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To create a band-pass filter using an RLC-circuit, i drew the following circuit (assuming R is given as 120):

RLC-circuit

The corresponding transfer function is $$ H(\mathrm{j}\,\omega)=\frac{120\, C\, \mathrm{j}\, \omega}{1 + 120\, C\,\mathrm{j}\,\omega + C\, L\, (\mathrm{j}\,\omega)^2} $$ I know it is possible to find the corresponding L and C values to create a band-pass filter at resonance, giving a peak at the resonance frequency and -40dB/dec decrease at higher and lower frequencies in the bode plot. In that case the denominator of the transfer function has 2 complex roots.

My question is: is it possible to find L and C values to create a band-pass filter without resonance. Thus with 2 real roots for the denominator. Since \$\omega=0\$ is a zero of the transfer function, I would assume there is an increase of 20dB/dec until reaching the first pole. Then a steady part and afterwards a decrease of -20dB/dec starting from the second pole.

I did some quick calculations and found example values for C and L (see the circuit picture). I drew the circuit in LTSpice, did an AC-analysis and got this bode plot:

Bode plot of circuit

This is exactly what I was expecting. However, since I can't find any resources about this configuration of RLC-circuits (on the internet and in my textbooks), I'm not sure this is actually correct. Is this circuit a real band-pass filter, or just some look-a-like? And is my statement that this circuit is not in resonance correct?

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My question is: is it possible to find L and C values to create a band-pass filter without resonance.

There is no such thing - even if the circuit is so over damped to be daft, there is still the natural resonant frequency of the circuit which is \$\dfrac{1}{2\pi\sqrt{LC}}\$.

Is this circuit a real band-pass filter, or just some look-a-like?

Yes it absolutely is.

And is my statement that this circuit is not in resonance correct?

No it isn't.

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  • \$\begingroup\$ Thanks for your fast response! So if I understand you correctly, the circuit I created is still in resonance, but it is over damped (that's why there is no peak)? And every RLC-circuit, no matter the values of R, L and C has always a resonant frequency? \$\endgroup\$
    – user1176420
    Jan 6, 2016 at 21:55
  • \$\begingroup\$ According to your graph (bode plot) there is a peak. And yes, when you start to learn about pole-zero diagrams you'll realize this and, just to whet your appetite, that resonant peak is at infinity. Now I bet you are confused. \$\endgroup\$
    – Andy aka
    Jan 6, 2016 at 21:57
  • \$\begingroup\$ Haha ok :p well that's quite confusing yes. But I think I got it more or less \$\endgroup\$
    – user1176420
    Jan 6, 2016 at 22:00

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