I am working with certain input-output maps that can be thought of as large RLC-networks. I thought maybe this might be a place to get some thoughts/ideas/answers.

My basic question is, given some large connected RLC network (all linear and ideal elements) and two ports in the network, say 'a' and 'b', can the transfer-function between the ports have non-minimum phase zeros?

More generally, is it true that every transfer function representing an RLC-circuit network is minimum phase?

I suspect the answer is true, but I am having a hard time proving it.


  • $\begingroup$ Very good question :-) I don't know enough about circuit analysis to answer it off the top of my head, but hopefully someone will. $\endgroup$ – David Z Oct 14 '11 at 17:18
  • $\begingroup$ For those of us ignorant of these things, what exactly is a "non minimum phase zero"? I guess it means that the AC voltage at the two points corresponding to the ports are locked to be in phase. Also what does "minimum phase" mean? If you define the terms, even if they are well known, you have more people who can potentially answer. $\endgroup$ – Ron Maimon Oct 14 '11 at 17:19
  • $\begingroup$ Minimum phase just means all zeros on the left hand side of the complex plane. In discrete time series analysis it means all zeros inside the unit circle. $\endgroup$ – user1631 Oct 14 '11 at 17:53
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    $\begingroup$ @user1631: I see. This is asking for a proof of causality. $\endgroup$ – Ron Maimon Oct 14 '11 at 20:49
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    $\begingroup$ Not really. Causality has to do with the location of the poles. You could have the poles on the left hand side, zeros on the right hand side, and it would be causal but non-minimum phase. The question is whether this could be realized with passive components. $\endgroup$ – user1631 Oct 14 '11 at 22:18

More generally, is it true that every transfer function representing an RLC-circuit network is minimum phase?

I suspect the answer is true, but I am having a hard time proving it.

It's not true because you can have an RLC all-pass filter. To see a more specific example, let's analyse a lattice phase equaliser topology:

Lattice phase equaliser

Writing the node equations:

$$(V_A - 1)Z^{-1} + (V_A - 0)Z'^{-1} = 0\quad{\rm (node\ A)}$$

$$(V_B - 1)Z'^{-1} + (V_B - 0)Z^{-1} = 0\quad{\rm (node\ B)}$$


$$V_A(Z^{-1} + Z'^{-1}) - Z^{-1} = 0\quad{\rm (node\ A)}$$

$$V_B(Z^{-1} + Z'^{-1}) - Z'^{-1} = 0\quad{\rm (node\ B)}$$

Subtracting the equations and reordering:

$$(V_A - V_B)(Z^{-1} + Z'^{-1}) - (Z^{-1} - Z'^{-1}) = 0$$

$$(V_A - V_B)(Z^{-1} + Z'^{-1}) = Z^{-1} - Z'^{-1}$$

$$V_A - V_B = \frac{Z^{-1} - Z'^{-1}}{Z^{-1} + Z'^{-1}}$$

By linearity and definition of transfer function:

$$ H(s) = \frac{Z(s)^{-1} - Z'(s)^{-1}}{Z(s)^{-1} + Z'(s)^{-1}}$$

If we use an inductor $L$ as impedance $Z$ and a capacitor as impedance $Z'$ we get:

$$Z(s) = sL$$

$$Z'(s) = (sC)^{-1}$$

$$H(s) = \frac{\frac{1}{sL} - sC}{\frac{1}{sL} + sC}$$

$$H(s) = \frac{\frac{1 - s^2LC}{sL}}{\frac{1 + s^2LC}{sL}}$$

$$H(s) = \frac{1 - s^2LC}{1 + s^2LC}$$

$H(s)$ has zeroes at $s = \pm(LC)^{-\frac{1}{2}}$, so it cannot be minimum phase.

[Added 10/15]

Zeroes in the right half-plane can be obtained even when limited to RC circuits. To see that, consider the transfer function of this filter:

RC non minimum phase filter

We can get the node voltages directly, because both branches are generalized voltage dividers:

$$\displaystyle V_A = \frac{(sC)^{-1}}{R + (sC)^{-1}}$$

$$\displaystyle V_B = \frac{R}{R + (sC)^{-1}}$$

$$\displaystyle H(s) = V_A - V_B = \frac{(sC)^{-1} - R}{R + (sC)^{-1}} = \frac{1 - sRC}{sRC + 1} = -\frac{s - (RC)^{-1}}{s + (RC)^{-1}}$$

The general restrictions in RC (and RL) transfer functions are:

  1. All poles are simple and on the negative real axis.
  2. All residues are real but can be positive or negative.
  3. Zeros can be anywhere in the s-plane, but complex zeros must be in conjugate pairs.
  4. Zero and infinite frequency cannot be poles.

(Extracted from p. 5 of The synthesis of voltage transfer functions, the best online reference I've been able to find.)

  • $\begingroup$ Great! A clear example demonstrating the point. Thank-you. Any ideas if no inductors are allowed, so only with an RC circuit? $\endgroup$ – 1yen Oct 15 '11 at 8:20
  • $\begingroup$ @1yen I added the analysis of an example of non minimum phase RC filter. $\endgroup$ – mmc Oct 16 '11 at 1:57
  • $\begingroup$ thanks again! The reference also looks very useful and interesting, so will be having a peek at that too! $\endgroup$ – 1yen Oct 17 '11 at 6:27

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