hwso-0401(new)

hwso-0401(new) - HW #1 Solutions ECE 178 WINTER 2004 B.S....

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HW #1 Solutions ECE 178 WINTER 2004 B.S. Manjunath TAs: Srivatsan Pallavaram, Evan Ruzanski and Christopher Utley Problem 2.1 The diameter, x, of the retinal image corresponding to the dot is obtained from similar triangles, as shown in Fig. P2.1. That is, which gives x = 0.07d. From the discussion in Section 2.1.1, and taking some liberties of interpretation, we can think of the fovea as a square sensor array having on the order of 337,000 elements, which translates into an array of size 580 X 580 elements. Assuming equal spacing between elements, this gives 580 elements and 579 spaces on a line 1.5 mm long. The size of each element and each space is then s = [(1.5mm)/1, 159] = 1.3 X 10 -6 m. If the size (on the fovea) of the imaged dot is less than the size of a single resolution element, we assume that the dot will be invisible to the eye. In other words, the eye will not detect a dot if its diameter, d, is such that 0.07(d) < 1:3 X 10 -6 m, or d < 18.6 X 10 -6 m. Problem 2.2 Brightness adaptation. Problem 2.3 λ = c/ ν = 2:998 X 10 8 (m/s)=60(1/s) = 4.99 X 10 6 m = 5000 Km. λ : wavelength of the alternating current, c: velocity of light, ν : frequency of the alternating current Problem 2.4 (a) From the discussion on the electromagnetic spectrum in Section 2.2, the source of
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hwso-0401(new) - HW #1 Solutions ECE 178 WINTER 2004 B.S....

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