Equivalent rectangular bandwidth
part 1/3 · 7.2 KB total
──────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────
top
The equivalent rectangular bandwidth or ERB is a measure used in psychoacoustics, which gives an approximation to the bandwidths of the filters in human hearing, using the unrealistic but convenient simplification of modeling the filters as rectangular band-pass filters, or band-stop filters, like in tailor-made notched music training (TMNMT).
Contents
• See also
──────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────
Approximations
For moderate sound levels and young listeners, Moore & Glasberg (1983) suggest that the bandwidth of human auditory filters can be approximated by the polynomial equation:cite-ref-mooreglasberg-1-0[1]
where F is the center frequency of the filter, in kHz, and ERB( F ) is the bandwidth of the filter in Hz. The approximation is based on the results of a number of published simultaneous masking experiments and is valid from 0.1–6500Hz.cite-ref-mooreglasberg-1-1[1]
Seven years later, Glasberg & Moore (1990) published another, simpler approximation:cite-ref-glasbergmoore-2-0[2]
where f is in Hz and ERB(f) is also in Hz. The approximation is applicable at moderate sound levels and for values of f between 100 and 10000Hz.cite-ref-glasbergmoore-2-2[2]
ERB-rate scale
The ERB-rate scale, or ERB-number scale, can be defined as a function ERBS(f) which returns the number of equivalent rectangular bandwidths below the given frequency f. The units of the ERB-number scale are known ERBs, or as Cams, following a suggestion by Hartmann.cite-ref-3[3] The scale can be constructed by solving the following differential system of equations:
{ E R B S ( 0 ) = 0 d f d E R B S ( f ) = E R B ( f ) {\displaystyle {\begin{cases}\mathrm {ERBS} (0)=0\\{\frac {df}{d\mathrm {ERBS} (f)}}=\mathrm {ERB} (f)\\\end{cases}}}
The solution for ERBS(f) is the integral of the reciprocal of ERB(f) with the constant of integration set in such a way that ERBS(0) = 0.cite-ref-mooreglasberg-1-2[1]
Using the second order polynomial approximation (Eq.1) for ERB(f) yields:
E R B S ( f ) = 11.17 ⋅ ln ( f + 0.312 f + 14.675 ) + 43.0 {\displaystyle \mathrm {ERBS} (f)=11.17\cdot \ln \left({\frac {f+0.312}{f+14.675}}\right)+43.0} cite-ref-mooreglasberg-1-3[1]
──────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────