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Complex Numbers converting from Polar form to Acos(wt + x)

  1. Sep 5, 2014 #1
    1. The problem statement, all variables and given/known data

    "Put each of the following into the form Acos(ωt+θ)..."

    (a.) 4ejt+4e-jt

    2. Relevant equations

    Euler's Identity: e = cos(θ)+jsin(θ)
    Phasor Analysis(?): Mcos(ωt+θ) ←→ Me
    j = ej π/2
    Trignometric Identities

    3. The attempt at a solution

    I attempted to use phasor analysis to simply have it be 4cos(ωt)+4cos(ωt), however, the given problem doesn't seem to follow the same form as the "examples" for phasor analysis.

    I also attempt to use Euler's identity (thus changing it into the cos()+jsin() form) but that still confused me because the example given wth Euler's is for a e not an ejt.

    I basically need a solid jump off point so I can correctly reason the answer out.
     
    Last edited: Sep 5, 2014
  2. jcsd
  3. Sep 5, 2014 #2

    Simon Bridge

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    No it doesn't. In the phasor form, the expression is the sum of two vectors - the resultant vector is what you want to express in cos form.
    Careful - the minus sign is important.

    ... put ##e^{j\theta}=e^{jt}## and solve for ##\theta##.
     
  4. Sep 5, 2014 #3
    So, basically ##j\theta=jt##, thus ##\theta=t##?

    So then we'd have ##4e^{j\theta}+4e^{-j\theta}##?

    How do I account for the negative then?

    If that's the case (and ##Mcos(ωt+\theta)## is interchangeable with ##Me^{j\theta}##) wouldn't the end result be:

    ##4cos(ωt+\theta) + 4cos(ωt+(-\theta))##?​

    I also know: ##Acos(ωt + θ)=\frac{1}{2}(X+X^∗)## where ##X=Ae^{j(ωt+θ)}=Ae^{jθ}e^{jωt}##.

    But I'm unsure how to use that info, is that even applicable in this situation (I feel a little ridiculous for not understanding this but I'm trying my best!)?
     
    Last edited: Sep 5, 2014
  5. Sep 5, 2014 #4

    Simon Bridge

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    Put ##e^{-j\theta}=e^{j(-\theta)}## ... what is special about negative angles?

    ... it isn't. ##Me^{j\theta}=M(\cos\theta + j\sin\theta)## ... you know this.
     
  6. Sep 6, 2014 #5
    Using that information we have:

    ##4e^{j\theta}+4e^{j(-\theta)}##​

    Expanding that out we have:

    ##4(\cos(\theta) + j\sin(\theta)) + 4(\cos(-\theta) + j\sin(-\theta))##​

    We know that ##\cos(\theta)+\cos(-\theta)## is ##2\cos(\theta)## due to trigonometric identities.

    We also know that the opposite is true for sine, rather that ##sin(-\theta) = -sin(\theta)##. So ##sin(\theta)+sin(-\theta) = 0##.

    If we use that knowledge we end up with:

    ##8cos(\theta)+0 → 8cos(\theta) → 8cos(t)##​

    Is that the correct thought process?
     
  7. Sep 6, 2014 #6

    rude man

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    We have a winner!

    You could also have gone

    cosθ = [e + e-jθ]/2

    so e + e-jθ = 2 cosθ

    so 4[e + e-jθ] = 8 cosθ
     
  8. Sep 6, 2014 #7
    Hah!

    It was so easy but for some reason I was battling myself the whole time thinking it was much more complicated.
    Thank you for the assistance Simon and Rude Man.
     
  9. Sep 6, 2014 #8

    Simon Bridge

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    The other approach was to draw the phasors.
    The "jt" version is an arrow, length 1, rotating anti-clockwise; while the "-jt" version is an arrow length 1 rotating clockwise. At t=0 they are both aligned on the real axis - so they add to 2.
    As the rotate in opposite directions, their imaginary components (the y-components) will always cancel.
    The real components (the x components) are the cosine of the angle.

    Well done though.
     
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