Phasors

Discussion in 'Homework Help' started by wolf12, Apr 29, 2011.

  1. wolf12

    Thread Starter New Member

    Apr 18, 2011
    3
    0
    i = 4 sin(10t + 10◦)
    Which phasor correct for the sinusoidal above?
    1. magnitude = 4 phase = -80◦
    2. maginitude = 4 phase = 100◦ (this is my answer, is it wrong ? why?)

    2nd problem

    dv/dt = −ωVm sin(ωt + φ) = ωVm cos(ωt + φ + 90◦)

    = Re(ωVm * e^jωt * e^jφ * e^j90◦ ) = Re(j * ω * V * e^jωt )

    how did j appear in Re(j ω V e^jωt )?
     
  2. jegues

    Well-Known Member

    Sep 13, 2010
    735
    43
    For the first portion of your question, phasors are usually written in the following form,

    V_{m}sin( \omega t + \phi)

    Where V_{m} is the magnitude, omega is the frequency (in rad/s) and \phi is the phase shift.

    For the second part of your question, are you trying to figure out what the derivative of this phasor is, like so,

    \frac{d}{dt}\left( -\omega V_{m}sin( \omega t + \phi) \right)

    ?
     
  3. wolf12

    Thread Starter New Member

    Apr 18, 2011
    3
    0
    1st question - Very sorry , its polar form, i cudnt find that angle symbol here, so i wrote magnitude = .. phase =..

    2nd question - Sorry Again,
    v(t) = Re(V * e^jωt ) = Vm cos (ωt+φ),

    dv/dt = −ωVm sin(ωt + φ) = ωVm cos(ωt + φ + 90◦)

    = Re(ωVm * e^jωt * e^jφ * e^j90◦ ) = Re(j * ω * V * e^jωt )

    how did j appear in Re(j ω V e^jωt )?
     
  4. jegues

    Well-Known Member

    Sep 13, 2010
    735
    43
    Okay so for the first question again, it will be of the form,

    V_{m} \quad \angle \phi

    For the second question, I don't see how you get that j down there.

    \frac{dv}{dt} = \omega V_{m}cos(\omega t + \phi + \beta), \quad \text{ Where, } \beta = 90^{o}

    So,

    Re \left{ V_{m}e^{j(\phi + \beta)}e^{jwt} \right}

    Note, e^{jwt} in simply a complex factor, the information we seek is representated by,

    \overline{V} = V_{m}e^{j(\phi - \beta)}} = V_{m} \angle (\phi - \beta)
     
    Last edited: Apr 29, 2011
  5. wolf12

    Thread Starter New Member

    Apr 18, 2011
    3
    0
    Its solved :) 1st one i have took it like 90-θ when i should take θ-90 ((ωt+10)-90)

    2nd one,
    e^{j90} = cos90 + jsin90 = 0+j = j

    Thanks :)
     
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