UNIT 1BOOLEAN ALGEBRA AND LOGIC GATES

Number Base Conversions :1. List the different number systems?

i) Binary Number system

ii) Octal Number systemiii) Decimal Number system

iv) Hexadecimal Number system2. Express the following in decimal:

a) (10110.0101)2, b) (16.5)16, c) (26.24)8, d) (FAFA.B)16, e) (1010.1010)2(a) (10110.0101)2 = (1x24) + (0x23) + (1x22) + (1x21) + (0x20) + (0x2-1) + (1x2-2) + (0x2-3) + (1x2-4)= 16 + 4 + 2 + 0.25 +0.0625 = (22.3125)10(b) (16.5)16 = (1x161) + (6x160) + (5x16-1) = 16 + 6 + (5 (0.0615))= (22.3125)10(c) (26.24)8 = (2x81) + (6x80) + (2x8-1) + (4x8-2)= 16 + 6 + 2/8 + 4/64 = (22.3125)10(d) (FAFA.B)16= (Fx163) + (Ax162) + (Fx161) + (Ax160) + (Bx16-1) = (15x163) + (10x162) + (15x161) + (10x160) + (11x16-1)= (64,250.6875)10(e) (1010.1010)2= (1x23) + (0x22) + (1x21) + (0x20) + (1x2-1) + (0x2-2) + (1x2-3) + (0x2-4)= 8 + 2 + 0.5 +0.125 = (10.625)103. Convert the following binary numbers to hexadecimal and to decimal:

a) 1.11010, b) 1110.10

Explain why the decimal answer in (b) is 8 times that of (a).To convert from binary to hexadecimal:

Each 4 binary digits are equal to 1 hexadecimal digit:

a) (0001.11010)2 = (1.D0)16

b) (1110.1000)2 = (E.8) 16

To convert from binary to decimal:

a) (1.11010)2 = (1x20) + (1x2-1) + (1x2-2) + (0x2-3) + (1x2-4) + (0x2-5)= (1) + (0.5+0.25+0.0625)

= (1.8125)10 b) (1110.10) 2 = (1x23) + (1x22) + (1x21) + (0x20) + (1x2-1) + (0x2-2) = (8+4+2) + (0.5)

= (14.5) 10The decimal answer in (b) is 8 times that of (a) because the binary number in (b) is the same as that in (a) except that the point is shifted to the right 3 digits and this means that it is multiplied by 23.4. Convert (9B2.1A) H to its decimal equivalent.

N = (9 x 16 2) + (B x 16 1) + (2 x 16 0) + (1 x 16 -1) + (A (10) x 16 -2) = 2304 + 176 + 2 + 0.0625 + 0.039

= (2482.1)10 5. Find the decimal equivalent of (346)7.

(May, 2004)

(346)7.= (3x72) + (4x71) + (6x70)

= (181)10.6. Convert 0.640625 decimal number to its octal equivalent.

0.640625 x 8 = 5.125

5 0.125 x 8 = 1.0

1 (0.640 625)10 = (0.51) 87. Convert the (153.513)10 to octal.

(May, 2010)Integer part:

8 153

8 19 --1

2 --3

= (231)8Fractional part:

0.513 x 8 = 4.104 4

0.104 x 8 = 0.832 0

0.832 x 8 = 6.656 6

0.656 x 8 = 5.248 5

0.248 x 8 = 1.984 1

0.984 x 8 = 7.872 7

= (0.406517)8 (approximate) (153.513)10 = (231. 406517)88. Convert 0.1289062 decimal number to its hex equivalent

0.1289062 x 16 = 2.06252 0.0625 x 16 = 1.0

1Ans = (0.21)169. Convert the following number from one base to other

(65.342)8 = ( ) 7.

(May, 2008)(65.342)8= (6x81) + (5x80) + (3x8-1) + (4x8-2) + (2x8-3)

= (53.4414062)10Now, convert this number to base 7.

Integer part:

7 53

7 7 --4

1 --0

= (104)7Fractional part

0.4414062x 7 = 3.0898434

3

0.0898434x 7 = 0.6289038

0

0.6289038 x 7 = 4.4023266

4

0.4023266x 7 = 2.8162862

2= (0.3042)7 (65.342)8 = (104.3042) 7 10. Convert 22.64 to hexadecimal number.Integer part:

16 22

1 --6

= (16)16Fractional Part:0.64 x 16 = 10.24

A0.24 x 16 = 3.84

30.84 x 16 = 13.44

D0.44 x 16 = 7.04

7Ans = (16. A3D7) 1611. Convert (231.3)4 to base 7.

(May, 2005)

Convert the given number to decimal,

(231.3)4 = (2x42) + (3x41) + (1x40) + (3x4-1)

= 32+ 12+ 1+ 0.75 = (45.75)10Now, convert this number to base 7.

Integer part:

7 45

6 --3

= (63)7Fractional part:0.75x 7 = 5.25

5

0.25x 7 = 1.75

1

0.75x 7 = 5.25

5

0.25x 7 = 1.75

1= (0.5151)7 (231.3)4 = (63.5151)712. Convert (634)8 to binary

6 3 4

110 011 100Ans = (110011100)213. Convert the following number from one base to other

(Nov, 2006) (a) (354.52) 6 = ( )10

(b) (100)10 = ( ) 16.(a) (354.52) 6 = (3x62) + (5x61) + (4x60) + (5x6-1) + (2x6-2) = 108+ 30+ 4 + 0.8333+ 0.0555

= (142.888) 10 (354.52) 6 = (142.888) 10(b) (100)10 16 100

6 -- 4

= (64)7 (100)10 = (64)16.14. A hexadecimal counter capable of counting upto atleast (10,000)10 is to be constructed. What is the minimum number of hexadecimal digits that the counter must have?

(May, 2004)Soln:

(10,000)10 = (2710)16Complements :15. What are the different types of number complements?

i) rs Complement

ii) (r-1)s Complement.

16. What is the range of values that can be represented using n-bit 2s complement form of representation? What is the corresponding range with n-bit 1s complement form?

(May, 2006)

The given number N in the base 2 having n digits. The 2s complement of N is defined as follows.

2s complement of N= + (2n-1- 1) to (2n-1),

Where, n is number of digits.

The given number N in the base r= 2 having n digits. The (r-1)s complement of N is defined as follows.

1s complement of N= + (2n-1- 1) to (2n-1-1) Where, N= given number or digit

17. Add (1 0 1 0)2 and (0 0 1 1)21 0 1 0

(+) 0 0 1 1---------

1 1 0 1

---------

Answer = (1 1 0 1)218. Substract (0 1 0 1)2 from (1 0 1 1)21 0 1 0

(-) 0 1 0 1--------

0 1 1 0

--------

Answer = (1 1 0)219. Why complementing a number representation is needed?

Complementing a number becomes as in digital computer for simplifying the

subtraction operation and for logical manipulation complements are used.20. Obtain the 1s and 2s complement of the following binary numbers:

a) 11101010 b) 01111110 c) 00000001 d) 10000000 e) 00000000

Soln:

1s complement : change every 1 to 0 and vice versa.

2s complement: change every 1 to 0 and vice versa, then add (1) to the LSB.

a) 11101010

1s complement: (00010101)22s complement : 0 0 0 1 0 1 0 1

(+) 1

--------------------

(0 0 0 1 0 1 1 0)2 --------------------- b) 01111110

1s complement: (10000001)2 2s complement : 1 0 0 0 0 0 0 1

(+) 1

---------------------

(1 0 0 0 0 0 1 0)2 ---------------------c) 00000001

1s complement: (01111110)2 2s complement : 0 1 1 1 1 1 1 0

(+) 1

----------------------

(1 1 1 1 1 1 1 1)2

-----------------------

d) 10000000

1s complement: (01111111)22s complement : 0 1 1 1 1 1 1 1

(+) 1

-----------------------

(1 0 0 0 0 0 0 0)2

-----------------------e) 00000000

1s complement: (11111111)2

2s complement : 0 1 1 1 1 1 1 1

(+) 1

-----------------------

(1 0 0 0 0 0 0 0 0)2

-----------------------21. Find 2s complement of (1 0 1 0 0 0 1 1) 2Soln:

0 1 0 1 1 1 0 0 1 - 1s Complement

(+) 0 0 0 0 0 0 1----------------------(0 1 0 1 1 1 0 1 0)2- 2s complement.----------------------

22. Substract (1 1 1 0 0 1)2 from (1 0 1 0 1 1)2 using 2s complement method.Soln:

0 0 0 1 1 0

- 1s Complement of (1 1 1 0 0 1)2+ 0 0 0 0 1--------------0 0 0 1 1 1

- 2s complement.--------------

1 0 1 0 1 1

+ 0 0 0 1 1 1

- 2s comp. of (1 1 1 0 0 1)2---------------

1 1 0 0 1 0 in 2s complement form---------------

To get the answer in true form , take the 2s complement and assign negative number to the answer.Answer in true form - ( 0 0 1 1 1 0 )223. Perform subtraction using 1s complement (11010)2 (10000)2. 1 1 1 1 1 1 0 1 0

(+) 0 1 1 1 1

1s complement of (10000)2 1 0 1 0 0 1

(+) 1

(Add carry to LSB)

(0 1 0 1 0)2 Y- X = (01010)224. Given two binary numbers X = 1010100 and Y = 1000011, perform subtraction.(a) X -Y and (b) Y - X using 2's complements.

a) X = 1 0 1 0 1 0 0

0 1 1 1 1 0 1

(2's complement of Y) ----------------

1 0 0 1 0 0 0 1

Discard end carry

Answer: X - Y = (0010001)2b) Y = 1 0 0 0 0 1 1

0 1 0 1 1 0 0

(2's complement of X) -----------------

1 1 0 1 1 1 1

There is no end carry, So, take 2s complement again for the above answer.Answer is Y-X = - (2's complement of 1101111) = - (0010001)225. Given two binary numbers X = 1010100 and Y = 1000011, perform subtraction.(a) X -Y and (b) Y - X using 1's complements.

a) X - Y = 1010100 - 1000011

X = 1 0 1 0 1 0 0

0 1 1 1 1 0 0

(1's complement of Y) -----------------

1 0 0 1 0 0 0 0

Discard end carry = + 1

Answer: X - Y = (0010001)2b) Y - X = 1000011 - 1010100

Y = 1 0 0 0 0 1 1

0 1 0 1 0 1 1

(1's complement of X) ----------------- 1 1 0 1 1 1 0

There is no end carry.

Answer is Y - X = - (1's complement of 1101110) = - (0010001)226. What are the different ways to represent a negative number?(Nov, 2006)

The different ways of representing a negative number are-

i. In ordinary arithmetic, the negative sign is indicated by a minus sign.

ii. In signed magnitude representation, in which MSB is indicated as 0 to represent negative number.

iii. In signed 1s complement representation, in which the negative number is indicated by its 1s complement.iv. In signed 2s complement representation, in which the negative number is indicated by its 2s complement.Binary Codes :27. Mention the different type of binary codes?

The various types of binary codes are,

i. BCD code (Binary Coded decimal).ii. Self-complementing code.iii. The excess-3 (Xs-3) code.iv. Gray code.v. Binary weighted code.vi. Alphanumeric code.vii. The ASCII code.viii. Extended binary-coded decimal interchange code (EBCDIC).ix. Error-detecting and error-correcting code.x. Hamming code.28. State the different classification of binary codes?

i. Weighted codesii. Non - weighted codesiii. Reflective codesiv. Sequential codes

v. Alphanumeric codesvi. Error Detecting and correcting codes.29. What is meant by bit?

A binary digit is called bit30. Define byte?

Group of 8 bits.31. State the steps involved in Gray to binary conversion?

The MSB of the binary number is the same as the MSB of the gray code number. So write it down. To obtain the next binary digit, perform an exclusive OR operation between the bit just written down and the next gray code bit. Write down the result.32. What are error detecting codes?

(Nov, 2007)

When the digital information in the binary form is transmitted from one circuit or system to another circuit or system an error may occur. To maintain the data integrity between transmitter and receiver, extra bit or more than one bit is added in the data. The data along with the extra bit/bits forms the code. Code which allow only error detection are called error detecting codes.

33. Convert gray code 101011 into its binary equivalent.

Gray Code : 1 0 1 0 1 1

Binary Code: 1 1 0 0 1 034. State the abbreviations of ASCII and EBCDIC code?

ASCII - American Standard Code for Information Interchange.

EBCDIC - Extended Binary Coded Decimal Information Code.35. What is advantage of gray codes over binary number sequence? (May, 2007)

The advantage of gray codes over the binary number is that only one bit in the code group changes when going from one number to the next.

The gray code is used in applications where the normal sequence of binary number may produce an error or ambiguity during the transition from one number to next.

Boolean Algebra & Theorems :36. What are basic properties of Boolean algebra?

The basic properties of Boolean algebra are commutative property, associative property and distributive property.

37. 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

38. State the commutative property of Boolean algebra.

The commutative property states that the order in which the variables are ORed makes no difference. The commutative property is:

(A+B) = (B+A)39. State the distributive property of Boolean algebra.

The distributive property states that ANDing several variables and ORing the result with a single variable is equivalent to ORing the single variable with each of the several variables and then ANDing the sums. The distributive property is:

A+BC= (A+B) (A+C)

40. 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.

(A.B)' = A'+B'

2) The complement of a sum term is equal to the product of the complements.

(A+B)' = A'.B'

41. State the absorption law of Boolean algebra.

The absorption law of Boolean algebra is given by,

A+AB=A, A (A+B) =A.

42. Define duality property.

Duality property states that, starting with a Boolean relation, you can derive another Boolean relation by

1. Changing each OR sign to an AND sign

2. Changing each AND sign to an OR sign

3. Complementing any 0 or 1 appearing in the expression

For Example: A+ A= 1 is A. A= 043. Show that A+A.B = A+B using the theorems of Boolean algebra.(Nov, 2005)

LHS= A+A.B

= A+ AB + AB

[ A+AB = A]

= A+ B (A+ A)

= A+ B (1)

[A+ A= 1]

= A+ BCanonical Form :44. What are minterms?

(May, 2008)

Each individual term in standard SOP form is called minterms

45. What are maxterms?

Each individual term in standard POS form is called maxterms.46. 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)47. 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] = m7 + m6 +m5 +m4

= m (4, 5, 6, 7)48. Convert the following function into sum of product form

(AB+C)(B+CD).

(May, 2008)= (AB.B+ B.C+ AB.CD+ C.CD)

= AB+ BC+ ABCD

[B. B= 1] [C.C= 0]AND each product term having missing literals, by ORing the literals and its complement

= AB (C+ C) (D+ D) + BC (A+ A) (D+ D) + ABCD

= (ABC+ ABC) (D+ D) + (ABC+ ABC) (D+ D) + ABCD

= ABCD+ ABCD+ ABCD+ ABCD+ ABCD+ ABCD+ ABCD+ ABCD+ ABCD= ABCD+ ABCD+ ABCD+ ABCD+ ABCD+ ABCD.= m15+ m14+ m13+ m12+ m7+ m6 F(A,B,C,D)= m( 6,7, 12,13,14,15)

49. 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)

50. 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')

51. Find the complements for the following functions

(a) F1= xy+ xy.(b) F2= (xy + yz + xz) x.

(Nov, 2007)

(a) F1= xy+ xy

F1= (xy+ xy)

= (xy). (xy)

= (x+y) (x+y)

= xx+ xy+ yx+ yy

= xy+ xy.(b) F2= (xy + yz + xz) x. F2 = ((xy + yz + xz) x)

= (xy + yz + xz) + x

= [(xy) (yz) (xz)] + x

= [(x+y) (y+z) (x+z)] + x

= [(xy+ xz+ 0+ yz) ( x+z)] + x

= xxy+ xxz+ xyz+ xyz+ xzz+ yzz+ x

= xy+ xz+ xyz+ xyz+ xz+ yz+ x

[x+ x = x], [x. x = x]

= xy+ xz+ xz (y+ y) + yz+ x

[x+ x= 1]

= xy+ xz+ xz (1) + yz+ x

= xy+ xz+ yz+ x

= xy+ x+ xz+ yz

= x(y+1) + xz+ yz

[y+1= 1]

= x (1+z) + yz

[y+1= 1]

= x+ yz

52. Obtain the complement of f = wxy + xy+ wxz using De Morgans theorem.

(May, 2006)

f = (wxy + xy+ wxz)

= (wxy) (xy) (wxz)

= (w+x+ y) (x+ y) (w+ x+ z)

= (wx+ wy+ xx+ xy+ xy+ yy) (w+ x+ z)

= (wx+ wy+ xy+ xy) (w+ x+ z)

= wx. w+ wy. w+ xy. w+ xy. w+ wx. x+wy. x+ xy. x+ xy. x+

wx. z+ wy. z+ xy. z+ xy.z

= wx+ wy+ wxy+ wxy+ wx+ wxy+ 0 + xy+ wxz+ wyz+ xyz+

xyz

= wx+ wy+ wxy+ wxy+ wxy+ xy+ wxz+ wyz+ xyz+ xyz

= wx( 1+ y+ y+ z)+ wy( 1+ x+ z)+ xy(1+ z)+ xyz

= wx(1)+ wy(1)+ xy(1)+ xyz

= wx+ wy+ xy+ xyz

Minimization of Boolean Expressions :53. Simplify the following using De Morgan's theorem [((AB)'C)'' D]'

L.H.S. = [((AB)'C)'' D]' = ((AB)'C)'' + D'

[(AB)' = A' + B']

= (AB)' C + D'

= (A' + B) C + D'

54. 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]

55. Reduce A.A'C

A.A'C = 0.C

[A.A' = 0]

= 0

56. Reduce AB + (AC)' + ABC (AB + C)

AB + (AC)' + ABC (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]

57. 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'

58. 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]

59. 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

60. Reduce A (A + B)

A (A + B) = A.A + A.B= A + A.B

= A (1 + B)

[1 + B = 1]

= A.

61. Simplify the following Boolean function:(a) x (x+y) (b) xy + xz + yz.

(May, 2007)(a) x (x+y)

= xx+ xy

= xy.

[x. x= 0]

(b) xy + xz + yz.

= xy + xz + yz( x+ x)

= xy + xz + xyz + xyz

= xy + xyz + xz +xyz

= xy (1+ z) + xz (1+y)

[1+y= 1]

= xy+ xz.

62. Simplify the following Boolean functions to a minimum number of literals

a) (x + y) (x + y)

b) xy + xz + yz

(May, 2010)a) (x+ y) (x+ y)

= x.x+ xy+ yx+ yy

= x+ xy+ xy+ 0

[ x. x= x]; [ y. y= 0]

= x (1+ y+ y)

= x (1)

[ 1+y= 1 ]

= x.

b) xy + xz + yz.

= xy + xz + yz (1)= xy + xz + yz (x+ x)

[x+ x= 1]

= xy + xz + xyz + xyz

Re-arranging,

= xy + xyz + xz +xyz

= xy (1+ z) + xz (1+y)

= xy (1) + xz (1)

[1+y= 1]; [1+z= 1] = xy+ xz.

63. 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

64. Find the minterm of xy+yz+xy'z

(Nov, 2008)

= xy+ yz+ xyz

= xy+ z (y+ xy)

= xy+ z (y+ x)

[x+ xy = x+ y]65. Simplify the following Boolean expression to a minimum number of literals:

AB+ ACD+ ABD+ ABCD

(May, 2009)

= AB (1+D) + ACD+ ABCD

= AB (1) + ACD+ ABCD

[1+ x = 1]

= AB+ ACD+ ABCD

= AB+ ABCD+ ACD

= AB (1+CD) + ACD

= AB (1) + ACD

[1+ x = 1]= AB+ ACD

= A (B+CD)

Simplification Of Boolean functions using K-map & Tabulation Methods :66. What are the methods adopted to reduce Boolean function?

i) Karnaugh map

ii) Tabular method or Quine Mc-Cluskey method

iii) Variable entered map technique.

67. 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, with each squares representing one minterm of the function.68. What is meant by three variable map?

Three variable map have 8 minterms for three variables, hence the map consists of 8 squares, one for each minterm.69. What is a prime implicant?

A prime implicant is a product term obtained by combining the maximum possible number of adjacent squares in the map.70. What is an essential implicant?

If a min term is covered by only one prime implicant, the prime implicant is said to be essential.71. What are called dont 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 ord in the truth tables and are called dont care conditions or incompletely specified functions.72. Simplify the following Boolean function by Karnaugh map method:

F (A, B, C, D) = m (1, 5, 9, 12, 13, 15)

(May, 2009)

Therefore,F= ABD+ CD+ ABC

73. What are the drawbacks of Karnaugh map?

(Nov, 2007)The drawbacks of the K-map method are

i. Generally it is limited to six variable map (i.e.) more than six variable involving expressions are not reduced.ii. The map method is restricted in its capability since they are useful for simplifying only Boolean expression represented in standard form.iii. It is not suitable for computer reduction.iv. Care must be taken to fill in every cell with the relevant entry, such as a 0, 1 (or) dont care terms.74. Explain or list out the advantages and disadvantages of K-map method?

The advantages of the K-map method are

i. It is a fast method for simplifying expression up to four variables.ii. It gives a visual method of logic simplification.iii. Prime implicants and essential prime implicants are identified fast.iv. Suitable for both SOP and POS forms of reduction.v. It is more suitable for class room teachings on logic simplification.

The disadvantages of the K-map method are

i. It is not suitable for computer reduction.ii. K-maps are not suitable when the number of variables involved exceed four.iii. Care must be taken to fill in every cell with the relevant entry,such as a 0, 1 (or) dont care terms.

75. Why we go in for tabulation method?

This method can be applied to problems with many variables and has the advantage of being suitable for machine computation.76. List out the advantages and disadvantages of Quine Mc-Cluskey method?

The advantages are,

i. This is suitable when the number of variables exceed four.ii. Digital computers can be used to obtain the solution fast.iii. Essential prime implicants, which are not evident in K-map, can be clearly seen in the final results.

The disadvantages are,

i. Lengthy procedure than K-map.ii. Requires several grouping and steps as compared to K-map.iii. It is much slower.iv. No visual identification of reduction process.v. The Quine Mc-Cluskey method is essentially a computer reduction method.Logic Gates :77. What is a Logic gate?

Logic gates are the basic elements that make up a digital system. The electronic gate is a circuit that is able to operate on a number of binary inputs in order to perform a particular logical function.78. Distinguish between positive logic and negative logic.

(Nov, 2003)

In binary logic, two voltage levels represent the two binary digits, 1 and 0. If the higher of the two voltages represents a 1 and the lower voltage represents a 0, the system is called positive logic system. On the other hand, if the lower voltage represents a 1 and the higher voltage represents a 0, then it is a negative logic system.79. What are the basic digital logic gates?

The three basic logic gates are

AND gate OR gate NOT gate

80. Which gates are called as the universal gates? What are its advantages?

The NAND and NOR gates are called as the universal gates. These gates are used to perform any type of logic application.81. Bubbled OR gate is equal to--------------

NAND gate82. Bubbled AND gate is equal to--------------

NOR gate83. How will you use a 4 input NOR gate as a 2 input NOR gate?(May, 2003)

By connecting unused inputs to logic 0, we can use 4-input NOR gate as a 2 input NOR gate.

84. How will you use a 4 input NAND gate as a 2 input NAND gate?(Nov, 2002)

By connecting unused inputs to logic 1, we can use 4-input NAND gate as a 2 input NAND gate.

85. What is meant by a functionally complete set of logic gates?(May, 2005)

A set of logic gates by which we can implement any logic function is called functionally complete set of logic gates.

86. Show that a positive logic NAND gate is the same as a negative logic NOR gate.

(May, 2003; Nov, 2004)

Logic expression for NAND gate is, Y= (A .B)

Y= (A .B)

= A +B

Y= A + B is the logic expression for negative logic NOR gate.

87. What happens when all the gates is a two level AND-OR gate network are replaced by NOR gate

(May, 2004; Nov, 2004, IT)

The output will change. We will get complemented output when all applied inputs are complemented.

88. Realize OR gate using NAND gate.

(Nov, 2005)

OR gate using AND gate

UNIT 2COMBINATIONAL LOGIC

1. Define Combinational circuit.

(May 2009)A combinational circuit consists of logic gates whose outputs at anytime are determined directly from the present combination of inputs, without regard to previous inputs.

2. What is a half-adder?

A half-adder is a combinational circuit that can be used to add two bits. It has two inputs that represent the two bits to be added and two outputs, with one producing the SUM output and the other producing the CARRY.

3. Give the truth table for a half adder.InputsOutputs

ABSum (S)Carry (C)

0000

0110

1010

1101

4. From the truth table of a half adder derive the logic equation

5. Draw the logic diagram of a half-adder.

(Nov, 05; Nov 09)

6. What is a full adder?

(May, 2007)

A full adder is a combinational circuit that forms the arithmetic sum of three input bits. It consists of 3 inputs and 2 outputs. Two of the input variables, represent the significant bits to be added. The third input represents the carry from previous lower significant position. The block diagram of full adder is given by,

7. Give the truth table for a full adder.InputsOutputs

ABCinSum (S)Carry (Cout)

00000

00110

01010

01101

10010

10101

11001

11111

8. From the truth table of a full adder derive the logic equation

9. Draw the circuit of a full-adder.

10. What is half-Subtractor?

A half-subtractor is a combinational circuit that can be used to subtract one binary digit from another to produce a DIFFERENCE output and a BORROW output. The BORROW output here specifies whether a 1 has been borrowed to perform the subtraction.

11. Give the truth table for a half Subtractor.

(Nov, 2005)InputsOutputs

ABDifference (D)Borrow (Bout)

0000

0111

1010

1100

12. From the truth table of a half-Subtractor derive the logic equation

13. Draw the circuit of a half-subtractor.

14. What is a full-subtractor?

A full subtractor performs subtraction operation on two bits, a minuend and a subtrahend, and also takes into consideration whether a 1 has already been borrowed by the previous adjacent lower minuend bit or not. As a result, there are three bits to be handled at the input of a full subtractor, namely the two bits to be subtracted and a borrow bit designated as Bin. There are two outputs, namely the DIFFERENCE output D and the BORROW output Bo. The BORROW output bit tells whether the minuend bit needs to borrow a 1 from the next possible higher minuend bit.

15. Give the truth table for a full-subtractor.

(Nov, 2004)InputsOutputs

ABBinDifference(D)Borrow(Bout)

00000

00111

01011

01101

10010

10100

11000

11111

16. From the truth table of a full-Subtractor derive the logic equation

17. Draw the circuit of a full-subtractor.

18. What is Binary parallel adder?

A binary parallel adder is a digital function that produces the arithmetic sum of two binary numbers in parallel.

19. What is BCD adder?

A BCD adder is a circuit that adds two BCD digits in parallel and produces a sum digit also in BCD.

20. What are Parity Generator/ Checker?

A parity bit is used for the purpose of detecting errors during transmission of binary information. A parity bit is an extra bit included in a binary message to make the number of 1s either odd or even.

The message, including the parity bit is transmitted and then checked at the receiving end for errors. An error is detected if the checked parity does not correspond with the one transmitted.

The circuit that generates the parity bit in the transmitter is called a parity generator and the circuit that checks the parity in the receiver is called a parity checker.

21. Draw the circuit diagram for 3-bit parity generator.

(Nov, 2007)

22. Draw the logic diagram of 4 bit even parity checker.

(Nov, 2008)

23. What is Magnitude Comparator?

A Magnitude Comparator is a combinational circuit designed primarily to compare the relative magnitude of two binary numbers.

It receives two n-bit numbers A and B as inputs and the outputs are A>B, A=B and A