Low pass LC filter - Question

281

06:31:20

The grey graph line is related to the phase, but I'd like to understand the blue line if someone can help here ;)
- @ below 120Hz it shows 20dB
- @ 1kHz it shows 24dB
- @ 1,5kHz it shows 73dB
- @ 5kHz it shows 0dB
where the higher the f the higher attenuation
Three questions :
1: why is there such a high gain @ 1,5kHz (the lower the resistance, the higher the current, the lower is the peak) - is this due to the resonance frequency ? How to get rid of it ?
2: why isn't the start / base not 0dB (it shifts up the higher the amplitude is set, but Im only interested in the relative value, hence question 3)
3: is it possible to just subtract 20dB from every value to bring the baseline to 0 so that eg 1kHz would become 4dB ?
Sorry if the question seems basic (or stupid) to you experts, just puzzled with this and eager to understand.

published 7 years ago

stanislav_maslovski

7 years ago

1. Yes, the high output voltage is due to the LC resonance. The resonant frequency is f = 1/(2 pi sqrt(LC)). The value of the voltage gain is determined by the quality factor Q of the loaded LC tank, which is in your case Q = R/rho, where rho = sqrt(L/C) is the characteristic impedance and R is the load resistance. If you want to get rid of the sharp resonance, you must have Q = 1. In your case, rho = 1 Ohm, and Q = 1 when R = 1 Ohm. http://everycircuit.com/circuit/5588730831437824

stanislav_maslovski

7 years ago

2. The 0 dB in EC corresponds to 1 V. I do not think there is any way to change this reference voltage.

stanislav_maslovski

7 years ago

3. Yes, you can either set the input voltage to 1 V instead of 10 V, or subtract 20 dB.

solivi

7 years ago

Great, may thanks for your answers (at least it confirms my way of thinking as a neophite 😉)

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