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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.

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  • \$\begingroup\$ 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 \$\endgroup\$ – JYelton Sep 10 '14 at 20:59
  • \$\begingroup\$ @JYelton thanks! I was having trouble doing that for somereason...although it's worked for me before. \$\endgroup\$ – codedude Sep 10 '14 at 21:00
  • \$\begingroup\$ Sometimes I've noticed a delay when loading Mathjax scripts. Perhaps it was just being slow. :) \$\endgroup\$ – JYelton Sep 10 '14 at 21:03
  • \$\begingroup\$ That may have been it. My connection is rather slow. \$\endgroup\$ – codedude Sep 10 '14 at 21:04
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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).

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Bandwidth is just a matter of definition, of how we label what's around us.

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.

A 100Hz cut off low pass filter has a band of 100Hz. Just think of it as a bandpass with the lowest frequency being zero...

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  • \$\begingroup\$ Hmmm. So you don't have to do anything with +-3dB...that's easy then I guess :) \$\endgroup\$ – codedude Sep 10 '14 at 21:05
  • \$\begingroup\$ Wait, of course: the cut off frequency is defined as the frequency where the amplitude is -3dB with respect to center band. \$\endgroup\$ – Vladimir Cravero Sep 10 '14 at 21:07

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