# Noise bandwidth/RMS value of 2nd order filter shortcut?

Discussion in 'Homework Help' started by cx1111, May 7, 2014.

1. ### cx1111 Thread Starter New Member

Nov 4, 2013
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0
Hello

I am trying to find the NOISE BANDWIDTH of second order filter.

I have a transfer function of a second order filter: 1/(as+1)(bs+1) which is obviously a cascade of two low pass filters with TF 1/(as+1) and 1/(bs+1).

To find the noise bandwidth, I need the area under the curve of the second order filter, which is equal to the filtered noise rms value given white noise input. I could get this by integrating the second order transfer function squared, but that is a giant pain to integrate and is very long.

I wonder if there is a simpler linear solution to this problem since the rms values/areas under the curves or bandwidths of the first order filters are relatively simple to find. Since this is not noise summation, but rather cascading filters, is there a way to find the solution by combining solutions of the first order filters?

ie mathematically, we have two functions of frequency. I can easily get the areas under each of the curves. I want the area under the curve obtained by multiplying the two individual functions. Can I get that area from the area of the individual curves?

Thank you very much.

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
Wolfram's online integrator makes short work of the problem - since one is presumably only integrating the square of the transfer function magnitude.

$|G (j\omega) | =\frac { 1}{\sqrt { (1-ab\omega^2)^2+(a+b)^2\omega^2} }$

If I'm correct the integral of this function squared turns out to be a simple closed form.

To round this out towards a final solution then requires one to determine the maximum value of the magnitude.

Last edited: May 8, 2014
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3. ### cx1111 Thread Starter New Member

Nov 4, 2013
9
0

Thank you!

I just wanted to know if there was a clever way of linearly combining the solutions to the individual filter responses. I would just feel dumb if I did a whole lot of integration for nothing but I guess that's not the case

4. ### daviddeakin Active Member

Aug 6, 2009
207
27
Actually you can do it that way. I happen to know the noise bandwidth of a first-order filter is $\pi$/2 times the 3dB bandwidth, which is not difficult to prove. Therefore, if you were to cascade two such filters, the noise bandwidth would be $\pi$/4 times the (single section) 3dB bandwidth. (This works out to be 1.22 times the 3dB bandwidth of the whole filter).

Last edited: May 8, 2014
5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
I'm not convinced this is true for the OP's actual example - in particular if a & b are different. It's a matter of checking the math I guess.

Last edited: May 9, 2014
6. ### daviddeakin Active Member

Aug 6, 2009
207
27
Hmm, I didn't think of that.