Question

With four Boolean inputs mentioned in the first four columns of the following table, report the outputs from the circuits described by the Boolean expressions in the topmost row of the table in the 5^th and 6^th columns, in the corresponding rows in the 5^th and 6^th columns of the table. The circuit in the 5^th column uses two AND gates, two NOT gates, and one OR gate. The circuit in the 6^th column uses two AND gates and one OR gate. You can draw the circuits or construct partial truth tables which have just the given combinations of input values instead of the 16 possible combinations, if you wish. You can also find the answers by simply using the meaning of AND, OR, and NOT gates.

X1	X2	X3	X4	NOT(X1) AND ((X2 OR X3) AND (NOT(X4)))	(X1 AND X2) OR (X3 AND X4)
0	0	0	0
1	1	0	0
0	1	1	0

Answer 1

Truth Table: 5^th column i.e NOT(X₁) AND ((X₂ OR X₃) AND (NOT(X₄)))

X1	X2	X3	X4	NOT(X1)	NOT(X4)	X2 OR X3	NOT(X1) AND ((X2 OR X3)	NOT(X1) AND ((X2 OR X3) AND (NOT(X4)))
0	0	0	0	1	1	0	0	0
0	1	1	0	1	1	1	1	1
1	1	0	0	0	1	1	0	0

The Boolean expression for 5^th column is F (X1, X2, X3, X4) = ​ M (0, 12)

Mo = 0000 = (X1+X2+X3+X4)

M12 = 1100 = (X1'+X2'+X3+X4)

F (X1, X2, X3, X4) = (X1+X2+X3+X4)(X1'+X2'+X3+X4)

= [X1+X2+(X3+X4)][X1'+X2'+(X3+X4)]
= [(X1+X2)(X1'+X2')+(X3+X4)] { By Distributive law P+QR = (P+Q)(P+R) }
= [X1(X1'+X2')+X2(X1'+X2')+(X3+X4)] { By Distributive law P+QR = (P+Q)(P+R) }
= [(X1X1'+X1X2')+(X1'X2+X2X2')+(X3+X4)] { By Distributive law P(Q+R) = PQ+PR }
= [(0+X1X2')+(X1'X2+0)+(X3+X4)] { We know that PP'=0 }
= [X1X2'+X1'X2+(X3+X4)]

The Simplified Boolean expression for 5^th column is F (X1, X2, X3, X4) = X1X2'+X1'X2+ X3+X4

Circuit for 5^th column

Truth Table: 6^th column i.e (X₁ AND X₂) OR (X₃ AND X₄)

X1	X2	X3	X4	X1 AND X2	X3 AND X4	(X1 AND X2) OR (X3 AND X4)
0	0	0	0	0	0	0
0	1	1	0	0	0	0
1	1	0	0	1	0	1

The Boolean expression for 6^th column is  F (X1, X2, X3, X4) = ​ M (0, 6)

Mo = 0000 = (X1+X2+X3+X4)

M12 = 0110 = (X1+X2'+X3'+X4)

F (X1, X2, X3, X4) = (X1+X2+X3+X4)(X1+X2'+X3'+X4)

= (X1+X2+X3+X4)(X1+X2'+X3'+X4)
= [((X1+X4)+X2+X3)((X1+X4)+X2'+X3')]
= [(X1+X4)+(X2+X3)(X2'+X3')] { By Distributive law P+QR = (P+Q)(P+R) }
= [(X1+X4)+(X2(X2'+X3')+X3(X2'+X3'))] { By Distributive law P+QR = (P+Q)(P+R) }
= [(X1+X4)+(X2X2'+X2X3')+(X2'X3+X3X3'))] { By Distributive law P(Q+R) = PQ+PR }
= [(X1+X4)+(0+X2X3')+(X2'X3+0))]  { We know that PP'=0 }
= X1+X4+X2X3'+X2'X3

The Simplified Boolean expression for 5^th column is F (X1, X2, X3, X4) = X1+X4+X2X3'+X2'X3

Circuit for 6^th column