# Simulating an RLC bandpass filter with LTspice

I have analyzed the filter in question both in practice and in simulation (with LTspice)

The calculations I performed to determine the high and low cutoff frequency and the resonance frequency are as follows:

• fres = 1/2pi * sqrt(LC)
• fch = R/2pi * L
• fcl = 1/2pi * RC

From these, I obtained:

• fres = ~10kHz
• fch = ~238kHz
• fcl = ~106Hz

The results I obtained coincide with those recorded during the practical test, but not in the simulated one.

Below the trend measured at the node between the inductor and the resistor:

Now I'm wondering why it behaves this way. Also, since this is a passive filter, shouldn't the maximum gain be 0 dB?

• The answer below is the correct one, but I have to wonder: why did you choose AC 5 in the first place? I'm interested in the reason. Dec 23, 2021 at 10:16
• the value has been chosen for no particular reason. Dec 23, 2021 at 17:57
• In your honest opinion, do you think that is a healthy attitude when searching for results? Don't forget about the GIGO. As it happens, this wasn't the case, but it did cause you to misinterpret the results. Dec 23, 2021 at 20:14
• I asked for this very reason, unfortunately I did not know that ltspice measured in dbV and I am new to the subject. However I understand the concept and I thank you for the advice, maybe if the situation had been slightly different I would have undoubtedly made a mistake by misinterpreting the results. Dec 23, 2021 at 21:17