5
\$\begingroup\$

For calculating the equivalent noise bandwidth of a non-brickwall filter, I can find two different sets of numbers, both of which claim they are similar things:


Order   EqNBW
1   1.5708
2   1.1107
3   1.0472
4   1.0262
5   1.0166
6   1.0115
7   1.0084
8   1.0065
9   1.0051
10  1.0041

 


1 1.57
2 1.22
3 1.16
4 1.13
5 1.12

Which is correct?

Or are they both correct; just used in different calculations?

Update

After figuring this out, I made a chart of the different factors and the types of filters they work for: ENBW Filter correction factors vs order

\$\endgroup\$
  • 2
    \$\begingroup\$ Good question +1 \$\endgroup\$ – Andy aka Jan 19 '17 at 8:45
2
\$\begingroup\$

The effective noise bandwidth depends on the shape of transfer function. It's easy to calculate it numerically.

See my Matlab script below that calculates the ENBW for a Butterworth lowpass filter. You can adapt it to your needs.

for N=1:10
  [b,a] = butter(N, 1, 's');
  f = @(x) (abs(freqs(b,a,x)).^2);
  bw = integral(f, 0, 1e6);
  fprintf('Order: %d, ENBW: %g\n',N, bw);
end 

In case you don't have Matlab, the output is given below

Order: 1, ENBW: 1.5708
Order: 2, ENBW: 1.11072
Order: 3, ENBW: 1.0472
Order: 4, ENBW: 1.02617
Order: 5, ENBW: 1.01664
Order: 6, ENBW: 1.01152
Order: 7, ENBW: 1.00844
Order: 8, ENBW: 1.00645
Order: 9, ENBW: 1.0051
Order: 10, ENBW: 1.00412
\$\endgroup\$
  • \$\begingroup\$ Ohhhhhhhhhhhhhhh... The 1.11 number is for a 2nd-order Butterworth filter, and the 1.22 number is for 2× 1st-order filters in cascade? \$\endgroup\$ – endolith Jan 19 '17 at 17:16
  • 1
    \$\begingroup\$ Yes, according to your reference (analog.intgckts.com/equivalent-noise-bandwidth). \$\endgroup\$ – Mario Jan 19 '17 at 17:24
  • \$\begingroup\$ Mario, could you clear something up? Is the \$\pi\$/2 figure (order = 1) applied at the signal level (i.e. the voltage or current) or does it apply to the power thus making the factor to be applied to signals \$\sqrt{\pi/2}\$? I would appreciate your guidance. @carloc - this is the post I mentioned. \$\endgroup\$ – Andy aka Jan 20 '17 at 9:19
  • \$\begingroup\$ @Andyaka It's applied to your bandwidth. So to calculate the total noise of a resistor connected to a capacitor (first order RC LPF) you'd use $$\sqrt{ 4 k_\text{B} T R \left(\frac \pi 2 \Delta f \right) }$$ So if \$f_c\$ of the RC filter were 10 kHz, you'd pretend it's a brickwall filter at 15.7 kHz instead. \$\endgroup\$ – endolith Jan 20 '17 at 14:47
  • \$\begingroup\$ @endolith so is that 1.57 or sqrt(1.57) for noise voltage? \$\endgroup\$ – Andy aka Jan 20 '17 at 14:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.