# Bandwidth of low pass filter

Say I have some low pass filter of the form:

$$G(s) = \frac{a}{s+b}$$

How do I find the bandwidth? I know for example that with bandpass filters it's the difference between +-3dB of the cutoff frequency but I'm not sure how that would work for this example.

• I converted your image to a formula using Mathjax, so it can be read by screen readers, etc. For more information see meta.math.stackexchange.com/q/5020/7195 – JYelton Sep 10 '14 at 20:59
• @JYelton thanks! I was having trouble doing that for somereason...although it's worked for me before. – codedude Sep 10 '14 at 21:00
• Sometimes I've noticed a delay when loading Mathjax scripts. Perhaps it was just being slow. :) – JYelton Sep 10 '14 at 21:03
• That may have been it. My connection is rather slow. – codedude Sep 10 '14 at 21:04

In your case $$G(s) = \frac{a}{s+b} = \left(\frac{a}{b}\right)\left( \frac {1}{1+\frac {s}{b}}\right)$$
The output will be -3dB compared to the passband when $s = jb$ (equivalent to $\frac{b}{2\pi}$ Hz) so the bandwidth is $b$ radians/second and the passband gain is (a/b).
A low pass filter is a box that let all the frequencies up to the cutoff pass almost untouched, so the width of the bunch of frequencies that can pass is from 0 to $f_C$, and we call this bandwidth.