Ae^{j\phi}\)

So what happens inbetween the time dependent exponential \(Ae^{j(wt+\phi)}\) and the phasor \(Ae^{j\phi}\)
 

Papabravo

Joined Feb 24, 2006
13,759
The information contained within the phasor has been separated from the time dependent information which was the original purpose of the exercise. What is left after dropping the time dependent part is useful in its own right since it transforms a complicated problem into a simpler one. Once the solution has been arrived at the time dependent portion can be reintroduced. It is very similar to the use of Laplace transforms for solving differential equations.

To answer your question, the original function is equal to a phasor transform times a complex exponential. The phasor transform can be manipulated on its own like a complex number and simplifies the solution of problem involving periodic waveforms. Believe it, or don't believe it I really can't help you there.

Would it make you happier if I said that the transform is equal to the original function multiplied by:

\(e^{-j\omega t}\)
 
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Thread Starter

jaydnul

Joined Apr 2, 2015
152
Ok, I understand this now; it's like separable solutions: \(V=v(t)v(\phi)\)

I'm still not sure how we got our voltage in exponential form. I fully understand: \(V=cos(wt+\phi)=Re(e^{j(wt+\phi)})\)

But how do we get from that, to this: \(V=cos(wt+\phi)=e^{j(wt+\phi)}\)

Because: \(e^{j(wt+\phi)}\neq Re(e^{j(wt+\phi)})\)

Right?
 

Papabravo

Joined Feb 24, 2006
13,759
As I see it you can approach the problem you are having in one of two ways:
  1. You can consider a portion of the expression of the complex exponential to represent the phasor transform, OR
  2. You can multiply the complex exponential by \(e^{-j \omega t}\) and what you are left with is the phasor transform, by definition
Now you can investigate the properties of the phasor transform without having to worry about the time dependency.
 

Papabravo

Joined Feb 24, 2006
13,759
As a side note, the concept of deriving and using a transform is not necessarily the same process as proving a trigonometric identity. What we did here was to isolate a portion of an expression so we could study just that portion of the expression. It might also help to realize that the magnitude of \(e^{- j \omega t}\) is just 1 for all t and all ω.
 
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