EC6302- Digital Electronics III semester ECE

EC6302- Digital Electronics III semester ECE

UNIT I
MINIMIZATION TECHNIQUES AND LOGIC GATES

1. What are basic properties of Boolean algebra?
The basic properties of Boolean algebra are 1.commutative property, 2.associative
property, 3.distributive property and 4.idem potency.

2. What is meant by bit?
A binary digit is called a bit.

3. Define byte.
Byte is defined as a group of 8 bits.

4. State the associative property of Boolean algebra.
The associative property of Boolean algebra states that the OR ing of several
variables results in the same regardless of the grouping of the variables. The
associative property is illustrated as follow:
A+ (B+C) = (A+B) +C

5. State the commutative property of Boolean algebra.
The commutative property states that the order in which the variables are
OR ed when changed does not make any difference in the characteristics. The
commutative property is illustrated as follow:
A+B=B+A

6. State the distributive property of Boolean algebra.
The distributive property states that AND ing several variables and OR ing the
result with a single variable is equivalent to OR ing the single variable with each of
the several variables and then AND ing the sums. The distributive property illustrated
as follow is:
A+BC= (A+B) (A+C)

7. Write the absorption law of Boolean algebra.
The absorption law of Boolean algebra is given by X+XY=X.

8. Simplify A.A'C.
A.A'C = 0.C
[A.A' = 0]
=0

9. Simplify A(A + B)
A(A + B) = AA + AB
= A(1 + B) [1 + B = 1]
= A.
10. Reduce A'B'C' + A'BC' + A'BC

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A'B'C' + A'BC' + A'BC = A'C'(B' + B) + A'B'C
= A'C' + A'BC
[A + A' = 1]
= A'(C' + BC)
= A'(C' + B)
[A + A'B = A + B]



11. State and prove Demorgan’s theorem.
i. (AB)’= A’+B’
ii. (A+B)’= A’B’

12. Reduce AB + (AC)' + AB'C(AB + C)
AB + (AC)' + AB'C(AB + C) = AB + (AC)' + AAB'BC + AB'CC
= AB + (AC)' + AB'CC
[A.A' = 0]
= AB + (AC)' + AB'C
[A.A = 1]
= AB + A' + C' + AB'C [(AB)' = A' + B']
= A' + B + C' + AB'C [A + AB' = A + B]
= A' + B'C + B + C' [A + A'B = A + B]
= A' + B + C' + B'C
=A' + B + C' + B'
=A' + C' + 1
=1
[A + 1 =1]

13. Simplify the following expression Y = (A + B)(A + C' )(B' + C' )
Y = (A + B)(A + C' )(B' + C' )
= (AA' + AC +A'B +BC )(B' + C')
[A.A' = 0]
= (AC + A'B + BC)(B' + C' )
= AB'C + ACC' + A'BB' + A'BC' + BB'C + BCC'
= AB'C + A'BC'

14. Show that (X + Y' + XY)( X + Y')(X'Y) = 0
(X + Y' + XY)(X + Y')(X'Y) = (X + Y' + X)(X + Y' )(X' + Y) [A + A'B = A + B]
= (X + Y' )(X + Y' )(X'Y) [A + A = 1]
= (X + Y' )(X'Y) [A.A = 1]
= X.X' + Y'.X'.Y
=0 [A.A' = 0]

15. Prove that ABC + ABC' + AB'C + A'BC = AB + AC + BC
ABC + ABC' + AB'C + A'BC=AB(C + C') + AB'C + A'BC
=AB + AB'C + A'BC

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=A(B + B'C) + A'BC
=A(B + C) + A'BC
=AB + AC + A'BC
=B(A + C) + AC
=AB + BC + AC
=AB + AC +BC
Hence proved.




16. Apply De Morgan’s theorem for the function . [Nov /Dec – 08]


= (A+B+C)' + D'
= A'. B'. C' + D'
17. Define – Duality Property
Duality property is defined as; every algebraic expression deducible from the
postulates of Boolean algebra remains valid, if the operators and identity elements are
interchanged. If the dual of an algebraic expression is desired, we have to simply
interchange OR and AND operators and replace 1's by 0's and 0's by 1's.

18. Find the complement of the functions F1 = x'yz' + x'y'z and F2 = x(y'z' + yz). By
applying De-Morgan's theorem.
F1 = x'yz' + x'y'z
Complement of F1 = F1' = (x'yz' + x'y'z)'
= (x'yz')'(x'y'z)'
= (x + y' + z)(x + y +z')

Complement of F2 = F2' = [x(y'z' + yz)]'
= x' + (y'z' + yz)'
= x' + (y'z')'(yz)'
= x' + (y + z)(y' + z')

19. Convert the given expression in canonical SOP form Y = AC + AB + BC
Y = AC + AB + BC
=AC(B + B' ) + AB(C + C' ) + (A + A')BC
=ABC + ABC' + AB'C + AB'C' + ABC + ABC' + ABC
=ABC + ABC' +AB'C + AB'C' [(A + A) =1]

20. Write the names of basic logical operators.
The names of the basic logic operators are
1. NOT / INVERT
2. AND
3. OR

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21.State the two canonical forms of Boolean algebra. [Apr. / May –
07]
The two canonical forms of Boolean algebra are:
i. Sum of products
Products of Sum

22.What are the methods adopted to reduce Boolean function?
There are four methods used to reduce the Boolean function. They are
i. Algebraic method
ii. Karnaug map
iii. Tabulation method or Quine Mc-Cluskey method
iv. Variable entered map technique.


23. What is a Karnaugh map?
A Karnaugh map or K map is a pictorial form of truth table, in which the map
diagram is made up of squares called cells, with each cell representing one minterm of
the function. The number of cells in the map depends upon the number of variable
(literal) in the minterm or maxterm of the functions.

24. State the limitations of Karnaugh map.
The limitations of Karnaugh map are:
i. If the number of variables increases it is difficult to make judgments
about grouping of cell and it is generally limited to six variables.
ii. The K – Map simplification is manual technique and linearly depends
upon human abilities.
iii. It can not be programmed to run on a computer as it is not algorithmic.

21. Find the minterms of the logical expression Y = A'B'C' + A'B'C + A'BC + ABC'
Y = A'B'C' + A'B'C + A'BC + ABC'
= m0 + m1 + m3 + m6
= m(0, 1, 3, 6)

22. Write the maxterms corresponding to the logical expression.
Y = (A + B + C' )(A + B' + C')(A' + B' + C)
= (A + B + C' )(A + B' + C')(A' + B' + C)
=M1.M3.M6
= M(1,3,6)

23. What are called don’t care conditions?
In some logic circuits certain input conditions never occur, therefore the
corresponding output never appears. In such cases the output level is not

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