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.

 

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.

 

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.

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

 

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).

 

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).

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.

 

I'm not convinced this is true for the OP's actual example - in particular if a & b are different.

Hmm, I didn't think of that.