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Bandwidth at resoance and peak dB level

by djstar
Tags: bandwidth, peak, resoance
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djstar
#1
Feb24-13, 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.
Attached Thumbnails
bode plot.jpg   schematic.jpg  
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jsgruszynski
#2
Feb24-13, 05:21 PM
P: 281
http://en.wikipedia.org/wiki/RLC_circuit
djstar
#3
Feb25-13, 04:10 AM
P: 11
Thanks for the reply but that still doesn't tell me how to calculate the -22 dB.

milesyoung
#4
Feb25-13, 09:46 AM
P: 482
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.
djstar
#5
Feb25-13, 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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