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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    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
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    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
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    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
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    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
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    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
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    Hmm, I didn't think of that.
     
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