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When the spot is at x, the maximum intensity is absorbed by all three retinula cells, and V is maximum. That is, Ie1 = Ie2 = Ie3 = Iemax:

emax 2n

-J 2rc5(u)[aTCexp(-a|u|)] (j°du = koloan 5.2-84 Thus,

V; = F exp{GKA ln[(koIo an)3/B3] = F[(ko Io a n)/B]3GKA = Vimax 5.2-86

Next, center the black spot of diameter 2r over the center receptor at x = 0. The effective intensity at retinula cell 1 is reduced:

1. = 1. -A1. = k l an-^ f" 2rsin(ru)[anexp(-a|u|)]du 5.2-87

Because the integrand is an even function, the integral can be written as k l r 2 sin(ru)

l„ = l.,„„-AL = kXarc—-^[arcexp(-au)] du 5.2-88

Jo u el elmax el o o

2n jo

The definite integral has the well-known solution:

Ie1 = akoIo [n - 2 tan-1(r/a)] 5.2-89 Now the offset DSFs have the Fourier transforms:

s2 (x) = 1 [l + (x + b)2/ a2 ]—^ S2 (u) = an exp(-a| u|) e+jub 5.2-90

s3 (x) = l/ [l + (x - b)2/ a2 ] ——^ S3 (u) = a n exp(-a| u|) e- jub 5.2-91

The Euler relation, e±jub = cos(ub) ± j sin(ub), is used to finally obtain:

S2(u) = an exp(-a |u|) [cos(ub) + j sin(ub)] 5.2-92

S3(u) = an exp(-a |u|) [cos(ub) - j sin(ub)] 5.2-93 Now Ie2 can be found:

le2 = koloan-^^ f sin(ru)[anexp(-a|u|)[cos(bu) + jsin(bu)]du 5.2-94 2n J-» u

The (odd) j sin(bu) term integrates to zero, so kolo _ r sin(ru)

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