What is the transfer function of this LC low-pass filter?

Discussion in 'General Electronics Chat' started by zhaojia, Nov 23, 2007.

1. zhaojia Thread Starter New Member

Nov 23, 2007
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I want to get a transfer function of the following LC low-pass filter.
And, I want to know how it's cut-off frequency is derived.

An LC low-pass filter:

http://www.williamson-labs.com/images/filters-lc.gif
(The first one, the Ui(jw) is at left for input and Uo(jw)) is the output at right.)

The first circuit.

What is the expression for G(jw) = Uo(jw) / Ui(jw)?
And What is the cut-off frequency wc?
{the value wc makes G(jwc) = 1/sqrt(2) * G(j0) }

Nov 23, 2007
7
0
3. zhaojia Thread Starter New Member

Nov 23, 2007
7
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Sorry.
And What is the cut-off frequency wc?
{the value wc makes G(jwc) = 1/sqrt(2) * G(j0) }
This is NOT the correct concept of cut-off frequency.

The cut-off frequency, in fact, is a corner frequency f0 at which the attenuation characteristics emerge.
It can be derived form the mathematically reform.

4. thingmaker3 Retired Moderator

May 16, 2005
5,072
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fc = 1/(2*$\Pi$*R*C)

5. zhaojia Thread Starter New Member

Nov 23, 2007
7
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Thanks. But this fc is the cut-off frequency of a RC (NOT LC) low-pass filter.

6. GS3 Senior Member

Sep 21, 2007
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http://en.wikipedia.org/wiki/Cutoff_frequency

7. The Electrician AAC Fanatic!

Oct 9, 2007
2,540
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Imagine that the inductor were replaced by a resistor, R1, and the capacitor were replaced by another resistor, R2. Then you would have a voltage divider, and Vo/Vi would be given by R2/(R1+R2). If you replace the resistance of R1 by the impedance of the inductor, j w L, and the resistance of R2 by the impedance of the capacitor, 1/(j w C), the formula for the voltage divider will become the transfer function you want, namely ZC/(ZL+ZC). You will have to do some algebraic simplification to get it to look like what you would find in a book.

8. zhaojia Thread Starter New Member

Nov 23, 2007
7
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Thank GS3 and The Electrician for the aproperate and nice reply.