BOOLEAN ALGEBRA. 1. State & Verify Laws by using :


 Rosalind Lester
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1 BOOLEAN ALGEBRA. State & Verify Laws by using :. State and algebraically verify Absorption Laws. (2) Absorption law states that (i) X + XY = X and (ii) X(X + Y) = X (i) X + XY = X LHS = X + XY = X( + Y) = X. [ + Y = ] = X = RHS. Hence proved. (ii) X(X + Y) = X LHS = X(X + Y) = X. X + XY = X + XY = X( + Y) = X. = X = RHS. Hence proved. [ mark for the statement] [ mark for proving it algebraically]
2 2. State and verify Distributive Laws algebraically. (2) Distributive law state that (a) X(Y +Z) = XY + XZ (b) X + YZ = (X + Y)(X + Z) now proof for st no. is as simple as we can see = XY + XZ = X(Y +Z) L.H.S=R.H.S. now proof for 2nd. law R.H.S. = (X + Y)(X + Z) = XX + XZ + XY + YZ = X + XZ + XY + YZ (XX = X Indempotence law) = X + XY + XZ + YZ = X( + Y) + Z(X + Y) = X. + Z(X + Y) ( + Y = property of and ) = X + XZ + YZ) (X. = X property of and ) = X( + Z) + YZ = X. + YZ ( + Z = property of and ) = X. + YZ (X. = X property of and ) = L.H.S. Hence proved. 3. State and verify Demorgan's Laws algebraically. (2) 4. Prove algebraically the third distributive law X + X Y = X + Y. L.H.S. = X + X Y = X. + X Y (X. = X property of and ) = X( + Y) + X Y ( + Y = property of and ) = X + XY + X Y = X + Y(X + X ) = X + Y. (X + X = complementarity law) = X + Y (Y. = Y property of and ) = R.H.S. Hence proved. 2
3 2. Verify the following algebraically: (2). (A +B ).(A +B)=A.B+A.B LHS (A + B ). (A+B) = A.A + A.B + A.B + B.B = + A.B + A.B + = A.B + A.B = RHS (Verified) 2. Verify the following using Boolean Laws. X + Y'= X.Y+X.Y'+X'.Y' L.H.S =X + Y =X.(Y+Y )+ (X + X ).Y =X.Y + X.Y + X.Y +X.Y =X.Y + X.Y + X.Y =R.H.S OR R.H.S =X.Y + X.Y + X.Y =X.(Y + Y )+ X.Y =X. + X.Y =X + X.Y =X + Y =L.H.S Verify the following using Truth Table. (2). U. (U' +V) = (U + V) 2. X+Y. Z=(X+Y).(X+Z) 3. X+ (Y+Z) = (X+Y) + Z 4. A(B + B C + B C ) = A 5. A + A B = A + B' 6. (x + y + z).(x + y + z) = y + z 7. A B C + A BC + AB C = A C + B C 5. Name the law shown below and verify it using a truth table. (2). A+B.C= (A+B).(A+C)  > Distributive Law 2. X+X.Y=X+Y > third distributive law 3
4 6. What does duality principle state? What is its usage in Boolean algebra? The principle of duality states that starting with a Boolean relation, another Boolean relation can be derived by :. Changing each OR sign(+) to an AND sign(.). 2. Changing each AND sign(.) to an OR sign(+). 3. Replacing each by and each by. Principle of duality is use in Boolean algebra to complement the Boolean expression. 7. Write dual of the following Boolean Expression : (a) (x + y ) (b) xy + xy + x y (c) a + a b + b (d) (x + y + z)(x + y) (a) xy (b) (x + y)(x + y )(x + y) (c) a. (a + b). b (d) xy z + xy 8. Write the equivalent expression for the following Logical Circuit: (2) 4
5 9. Draw a logical Circuit Diagram for the following Boolean Expression: 2. (U + V').W' + Z 2. Draw the logic circuit for F=AB' + CD'. Draw a logical Circuit Diagram for the following Boolean Expression by using NOR gate and NAND Gate 5
6 . CANONICAL SOP AND POS. What do you mean by canonical form of a Boolean expression? Which of the following are canonical? (i) ab + bc (ii) abc + a bc + ab c (iii) (a + b)(a +b ) (iv) (a + b + c)(a + b + c)(a + b +c ) (v) ab + bc + ca Boolean Expression composed entirely either of Minterms or maxterms is referred to as canonical form of a Boolean expression. (i)non canonical (ii) canonical (iii) canonical (iv) canonical (v) Non canonical 6
7 Type : Q:.(i) Express P+Q R in canonical SOP form. () (ii) X + X Y + X Z = X(Y + Y )(Z + Z ) + X Y(Z + Z ) + X Z ( (Y + Y ) = (XY + XY )(Z + Z ) + X YZ + X YZ + X YZ + X Y Z = Z(XY + XY ) + Z (XY + XY ) + X YZ + X YZ + X YZ + X Y Z = XYZ + XY Z + XYZ + XY Z + X YZ + X YZ + X YZ + X Y Z By removing duplicate terms we get canonical Sum of=product form : XYZ + XY Z + XYZ + XY Z + X YZ + X YZ + X Y Z Q: 2.(i) Express P+Q R in POS form. () : P+Q R = (P + Q ) ).(P + R) [Distributive Law] = (P + Q + RR )(P + R + QQ ) = (P + Q + R)(P + Q + R ) (P + R + Q)(P + R + Q ) [Distributive Law] = (P + Q + R)(P + Q + R ) (P + R + Q) [ By Removing the duplicate terms] (ii) (X + Y)(Y + Z)(X + Z) = (X + Y + ZZ )(XX + Y + Z)(X + YY + Z) = (X + Y + Z)(X + Y + Z )(X + Y + Z)(X + Y + Z)(X + Y + Z)(X + Y + Z) By removing duplicate terms we get canonical Product of Sum form: (X + Y + Z)(X + Y + Z )( X + Y + Z)( X + Y + Z) Type 2: Q: 3 7
8 Type 3: Q: 4 Convert the following Boolean expression into its equivalent Canonical Product of Sum form: X.Y. Z + X.Y.Z + X.Y.Z () X.Y.Z + X.Y.Z + X.Y.Z X Y Z F POS X + Y + Z X + Y + Z X + Y + Z X + Y + Z X + Y + Z F( (X,Y,Z) = (X + Y + Z)( X + Y + Z )( X + Y + Z) ( X + Y + Z)( X + Y + Z ) Type 4: Q.5 2. K MAP Following each question carry 3 marks: ) 8
9 2) ) Reduce the following Boolean Expression using KMap: F(A,B,C,D)= (,,2,4,5,6,8,) 2) Reduce the following Boolean Expression using KMap: F(U,V,W,Z)= (,,2,4,5,6,8,) 9
10 3) Reduce the following Boolean Expression using KMap: F(A,B,C,D)= π(,,2,4,5,6,8,) 4) Reduce the following Boolean Expression using KMap: F (X, Y, Z, W) = (,,3,4,5,7,9,,,3,5) 5) Reduce the following Boolean expression using K map: F(M,N,O,P)= (,,3,4,5,6,7,9,,,3,5) 6) Obtain simplified form for a Boolean expression F(x,y,z,w) = Σ(,3,4,5,7,9,,2,3,5) using K Map 7) F(P,Q,R,S)= (,3,5,6,7,,2,5) 8) F(P, Q,R, S) = Σ (3,5,7,,,3,5) 9) F(X,Y,Z,W) = Σ (,3,4,5,7,,3,5) ) F(P,Q,R,S)= π (,3,5,6,7,,2,5) ) F(P, Q, R, S) = Σ(, 2, 3, 5,6, 7, 9,, 2, 3, 5) 2) F (A,B,C,D) = (,,3,4,5,7,8,,2,4,5) 3) F (A,B,C,D) = π(5,6,7,8,9,2,3,4,5) 4) F (X,Y,Z,W) = (,,4,5,7,8,9,2,3,5) 5) F (A,B,C,D) = (,2,3,4,6,7,8,,2) 6) F(U,V,W,Z)= π 7) F(P,Q,R,S)=Σ(l,3,5,8,,2,5) 8) F(U,V,W,Z) = (,, 2, 3, 4,, ) 9) F(P, Q, R, S) = (,,2,4,5,6,8, 2) 2) F(A,B,C,D)= (,3,4,5,6,7,2,3) 2) F (U, V, W, Z) = II (,, 3, 5, 6, 7, 5) 22) F(X,Y,Z,W) = Σ(,,6,8,9,l,,2,5)
X Y Z F=X+Y+Z
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