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- 2nd November 2010, 23:16 #1

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## derivation of cutoff frequency of low pass filter

how can i derive the formula of cutoff frequency of a second order low pass filter?

it is a simple two stage RC filter, but when it comes to derivation of cutoff frequency the equation i face with extraordinariliy complex (at least for me)

can anyone give a suggestion?

(may be a link to such a derivation?)

thanks in advance

- 2nd November 2010, 23:16

- 2nd November 2010, 23:35 #2

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## Re: derivation of cutoff frequency of low pass filter

Yes, it's not easy - however, rather straightforward. Use the complex transfer function (2nd order) and calculate the magnitude.

Then, set the magnitude equal to 1/sqrt(2) and solve for w. But I do not recommend this procedure cause you will learn not too much. Therefore, the normal and classical approach is to use tables in textbooks (computer derived based on pole data) or filter simulation programs.

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- 2nd November 2010, 23:35

- 2nd November 2010, 23:49 #3

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## Re: derivation of cutoff frequency of low pass filter

thank you very much LvW

unfortunately i have to derive the formula manually, and i'll be grateful if you can help me in this respect.

i derived the transfer function, but i can't solve it for w (it becomes too complex)

do you have any recommendation?

where can i find an example derivation?

thanks in advance

- 2nd November 2010, 23:49

- 3rd November 2010, 10:33 #4

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## Re: derivation of cutoff frequency of low pass filter

If both RC stages are decoupled from each other it is not too complicated (decoupling means that the first RC stage is not loaded by the 2nd stage because a buffer is in between).

In this case, you simply multiply two transfer functions of 1st order like:

H(jw)=1/(1+jwT1)*1/(1+jwT2).

So you get a transfer function of 2nd order. Then, the magnitude can be derived by separating the real and imaginary parts, respectively. I assume you are familiar with writing magnitudes of a complex expression.

As a next step set this magnitude equal to 1/sqrt(2) which leads to a 4th order equation.

However, you are lucky because this equation includes only even exponents (w^2 and w^4). So you can substitute w^2=x leading to a simple 2nd order equation which can be solved using classical approaches.Last edited by LvW; 3rd November 2010 at 10:39.

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- 19th August 2012, 04:34 #5
## Re: derivation of cutoff frequency of low pass filter

is it 1/sqrt(2) or 1/sqrt(2) times the pass band gained.Have you assumed that the gain is 1?

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