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Differentiating a sinusoid

  1. Apr 23, 2013 #1
    The textbook says that differentiating a sinusoid is the same as multiplying the phasor by jω. Shouldn't jω be jωejωt?
     

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  3. Apr 23, 2013 #2

    tiny-tim

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    Hi Turion! :smile:
    No, you start with Re(Vejωt),

    and differentiating gives you Re(jωVejωt). :wink:
     
  4. Apr 23, 2013 #3
    [tex]So:\\ v(t)=Re(\overrightarrow { V } { e }^{ jωt })\\ \frac { dv(t) }{ dt } =Re(j\omega \overrightarrow { V } { e }^{ jωt })\\ \frac { dv(t) }{ dt } =j\omega v(t)\\ \\ But\quad the\quad book\quad asserts\quad that:\\ \frac { d\overrightarrow { V } }{ dt } =\overrightarrow { V } j\omega [/tex]
     
  5. Apr 23, 2013 #4
    The book asserts that if V1 and V2 are phasors that represent v(t) and dv(t)/dt, respectively, then they are related by:

    V2 = jωV1
     
  6. Apr 23, 2013 #5
    Isn't that what I just posted?
     
  7. Apr 23, 2013 #6

    tiny-tim

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    No, it asserts that [tex]\frac { dv}{ dt } \leftrightarrow \overrightarrow { V } j\omega [/tex]

    (You can't have dV/dt anyway, since V (the "complex voltage") is a constant! :wink:)

    See eg https://www.physicsforums.com/library.php?do=view_item&itemid=303
     
  8. Apr 23, 2013 #7
    You wrote:
    Re(jωVejωt) = jωv(t)

    which isn't true. The LHS is real, the RHS is complex.

    V is a constant, so the last line is:
    0 = jωV

    Edit:
    tiny-tim beat me to it.
     
  9. Apr 23, 2013 #8
    [tex]So\quad \frac { dv }{ dt } =\overrightarrow { V } j\omega \quad is\quad right\quad and\quad \frac { dv }{ dt } =vj\omega \quad is\quad wrong.\\ How\quad did\quad the\quad author\quad derive\quad that\quad \frac { dv }{ dt } =\overrightarrow { V } j\omega ?[/tex]
     
  10. Apr 23, 2013 #9

    f95toli

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    You need to keep track of V and v.
    They represent the voltage in different domains: one is real and the other complex, and they never "live" in the same space. Hence, they are not the same thing.

    Phasors is just shorthand. What you are "really" doing is applying a Fourier transform to a differential equation. This is not something you have to worry about as long as you keep the usual rules for phasors in mind, but it helps explain why V and v are fundamentally very different.
     
  11. Apr 23, 2013 #10

    tiny-tim

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    No!!

    As both milesyoung :smile: and I have said, V is constant.

    dv/dt is not constant, it's sinusoidal, it can't equal jωV.

    Your book doesn't say =, it says ##\leftrightarrow##.

    The v and V methods are completely different and incompatible (see the link above).
     
  12. Apr 23, 2013 #11
    There's a difference? I wasn't aware.
     
  13. Apr 23, 2013 #12

    tiny-tim

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    erm … yes!!!!

    Books don't use fancy symbols like that if they don't need to!

    = means that the LHS and RHS (as f95toli :smile: says) "live" in the same space.

    ##\leftrightarrow## is used to show that the LHS and RHS are completely different, but that they represent the same thing.

    Turion, you need to read all about complex voltage and current again, from the start.
     
  14. Apr 23, 2013 #13
    I don't see a derivation in your library entry as attached.

    The issue is that I don't see how the author derived ##\frac { dv }{ dt } \leftrightarrow\overrightarrow { V } j\omega##.
     

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  15. Apr 23, 2013 #14

    tiny-tim

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    I don't see how you don't see it, both in the book and in the library entry. :confused:

    You really need to spend an hour studying this, from the start.​
     
  16. Apr 24, 2013 #15
    Normally this type of problem is encountered in transmission line theory.Here the derivative must be replaced by partial derivative.V is being converted to frequency domain in this problem.This is called the phasor form.This type of conversion reduces the computation steps involved in the time domain form.If you again want to convert to time domain form then multiply the right hand expression of the phasor form with exp(jwt) and then take the real part i.e. only cosine part.
     
  17. Apr 24, 2013 #16
    [itex]\partial[/itex]V/[itex]\partial[/itex]t=Re(jwVsexp(jwt)).
    This means taking the real part ,of phasor form multiplied with exp(jwt), gives you back the time domain form of V.
    Here phasor form is jwVs.
     
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