# Why are there two sets of ENBW correction factors?

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

• Good question +1 – Andy aka Jan 19 '17 at 8:45

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

• 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? – endolith Jan 19 '17 at 17:16
• Yes, according to your reference (analog.intgckts.com/equivalent-noise-bandwidth). – Mario Jan 19 '17 at 17:24
• 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. – Andy aka Jan 20 '17 at 9:19
• @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. – endolith Jan 20 '17 at 14:47
• @endolith so is that 1.57 or sqrt(1.57) for noise voltage? – Andy aka Jan 20 '17 at 14:48