Electronic – Is the resonance frequency actually at the center of the passband of a bandpass filter

band passcircuit analysisfilter

Assume we have the following circuit

enter image description here

It is a band-pass filter.

There is something I don't get.

We know the the resonance frequency is given by \$ f_{Resonance}=\frac{1}{2\pi\sqrt{LC}} \$ and we know that the lowpass and highpass cutoff frequencies given are given by

$$ f_{1}=\frac{RC+\sqrt{\left(RC\right)^{2}+4LC}}{4\pi LC},\thinspace\thinspace\thinspace\thinspace\thinspace f_{2}=\frac{-RC+\sqrt{\left(RC\right)^{2}+4LC}}{4\pi LC} $$

The bandwidth is given by \$ BW=\frac{1}{2\pi}\frac{R}{L} \$

Now we also know that the resonance frequency is in the middle of the bandwidth. So we are supposed to have:

$$ \begin{cases}
\end{cases} $$

I have built a circuit as described in the picture, up to a change in the value of the resistor. My RLC circuit is the same, but with $ R= 1491 \varOmega $.

But for my values:

$$ f_{Resonance}=\frac{1}{2\pi\sqrt{84\cdot10^{-3}\cdot8.3\cdot10^{-9}}}=6027.558Hz $$

$$ BW=\frac{1}{2\pi}\frac{1491}{84\cdot10^{-3}}=2825 $$

$$ f_{1}=\frac{\left(1491\cdot8.3\cdot10^{-9}\right)+\sqrt{\left(1491\cdot8.3\cdot10^{-9}\right)^{2}+4\cdot84\cdot10^{-3}\cdot8.3\cdot10^{-9}}}{4\pi\cdot84\cdot10^{-3}\cdot8.3\cdot10^{-9}}=7603.349Hz $$

$$ f_{2}=\frac{-\left(1491\cdot8.3\cdot10^{-9}\right)+\sqrt{\left(1491\cdot8.3\cdot10^{-9}\right)^{2}+4\cdot84\cdot10^{-3}\cdot8.3\cdot10^{-9}}}{4\pi\cdot84\cdot10^{-3}\cdot8.3\cdot10^{-9}}=4778.349 $$

$$ f_{Resonance}+\frac{2825}{2}=7440.058Hz\neq7603.349 $$

$$ f_{Resonance}-\frac{2825}{2}=4615.058\neq4778.349 $$

So I get that the resonance frequency is not centered in the bandwitch. What went wrong here?

Best Answer

I remember learning in Circuit Theory that the resonance frequency of a series RLC circuit is the geometric mean (or geometric middle) of the half-power (cutoff) frequencies:

enter image description here

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I hope this helps.