we saw that multiplying the phase angle by 1 flipped the image with respect to both coordinate axes. Suppose that instead we multiplied the magnitude of the transform by 1 and then took the inverse DFT using the equation: g(x, y) = 1 { F(u,v) ejf(u,v)}. (a) * What would be the difference between the two images g(x, y) and f (x, y)? [Remember, F(u,v) is the DFT of f (x, y).] (b) Assuming that they are both 8-bit images, what would g(x, y) look like in terms of f (x, y) if we scaled the intensity values of g(x, y) using Eqs. (2-31) and (2-32), with K = 255?