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)

| improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.