Unit-1(2-MARKS)

1) What are basic properties of Boolean algebra? 
The basic properties of Boolean algebra are commutative property, associative Property
and distributive property.

2) 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 stated
as follows:
 A+ (B+C) = (A+B) +C

3) State the commutative property of Boolean algebra.
The commutative property states that the order in which the variables are OR ed makes
no difference. The commutative property is:
 A+B=B+A

4) 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 the several
Variables and then AND ing the sums. The distributive property is:
 A+BC= (A+B) (A+C)

5) State the absorption law of Boolean algebra. 
The absorption law of Boolean algebra is given by X+XY=X, X(X+Y) =X.

6) State De Morgan's theorem. 
De Morgan suggested two theorems that form important part of Boolean algebra. They
are,
 1) The complement of a product is equal to the sum of the complements.
 (AB)' = A' + B'
 2) The complement of a sum term is equal to the product of the complements.
 (A + B)' = A'B'

7) Reduce A (A + B) 
 A (A + B) = AA + AB
 = A (1 + B) [1 + B = 1]
= A.

8) Reduce A'B'C' + A'BC' + A'BC 

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]

9) 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]


10) 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'
11) 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]


12) 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
=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 ...Proved


13) 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]

14) Define duality property. 
Duality property states that 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 simply interchange OR and AND operators and
replace 1's by 0's and 0's by 1's.


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

16) Simplify the following expression 
Y = (A + B) (A = C) (B + C)
= (A A + A C + A B + B C) (B + C)
= (A C + A B + B C) (B + C)
= A B C + A C C + A B B + A B C + B B C + B C C
= A B C

17) Given the two binary numbers X = 1010100 and Y = 1000011, perform the subtraction (a) X -Y and (b) Y - X using 2’s complements.

a) X = 1010100

 2’s complement of Y = + 0111101 --------------

 Sum = 10010001

 Discard end carry

Answer: X - Y = 0010001

b) Y = 1000011

 2’s complement of X = + 0101100 ---------------

Sum = 1101111

 There is no end carry,

 Therefore the answer is Y-X = -(2’s complement of 1101111) = -0010001

18). Given the two binary numbers X = 1010100 and Y = 1000011, perform the subtraction (a) X -Y and (b) Y - X using 1’s complements.

a). X - Y = 1010100 – 1000011

 X = 1010100

1’s complement of Y = + 0111100 --------------

Sum = 10010000

End -around carry = + 1 --------------

Answer: X - Y = 0010001

b). Y - X = 1000011 – 1010100

 Y = 1000011

1’s complement of X = + 0101011 -----------

 Sum = + 1101110

There is no end carry.

Therefore the answer is Y - X = -(1’s complement of 1101110) = -0010001

19). what is meant by parity bit?

A parity bit is an extra bit included with a message to make the total number of 1’s either even or odd. Consider the following two characters and their even and odd parity: With even parity with odd parity ASCII A = 1000001 01000001 11000001, ASCII T = 1010100 11010100 01010100. In each case we add an extra bit in the left most position of the code to produce an even number of1’s in the character for even parity or an odd number of 1’s in the character for odd parity. The parity bit is helpful in detecting errors during the transmission of information from one location to another.

20).What are registers?

Register is a group of binary cells. A register with n cells can store any discrete quantity of information that contains n bits. The state of a register is an n-tuple number of 1’s and 0’s, with each bit designating the state of one cell in the register.

21). What is meant by register transfer?

A register transfer operation is a basic operation in digital systems. It consists of transfer of binary information from one set of registers into another set of registers. The transfer may be direct from one register to another, or may pass through data processing circuits to perform an operation.

22). Define binary logic?

Binary logic consists of binary variables and logical operations. The variables are designated by the alphabets such as A, B, C, x, y, z, etc., with each variable having only two distinct values: 1 and 0. There are three basic logic operations: AND, OR, and NOT.

23). Define logic gates?

Logic gates are electronic circuits that operate on one or more input signals to produce an output signal. Electrical signals such as voltages or currents exist throughout a digital system in either of two recognizable values. Voltage- operated circuits respond to two separate voltage levels that represent a binary variable equal to logic 1 or logic 0.

24).Define duality property.

Duality property states that 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 simply interchange OR and AND operators and replace 1’s by 0’s and 0’s by 1’s.

25.Find the complement of the functions F₁= x’yz’ + x’y’z and F₂= x(y’z’ + yz) by applying De Morgan’s theorem as many times as necessary.

F₁’ = (x’yz’ + x’y’z)’ = (x’yz’)’(x’y’z)’ = (x + y’ + z)(x + y +z’)

F₂’ = [x(y’z’ + yz)]’ = x’ + (y’z’ + yz)’ = x’ + (y’z’)’(yz)’

      = x’ + (y + z)(y’ + z’)

26).Find the complements of the functions F1 = x’yz’ + x’y’z and F2 = x(y’z’ + yz)  by taking their duals and complementing each literal.

F1= x’yz’ + x’y’z. The dual of F1 is (x’ + y + z’)(x’ + y’ + z).

 Complementing each literal: (x + y’ + z)(x + y + z’)

 F2= x(y’z’ + yz). The dual of F2 is x + (y’ + z’)(y + z).

Complement of each literal: x’ + (y + z)(y’ + z’)

27).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]

28).Convert the given expression in canonical POS form Y = ( A + B)(B + C)(A + C)

Y = ( A + B)(B + C)(A + C)

 = (A + B + C.C’ )(B + C + A.A’ )(A + B.B’ + C)

= (A + B + C)(A + B + C’ )(A + B +C)(A’ + B +C)(A + B + C)(A + B’ + C) [A + BC = (A + B)(A + C) Distributive law]

= (A + B + C)(A + B + C’)(A’ + B + C)(A’ + B + C)(A + B’ + C)

29). 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 =ôP____________

30).Write the maxterms corresponding to the logical expression Y = (A + B + C’ )(A + B’ + C’)(A’ + B’ + C)

Y = (A + B + C’ )(A + B’ + C’)(A’ + B’ + C) =M1.M3.M6 =ö0_______

31).Convert (4021.2)₅to its equivalent decimal.

(4021.2)₅= 4 x 5³+ 0 x 5²+ 2 x 5¹+ 1 x 5⁰+ 2 x 5^-1

= (511.4)₁₀

32) Using 10’s complement subtract 72532 – 3250

M = 72532 10’s complement of N = + 96750 -----------

 Sum = 169282

Discard end carry

Answer = 69282

33) 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 defined, it can be either high or low. These output levels are indicated by ‘X’ or‘d’ in the truth tables and are called don’t care conditions or incompletely specified functions.

34) Write down the steps in implementing a Boolean function with levels of NAND Gates?

Simplify the function and express it in sum of products. Draw a NAND gate for each product term of the expression that has at least two literals. The inputs to each NAND gate are the literals of the term. This constitutes a group of first level gates. Draw a single gate using the AND-invert or the invert-OR graphic symbol in the second level, with inputs coming from outputs of first level gates. A term with a single literal requires an inverter in the first level. How ever if the single literal is complemented, it can be connected directly to an input of the second level NAND gate.

35) Give the general procedure for converting a Boolean expression in to multilevel NAND diagram?

Draw the AND-OR diagram of the Boolean expression. Convert all AND gates to NAND gates with AND-invert graphic symbols. Convert all OR gates to NAND gates with invert-OR graphic symbols. Check all the bubbles in the same diagram. For every bubble that is not compensated by another circle along the same line, insert an inverter or complement the input literal.