enter image description here

I have a parallel circuit and I need to find its resonant frequency. However, there is only a capacitor paralleled with a resistor and there is no inductor included. So is it correct that in using the formula \$f_{res}=\frac{1}{2\pi\sqrt{LC}}\$ I should just assume that L is non-existent?

I am just learning about RLC circuits and I am struggling when simplifying and solving these types of circuits.

  • \$\begingroup\$ What you actually drew is an RC filter. It will have a cut-off frequency, but it won't have a resonance. electronics.stackexchange.com/questions/350482/… \$\endgroup\$
    – Kyle B
    Commented Nov 22, 2022 at 5:58
  • \$\begingroup\$ @char can you ask a specific question? \$\endgroup\$
    – Voltage Spike
    Commented Nov 22, 2022 at 6:14
  • \$\begingroup\$ @VoltageSpike I have to solve for the resonant frequency given that circuit. However, I am not sure how to solve it because base on the formula it needed the value for L or the inductance, my problem is, there is no inductor in the circuit so how do I solve for the resonant frequency. \$\endgroup\$
    – char
    Commented Nov 22, 2022 at 6:33
  • 1
    \$\begingroup\$ Because we don't know what transfer function TF is needed (a voltage gain, an input impedance?), I will assume that the stimulus is the input source implying a voltage gain. In this case, the denominator is \$D(s)=1+s(R_1C_1+C_2(R_1+R_2))+s^2R_1C_1R_2C_2\$ which you can put under a normalized form \$D(s)=1+\frac{s}{\omega_0Q}+(\frac{s}{\omega_0})^2\$. You see a resonant frequency and identification between the terms is easy. But considering the very low \$Q\$, you can factor \$D\$ as two cascaded poles. \$\endgroup\$ Commented Nov 22, 2022 at 7:54
  • 1
    \$\begingroup\$ Your circuit is a passive RC-bandpass. At a certain frequency (midfrequency) the transfer function has maximum gain and a zero phase shift. This frequency could be called also "resonant frequency" because the definition for resonance is fulfilled (real transfer function, zero phase shift) \$\endgroup\$
    – LvW
    Commented Nov 22, 2022 at 8:13

3 Answers 3


Since the difference between the two RC time constants is quite large, separate answers are given for the two as a first approximation. Thus, the lower cut-off frequency will be 1 kHz, so 1/(2πR1C1). And the upper cut-off frequency will be 528 kHz, so 1/(2πR2C2). enter image description here The amplitude decreases in proportion to the two capacitors if we look at the circuit together. The frequencies are slightly shifted, but not significantly. enter image description here At zero phase shift, the "resonant frequency" is 23.7 kHz. enter image description here


I am disappointed by the comments so far. OP states that he is completely new to the subject and I'm certain that the comments (except for Verbal Kint's) has done nothing but confuse OP even more.

A second order low pass filter will have a transfer function of the general form

$$H(s) = \frac{V_o(s)}{V_{in}(s)} = \frac{\omega_0^2}{s^2 + 2\zeta\omega_0s + \omega_0^2} \tag1$$

where \$\omega_0 \$ is called the undamped resonance frequency and \$\zeta \$ is called the damping ratio.

The reason there may arise confusion in this question, is because a transfer function is the ratio between an input and an output. In your schematic, it is not clear what the output is, but I think it's very likely that the voltage across \$C_2\$ is the output.

Your task

You need to find the ratio \$H(s) = \frac{V_{C_2}(s)}{V_{in}(s)} \$ and obtain it in the form as shown above. There are many ways of doing this - one way is with node analysis. I can set the node equations up for you, but you will have to solve them yourself. Call the voltage across \$R_1\$ for \$V_1\$ and the voltage across \$C_2\$ for \$V_{C_2}\$ and ground the bottom terminal of the voltage source:

$$\begin{cases} \frac{V_1-V_{C_2}}{R_2} + (V_1-V_{in})sC_1 + \frac{V_1}{R_1} = 0 \\\\ V_{C_2}sC_2 + \frac{V_{C_2} - V_1}{R_2} = 0 \end{cases} $$

Solving these should bring you close to (1).

Answers to Lwv's questions in the comments

Do you agree that the classical definition of "resonance" is based on a phase shift of zero deg ?

No. I use this definition of resonance from Lathi: "resonance is observed when the input signal is identical to or very similar to a characteristic mode of the system".

how such a lowpass can show a resonance effect?

Consider a completely undamped low pass filter (impossible to realize in real life)

$$H(s) = \frac{100}{s^2+100} $$

This system has an undamped resonance frequency of \$10 \: \text{rad/s} \$ which can be verified by inspecting the impulse response: -

enter image description here

$$\omega_0 = \frac{2\pi}{T} = \frac{2\pi}{(0.788-0.158) \: \text{s}} = 10 \: \text{rad/s} $$

Now, what happens if I feed the system a signal with a frequency that is equal to its own resonance frequency? As Lathi says it himself "this would be like hiring an alcoholic to taste liquor. An alcoholic would gladly do the job without pay. Think what will happen if he were paid by the amount of liqour he tasted! The system response will be very high".

This is the response I get if I feed a 10 rad/s sine-wave into the system: -

enter image description here

  • \$\begingroup\$ Best add the bit about \$\omega_0=1/(R_1R_2C_1C_2)\$ to differentiate between what will be vs what OP expects (e.g. OP expects an L, that's his confusion). \$\endgroup\$ Commented Nov 22, 2022 at 10:45
  • \$\begingroup\$ @Carl Because you have mentioned the term "confusion": According to your explanation, the frequency ωo is called the undamped resonance frequency. To me, this sounds a bit strange (and really will confuse the reader). In nearly all relevant books and contributions this quantity (wo) is called "pole frequency". Do you remember that the classical definition of resonance is based on a phase shift of zero deg? Only in case of a bandpass the pole frquency is identical to the resonant frequency. But you were speaking of a lowpass only!! Why? \$\endgroup\$
    – LvW
    Commented Nov 22, 2022 at 11:05
  • \$\begingroup\$ @LvW I don't find it strange at all to call \$\omega_0\$ the undamped resonance frequency. Have you tried plotting the impulse response of an undamped (\$\zeta = 0 \$) second order circuit? To me, calling \$\omega_0 \$ for "pole frequency" does not make sense. I think that term will definitely confuse OP more. The only book I've used in my electronics courses that wasn't completely garbage was "Electrical Engineering - Principles and Applications" By Hambley, and he calls \$\omega_0\$ for undamped resonance frequency. \$\endgroup\$
    – Carl
    Commented Nov 22, 2022 at 12:06
  • \$\begingroup\$ @Carl At the beginning of your contribution you show the classical 2nd-order lowpass transfer function containing the quantity wo (you call it "resonant frequency"). May I ask you (a) how such a lowpass can show a resonance effect and (b) if you agree that the classical definition of "resonance" is based on a phase shift of zero deg ? This frequency wo is called either "undamped natural frequency" or "pole frequency" because its value desribes - together with a characteristic angle - the pole position in the complex frequency plane. \$\endgroup\$
    – LvW
    Commented Nov 22, 2022 at 15:42
  • \$\begingroup\$ Addendum: For example, the standard textbook book "Microelectronic Circuits" (Sedra Smith) uses the term "pole frequency" for the frequency which describes the position of the poles in th s-plane. \$\endgroup\$
    – LvW
    Commented Nov 22, 2022 at 15:56

If we assume that the response is observed across capacitor \$C_2\$, we can use the fast analytical circuits techniques or FACTs and show how quick it is to determine this transfer function (TF). This is a good example for my on-going crash-course that I currently write as a follow-up to the first one.

First you start by zeroing the excitation and replace the input source by a short circuit. Then, you temporarily remove each capacitor and "look" through its respective terminals to infer the resistance \$R\$ you see. Once you have \$R\$ for each capacitor, you can assemble the time constant as \$\tau=RC\$. And what is cool is that it is all done by inspecting the circuit - you read it in your head - without a line of algebra. The below picture shows the process:

enter image description here

Then you proceed with high-frequency gains - instead of going through an NDI, null double injection - you use the generalized transfer function documented by Ali Hajimiri and it is quite straightforward here. Assemble all the time constants to obtain the transfer function. Now, you can rearrange it to reveal the mid-band gain because we are always determining transfer functions for an upcoming design exercise: what components values will lead me to the response I want. This is what is called design-oriented analysis or D-OA dear to the late Dr. Middlebrook. The equation you have must serve a design purpose and reveal the salient points of the circuit. This is what I've done in the lower side of the above picture in which the mid-band gain appears and is clearly expressed in terms of the components values.

Then test the results in a Mathcad sheet, comparing the response from a brute-force expression with the low-entropy formula: they are identical:

enter image description here

There is one zero at the origin - a dc block - and a pole which form the bandpass response.


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