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CS8351- DIGITAL PRINCIPLES AND SYSTEM DESIGN

UNIT I

Short Question and Answers

1. Define : system

It is defined by base or radix which gives the number of digits the system uses to represent the in the system Eg: Decimal Number system : Uses 0 to 9 – Total nos. required 10, Base of Decimal N.S is 10. 2. What is logic? It tells you a certain preposition (declarative statement) is true if certain conditions are true.

3. What is the use of Logic? Logic is applied to digital circuit to implement logic functions.

4. What is logic Gate? The logic gate is a digital circuit with one or more inputs and generates only one output.

5. What is and what’s the use of Truth Table? The of logic gate can be easily understood with the help of truth table. A truth table is a table that shows all the input and output possibilities of logic circuit (ie) it indicates the output for different possibilities of the inputs.

6. What are Universal Gates? Why it is called as Universal Gates? NAND and NOR gates are called Universal Gates since it is used to perform AND, OR, Inverter operations.

7. Gate operations OR: For 2 input OR gate, output is HIGH if anyone input either A or B is high. Output is Low if both A and B are LOW.

AND: For 2 input AND gate, the output is HIGH if both A and B are HIGH. output is LOW if anyone input either A or B is Low.

NOT: When the input is Low, the output is HIGH. When the input is high, the output is LOW.

NAND: For 2 input NAND gate, the output is LOW if inputs A and B are HIGH. Output is HIGH if either A or B is LOW, or both are LOW.

NOR: For 2 input NOR gate, the output is LOW if either A or B is HIGH or if both A and B are HIGH. Output is HIGH if both A and B are low.

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EX-OR: For 2 input EX-OR gat, the output is HIGH if anyone input is High. Output is LOW if both A and B are equal.

EX-NOR: For 2 input EX-NOR gat, the output is HIGH if both A and B are Equal. Output is LOW if anyone input is if anyone input is HIGH. 8. Gate and IC number

GATE IC NO OR ( 3 Inputs ) 7432 AND ( 3 Inputs ) 7408 NOT ( 3 Inputs ) 7404 NOR ( 3 Inputs ) 7402 NAND ( 3 Inputs ) 7400 EX OR ( 3 Inputs ) 7486 AND( 3 Inputs ) 7411

9. What is Boolean ? It is a convenient and systematic way of expressing and analyzing the operation of logic circuits.

10. Define Variable, Complement, and Literal. Variable: It is a used to represent a logical quantity. Complement: It is the inverse of a variable. Denoted by bar over the variable. Literal: It is a variable or the complement of a variable.

11. Demorgan’s Theorems: First theorem: The complement of a of variables is equal to the sum of the complements of the variables. Second theorem: The complement of a sum of variables is equal to the product of complements of the variables.

12. What is the use of standard forms of Boolean Expressions? It makes the evaluation, implication and implementation of Boolean expressions much more systematic and easier.

13. Define SOP and POS: SOP: Two or more product terms are summed by Boolean . (or) of product terms ORed together. POS: When two or more sum terms are multiplied the resulting Expression is a POS.

14. What is domain of a Boolean expression? It is the of variables in the expression in either complemented or uncomplemented form.

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15. What is standard or canonical SOP and std POS? A standard SOP is one in which all the variables in the domain should appear in each product term in the expression. A standard POS in one in which all the variables in the domain should appear in each sum term in the expression.

16. Define Minterm and Maxterm: Minterm: Each individual term in canonical SOP is called Minterm. Maxterm: Each individual term in canonical POS is called Maxterm.

17. Short note on Karnaugh Map: It is similar to truth table and it presents all of the possible values of input variables and the resulting output for each value. It’s an array of cells in which each cell represents a binary value of the input variables. It’s used to minimize the Boolean Expression.

18. Short note on don’t care condition or incompletely specified functions. In some logic circuits certain input conditions never occur, so the corresponding output never appears. In such cases, the output level is not defined, it can be either ‘0’ or ‘1’ and are indicated by ‘X’ or ‘d’ in the truth table and are called don’t care output or don’t care condition.

19. What is the largest decimal number that can be represented in binary with eight bits? 256-1 = 255

20. Determine the weight of the 1 in the binary number 10000. 16 (2 power of 4)

21. What is the use of 2’s complement numbers? 2’s complement is commonly used in computers to handle negative numbers.

22. Why we need alphanumeric codes and give an example? For better communication we need not only numbers, but also letters and other symbols. E.g.: ASCII – American Standard Code for Information Interchange.

23. What is the difference between Parity and Hamming code? Parity code: It is used only for detecting the error and also if error occurs more than one bit position it is invalid to detect the error. Hamming Code: It is used for detection and correction of error in digital transmission.

24. CODES

Weighted Code Ex: Binary and BCD

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UnweightedCode Ex: Gray code and Excess – 3 code.

Reflective code Ex: Excess 3, 8421 code. Sequential Code Ex: Excess 3 and 8421 Code.

Alphanumeric Code: Ex: ASCII Code.

Error Detecting Code Ex: Parity code and Biquinary code.

Error Correction Code Ex: Hamming Code.

25. What are basic properties of Boolean algebra? The basic properties of Boolean algebra are commutative property, and .

26. 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

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

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

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

30. Reduce A (A + B) A (A + B) = AA + AB = A (1 + B) [1 + B = 1] = A. 31. Reduce A'B'C' + A'BC' + A'BC A'B'C' + A'BC' + A'BC = A'C'(B' + B) + A'B'C

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= A'C' + A'BC [A + A' = 1] = A'(C' + BC) = A'(C' + B) [A + A'B = A + B] 32. 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]

33. 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'

34. 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]

35. 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

36.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]

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37.Define duality property. Duality property states that every deducible from the postulates Of Boolean algebra remains valid if the operators and elements are interchanged. If the dual of an algebraic expression is desired, we simply interchange OR and AND operators and replace 1's by 0's and 0's by 1's.

38.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')

39. 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

40.What are the methods adopted to reduce Boolean ? i) Karnaug map ii) Tabular method or QuineMc-Cluskey method iii) Variable entered map technique.

41.State the limitations of karnaugh map. i) Generally it is limited to six variable map (i.e) more then six variable involving expression are not reduced. ii) The map method is restricted in its capability since they are useful forsimplifying only Boolean expression represented in standard form.

42. 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 unction.

43. 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)

44. 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.

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45. State the postulates and theorems of Boolean algebra.

X + 0 = X X · 1 = X X + X' = 1 X · X' = 0 X + X = X X · X = X X + 1 = 1 X · 0 = 0 (X')' = X X + Y = Y + X XY = YX X + (Y + Z) = (X + Y) + Z X (YZ) = (XY) Z X(Y + Z) = XY + XZ X + YX = (X + Y) (X + Z) (X + Y)' = X'Y' (XY)' = X' + Y' X + XY = X X(X + Y) = X

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. 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]

48. 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]

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