2 Marks Q&A - Weeblyshanthirevathi.weebly.com/uploads/1/2/8/5/12854997/dpsd-2-marks.pdf · CHAPTER 1 BOOLEAN ALGEBRA AND LOGIC GATES Number Base Conversions : 1. List the different

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2 Marks Q&A

Digital Principles&

Systems Design

3rd SEM CSE & IT

PANIMALAR INSTITUTE OF TECHNOLOGY(JAI SAKTHI EDUCATIONAL TRUST)CHENNAI- 600 123

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CHAPTER 1

BOOLEAN ALGEBRA AND LOGIC GATES

Number Base Conversions :

1. List the different number systems?

i) Binary Number system

ii) Octal Number system

iii) Decimal Number system

iv) Hexadecimal Number system

2. 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)



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= (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)10

3. 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) 10

The 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



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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) 8

7. Convert the (153.513)10 to octal. (May, 2010)

Integer part:

8 153

8 19 -- 1

2 -- 3

= (231)8

Fractional 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)8

8. Convert 0.1289062 decimal number to its hex equivalent

0.1289062 x 16 = 2.0625 2

0.0625 x 16 = 1.0 1



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Ans = (0.21)16

9. 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)10

Now, convert this number to base 7.

Integer part:

7 53

7 7 -- 4

1 -- 0

= (104)7

Fractional 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)16

Fractional Part:

0.64 x 16 = 10.24 A

0.24 x 16 = 3.84 3

0.84 x 16 = 13.44 D

0.44 x 16 = 7.04 7



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Ans = (16. A3D7) 16

11. 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)10

Now, convert this number to base 7.

Integer part:

7 45

6 -- 3

= (63)7

Fractional 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)7

12. Convert (634)8 to binary

6 3 4

110 011 100

Ans = (110011100)2

13. 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



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(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)16

Complements :15. What are the different types of number complements?

i) r’s Complement

ii) (r-1)’s Complement.



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16. What is the range of values that can be represented using n-bit 2’s complement

form of representation? What is the corresponding range with n-bit 1’s

complement form? (May, 2006)

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

defined as follows.

2’s 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.

1’s 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)2

1 0 1 0

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

Answer = (1 1 0 1)2

18. Substract (0 1 0 1)2 from (1 0 1 1)2

1 0 1 0

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

Answer = (1 1 0)2

19. 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 1’s and 2’s complement of the following binary numbers:

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



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Soln:

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

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

a) 111010101’s complement: (00010101)2

2’s complement : 0 0 0 1 0 1 0 1

(+) 1--------------------(0 0 0 1 0 1 1 0)2

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

b) 011111101’s complement: (10000001)2

2’s complement : 1 0 0 0 0 0 0 1

(+) 1---------------------(1 0 0 0 0 0 1 0)2

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

c) 000000011’s complement: (01111110)2

2’s complement : 0 1 1 1 1 1 1 0

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

(1 1 1 1 1 1 1 1)2-----------------------

d) 100000001’s complement: (01111111)2

2’s complement : 0 1 1 1 1 1 1 1

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

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

e) 000000001’s complement: (11111111)2

2’s complement : 0 1 1 1 1 1 1 1



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(+) 1-----------------------(1 0 0 0 0 0 0 0 0)2-----------------------

21. Find 2’s complement of (1 0 1 0 0 0 1 1) 2

Soln:

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

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

Soln:

0 0 0 1 1 0 - 1’s Complement of (1 1 1 0 0 1)2+ 0 0 0 0 1--------------0 0 0 1 1 1 - 2’s complement.--------------

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

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

To get the answer in true form , take the 2’s complement and assign negative

number to the answer.

Answer in true form - ( 0 0 1 1 1 0 )2

23. Perform subtraction using 1’s complement (11010)2 – (10000)2.

1 1 1 11 1 0 1 0

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

1 0 1 0 0 1

(+) 1 (Add carry to LSB)

(0 1 0 1 0)2

Y- X = (01010)2



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24. 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 00 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)2

b) Y = 1 0 0 0 0 1 10 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 2’s complement again for the above answer.

Answer is Y-X = - (2's complement of 1101111)

= - (0010001)2

25. 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)2

b) 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)



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= - (0010001)2

26. 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 1’s complement representation, in which the negative number is

indicated by its 1’s complement.

iv. In signed 2’s complement representation, in which the negative number is

indicated by its 2’s 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 (X’s-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 codes

ii. Non - weighted codes

iii. Reflective codes

iv. Sequential codes



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v. Alphanumeric codes

vi. Error Detecting and correcting codes.

29. What is meant by bit?

A binary digit is called bit

30. 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 0

34. 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.



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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.



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(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’= 0

43. Show that A+A’.B = A+B using the theorems of Boolean algebra. (Nov, 2005)

LHS= A+A’.B

= A+ AB + A’B [ A+AB = A]

= A+ B (A+ A’)

= A+ B (1) [A+ A’= 1]

= A+ B

Canonical 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'



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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+C’D). (May, 2008)

= (AB.B+ B.C+ AB.C’D+ C.C’D)

= AB+ BC+ ABC’D [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’) + ABC’D

= (ABC+ ABC’) (D+ D’) + (ABC+ A’BC) (D+ D’) + ABC’D

= ABCD+ ABCD’+ ABC’D+ ABC’D’+ ABCD+ ABCD’+ A’BCD+ A’BCD’

+ ABC’D

= ABCD+ ABCD’+ ABC’D+ ABC’D’+ A’BCD+ A’BCD’.

= 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

= ПM (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.



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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’+ x’y.

(b) F2= (xy + y’z + xz) x. (Nov, 2007)

(a) F1= xy’+ x’y

F1’= (xy’+ x’y)’

= (xy’)’. (x’y)’

= (x’+y) (x+y’)

= x’x+ x’y’+ yx+ yy’

= x’y’+ xy.

(b) F2= (xy + y’z + xz) x.

F2’ = ((xy + y’z + xz) x)’

= (xy + y’z + xz)’ + x’

= [(xy)’ (y’z)’ (xz)’] + x’

= [(x’+y’) (y+z’) (x’+z’)] + x’

= [(x’y+ x’z’+ 0+ y’z’) ( x’+z’)] + x’

= x’x’y+ x’x’z’+ x’y’z’+ x’yz’+ x’z’z’+ y’z’z’+ x’

= x’y+ x’z’+ x’y’z’+ x’yz’+ x’z’+ y’z’+ x’ [x+ x = x], [x. x = x]

= x’y+ x’z’+ x’z’ (y’+ y) + y’z’+ x’ [x+ x’= 1]

= x’y+ x’z’+ x’z’ (1) + y’z’+ x’

= x’y+ x’z’+ y’z’+ x’

= x’y+ x’+ x’z’+ y’z’

= x’(y+1) + x’z+ y’z’ [y+1= 1]

= x’ (1+z) + y’z’ [y+1= 1]



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= x’+ y’z’

52. Obtain the complement of f = wx’y + xy’+ wxz using De Morgan’s theorem.(May, 2006)

f’ = (wx’y + xy’+ wxz)’

= (wx’y)’ (xy’)’ (wxz)’

= (w’+x+ y’) (x’+ y) (w’+ x’+ z’)

= (w’x’+ w’y+ xx’+ xy+ x’y’+ yy’) (w’+ x’+ z’)

= (w’x’+ w’y+ xy+ x’y’) (w’+ x’+ z’)

= w’x’. w’+ w’y. w’+ xy. w’+ x’y’. w’+ w’x’. x’+w’y. x’+ xy. x’+ x’y’. x’+

w’x’. z’+ w’y. z’+ xy. z’+ x’y’.z’

= w’x’+ w’y+ w’xy+ w’x’y’+ w’x’+ w’x’y+ 0 + x’y’+ w’x’z’+ w’yz’+ xyz’+

x’y’z’

= w’x’+ w’y+ w’xy+ w’x’y’+ w’x’y+ x’y’+ w’x’z’+ w’yz’+ xyz’+ x’y’z’

= w’x’( 1+ y’+ y+ z’)+ w’y( 1+ x+ z’)+ x’y’(1+ z’)+ xyz’

= w’x’(1)+ w’y(1)+ x’y’(1)+ xyz’

= w’x’+ w’y+ x’y’+ 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]



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= 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)' + 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]

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]



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= 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 + x’z + yz. (May, 2007)

(a) x (x’+y)

= xx’+ xy

= xy. [x. x’= 0]

(b) xy + x’z + yz.

= xy + x’z + yz( x+ x’)

= xy + x’z + xyz + x’yz

= xy + xyz + x’z +x’yz



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= xy (1+ z) + x’z (1+y) [1+y= 1]

= xy+ x’z.

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

a) (x + y) (x + y’)

b) xy + x’z + 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 + x’z + yz.

= xy + x’z + yz (1)

= xy + x’z + yz (x+ x’) [x+ x’= 1]

= xy + x’z + xyz + x’yz

Re-arranging,

= xy + xyz + x’z +x’yz

= xy (1+ z) + x’z (1+y)

= xy (1) + x’z (1) [1+y= 1]; [1+z= 1]

= xy+ x’z.

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+ xy’z



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= xy+ z (y+ xy’)

= xy+ z (y+ x) [x+ x’y = x+ y]

65. Simplify the following Boolean expression to a minimum number of literals:

A’B’+ A’C’D’+ A’B’D+ A’B’CD’ (May, 2009)

= A’B’ (1+D) + A’C’D’+ A’B’CD’

= A’B’ (1) + A’C’D’+ A’B’CD’ [1+ x = 1]

= A’B’+ A’C’D’+ A’B’CD’

= A’B’+ A’B’CD’+ A’C’D’

= A’B’ (1+CD’) + A’C’D’

= A’B’ (1) + A’C’D’ [1+ x = 1]

= A’B’+ A’C’D’

= A’ (B’+C’D’)

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.



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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 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.

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+ C’D+ 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.



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iv. Care must be taken to fill in every cell with the relevant entry, such as a 0, 1 (or)

don’t 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)

don’t 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.



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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 gate

82. Bubbled AND gate is equal to--------------

NOR gate



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83. 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



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CHAPTER 2

COMBINATIONAL 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.

Inputs OutputsA B Sum (S) Carry (C)0 0 0 00 1 1 01 0 1 01 1 0 1

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



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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.

Inputs OutputsA B Cin Sum (S) Carry (Cout)0 0 0 0 00 0 1 1 0



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0 1 0 1 00 1 1 0 11 0 0 1 01 0 1 0 11 1 0 0 11 1 1 1 1

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.



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11. Give the truth table for a half Subtractor. (Nov, 2005)

Inputs OutputsA B Difference (D) Borrow (Bout)0 0 0 00 1 1 11 0 1 01 1 0 0

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.



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15. Give the truth table for a full-subtractor. (Nov, 2004)

Inputs OutputsA B Bin Difference(D) Borrow(Bout)0 0 0 0 00 0 1 1 10 1 0 1 10 1 1 0 11 0 0 1 01 0 1 0 01 1 0 0 01 1 1 1 1

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

17. Draw the circuit of a full-subtractor.



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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 1’s 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)



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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<B. Depending upon the relative magnitudes of the two numbers, one of the

outputs will be high.

24. List out the applications of comparators?

The following are the applications of comparator

Comparators are used as a part of the address decoding circuitry in computers to

select a specific input/output device for the storage of data.

They are used to actuate circuitry to drive the physical variable towards the

reference value.

They are used in control applications.

25. What is the need for code conversion? (May, 2009)

If two systems working with different binary codes are to be synchronized in

operation, then we need digital circuit which converts one system of codes to the other.

The process of conversion is referred to as code conversion.

26. What is code converter?

It is a circuit that makes the two systems compatible even though each uses a

different binary code. It is a device that converts binary signals from a source code to its

output code. One example is a BCD to Xs3 converter.



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27. Construct a 4-bit binary to gray code converter circuit and discuss its

operation. (May, 2006)

The gray code is often used in digital systems because it has the advantage that

only bit in the numerical representation changes between successive.

28. What is logic synthesis in HDL? (Nov, 2006; Nov, 2007)

Logic Synthesis is the automatic process of transforming a high level language

description such as HDL into an optimized netlist of gates that perform the operations

specified by the source code.

It is the process of deriving a list of components and their interconnections from

the model of a digital system described in HDL.

29. List the important features of HDL. (Nov, 2006; May 2010)

1. It is specifically oriented to describe hardware structures and behaviors.

2. It can be used to represent logic diagrams, Boolean expressions and other

complex digital circuits.

3. It is used to represent and document digital systems in a form that can be read

by both humans and computers.

30. Mention any two uses of HDL. (May, 2006)

1. HDL is a language that describes the hardware of digital systems in textural form.

2. It can be used to represent logic diagrams, Boolean expressions and other more

complex digital circuits.



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3. It is used to represent and document digital systems in a form that can be read by

both humans and computers.

4. The language content can be stored and retrieved easily and processed by

computer software in an efficient manner.

CHAPTER 3

DESIGN WITH MSI DEVICES

Decoders & Encoders1. What do you mean by analyzing a combinational circuit?

The reverse process for implementing a Boolean expression is called as analyzing

a combinational circuit. (ie) the available logic diagram is analyzed step by step and

finding the Boolean function.

2. What is decoder? (May, 09)

A decoder is a combinational circuit that decodes the binary information on ‘n’

input lines to a maximum of 2n unique output lines. The general structure of decoder

circuit is –



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3. What is encoder? (May, 10)

An encoder is a combinational circuit that converts binary information from 2n

input lines to a maximum of ‘n’ unique output lines. The general structure of encoder

circuit is –

4. List out the applications of decoder?

1. Decoders are used in counter system.

2. They are used in analog to digital converter.

3. Decoder outputs can be used to drive a display system.

5. What are the functions of encoders and decoders? (Nov, 2006)

An encoder is a combinational circuit that converts binary information from 2n

input lines to a maximum of ‘n’ unique output lines.

A decoder is a combinational circuit that decodes the binary information on ‘n’

input lines to a maximum of 2n unique output lines.

6. Distinguish between decoder and encoder

S.No Decoder Encoder



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1 One of the input lines is activated corresponding

to the binary input

The input lines generate the binary code,

corresponding to the input value

2 Input of the decoder is an encoded information

presented as ‘n’ input producing 2n possible

outputs.

Input of the encoder is a decoded

information presented as ‘2n’ inputs

producing ‘n’ outputs.

3 The input code generally has a fewer bits than

the output code.

The input code generally has more bits

than the output code.

4

7. Implement the logic function f= ∑m (0, 2, 3, 6) using a decoder (May, 2006)

8. What is priority encoder? (May, 2008, May, 2007)

A priority encoder is an encoder that includes the priority function. The operation

of the priority encoder is such that if two or more inputs are equal to 1 at the same time,

the input having the highest priority will take precedence.

Inputs Outputs



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D0 D1 D2 D3 Y1 Y0 V01xxx

001xx

0001x

00001

x0011

x0101

01111

Multiplexers & Demultiplexers9. What is a multiplexer? (Nov, 06; May 10)

A multiplexer is a digital switch which allows digital information from several

sources to be routed into a single output line.

The basic multiplexer has several data-input lines and a single output line. The

selection of a particular input line is controlled by a set of selection lines. Normally there

are 2n input lines and n selection lines.

10. What is the function of the enable input in a Multiplexer?

The function of the enable input in a MUX is to control the operation of the unit.

11. Implement the logic function f= AB + A’.B’ using a suitable multiplexer.

(Nov, 2005)

f= AB + A’B’

= ∑m (3, 0)



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12. How can a multiplexer be used to convert 8-bit parallel data into serial form?

Draw the circuit and briefly explain. (May, 2006)

Here, binary counter is used to derive the select inputs of the multiplexer so that

as the binary counter increments its count, the next bit is available at the output of the

multiplexer. The binary counter counts from 000 to 111, therefore D0 through D7 bits are

available at the output of the multiplexer as serial output.

13. Mention the uses of multiplexer.

1. It can be used to realize a Boolean function

2. It can be used in communication systems e.g., time division multiplexing.

14. Mention any two applications of multiplexers. (May, 07; May, 09)

1. Data routing



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2. Logic function generator

3. Control sequencer

4. Parallel-to-serial converter.

15. Construct a 16×1 multiplexer with two 8×1 multiplexer and 2×1 multiplexer.

(Nov, 2008)

16. What is a demultiplexer? (May 2008)

A demultiplexer is a combinational logic circuit with an input line, 2n output lines

and n select lines. It routes the information present on the input line to any of the output

lines. The output line that gets the information present on the input line is decided by the

bit status of the selection lines.



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17. Mention the uses of Demultiplexer.

Demultiplexer is used in computers when a same message has to be sent to

different receivers. Not only in computers, but any time information from one source can

be fed to several places.

18. How can a decoder be converted into a demultiplexer? (Nov, 2005)

Decoder is a circuit which converts one form of code into another. Demultiplexer

is a circuit which converts one input to many outputs. If the enable line E is taken as a

data input line A and B are taken as selection lines, then it is a demultiplexer.

19. Can a decoder function as a Demultiplexer?

Yes. A decoder with enable can function as a Demultiplexer if the enable line E is

taken as a data input line A and B are taken as selection lines.

20. Give the applications of Demultiplexer.

1. It finds its application in Data transmission system with error detection.

2. One simple application is binary to Decimal decoder.

21. Give other name for Multiplexer and Demultiplexer.

Multiplexer is otherwise called as Data selector.

Demultiplexer is otherwise called as Data distributor.



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22. Distinguish between decoder and demultiplexer. (May, 04; Nov, 09)

S.No Decoder Demultiplexer

1Decoder is a many input to many

output device.

Demultiplexer is a one input to many

output devices.

2There are no selection lines. The selection of specific output line is

controlled by the value of selection lines.

23. Design 8: 1 multiplexer using two 4:1 multiplexers.

24. Design 1: 8 demultiplexer using two 1: 4 demultiplexers.



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Memory & Programmable logic25. Define address and word

Each bit combination of the input variable is called on address. Each bit combination that

comes out of the output lines is called a word.

26. What is RAM?

A memory unit is a collection of storage cells together with associated circuits

needed to transfer information in and out of the device. The time it takes to transfer

information to or from any desired random location is always the same. Hence, the name

random-access memory (RAM).

27. List the types of RAM.

1. Static RAM

2. Dynamic RAM

28. Explain SRAM?

1. Static RAM (SRAM) consists of internal latches that store the binary

information. The stored information remains valid as long as the power is applied to the

unit.

2. SRAM is easier to use and has shorter read and write cycle.

3. The memory capacity of a static RAM varies from 64 bit to 1 mega bit.

29. Explain DRAM?

1. The dynamic RAM (DRAM) stores the binary information in the form of

electric charges on capacitors. The capacitors are provided inside the chip by MOS

transistors.

Dynamic RAM



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2. The stored charges on the capacitors tend to discharge with time and the

capacitors must be tending to discharge with time and the capacitors must be periodically

recharged by refreshing the dynamic memory.

3. DRAM offers reduced power consumption and larger storage capacity in a

single memory chip.

30. Differentiate static RAM and dynamic RAM.

S.No Static RAM Dyanamic RAM

1It contains less memory cells per

unit area.

It contains more memory cells per unit

area.

2Its access time is less, hence faster

memories.Its access time is greater than static RAM

3It consists of number of flip-flops.

Each flip-flop stores one bit.

It stores the data as a charge on the

capacitor. It consists of MOSFET and

capacitor for each cell.

4 Refreshing circuitry is not required.

Refreshing circuitry is required to maintain

the charge on the capacitors every time

after every few milliseconds. Extra

hardware is required to control refreshing.

This makes system design complicated.

5 Cost is more Cost is less.

31. Differentiate volatile and non-volatile memory?

S. No. Volatile memory Non-volatile memory

1

They are memory units which

loses stored information when

power is turned off.

It retains stored information when power is

turned off.

2 E.g. SRAM and DRAM E.g. Magnetic disc and ROM



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32. What are the advantages of RAM?

The advantages of RAM are

1. Non-destructive read out

2. Fast operating speed

3. Low power dissipation

4. Compatibility

5. Economy.

33. In what ways memory expansion can be achieved?

The memory expansion can be achieved in two ways:

1. By expanding word size,

2. By expanding memory capacity.

34. How memory expansion can be achieved by expanding word size?

The word size of the memory IC can be increased by connecting two memory

IC’s in such a way that their data bus is in series and address bus in parallel. Both

memory IC’s are selected simultaneously by common chip select signal to access entire

expanded word at time.

35. How memory expansion can be achieved by expanding memory capacity?

The memory capacity can be increased by connecting two or more memory IC’s

in parallel ie., the address, data and control lines are connected in parallel to all memory

IC’s. Each IC is selected by the separate chip select signal generated by the address

decoder.

36. What is ROM?

A Read-only memory (ROM) is essentially a memory device in which permanent

binary information is stored. The binary information must be specified by the designer

and is then embedded in the unit to form the required interconnection pattern. Once the

pattern is established, it stays within the unit even when power is turned OFF and ON

again.



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37. List the types of ROM.

1. Masked ROM,

2. Programmable ROM (PROM)

3. Erasable ROM (EPROM)

4. Electrically Erasable ROM (EEROM)

38. Explain masked ROM.

In masked ROM, mask programming is done by the manufacturer during the last

fabrication process of the unit. The procedure for fabricating a ROM requires that the

customer fill out the truth table, the ROM is to satisfy.

39. Explain PROM.

The PROM (Programmable Read-only memory), allows user to store data/

program. It uses the fuses with material like nichrome and polycrystalline. The user can

blow the fuses by passing around 20 – 50 mA of current for a period of 5 – 20 µsec. The

blowing of fuses according to the truth table is called programming of ROM.

The PROM programmer selectively burns the fuses according to the bit pattern to

be stored.

The PROM’s are one-time programmable, once programmed, the information

stored is permanent.

40. Explain EPROM.

The EPROM (Erasable PROM), uses MOS circuitry. They store 1’s and 0’s as

packets of charge in a buried layer of IC chip. It can be programmed by the user by a

special EPROM programmer.

The information stored can be erased by exposing the chip to Ultraviolet through

its quartz window for 15 to 20 minutes.

In EPROM’s, it is not possible to erase selective information, when erased the

entire information is lost. The chip can be reprogrammed.

It is ideally suited for product development, college laboratories, etc.



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41. Explain EEPROM.

The EEPROM (Electrically Erasable PROM), also uses MOS circuitry. Data is

stored as charge or no charge on an insulating layer, which is made very thin (< 200Å).

Therefore a voltage as low as 20- 25V can be used to move charges across the thin barrier

in either direction for programming or erasing ROM.

It allows selective erasing at the register level rather than erasing all the

information, since the information can be changed by using electrical signals.

It has chip erase mode by which the entire chip can be erased in 10 msec. Hence

EEPROM’s are most expensive.

42. What is a Programmable logic device (PLD)?

A Programmable logic device (PLD) is an integrated circuit with internal logic

gates that are connected through electronic fuses. Programming the device involves

blowing the fuses along the paths that must be disconnected so as to obtain a particular

configuration. It is divided into an AND array and an OR array to provide an AND-OR

sum of product implementation.

43. List basic types of programmable logic devices.

1. PROM: Programmable Read only memory,

2. PLA: Programmable logic Array,

3. PAL: Programmable Array Logic,

4. FPGAs: Field programmable Gate Arrays,

5. CPLDs: Complex programmable Logic Devices.

44. Explain PROM?

The Programmable read-only memory has a fixed AND array constructed as a

decoder and programmable OR array. The programmable OR gates implement the

Boolean function in sum of minterms.



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45. Explain PAL?

It has a programmable AND array and a fixed OR array. The AND gates are

programmed to provide the product terms for the Boolean functions, which are logically

summed in each OR gate.

46. Explain PLA

The most flexible PLD is the programmable logic array (PLA), where both the

AND and OR arrays can be programmed. The product term in the AND array may be

shared by any OR gate to provide the required sum of product implementation.

47. Differentiate ROM & PLD’s.

S.No ROM (Read Only Memory) PLD’s (Programmable Logic Array)

1

1.It is a device that includes both

the decoder and the OR gates with

in a single IC package

It is a device that includes both AND

and OR gates with in a single IC package

2

ROM does full decoding of the

variables and generates all the

minterms

PLD’s does not provide full decoding of

the variable and does not generate all the

minterms.



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48. What are the difference between PLA and PAL? (Nov, 2006)

S.No PLA PAL

1

PLA is a device with a

programmable AND array and

programmable OR array

PAL is a programmable logic device with a

fixed OR array and programmable AND

array.

2

PLA is comparatively difficult to

program, as both AND and OR

array are programmable.

PAL is easier to program as only AND gates

are programmable.

3 It is flexible. It is less flexible than PLA.

49. Give differences between PROM, PLA, and PAL. (Nov, 09)

S.No PROM PLA PAL

1

AND array is fixed and

OR array is

programmable

Both AND and OR arrays

are programmable

OR array is fixed and

AND array is

programmable

2Cheaper and simpler to

useCostliest and complex Cheaper and simpler

3All minterms are

decoded

AND array can be

programmed to get desired

minterms

AND array can be

programmed to get

desired minterms

4

Only Boolean functions

in standard SOP form

can be implemented

using PROM

Any Boolean functions in

SOP form can be

implemented using PLA

Any Boolean functions

in SOP form can be

implemented using PLA

50. What are the terms that determine the size of a PAL?

The size of a PAL is specified by the

a. Number of inputs

b. Number of products terms

c. Number of outputs



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51. What is meant by memory decoding?

The memory IC used in a digital system is selected or enabled only for the range

of addresses assigned to it.

52. What is access and cycle time?

The access time of the memory is the time to select word and read it. The cycle

time of a memory is a time required to complete a write operation.

53. When an overflow condition will encounter in an accumulator register?

(Nov, 2007)

When the output exceeds the capacity of accumulator.

54. What is the maximum range of a memory that can be accessed using 10

address lines? (May, 2008)

Maximum range of memory = 2 address lines.

= 210

= 1024 bytes.

HDL for Combinational Circuits

55. What is VHDL?

VHDL is a hardware description language that can be used to model a digital

system at many level of abstraction, ranging from the algorithmic level to the gate level.

The VHDL language has a combination of the following language.

1. Sequential language

2. Concurrent language

3. Net-list language

4. Timing specification

5. Waveform generation language.



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56. What are the features of VHDL?

The features of VHDL are

1. VHDL has powerful constructs.

2. VHDL supports design library.

3. The language is not case sensitive.

57. What is gate level modelling? (Nov, 2007)

This technique uses primitive gates and user-defined modules. It describes a

schematic diagram in a textural form.

58. What are the modelling techniques available to build HDL module?

(May, 2007)

1. Gate level modeling using instantiation of primitive gates and user-defined

modules.

2. Data flow modeling, using continuous assignment statements with keyword

‘assign’.

3. Behavioral modeling using procedural assignment with keyword ‘always’.

59. Define entity?

Entity gives the specification of input/output signals to external circuitry. An

entity is modeled using an entity declaration and at least one architecture body. Entity

gives interfacing between device and others peripherals.

60. List out the different elements of entity declaration?

The different elements of entity declaration are:

1. entity_name

2. signal_name

3. mode

4. in:

5. out:

6. input



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7. buffer

8. signal_type

61. What do you meant by concurrent statement?

Architecture contains only concurrent statements. It specifies behavior,

functionality, interconnections or relationship between inputs and outputs.

62. What are operators used in VDHL language?

There are different types of operators used in VHDL language

Logical operators : AND, OR, NOT, XOR, etc.,

Relational operator : equal to, <less than etc.,

Shift operators : SLL- Shift Left Logical, ROR- Rotate Right Logical etc.,

Arithmetic operators : Addition, subtraction etc.,

Miscellaneous operators: <= assign to etc.,

63. Define VHDL package?

A VHDL, package is a file containing definitions of objects which can be used in

other programs. A package may include objects such as signals, type, constant, function,

procedure and component declarations.



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CHAPTER 4SYNCHRONOUS SEQUENTIAL

CIRCUITS

1. What is sequential circuit?

The circuits in which the output variables depend not only on the present input but

they also depend upon the past history of these input variables are known as sequential

circuits.

Block diagram of sequential circuit

The memory elements are connected to the combinational circuit as a feedback

path. The present state and the external inputs determine the outputs and the next state of

the sequential circuit.

2. What are the differences between sequential and combinational logic circuits?

(Nov, 2004; Nov, 2007; May 2010)

S.No Combinational logic Sequential logic

1Output depends on present input Output depends not only on present input

but also depend upon the past inputs.

2Memory unit is not required Memory unit is required to store past input

variables

3 Faster in speed Slower

4 Easy to design Hard

5Ex: Adders, Subtractor, MUX,

DEMUX, Encoder, Decoder etc..,

Ex: Shift Registers, Counters

3. List the classifications of sequential circuit.

1. Synchronous sequential circuit.

2. Asynchronous sequential circuit.



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4. What is Synchronous sequential circuit?

A Synchronous sequential circuit is a system whose behavior can be defined from

the knowledge of its signal at discrete instants of time.

5. Mention one advantage and disadvantage of Asynchronous sequential circuit.

(Nov, 2005)

Advantage:

Because of the absence of clock it can operate faster than synchronous sequential

circuits.

Disadvantage:

The charge in input signal can affect memory elements at any instant of time and

it is more difficult to design.

6. Distinguish between synchronous and asynchronous sequential logic circuits.

(Nov, 2002; Nov, 2003; May, 2005)

S.No Synchronous sequential circuits Asynchronous sequential circuits

1Memory elements are clocked flip-

flops

Memory elements are either unclocked

flip-flops or time delay elements.

2

The change in input signals can

affect memory element upon

activation of clock signal.

The change in input signals can affect

memory element at any instant of time.

3

The maximum operating speed of

clock depends on time delays

involved.

Because of the absence of clock, it can

operate faster than synchronous circuits.

4 Easier to design More difficult to design

7. What is a clocked sequential circuit?

Synchronous sequential circuit that use clock pulses in the inputs of memory

elements are called clocked sequential circuit. One advantage as that they don’t cause

instability problems.



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Flip-flops8. What is called latch?

Latch is a simple memory element, which consists of a pair of logic gates with

their inputs and outputs inter connected in a feedback arrangement, which permits a

single bit to be stored.

Enable signal is provided with the latch. When enable signal is active, output

changes with output. When enable signal is not activated, input changes does not affect

output.

9. Draw the internal circuit of a NOR gate latch and derive the truth table.

(May, 2006)

The SR latch is a digital circuit with two inputs S and R and two cross-coupled

NOR gates.

Truth table:

S R Qn Qn+1 State

0

0

0

0

0

1

0

1

No Change

(NC)

0

0

1

1

0

1

0

0Reset

1

1

0

0

0

1

1

1Set

1

1

1

1

0

1

x

x

Indeterminate

*



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10. Draw the logic diagram of D-type latch. (Nov, 2007)

11. What is flip-flop?

Flip-Flops are synchronous bistable devices (has two outputs Q and Q’).

An edge-triggered Flip-Flop changes state either at the positive edge (rising edge)

or at the negative edge (falling edge) of the clock pulse and is sensitive to its

inputs only at this transition of the clock.

12. Differentiate Flip-flops from Latches. (May, 2010)

Latch is a sequential device that checks all of its inputs continuously and changes

its outputs according to any time, independent of a clocking signal.

Flip-flop is a sequential device that samples its inputs and changes its outputs

only at times determined by clocking signal.

13. List different types of flip-flops.

i) SR flip-flop

ii) JK flip-flop

iii) D flip-flop

iv) T flip-flop

14. What do you mean by triggering of flip-flop?

The state of a flip-flop is switched by a momentary change in the input signal.

This momentary change is called a trigger and the transition it causes is said to trigger the

flip-flop.



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15. Draw the diagram of a clocked SR flip-flop using four NAND gates.

(Nov, 2004)

16. Draw the logic diagram for clocked D Flip-Flop.

17. Draw the logic diagram for clocked JK Flip-Flop.

18. What is a characteristic table?

A characteristic table defines the logical property of the flip-flop by describing its

operation in tabular form.



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19. Draw the logic diagram for T Flip-Flop. (May, 2008)

20. Draw the logic diagram for Master-slave SR Flip-Flop.

21. What is race around condition? How can it be avoided? (May, 09; Nov, 09)

In a JK latch, when J and k are both high, then the output toggles continuously.

This condition is called a race around condition.



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Due to this, in the positive half cycle of the clock pulse (Enable), if J and K both

are HIGH, then the output toggles continuously.

To avoid this condition, an edge triggered or pulse triggered JK flip-flop is

created. In this flip-flop, the output changes only at the positive edge or a negative edge

of the clock.

22. Write down the characteristic table of SR flip flop.

Present State Inputs Next StateQn S R Qn+1

00001111

00110011

01010101

001x101x

23. Write down the characteristic table of JK flip flop. (Nov, 2008)

Present State Inputs Next StateQn J K Qn+1

00001111

00110011

01010101

00111010

24. Write down the characteristic table of D flip flop.

Present State Input Next StateQn D Qn+1

0011

0101

0101



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25. Write down the characteristic table of T flip flop.

Present State Input Next StateQn T Qn+1

0011

0101

0110

26. What is an excitation table?

During the design process we usually know the transition from present state to

next state and wish to find the flip-flop input conditions that will cause the required

transition. A table which lists the required inputs for a given chance of state is called an

excitation table.

27. Give the excitation table of a SR flip-flop.

Present State Next State InputsQn Qn+1 S R0011

0101

010x

x010

28. Give the excitation table of a JK flip-flop.

Present State Next State InputsQn Qn+1 J K0011

0101

01xx

xx10

29. Give the excitation table of a T flip-flop.

Present State Next State InputQn Qn+1 T0 0 00 1 11 0 11 1 0



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30. Give the excitation table of a D flip-flop.

Present State Next State InputQn Qn+1 D0 0 00 1 11 0 01 1 1

31. Derive the characteristic equation of a JK flip-flop. (May, 2003)

Characteristic Equation:

Qn+1= JQ’+ K’Q.

32. Derive the characteristic equation of a D flip-flop. (Nov, 2002, IT)

Characteristic Equation:

Qn+1= D.



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33. Derive the characteristic equation of a T flip-flop. (May, 2004, IT)

Characteristic equation:

Qn+1= TQn’+ T’Qn

34. What is the difference between truth table and excitation table?

i) An excitation table is a table that lists the required inputs for a given change of

state.

ii) A truth table is a table indicating the output of a logic circuit for various input

states.

35. How will you convert a SR flip-flop into D flip-flop?

Input Present state Next state Flip-Flop InputsD Qn Qn+1 S R0011

0101

0011

001x

x100

SR to D Flip-Flop



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36. How will you convert a SR flip-flop into JK flip-flop?

Inputs Present state Next state Flip-Flop InputsJ K Qn Qn+1 S R00001111

00110011

01010101

01001110

0x001x10

x0x10001

SR to JK Flip-Flop

37. How will you convert a SR flip-flop into T flip-flop?

Input Present state Next state Flip-Flop InputsT Qn Qn+1 S R0011

0101

0110

0x10

x001



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SR to T Flip-Flop

38. How will you convert a JK flip-flop into T flip-flop?

Input Present state Next state Flip-Flop InputsT Qn Qn+1 J K0011

0101

0110

0x1x

x0x1

JK to T Flip-Flop

39. How will you convert a JK flip-flop into D flip-flop?

Input Present state Next state Flip-Flop InputsD Qn Qn+1 J K0011

0101

0011

0x1x

x1x0



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JK to D Flip-Flop

40. How will you convert a D flip-flop into T flip-flop?

Input Present state Next state Flip-Flop InputT Qn Qn+1 D0011

0101

0110

0110

D to T Flip-Flop

41. How will you convert a T flip-flop into D flip-flop?

Input Present state Next state Flip-Flop Input

D Qn Qn+1 T0011

0101

0011

0110



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T to D Flip-Flop

42. What are the models used to represent clocked sequential circuits? (Nov, 2006)

The clocked sequential circuits are represented by two models as

1. Moore circuit,

2. Mealy circuit.

43. What is the difference between a Mealy machine and Moore machine?

(Nov, 2008; May, 2005)

S.No Moore model Mealy model

1 Its output is a function of present

state only.

Its output is a function of present state as

well as present input.

2 An input change does not affect the

output.

An input change may affect the output of

the circuit

3 It requires more number of states for

implementing same function.

It requires less number of states for

implementing same function.

44. Define state of sequential circuit?

The binary information stored in the memory elements at any given time defines

the “state” of sequential circuits.

45. Define state diagram.

A graphical representation of a state table is called a state diagram.



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46. What is the use of state diagram?

i) Behavior of a state machine can be analyzed rapidly.

ii) It can be used to design a machine from a set of specification.

47. What is state table?

A table, which consists time sequence of inputs, outputs and flip-flop states, is

called state table. Generally it consists of three section present state, next state and

output.

48. A reduced state table has 14 rows. What is the minimum number of flip-flops

needed to build the sequential circuit? (Nov, 2004)

24 ≥ 14.

Therefore, 4 flip-flops.

49. What is a state equation?

A state equation also called, as an application equation is an algebraic expression

that specifies the condition for a flip-flop state transition. The left side of the equation

denotes the next state of the flip-flop and the right side; a Boolean function specifies the

present state.

Counters

50. What is counter?

A counter is used to count pulse and give the output in binary form.

51. What is synchronous counter?

In a synchronous counter, the clock pulse is applied simultaneously to all flip-

flops. The output of the flip-flops changes state at the same instant. The speed of

operation is high compared to an asynchronous counter.



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52. What is an Asynchronous/ ripple counter?

In an Asynchronous counter, the clock pulse is applied to the first flip-flops. The

change of state in the output of this flip-flop serves as a clock pulse to the next flip-flop

and so on. Here all the flip-flops do not change state at the same instant and hence speed

is less.

53. What is the difference between synchronous and asynchronous counter?

S.No Synchronous counter Asynchronous counter

1

Clock pulse is applied

simultaneously

Clock pulse is applied to the first flip-flop;

the change of output is given as clock to

next flip-flop.

2All the flip-flops are clocked

simultaneously.

All the flip-flops are not clocked

simultaneously.

3Design involves complex logic

circuit as number of state increases

Logic circuit is very simple even for more

number of states.

4 Speed of operation is high Speed of operation is low.

54. What is up counter?

A counter that increments the output by one binary number each time a clock

pulse is applied.

55. What is down counter?

A counter that decrements the output by one binary number each time a clock

pulse is applied.

56. What is up/down counter?

A counter, which is capable of operating as an up counter or down counter,

depending on a control lead (Up/ down).

57. What is a ripple counter?

A ripple counter is nothing but an asynchronous counter, in which the output of

the flip-flop changes state like a ripple in water.



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58. What are the uses of a counter?

1. The digital clock

2. Auto parking control

3. Parallel to serial data conversion.

59. What is meant by modulus of a counter?

By the term modulus of a counter we say it is the number of states through which

a counter can progress.

60. What is meant by natural count of a counter?

By the term natural count of a counter we say that the maximum number of states

through which a counter can progress.

61. A ripple counter is a ------------ sequential counter.

Ans: Asynchronous.

62. What is modulo-N counter? (May, 2008)

A modulo-N counter is a counter of that goes through a repeated sequence of N

counts.

63. The number of flip-flops required for modulo-18 counter is -------

2n ≥ N

25 ≥ 18

Therefore, five flip-flops are required.

64. What is the minimum number of flip-flops needed to build a counter of

modulus- 8? (May, 2004)

2n ≥ N

23 ≥ 8

Therefore, three flip-flops are required.



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65. How many flip-flops are required for designing synchronous MOD50 counter?

(May, 2009)

2n ≥ N

26 ≥ 8

Therefore, six flip-flops are required.

66. Form the truth table for 3-bit binary down counter.

Clk Q2 Q1 Q0

1

1

1

1

1

1

1

1

1

1

1

1

1

0

0

0

0

1

1

1

0

0

1

1

0

0

1

1

0

1

0

1

0

1

0

1

67. What is a ring counter?

A counter formed by circulating a ‘bit’ in a shift register whose serial output has been

connected to its serial input.

68. What is BCD counter?

A BCD counter counts in binary coded decimal from 0000 to 1001 and back to

0000. Because of the return to 0000 after a count of 1001, a BCD counter does not have a

regular pattern as in a straight binary counter.

69. What is a binary counter? (Nov, 2006)

A counter that follows the binary sequence is called binary counter. An n-bit

binary counter consists of n flip-flops and can count in binary from 0 to 2n-1.



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70. What are the uses of a ring counter?

i) Control section of a digital system.

ii) Controlling events, which occur in strict time sequence.

71. State the relative merits of series and parallel counters. (May, 2003)

In comparison with parallel counters the serial counters have simple logic circuits;

however, serial counters are low speed counters as the clock is propagated through

number is flip-flops before it reaches the last flip-flop.

72. What is Johnson counter?

It is a ring counter in which the inverted output is fed into the input. It is also

know as a twisted ring counter.

73. What is a cycle counter?

A cycle counter is a counter that outputs a stated number of counts and then stops.

Shift Register

74. What is a register?

Memory elements capable of storing one binary word. It consists of a group of

flip-flops, which store the binary information.

75. What is a shift register? (Nov, 2003)

A register capable of shifting its binary information in one or both directions is

called shift register. The logical configuration of a shift register consists of a chain of

flip-flops in cascade, with the output of one flip-flop connected to the input of the next

flip-flop. All flip-flops receive common clock pulses, which activate the shift from one

stage to the next.



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76. What is serial shifting?

In a shift register, if the data is moved 1 bit at a time in a serial fashion, then the

technique is called serial shifting.

77. What is parallel shifting?

In a shift register all the data are moved simultaneously and then the technique is

called parallel shifting.

78. How many flip-flops are needed to build an 8-bit register? (Nov, 2002)

8 -flops are needed to build an 8-bit register.

79. Write the uses of a shift register.

i) Temporary data storage

ii) Bit manipulations.

80. What are the applications of shift registers? (May, 2005)

1. A serial-in-serial-out shift register can be used to introduce time delay in digital

signals.

2. A serial-in-parallel-out shift register can be used to convert data in the serial form

to the parallel form.

3. A parallel-in-serial-out shift register can be used to convert data in the parallel

form to the serial form.

4. A shift register can also be used as a counter.

81. A shift register comprises of JK flip-flops. How will you complement the

contents of the register? (May, 2003)

In shift register outputs J and K of previous flip-flop is connected to the inputs of

the next flip-flop. If these lines are connected through OR gate, we can complement the

contents of flip-flop. When complement line is high all J and K inputs will be high and

flip-flops will complement the output.



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82. How many states are there in a 3-bit ring counter? What are they? (May, 2007)

The number of states in a 3-bit counter is three- 001, 010, 100.



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CHAPTER 5ASYNCHRONOUS SEQUENTIAL CIRCUITS

1. What are the problems involved in asynchronous circuits?

The asynchronous sequential circuits have three problems namely,

a. Cycles

b. Races

c. Hazards

2. Define cycles.

If an input change includes a feedback transition through more than unstable state

then such a situation is called a cycle.

3. What is meant by race? (May, 2004; Nov, 2003)

When two or more binary state variables change their value in response to a

change in an input variable, race condition occurs in asynchronous sequential circuits.

In case of unequal delays, a race condition may cause the state variables to change in an

unpredictable manner.

4. Define critical & non-critical race.

The final stable state that the circuit reaches does not depend on the

order in which the state variables change, the race is called non-critical race.

The final stable state that the circuit reaches depends on the order in which the

state variables change, the race is called critical race.

5. What is critical race? Why should it be avoided? (Nov, 2005)

Race exists in synchronous sequential circuits when two or more binary state

variables charge during a state transition.

A race becomes critical if the correct next value in not reached during a state

transition. For the proper operation of the circuits, the critical races must be avoided.



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6. What is meant by a non-critical race? What is its cause? (May, 2006)

A race condition is said to exist in an asynchronous sequential circuit when two or

more binary state variables changes value in response to a change in an input variable.

The order by which the state variables change may not be known in advance if the

final stable state that the circuit reaches does not depend on the order in which the state

variable change, the race is called a non-critical race.

7. How can a race be avoided?

Races can be avoided by directing the circuit through intermediate unstable states

with a unique state – variable change.

8. What is a hazard? (May, 04; May, 09)

Hazards are unwanted switching transients that may appear at the output of

a circuit because different paths exhibit different propagation delays.

9. What is a hazard in combinational circuits? (May, 2007; May, 2008)

The unwanted switching transients that may appear at the output of a circuit are

called hazards. The hazards cause the circuit to malfunction. The main cause of hazards is

the different propagation delays at different paths.

Hazards occur in the combinational circuits, where they may cause a temporary

false output value. When such combinational circuits are used in the asynchronous

sequential circuits, they may result in a transition to a wrong stable state.

10. What are the types of hazards?

The 3 types of hazards are

1. Static – 0 hazards

2. Static – 1 hazard

3. Dynamic hazards.

11. Define static 0-hazard, static 1-hazard and dynamic hazard. (May, 2005)

In a combinational circuit, if output goes momentarily 0 when it should remain a

1, the hazard is known as static-1 hazard.



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If the output goes momentarily 1 when it should remain a 0, the hazard is known

as static-0 hazard.

When the output changes three or more times when it should change from 1 to 0

or from 0 to 1 is known as dynamic hazard.

12. Does Hazard occur in sequential circuit? If so what is the problem caused?

Yes, Hazards occur in sequential circuit that is Asynchronous sequential circuit.

It may result in a transition to a wrong state.

13. Describe how to detect and eliminate hazards from an asynchronous network.

(May, 2005)

Hazards can be eliminated by enclosing two minterms or maxterms.

14. How can the hazards in combinational circuit be removed?

Hazards in the combinational circuits can be removed by covering any two

min terms that may produce a hazard with a product term common to both. The

removal of hazards requires the addition of redundant gates to the circuit

15. How does an essential hazard occur?

An essential hazard occurs due to unequal delays along two or more paths that

originate from the same input. An excessive delay through an inverter circuit in

comparison to the delay associated with the feedback path causes essential hazard.

16. What are the two types of asynchronous circuits? How do they differ?

(May, 2006)

Two types of asynchronous circuits are –

1. Fundamental mode circuit,

2. Pulse mode circuit.



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According to the characteristics of the input,

Fundamental mode circuit:

The input is allowed to charge after the steady state condition. Here the inputs are

levels and not pulses.

Pulse mode circuit:

Here the inputs are pulses. The pulses width must not be so long that it is still

present after the new state is reached.

Generally Fundamental mode circuit is preferred over pulse mode circuits because

it is very difficult to fix the pulse width.

17. What are the assumptions that must be made for fundamental mode circuit?

(May 2008)

1. The input variables change only when the circuit is stable.

2. Only one input variable can change at a given time.

3. Inputs are levels and not pulses.

18. What are the assumptions made for pulse mode circuit?

(Nov 2006; May, 2007; Nov, 2007)

1. The input variables are pulses instead of levels.

2. The width of the pulses is long enough for the circuit to respond to the input.

3. The pulse width must not be so long that it is still present after the new state is

reached.

19. What is meant by flow table?

During the design of asynchronous sequential circuits, it is more convenient to

name the states by letter symbols without making specific reference to their binary

values. Such a table is called a flow table.

20. Define primitive flow table?

A primitive flow table is a flow table with only one stable total state in each row.

Remember that a total state consists of the internal state combined with the input.



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21. Define merging?

The primitive flow table has only one stable state in each row. The table can be

reduced to a smaller numbers of rows if two or more stable states are placed in the same

row of the flow table. The grouping of stable states from separate rows into one common

row is called merging.

22. Give the procedural steps for determining the compatibles used for the

purpose of merging a flow table.

The purpose that must be applied in order to find a suitable group of compatibles

for the purpose of merging a flow table can be divided into 3 procedural steps.

i. Determine all compatible pairs by using the implication table.

ii. Find the maximal compatibles using a Merger diagram

iii. Find a minimal collection of compatibles that covers all the states and is closed.

23. What are the steps for the design of asynchronous sequential circuit? (Nov, 09)

1. Construction of a primitive flow table from the problem statement.

2. Primitive flow table is reduced by eliminating redundant states using the state

reduction.

3. State assignment is made.

4. The primitive flow table is realized using appropriate logic elements.

24. What is the advantage of debounce circuit? (Nov, 2008)

A debounce circuit is a circuit which removes the series of pulses that result from

a contact bounce and produces a single smooth transition of the binary signal from 0 to 1

or from 1 to 0.



2 Marks Q&A - Weeblyshanthirevathi.weebly.com/uploads/1/2/8/5/12854997/dpsd-2-marks.pdf · CHAPTER 1 BOOLEAN ALGEBRA AND LOGIC GATES Number Base Conversions : 1. List the different

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