# How to calculate the gain of a RLC tank filter?

I'm supposed to create a passive bandpass filter to select a reasonably sinusoidal 750 Hz signal from a 250 Hz square wave.

Since we have very limited choices for the inductor (0.03 H) and capacitor (0.22, 0.47 and 1 uF) in the lab, and also very limited space to put those components on, I would need a really low resistance for a small bandwidth if I were to use a series RLC circuit. So I decided to use a parallel circuit instead, which I simulated on qucs and went on to the lab and measured the gain as a function of input frequency. And it worked, kinda of, but I don't know how to fit the data I gathered. Also, the gain was really low, and the filtered signal needs to have an amplitude of at least 10% of the original wave. Is the parallel RLC filter not the answer?

• can U compute X(f ) /R and R / X(f) for L or C? Apr 19 '18 at 1:26
• Using the same setup in LTspice, I get ~0dB@750Hz. Using a 250Hz square wave as input, for 1s, I get the 3rd harmonic, wobbly (about 90mV max difference between ripples), and with a general 400mV peak amplitude. The source has 1Ohm resistance. Did your lab setup involve additional elements? Maybe the supply/sink had significant I/O resistances/impedances? Apr 19 '18 at 6:05
• My lab setup was exactly the same as in the picture. The generator has an output impedance of 50 Ohm, and the inductor has an internal resistance of 7.8 Ohm but I aren't those irrelevant since they are small compared to the 1k resistor I used?
– user186245
Apr 19 '18 at 12:07
• @John Here's what I see: i.stack.imgur.com/QJitS.png . For 2V input there's well over 1V ouput, with the included 7.8 resistance. Only if there's a resistance at the output does the amplitude drop, but then the Q will lower itself, so the bandwidth will "fatten". Maybe your input signal was not 2V? Apr 19 '18 at 13:19

You can solve the Q = Rs/Xs(f) or Q= Xp(f)/Rp but if limited to these 3 cap values. Then choose the best combination. I don’t expect you to do as follows but this is what is possible to get voltage gain BT raising the output impedance and lowering source reactance.

Here you can obtain a gain of 17dB with this 4th Order BPF again by impedance ratios by raising source impedance with a smaller series cap then lowering source impedance with a larger shunt cap such that the parallel resonance of (0.22//1 in series with 1uF) // (0.47+0.47) to resonate at 750 with 30mH.

An impedance RLC nomograph makes it easier, as well as a Falstad Simulator. Here I show how yours works from a graph. • Any questions?? Apr 19 '18 at 11:42
• Hi, how can I find a formula for the gain? Do I need to find the output and input voltages using Kirchoff's laws? Also, I understand how the ressonance frequency of the circuit you designed is 750 Hz, but I don't understand how it's a bandpass filter. Our teacher didn't even teach us about anything other than LP filters, nor gave us any references, do you know of any good one?
– user186245
Apr 19 '18 at 12:15
• sim.okawa-denshi.jp/en/RLCbpkeisan.htm and sim.okawa-denshi.jp/en/Fkeisan.htm ( consider ω = 2πf for s ) Apr 19 '18 at 12:23
• for advanced graphical solutions find a good RLC nomograph, then you get ballpark solutions of Q from impedance ratio right on graph Apr 19 '18 at 12:30
• If you understand impedance, reactance ratios, then gain is easy, for series vs parallel but for complex i.e. 10th order LC filters better tools are used. like Falstad or other to make steep LPF HPF, PBF etc Apr 19 '18 at 12:52