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Homework Help: Convert trigo to euler form

  1. Jun 20, 2011 #1
    1. I have a complex Ohm question in which u(t) is given as Umax*[itex]\sqrt{2}[/itex]cos([itex]\varpi[/itex]+[itex]\varphi[/itex]), i know how to convert from trigo to euler form if i have both sin and cos but this doesn't. Is it possible to convert just a cos to Euler form ?





    3. Since it is a complex impedance i tried to reason that if we calculate the rms intensity we simply take the real part and leave out the imaginary part, so i by-part the conversion. Is it ok to do that ?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 21, 2011 #2

    ehild

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    cos(ωt +φ) = Re[ei(ωt +φ)]. One calculates with the exponential form, then takes the real part of the result to get the average power, and it comes out that the rms voltage or current is 1/√2 times of the amplitude of the exponential form.

    ehild
     
  4. Jun 21, 2011 #3
    Thank you, i understand now, so the given formula gives the numerical result of the Urms and the imaginary part is implied from the angle?
     
  5. Jun 21, 2011 #4

    ehild

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    I do not get you. In the original formula, the time dependence should be U(t) = Umax cos (ωt+φ), and Umax =√2*Urms. The Euler form is U0ei(ωt+φ), with U0=Umax.

    ehild
     
  6. Jun 21, 2011 #5
    Oh i meant i(t) = cos(ωt+φ) + j*sin(ωt+φ) so the imaginary part of i(t) is j*sin(ωt+φ) so for U=Ri(t), we take the real part and get U = R*cos(ωt+φ) and hence the imaginary part is implied by (ωt+φ), i.e, it is not written out but we can easily get it to be j*sin(ωt+φ). Is that not correct ?
     
  7. Jun 21, 2011 #6

    ehild

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    It is about right, but the imaginary part is defined as sin(ωt+φ), without j. But it is multiplied by j when you write the trigonometric form of the complex number. Any complex number is of the form

    z=x + j*y, x is called the real part and y the imaginary part.

    ehild
     
  8. Jun 22, 2011 #7
    Thank you i handed in the work yesterday and it seems i somehow got it right, if only the numerical values. Cheers :D
     
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