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The following circuit:

enter image description here

can be described by:

$$\frac{d^2v_c}{dt^2} + \frac{R}{L}\frac{dv_c}{dt} + \frac{1}{LC}v_c = \frac{R}{LC}i_S$$

From there, it's clear that the undamped natural frequency, or undamped resonance frequency, or simply resonance frequency, is:

$$\omega_0 = \frac{1}{\sqrt{LC}}$$

\$\omega_0\$ indicates the frequency at which the circuit resonates, right? In other words, if the circuit is driven by a sinusoidal signal of frequency \$\omega_0\$, it will resonate, correct?

The characteristic equation:

$$s^2 + \frac{R}{L}s + \frac{1}{LC}$$

Has 2 (potentially complex) roots, \$s_1\$ and \$s_2\$, which are called natural frequencies.

As far as I understand, the natural frequencies are the frequencies at which the circuit will oscillate if it is excited and no drive force is present. For instance, if \$i_S\$ produces a pulse of 1 Amp. for 1 second at \$t = 0\$. Then, the circuit will oscillate at these frequencies from \$t = 0^+\$ onwards, until \$R\$ dissipates all the energy, or for ever in case \$R\$ is not present and all elements are ideal. Is this correct?

If so, how can it have 2 natural frequencies? At which one of those will the circuit oscillate, \$s_1\$ or \$s_2\$?

Finally, let's suppose the inductor is removed from the circuit. Then, it becomes a first order circuit, and its characteristic equation has only one root. This root is called natural frequency. What does this natural frequency indicate, given that first order circuits can't oscillate?

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  • \$\begingroup\$ Notice that we can use Laplace transform to write: $$\text{V}_\text{C}\left(t\right)=\mathscr{L}_\text{s}^{-1}\left[\frac{\displaystyle1}{\displaystyle\text{sC}}\cdot\frac{\displaystyle\text{R}}{\displaystyle\text{R}+\text{sL}+\frac{\displaystyle1}{\displaystyle\text{sC}}}\cdot\text{i}_\text{source}\left(\text{s}\right)\right]_{\left(t\right)}\tag1$$ \$\endgroup\$ Commented Dec 10, 2023 at 21:34
  • \$\begingroup\$ I haven't studied anything about transforms yet I'm afraid. \$\endgroup\$
    – Dan
    Commented Dec 10, 2023 at 21:47
  • \$\begingroup\$ @Dan If \$\left(\frac{\text{d}}{\text{d}t}-\alpha\right)\$ is an operator that can be applied to \$w\$, as in the homogeneous equation \$\left(\frac{\text{d}}{\text{d}t}-\alpha\right)w=0\$, then solutions are of the form \$w_t=A_0\exp\left(\alpha\,t\right)\$. Here, \$\alpha\$ is a frequency. It must be true that \$\alpha\,t\$ is unitless. Does the fact that \$\alpha\$ is a frequency mean oscillation??? Of course not. Yet it is still a frequency. This answers the part of your question about 1st order and the confusion of it with the word: oscillation. \$\endgroup\$ Commented Dec 11, 2023 at 1:19
  • \$\begingroup\$ @Dan The term spin was applied to describe a specific set of observed behaviors in experiments. But it is the math that describes its meaning. Not the word spin. And if you lose yourself trying to rework spin in many different macro-scale ways, you'll only be lost to reality. Don't waste time over-parsing words that humans associate with some math to communicate between each other. Look at the math and/or geometry when you want to understand something. \$\endgroup\$ Commented Dec 11, 2023 at 1:20
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    \$\begingroup\$ @Dan I am as well. Understood. Just as a final note: The \$\alpha\$ I wrote above in that general form solution is a frequency. So is the \$\beta\$. But the \$\alpha\$ is involved in the rate of the declining nature (it's negative here) of the last factor there, which may of course oscillate on \$\beta\$. Anyway, there is so much more to all this. See if you can find a meetup of some kind. I learned a great deal from a very good mathematician when my wife met his wife and we then started visiting each other once or twice a month. It can happen like that, too. \$\endgroup\$ Commented Dec 13, 2023 at 3:53

3 Answers 3

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Then, the circuit will oscillate at t̲h̲e̲s̲e̲ f̲r̲e̲q̲u̲e̲n̲c̲i̲e̲s̲ from t=0+ onwards, until R dissipates all the energy, or for ever in case R is not present and all elements are ideal. Is this correct?

Correct, but partially. I marked the problematic section in the quote above. I'll explain later.

If so, how can it have 2 natural frequencies? At which one of those will the circuit oscillate, s1 or s2?

You first write the transfer function in the following form:

$$ H(s)=a \ \frac{\omega_n^2}{s^2+2\zeta \omega_n \ s + \omega_n^2} $$

Here \$a\$ is the gain (in your case it's 1 as there's no amplification), \$\omega_n\$ is the natural (resonance) frequency, and \$\zeta\$ is the damping ratio.

Now you'll see that the transfer function has two roots:

$$ s_{1,2}=-\zeta \omega_n ±j\omega_n\sqrt{1-\zeta^2} $$

The imaginary part of the roots' equation above is related to oscillation. And as you can see, there's only one natural (oscillation) frequency and it's \$\omega_n\$. For oscillation, the damping ratio should be \$0\leq\zeta<1\$ otherwise the circuit can't oscillate. In your case, having R removed from the circuit (i.e. infinite resistance) corresponds to \$\zeta=0\$ meaning pure oscillation without any damping i.e. constant amplitude.

Finally, let's suppose the inductor is removed from the circuit. Then, it becomes a first order circuit, and its characteristic equation has only one root.

Correct.

This root is called natural frequency.

No, it is not. I don't know where you got this info from, but it's wrong. The single root of a first-order system is NOT called "natural frequency". I'll explain this below.

What does this natural frequency indicate, given that first order circuits can't oscillate?

There's no oscillatory behaviour, so maybe we shouldn't use "frequency" here.

You may remember, that the analysis of a system's response can be done in two parts: forced response (zero "initial" state e.g. you apply a step input but all the initial conditions are zero), and natural response (zero input e.g. you removed the input). The total response is the sum of these two (remember superposition).

Now assume you applied a step input to the circuit (turned the voltage source on whilst all other initial conditions were zero), and after some time you removed the input. So the natural response of a simple RC circuit in which the output is the capacitor voltage becomes

$$ \mathrm{V_O(t)=V_C(t)=V_{C0} \ e^{-t/RC}} $$

where \$\mathrm{V_{C0}}\$ is the capacitor's voltage just before the input is removed. The pole of the transfer function is \$\omega_p=1/RC\$, yes, but the \$RC\$ here is the time constant of the natural response, so \$\omega_p\$ is the reciprocal of this time constant, not "natural frequency".

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  • \$\begingroup\$ Thanks. Sorry, let me see if I understand: then wd (damped natural frequency) is the frequency at which the circuit will oscillate. This is true for the time domain, when the circuit is fed with a step input for instance. Also, w0 is the resonant frequency, which is strictly different than wd. This is the frequency a resonant circuit must be fed with to resonate, that is, to reach its peak gain. \$\endgroup\$
    – Dan
    Commented Dec 12, 2023 at 11:06
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    \$\begingroup\$ @Dan \$\omega_d\$ is the damped frequency which can be obtained from the imaginary part of the roots' equation above i.e. \$\omega_d=\omega_n \ \sqrt{1-\zeta^2}\$. And yes, \$\omega_d\$ and \$\omega_n\$ are different, but can be the same when there's no damping i.e. \$\zeta=0\$ because damped natural frequency becomes equal to the natural frequency, at which the tank tends to oscillate. If you apply the resonance frequency to an RLC tank the L and C will simply cancel each other (e.g. -A + (+A) = 0) so you'll get the maximum amplitude possible. \$\endgroup\$ Commented Dec 12, 2023 at 14:25
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You must make your scope wider. Let's rewrite the differential equation in case current Is=0 i.e. the natural oscillation can be the major phenomena:

enter image description here

This means that there's a mechanism which recreates Vc by applying to Vc a linear operation (derivatives, multiplications by numbers and summing the results). The functionality is the same as in a linear system where the input is connected directly to the output.

The possible functions Vc(t) are called natural responses.

Which of the possible natural responses actually happen depend on boundary conditions and, of course, they vary widely depending on L, C and R. The natural response Vc is a decaying exponential when L=0; the resistor discharges the capacitor if someone had put there some voltage just before the starting point of the time.

When L and C are both >0, the natural response Vc is a decaying sinusoidal oscillation (assumed R is small enough). It never decays if R=0, it oscillates infinitely. The oscillation frequency is called "natural frequency". There's 2 possible natural frequencies which have only different signs, the absolute value is the same for both.

You may say: "What the f**k, its not a linear system with direct wire feedback, it's an LRC circuit!" Yes it is an LRC circuit, but in math it behaves like the mentioned linear system and at least I have found it to be an useful way to think it. It works also when current Is is not zero. Current Is can be considered to be added to the input in a summing junction, but as fed through a multiplication block (an amplifier). I'm sure one day you are forced to accept my approach if you start to work with feedback control theory or its applications in electronics, say sinewave oscillators for a start.

Thinking the fundamental operating law of a system as "output must be the same as the input" will have much more applications than the mentioned ones. For example you can construct with it the numerical simulation of the system. It would much more intuitive if you didn't use a 2nd degree differential equation for a single state variable, but instead of it use a group of two 1st degree equations for two state variables, namely for the capacitor voltage and the inductor current.

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With regard to a generalized stable second order system which has some damping (damping ratio > 0)....

There are three particular frequencies of interest.....

  1. The undamped natural frequency. This is sometimes referred to as the pole frequency and is denoted wp, wn or w0. In the frequency domain it is the frequency where the lag is -90 degrees and the magnitude, Av(wn)/Av(dc) is equal to Q (the quality factor) which is equal to 1/(2 * zeta) where zeta is the damping ratio. In the time domain it is the frequency at which the system will theoretically oscilate forever (at a constant pk to pk amplitude of twice the amplitude of the step input) if there is no damping (zeta = 0), when subject to a step input.

  2. The damped natural frequency. This is denoted wd. This frequency is a bit lower than wn and, in the time domain, it is the frequency at which the system will oscilate at when subject to a step input. Because there is damping in the system (zeta > 0) the system will oscillate at wd with decaying amplitude oscillations.

  3. The resonant frequency sometimes referred to as the natural frequency or peaking frequency. Denoted wpk. In the frequency domain this is the frequency at which the peak in the magnitude response occurs. It is a little lower in frequency than wd.

So we have 3 frequencies of interest and when there is damping in the system wpk < wd < wn. As the damping ratio is reduced wpk and wd move up in frequency towards wn until when the damping ratio (zeta) becomes zero (no damping), wpk = wd = wn and the resonant frequency (peaking frequency) occurs at wn.

So in the time domain, when zeta is zero, the system theoretically oscillates for ever at constant amplitude (when subject to a step input) and in the frequency domain the system theoretically peaks with an infinite magnitude resonant peak at wn. In this situation with no damping, the system could be referred to as marginally stable with its poles on the imaginary axis of the s-plane.

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