Question

please answers all of them!

1. Simplify, using algebraic manipulations, the following Boolean expressions to a mini- mum number of terms and factors. (a) XYZ + XY + XYZ (b) XYZ + XZ 2. Find the complement of the following expression: (a) XY + XY 3. Using DeMorgan's Theorem, express the following function .... (a) F= XY+XY + ÝZ ... with only OR and complement operations. 4. Propose and solve your own logic simplification problem using logic theorems 5. Simplify the following expressions using K-Map (a) XY+YŻ + XYZ 6. Simplify the following expressions by means of a four-variable K-Map (a) AD + BD + BC + ABD (b) ABC + CD + BCD + BC 7. Propose and solve your own K-Map simplification problem (with do not care conditions)

Answer 1

1) a) Given F = XYZ+X'Y+XYZ'

F = XYZ+XYZ'+X'Y

F = XY(Z+Z')+X'Y { By Distributive law PQ+PR = P(Q+R) }

F = XY(1)+X'Y  { We know that P+P'= 1 }

F = XY+X'Y

F = Y(X+X') { By Distributive law PQ+PR = P(Q+R) }

F = Y(1)  { We know that P+P'= 1 }

F = Y

b) Given F = X'YZ+XZ

F = Z(X'Y+X)

F = Z(X'+X)(Y+X)  { By Distributive law P+QR = (P+Q)(P+R) }

F = Z(1)(Y+X)  { We know that P+P'= 1 }

F = Z(Y+X)

F = YZ+XZ { By Distributive law P(Q+R) = PQ+PR }

F = YZ+XZ

2) a) Given F = XY'+X'Y

F' = [XY'+X'Y]'

F' = [XY']' [X'Y]'  { We know that (P+Q)'= P' Q' }

F' = [X'+(Y')'] [(X')' + Y']   { We know that (PQ)'= P' + Q' }

F' = [X'+Y] [X + Y']  { We know that (P')'= P }

F' = (X'+Y)(X + Y')

3) a) Given F = XY+X'Y'+Y'Z

F = XY+Y'(X'+Z) { By Distributive law PQ+PR = P(Q+R) }

F = [X+Y'(X'+Z)][Y+Y'(X'+Z)]  { By Distributive law P+QR = (P+Q)(P+R) }

F = [(X+Y')(X+(X'+Z))][(Y+Y')(Y+(X'+Z))]  { By Distributive law P+QR = (P+Q)(P+R) }

F = [(X+Y')((X+X')+Z)][(Y+Y')(Y+(X'+Z))]

F = [(X+Y')((1)+Z)][(1)(Y+(X'+Z))]  { We know that P+P'= 1 }

F = [(X+Y')(1+Z)][(Y+X'+Z)]

F = [(X+Y')(1)][(Y+X'+Z)]  { We know that 1+P= 1 }

F = (X+Y')(X'+Y+Z)

4)

Given Equation

F = A'B + AB'

G = A'B' + AB

Consider

G' = (A'B' + AB)'

G' = (A'B')' (AB)'  { We know that (P+Q)'= P' Q' }

G' = [(A')'+ (B')'] [A'+B']   { We know that (PQ)'= P' + Q' }

G' = [A+B] [A'+B']   { We know that (P')'= P }

G' = [A(A'+B')+B(A'+B')] { By Distributive law P(Q+R) = PQ+PR }

G' = [(AA'+AB')+(A'B+BB')] { By Distributive law P(Q+R) = PQ+PR }

G' = [(0+AB')+(A'B+0)] { We know that PP'= 0 }

G' = AB'+A'B { We know that P+0= P }

G' = A'B + AB'

given F = A'B + AB'

From the above Equation we can say that F = G'

5) a) Given F = XY+YZ'+X'Y'Z'

Above Function in K-map as follows

K-map XY YZ ΥΖ 00 01 11 ΧΎ7'

Simplified K-map as follows

K-map YZ 00 0 1 11 10 X'Z'

The Simplified SOP of F (X, Y, Z) = F = XY+X’Z’

6) a) Given F = A'D+BD+B'C+AB'D

Above Function in K-map as follows

K-map A'D B'C CD 00 01 1 1 BD 11 13 15 AB'D 10

Simplified K-map as follows

K-map B'c CD 00 01 Τ.

The Simplified SOP of F (A, B, C, D) = D + B’C

6) b) Given F = ABC+CD+BC'D+B'C

Above Function in K-map as follows

K-map со в'c 10 CD 00 01 (11 BC'D ABC 11 12

Simplified K-map as follows

K-map CD 00 01 11 BD 11 13 10

The Simplified SOP of F (A, B, C, D) = BD+CD+AC