How do you calculate the noise equivalent bandwidth for an IIR or FIR filter?


Hm, isn't it the same as for any filter?

$$ B_\text{eq} = \frac{1}{|H(e^{j\omega_0})|^2} \int_0^\infty |H(e^{j\omega})|^2 d\omega $$

where $H(e^{j\omega})$ is the frequency response, and $\omega_0$ is the max-abs frequency. For FIR filters, you can just use e.g. freqz in MATLAB/Octave. For IIR filters, you'd have to do analysis or take measurements (or see if you can get away with truncating)

  • $\begingroup$ Shouldn't the denominator term be $|H(e^{j\omega_0})|^2$ ? onmyphd.com/?p=enbw.equivalent.noise.bandwidth $\endgroup$ – Peter K. Nov 11 '13 at 22:15
  • $\begingroup$ Yes you are correct, I'll fix it $\endgroup$ – toast Nov 11 '13 at 23:55
  • $\begingroup$ Shouldn't it be in discrete frequency, not continuous? $\endgroup$ – random_dsp_guy Nov 12 '13 at 0:04
  • $\begingroup$ I do not think so. The DTFT is continuous (actually a complex Fourier series). $\endgroup$ – toast Nov 12 '13 at 0:16

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