## J0

Because the -j sin(bu) term integrates to zero, it is clear that Ie2 = Ie3. When the trigonometric identity, sin(A) cos(B) = (l/2)[sin(A + B) + sin(A B)], is used, the definite integrals become

Ie2 = Ie3 = akoIo {n - tan^ [(r + b)/a] - tan^ [(r - b)/a]} 5.2-96

Substituting Equations 5.2.89 and 5.2.96 into 5.2.81, 5.2.82, and 5.2.83 yields

Vimin = F exp{GKA ln {(akoIo/B)3 [n - 2 tan-1(r/a)][n - tan-1[(r + b)/a]

Vimin = F exp{ ln {(akoIo/B)3GKA [n - 2 tan-1(r/a)]GKA[n - tan-1[(r + b)/a]

Vimin = F (akoIo/B)3GKA [n - 2 tan-1(r/a)]GKA {[n - tan-1[(r + b)/a]

At this point in this example, it is useful to assume reasonable numerical values for the parameters, and to calculate CVi = (Vimax - Vimin)/(Vimax + Vimin) numerically and compare it to CV1 for just one receptor. Let K = 3, A = 0.01, G = 100, r = 0.1°, b = 0.2°, a = 1°. The other parameters (kg, Io, F, B) cancel in calculating Cvi. CVi is found:

F(akolo/B)9n9 - F(akolo/B)9[n - 0.4]3[n - 0.4]6 = 1.3318 F(akolo/B)9 n9 + F(akolo/B)9[ - 0.4]3 [n - 0.4]6 4.6300

But for just one receptor:

C = F(k0l0a nB)GKA - F((lkoloaB)[n - 2 tan- 1(r/a)]} V1 F(koloanB)GKA + F((]koloaB)^ 2tan- l(r/a)]^ 52100

The ratio of contrast improvement is R = CVi/CV1 = 2.93.

Thus, it is clear that this simple mathematical model of the synthetic aperture/MSP system indicates that there will be a significant improvement in contrast in the hyper-polarizing response of the medullar interneuron to a moving, contrasting object.

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