Talk:Impedance

I changed two bits saying, essentially, "with constant voltage, capacitors in the circuit prevent current from flowing", since the capacitor could be in parallel with a resistor, for instance, and then a current would flow. Also, my statement that impedance in DC circuits is redundant is false: DC doesn't mean constant voltage, and even if the voltage is normally constant, the behavior during switch on/off is analyzed using the impedance. AxelBoldt 17:36 Feb 23, 2003 (UTC)

Nominal impedance

I am not sure what the nominal impedance article is supposed to show. I tried to update it a little, but couldn't really get it into a workable form without being too heavy on physics/math. Simplifying it too much will just give people ideas of impedance that are wrong, and inhibit further understanding. We should get a regular old guitar player to read through it and mark it up with questions they have.

I'm thinking maybe it should just be absorbed into this article, and the first section or so of this article should be a layman's introduction. Opinions? - Omegatron 21:48, Sep 13, 2004 (UTC)

One thing that confuses me

i got confused while trying to write the nominal impedance article (teaching something is the best way of learning it yourself). maybe this should be posted in the thevenin's theorem article instead.

you can create a filter out of passive components, say a 5th order chebyshev bandpass filter (if that's a possible filter). anyway, if you plotted the frequency response of this filter with a resistive load, the graph goes all over the place and makes lots of squigglies. and yet you can collapse the entire circuit down into a single voltage divider with a voltage source, R+jX source impedance, and R+jX load impedance? That seems like more complication than can fit into just those two terms. maybe I don't grasp something. - Omegatron 01:29, Sep 14, 2004 (UTC)

I just took a brief look, but I think the confusion arises because you may have forgotten that the R+jX term only applies at one frequency. That is this is another simplification. The filter is commonly specified in the frequency domain (at steady state - another simplification) and the impedance will, in general, be a function of frequency. By the way, for some filters the transient response is important too. --AJim 02:57, 14 Sep 2004 (UTC)

Hmm. For a load that is simply a capacitor, the impedance would be

<math>

Z = R + j X = 0 + j {-1 \over 2 \pi f C} = {-j \over 2 \pi f C} <math>

Right? So the R + jX is not an approximation at a particular frequency, it contains the magnitude of resistance and phase change for all frequencies (because of the f term). And somehow any combination of inductors capacitors and resistors can apparently be collapsed down into a single R + jX term (which seems suspicious to me). Perhaps during the thevenin conversion the voltage source magnitude is the part that is only valid at a specific frequency?

I'm pretty sure transient (step) response, impulse response, and frequency response are just different ways of looking at the same thing. - Omegatron 03:43, Sep 14, 2004 (UTC)

In Thevenin's theorem, the equivalent impedance is only half the story. There is also the equivalent voltage source, which can be arbitrarily complex. I believe you will find that that is where the complexity goes when you do the reduction. Gazpacho 03:57, 14 Sep 2004 (UTC)

Aha! So is it true that everything can be collapsed into a single R+jX? (I mean one for source and one for load.) - Omegatron 04:40, Sep 14, 2004 (UTC)

Yes. The input can produce a complex signal at the load, but the coupling between that signal and the load is linear. Gazpacho 05:35, 14 Sep 2004 (UTC)