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:

$$ \begin{cases} f_{1}=f_{Resonance}+\frac{BW}{2}\\ f_{2}=f_{Resonance}-\frac{BW}{2} \end{cases} $$

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?

  • \$\begingroup\$ Use a simulator to plot the transfer curve and carefylly look at the shape of the curve around the centre frequency, is \$f_{resonance}\$ exactly in the middle between the bandwidth's -3 dB points: \$f_{-3dB} = f_{resonance} +/- xxx kHz (absolte value) or is it relative, like: \$f_{-3dB,low}\$ = 0.9 * \$f_{resonance}\$, \$f_{-3dB,high}\$ = 1.1 * \$f_{resonance}\$? \$\endgroup\$ – Bimpelrekkie May 7 at 12:32
  • \$\begingroup\$ @Bimpelrekkie I guess that in reality the resonance frequency is not exactly in the middle. But in my theoretical calculation isnt it supposed to work? \$\endgroup\$ – FreeZe May 7 at 12:37
  • 1
    \$\begingroup\$ Why would it, the behavior of the components is relative to frequency therefore so are their impedance and as a result the transfer function. So you make an error, this error does get smaller as Q = Bandwidth / \$f_{res}\$ gets smaller. But your Q is only about 3 so if you do not compensate for the error your calculations will result in numbers with large errors. In the end it is easier not to assume \$f_{res}\$ is in the absolute middle but in the relative (geometric) middle. \$\endgroup\$ – Bimpelrekkie May 7 at 12:41
  • \$\begingroup\$ @Bimpelrekkie, almost correct: Q = fres / Bandwidth \$\endgroup\$ – Bart May 7 at 12:50
  • \$\begingroup\$ @Bart Of course you're correct, \$Q = \frac{f_{res}}{BW}\$, at least someone is paying attention ;-) Mysteriously I did come to the right value of Q = 3. (Q = 1/3 would not make sense). \$\endgroup\$ – Bimpelrekkie May 7 at 15:24

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

enter image description here

I hope this helps.

  • 1
    \$\begingroup\$ Yes, geometric middle, that is correct. \$\endgroup\$ – Bimpelrekkie May 7 at 12:44
  • \$\begingroup\$ Thank you for confirming @Bimpelrekkie \$\endgroup\$ – Prathik Prashanth May 7 at 12:44
  • \$\begingroup\$ Thank you, indeed, this was very helpful! \$\endgroup\$ – FreeZe May 7 at 15:04
  • 1
    \$\begingroup\$ You can use the embedded Latex editor to write and format equations easily. \$\endgroup\$ – Mitu Raj May 8 at 5:03

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