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Published on: 30/07/2018
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1.
A boolean function F defined on three input variables X,Y and Z is if and only if number od 1 input is odd. Draw the truth table for the above function and express it in canonical SOP form.
2.
Convert \((\overset { \_ \_ \_ \_ \_ \_ \_ }{ \bar { X } .Y } ).(\overset { \_ \_ \_ \_ \_ \_ \_ \_ }{ \bar { X } .\bar { Z } } )\) into canonical Sum of Products form
3.
Given the following truth table, device a Sum of Product(SOP) and Product of Sum(POS) form of boolean expression from it:
| A | B | C | F(A,B,C) |
| 0 | 0 | 0 | 1 |
| 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 1 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 0 |
4.
State and verify distributive law in boolean algebra.
5.
State and verify absorption law using truth table.
6.
Name the law shown below and verify it using a truth table: \(X+\overset { - }{ X } .Y=X+Y\)
7.
Verify the following using boolean laws:
\(A+C=A+\overset { - }{ A } .C+B.C\)
8.
Express \(P+\bar { Q } .R\) in canonical SOP form
9.
How many input combinations can be there in the truth table of a logic system having(N) input binary variables?
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 Sum of Product(SOP) form of the function F(A,B,C) for the following truth table representation of F:
| A | B | C | F |
| 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 1 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 1 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 0 |
| 1 | 1 | 1 | 1 |
13.
Write the product of Sum (POS) form of the function G(U,V,W) for the following truth table representation of G:
| U | V | W | G |
| 0 | 0 | 0 | 1 |
| 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 1 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 0 | 1 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 0 |
| 1 | 1 | 1 | 1 |
14.
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 |
15.
Write the Sum of Product(SOP) form of the function F(P,Q, R) for the following truth table representation of F:
| P | Q | R | 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 |
1.
Truth table
| X | Y | Z | F |
| 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 1 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 0 | 1 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 0 |
| 1 | 1 | 1 | 1 |
Boolean expression: \(F=\bar { X } .\bar { Y } .Z+\bar { X } .Y\bar { Z } +X.\bar { Y } .\bar { Z } +X.Y.Z\)
Canonical form \(F=\sum (1,2,4,7)\)
2.
Step 1 Simplify the given expression using appropriate theorems/ rules.
\((\overset { \_ \_ \_ \_ \_ \_ \_ }{ \bar { X } .Y } ).(\overset { \_ \_ \_ \_ \_ \_ \_ \_ }{ \bar { X } .\bar { Z } } )=(\overset { \_ \_ \_ }{ \bar { X } } +\bar { Y } ).(\overset { \_ \_ \_ }{ \bar { X } } +\overset { \_ \_ \_ }{ \bar { Z } } )\\ =(X+\bar { } ).(X+Z)\\ =X.X+X.Z+X.\bar { Y } +\bar { Y } .Z\\ =X+X.Z+X.\bar { Y } +\bar { Y } .Z\\ =X.(1+Z)+X.\bar { Y } +\bar { Y } .Z\\ =X+X.\bar { Y } +\bar { Y } .Z\\ =X(1+\bar { Y } )+\bar { Y } .Z=X+\bar { Y } .Z\)
Since it is a 3 variables expression, a product term must have all 3 variables.
Step 2 Wherever a variable is missing, multiply that term with (missing variable+ compliment of missing variable)
\(=X.(Y+\bar { Y } ).(Z+\bar { Z } )+(X+\bar { X } ).\bar { Y } .Z\\ =(X.Y+X.\bar { Y } ).(Z+\bar { Z } )+X.\bar { Y } .Z+\bar { X } .\bar { Y } .Z\\ =X.Y.Z+X.Y.\bar { Z } +X.\bar { Y } .Z+X.\bar { Y } .\bar { Z } +X.\bar { Y } .Z+\bar { X } .\bar { Y } .Z\)
Step 3 Remove all duplicate term, i.e.,
\(=X.Y.Z+X.Y.\bar { Z } +X.\bar { Y } .Z+X.\bar { Y } .\bar { Z } +\bar { X } .\bar { Y } .Z\)
This is the desired canonical Sum of Products form.
3.
| A | B | C | F | MInterms | Maxterm |
| 0 | 0 | 0 | 1 | \(\bar { A } .\bar { B } .\bar { C } \) | |
| 0 | 0 | 1 | 1 | \(\bar { A } .\bar { B } .C\) | |
| 0 | 1 | 0 | 0 | \(A+\bar { B } +C\) | |
| 0 | 1 | 1 | 1 | \(\bar { A } .B.C\) |
| A | B | C | F | Minterms | Maxterm |
| 1 | 0 | 0 | 0 | \(\bar { A } +B+C\) | |
| 1 | 0 | 1 | 1 | \(A.\bar { B } .C\) | |
| 1 | 1 | 0 | 1 | \(A.B.\bar { C } \) | |
| 1 | 1 | 1 | 0 | \(\bar { A } +\bar { B } +\bar { C } \) |
Sum of Products(SOP) expression contain sum of all minterms for output as 1.
\(=\bar { A } .\bar { B } .\bar { C } +\bar { A } .\bar { B } .C+\bar { A } .B.C+A.\bar { B } .C+A.B.\bar { C } \)
\(=\sum (0,1,3,5,6)\)
Product of Sum(POS) expression contains product of all maxterms for output as 0.
\(=(A+\bar { B } +C).(\bar { A } +B+C).(\bar { A } +\bar { B } +\bar { C } )\quad \)
\(=\prod (2,4,7)\)
4.
Distributive law state that:
(i)X.(Y+Z)=X.Y+X.Z
(ii) X+Y.Z=(X+Y).(X+Z)
Verification:
(i) Truth table of : X.(Y+Z)=X.Y+X.Z
| X | Y | Z | Y+Z | X.(Y+Z) | X.Y | X.Z | X.Y+X.Z |
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 |
| 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 |
| 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 |
| 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 1 | 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 1 | 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
Both columns X.(Y+Z) and X.Y+X.Z are identical.
\(\therefore \) X.( Y+Z)=X.Y+X.Z . Hence Proved.
(ii) Truth table for X+Y.Z=(X+Y).(X+Z)
| X | Y | Z | Y.Z | X+Y.Z | X+Y | X+Z | (X+Y).(X+Z) |
| 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 |
Both columns X+Y.Z and (X+Y).(X+Z) are identical.
\(\therefore \) X+Y.Z=(X+Y).(X+Z) . Hence Proved.
5.
Absorption Law It states that for boolean variable X and Y,
(i) X+X.Y=X (ii) X.(X+Y)=X
Verification:
(i) X+X.Y=X
| X | Y | X.Y | X+X.Y |
| 0 | 0 | 0 | 0 |
| 0 | 1 | 0 | 0 |
| 1 | 0 | 0 | 1 |
| 1 | 1 | 1 | 1 |
Both columns X and X+X.Y are identical.
So, X+X.Y=X Hence Proved.
(ii) X.(X+Y)=X
| X | Y | X+Y | X.(X+Y) |
| 0 | 0 | 0 | 0 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 1 | 1 |
Both columns X and X.(X+Y) are identical.
So, X.(X+Y)=X Hence Proved.
6.
Name of a given law is redundant literal law.
| X | Y | \(\overset { - }{ X } \) | \(\overset { - }{ X } .Y\) | \(X+\overset { - }{ X } .Y\) | X+Y |
| 0 | 0 | 1 | 0 | 0 | 0 |
| 0 | 1 | 1 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 | 1 | 1 |
| 1 | 1 | 0 | 0 | 1 | 1 |
Since, last two columns of the above truth table are identical, hence, \(X+\overset { - }{ X } .Y=X+Y\) is verified.
7.
\(A+C=A+\overset { - }{ A } .C+B.C\)
RHS=\(A+\overset { - }{ A } .C+B.C\)
\(A+C=A+\overset { - }{ A } .C+B.C\\ \quad \quad \quad \quad =(A.\overset { - }{ A } ).(A+C)+B.C\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad [By\quad using\quad distributive\quad law]\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad [By\quad using\quad A+\overset { - }{ A } =1]\\ \quad \quad \quad \quad =1.(A+C)+B.C\\ \quad \quad \quad \quad =A+C+B.C=A+C.(1+B)\\ \quad \quad \quad \quad =A+C.(1)=A+C\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad [\therefore 1+B=1]\\ \quad \quad \quad \quad =LHS\)
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.
( )
2N input combinations can be there in the truth table of a logic system.
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.
( )
| A | B | C | F | Minterms |
| 0 | 0 | 0 | 0 | \(\bar{A}.\bar { B } .\bar { C }\) |
| 0 | 0 | 1 | 0 | \(\bar{A}.\bar { B } .C\) |
| 0 | 1 | 0 | 1 | \(\bar { A } .B.\bar { C }\) |
| 0 | 1 | 1 | 1 | \(\bar { A } .B.C\) |
| 1 | 0 | 0 | 1 | \(A.\bar { B } .\bar { C }\) |
| 1 | 0 | 1 | 0 | \(A.\bar { B } .C\) |
| 1 | 1 | 0 | 0 | \(A.B .\bar { C }\) |
| 1 | 1 | 1 | 1 | A.B.C |
SOP form of function F(A, B, C) is: \(\bar { A } .B.\bar { C } +\bar { A } .B.C+A.\bar { B } .\bar { C } +A.B.C\)
13.
( )
| U | V | W | G | Maxterms |
| 0 | 0 | 0 | 1 | U+V+W |
| 0 | 0 | 1 | 0 | \(U+V+\bar { W } \) |
| 0 | 1 | 0 | 1 | \(U+\bar {V}+ W \) |
| 0 | 1 | 1 | 1 | \(U+\bar {V}+\bar { W } \) |
| 1 | 0 | 0 | 1 | \(\bar {U}+V+W \) |
| 1 | 0 | 1 | 0 | \(\bar{U}+V+\bar { W } \) |
| 1 | 1 | 0 | 0 | \(\bar {U}+\bar {V}+W \) |
| 1 | 1 | 1 | 1 | \(\bar {U}+\bar {V}+\bar { W } \) |
POS form of function G(U,V,W) is : \((U+V+\bar { W } ).(U+\bar { V } +\bar { W } ).(\bar { U } +V+\bar { W } ).(\bar { U } +\bar { V } +W)\)
14.
( )
| 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: \((X+Y+\bar { Z } ).(X+\bar { Y } +Z).(\bar { X } +Y+Z).(\bar { X } +Y+\bar { Z } )\)
15.
( )
| P | Q | R | F | Minterms |
| 0 | 0 | 0 | 1 | \(\bar { P } .\bar { Q } .\bar { R } \) |
| 0 | 0 | 1 | 0 | \(\bar { P } .\bar { Q } .R \) |
| 0 | 1 | 0 | 0 | \(\bar { P } .Q .\bar { R } \) |
| 0 | 1 | 1 | 1 | \(\bar { P } . Q .R \) |
| 1 | 0 | 0 | 0 | \(P .\bar { Q } .\bar { R } \) |
| 1 | 0 | 1 | 0 | \(P .\bar { Q } .R \) |
| 1 | 1 | 0 | 1 | \(P.Q .\bar { R } \) |
| 1 | 1 | 1 | 1 | P.Q.R |
SOP form of function F(P, Q, R) is: \(=\bar { P } .\bar { Q } .\bar { R } +\bar { P } .Q.R+P.Q.\bar { R } +P.Q.R\)
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