Let's say we have any circuit, with two available terminals, consisting exclusively of ideal constant-resistance resistors, ideal constant-capacitance capacitors, and ideal constant-inductance inductors. We connect the two terminals of the circuit to a sinusoidal independent voltage source. Is there a frequency (or a set of frequencies) for the voltage source where in steady-state resonance occurs (the equivalent reactance of the circuit is exactly zero)?

This is a theoretical question. The frequency can be any (except infinite since we can't achieve exactly that, only approach it, and at infinite frequency capacitors acts as short-circuits and inductors as open-circuits, reducing the circuit to only resistors), and the value of the resistances, inductances and capacitances can be any constant value.

I'm not sure how to proceed to proof the above, i.e. whether there exists or not resonant frequencies for any circuit. Perhaps one way is to set the imaginary part of the equivalent impedance (seen by the source from the circuit) to zero: \$\Im [{\hat Z}_\text{eq}(\omega_0)] = 0\$ where \$\omega_0\$ is the resonant angular frequency. But what guarantees that there exists or not a particular value for the frequency that cancels the equivalent reactance? In other words, what guarantees that the equation \$\Im [{\hat Z}_\text{eq}(\omega_0)] = 0\$ has real solutions for \$\omega_0\$?

Another way is to simply find a counter-example, since that's what it takes to answer "do all circuits". So, in LTspice, I invented an arbitrary circuit, specifying the value of resistances, inductances and capacitances, I connected it to a sinusoidal voltage source with 0° of phase angle, and performed an AC Analysis sweeping the source frequency through a large range (from 1 Hz to 100 kHz). Then, I measured the negative of the current through the voltage source (since that's the current that enters to the circuit satisfying the passive sign convention in the passive circuit, and LTspice measures currents of voltage sources such that they enter into the positive terminal), and saw the plot of its amplitude and phase angle as a function of frequency. If there is a frequency where the phase angle of the current is exactly zero (and the amplitude is not zero), it means the current through the circuit is in phase with the voltage across the circuit, which happens only if the equivalent impedance of the circuit is purely resistive, so the equivalent reactance is zero.

But after doing all of that, surprisingly there was a frequency where the current phase angle was zero (so resonance occurred at that frequency). This is shown in the following figure.

Figure 1

Figure 1.

Figure 2

Figure 2.

As we can see, at \$f \approx\$ 361.415 Hz the phase angle of the current is zero and so resonance occurs as explained above. We can check this by running a phasor analysis at that frequency:

Figure 3

Figure 3.

Or by running a transient analysis at that frequency (notice the circuit voltage is in phase with the circuit current):

Figure 4

Figure 4.

  • \$\begingroup\$ Although you've gone to some effort (thanks) in writing this out, I want to make sure that you are also including a resistance-only black box. In this case, there is no meaningful resonance frequency. Perhaps DC would be your call here? \$\endgroup\$
    – jonk
    Commented Dec 24, 2021 at 20:46
  • 2
    \$\begingroup\$ What about a simple RC lowpass? It always has reactance no matter the frequency. \$\endgroup\$ Commented Dec 24, 2021 at 20:46
  • \$\begingroup\$ @jonk The arbitrary circuit must contain at least one resistor, one inductor and one capacitor, and the input voltage must be sinusoidal in time (DC is not sinusoidal even though it can be regarded as zero frequency). \$\endgroup\$
    – alejnavab
    Commented Dec 24, 2021 at 20:51
  • 2
    \$\begingroup\$ @JonathanS. True, but I meant that the arbitrary circuit must contain at least one resistor, one inductor and one capacitor. \$\endgroup\$
    – alejnavab
    Commented Dec 24, 2021 at 20:53
  • 1
    \$\begingroup\$ If the circuit contains only R, L and C components and also contains at least one each of an L and a C then there must be a frequency where the net L and net C have equal reactance and thus are in resonance. This is trivially true since the capacitive reactance is infinite at 0 Hz and zero at infinite hertz and because it varies continuously and so will take all possible values. \$\endgroup\$ Commented Dec 24, 2021 at 20:57

2 Answers 2


What you want to know can be reduced to: is it true for all random (realizable) transfer functions that there exists a frequency where the imaginary part becomes zero? The condition is for the transfer function to be made of inductors and capacitors in equal parts (e.g. for every N inductors there are N capacitors).

The answer is no. And the proofs are the lowpass (or @Jonathan S.'s comment, generalized) and the highpass (@Tesla23's answer), at least.

Both of these can be made with equal parts Ls and Cs, both of them will have either no \$s\$ in the numerator, or the same power as the denominator. That will ensure either complete, or no cancellation, whatsoever, of \$s\$, leaving the imaginary part converge to zero only at 0 (no frequency) or infinity (and beyond). Any other transfer function will have an \$s\$ that is either of power of at least 1 (greater than the 0th power of the denominator), or at least 1 less than the highest power of the denominator. For all these cases there may be a case where the imaginary part is zero at some frequency, but it's not necessarily true. E.g. a pure zero, \$s^2+\omega_z^2\$, will give a discontinuity, in which case it's a matter of limits, from \$+\infty\$ or \$-\infty\$, but this case is ideal. Or, if active elements are involved, RHP zeroes.


Looks like you are after a form of Foster's reactance theorem (https://en.wikipedia.org/wiki/Foster%27s_reactance_theorem) for lossy circuits, well the bad news is that there isn't one. Consider:


simulate this circuit – Schematic created using CircuitLab

If my quick analysis is correct, then if \$R < \sqrt{\frac{L}{C}}\$ , the input reactance doesn't vanish at any frequency.

  • \$\begingroup\$ Thanks! Interesting, I hadn't heard of that theorem. I'll wait some time before choosing this as the accepted answer, just in case someone adds their own. \$\endgroup\$
    – alejnavab
    Commented Dec 24, 2021 at 23:18
  • \$\begingroup\$ I solved the equations for this circuit too and I always got a resonant frequency where the impedance is entirely resistive for all values of R (though I can't rule out that I made a mistake somewhere). However, there is a circuit that doesn't have a resonance for certain component values: (R+L)||C. See: en.wikipedia.org/wiki/RLC_circuit#Other_configurations \$\endgroup\$ Commented Dec 25, 2021 at 0:43
  • \$\begingroup\$ @JonathanS. I also solved the equation, and agree with Tesla23 that if R^2<L/C, then the input reactance is never 0, no matter what the frequency. However, if R=0, which is less than L/C, then there is a frequency where the impedance is infinite. \$\endgroup\$ Commented Dec 25, 2021 at 1:25

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