5
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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

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1
  • 2
    \$\begingroup\$ Good question +1 \$\endgroup\$
    – Andy aka
    Commented Jan 19, 2017 at 8:45

1 Answer 1

2
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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
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6
  • \$\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
    Commented Jan 19, 2017 at 17:16
  • 1
    \$\begingroup\$ Yes, according to your reference (analog.intgckts.com/equivalent-noise-bandwidth). \$\endgroup\$
    – Mario
    Commented Jan 19, 2017 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
    Commented Jan 20, 2017 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
    Commented Jan 20, 2017 at 14:47
  • \$\begingroup\$ @endolith so is that 1.57 or sqrt(1.57) for noise voltage? \$\endgroup\$
    – Andy aka
    Commented Jan 20, 2017 at 14:48

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