5.8. Of interest is the motion sensitivity of a photoreceptor when a one-dimensional DSF, s(x) = exp[-x2/(2a2)], views a one-dimensional, spatial sinewave object given a small displacement, 8x. Mathematically this object can be written:
For algebraic ease, this sinewave can be written as a shifted cosine wave when Fourier transforming. Thus, F(u, 8x) = (Io/2)[2n 8(u) + 1/2 8(u + uo) + 1/2 8(u - uo]) exp[-ju(X/4 + 8x)]. Note that uo = 2n/X, where x = X is the spatial period of the sinewave. The right-hand exponential term shifts the cosine wave by 90° to make a -sinewave, and gives an additional displacement, 8x, required to give a contrast change. The contrast is defined here as C(8x) = AIe/Ie(0). Ie(0) is the effective absorbed light intensity with 8x = 0. AIe = Ie(0) - Ie(8x).
a. Use the Fourier transform approach to find an expression for Ie(0).
c. Use the expression for the contrast, C(8x), to calculate the 8x required to produce C(8x) = 0.01 as a function of the receptor DSF half-angle,
6m/2. Note that ct = 9m/2^ln(4) , let X = 5°. Consider 0.25 9 6m/2 9 2.5°. (Be careful to use proper angle dimensions, degrees or radians, when solving this problem.)
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