Question

BC z 0 0 1 ololo 1 0 0 0 0 1 10 110

1.Show the circuit for the truth table shown above
2.Extract the Boolean equation to describe the circuit you found. Don’t simplify it yet!!!

3,Reduce the Boolean equation to its simplest form. Cite any Boolean identities used

4.Draw the logic circuit corresponding to the simplified expression and its truth table.

Answer 1

1) CIRCUIT:

......OV . . . . . . A'BC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . BO CO K- A'BC . . . . A AB'C' D V . . . . . . . . . . . KI N . . . . . . . . . . . (0) . . . . . . . . . AB'C . . . . . . . . . . . . . . . . . . . . . . . . . ... ...ABC ABC ...........

2) Given Z (A, B, C) = ​ m (0, 2, 3, 4, 5, 6)

m0 = 000 = A'B'C'

m2 = 010 = A'BC'

m3 = 011 = A'BC

m4 = 100 = AB'C'

m5 = 101 = AB'C

m6 = 110 = ABC'

The Boolean equation for Z = A'B'C'+A'BC'+A'BC+AB'C'+AB'C+ABC'

3) Given Z = A'B'C'+A'BC'+A'BC+AB'C'+AB'C+ABC'

Z = A'B'C'+ A'B(C'+C)+AB'(C'+C)+ABC'  { By Distributive law PQ+PR= P(Q+R) }

Z = A'B'C'+ A'B(1)+AB'(1)+ABC' { We know that P+P'=1 }

Z = A'B'C'+A'B+AB'+ABC'

Z = A'B'C'+A'B+A(B'+BC')

Z = A'B'C'+ A'B+A(B'+B)(B'+C') { By Distributive law P+QR = (P+Q)(P+R) }

Z = A'B'C'+A'B+A(1)(B'+C') { We know that P+P'=1 }

Z = A'B'C'+A'B+A(B'+C')

Z = A'B'C'+A'B+AB'+AC' { By Distributive law P(Q+R) = PQ+PR }

Z =   A'B+AB'+A'B'C'+AC'
Z =   A'B+AB'+C'(A'B'+A)
Z =   A'B+AB'+C'(A'+A)(B'+A) { By Distributive law P+QR = (P+Q)(P+R) }
Z =   A'B+AB'+C'(1)(B'+A)  { We know that P+P'=1 }
Z =   A'B+AB'+C'(B'+A)

Z = A'B+AB'+AC'+B'C { By Distributive law P(Q+R) = PQ+PR }

Which is Required Simplified Boolean Equation

4) CIRCUIT:

. . . . . . . AB R . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . AB' A B . . . . . . . . . . . . . . OK . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . D B'C I . . . . . . . . . . . . . . . . . . . . . . ..... ... A L ...........