# 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

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?