Modeling neuronal response to disparity gradient

Abstract

There is a rich literature of physiological studies that a subset of neurons in visual cortices is discriminative of 3-D surface orientation using only the disparity gradient information. One of the physiological models to account for this sensibility to surface slant is the dif-frequency disparity model. Although this model is physiologically plausible, no computational analysis is available to explain how first-order-disparity sensitive neurons detect slanted surface. In this paper a computational model based on the dif-frequency disparity model is presented. In particular, analytical expressions that fit well with neuronal responses to broadband stimuli are obtained when simple cell receptive field is described by log-Gabor filters. It is shown with mathematical analysis and numerical simulations that our proposed model can not only account for physiological data of neuronal response to surface slant but also detect disparity gradient from random dot and sinusoidal grating stereograms.

Appendices

Appendix A: Disparity gradient tuning curve to sinusoidal gratings stimulus

In this appendix, the derivation of the response function of a complex cell to sinusoidal gratings is provided.

The receptive field profile of a simple subunit in the disparity energy model can be represented by a one-dimensional (1-D) Gabor function

$$ \begin{gathered} f(x;x_{0} ,\omega_{0} ,\phi ) = \frac{1}{{\sqrt {2\pi } \sigma }}\exp \left( { - \frac{{(x + x_{0} )^{2} }}{{2\sigma^{2} }}} \right) \hfill \\ \exp \left[ { - i\left( {\omega_{0} (x + x_{0} ) + \phi } \right)} \right] \hfill \\ \end{gathered} $$

(A1)

where $\sigma ,\omega_{0} ,\phi$ are, respectively, Gaussian scale, preferred angular frequency and phase of the cell. $x_{0}$ is the position of Gaussian peak and set to be zero $\left( {x_{0} = 0} \right)$ without loss of generality.

The receptive field profiles of left and right eye can be described as:

$$ f_{L} (x) = f(x;0,\omega_{0} ,\phi ) $$

(A2)

$$ f_{R} (x) = f(x;\Updelta x,k\omega_{0} ,\phi + \Updelta \phi ) $$

(A3)

where $k$ is the spatial frequency ratio between right and left receptive field. $\Updelta x$ and $\Updelta \phi$ are interocular position and phase difference, respectively.

Given Eq. 3, the corresponding image patch within the receptive field is

$$ I(x) = I_{L} (x) $$

$$ I(x) = I_{R} (x + d(x)) = I_{R} (x + {{\Updelta}}dx + d_{0} ) $$

which is equivalent to

$$ I_{L} (x) = I(x) $$

$$ I_{R} (x) = I\left( {\frac{{x - d_{0} }}{{1 + {{\Updelta}}d}}} \right) $$

If $\hat{I}\left( \omega \right)$ is the Fourier transform of $I\left( x \right),$ then we have

$$ I_{L} (x) = \int\limits_{ - \infty }^{ + \infty } {\hat{I}\left( \omega \right)\exp \left( {i\omega x} \right)d\omega } $$

(A4)

$$ I_{R} (x) = \int\limits_{ - \infty }^{ + \infty } {\hat{I}\left( \omega \right)\exp \left( {i\frac{{\omega \left( {x - d_{0} } \right)}}{1 + \Updelta d}} \right)d\omega } $$

(A5)

According to the disparity energy model, the response of a complex cell is the linear summation of the responses of its monocular unit:

$$ R_{C} = \left| C \right|^{2} ,\quad C = \int\limits_{ - \infty }^{ + \infty } {\left( {f_{L} (x)I_{L} (x) + f_{R} (x)I_{R} (x)} \right)dx} $$

Substitute Eq. A2, A3, A4 and A5 into the above equation yields

$$ \begin{aligned} C = & \int\limits_{ - \infty }^{ + \infty } {\rho \left( \omega \right)d\omega \exp \left( { - \frac{{\sigma^{2} }}{2}\left( {\omega - \omega_{0} } \right)^{2} } \right)\exp \left[ {i\left( {\theta \left( \omega \right) - \phi } \right)} \right]} \\ & \quad + \int\limits_{ - \infty }^{ + \infty } {\rho \left( \omega \right)d\omega \exp \left( { - \frac{{\sigma^{2} }}{2}\left( {\frac{\omega }{{1 + {{\Updelta}}d}} - k\omega_{0} } \right)^{2} } \right)} \exp \left[ {i\;\left( {\theta \left( \omega \right) - \frac{{\omega \left( {{{\Updelta}}x + d_{0} } \right)}}{{1 + {{\Updelta}}d}} - \phi - \Updelta \phi } \right)} \right] \\ \end{aligned} $$

(A6)

In the deduction, the following equality is used:

$$ \int\limits_{ - \infty }^{ + \infty } {\exp \left( { - \frac{{x^{2} }}{{2\sigma^{2} }}} \right)\exp (i\omega x)dx = \sqrt {2{{\Uppi}}} \sigma \exp } \left( { - \frac{{\sigma^{2} \omega^{2} }}{2}} \right) $$

For a sinusoidal grating stimulus, its transfer function contains only two frequency components and its Fourier phase changes linearly in related to the frequency.

$$ \rho \left( \omega \right) = \delta \left( {\Upomega + \omega } \right) + \delta \left( {\Upomega - \omega } \right) $$

(A7)

$$ \theta \left( \omega \right) = a\omega $$

(A8)

where $\delta \left( \omega \right)$ is the Dirac’s delta function and $a$ is a constant.

By making use of Eqs. A7 and A8, and let mean disparity, interocular position and phase difference to be zero, Eq. A6 can be simplified as

$$ R_{C}^{\sin } \approx \left[ \begin{aligned} & {\exp \left( { - \frac{{\sigma^{2} }}{2}\left( {\frac{\Upomega }{{1 + {{\Updelta}}d}} - k\omega_{0} } \right)^{2} } \right)} \\ & + {\exp \left( { - \frac{{\sigma^{2} }}{2}\left( {\Upomega - \omega_{0} } \right)^{2} } \right)} \end{aligned} \right]^{2} $$

(A9)

Appendix B: Disparity gradient tuning curve to broadband stimulus

In this appendix, we give the derivation of the response function of our model cell to broadband stimuli such as random dot patterns and illumination bars.

The log-Gabor filter can be expressed in the spatial frequency domain as

$$ \hat{g}\left( {\omega ;x_{0} ,\omega_{0} ,\phi } \right) = \exp \left( { - \frac{{\sigma_{\omega }^{2} }}{2}\left( {\log {\omega \mathord{\left/ {\vphantom {\omega {\omega_{0} }}} \right. \kern-\nulldelimiterspace} {\omega_{0} }}} \right)^{2} } \right)\exp \left[ { - i\left( {x_{0} \left( {\log {\omega \mathord{\left/ {\vphantom {\omega {\omega_{0} }}} \right. \kern-\nulldelimiterspace} {\omega_{0} }}} \right) - \phi } \right)} \right] $$

(B1)

where $\sigma_{\omega }$ is the Gaussian scale that controls the spatial frequency bandwidth of the filter. $\omega_{0}$ is the preferred angular spatial frequency of the cell. $x_{0}$ is the retinal position of the receptive field center and $\phi$ is the carrier phase.

With Eq. B1, the left and right receptive field profile in the spatial frequency domain is

$$ \hat{g}_{L} \left( \omega \right) = \hat{g}\left( {\omega ;x_{c} ,\omega_{0} ,\phi } \right) $$

(B2)

$$ \hat{g}_{R} \left( \omega \right) = \hat{g}\left( {\omega ;x_{c} ,k\omega_{0} ,\phi } \right) $$

(B3)

where $x_{c}$ is the retinal position of the receptive field center (we let $x_{c} = 1$ throughout this paper), $k$ is the spatial frequency ratio between the right and left receptive field (see Eq. 2).

For a stimulus with a mean disparity $d_{0}$ and a constant disparity gradient $\Updelta d,$ its left and right images in logarithmic scale can be expressed as

$$ \begin{aligned} I_{L} (\log x)& = I\left[ {\log x} \right] \\ I_{R} (\log x)& = I\left[ {\log \left( {x - d_{0} } \right) - \log \left( {1 + \Updelta d} \right)} \right] \\ \end{aligned} $$

If the mean disparity is known beforehand, it can be nullified by shifting the left retinal image by $d_{0} .$ Therefore without loss of generality, we assume $d_{0} = 0.$ Given the above equations, the Fourier transform of left and right retinal image in logarithmic scale can be written as

$$ \hat{I}_{L} \left( {\log \omega } \right) = \hat{I}\left( {\log \omega } \right) $$

(B4)

$$ \hat{I}_{R} \left( {\log \omega } \right) = \exp \left( { - i\log \omega \log \left( {1 + \Updelta d} \right)} \right)\hat{I}\left( {\log \omega } \right) $$

(B5)

The response of a complex cell is the power of the sum of the left and right monocular response, which can be written in logarithmic spatial frequency domain with Parseval’s theorem as

$$R=\left|{\begin{aligned}& \int\limits_{0}^{ + \infty } \hat{g}_{L}(\text{log}\;\omega)\hat{I}_{L}(\text{log}\;\omega) d(\text{log}\;\omega)\\ & + \int\limits_{0}^{ + \infty } \hat{g}_{R}(\text{log}\;\omega)\hat{I}_{R}(\text{log}\;\omega) d(\text{log}\;\omega) \end{aligned}}\right|^2$$

(B6)

Let

$$ C \equiv \int_{0}^{ + \infty } {\hat{g}_{L} \left( {\log \omega } \right)\hat{I}_{L} \left( {\log \omega } \right)d\log \omega } + \int_{0}^{ + \infty } {\hat{g}_{R} \left( {\log \omega } \right)\hat{I}_{R} \left( {\log \omega } \right)d\log \omega } $$

(B7)

Substituting Eqs. B2, B3, B4 and B5 into Eq. B7 and separating the magnitude and phase of the stimulus with

$$ \hat{I}\left( {\log \omega } \right) = \rho \left( {\log \omega } \right)\exp \left[ { - i\theta \left( {\log \omega } \right)} \right] $$

We have

$$ \begin{aligned} C = & \int\limits_{0}^{ + \infty }\rho \left( {\log \omega } \right)d\log \omega \exp {\left\{ { - \frac{{\sigma_{\omega }^{2} }}{2}\left[ {\log \left( {\frac{\omega }{{\omega_{0} }}} \right)} \right]^{2} } \right\}}\\ & \quad \times \exp {\left[ { - i\left( {\log \left( {\frac{\omega }{{\omega_{0} }}} \right) - \phi + \theta \left( {\log \omega } \right)} \right)} \right]} \\ & \quad + \int\limits_{0}^{ + \infty } {\rho \left( {\log \omega } \right)d\log \omega \exp \left\{ { - \frac{{\sigma_{\omega }^{2} }}{2}\left[ {\log \left( {\frac{\omega }{{k\omega_{0} }}} \right)} \right]^{2} } \right\}}\\ &\quad \times \exp \left[ { - i\left( {\log \left( {\frac{\omega }{{k\omega_{0} }}} \right) - \phi + \theta \left( {\log \omega } \right)} \right)} \right]\\ & \quad \times\exp \left[ { - i\log \omega \log \left( {1 + {{\Updelta}}d} \right)} \right] \\ \end{aligned} $$

According to the physiological data of Pollen and Ronner (1981), the spatial frequency ratio $(k)$ is within one-fourth octave. This means $0.84 \le k \le 1.19.$ In addition, it is shown in Fig. 9 of Sanada and Ohzawa (2006) that the frequency ratios of most cells concentrate around unit. Therefore, it can be reasonably assumed that the frequency ratio $k$ is not too far away from unit. In this case, the response of a model complex cell $\left( R \right)$ can be approximated as

$$R \approx \left[{\begin{array}{*{20}l} & \int\limits_{0}^{ + \infty } {\rho \left( {\log \omega } \right)d\log \omega \exp \left[ { - \frac{{\sigma_{\omega }^{2} }}{2}\left( {\log \frac{\omega }{{\omega_{0} }}} \right)^{2} } \right]} \\& \quad \times\cos \left( {\log \frac{\omega }{{\omega_{0} }} - \phi + \theta \left( {\log \omega } \right)} \right) \\ & \quad +\int\limits_{0}^{ + \infty } {\rho \left( {\log \omega } \right)d\log \omega \exp \left[ { - \frac{{\sigma_{\omega }^{2} }}{2}\left( {\log \frac{\omega }{{\omega_{0} }}} \right)^{2} } \right]}\\&\quad \times\cos \left( {\log \frac{\omega }{{k\omega_{0} }} - \phi + \log \omega \log \left( {1 + \Updelta d} \right) + \theta \left( {\log \omega } \right)} \right)\end{array} } \right] ^{2} \\+\left[{\begin{array}{*{20}l}& \int\limits_{0}^{ + \infty } {\rho \left( {\log \omega } \right)d\log \omega \exp \left[ { - \frac{{\sigma_{\omega }^{2} }}{2}\left( {\log \frac{\omega }{{\omega_{0} }}} \right)^{2} } \right]}\\ &\quad\times { \sin \left( {\log \frac{\omega }{{\omega_{0} }} - \phi + \theta \left( {\log \omega } \right)} \right)} \\ & \quad +\int\limits_{0}^{ + \infty } \rho \left( {\log \omega } \right)d\log \omega \exp \left[ { - \frac{{\sigma_{\omega }^{2} }}{2}\left( {\log \frac{\omega }{{\omega_{0} }}} \right)^{2} } \right]\\ &\quad\times\sin \left( {\log \frac{\omega }{{k\omega_{0} }} - \phi + \log \omega \log \left( {1 + \Updelta d} \right) + \theta \left( {\log \omega } \right)} \right)\end{array} }\right]^{2} $$

With sum-to-product trigonometric identity, we have

$$ R = \left[{\begin{array}{*{20}l} & [2\rho \left( {\log \omega } \right)d\log \omega \exp \left[ { - \frac{{\sigma_{\omega }^{2} }}{2}\left( {\log \frac{\omega }{{\omega_{0} }}} \right)^{2} } \right] \\ & \quad \int\limits_{0}^{ + \infty } \cos \left( {\log \frac{\omega }{{\sqrt k \omega_{0} }} - \phi + \frac{{\log \omega \log \left( {1 + {{\Updelta}}d} \right)}}{2} + \theta \left( {\log \omega } \right)} \right)\\&\quad \times\cos \left( {\frac{1}{2}\left( {\log \omega \log \left( {1 + {{\Updelta}}d} \right) - \log k} \right)} \right)\end{array} } \right]^{2} \\ + \left[{\begin{array}{*{20}l}& 2\rho \left( {\log \omega } \right)d\log \omega \exp \left[ { - \frac{{\sigma_{\omega }^{2} }}{2}\left( {\log \frac{\omega }{{\omega_{0} }}} \right)^{2} } \right] \\ & \quad \int\limits_{0}^{ + \infty} \sin \left( {\log \frac{\omega }{{\sqrt k \omega_{0} }} - \phi + \frac{{\log \omega \log \left( {1 + {{\Updelta}}d} \right)}}{2} + \theta \left( {\log \omega } \right)} \right)\\&\quad \times\cos \left( {\frac{1}{2}\left( {\log \omega \log \left( {1 + {{\Updelta}}d} \right) - \log k} \right)} \right)\end{array}}\right]^{2} $$

By the theorem that the square of a single integration is equivalent to a double integration

$$ \left( {\int {f(x)dx} } \right)^{2} = \iint {f(x)f(x')dxdx'} $$

The previous result can be simplified as

$$ \begin{aligned}R &= 4 \iint \limits_{{0 < \omega ,\omega^{\prime} < + \infty }} \exp \left[ { - \frac{{\sigma_{\omega }^{2} }}{2}\left( {\log \frac{\omega }{{\omega_{0} }}} \right)^{2} } \right]\\& \quad \times\cos \left( {\frac{1}{2}\left( {\log \omega \log \left( {1 + {{\Updelta}}d} \right) - \log k} \right)} \right) \hfill \\&\quad\times\exp \left[ { - \frac{{\sigma_{\omega }^{2} }}{2}\left( {\log \frac{{\omega^{\prime}}}{{\omega_{0} }}} \right)^{2} } \right]\\& \quad \times \cos \left( {\frac{1}{2}\left( {\log \omega^{\prime}\log \left( {1 + {{\Updelta}}d} \right) - \log k} \right)} \right)\\& \quad\times\alpha \rho \left( {\log \omega } \right)d\log \omega \rho \left( {\log \omega^{\prime}} \right)d\log \omega^{\prime}\end{aligned} $$

(B8)

where

$$\alpha = \cos \left( \begin{gathered}\theta ( \log {\omega } ) - \theta ( \log {\omega^{\prime}} ) + \log \frac{\omega }{{\omega^{\prime}}} \\ + {\frac{1}{2}} \log ( {1 + {{\delta}}d} ) \log \frac{\omega }{{\omega^{\prime}}} \end{gathered}\right)$$

(B9)

Equation B8 is a general expression of the response of our model complex cell and can be further specialized for a particular stimulus if the property of that stimulus in frequency domain is available.

For a broadband stimulus, its Fourier magnitude remains approximately constant across the whole spatial frequency domain and is independent of its phase. Therefore we let

$$ \rho \left( {\log \omega } \right) = \rho $$

Also note that $\alpha$ in Eq. B9 is usually not a constant since the Fourier phase is random. Therefore Eq. B10 is an approximated specialization of Eq. B8 by letting $\alpha = 1.$

$$ R_{\text{noise}} = \frac{{8\pi \rho^{2} }}{{\sigma_{\omega }^{2} }}\exp \left[ { - \frac{1}{{4\sigma_{\omega }^{2} }}\log^{2} \left( {1 + {{\Updelta}}d} \right)} \right]\cos^{2} \left[ {\frac{1}{2}\left( {\log \omega_{0} \log \left( {1 + {{\Updelta}}d} \right) - \log k} \right)} \right] $$

(B10)

Bar stimulus can be represented by a delta function

$$ I(x) = \left\{ \begin{gathered} 1,\quad x = x_{b} \hfill \\ 0,\quad x \ne x_{b} \hfill \\ \end{gathered} \right. $$

And its Fourier transform is

$$ \rho \left( {\log \omega } \right) = 1,\theta \left( {\log \omega } \right) = x_{b} \log \omega $$

Thanks to its linear Fourier phase, $\alpha$ of Eq. B9 is as a constant. Substitute this result into Eq. B8 yields Eq. B10 but with $\rho = 1.$