Figure P54

53 (x) = exp[- (x - b)2/(2o2)) Note: 0 = 9^/7^(4), so Si (±9^ ) = 0.5

Assume

Let A = 0.01°, 9m/2 = 1.4°, K = 10, G = 100, and F = 10-3. _

Derive an expression for the limiting contrast: CVi = Vi/Vi - Vi (as A/0m/2 ^ 0) as a function of 0m/2/A and b. Show how CVi behaves as b is decreased to zero. Assume the square-wave object is moved A/2 so a black stripe is centered on x = 0. Note that at limiting resolution the square wave can be represented by the first harmonic in its Fourier series. A similar development can be used as that described in the Equations 5.2-77 to 5.2-108. 5.5. A one-dimensional, linear, LI system can be modeled by the integral equation:

Assuming the instantaneous frequency, r(x) S 0 everywhere. The spatial inhibition function is given by k k(x) =-^^

a. Find an expression for K(u), the Fourier transform of k(x). Assume the input e(x) is a spatial sinewave riding on a dc bias so r(x) S 0. That is, e(x) = A + B sin(ux) over all x. A > B.

b. Calculate and plot the one-dimensional, LI system SS frequency response, (R/E) (u), where u is the spatial frequency in rad/mm. Let ko = 2, and a = 1 mm.

5.6. The spatial inhibition function, k(x), of a one-dimensional, linear LI system is shown in Figure P5.6. k(x) is the product of a cosine wave, cos(uox), and a rectangular gating function, g(x) = ko for |x| 3 3n/(2uo), and g(x) = 0 for |x| > 3n/(2uo). That is, k(x) = g(x) cos(uox). Note that this k(x) is inhibitory in its center, and excitatory in its periphery.

a. Use the Fourier transform theorem, g(t) cos(raot) ^ 1/2 (G(ra + rao) + G(ra - rao)}, to find K(u).

b. Again, let the LI system input be dc plus a spatial sine wave: e(x) = A + B sin(uox). Calculate and plot (R/E) (u). Let ko = 2, uo = 1 r/mm.

5.7. A photoreceptor has a one-dimensional DSF given by