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Complex Impedance Derivation

  1. Oct 27, 2008 #1
    I have a quick question regarding how the complex impedance of an AC circuit is derived.

    The voltage and current in an AC circuit are given by the equations
    [tex]
    V=V_{m}cos(\omega t) = V_{m}Re(e^{i\omega t})
    [/tex]

    [tex]
    I=I_{m}cos(\omega t-\phi) = I_{m}Re(e^{i(\omega t-\phi)})
    [/tex]

    The impedance is given by
    [tex]
    Z = \frac{V}{I}} = \frac{V_{m}Re(e^{i\omega t})}{I_{m}Re(e^{i(\omega t - \phi)})}
    [/tex]
    Some online resources state that
    [tex]
    Z=\frac{V_{m}}{I_{m}}e^{i\phi}
    [/tex]

    Do they actually mean that
    [tex]
    Z=\frac{V_{m}}{I_{m}}Re(e^{i\phi})
    [/tex]?

    if so...
    [tex]
    \frac{Re(e^{i\omega t})}{Re(e^{i(\omega t - \phi)})}\neq Re(\frac{e^{i\omega t}}{e^{i(\omega t - \phi)}})
    [/tex]

    What am I failing to see?
     
    Last edited: Oct 27, 2008
  2. jcsd
  3. Oct 27, 2008 #2
    Ok.. I think I've figured it out. For the benefit of others stumbling on this or similar problems, here's my attempt at a solution. I would be happy if anyone could confirm this

    The voltage and current given by
    [tex]
    V=V_{m}cos(\omega t) = V_{m}Re(e^{i\omega t})
    [/tex]

    [tex]
    I=I_{m}cos(\omega t-\phi) = I_{m}Re(e^{i(\omega t-\phi)})
    [/tex]

    have the Phasor Representation
    [tex]
    V \rightarrow V_{m}
    [/tex]​
    and

    [tex]
    I \rightarrow I_{m}e^{-i\phi}
    [/tex]

    Though the impedance [tex]Z[/tex] is given by
    [tex]
    Z = \frac{V}{I}} = \frac{V_{m}Re(e^{i\omega t})}{I_{m}Re(e^{i(\omega t - \phi)})}
    [/tex]
    its phasor representation is given by
    [tex]
    Z = \frac{V}{I}} \rightarrow \frac{V_{m}}{I_{m}e^{-i\phi}}
    [/tex]
     
  4. Oct 27, 2008 #3

    rbj

    User Avatar

    this should probably go to the Electrical Engineering forum, but makes no difference to me.

    now you realize that the 2nd or right-hand equal signs in both equations are wrong, do you?

    edit: oooops, i didn't see the "Re" in there. so the equations are not wrong, but expressing a real sinusoid as the real part of the complex exponential is a bit clunky (and i take issue with the pedagogical reasoning behind it in beginner's textbooks).

    i would express it as this:

    [tex] v(t) = V \cos(\omega t + \phi) = \frac{1}{2} \left( V e^{i (\omega t + \phi)} + V e^{-i (\omega t+\phi)} \right) [/tex]

    [tex] = \frac{1}{2} \left( (Ve^{i \phi}) e^{i \omega t} + (Ve^{-i \phi}) e^{-i \omega t} \right) [/tex]


    [tex] = \frac{1}{2} \left( \mathbf{V} e^{i \omega t} + \mathbf{V}^* e^{-i \omega t} \right) [/tex]


    so that the understanding is that for every real sinusoid, there is a positive-frequency component and a negative-frequency component. it's not hard to see that whatever happens to the positive-frequency component, precisely the same happens to the negative-frequency component except it's the complex conjugate of what was done to the positive-frequency component. to get the real result, in a linear electric circuit, we apply superposition and ask what happens if we just send the positive-frequency component through the circuit (what happens with the negative is not different, but a "mirror image" so nothing is new when investigating that).

    [tex] v(t) = \mathbf{V} e^{i \omega t} [/tex]

    you will notice that for every R, L, and C in the circuit that both the current and the voltage on every element is some constant times the same complex exponential [itex]e^{i \omega t}[/itex], so that factor (which is never zero) is factored out of all of the R, L, C volt-amp equations and it's factored out of the Kirchofff's Current Law equation for every node and out of the Kirchoff's Voltage Law for every loop. then you have simpler algebraic (with some complex terms) equations that have no function of time in them.




    what you need to do is play a little less "fast and loose" with the math. there are time-domain expressions of voltage and current (both as a function of time), and then there are "phasor" or frequency-domain representations of sinusoidal voltage and current signals. usually we use small-case letters for the time-domain functions (like [itex]v(t)[/itex]) and the corresponding large-case letters for their frequency-domain counterparts (like [itex]\mathbf{V}[/itex] or [itex]V(i \omega)[/itex]). they map to each other, but they are not equal. there's a transformation in between in there.
     
    Last edited: Oct 27, 2008
  5. Oct 27, 2008 #4
    Thank you kindly for your reply rb-j.
    I did not realize that the equations
    [tex]
    V_{m}cos(\omega t) = V_{m}Re(e^{i\omega t})
    [/tex]

    [tex]
    I_{m}cos(\omega t-\phi) = I_{m}Re(e^{i(\omega t-\phi)})
    [/tex]
    are wrong. In fact I'm not sure why the would be, given that [tex]V_{m}[/tex] and [tex]I_{m}[/tex] are just real numbers.

    Also, the transformation that you alluded to, is it
    [tex]v(t)=Re(Ve^{i \omega t})[/tex].

    Knowing that small-case letters represent time-domain functions and upper-case their frequency-domain counterparts saves me a lot confusion. Thank you.
     
  6. Oct 27, 2008 #5
    Looks like you were already ahead of me.
    Thank you again for your help.
    :)
     
  7. Oct 27, 2008 #6

    rbj

    User Avatar

    well, now i think i corrected all of my mistakes, so i hope the confusion is reduced.
     
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