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20085eeM16_2_Homework2_Soln

20085eeM16_2_Homework2_Soln - 2 7-Seqment Decoder You are...

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2. 7-Seqment Decoder You are designing a combinational system for displaying binary coded decimal (BCD) digits. F b a. Write down the truth table for z b = F(x b ) X 3 X 2 X 1 X 0 Z 6 Z 5 Z 4 Z 3 Z 2 Z 1 Z 0 0 0 0 0 0 1 1 1 1 1 1 0 0 0 1 0 0 0 0 0 1 1 0 0 1 0 1 1 0 1 1 0 1 0 0 1 1 1 1 0 0 1 1 1 0 1 0 0 1 0 1 0 0 1 1 0 1 0 1 1 1 1 0 1 1 0 0 1 1 0 1 0 1 1 1 1 0 0 1 1 1 0 1 0 0 0 1 1 1 0 0 0 1 1 1 1 1 1 1 1 0 0 1 1 1 1 0 0 1 1 b. Write the switching expressions for z b in a sum-of-product form. Z 6 = x3’x2’x1x0’ + x3’x2’x1x0 + x3’x2x1’x0’ + x3’x2x1’x0 + x3’x2x1x0’ + x3x2’x1’x0’ + x3x2’x1’x0 Z 5 = x3’x2’x1’x0’ + x3’x2’x1x0’ + x3’x2’x1x0 + x3’x2x1’x0 + x3’x2x1x0 + x3x2’x1’x0’ + x3x2’x1’x0 Z 4 = x3’x2’x1’x0’ + x3’x2x1’x0’ + x3’x2x1’x0 + x3’x2x1x0’ + x3x2’x1’x0’ + x3x2’x1’x0 Z 3 = x3’x2’x1’x0’ + x3’x2’x1x0’ + x3’x2x1x0’ + x3x2’x1’x0’ Z 2 = x3’x2’x1’x0’ + x3’x2’x1x0’ + x3’x2’x1x0 + x3’x2x1’x0 + x3’x2x1x0’ + x3x2’x1’x0’ Z 1 = x3’x2’x1’x0’ + x3’x2’x1’x0 + x3’x2’x1x0 + x3’x2x1’x0’ + x3’x2x1’x0 + x3’x2x1x0’ + x3’x2x1x0 + x3x2’x1’x0’ + x3x2’x1’x0 Z 0 = x3’x2’x1’x0’ + x3’x2’x1’x0 + x3’x2’x1x0’ + x3’x2’x1x0 + x3’x2x1’x0’ + x3’x2x1x0 + x3x2’x1’x0’ + x3x2’x1’x0 BCD coder 5 x b control signal decoder 4 3 6 x in {0,1,2 …,9} z b 1 0 2
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c. Use Boolean algebra to minimize switching expressions. Z 6 = x3’x2’x1x0’ + x3’x2’x1x0 + x3’x2x1’x0’ + x3’x2x1’x0 + x3’x2x1x0’ + x3x2’x1’x0’ + x3x2’x1’x0 = x3’x2’x1(x0’ + x0) + x3’x2x1’(x0’ + x0) + x3’x2x1x0’ + x3x2’x1’(x0’ + x0) By Distributive = x3’x2’x1 + x3’x2x1’ + x3’x2x1x0’ + x3x2’x1’ By Complement Z 5 = x3’x2’x1’x0’ + x3’x2’x1x0’ + x3’x2’x1x0 + x3’x2x1’x0 + x3’x2x1x0 + x3x2’x1’x0’ + x3x2’x1’x0 = x3’x2’x0’(x1’ + x1) + x3’x2’x1x0 + x3’x2x0(x1’ + x1) + x3x2’x1’(x0’ + x0) By Distributive
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