Electronic – Resonance frequency of filter independent of resistance

circuit analysisfilterfrequencyresonance

I have the following circuit and problem.

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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?

Best Answer

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

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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:

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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:

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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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