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Going from phasor form to instantaneous time sinusoidal functions

  1. Jan 16, 2012 #1
    So I'm currently reviewing for my electromagnetism class, but I do not remember the techniques from going to phasor to instantaneous time. Maybe someone can explain to me what is going on.

    1. The problem statement, all variables and given/known data

    Find the instantaneous time sinusoidal functions corresponding to the following phasors:


    My question is specifically on Part A and Part B

    2. Relevant equations

    Relevant equations includes Eulers:

    c9f2055dadfb49853eff822a453d9ceb.png

    Which let you go from polar to rectangular.

    3. The attempt at a solution

    For Part A, we are given V~ = -5exp(j*pi/3)

    I just went straight from phasor to instantaneous time:

    V~ = v(t) -> -5exp(j*pi/3) = Re{V~ * exp(jwt) = -5exp(jwt + pi/3)

    But for some strange reason, in the solution manual they subtract pi from the exponential to remove the negative sign. I'm not exactly sure why they do this. It is the same for Part B, why do they subtract pi/2. Can anyone explain this to me? I've also attached the solution to part A and B below.
     
    Last edited by a moderator: Apr 19, 2017
  2. jcsd
  3. Jan 17, 2012 #2

    rude man

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    First, you need to multiply all your coefficients (voltages and currents) by √2. The transformed voltages and currents represent rms values. The time-domain ones obviously represent peak voltages.

    OK, now realize that -1 = +1*exp(jπ) = +1*exp(-jπ). You don't want a negative sign in front of your time-domain voltage expression, it's meaningless. What counts is phase. -1V = +1V shifted 180 degrees.

    Use Euler to convince yourself that -1 = exp(jπ) = exp(-jπ). Also that j = exp(jπ/2).

    Subtracting π/2 from the exponent is subtracting 90 degrees from the phase shift. Multiplying any transformed voltage by j is equivalent to adding (or subtracting) π from the exponent. So you need to subtract π/2, then add π, to figure out part b.
     
    Last edited by a moderator: Apr 19, 2017
  4. Jan 18, 2012 #3
    Thanks for the reply, so you are saying that the the subtraction of the ∏ phase in part is to get rid of the negative. Meaning if I add ±∏, I should get the same answer since it shifts the negative to positive phase either way. That means:

    5exp(j(4∏/3 + wt)) should also be correct? Yes?
     
  5. Jan 19, 2012 #4

    rude man

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    Yes, but convention is to never exceed pi/2 in magnitude.

    So exp(-2pi/3) is more conventionally expressed as exp(pi/3) etc.
     
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