1  

BOOLEANALGEBRA1.State&VerifyLawsbyusing:‐

1. State and algebraically verify Absorption Laws.           (2) 

Ans: Absorption law states that (i) X + XY = X and (ii) X(X + Y) = X (i) X + XY = X LHS = X + XY = X(1 + Y)         = X . 1 [∵ 1 + Y = 1]         = X = RHS. Hence proved. (ii) X(X + Y) = X LHS = X(X + Y) = X . X + XY         = X + XY         = X(1 + Y)         = X . 1        = X = RHS. Hence proved. [1 mark for the statement] 

[1 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 1st 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(1 + Y) + Z(X + Y) = X.1 + Z(X + Y) (1 + Y = 1 property of 0 and 1) = X + XZ + YZ) (X . 1 = X property of 0 and 1) = X(1 + Z) + YZ = X.1 + YZ (1 + Z = 1 property of 0 and 1) = X.1 + YZ (X . 1 = X property of 0 and 1) = 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.1 + X’Y        (X . 1 = X property of 0 and 1)            = X(1 + Y) + X’Y   (1 + Y = 1 property of 0 and 1)            = X + XY +  X’Y                                                                    = X + Y(X + X’)            = X + Y.1          (X + X’ =1 complementarity law)            = X + Y            (Y . 1 = Y property of 0 and 1)            = R.H.S.           Hence proved. 

 

3  

2.Verifythefollowingalgebraically:‐(2)

1. (A’+B’).(A +B)=A’.B+A.B’ Ans:

LHS (A’ + B’) . (A+B) = A’.A + A’.B + A.B’ + B’.B = 0 + A’.B + A.B’+ 0 = A’.B + A.B’ = RHS (Verified)

2. Verify the following using Boolean Laws. X + Y'= X.Y+X.Y'+X'.Y' Ans: 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.1 + X’.Y’ =X + X’.Y’ =X + Y’

=L.H.S 3.

4.VerifythefollowingusingTruthTable.(2)

1. 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.Namethelawshownbelowandverifyitusingatruthtable.(2)

1. A+B.C= (A+B).(A+C) - > Distributive Law 2. X+X’.Y=X+Y -> third distributive law

4  

6.Whatdoesdualityprinciplestate?WhatisitsusageinBooleanalgebra?The principle of duality states that starting with a Boolean relation, another Boolean relation can be derived by : 1. Changing each OR sign(+) to an AND sign(.).   2. Changing each AND sign(.) to an OR sign(+). 3. Replacing each 0 by 1 and each 1 by 0. Principle of duality is use in Boolean algebra to complement the Boolean expression.  

7.WritedualofthefollowingBooleanExpression:

(a) (x + y’) (b) xy + xy’ + x’y (c) a + a’b + b’ (d) (x + y’ + z)(x + y) Ans:

(a) xy’     (b) (x + y)(x + y’)(x’ + y) (c) a . (a’ + b) . b’     (d) xy’z + xy 

8.WritetheequivalentexpressionforthefollowingLogicalCircuit:(2)

5  

9.DrawalogicalCircuitDiagramforthefollowingBooleanExpression:2

1. (U + V').W' + Z

2. Draw the logic circuit for F=AB' + CD'   Ans: 

  

10.DrawalogicalCircuitDiagramforthefollowingBooleanExpressionbyusingNORgateandNANDGate

6  

11.CANONICALSOPANDPOS

1. 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 

 

Type 1:

Q: 1.(i)E

(ii) X + X= X(Y + Y= (XY + X= Z(XY + = XYZ + XBy remoXYZ + XY

Q:2.(i)E

Ans::

P====

(ii) (X + = (X + Y += (X + Y +By remo(X + Y + Z

Type 2:

Q:3

Express

X’Y + X’Z’Y’)(Z + Z’) +XY’)(Z + Z’XY’) + Z’(XXY’Z + XYZving duplY’Z + XYZ’ 

Express

P+Q ’R = (P + Q’)= (P + Q’ = (P + Q’ = (P + Q’

Y)(Y + Z+ ZZ’)(XX’ + Z)(X + Y ving duplZ)(X + Y + 

P+Q’Rin

+ X’Y(Z + ) + X’YZ +XY + XY’) Z’ + XY’Z’ licate term+ XY’Z’ + 

P+Q’Rin

).(P + R) + RR’)(P+ R)(P + + R)(P +

Z)(X + Z)  + Y + Z)(X+ Z’)(X + licate termZ’)( X’ + Y

ncanonic

Z’) + X’Z’( X’YZ’ + X+ X’YZ + X+ X’YZ + Xms we getX’YZ + X’Y

nPOSfor

[DistribP + R + Q

Q’ + R’)Q’ + R’)

X + YY’ + ZY + Z)(X’ +ms we getY + Z)( X +

calSOPfo

(Y + Y’) X’YZ’ + X’YX’YZ’ + X’YX’YZ’ + X’Yt canonicaYZ’ + X’Y’Z

rm.

butive LaQ’) (P + R + (P + R +

Z) + Y + Z)(X t canonica+ Y’ + Z) 

7

orm.

Y’Z’ YZ’ + X’Y’YZ’ + X’Y’Zal Sum‐ofZ’ 

aw]

Q)(P + RQ) [ By R

+ Y + Z)(Xal Produc

Z’ Z’ f=Product

R + Q’) [DRemoving

X + Y’ + Z)ct‐of‐Sum 

t form : 

Distributig the dup

 form: 

ive Law] licate ter

(1)

(1)

rms]

)

)

 

Type 3:

Q:4Conform:

Type4:Q.5

12.K‐M

Followin1)

nvertthe

F(

AP

ngeachq

efollowinX.Y’.

X00001111

(X,Y,Z)=

question

ngBooleZ+X’.Y.Z

Y00110011

(X+Y+

ncarry3

eanexpreZ+X’.Y.Z

X.Y’.Z+101

Y00110011

Z)(X+Y

marks:

8

essioninZ’

+X’.Y.Z+011

Z01010101

Y+Z’)(X’

ntoitseq

+X’.Y.Z’010

F00110100

’+Y+Z)

uivalent

PX+X+

X’+

X’+X’+

(X’+Y’+

tCanonic

POS+Y+Z+Y+Z’

+Y+Z

+Y’+Z+Y’+Z’

+Z)(X’+

calProdu

+Y’+Z’)

uctofSum(1)

m

9  

2)

1) Reduce the following Boolean Expression using K-Map: F(A,B,C,D)=(0,1,2,4,5,6,8,10)

2) Reduce the following Boolean Expression using K-Map: F(U,V,W,Z)=(0,1,2,4,5,6,8,10)

 

10  

3) Reduce the following Boolean Expression using K-Map: F(A,B,C,D)= π(0,1,2,4,5,6,8,10)  

4) Reduce the following Boolean Expression using K-Map: F (X, Y, Z, W) = (0,1,3,4,5,7,9,10,11,13,15)

 

5) Reduce the following Boolean expression using K map: F(M,N,O,P)= (0,1,3,4,5,6,7,9,10,11,13,15)

 

6) Obtain simplified form for a Boolean expression F(x,y,z,w) = Σ(1,3,4,5,7,9,11,12,13,15) using K Map 

7) F(P,Q,R,S)=  ∑(0,3,5,6,7,11,12,15) 

 

8) F(P, Q ,R, S) = Σ (3,5,7,10,11,13,15)   

 

9) F(X,Y,Z,W) = Σ (0,3,4,5,7,11,13,15)   

10) F(P,Q,R,S)=  π (0,3,5,6,7,11,12,15)   

11) F(P, Q, R, S) = Σ(1, 2, 3, 5,6, 7, 9, 11, 12, 13, 15) 

12) F (A,B,C,D) = ∑(0,1,3,4,5,7,8,10,12,14,15) 

13) F (A,B,C,D) = π(5,6,7,8,9,12,13,14,15) 

14) F (X,Y,Z,W) = ∑(0,1,4,5,7,8,9,12,13,15) 

15) F (A,B,C,D) = ∑(0,2,3,4,6,7,8,10,12)   

16) F(U,V,W,Z)= π 17) F(P,Q,R,S)=Σ(l,3,5,8,11,12,15) 

18) F(U,V,W,Z) = (0, 1, 2, 3, 4, 10, 11) 

19) F(P, Q, R, S) = (0,1,2,4,5,6,8, 12) 

20) F(A,B,C,D)=(1,3,4,5,6,7,12,13) 21) F (U, V, W, Z) = II (0, 1, 3, 5, 6, 7, 15) 

 

22) F(X,Y,Z,W) = Σ(0,1,6,8,9,l0,11,12,15) Ans: