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[Note that sin(au + n/4) reaches its maximum for au = n/4 rad; this broadens the low-frequency peak of Ssa(u).]

Compare the expression for Ssa(u) above with the Fourier transform of the DSF of a single ommatidium: Sj(u) = an exp(-au), for u S 0. Figure 5.2-10 plots Sj(u) and Ssa(u) vs. u. For simplicity, let a = 1° = b. Note that Ssa(u) has less low-frequency attenuation than a single ommatidium DSF, but has a zero at about u = 2.3.

FIGURE 5.2-10 Plot of the Fourier transform of a single receptor DSF (trace 2) and the two-receptor, synthetic aperture DSF (trace 1) vs. spatial frequency, u. Trace 3 is the zero reference.

To verify that the SA architecture leads to increased intensity contrast, first calculate the intensity contrast for a one-dimensional black spot of "diameter" 2r, centered on x = 0. p(x) for the spot is p(x) = I0 {1 [U(x + r) - U(x - r)]} The Fourier transform of this spot is

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