
#1
Feb2413, 02:27 PM

P: 11

Hi,
I'm investigating what happens to bandwidth at resonance of a series RLC circuit when each component is doubled. i have simulated the circuits and created a bode plot. You can see from the photos that at resonance the gain peaks at 20.82 dB. I would like to know how to calculate this value. i have an input voltage of 10Vp and the voltage at resonance across the resistor is 6.428V. i understand that the voltage across both resistors is 7.07V because the multimeter is measuring RMS but how do i calculate the db level. 



#2
Feb2413, 05:21 PM

P: 272




#3
Feb2513, 04:10 AM

P: 11

Thanks for the reply but that still doesn't tell me how to calculate the 22 dB.




#4
Feb2513, 09:46 AM

P: 427

bandwidth at resoance and peak dB level
You have a voltage divider with impedances:
Z1 = R1 + j*omega*L + 1/(j*omega*C) = R1 + j*(omega*L  1/(omega*C) ) Z2 = R2 Your measured voltage, call it Vout, is then given by: Vout = Z2/(Z1 + Z2) = R2/(R1 + R2 + j*(omega*L  1/(omega*C) ) ) with magnitude: Vout = R2/sqrt( (R1 + R2)^2 + (omega*L  1/(omega*C) )^2 ) Find peak value by finding roots of the derivative of Vout with respect to omega and substitute into Vout, you probably know the drill. My solution gives: Peak value = R2/sqrt( (R1+R2)^2 + (L/sqrt(L*C)  sqrt(L*C)/C)^2 ) = 0.0909 V = 20.8279 dBV for R1 = 10 ohm, R2 = 1 ohm, L = 70 mH, C = 110 uF. 



#5
Feb2513, 10:12 AM

P: 11

Thanks,
Once you're broke down the equations its made it easier to see what is going on. Thanks for the help :) 


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