[sdiy] Passive filters and impedances
Tom Wiltshire
tom at electricdruid.net
Tue Nov 3 23:38:15 CET 2015
Hi Don,
Ok, you've picked the other approach - Kirchoff's law. I haven't tried that yet, but I'm willing to give it a go. Was there any particular reason you chose to do it that way rather than the simplify-the-impedances method I tried?
> Instead of turning to a symbolic math program, I think it's a lot more educational, self-reliant, and true to the DIY spirit to just do it yourself. It's not difficult, and it's an important and useful skill to have.
Agreed. That was my motivation.
> Here, I'll help get you started. For the two pole circuit, let Va be the voltage at the junction of R1, R2, and C1. The sum of the currents going into this junction must be zero.
>
> (1/R1)(Vin - Va) - (1/sC1) Va + (1/R2)(Vo - Va) = 0
This is just the application of Ohm's law with impedances around the junction, right? I get this part.
> Rearrange that into the junction order, by the V's:
>
> (1/R1)Vin + (-1/R1 - 1/R2 - 1/sC1)Va + (1/R2)Vo = 0
Ok, I see what you've done here, and now you've done it, I see what you meant by "in junction order, by the V's", although I found it unintelligible on it's own!
> Next, do the same for junction Vo, and you're well on your way.
I think I can see how I'd do the same summing-the-currents process around Vo, which will presumably give me another equation somewhat similar to the one you've just derived. But then what?
My guess - they're both equal to zero, so I can set them either side of an equals sign (e.g. they're equal to each other). Then I start rearranging them until I finish up with Vo/Vin on one side and everything else (the transfer function) on the other side?
Thanks,
Tom
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