# Is the resonance frequency actually at the center of the passband of a bandpass filter?

Assume we have the following circuit

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?

• 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}$? – Bimpelrekkie May 7 at 12:32 • @Bimpelrekkie I guess that in reality the resonance frequency is not exactly in the middle. But in my theoretical calculation isnt it supposed to work? – FreeZe May 7 at 12:37 • Why would it, the behavior of the components is relative to frequency therefore so are their impedance and as a result the transfer function. So you make an error, this error does get smaller as Q = Bandwidth /$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}$is in the absolute middle but in the relative (geometric) middle. – Bimpelrekkie May 7 at 12:41 • @Bimpelrekkie, almost correct: Q = fres / Bandwidth – Bart May 7 at 12:50 • @Bart Of course you're correct,$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). – Bimpelrekkie May 7 at 15:24