Info

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)

Was this article helpful?

0 0
Peripheral Neuropathy Natural Treatment Options

Peripheral Neuropathy Natural Treatment Options

This guide will help millions of people understand this condition so that they can take control of their lives and make informed decisions. The ebook covers information on a vast number of different types of neuropathy. In addition, it will be a useful resource for their families, caregivers, and health care providers.

Get My Free Ebook


Post a comment