Digital Logic HW 3 Solutions

# Digital Logic HW 3 Solutions - 8 + 1 10 = 48 Problem 4 3.6...

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Solution. Problem 1 2.39 Problem 2 2.51a module prob2 51 (x1, x2, x3, x4, f1, f2); input x1, x2, x3, x4; output f1, f2; assign f1 = (x1 & x3) | (~x1 & ~x3) | (x2 & x4) | (~x2 & ~x4); assign f2 = (x1 & x2 & ~x3 & ~x4) | (~x1 & ~x2 & x3 & x4) | (x1 & ~x2 & ~x3 & x4) | (~x1 & x2 & x3 & ~x4); endmodule Problem 3 3.1 a.

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b. number of transistors = NOT gates × 2 + AND gates 8 + OR gate 10 = 3 2 + 4
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Unformatted text preview: 8 + 1 10 = 48 Problem 4 3.6 a. b. The canonical SOP expression is f = x 1 2 3 + 1 2 3 + 1 2 3 + 1 2 3 + 1 2 3 The number of transistors required using only AND, OR, and NOT gates is number of transistors = NOT gates 2 + AND gates 8 + OR gates 12 = 3 2 + 5 8 + 1 12 = 58 Problem 5 3.9 Pull down network...
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## Digital Logic HW 3 Solutions - 8 + 1 10 = 48 Problem 4 3.6...

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