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Given
using rule that Z'+Z=1
now using Y+Y=Y
is the answer
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Reduce Boolean equation X=BC+A'B'C+BC'+AB'C to a minimum SOP form. A/
Using Boolean algebra, the complete simplification of A'B'C+ABC+AB'C'+A'BC' gives us: ABC+A'B'C' (ABC (ABC) eB о Авес AB+C' (AEB).(BEC)
Reduce the following equation using Boolean algebra and show all of your steps. 0 - A'B'C + A'BC' + A'BC + ABC
simplify the following boolean expression using boolean identities(A' means NOT A): X=(AB'C')+(AB'C)+(ABC)
#4 Given the Boolean function F(A,B,C) = A'C + A'B + AB'C + BC, a) construct the truth table. b) Simplify the expression and draw the resulting combinational circuit (AND, OR, NOT).
(c) Use a Karnaugh map to reduce the following 4-variable expression to a minimum SOP form. CD(AB+AB)+CD(AB+AB)+CD(AB+ AB)+ ABCD+ABCD+ACD+ACD (5 marks)
(10) Applying DeMorgan, simplify logical equivalent of X=(A[BC]+ [ABC] + A'B'C') Draw the equivalent circuit utilizing minimum number of logic gates. Verify attachment for X
1. Use K-maps to reduce each of the following to a minimized SOP form: (a) A + BC + CD (b) ABCD + ABCD + ABCD + ABCD (c) ABCD + CD) + ABCD + CD) + ABCD (d) (AB + ABXCD + CD) (e) AB + AB + CD + CD 2. Use K-maps to find the minimum SOP expression for the logic function shown in the table to the right. Implement the circuit using NAND gates only. Inputs...
1. Convert the following expressions to standard SOP form (a) BC + DE(BC + DE) (b) BC(CD + CE) (c) B + C[BD + C + DE] 2. Develop a truth table for each of the following standard SOP expressions (a) ABCD + ABCD + ABCD + ABCD (b) WXYZ + WXYZ + WXYZ + WXYZ + WXYZ
answer asap and show work: 1) Using Boolean Algebraic Properties, reduce the following expressions to a minimum SOP form. Show each step. a) ? = ?’?’?’ + ?’?’? + ?’?? + ??? + ???’: b) ? = ?’?’?’ + ?’?’? + ??? + ??’? : c)? = ?? + ?’?’? + ?’??’ + ????’ + ?’?? + ??: d) ? = ??? + ??? + ?’?’:
[8] Using properties of Boolean algebra, simplify the following Boolean expressions so they could be built with the minimum number of gates. a. X= A + BC + AB + ABC + B b. Y = AB + B(AC + BC + ABC' + A) C. W = ABC' + AB'C' + B'CD + A'C + BC d. Z = (A + B')' + (ABC')' +A(B + A'C)'