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:: EPE Chat Zone ­:: ­Radio Bygones Message Board :: » EPE Forum Archives 2010 - » Archive through 28 May, 2013 » LC FILTER « Previous Next »

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Frequent Contributor
Username: james

Post Number: 232
Registered: 02-2007

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Posted on Monday, 27 May, 2013 - 10:47 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)


Cut off frequency,Fc = 1/(2PRC)

Cut off frequency,Fc = R/(2PL)


My question - What's the formula to calculate the cut off frequency of a LC filter?

Mike & Richard Tooley use the low pass form of this filter in this month's Jump Start in their Radio's AM demodulator.
They use -
L = 3.3mH
C = 100nF
Which I would assume gives a cut off frequency of somewhere around 20KHz.

I know that the resonant frequency is given by -

Fres = 1/(2P * SQRT(LC))


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Username: alexr

Post Number: 247
Registered: 02-2008

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Posted on Tuesday, 28 May, 2013 - 08:38 am:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

The cut-off frequency is the frequency at which the output falls to 1/2 of the input. In this case this will be when XL = XC which also happens to the condition that defines resonance.

So just change Fres to Fc in your formula and you have the answer.

The values you quote give a Fc of about 8.76KHz which, since AM audio bandwidth is usually limited to about 10KHz, sounds about right.

(Message edited by alexr on 28 May, 2013)

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