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This is a silly question to ask because the answer might be obvious, but I still have my doubts.

If given a circuit composed only of resistors, inductors, and capacitors, if its impulse response decays over time, there should be negative exponents in that response. But that means that the complex roots of the characteristic polynomial should have negative (or zero) real parts. I assumed this was because the characteristic polynomial has positive coefficients but I later found out that not all polynomials with positive coefficients have no positive real parts in their roots. For instance, take a look at \$s^5+s^4+s^3+s^2+s+1\$.

I know there might be some physical explanation to this even though I'm looking from a mathematical standpoint.

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  • \$\begingroup\$ Interesting question. +1 \$\endgroup\$
    – jonk
    Commented May 17, 2018 at 6:07
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    \$\begingroup\$ Just a quibble about the question title - the question is really about circuits containing only passive components, whether or not they are linear. There are linear active circuits where the impulse response does not decay (at least, until the signal amplitude in a real circuit increases to the point where the response is no longer linear) - for example many oscillator circuits function using linear circuit behaviour. \$\endgroup\$
    – alephzero
    Commented May 17, 2018 at 9:26

4 Answers 4

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In a circuit that is composed of only Ls, Cs and Rs ...

An input impulse will store energy in the Ls and Cs. The stored energy will be dissipated in the Rs, and will tend to zero over time.

Conversely, if there are no Rs, no means of dissipation, then the energy will remain stored, and the impulse response will last indefinitely.

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    \$\begingroup\$ The best answer, IMHO. Mathematics is a very useful tool, but it's not the only tool in the box. When a simple argument based on energy conservation answers a question about electrical engineering, advanced maths is overkill. \$\endgroup\$
    – alephzero
    Commented May 17, 2018 at 9:21
  • \$\begingroup\$ This conservation argument pairs nicely with jonk's answer; it gives a bulletproof theoretical reason why it should be impossible to form a circuit with certain transfer functions using passive components only. \$\endgroup\$
    – zwol
    Commented May 17, 2018 at 16:23
  • \$\begingroup\$ @alephzero It was the first answer that crossed my mind, too, and I agree that Neil's answer is a good one. The OP specifically wrote: "I know there might be some physical explanation to this even though I'm looking from a mathematical standpoint." Because of that statement, suggesting the OP was searching for a mathematical inconsistency with reality, I made the decision to avoid that argument. \$\endgroup\$
    – jonk
    Commented May 17, 2018 at 18:10
  • \$\begingroup\$ @jonk Exactly, but it turns out that sometimes when the laws of physics answer a question then proving through maths is no longer the best choice (in other words, I've given up). \$\endgroup\$
    – mjtsquared
    Commented May 18, 2018 at 13:37
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I suspect the problem may arrive from the requirement of an impossible domain for passive component values. Your characteristic equation breaks down (as I'm sure you already know) into: \$\left(s+1\right)\left(s^2+s+1\right)\left(s^2-s+1\right)\$. The first two factors certainly can be formed with passive components. But the last term appears to require a negative-valued component.

Can you can find any passive circuit arrangement where the characteristic equation is \$s^2-s+1\$?

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  • \$\begingroup\$ Okay, so does that mean that all characteristic polynomials derivable from passive circuit arrangements are somehow unique such that they don't take this impossible form? \$\endgroup\$
    – mjtsquared
    Commented May 17, 2018 at 5:44
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    \$\begingroup\$ @mjtsquared If you come up with an instance illustrating this form, I'd certainly be interested. If you review an answer I wrote here you can see the extraction of one such characteristic equation. Work out how I developed it and see if you can see some possible re-arrangement that could achieve such a characteristic equation -- even in theory. I would be interested. \$\endgroup\$
    – jonk
    Commented May 17, 2018 at 5:56
  • \$\begingroup\$ @mjtsquared I could trivially develop such a circuit if I could find a negative resistor value. In fact, if you were to set \$L=1\$, \$C=1\$, and \$R=-1\$ here then your characteristic equation would be demonstrated. But the domain range of component values tends to be limited to non-negative values for a passive system. (Certainly, with an active system it could be achieved, though.) \$\endgroup\$
    – jonk
    Commented May 17, 2018 at 6:06
  • \$\begingroup\$ While it will lead to the same result, the rigorous version of your answer is actually more complicated. Can you design a circuit using only RLC components (with or without negative values) with the transfer function \$1/((s+1)(s^2+s+1)(s^2-s+1))\$? Every component in the system will influence every pole or zero in the system, and it may not be obvious that poles stay on the left side of the complex plane (they do) when adding a component with positive value. \$\endgroup\$
    – Sven B
    Commented May 17, 2018 at 7:55
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The simple answer is related to the Routh-Hurwitz stability criterion. This means that it simply doesn't matter how the terms of the polynomial end up as long as all the roots have negative real part. Only by satisfying this criterion, the impulse response will decay in time.

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  • \$\begingroup\$ That's interesting. It looks like to answer this question is something I would have to study control theory for. \$\endgroup\$
    – mjtsquared
    Commented May 17, 2018 at 8:49
  • \$\begingroup\$ @mjtsquared Not necessarily, it could be just signal processing (analog or digital, but analog in this case), but not necessarily all of it (huge domain), mostly basic filtering. \$\endgroup\$
    – a concerned citizen
    Commented May 17, 2018 at 8:52
  • \$\begingroup\$ I've been doing a lot of examples lately and it seems like every one of them pass the criterion (as expected), although I still can't find any generality. I'm starting to hypothesize that there's gotta be something in KCL and KVL that allows this to happen. \$\endgroup\$
    – mjtsquared
    Commented May 17, 2018 at 16:53
  • \$\begingroup\$ Neil_UK and jonk have good answers: with only passive elements it's impossible to have positive real parts in the roots. With active elements, or digital, things change. For example it's easy to make negative resistors (analog), or force right-hand side poles directly in the transfer function (digital), and whatnot. Unless someone comes with the equivalence of antimatter as antiinductance, anticapacitance, or antiresistance, I'm afraid that polynomial you show in your question is impossible to achieve (again, with passives, only). \$\endgroup\$
    – a concerned citizen
    Commented May 17, 2018 at 17:04
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"I assumed this was because the characteristic polynomial has positive coefficients but I later found out that not all polynomials with positive coefficients have no positive real parts in their roots."

On a side note, I believe this is referring to Descartes rule of signs.

  1. When a polynomial is ordered by descending exponent, then the number of positive roots is either equal to the number of sign differences between consecutive nonzero coefficients, or is less than it by an even number.

  2. The number of negative roots can be found by changing the real axis. This is done by inverting the sign of all odd-powered terms.

The catch here is that the roots have to be real. So this means that the rule actually is:

All coefficients have the same sign \$\Leftrightarrow\$ All real roots (if any) are negative

The example that was given has complex roots, and so the rule doesn't apply to them. The real roots in the example are negative though (\$s=-1\$).

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  • \$\begingroup\$ Descartes's rule of signs only work for real roots and that's the easy part. I'm referring to the real parts of potentially complex roots of the characteristic polynomial. \$\endgroup\$
    – mjtsquared
    Commented May 17, 2018 at 8:33
  • \$\begingroup\$ My answer was intended to accompany your statement: "I assumed this was because the characteristic polynomial has positive coefficients but I later found out that not all polynomials with positive coefficients have no positive real parts in their roots." It may help other people that stumble on this question. \$\endgroup\$
    – Sven B
    Commented May 17, 2018 at 8:42

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