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I have the following circuit and problem.

enter image description here

What is the resonance frequency of the filter, \$ \frac{\mathbf{I}_{out}}{\mathbf{I_{in}}} \$?

I found this rather straight forward formula for the resonance frequency for a parallel RLC circuit.

\$f_{resonance}=\frac{1}{2\pi \sqrt{LC}} =\frac{1}{2\pi \sqrt{62\text{nF} \cdot 63\text{nH}}}=2.55 \text{MHz}\$

Okay, so from the formula it's seems like the the resonant frequency is about 2.55 MHz, but is this right? Is the resonant frequency independent of the resistance? Why even include it then?

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4 Answers 4

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The study of this simple filter can be undertaken applying the current division law and then you develop and rearrange the expression:

enter image description here

Or you apply the fast analytical circuits techniques or FACTs and obtain the result in a few minutes, the time to write this answer: reduce the excitation to 0 A (open-circuit the left-side source) and "look" at the resistance offered by the connecting terminals of the capacitor and the inductor. Collect the time constants in this mode, assemble them to form the denominator and there you go for this simple circuit:

enter image description here

Considering a low quality factor, you can even apply the low-\$Q\$ approximation and obtain a new expression with two cascaded poles. The resulting transfer function plot is there:

enter image description here

No need for KVL or KCL, the FACTs get you to the well-ordered transfer function by inspecting the circuit without writing a line of algebra.

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  • \$\begingroup\$ Ahh okay, it's good to see that you also get 2.55 MHz resonant frequency. Thank you. I know it's not nice to intrude you like this, but would you help me with another problem of mine? I really need some help with it. That is of course, if you have the time to do it, sir. Here is the link: electronics.stackexchange.com/questions/492444/… \$\endgroup\$
    – Carl
    Apr 14, 2020 at 7:05
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    \$\begingroup\$ I saw it last time but this is 6th-order circuit with up to 20 terms for some of the \$s\$ coefficients - see this link for the FACTs at work with a 6th-order network. Furthermore, you have 3 independent generators, correct? Is this a school exercise of some sort or a real problem to solve? \$\endgroup\$ Apr 14, 2020 at 7:21
  • \$\begingroup\$ There are actually 4 independent sources. And yes this is an exercise made by instructor at my university. I have tried simulating the program with LT-spice, but I'm not certain that I get a correct result. \$\endgroup\$
    – Carl
    Apr 14, 2020 at 7:29
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Is the resonant frequency independent of the resistance?

The natural resonant frequency is independent of the series resistance.

Why even include it then?

The inductor has series winding resistance so you can't ignore it. If you were doing a more complex analysis of the circuit like trying to find the magnitude of the transfer function (for instance), you need to include it.

What is the resonance frequency of the filter, \$\dfrac{I_{out}}{I_{in}}\$?

No, that's a transfer function and not the resonant frequency.

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  • \$\begingroup\$ Hmm alright, I guess I'm on the right track then. Although I wonder my instructor wrote the question like that, and why he included the current-arrows in the circuit schematic. \$\endgroup\$
    – Carl
    Apr 13, 2020 at 15:58
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    \$\begingroup\$ It's a very poor resonant tuned circuit by the way - are you sure the inductance is 63 nH and not 63 uH? I say this because 63 uH gives a very nice frequency response whereas with 63 nH the response is almost the same as if the inductor were zero in value. \$\endgroup\$
    – Andy aka
    Apr 13, 2020 at 16:00
  • \$\begingroup\$ My guess is as good as yours, since this is a problem designed by my electrical circuits 2 instructor, although I'm quite sure it is meant to be 63nH. I have inserted a picture of the full problem, to show you exactly what I myself am seeing. \$\endgroup\$
    – Carl
    Apr 13, 2020 at 16:17
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    \$\begingroup\$ Hmm it doesn't sound right to me! \$\endgroup\$
    – Andy aka
    Apr 13, 2020 at 16:20
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    \$\begingroup\$ The value of inductance is so low that the circuit performs pretty much like an RC network i.e. there's no resonant peak to speak of and Q = 0.0246 i.e. a pretty poor example of a resonant filter. But, you are at the mercy of the instructors. \$\endgroup\$
    – Andy aka
    Apr 13, 2020 at 16:29
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Well, we have the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

When we use and apply KCL, we can write the following set of equations:

$$\text{I}_\text{x}=\text{I}_1+\text{I}_2\tag1$$

When we use and apply Ohm's law, we can write the following set of equations:

$$ \begin{cases} \text{I}_1=\frac{\text{V}_1}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_1-\text{V}_2}{\text{R}_2}\\ \\ \text{I}_2=\frac{\text{V}_2}{\text{R}_3} \end{cases}\tag2 $$

Substitute \$(2)\$ into \$(1)\$, in order to get:

$$ \begin{cases} \text{I}_\text{x}=\frac{\text{V}_1}{\text{R}_1}+\text{I}_2\\ \\ \text{I}_2=\frac{\text{V}_1-\text{V}_2}{\text{R}_2}=\frac{\text{V}_2}{\text{R}_3} \end{cases}\tag3 $$

Solving \$(3)\$ for \$\text{I}_2\$, gives:

$$\text{I}_2=\frac{\text{I}_\text{x}\text{R}_1}{\text{R}_1+\text{R}_2+\text{R}_3}\tag4$$

So, when we have the transfer function we get:

$$\mathcal{H}:=\frac{\text{I}_2}{\text{I}_\text{x}}=\frac{1}{\text{I}_\text{x}}\cdot\frac{\text{I}_\text{x}\text{R}_1}{\text{R}_1+\text{R}_2+\text{R}_3}=\frac{\text{R}_1}{\text{R}_1+\text{R}_2+\text{R}_3}\tag5$$


Now, applying this to your circuit we need to use (from now on I use the lower case letters for the function in the 'complex' s-domain where I used Laplace transform) the fact that the resistor \$\text{R}_1\$ is replaced by a capacitor and \$\text{R}_2\$ is replaced by an inductor, so:

  • $$\text{R}_1=\frac{1}{\text{sC}}\tag6$$
  • $$\text{R}_2=\text{sL}\tag7$$

So, we get as the transfer function:

$$\mathcal{H}\left(\text{s}\right)=\frac{\text{i}_2\left(\text{s}\right)}{\text{i}_\text{x}\left(\text{s}\right)}=\frac{1}{\text{sC}}\cdot\frac{1}{\frac{1}{\text{sC}}+\text{sL}+\text{R}_3}=\frac{1}{\text{s}^2\text{CL}+\text{sCR}_3+1}\tag8$$

Now, we can use \$\text{s}=\text{j}\omega\$ (where \$\text{j}^2=-1\$):

$$\underline{\mathcal{H}}\left(\text{j}\omega\right)=\frac{1}{\left(\text{j}\omega\right)^2\text{CL}+\text{j}\omega\text{CR}_3+1}=\frac{1}{1-\omega^2\text{CL}+\omega\text{CR}_3\text{j}}\tag9$$

Now, we need to find \$\left|\underline{\mathcal{H}}\left(\text{j}\omega\right)\right|\$:

$$\left|\underline{\mathcal{H}}\left(\text{j}\omega\right)\right|=\left|\frac{1}{1-\omega^2\text{CL}+\omega\text{CR}_3\text{j}}\right|=\frac{1}{\left|1-\omega^2\text{CL}+\omega\text{CR}_3\text{j}\right|}=$$ $$\frac{1}{\sqrt{\left(1-\omega^2\text{CL}\right)^2+\left(\omega\text{CR}_3\right)^2}}\tag{10}$$

Solving for which \$\omega\$, \$\left|\underline{\mathcal{H}}\left(\text{j}\omega\right)\right|\$ is at a maximum gives the resonant frequency:

$$\frac{\partial\left|\underline{\mathcal{H}}\left(\text{j}\hat{\omega}\right)\right|}{\partial\hat{\omega}}=0\space\Longleftrightarrow\space\hat{\omega}=\frac{1}{\text{L}}\cdot\sqrt{\frac{1}{2}\cdot\left(\frac{2\text{L}}{\text{C}}-\text{R}_3^2\right)}\tag{11}$$

Applying this to your circuit gives an imaginary resonant frequency. Therefore I think that the other person (@Andyaka) who answered this question is maybe right. The values of the components do not add up to a possible answer.

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  • \$\begingroup\$ Huh, weird that you get an imaginary result. Your calculations are usually spot on though, so it’s a little odd. Anyway, thank you for the answer, Jan. \$\endgroup\$
    – Carl
    Apr 13, 2020 at 20:09
  • \$\begingroup\$ @Carl no it is not weird, my calculation shows that for your component values there is no resonance frequency. Changing the values of the components influences if there is such a frequency. And as I showed the resistance does matter in your circuit. Only when you have a pure series RLC circuit the resistance can be ignored. And I am glad that I could help you, you're welcome. \$\endgroup\$ Apr 13, 2020 at 20:13
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Equation

https://en.wikipedia.org/wiki/RLC_circuit

https://nptel.ac.in/content/storage2/courses/108105053/pdf/L-17(NKD)(ET)%20((EE)NPTEL).pdf

In the second link, the derivation is correct apart from the last step where the final equation is incorrect.

EDIT

Considering the above equation.

Adding resistance in series with the inductor reduces the oscillation frequency below that frequency obtained with the ideal zero resistance.

The 40R resistance value in the op's circuit is so large that it damps away any oscillation.

Transposing the above equation to make R the subject ant setting Wo to zero it is possible to obtain the maximum value of resistance which will still enable oscillation, all be it at a very low frequency. For the op's circuit the max value of resistance calculates to be just over 1.008 Ohms, well below the shown 40 Ohms. Any value for resistance larger than about 1.008 Ohms will give a negative frequency result when using the above equation.

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