Computer Architecture HW Assignment1 Result.pdf

A b c f a babc a b c abc 0 0 0 0 0 0 1 0 1 0 0 1 1 1

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A B C F= ( A + B`+A`B`C`) (A + B + C) + (ABC) 0 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1 ii- Check the correct logic diagram of the original Boolean expression for F from the following a = b = = = = B C C A B F A B C B A A
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b- c- d- e- f- None of the above B C C A B F A B C B A A B C C A B F A B C B A A B C C A B F A B C B A A B C C A B F A B C B A A
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iii- Which one of the following represents the simplified algebraic expression for F: a- A+ B`+C b- A+ B`C c- A`+ B + C d- AB` +C e- A`B` +AC f- None of the above iv- Draw the logic diagram for the simplified expression and match it with one of the following diagrams: a- b- c- d- e- f- None of the above v- Insert the total number of 2-input gates used for the simplified and original Boolean experssions in the following table (Hint; NOT gate is also considered 2-input gate): Original Boolean function Simplified Boolean function 4- Using four-variable k-maps. i- Use the following k-map to insert the logic values that corresponds to 0 or 1 for; F(A,B,C,D) = ABCD`+A`BCD`+ A`B`C`D`+ A`B`C D`+ AB C`D`+ ABC`D`+ ABC D` ii- The K-map given below is for F(A, B, C, D) = ABCD` + AB`C`D`+A`B`C`D`+ A`BC`D`+ AB`C`D + AB`CD + ABC`D + ABC`D`+ ABCD = = AB CD 00 01 11 10 00 0 4 12 8 01 1 5 13 9 11 3 7 15 11 10 2 6 14 10 AB CD 00 01 11 10 00 1 1 1 1 01 1 1 11 1 1 10 1 B A F C B C F A B C F A A C F B F B A C A
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Which one of the following expressions represents the simplified Boolean function for F? a- F=C`D`+AB+AB`D b- F=C`D`+AB+AD c- F=CD+AB+AD d- F=A`C`D`+AB+AC`+AD e- F= AB +C`D` +AC`D +ACD f- None of the above iii- Which one of the following expressions represents the simplified Boolean function for F(A, B, C, D) = A`B`CD + A`B`CD`+ AB`CD`+AB`CD+AB`C`D+ ABC`D + A`BCD` + ABC D`+ A`BCD+ ABCD a- F= CD +CD`+AC`D b- F= CD+CD`+AD c- F= C+AC`D c- F= AC`D +A`C+AC e- F=AD+C f- None of the above iv- Which one of the following expressions represents the simplified Boolean function for F(A, B, C, D) = A`B`C`D+ A`BC`D+ ABCD + A`B`C`D` + A`B`C D + A`B C`D` + A`B C D`+ A`B C D + ABC`D` + A B`C D`+ A B`C D +AB`C`D` + A B C D` + A` B` C D` AB CD 00 01 11 10 00 1 1 1 1 01
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  • Spring '10
  • DR.WEEKS
  • Boolean Algebra, +ABC

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