An analysis of the basal basilar membrane response to tones
A comparison of BM displacement and Neural excitation patterns basal to CF

J. B. Allen1
AT&T Labs-Research
Florham Park, NJ 07932-0971

Abstract

Abstract
We provide an analysis of the basal response of basilar membrane (BM) and cilia response for low frequency tonal stimulation. In the base, the BM impedance is K0 e-2ax/i w while the scala impedance is i wrc A(x). Thus the pressure is very close to that in a ridged-walled box, namely P(x) = i wrc (L-x) ust, since the BM impedance is much greater than the scala impedance. It follows from Hooke's Law that the BM displacement must vary in an exponential manner with place: x(x,w) = - w2 rc (L-x) ust e2ax/K0.

Hair cells are known to be displacement detectors (Hudspeth and Corey, 1977). Above 1 kHz Dallos has found that the inner hair cell (IHC) responds to the shear displacement of the sub-tectorial space. Direct measurements of the neural population by Kim et al., as well as transformations of populations of neural tuning curves from frequency to place (Allen, 1991), have shown that the basal neural response is independent of x. These results are consistent with two-tone suppression (Fahey and Allen, 1979) (and upward spread of masking) data which show that the low frequency suppression threshold is approximately independent of frequency.

There are two possible explanations for the above contradiction. Either the inner and outer hair cell sensitivity must vary as e-2ax, or the micromechanics of the tectorial membrane must transform the BM displacement shear signal from exponential to a constant place dependence. The evidence for a gradient in cilia sensitivity is in the direction of increased sensitivity for lower frequencies, because the cilia increase in length with place. This leaves us with the option that the TM to cilia stiffness ratio must vary as e-2ax to compensate for the e2ax dependence of the BM displacement.

A simple and natural solution to all these requirements is to assume that the linear component of the partition stiffness is dominated by the TM radial stiffness. Model results support these conclusions.

Transmission line model

The place to frequency map

Neural excitation pattern

Stiffness dominated tail region

  • For each pure tone stimulus, the partition impedance is compliance dominated from the stapes to Xz(f) (i.e., a few mm basal to the CF)
    Zp(x,w) » K0 e-2ax/i w

    figs/expat2.gif
  • Hooke's Law relates partition pressure P(x) and displacement D(x)
    P(x) = K0 e-2ax     D(x)

Cochlear Pressure for a tone

  • From the WKB solution method, the spatial pressure distribution of a tone stimulus in the base of the cochlea is given by
    P(x,w)/P(0,w)
    =
      æ
    Ö

    c(x)/c(0)
     
    e-iwòx = 0x dx/c(x)
    =
    e -ax/2 e -iwt(x,w)
    where c(x) = Ö{K(x) A/r} is the local wave speed

  • Conclusion:

    • The partition pressure magnitude decays as
      |P(x)| µ e-ax/2

    • The partition displacement magnitude increases as
      |D(x)| µ e3ax/2

  • Since the inner haircell is a displacement detector above about 1 kHz

    • For the cat (where 2 = ea 0.3), the cilia (neural) response should grow as
      |D(x)| µ e3ax/2

  • This gives a partition displacement growth of 3 dB/mm

Estimation of basal EP slope

  • Neural excitation pattern estimated from FTC

    figs/ep1.gif

  • Slopes S1, S2, and S3 (dB/mm)

    CF S1 S2 S3
    kHz SLOPE* (dB/mm)
    5.0 9.3 32.7 -66.1
    4.0 5.0 26.3 -69.3
    2.0 1.3 15.2 -34.5
    1.0 1.2 17.4 -25.6
    0.5 0.3 14.8 -34.5
    0.25 0.3 17.1 -11.0

     * Mult by 3 mm/oct to convert to dB/oct

  • FINAL CONCLUSION:

  • There must be a transduction filter H(x,f) to account for the slope difference of 3 dB/mm for D(x,f) and 0.3-1.3 dB/mm for the cilia EP C(x,f)

A natural solution to the transduction filter problem

  • I propose that the partition stiffness is dominated by the tectorial membrane stiffness Kt(x)

    figs/rectamodel.gif
  • The partition impedance is: Kp = [(Kt Kc)/( Kt + Kc)] + Kbm

    • Assume: Kc >> Kt µ e-2ax and Kbm » Kt

  • It follows that
    fz(x) º 1
    2 p
      _____
    ÖKt/Mt
     
    = fcf(x)/Ö2
    and at 0.5 kHz
    H º C
    D
    = Kt/Kc » eax/10-3ax/2


Footnotes:

1 jba@research.att.com; TEL: 973/360-8545


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On 20 Feb 1999, 07:24.