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(2 points) Given four inputs: a, b, c & d, where (a, b) represents a 2-bit unsigned binary number X; and (c, d) represents a 2-bit unsigned binary number Y (so both X and Y are in the range #0 to #3) - i. e. same as q. 2 This time, there are 4 outputs - in effect, you are building 4 separate truth tables - o3, o2, o1, o0. The resulting number [o3:o0] corresponds to the unsigned 4-bit product of the two numbers X, Y E. g., if a = 1, b = 0 (i. e. X = b10 => #2); c = 1, d = 1 (i. e. Y = b11 => #3); then [o3:o0] = {0, 1, 1, 0}, since #2 * #3 = #6.

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(2 points) Given four inputs: a, b, c & d, where (a, b) represents a 2-bit unsigned binary numbe...
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