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Published on: 31/07/2019
Boolean Algebra
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Questions + Answers key
Take MCQ Computer Science Test
1.
Prove that (a’+b’)(a’+b)(a+b’)=a’b’
2.
Write the equivalent Boolean Expression for the following Logic Circuit
3.
Draw a Logical Circuit Diagram for the following Boolean Expression: A.(B+C’)
4.
Convert the following Boolean expression into its equivalent Canonical Sum of Product Form((SOP) (X’+Y+Z’).(X’+Y+Z).(X’+Y’+Z).(X’+Y’+Z’)
5.
Convert the following boolean expression into the equivalent canonical Product of Sum(POS) form \(A.\bar { B } .C+\bar { A } .B.C+\bar { A } .B.\bar { C } \)
6.
Verify \(\bar { X } .Y+X.\bar { Y } +\bar { X } .\bar { Y } =\bar { X } +\bar { Y } \) using truth table.
7.
Name the law shown below and verify it using a truth table
A+B.C=(A+B).(A+C)
8.
Express \(P+\bar { Q } .R\) in canonical SOP form
9.
Why are NAND and NOR gates called Universal gates?
10.
Find the complement of the following boolean function \({ F }_{ 1 }=A.B+\bar { C } .\bar { D } \) :
11.
Write the dual of the boolean expression \((A+0).(A.1.\bar { A } )\)
12.
Write the Product of Sum(POS) form of the function F(X, Y, Z) for the following truth table representation of F:
| X | Y | Z | F |
| 0 | 0 | 0 | 1 |
| 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 |
13.
Give the logic function F=AB+A’B’+A’B. Assuming the complements are available. Simplify the function using De Morgan’s Theorem. Synthesize F by using NOR gates only.
14.
Draw the truth table for a full adder and then simplify the Sum and Carry by using K-map. After that draw its circuit diagram
15.
Reduce the following Boolean Expression with the help of Karnnaugh Map F(U,V,W,Z)=S(0,1,2,3,12,13,14,15)
16.
Solve the following F(a,b,c,d)=∑(0,1,3,4,5,7,8,9,11,12,13,15). Obtain the simplified form using K-Map.
17.
If F(a,b,c,d)=(0,2,4,5,7,8,10,12,13,15), obtain the simplified form using K-Map
18.
Reduce the following Boolan Expression using K-Map:
F(U,V,W,Z)=\(\pi\) (0,1,2,3,4,5,6,810,1)
19.
Reduce the following Boolean expression using
K – Map F (A, B, C, D) =Σ (0,2,3,4,6,7,8,10,12)
20.
Reduce the following Boolean expression using K-Map:
F(A,B,C,D)=(0,1,2,4,5,8,9,10,11)
1.
LHS=(a’+b’)(a’+b)(a+b’)
=(a’a’+a’b+a’b’+b’b)(a+b’)
=(a’+a’b+a’b’+0)(a+b’)
=aa’+a’b’+aa’b+a’bb’+a’ab’+a’b’b’
=0+a’b’+0+)+0+0+a’b’=a’b’=RHS
2.
F(P,Q)=(P'+Q).(P+Q')
3.
4.
F( X , Y , Z ) = ∏ (4 , 5 , 6 , 7)
= Σ (0 , 1 , 2 , 3)
= X’. Y’. Z’ + X’. Y’. Z + X’. Y. Z’ + X’. Y. Z
5.
F= \(A.\bar { B } .C+\bar { A } .B.C+\bar { A } .B.\bar { C } \)
i.e., (101)+(011)+(010)=m5+m3+m2=\(\sum (2,3,5)\)
POS expression will be :
\(=\prod (0,1,4,6,7)\)
= M0.M1.M4.M6.M7
=(000).(001).(100).(110).(111)
=\((A+B+C).(A+B+\bar { C } ).(\bar { A } +B+C).(\bar { A } +\bar { B } +C).(\bar { A } +\bar { B } +\bar { C } )\)
6.
\(\bar { X } .Y+X.\bar { Y } +\bar { X } .\bar { Y } =\bar { X } +\bar { Y } \)
| X | Y | \(\bar { X } \) | \(\bar { Y } \) | \(\bar { X } .Y\) | \(X.\bar { Y } \) | \(\bar { X } .\bar { Y } \) | \(\bar { X } .Y+X.\bar { Y } +\bar { X } .\bar { Y } \) | \(\bar { X } +\bar { Y } \) |
| 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 |
| 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 1 |
| 1 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
Both columns \(\bar { X } .Y+X.\bar { Y } +\bar { X } .\bar { Y } \) and \(\bar { X } +\bar { Y } \) are identical.
So, \(\bar { X } .Y+X.\bar { Y } +\bar { X } .\bar { Y } =\bar { X } +\bar { Y } \) Hence proved.
7.
A+B.C=(A+B).(A+C)
The above stated law is called distributive law.
| A | B | C | B.C | A+B.C | A+B | A+C | (A+B).(A+C) |
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 |
| 1 | 0 | 1 | 0 | 1 | 1 | 1 | 1 |
| 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
Since, column five and column eight are identical, hence, the given law A+B.C=(A+B).(A+C) is verified
8.
( )
\(\Longrightarrow P+\bar { Q } .R\)
\(\Longrightarrow P.(Q+\bar { Q } ).(R+\bar { R } )+(P+\bar { P } ).\bar { Q } .R\)
\(\Longrightarrow (P.Q+P.\bar { Q } ).(R+\bar { R } )+P.\bar { Q } .R+\bar { P } .\bar { Q } .R\)
\(\Longrightarrow R.(P.Q+P.\bar { Q } ).\bar { R } .(P.Q+P.\bar { Q } )+P.\bar { Q } .R+\bar { P } .\bar { Q } .R\)
\(\Longrightarrow P.Q.R+P.\bar { Q } .R+P.Q.\bar { R } +P.\bar { Q } .\bar { R } +P.\bar { Q } .R+\bar { P } .\bar { Q } .R\)
Now, remove duplicate term,
\(\Longrightarrow P.Q.R+P.\bar { Q } .R+P.Q.\bar { R } +P.\bar { Q } .\bar { R } +\bar { P } .\bar { Q } .R\)
9.
( )
NAND and NOR gates are easier to design and basic functions like AND, OR and NOT, etc., can be easily implemented using NAND/NOR gates. So, these gates are called Universal gates.
10.
( )
\(\overset { \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ }{ (A.\bar { B } +\bar { C } .\bar { D } ) } \) \(=\overset { \_ \_ \_ \_ \_ \_ \_ \_ \_ \quad \quad \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ }{ (A.\bar { B } ).(\bar { C } +\bar { D } ) } \) [DeMorgan's law]
\(=(\bar { A } +\overset { \_ \_ \_ \_ }{ \bar { B } } ).(\overset { \_ \_ \_ }{ \bar { C } } +\overset { \_ \_ \_ \_ }{ \bar { D } } )\) [Demorgan's law]
\(=(\bar { A } +B).(C+D)\)
11.
( )
Using duality principle, changing (+) to (.) and vice-versa and by replacing 0's with 1's and 1's with 0's, the dual for the given expression is as follows:\((A.1)+(A+0+\bar { A } )\)
12.
( )
| X | Y | Z | F | Maxterms |
| 0 | 0 | 0 | 1 | X+Y+Z |
| 0 | 0 | 1 | 0 | \(X+Y+\bar { Z } \) |
| 0 | 1 | 0 | 0 | \(X+\bar {Y}+ { Z } \) |
| 0 | 1 | 1 | 1 | \(X+ \bar {Y}+\bar { Z } \) |
| 1 | 0 | 0 | 0 | \(\bar {X}+ Y+Z \) |
| 1 | 0 | 1 | 0 | \(\bar {X}+ Y+\bar { Z } \) |
| 1 | 1 | 0 | 1 | \(\bar {X}+ \bar {Y}+Z \) |
| 1 | 1 | 1 | 1 | \(\bar {X}+ \bar {Y}+\bar { Z } \) |
POS form of function F(X, Y, Z) is: \((X+Y+\bar { Z } ).(X+\bar { Y } +Z).(\bar { X } +Y+Z).(\bar { X } +Y+\bar { Z } )\)
13.
14.
15.
16.
17.
18.
F(U,V,W,Z)= (0,1,2,4,5,6,8,10)
=\(\Sigma\)(3,7,9,11,12,13,14,15)
19.
F= C’.D’ + A’.C + B’.D’
20.
F(A,B,C,D)= A’.C’+A.B’ + B’.D’
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