2
\$\begingroup\$

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?

\$\endgroup\$
5
  • \$\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, 2021 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, 2021 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, 2021 at 12:41
  • \$\begingroup\$ @Bimpelrekkie, almost correct: Q = fres / Bandwidth \$\endgroup\$
    – Bart
    May 7, 2021 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, 2021 at 15:24

1 Answer 1

Reset to default
7
\$\begingroup\$

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.

\$\endgroup\$
4
  • 1
    \$\begingroup\$ Yes, geometric middle, that is correct. \$\endgroup\$
    – Bimpelrekkie
    May 7, 2021 at 12:44
  • \$\begingroup\$ Thank you for confirming @Bimpelrekkie \$\endgroup\$
    – Prathik Prashanth
    May 7, 2021 at 12:44
  • \$\begingroup\$ Thank you, indeed, this was very helpful! \$\endgroup\$
    – FreeZe
    May 7, 2021 at 15:04
  • 1
    \$\begingroup\$ You can use the embedded Latex editor to write and format equations easily. \$\endgroup\$
    – Mitu Raj
    May 8, 2021 at 5:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.