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I'm currently considering this circuit:

enter image description here

Which is really just two RL filters added together. A single RL-filter looks like this:

enter image description here

I want to be able to multiply the transfer functions of the left and right circuit to be determine the net transfer function.

Now the literature I'm using tells me that I must make sure that my right circuit has a high input impedance, and that my left circuit has a low output impedance. The concept of impedance isn't new to me, but input and output impedances are. Luckily the literature provided some definitions.

First, it defined the input impedance of my right circuit as the impedance seen by a voltage source attached to its input port. I intuitively understand why this quantity must be high: the current at the R1-L1 junction must be split up, and the higher the input impedance, the less current flows into my right circuit. And that's exactly what I want!

But I cannot for the life of me figure out what output impedance means in this context. The literature I'm using tells me that it is "the impedance a voltage source would see if it was attached to the output of the circuit, when the input of the circuit is short circuited". But how is this applicable to the situation of the two linked circuits? Surely, if Vin were applied in between L1 and L2, and "Vin" in the picture were short circuited, then this definition of output impedance would be applicable, and I would want it to be very low so that little voltage and a lot of current is given to that part of the system. But that is not the case; there is a seperate voltage supply at the left hand side of the circuit, meaning that I could never replace the left circuit with a component of fixed impedance--whereas I could do that with the right circuit.

So what would be a good defintion for output impedance in this context, and why would I want it to be low? And what does it have to do with the quoted circuit? (I'm assuming the definition is correct, or at least applicable, but I'm just not seeing it)

Heatherfield
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  • hint: what is the internal impedance of an ideal voltage source that would be driving the circuit at Vin? – hyportnex Feb 06 '19 at 22:43
  • It's zero--although adding an extra resistor doesn't make much difference at this state, haha. I think I understand what you're getting at but that's not the problem I'm seeing; the problem is that wheras I can model the right circuit as a "black box" with a set impendance, I feel like I cannot do the same to the left circuit. After all, the voltage source, ideal or not, is keeping the series circuit at a certain voltage. Wouldn't I lose that detail if I'd exchange the whole left circuit for a "black box", thereby erasing the voltage source? – Heatherfield Feb 06 '19 at 22:51
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    Calculate the transfer function of two resistors in series. You'll find that the result is only the product of the two resistors' individual transfer functions under exactly the conditions you're asking about here. By doing the explicit calculation, you'll also understand why. – DanielSank Feb 06 '19 at 23:17

1 Answers1

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But I cannot for the life of me figure out what output impedance means in this context.

It's the same thing here as anywhere:

$$Z_{out}=\frac{dv_{out}}{di_{out}}$$

(where we use the passive current convention and define $i_{out}$ as positive when it flows in to the output port).

But how is this applicable to the situation of the two linked circuits?

It's applicable to the two cascaded circuits, because the output of stage 1 has to drive the input of stage 2. And if we don't have $Z_{out(1)}\ll Z_{in(2)}$ then the loading provided by stage 2 (current flowing into stage 2's input rather than through stage 1's inductor) will change the behavior of stage 1.

Surely, if Vin were applied in between L1 and L2, and "Vin" in the picture were short circuited,

What you're really interested in is $\frac{dv_{out}}{di_{out}}$. What you quoted about connecting a voltage source is just an easy way to calculate this derivative in a model of the circuit.

The Photon
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  • That's a much nicer definition, thank you very much! You write that the circuit I described is a method of calculating that derivative; do you have any resources explaining why that is so? – Heatherfield Feb 07 '19 at 00:13
  • Because it's a linear circuit, you know if the output voltage is 0, then output current must also be 0. So you only need one more point on the i-v curve to determine $\frac{dv_{out}}{di_{out}}$. – The Photon Feb 07 '19 at 00:20
  • Ah, that explains why I'm adding a voltage source in the "middle" but it still feels naughty to short the original Vin; isn't that still a component that will in the end determine the $i_{out}$ I'm looking for (as second point on the curve)? – Heatherfield Feb 07 '19 at 00:29
  • You're not inserting a source in the middle. You're separating your two stages and evaluating each one separately. One way to do evaluate your filter stage (or any other linear 2-port network) is to model it with voltage or current sources attached to the ports and calculate the Z-parameters. You wouldn't attach a source in the middle if you were evaluating your 2-stage filter as a whole. – The Photon Feb 07 '19 at 00:45
  • My apologies, what I meant was: normally the input port of my filter is not shorted, and neither is the output port. When I use my filter in practise, I can apply a voltage to the input-port and measure a voltage at the output-port. The question I meant to ask was: I understand that I can just as well apply a voltage at the output port to model the output impedance response, but doesn't shorting the input port drastically change my circuit? – Heatherfield Feb 07 '19 at 00:51
  • If you're driving the input with a low-impedance source (something approximating an ideal voltage source), then that will look like a short circuit relative to any stimulus provided on the output port. If you are working in audio frequencies, this is a reasonable way to model your circuit. If you were actually calculating the Z parameters of your 2-port, you'd want to terminate the input with an open when calculating the effects of stimulus at the output. Because you're calculating the voltage response to current inputs, you want the current on the port you're not "testing" to be 0. – The Photon Feb 07 '19 at 01:33
  • But you might not really want the canonical Z-parameters here. Just knowing the effective output impedance when driven by a low impedance generator may be more useful for the way you're using this filter. – The Photon Feb 07 '19 at 01:35