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Three phase phasor proof or simplification

  1. Sep 20, 2015 #1
    So I'm really rusty on phasors, I was reading that a space vector current was

    i(t) = I(cos(wt)<0 + cos(wt - 120)<120 + cos(wt - 240)<240 ) = 3/2 * I < wt

    and I couldn't figure out how that could be so (still can't, please help)

    So I tried to go back to basics and I went back and read:

    A = CosΘ.Cos(wt) + Cos(Θ - 120).Cos(wt - 120) + Cos(Θ + 120).Cos(wt + 120)
    = 3/2 * Cos(Θ - wt)

    so I went back to Trig identities.....and pages of scribbling later still couldn't figure it out.
    Does anyone have a proof for either of these?

    Thanks!
     
  2. jcsd
  3. Sep 20, 2015 #2
    A note: phasors and space vectors are two very different concepts.
    - A phasor is a representation of a sinusoidal function with parameters that are time-invariant.
    - A space vector is a transformation that maps a set of real-valued functions to a complex-valued function, which usually has some type of spatial interpretation (used a lot in electric machine theory).

    Have you tried converting all the terms to their complex representation? Then do a bit of algebra.
     
  4. Sep 20, 2015 #3
    Yeah as far as I remember a space vector is the addition of phasors through feeding a sineusoidally current to a sineusoidally distributed winding in space.
    Yeah I tried that thanks.
    As you can see in what you quoted the angle term becomes the frequency * time. It should be simple, I'm just not doing something right.

    What about:
    no idea how it reduces to that?
     
  5. Sep 21, 2015 #4
    No, phasors aren't involved at all when you're working with space vectors, which is a good thing, since you'd otherwise be limited to steady-state analysis.

    If you have a function ##i(t) = A\sin(\omega t + \theta)##, where ##A, \omega, \theta## are constants, then its phasor representation is ##I = Ae^{j\theta}##.

    If this:
    $$ \mathbf{i_s} = i_a(t) + i_b(t) \mathbf{a} + i_c(t) \mathbf{a}^2, \mathbf{a} = e^{2\pi/3}$$
    is the space-vector transformation, then, for instance, the term ##i_b(t) \mathbf{a}## is not a phasor, even though it is complex. ##i_b## is just an arbitrary real-valued function of ##t##, and ##\mathbf{a}## isn't a phasor representation of a sinusoidal function.

    ##i_a, i_b, i_c## aren't restricted to sinusoidal functions, and the space-vector transformation is just as useful for windings that aren't sinusoidally distributed.

    I can help you with the algebra if you show your work.

    Try doing the complex algebra first.
     
  6. Sep 21, 2015 #5
    Could please you elaborate on that?

    Anyway isn't what I've posted just phasors?
     
  7. Sep 21, 2015 #6
    That was the purpose of the rest of my post. Can you be a bit more specific? :smile:

    No, a phasor is a complex constant, i.e. they have no time dependency.

    None of these terms are phasors.

    Edit:
    Just to reiterate: a phasor is a complex constant, but not all complex constants are phasors. As I wrote, a phasor encodes the time-invariant amplitude and phase of a sinusoidal function.

    You can have voltage and current phasors, but their complex ratio, the impedance, isn't a phasor, since it doesn't encode the parameters of a sinusoidal function.
     
    Last edited: Sep 21, 2015
  8. Sep 21, 2015 #7
    Oups, that should be ##\mathbf{a} = e^{j2\pi/3}##.
     
  9. Sep 21, 2015 #8
    ok thanks, found the paper:

    i(t) = I(cos(wt)<0 + cos(wt - 120)<120 + cos(wt - 240)<240 ) = 3/2 * I < wt

    =[ei(wt)+e-i(wt)]/2 + [ei(wt +120)+ei(wt + 120)]/2 * (-1/2 + i√3/2) + [ei(wt -120)+ei(wt - 120)]/2 * (-1/2 + i√3/2) = ...

    And

    A = CosΘ.Cos(wt) + Cos(Θ - 120).Cos(wt - 120) + Cos(Θ + 120).Cos(wt + 120)
    = 3/2 * Cos(Θ - wt)

    using: sin(wt).sin(Θ) = [Cos(Θ - wt) - Cos(Θ + wt)] / 2
    and Cos(a).Cos(b) = Cos(a - b) + sin(a).Sin(b)

    cos(Θ - wt) + sin(wt).sin(Θ) + cos(Θ - 120 -(wt - 120)) + sin(Θ - 120).sin(wt - 120) + cos(Θ + 120 - (wt + 120) )+ sin(Θ + 120).sin(wt + 120)

    cos(Θ - wt) + [Cos(Θ - wt) - Cos(Θ + wt)] / 2 + cos(Θ - wt) + [cos(Θ - wt) - cos(Θ + wt + 240)] / 2 + cos(Θ - wt) + [cos(Θ - wt) - cos(Θ + wt + 240)/2 = ...

    cheers!
     
  10. Sep 21, 2015 #9
    hmm,
    so are you saying the thing that makes it a phasor is that there is not 't', the 'wt' is implicit, you have to keep it in the back of your mind? Like a phasor has an invisible frequency, but the phasor itself just a complex value?
     
  11. Sep 22, 2015 #10
    Not exactly. What makes a complex constant a phasor is the mapping it represents. It's just a compact way of representing a sinusoidal function with time-invariant parameters, which, in the framework of steady-state AC analysis, is very convenient mathematically to work with. But this is really getting away from the point I was trying to make:

    Space vectors have nothing to do with phasors. There are no complex constants, like ##Ae^{j\theta}##, which maps to a sinusoidal function, present in this expression:
    $$
    i_a(t) + i_b(t) \mathbf{a} + i_c(t) \mathbf{a}^2, \mathbf{a} = e^{j2\pi/3}
    $$
    You're just working with complex numbers here. Phasors belong in a completely different framework.

    There are similarities between space vectors and symmetrical components, in which you actually do work with phasors, but the theory of symmetrical components only deal with steady-state analysis, which is the crucial difference.
     
  12. Sep 22, 2015 #11
    I sort of hear what you're saying, I'll endevour to think about it more, I think some specific maths would help. What did you think of the track I went down with my working for the initial two expression derivations?
    Cheers
     
  13. Sep 22, 2015 #12
    Focusing on the first bit, I'd start with polar form. It'll make the algebra lot easier:
    $$
    \cos(\omega t) + \cos(\omega t + 2\pi/3)e^{j2\pi/3} + \cos(\omega t - 2\pi/3)e^{-j2\pi/3}
    $$
     
  14. Sep 22, 2015 #13
    I can see in your script the last negative is there, but for some reason it didn't display on my browser.

    Right, well, I suppose I could factor out the ej2pi/3

    Cos(wt) + ej2pi/3[cos(wt + 2pi/3) + cos(wt - 2pi/3).e^-1]
    ...actually that didn't help
    So I suppose those trig relations were of no relevance?

    Sorry, this is embarrassing
     
  15. Sep 22, 2015 #14
    You were on the right track before by converting the cosines to their complex form. I just wanted you to start out with:
    $$
    \cos(\omega t) + \cos(\omega t + 2\pi/3)e^{j2\pi/3} + \cos(\omega t - 2\pi/3)e^{-j2\pi/3}
    $$
    instead.

    No it's not :smile:.
     
  16. Sep 22, 2015 #15
    I appreciate your tact.

    So If I leave the Cosines in their current form, how can I simplify them?

    Thank you

    P.S also what about:
     
    Last edited: Sep 22, 2015
  17. Sep 22, 2015 #16
    Convert the complex exponentials to their rectangular form (like you did), i.e. ##e^{j2\pi/3} = -\frac{1}{2} + j\frac{\sqrt3}{2}, e^{j4\pi/3} = -\frac{1}{2} - j\frac{\sqrt3}{2}##, expand terms, and see what you get.

    There's probably a more elegant way to show it, but I'd just use the angle sum identities, so you have, for instance:
    $$
    \cos\left(\theta - \frac{2\pi}{3}\right) = \cos(\theta)\cos\left(\frac{2\pi}{3}\right) + \sin(\theta)\sin\left(\frac{2\pi}{3}\right) = -\frac{1}{2}\cos(\theta) + \frac{\sqrt{3}}{2}\sin(\theta)
    $$
    and expand terms.
     
  18. Sep 22, 2015 #17
    Very well, here goes:
    i(t) = I(cos(wt)<0 + cos(wt - 120)<120 + cos(wt - 240)<240 ) = I[ cos(wt) + cos(wt + 120)*(-0.5 + j√3/2) + cos(wt + 240)*(-0.5 - j√3/2)]
    = .... what am I not seeing?
    = 3/2 * I < wt
     
  19. Sep 22, 2015 #18
    Did you have a look at the angle sum identities?
     
  20. Sep 22, 2015 #19
    Ah, I see:
    i(t) = I(cos(wt)<0 + cos(wt - 120)<120 + cos(wt - 240)<240 ) = I[ cos(wt) + cos(wt + 120)*(-0.5 + j√3/2) + cos(wt + 240)*(-0.5 - j√3/2)]
    = I[ cos(wt) + (j√3/2-0.5)*(-0.5*cos(wt) + √3/2*sin(wt) - 0.5*cos(wt) - √3/2*sin(wt))]
    =I[ 1.5*cos(wt) - j√3/2*cos(wt)] ....hmm I don't have a calculator on me atm, but if we just took the real part that'd be right?
    = 3/2 * I < wt
     
  21. Sep 26, 2015 #20
    do you just take the real part?
     
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