DEPARTMENT OF COMPUTER SCIENCE AND ENGINEERING
Subject Name: Digital design and system architecture
Subject Code: U23ITTC01
Prepared by:
Mr.B.THIYAGARAJAN / AP /CSE
Ms.V.SWATHILAKSHMI / AP /CSE
Verified by:
Approved by:
UNIT I
REVIEW OF NUMBER SYSTEMS
Review of Number systems – Conversion of Number systems – Binary addition and subtractions – Binary
representation: Signed magnitude representation and Compliment representations – Binary codes – Boolean
Algebra – Boolean functions – Canonical forms – Simplifications of Boolean function: Theorems and laws,
Karnaugh Map and Quine McCluskey method.
Two Marks
1. Convert (101.01)2 to decimal number.
101.01 = [(1 × 22 ) + (0 × 21 ) + (1 × 20 )]. [(0 × 2−1 ) + (1 × 2−2 )]
= 5.25
(101.01)2 = 5.2510
2. Determine (377)10 in Octal and Hexa-Decimal equivalent.
3. Convert [115]10 and [235]10 to hexadecimal numbers.
4. Convert:
a) (475.25)8 to its decimal equivalent.
b) (549.B4)16 to its binary equivalent.
a). (475.25)8
(475.25)8 = [(4 × 82 ) + (7 × 81 ) + (5 × 80 )]. [(2 × 8−1 ) + (5 × 8−2 )]
= 317.328125
(475.25)8 = [317.328125]10
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b). (549.B4)16
(549.B4)16 = [10101001001.10110100]2
5. Convert each of the following decimal numbers to excess – 3 code. i). [18]10 ii). [56]10
i) [18]10
[18]10= (0100 1011) XS-3
ii) [56]10
[56]10= (1000 1001) XS-3
6. Convert 14310 into its binary and binary coded decimal equivalent.
Decimal to binary:
[143]10 = [10001111]2
Decimal to BCD:
[143]10 = [0001
0100
SMVEC/CSE/U23ITIC01/DDSA/UNIT 1
0011] BCD
2
7. Give the classification of binary codes.
8. What are weighted codes? Give example.
• In weighted codes, each digit is assigned a specific weight according to its position.
• In 8421 BCD code -1001 the weights of 1, 0, 0, 1 (from left to right) are 8, 4, 2 and 1 respectively.
• The codes 8421 BCD, 2421 BCD, 5211 BCD are all weighted codes.
9. What are non-weighted codes? Give example.
• The non-weighted codes are not positional weighted.
• Each digit position within the number is not assigned a fixed value (or weight).
• Excess-3 and gray code are non-weighted codes.
10. What is a unit distance code? Give an example.
• Unit distance code is a non – weighted code in which next increment or decrement causes the bit –
transition only at one place.
• Example: Gray Code.
11. What are reflective codes? Give example.
• A code is reflective when the code is self-complementing.
• In other words, when the code for 9 is the complement of 0, 8 for 1, 7 for 2, 6 for 3 and 5 for 4.
• 2421BCD, 5421BCD, 5211BCD and Excess-3 code are reflective codes.
12. What are self-Complement Codes?
•
Self-complement codes are whose arithmetic and logic complements are same.
•
Example: 2421, 631-1.
13. What are sequential codes? Give example.
In sequential codes, each succeeding code is one binary number greater than its preceding code.
This property helps in manipulation of data. 8421 BCD and Excess-3 are sequential codes.
14. What are alphanumeric codes? Give example.
Codes used to represent numbers, alphabetic characters, symbols and various instructions necessary for
conveying intelligible information. ASCII, EBCDIC, UNICODE are the most-commonly used alphanumeric
codes.
15. What are Error detecting and correcting codes? Give example.
•
Codes which allow error detection and correction are called error detecting and correcting codes.
•
Hamming code is the mostly commonly used error detecting and correcting code.
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16.
What is Error detection and Error correction?
Error detection is the ability to detect errors. Error correction has an additional feature that enables
identification and correction of the errors. Error detection always precedes error correction. Both can be
achieved by having extra/redundant/check bits in addition to data deduce that there is an error.
17.
Give each one example for error detecting code and error correcting code.
• Error detecting Codes :
• It is used to detect the errors present in the received data (bit stream).
• Example – Parity code and hamming code.
• Error Correcting Codes:
• It is used to correct the errors present in the receiving data (bit stream).
• Example – Hamming code.
18.
What is Excess-3 code?
•
Excess-3 [XS3], is a non-weighted code used to express decimal numbers.
•
It is particularly significant for arithmetic operations as it overcomes the shortcomings
encountered while using the 8421 BCD code to add two decimal digits whose sum exceeds 9.
•
This code is used in some old computers.
19.
What is Gray code or cyclic code? [OR] What is a gray code and mention its advantages.
Gray code is a type of unit distance code. In this code bit patterns for two consecutive numbers differ in
only one bit position. This property makes it very useful in Karnaugh Map [K-Map].
Advantages:
In Gray code, if we go from one decimal number to next, only one bit of the gray code changes.
Because of this feature, an amount of switching is minimized and the reliability of the switching systems is
improved.
20.
What is Binary Coded Decimal (BCD)?
The binary coded decimal (BCD) is a type of binary code used to represent a given decimal number in an
equivalent binary form.The most common BCD code is the 8421 BCD code. 8, 4, 2 and 1 represent the
weights of different bits in the four-bit groups, starting from the (MSB) most significant bit (to extreme
left) and proceeding towards the least significant (LSB) bit.
Advantage:
•
It allows easy conversion to decimal digits for printing or display and faster calculations.
21. Convert binary number 10110 to gray code.
The gray code is 11101
22. Convert gray code word 11011 to binary.
The binary number is 100102
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23. Convert the following binary code into a Gray code, 10101110002.
The gray code is 1111100100.
24. Give the binary, BCD, Excess–3, gray code representation of decimal numbers: 5, 8, 14.
Decimal Numbers
Binary Code
BCD Code
Excess – 3
Gray code
5
0101
0000 0101
1000
0111
8
1000
0000 1000
1011
1100
14
1110
0001 0100
0100 0111
1001
25. Convert gray number 1110 to its BCD equivalent
The it’s equivalent Binary number is [1011]2
BCD:
[1011]2 = (1 × 23 ) + (0 × 22 ) + (1 × 21 ) + (1 × 20 ) = 8 + 0 + 2 + 1
[1011]2 = [11]10
[11]10 = [0001 0001] BCD
26. Convert the following Excess-3 numbers into decimal numbers.
(a) [1011]XS-3
Step 1: Convert excess -3 into binary
Step 2: Binary to decimal
[1000]2 = (1 × 23 ) + (0 × 22 ) + (0 × 21 ) + (0 × 20 ) = 8 + 0 + 0 + 0
[1011]2 = [8]10
(b) [1001 0011 0111] XS-3
Step 1: Convert excess -3 into binary
Step 2: Binary to decimal
[011000000100]2 = (0 × 211 ) + (1 × 210 ) + (1 × 29 ) + (0 × 28 ) +
(0 × 27 ) + (0 × 26 ) + (0 × 25 ) + (0 × 24 ) +
(0 × 23 ) + (1 × 22 ) + (0 × 21 ) + (0 × 20 )
[1001 0011 0111] XS-3= [1540]10
27. What is a Parity code?
Some of the codes are used for detecting errors while transferring binary data from one digital system
to other. The data transferred from transmitter and the data received by the receiver should match.
Parity is counting the number of ones in a binary data before the transmission.
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28. What is even parity?
• An error-checking system that requires a binary number to have an even number of 1’s.
• Generating even parity bit is just an XOR function.
• Example:
000 1001 →0 correct
000 0001 →1 incorrect
29. What is odd parity?
• An error-checking system that requires a binary number to have an odd number of 1’s.
• Odd parity is generated using a XNOR function.
• Example:
000 0001 →0 correct
000 0011 →1 incorrect
30. What is hamming code?
Sender adds the parity bit to existing data bits before transmission and the Receiver checks for the
expected parity. If wrong parity is found, the received data is discarded and re-transmission is
requested.Hamming code is a code to detect error and also to correct it.It detects a single bit error and
also identifies the bit that is in error.
32. What is Signed-magnitude Representation?
In signed numbers it is necessary to represent negative as well as positive numbers. In sign-magnitude
representation the most significant bit (leftmost bit) is used to represent sign of the number. If the most
significant bit is 0, the number is positive, and if the most significant bit is 1, the number is negative. The
remaining bits of the number represent magnitude.
Sign magnitude format
Here are some examples of sign-magnitude numbers.
+6
=
0 000
0110
- 14
=
1 000
1110
+ 24
=
0 001
1000
- 64
=
1 100
0000
33. What are the Binary system complements?
In digital computers, to simplify the subtraction operation and for logical manipulation complements are
used. Two types of complements:
• 1’s complement
• 2’s complement
34.What is 1's complement?
The 1's complement of a number is found by changing all 1's to 0's and all 0's to 1's. This is called as
taking complement or 1's complement. Example of 1's Complement is as follows.
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35.What is 2's complement?
The 2's complement of binary number is obtained by adding 1 to the Least Significant Bit (LSB) of 1's
complement of the number.
2's complement = 1's complement + 1
Example of 2's Complement is as follows.
36. State the absorption law of Boolean algebra.
The absorption law of Boolean algebra is given by
X+XY=X and X (1+Y) = X
37. Explain consensus theorem in Boolean algebra.
In simplification of Boolean expression the redundant term in an expression can be eliminated to form
the equivalent expression.
𝐴𝐵 + 𝐴̅ 𝐶 + 𝐵𝐶 = 𝐴𝐵 + 𝐴̅ 𝐶
38. State the associative property of Boolean algebra. (Oct 22)
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.
A+ (B+C) = (A+B) +C
39. State De Morgan's theorem. (oct 22,aug 22)
De Morgan suggested two theorems that form important part of Boolean algebra. They are,
a. The complement of a product is equal to the sum of the complements.
̅̅̅̅̅̅̅
[𝐴. 𝐵] = 𝐴̅ + 𝐵̅
b. The complement of a sum term is equal to the product of the complements.
̅̅̅̅̅̅̅̅̅̅
[𝐴 + 𝐵] = 𝐴̅. 𝐵̅
40. Apply De Morgan’s theorem to the following expression (A+B+C+D).
̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅
(𝐴 + 𝐵 + 𝐶 + 𝐷)
̅. 𝑩
̅. 𝑫
̅. 𝑪
̅
= 𝑨
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41. What are the applications of De Morgan’s Theorem?
• Convert max term to min term or min term to max term form.
• Examine two different logic diagrams that perform same logic function.
42. Simplify the following using De Morgan's theorem [((AB)'C)'' D]'
̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅
̅̅̅̅̅̅̅̅̅
̿̿̿̿̿̿̿̿̿
̿̿̿̿̿̿̿̿̿
̅̅̅̅̅̅̅
̅̅̅̅ )𝐶 ) 𝐷] = ((𝐴𝐵
̅̅̅̅ )𝐶 ) + 𝐷
̅
[((𝐴𝐵
∵ [[𝐴.
𝐵] = 𝐴̅ + 𝐵̅
̅
= ((𝐴𝐵)𝐶) + 𝐷
̅
= ̿̿̿̿
𝐴𝐵 + 𝐶̅ + 𝐷
̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅
̿̿̿̿̿̿̿̿̿
̅̅̅̅ )𝑪) 𝑫] = 𝑨𝑩 + 𝑪
̅+ 𝑫
̅
[((𝑨𝑩
∵ [𝐴̅̿] = 𝐴̅
∵ 𝐴̿ = 𝐴
43. What are the methods adopted to reduce Boolean function?
• Algebraic Method using Boolean Postulates and laws
• Karnaugh map
• Tabulation method or Quine Mc-Cluskey method
44. Simplify A+AB+A’+B
𝐴 + 𝐴𝐵 + 𝐴̅ + 𝐵 = 𝐴(1 + 𝐵) + 𝐴̅ + 𝐵
= (𝐴. 1) + 𝐴̅ + 𝐵
= 𝐴 + 𝐴̅ + 𝐵
=1 +𝐵
45. Reduce A.A'C
𝐴. 𝐴̅ 𝐶 = 0. 𝐶
̅𝑪=𝟎
𝑨. 𝑨
46. Reduce A (A + B) [Apr’18]
𝐴(𝐴 + 𝐵) = 𝐴. 𝐴 + 𝐴. 𝐵
= 𝐴 + 𝐴. 𝐵
= 𝐴(1 + 𝐵)
= 𝐴. 1
𝑨(𝑨 + 𝑩) = 𝑨
∵ [(1 + 𝐴) = 1]
∵ [(𝐴. 1) = 𝐴]
∵ [(𝐴 + 𝐴̅) = 1]
̅ +𝑩=𝟏
𝑨 + 𝑨𝑩 + 𝑨
∵ [(𝐴. 𝐴̅) = 0]
∵ [(𝐴. 0) = 0]
∵ [(𝐴. 𝐴) = 𝐴]
∵ [(1 + 𝐴) = 1]
∵ [(𝐴. 1) = 𝐴]
47. Reduce 𝑨 𝑩 𝑪 + 𝑨 𝑩 𝑪 + 𝑨 𝑩𝑪
𝐴 𝐵 𝐶 + A B C + A BC = 𝐴 𝐶(B + B) + A BC
= [𝐴 𝐶]. 1 + ABC
= 𝐴 𝐶 + ABC
= A[C + BC]
𝑨 𝑩 𝑪 + 𝐀 𝐁 𝐂 + 𝐀 𝐁𝐂 = 𝐀[𝐂 + 𝐁]
∵ [(𝐴 + 𝐴̅) = 1]
∵ [(𝐴. 1) = 𝐴]
48. Reduce a (b + bc’)+ ab’.
𝑎(𝑏 + 𝑏 𝑐) + a b = 𝑎𝑏 + 𝑎𝑏 𝑐 + a b
= 𝑎𝑏[1 + 𝑐 ] + a b
= [𝑎𝑏 . 1 ] + a b
= 𝑎𝑏 + a b
= 𝑎(𝑏 + b)
= 𝑎(1)
𝒂(𝒃 + 𝒃 𝒄) + 𝐚 𝐛 = 𝒂
∵ [(1 + A) = 1]
∵ [(𝐴. 1) = 𝐴]
SMVEC/CSE/U23ITIC01/DDSA/UNIT 1
∵ [(A + AB) = (A + B)]
∵ [(𝐴 + 𝐴̅) = 1]
8
49. Find the result of A + A’D + AC’.
̅ D + AC̅ = A + AC̅ + A
̅D
A+ A
̅
̅D
= A [1 + C] + A
̅
= A [1] + AD
̅D
= A + A
̅
𝐀 + 𝐀𝐃 + 𝐀𝐂̅ = 𝐀 + 𝐃
∵ [(1 + A) = 1]
∵ [(1. A) = A]
̅ B] = [A + B]
∵ [A + A
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 = ̅̅̅̅̅̅̅̅̅̅̅̅
x̅yz̅ + x̅y̅z
= ̅̅̅̅̅̅̅
(x̅yz̅) ̅̅̅̅̅̅̅
(x̅y̅z)
= (x̿ + y̅ + z̿) (x̿ + y̿ + z̅)
̅̅̅
𝐅𝟏 = (𝐱 + 𝐲̅ + 𝐳) (𝐱 + 𝐲 + 𝐳̅)
̅̅̅
F2 = ̅̅̅̅̅̅̅̅̅̅̅̅̅̅
x(y̅z̅ + yz)
= x̅ + ̅̅̅̅̅̅̅̅̅̅̅̅̅
(y̅z̅ + yz)
(y̅z̅). ̅̅̅̅̅̅
(yz)
= x̅ + ̅̅̅̅̅̅
= x̅ + (y̿ + z̿). ( ̅y + z̅)
= x̅ + (y + z). ( ̅y + z̅)
= x̅ + y ̅y + y z̅ + ̅z
y + zz̅
= x̅ + 0 + y z̅ + ̅z
y + 0
= x̅ + y z̅ + ̅z
y
̅̅̅
𝐅𝟐 = 𝐱̅ + [𝐲 ⨁𝐳]
∵ [(𝐴. 𝐴̅) = 0]
∵ [𝐴 + 0 = 𝐴]
̅B ]
̅+A
∵ [A ⨁B] = [A B
51. Simplify the following expression Y = (A+B) (A+C) (B+C)
(A + B)(A + C)(B + C) = (AA + AC + AB + BC)(B + C)
= (A + AC + AB + BC)(B + C)
∵ [A. A] = A
= AB + ABC + ABB + BBC + AC + ACC + ABC + BCC
= AB + ABC + AB + BC + AC + AC + ABC + BC
∵ [A. A] = A
= AB + ABC + BC + AC
∵ [A + A] = A
= AB + BC [A + 1] + AC
= AB + BC [1] + AC
∵ [A + 1] = 1
(𝐀 + 𝐁)(𝐀 + 𝐂)(𝐁 + 𝐂) = 𝐀𝐁 + 𝐁𝐂 + 𝐀𝐂
∵ [A. 1] = A
52. If A and B are the Boolean variables & if A=1 & (A+B)’=0. Find B.
̅̅̅̅̅̅̅̅̅̅
[A + B] = ̅̅̅
A . ̅B
̅̅̅
when A = 1,
A =0
̅̅̅̅̅̅̅̅̅̅
[A + B] = 0. ̅B
̅̅̅̅̅̅̅̅̅̅
[A + B] = 0
∵ [(0. 𝐴̅) = 0]
∵ 𝐁 = 𝟏 𝐨𝐫 𝟎
53. Simplify the expression Z = AB + AB’. (A’C’)’.
̅ C̅)
̅̅̅̅̅̅̅̅
(A
Z = AB + AB
̅ [ 𝐴̿ + 𝐶̿ ]
= AB + AB
̅[𝐴+ C]
= AB + AB
̅ + AB
̅C
= AB + 𝐴AB
̅
̅
= AB + AB + ABC
̅ [1 + C]
= AB + AB
̅ [1]
= AB + AB
̅
= AB + AB
̅
= A [B + B]
SMVEC/CSE/U23ITIC01/DDSA/UNIT 1
∵ 𝐵𝑦 𝐷𝑒 𝑀𝑜𝑟𝑔𝑎𝑛′𝑠 𝑙𝑎𝑤
∵ [𝐴̿ = 𝐴]
∵ [𝐴. 𝐴 = 𝐴]
∵ [A + 1] = 1
∵ [A. 1] = A
9
= A [1]
𝐙=𝑨
54. Show that a) A+A’B=A+B and b) X’Y’Z+X’YZ+XY’ = X’Z+XY’
̅ B = A + AB + A
̅B
a).
A+ A
̅]
= A + B [A + A
= A + B [1]
̅𝐁=𝐀+𝐁
𝐀+ 𝐀
b).
̅
̅= ̅
̅
X̅
YZ+ ̅
X YZ + XY
X Z[̅
Y + Y] + XY
̅ Z [ 1] + XY
̅
= X
̅𝐘
̅𝐙+ 𝐗
̅ 𝐘𝐙 + 𝐗𝐘
̅= 𝐗
̅ 𝐙 + 𝐗𝐘
̅
𝐗
∵ [(𝐴 + 𝐴̅) = 1]
∵ [A + AB = 𝐴]
∵ [(𝐴 + 𝐴̅) = 1]
∵ [[A. 1] = A]
∵ [(𝐴 + 𝐴̅) = 1]
∵ [[A. 1] = A]
55. Define Minterm and Maxterm .(Dec 22)
Minterm:
• The products of Boolean expression where all possible variables appearance incomplement to runcomplement variables.
̅B
• Example ∶ AB + A
Maxterm
• A sum terms in a Boolean expression where all possible variables appearance, incomplement or
un-complement form
̅ + B)
• Example ∶ (A + B)( A
56. What is meant by karnaugh map or K-Map method?
• A Karnaugh map [k-map] is a Pictorial form of truthtable, in which the map diagram is made up of
cells, with each cell representing one minterm or maxterm of the function.
• It provides a simple straight forward procedure for minimizing Boolean function.
57. What are don’t care conditions? How does it help K-map for circuit simplification?
• 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
58. What is tabulation method?
• A method involving an exhaustive tabular search method for the minimum expression to solve a
Boolean equation for more variables is called as a tabulation method.
59. 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.
• They cannot be reduced further.
• A prime implicant is a group of minterms which cannot be combined with anyother minterm or
groups.
60. What is an essential prime implicant? (Dec 22)
• The Essential Prime Implicant is a prime implicant in which one or more minterms are unique.
• It contains atleast one minterm which is not contained in anyother prime implicant.
61. Define Duality Theorem. [OR] Define Duality Property.
• The Duality theorem states that starting with a Boolean relation we can derive another Boolean
relation by:
i) Changing OR (operation) i.e., + (Plus) sign to an AND (operation) i.e., · (dot) and Vice-versa.
ii) Complementary 0 or 1 appearing in the expression i.e. replacing contains 0 and 1 by 1 and 0
respectively.
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62. Convert the given expression in canonical SOP form Y = AC + AB + BC.
𝑌 = AC + AB + BC
̅ ]BC
̅]C + AB[C + C̅] + [A + A
= A[B + B
̅ BC
̅ C + ABC + ABC̅ + ABC + A
= ABC + AB
̅
̅
̅
𝒀 (𝑨, 𝑩, 𝑪) = 𝐀𝐁𝐂 + 𝐀𝐁𝐂 + 𝐀𝐁𝐂 + 𝐀𝐁𝐂
63. Convert the given expression in canonical SOP form Y =AB+A’C+ BC’.
̅ C + BC̅
𝑌 = AB + A
̅ [B + B
̅ ]BC̅
̅]C + [A + A
= AB [C + C̅] + A
̅ BC + A
̅B
̅ BC̅
̅C + ABC̅ + A
= ABC + ABC̅ + A
̅
̅
̅
̅
̅
̅
𝒀 (𝑨, 𝑩, 𝑪) = 𝐀𝐁𝐂 + 𝐀𝐁𝐂 + 𝐀𝐁𝐂 + 𝐀𝐁𝐂 + 𝐀𝐁𝐂
64. Express F=A+BC’ as sum of minterms.
𝐹 = A + BC̅
̅ ] BC̅
̅][C + C̅] + [A + A
= A[B + B
̅ BC̅
̅]] + ABC̅ + A
= [AB + AB
̅
̅ BC̅
̅
̅C̅ + ABC̅ + A
= ABC + ABC + ABC + AB
̅
̅
̅
̅
̅
̅
= ABC + ABC + ABC + ABC + ABC
= 111 + 110 + 101 + 100 + 010
= m7 + m6 + m5 + m4 + m2
𝑭 (𝑨, 𝑩, 𝑪) = 𝚺𝒎(𝟐, 𝟒, 𝟓, 𝟔, 𝟕)
65. Given F = B’ + A’B +A’C’ identify the redundant term using K-Map.
̅B + A
̅ C̅
̅ + A
𝐹= B
̅
̅ B[C + C̅ ] + A
̅ [B + B
̅
̅ ]C̅
= [A + A ]B [C + C̅ ] + A
̅
̅
̅
̅
̅
̅
̅
̅B
̅
̅
̅ C̅
= [AB + AB ][C + C ] + ABC + ABC + ABC + A
̅B
̅B
̅ BC + A
̅ BC̅ + A
̅ BC̅ + A
̅B
̅C + AB
̅C̅ + A
̅C+A
̅ C̅ + A
̅ C̅
= AB
̅
̅
̅
̅
̅
̅
̅
̅
̅
̅
̅
= ABC + ABC + AB C + AB C + ABC + ABC
= 101 + 100 + 001 + 000 + 011 + 010
= 𝑚5 + m4 + m1 + m0 + m3 + m2
𝐹 (𝐴, 𝐵, 𝐶) = Σ𝑚(0, 1, 2, 3, 4, 5)
K-Map:
66. Draw the logic diagram for X = AB+BC
67. Write the Boolean expression for the output of the system shown.
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C = (A + B) . B
C = [A. B]B
𝐂=𝟎
68. Write the POS representation of the following SOP function F(x, y, z) =∑m (0, 1, 3, 5, 7).
Let the SOP functions will ∑m (0, 1, 3, 5, 7)
Now taking left out values πM (2, 4, 6)
𝐹 (𝑥, 𝑦, 𝑧) = 𝜋𝑀(2, 4, 6)
= (1 + 0 + 1)(0 + 1 + 1)(0 + 0 + 1)
𝑭 (𝒙, 𝒚, 𝒛) = (𝒙 + 𝒚̅ + 𝒛)(𝒙̅ + 𝒚 + 𝒛)(𝒙̅ + 𝒚̅ + 𝒛)
̅𝒚𝒛 + 𝒙𝒚𝒛̅ + 𝒙𝒚
̅𝒛
69.Write the POS representation of the following SOP function𝑭(𝒙, 𝒚, 𝒛) = 𝒙
𝐹(𝑥, 𝑦, 𝑧) = 𝑥̅ 𝑦𝑧 + 𝑥𝑦𝑧̅ + 𝑥𝑦̅𝑧
= 011 + 110 + 101
= 𝑚3 + 𝑚6 + 𝑚5
𝐹(𝑥, 𝑦, 𝑧) = ∑m (3, 6, 5)
Let the remaining term for POS,
𝐹(𝑥, 𝑦, 𝑧) = 𝜋M (0, 1, 2, 4, 7)
= (1 + 1 + 1) (1 + 1 + 0) (1 + 0 + 1) (0 + 1 + 1) (0 + 0 + 0)
̅ + 𝒛)(𝒙
̅ + 𝒚 + 𝒛)(𝒙
̅+𝒚
̅ + 𝒛̅)
𝑭(𝒙, 𝒚, 𝒛) = (𝒙 + 𝒚 + 𝒛)(𝒙 + 𝒚 + 𝒛̅)(𝒙 + 𝒚
69. find the 2s complement of 1101 and 1001 (May 22)
(a) 1101
(b) 1001
1s comp-0010
2s comp-0011
1s comp-0110
2s comp -0111
70.State Commutative law (may 22)
A binary operator * on a set S is said to be commutative if,
for all x, y Є S
x*y=y*x
Eg:
0+ 1 = 1+ 0 = 1
a) x+ y= y+ x
0.1=1.0 =0
b) x. y= y. x
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5 Marks and 10 Marks
1. NUMBER SYSTEM:
• A numeral system (or system of numeration) is a writing system for expressing numbers, that
is, a mathematical notation for representing numbers of a given set, using digits or other
symbols in a consistent manner.
• The values of each digit are determined by the digit, the position of the digit in the number and
the base or radix of the number system.
• The number system will be of 4 types.
• Decimal
• Binary
• Octal
• Hexadecimal
Decimal:
• Decimal is said to be Base 10.
• When we count in Base Ten, we count starting with zero and going to nine in the order.
• It is a human language one.
0, 1, 2, 3, 4, 5, 6, 7, 8, 9
• Each additional digit represents a higher and higher quantity.
• This is applicable for base 10 as well as to other bases.
Binary:
• In a Binary number system, there are only two symbols used to represent numbers: 0 and 1.
• It is a circuit language one.
0, 1
Applications:
• The binary system is useful in computer science and electrical engineering.
• Transistors operate from the binary system and transistors are found in practically all electronic
devices.
Octal:
• Octal is Base Eight meaning eight symbols are used to represent all the quantities.
• When we count up one from the 7, we need a new placement to represent what we call 8 since
an 8 does not exist in Octal. So after 7 is 10.
• It is a machine language one.
0, 1, 2, 3, 4, 5, 6, 7
Hexadecimal:
• The hexadecimal system is Base Sixteen; this number system uses sixteen symbols to
represent numbers.
• Unlike binary and octal, hexadecimal has six additional symbols that it uses beyond the
conventional ones found in decimal.
• It is a machine language one.
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Decimal (10)
00
01
02
03
04
05
06
07
08
09
10
11
12
13
14
15
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F
Binary (2)
Octal (8)
Hexadecimal (16)
0000
0
0
0001
1
1
0010
2
2
0011
3
3
0100
4
4
0101
5
5
0110
6
6
0111
7
7
1000
10
8
1001
11
9
1010
12
A
1011
13
B
1100
14
C
1101
15
D
1110
16
E
1111
17
F
2. CONVERSION OF NUMBER SYSTEMS
2.1 Decimal to Other Base System
Steps
•
Step 1 − Divide the decimal number to be converted by the value of the new base.
•
Step 2 − Get the remainder from Step 1 as the rightmost digit (least significant digit) of new base
number.
•
Step 3 − Divide the quotient of the previous divide by the new base.
•
Step 4 − Record the remainder from Step 3 as the next digit (to the left) of the new base number.
Repeat Steps 3 and 4, getting remainders from right to left, until the quotient becomes zero in Step 3.
The last remainder thus obtained will be the Most Significant Digit (MSD) of the new base number.
1. Convert the following decimal to binary/ Octal/ Hexadecimal.
i). 26710
ii). 1476810
iii). 34.4510
i). (267)10
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ii). (14768)10
iii). (34.45)10
2.2 Other Base System to Decimal System
Steps
•
Step 1 − Determine the column (positional) value of each digit (this depends on the position of the
digit and the base of the number system).
•
Step 2 − Multiply the obtained column values (in Step 1) by the digits in the corresponding
columns.
•
Step 3 − Sum the products calculated in Step 2. The total is the equivalent value in decimal.
2.3 Binary to Octal
Steps
•
Step 1 − Divide the binary digits into groups of three (starting from the right).
•
Step 2 − Convert each group of three binary digits to one octal digit.
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2.4 Binary to Hexadecimal
Steps
•
Step 1 − Divide the binary digits into groups of four (starting from the right).
•
Step 2 − Convert each group of four binary digits to one hexadecimal symbol.
2.5 Octal to Binary
Steps
•
Step 1 − Convert each octal digit to a 3 digit binary number (the octal digits may be treated as
decimal for this conversion).
•
Step 2 − Combine all the resulting binary groups (of 3 digits each) into a single binary number.
2.6 Hexadecimal to Binary
Steps
•
Step 1 − Convert each hexadecimal digit to a 4 digit binary number (the hexadecimal digits may be
treated as decimal for this conversion).
•
Step 2 − Combine all the resulting binary groups (of 4 digits each) into a single binary number.
2. Convert the Binary number 111001012 to its decimal, octal and hexadecimal equivalent.
Decimal: - 111001012
= (1 × 27 ) + (1 × 26 ) + (1 × 25 ) + (0 × 24 ) + (0 × 23 ) + (1 × 22 ) + (0 × 21 ) + (1 × 20 )
= 128+ 64 + 32 + 0 + 0 + 4 + 0 + 1
[11100101]2 = [229]10
3. Convert the Octal number 24368 to its decimal, binary and hexadecimal equivalent.
Decimal:
[2436]8 = (2 × 83 ) + (4 × 82 ) + (3 × 81 ) + (6 × 80 )
= 1024 + 256 + 24 + 6
[2436]8= [1310]10
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4. Convert the hexadecimal number 28D16 to its decimal, binary and octal equivalent.
Decimal:
[28D]16 = 2 8 13
= (2 × 162 ) + (8 × 161 ) + (13 × 160 )
= 512 + 128+ 13
[28D]16 = [653]10
5. Convert FACE16 into its binary, octal and decimal equivalent.[May’17]
(i) Hexadecimal to binary:
(ii) Hexadecimal to octal:
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(iii) Hexadecimal to decimal
[FACE]16 = (1 × 215 ) + (1 × 214 ) + (1 × 213 ) + (1 × 212 ) + (1 × 211 ) + (0 × 210 ) + (1 × 29 )
+ (0 × 28 ) + (1 × 27 ) + (1 × 26 ) + (0 × 25 ) + (0 × 24 ) + (1 × 23 ) + (1 × 22 )
+ (1 × 21 ) + (0 × 20 )
[FACE]16 = [64206]10
6. Convert 23.62510 to octal (base 8). [Nov’16]
[23]10 = [27]8
0.625 x 8 = 5.0 5
(0.625)10 = (0.5)8
(23.625)10 = (27.5)8
7. Convert [1010111011101100]2 into its octal, decimal and hexadecimal equivalent. [Apr’16]
(i) Binary to octal:
(ii) Binary to decimal:
10101110111011002
[1010111011101100]2
= (1 × 215 ) + (0 × 214 ) + (1 × 213 ) + (0 × 212 ) + (1 × 211 ) + (1 × 210 )
+ (1 × 29 ) + (0 × 28 ) + (1 × 27 ) + (1 × 26 ) + (1 × 25 ) + (0 × 24 ) + (1 × 23 )
+ (1 × 22 ) + (0 × 21 ) + (0 × 20 )
[1010111011101100]2 = [44780]10
(iii) Binary to hexadecimal:
1010 1110 1110 1100
[1010111011101100]2 = [AEEC]16
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3. Binary addition and subtractions
Addition:
0+0=0
0+1=1
1+0=1
1 + 1 = 10
1 + 1 + 1 = 11
Subtraction:
0-0=0
0 - 1 = 11
(Borrow – 1, difference – 1)
1-0=1
1 -1=0
8. Perform the binary addition and subtraction for the following binary numbers.
(a) [10001]2 + [11101]2
(b) [10111]2 + [110101]2
(c) [1011011]2 - [10010]2
(d) [1000101] 2 – [101100] 2 .
(a). [10001]2 + [11101]2
10001
+ 11101
101110
[10001]2 + [11101]2 = [101110]2
(b) [10111]2 + [110101]2
10111
+ 110101
1001100
[10111]2 + [110101]2 = [101100]2
(c) [1011011]2 - [10010]2
1011011
- 10010
1001001
[1011011]2 – [10010]2 = [1001001]2
(d) [1000101] 2 – [101100] 2
1000101
- 101100
0011001
[1000101] 2 – [101100] 2 = [0011001]2
4. Binary system complements
In digital computers, to simplify the subtraction operation and for logical manipulation complements are
used. Two types of complements:
•
•
1’s complement
2’s complement
1's complement
The 1's complement of a number is found by changing all 1's to 0's and all 0's to 1's. This is called as
taking complement or 1's complement. Example of 1's Complement is as follows.
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2's complement
The 2's complement of binary number is obtained by adding 1 to the Least Significant Bit (LSB) of 1's
complement of the number.
2's complement = 1's complement + 1
Example of 2's Complement is as follows.
9. Do the 1’s complement for [1110]2, [1010]2 and 2’s complement for [1110]2, [101011]2.
1’s complement:
(i). [1110]2
11102
Change 0 to 1 and 1 to 0,
00012
[1110]2 = [0001]2
(ii). [1010]2
10102
Change 0 to 1 and 1 to 0.
01012
[1010]2 = [0101]2
2’s complement:
(i). [1110]2
11102
Converting the given number to 1’s complement.
Change 0 to 1 and 1 to 0,
00012
Adding ‘1’ at the end.
0001
1
0010
[1110]2 = [0010]2
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(ii). [101011]2
1010112
Converting the given number to 1’s complement.
Change 0 to 1 and 1 to 0,
0101002
Adding ‘1’ at the end.
010100
1
010101
[101011]2 = [010101]2
4.1 1’s complement subtraction
(i) Subtraction of smaller number from larger number
• Determine the 1’s complement of the smallest no
• Add the 1’s complement to the larger no
• Remove the carry and add it to the result. This is called end around carry.
(i) Subtraction of larger number from smaller number
• Determine the 1’s complement of the larger no
• Add the 1’s complement to the smaller no
• Answer is in 1’s complement form. To get the answer in true form take the 1’s
complement and assign negative sign to the answer.
10. Perform the subtraction for the following binary number using 1’s complement. (a) [110101] 2 –
[100101]2 (b) [101011]2 – [111001]2.
(a) [110101]2 – [100101]2
Subtrahend is 100101
Converting the given number to 1’s complement.
Change 0 to 1 and 1 to 0,
011010
Minued = 110101
1’s complemented Subtrahend = 011010
Adding = 1001111
Carry over is 1, so add that to the answer,
001111
1
010000
[110101]2 – [100101]2 = [010000]2
(b) [101011]2 – [111001]2
Subtrahend is 111001
Converting the given number to 1’s complement.
Change 0 to 1 and 1 to 0,
000110
Minued = 101011
1’s complemented Subtrahend = 000110
Adding = 110001
Since no Carry over, so convert the answer by 1’s complement and put ‘– ‘sign before it,
[110101]2 – [100101]2 = [-001110]2
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4.2 2’s complement subtraction
(i) Subtraction of smaller number from larger number
• Determine the 2’s complement of the smallest no
• Add the 2’s complement to the larger no
• Discard the carry.
(i) Subtraction of larger number from smaller number
• Determine the 2’s complement of the larger no
• Add the 2’s complement to the smaller no
• Answer is in 2’s complement form. To get the answer in true form take the 2’s
complement and assign negative sign to the answer.
11. Perform the subtraction for the following binary number using 2’s complement.
(a) [110110]2 – [10110]2 (b) [10110]2 – [11010]2.
(a) [11010]2 – [10110]2
Subtrahend is 010110
Converting the given number to 1’s complement.
Change 0 to 1 and 1 to 0,
101001
Adding ‘1’ at the end.
101001
1
101010
Minued = 110110
2’s complemented Subtrahend = 101010
1100000
Carry over is 1, so drop the carry over,
[11010]2 – [10110]2 = [100000]2
(b) [10110]2 – [11010]2
Subtrahend is 11010
Converting the given number to 1’s complement.
Change 0 to 1 and 1 to 0,
00101
Adding ‘1’ at the end.
00101
1
00110
Minued = 10110
2’s complemented Subtrahend = 00110
11100
Since no Carry over, so convert the answer by 2’s complement and put ‘– ‘sign before it,
Converting the given number to 1’s complement.
Change 0 to 1 and 1 to 0,
00011
Adding ‘1’ at the end.
00011
1
00100
[10110]2 – [11010]2 = [-00100]2
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12. Perform the following operation (756)8 – (437) 8 + (725) 16. Express the answer in octal
form.[Nov’15]
Since (437) 8 is in negative, taking 2’s complement,
100 011 111
Converting the given number to 1’s complement.
Change 0 to 1 and 1 to 0,
011 100 000
Adding ‘1’ at the end.
011 100 000
1
011 100 001
2’s complement of [100 011 111] = [011 100 001]
Adding:
Converting octal value to hexa value:
Adding:
Hexadecimal to Octal conversion:
(756)8 – (437) 8 + (725) 16 = [4764] 8
4.3 BCD addition:
BCD addition procedure as follows:
1 . Add two BCD numbers using ordinary binary addition.
2 . If four-bit sum is equal to or less than 9, no correction is needed. The sum is in proper BCD form.
3 . If the four-bit sum is greater than 9 or if a carry is generated from the four-bit sum, the sum is
invalid.
4 . To correct the invalid sum, add 01102 to the four-bit sum. If a carry results from this addition, add
it to the next higher-order BCD digit.
13. Convert the following number to Binary Coded Decimal [BCD] (a). 581 (b). 637
(a). 581
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(b). 637
14. Add the following numbers using BCD (a). 234 + 152 (b). 623 +137 (c). 599 + 984.
(a). 234 + 152
234 + 152 = [0011 1000 0110]
(b). 623 + 137
Since 1010 is greater than 9, its invalid condition.
Add 0110 to that value.
623 + 137 = [0111 0110 0000]
(c). 599 + 984
Since 1110, 1101 are greater than 9, its invalid condition.
Add 0110 to those values.
Carry over occurs at second number, add the carry to third one.
599
984 = [0001 0101 1000 0011]
15. Perform the following addition using BCD and Excess-3 addition (205+569) [Apr’15]
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BCD:
Since 1110 is greater than 9, its invalid condition.
Add 0110 to that value.
(205+569) = [0111 0111 0100]2
Excess-3:
Excess-3 addition:
Add 0011 to carry over takes places and subtract 0011 to non carry,
(205+569) = (1010 1010 0111)
5. Signed-magnitude Representation
In signed numbers it is necessary to represent negative as well as positive numbers. In sign-magnitude
representation the most significant bit (leftmost bit) is used to represent sign of the number. If the most
significant bit is 0, the number is positive, and if the most significant bit is 1, the number is negative. The
remaining bits of the number represent magnitude.
Sign magnitude format
Here are some examples of sign-magnitude numbers.
+6
=
0 000
0110
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- 14
=
1 000
1110
+ 24
=
0 001
1000
- 64
=
1 100
0000
In case of unsigned 8-bit binary numbers the decimal range is 0 to 255. For sign-magnitude 8bit numbers the largest magnitude is reduced from 255 to 127 because we need to represent both positive
and negative numbers.
Maximum positive number 0 1 1 1
1 1 1 1 + 127
Maximum negative number 1 1 1 1
1 1 1 1 = - 127
6. CLASSIFICATION OF BINARY CODES
The codes are broadly categorized into following four categories.
•
Weighted Codes
•
Non-Weighted Codes
•
Binary Coded Decimal Code
•
Alphanumeric Codes
•
Error Detecting Codes
•
Error Correcting Codes
6.1 Weighted Codes
Weighted binary codes are those binary codes which obey the positional weight principle. Each position of
the number represents a specific weight. Several systems of the codes are used to express the decimal
digits 0 through 9. In these codes each decimal digit is represented by a group of four bits.
6.2 Non-Weighted Codes
In this type of binary codes, the positional weights are not assigned. The examples of non-weighted codes
are Excess-3 code and Gray code.
6.2.1 Excess-3 code
The Excess-3 code is also called as XS-3 code. It is non-weighted code used to express decimal
numbers. The Excess-3 code words are derived from the 8421 BCD code words adding (0011)2 or (3)10
to each code word in 8421. The excess-3 codes are obtained as follows −
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Example
6.2.2 Gray Code
It is the non-weighted code and it is not arithmetic codes. That means there are no specific weights
assigned to the bit position. It has a very special feature that, only one bit will change each time the
decimal number is incremented as shown in fig. As only one bit changes at a time, the gray code is called
as a unit distance code. The gray code is a cyclic code. Gray code cannot be used for arithmetic
operation.
Application of Gray code
•
Gray code is popularly used in the shaft position encoders.
•
A shaft position encoder produces a code word which represents the angular position of the shaft.
6.3 Binary Coded Decimal (BCD) code
In this code each decimal digit is represented by a 4-bit binary number. BCD is a way to express each of
the decimal digits with a binary code. In the BCD, with four bits we can represent sixteen numbers (0000
to 1111). But in BCD code only first ten of these are used (0000 to 1001). The remaining six code
combinations i.e. 1010 to 1111 are invalid in BCD.
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Advantages of BCD Codes
•
It is very similar to decimal system.
•
We need to remember binary equivalent of decimal numbers 0 to 9 only.
Disadvantages of BCD Codes
•
The addition and subtraction of BCD have different rules.
•
The BCD arithmetic is little more complicated.
•
BCD needs more number of bits than binary to represent the decimal number. So BCD is less
efficient than binary.
6.4 Alphanumeric codes
A binary digit or bit can represent only two symbols as it has only two states '0' or '1'. But this is not
enough for communication between two computers because there we need many more symbols for
communication. These symbols are required to represent 26 alphabets with capital and small letters,
numbers from 0 to 9, punctuation marks and other symbols.
The alphanumeric codes are the codes that represent numbers and alphabetic characters. Mostly such
codes also represent other characters such as symbol and various instructions necessary for conveying
information. An alphanumeric code should at least represent 10 digits and 26 letters of alphabet i.e. total
36 items. The following three alphanumeric codes are very commonly used for the data representation.
•
American Standard Code for Information Interchange (ASCII).
•
Extended Binary Coded Decimal Interchange Code (EBCDIC).
•
Five bit Baudot Code.
ASCII code is a 7-bit code whereas EBCDIC is an 8-bit code. ASCII code is more commonly used
worldwide while EBCDIC is used primarily in large IBM computers.
6.5 Error Codes
There are binary code techniques available to detect and correct data during data transmission.
7. Boolean algebra: (Apri 22 /10 mark)
7.1
Introduction:
In 1854, George Boole, an English mathematician, proposed algebra for symbolically representing
problems in logic so that they may be analyzed mathematically. The mathematical systems founded upon
the work of Boole are called Boolean algebra in his honor.
The application of a Boolean algebra to certain engineering problems was introduced in 1938 by
C.E. Shannon. For the formal definition of Boolean algebra, we shall employ the postulates formulated by
E.V. Huntington in 1904.
7.2
Fundamental postulates of Boolean algebra:
The postulates of a mathematical system forms the basic assumption from which it is possible to
deduce the theorems, laws and properties of the system. The most common postulates used to formulate
various structures are:
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i)
Closure:
A set S is closed w.r.t. a binary operator, if for every pair of elements of S, the binary operator
specifies a rule for obtaining a unique element of S. The result of each operation with operator (+)
or (.) is either 1 or 0 and 1, 0 ЄB.
ii) Identity element:
A set S is said to have an identity element w.r.t a binary operation * on S, if there exists an element
e Є S with the property,
e* x = x * e = x
Eg:
0+ 0 = 0
0+ 1 = 1+ 0 = 1
1.1=1
1.0=0.1=1
a) x+ 0= x
b) x. 1 = x
iii) Commutative law:
A binary operator * on a set S is said to be commutative if,
for all x, y Є S
x*y=y*x
Eg:
0+ 1 = 1+ 0 = 1
a) x+ y= y+ x
0.1=1.0 =0
b) x. y= y. x
iv) Distributive law:
If * and • are two binary operation on a set S, • is said to be distributive over + whenever,
x . (y+ z) = (x. y) + (x. z)
Similarly, + is said to be distributive over • whenever,
x + (y. z) = (x+ y). (x+ z)
v) Inverse:
A set S having the identity element e, w.r.t. binary operator * is said to have an inverse, whenever
for every x Є S, there exists an element x’ Є S such that,
x. x’ Є e
a) x+ x’ = 1, since 0 + 0’ = 0+ 1 and 1+ 1’ = 1+ 0 = 1
b) x. x’ = 1, since 0 . 0’ = 0. 1 and 1. 1’ = 1. 0 = 0
Summary:
Postulates of Boolean algebra:
POSTULATES
(a)
(b)
Postulate 2 (Identity)
x+0=x
x.1=x
Postulate 3 (Commutative)
x+ y = y+ x
x . y = y. x
Postulate 4 (Distributive)
x (y+ z) = xy+ xz
x+ yz = (x+ y). (x+ z)
x+x’ = 1
x. x’ = 0
Postulate 5 (Inverse)
SMVEC/CSE/U23ITIC01/DDSA/UNIT 1
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7.3
Basic theorems and properties of Boolean algebra:
Basic Theorems:
The theorems, like the postulates are listed in pairs; each relation is the dual of the one paired with
it. The theorems must be proven from the postulates. The proofs of the theorems with one variable are
presented below. At the right is listed the number of the postulate that justifies each step of the proof.
1) a) x+ x = x
x+ x = (x+ x) . 1
= (x+ x). (x+ x’)
= x+ xx’
= x+ 0
=x
b) x. x = x
x. x = (x. x) + 0
= (x. x) + (x. x’)
= x ( x+ x’)
= x (1)
=x
-------------------------------------------------------------------------------------------
by postulate 2(b) [ x. 1 = x ]
5(a) [ x+ x’ = 1]
4(b) [ x+yz = (x+y)(x+z)]
5(b) [ x. x’ = 0 ]
2(a) [ x+0 = x ]
-------------------------------------------------------------------------------------------
by postulate 2(a) [ x+ 0 = x ]
5(b) [ x. x’ = 0]
4(a) [ x (y+z) = (xy)+ (xz)]
5(a) [ x+ x’ = 1 ]
2(b) [ x.1 = x ]
-------------------------------------------------------------------------------------------
by postulate 2(b) [ x. 1 = x ]
5(a) [ x+ x’ = 1]
4(b) [ x+yz = (x+y)(x+z)]
2(b) [ x. 1 = x ]
5(a) [ x+ x’= 1]
2) a) x+ 1 = 1
x+ 1 = 1 . (x+ 1)
= (x+ x’). (x+ 1)
= x+ x’.1
= x+ x’
=1
b) x .0 = 0
3) (x’)’ = x
From postulate 5, we have x+ x’ = 1 and x. x’ = 0, which defines the complement of x. The
complement of x’ is x and is also (x’)’. Therefore, since the complement is unique,
(x’)’ = x.
4) Absorption Theorem:
a) x+ xy = x
x+ xy = x. 1 + xy
= x (1+ y)
= x (1)
= x.
b) x. (x+ y) = x
-------------------------------------------------------------------------
by postulate 2(b) [ x. 1 = x ]
4(a) [ x (y+z) = (xy)+ (xz)]
by theorem 2(a)
[x+ 1 = x]
by postulate 2(a)
[x. 1 = x]
x. (x+ y) = x. x+ x. y
= x + x.y
= x.
c) x+ x’y = x+ y
-------------------------------------------------------
4(a) [ x (y+z) = (xy)+ (xz)]
by theorem 1(b)
[x. x = x]
by theorem 4(a)
[x+ xy = x]
-------------------
by theorem 4(a)
x+ x’y = x+ xy+ x’y
[x+ xy = x]
= x+ y (x+ x’)
= x+ y (1)
= x+ y
d) x. (x’+y) = xy
--------------- by postulate 4(a) [ x (y+z) = (xy)+ (xz)]
------------------5(a)
[x+ x’ = 1]
------------------2(b)
[x. 1= x]
x. (x’+y) = x.x’+ xy
= 0+ xy
= xy.
--------------- by postulate 4(a) [ x (y+z) = (xy)+ (xz)]
------------------5(b)
[x. x’ = 0]
------------------2(a)
[x+ 0= x]
SMVEC/CSE/U23ITIC01/DDSA/UNIT 1
30
Properties of Boolean algebra:
1. Commutative property:
Boolean addition is commutative, given by
x+ y = y+ x
According to this property, the order of the OR operation conducted on the variables makes no
difference. Boolean algebra is also commutative over multiplication given by,
x. y = y. x
This means that the order of the AND operation conducted on the variables makes no difference.
2. Associative property:
The associative property of addition is given by,
A+ (B+ C) = (A+B) + C
The OR operation of several variables results in the same, regardless of the grouping of the
variables.
The associative law of multiplication is given by,
A. (B. C) = (A.B) . C
It makes no difference in what order the variables are grouped during the AND operation of several
variables.
3. Distributive property:
The Boolean addition is distributive over Boolean multiplication, given by,
A+ BC = (A+B) (A+C)
The Boolean addition is distributive over Boolean addition, given by,
A. (B+C) = (A.B)+ (A.C)
4. Duality:
It states that every algebraic expression deducible from the postulates of Boolean algebra remains valid
if the operators and identity elements are interchanged. If the dual of an algebraic expression is desired,
we simply interchange OR and AND operators and replace 1’s by 0’s and 0’s by 1’s.
x+ x’ = 1 is x. x’ = 0
Duality is a very important property of Boolean algebra.
Summary:
Theorems of Boolean algebra:
x+x=x
1.
Idempotency
x+1=1
SMVEC/CSE/U23ITIC01/DDSA/UNIT 1
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2.
Involution
3.
Absorption
4.
5.
(x’)’ = x
x+ xy = x
x (x+ y) = x
x+ x’y = x+ y
x. (x’+ y)= xy
Associative
x+(y+ z)= (x+ y)+ z
x (yz) = (xy) z
DeMorgan’s Theorem
(x+ y)’= x’. y’
(x. y)’= x’+ y’
DeMorgan’s Theorems:
Two theorems that are an important part of Boolean algebra were proposed by DeMorgan.
The first theorem states that the complement of a product is equal to the sum of the complements.
(AB)’ = A’+ B’
The second theorem states that the complement of a sum is equal to the product of the
complements.
(A+ B)’ = A’. B’
Consensus Theorem:
In simplification of Boolean expression, an expression of the form AB+ A’C+ BC, the term BC is
redundant and can be eliminated to form the equivalent expression AB+ A’C. The theorem used for this
simplification is known as consensus theorem and is stated as,
AB+ A’C+ BC = AB+ A’C
The dual form of consensus theorem is stated as,
(A+B) (A’+C) (B+C) = (A+B) (A’+C)
7.4 Minimization of Boolean Expressions:
The Boolean expressions can be simplified by applying properties, laws and theorems of Boolean
algebra.
Problems: Simplify the following Boolean functions to a minimum number of literals:
1. x (x’+y)
= xx’+ xy
[ x. x’= 0 ]
= 0 + xy
[ x+ 0 = x ]
= xy.
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2.
x+ x’y
= x + xy + x’y
[ x+ xy= x]
= x+ y (x+x’)
= x+ y (1)
[ x+ x’ = 1]
= x+ y.
3. xy+ yz+ y’z
= xy+ z ( y+ y’)
= xy+ z ( 1 )
[ y+ y’ = 1]
= xy+ z.
4. (x+ y) (x+ y’)
= x.x+ xy’+ xy+ yy’
= x+ xy’+ xy+ 0
[ x. x= 0]; [ y. y’= 0]
= x (1+ y’+ y)
= x (1)
[ 1+y= 1 ]
= x.
5. xy + x’z + yz.
= 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)
[1+y= 1]
= xy+ x’z.
6. (x+ y) (x’+ z) (y+ z)
= (x+ y) (x’+ z)
[dual form of consensus theorem,
(A+ B) (A’+ C) (B+ C) = (A+ B) (A’+ C)]
7. x’y+ xy+ x’y’
= y ( x’+ x) + x’y’
[ x (y+ z) = xy+ xz ]
= y ( 1 ) + x’y’
[ x+ x’ = 1]
= y+ x’y’
[ x+ x’y’ = x+ y’ ]
= y+ x’.
8. x+ xy’+ x’y
= x (1+ y’)+ x’y
= x (1) + x’y
[ 1+ x = 1 ]
= x+ x’y
[ x+ x’y = x+ y ]
= x+ y.
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9. AB + (AC)' + AB’C (AB + C)
= AB + (AC)' + AAB'BC + AB'CC
= AB + (AC)' + 0+ AB'CC
[B.B' = 0]
= AB + (AC)' + AB'C
[C.C = 1]
= AB + A' + C' +AB'C
[(AC)' = A' + C']
= AB + A’ + C' + AB'
[C’ + AB’C = C’ + AB’]
= A' + B+ C’+ AB’
[A’ + AB = A’ + B]
Re- arranging,
= A' + AB’+ B+ C'
[A’ + AB = A’ + B]
= A' + B’+ B+ C'
[ B’+ B= 1]
= A' +1+ C’
[ A+ 1= 1]
=1
10. (x’+ y) (x+ y)
= x’.x+ x’y+ yx+ y.y
= 0+ x’y+ xy+ y
[ x.x’= 0]; [ x. x= x]
= y ( x’+ x+ 1)
= y( 1 )
[ 1+ x = 1 ]
= y.
11. xy+ xyz+ xy (w+ z)
= xy ( 1+ z+ w+ z)
= xy ( 1 )
[ 1+ x = 1 ]
= xy.
12. xy+ xyz+ xyz’+ x’yz
= xy ( 1+ z+ z’)+ x’yz
= xy ( 1 ) + x’yz
[ 1+ x = 1 ]
= xy+ x’yz
= y ( x+ x’z )
[ x+ x’y = x+ y]
= y ( x+ z ).
13. xyz+ xy’z+ xyz’
= xy (z+ z’) + xy’z
= xy+ xy’z
[ x+ x’= 1]
= x(y+ y’z)
[ x+ x’y = x+ y]
= x(y+ z)
14. x’y’z’+ x’yz’+ xy’z’+ xyz’
= x’z’ (y’+ y) + xz’ (y’+ y)
= x’z’+ xz’
[ x+ x’= 1]
= z’ (x’+ x)
= z’
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[ x+ x’= 1]
34
15. w’xyz’+ xyz’+ xy’z’+ xy’z
= xyz’ (w’+ 1) + xy’z’+ xy’z
= xyz’+ xy’z’+ xy’z
[ 1+ x = 1 ]
= xz’ (y+ y’) + xy’z
= xz’+ xy’z
[ x+ x’= 1]
= x (z‘+ y’z)
= x (z’+ y’).
[ x’+ xy’ = x’+ y’]
16. w’xy’z+ w’xyz+ wxz
= w’xz (y’+ y)+ wxz
= w’xz (1)+ wxz
[ x+ x’= 1]
= w’xz+ wxz
= xz (w’+ w)
= xz.
[ x+ x’= 1]
17. xy+ x (y+ z) + y (y+ z)
= xy+ xy+ xz+ yy+ yz
= xy+ xz+ y+ yz
[x+ x= x]; [x. x= x]
= xy+ xz+ y
[x+ xy= x]
= y+ xz
[x+ xy= x]
18. x’y’z’+ x’y’z+ x’yz’+ x’yz+ xy’z’
= x’y’ (z’+z) + x’y (z’+z)+ xy’z’
= x’ y’ (1) + x’y (1)+ xy’z’
[ x+ x’= 1]
= x’y’ + x’y + xy’z’
= x’(y’+y) + xy’z’
= x’ (1) + xy’z’
[ x+ x’= 1]
= x’ + xy’z’
= x’+ y’z’.
[ x’+ xy’ = x’+ y’]
19. [ xy’ (z+ wy) + x’y’] z
= [ xy’z+ xy’wy+ x’y’] z
= [ xy’z+ 0+ x’y’] z
[x. x’= 0]
= xy’z. z+ x’y’z
= xy’z+ x’y’z
[x. x= x]
= y’z (x+ x’)
= y’z (1) = y’z.
[ x+ x’= 1]
20. w’y (w’xz)’ + w’xy’z’ + wx’y
= w’y (w’’+ x’+ z’) + w’xy’z’ + wx’y
= w’y (w+ x’+ z’) + w’xy’z’ + wx’y
SMVEC/CSE/U23ITIC01/DDSA/UNIT 1
[ x’’ = x]
35
= w’yw+ w’y x’+ w’y z’ + w’xy’z’ + wx’y
= 0 + w’x’y+ w’y z’ + w’xy’z’ + wx’y
[x. x’= 0]
Re-arranging,
= w’x’y+ wx’y + w’y z’ + w’xy’z’
= x’y (w’+ w) + w’z’ (y+ xy’)
= x’y (1) + w’z’ (y+ xy’)
[ x+ x’= 1]
= x’y+ w’z’ (y+x)
[ x+ x’y = x+ y]
21. x’yz+ xy’z’+ x’y’z’+ xy’z+ xyz
= yz (x’+x) + xy’z’+ x’y’z’+ xy’z
= yz (1) + y’z’ (x+ x’) + xy’z
[ x+ x’= 1]
= yz+ y’z’ (1) + xy’z
[ x+ x’= 1]
= yz+ y’z’+ xy’z
= yz+ y’ (z’+ xz)
= yz+ y’ (z’+ x)
[ x’+ xy = x’+ y]
= yz+ y’z’+ xy’
22. [ xy+ xz]’+ x’y’z
= (xy)’. (xz)’+ x’y’z
= (x’+ y’). (x’+ z’)+ x’y’z
= x’x’+ x’z’+ x’y’+ y’z’+ x’y’z
= x’+ x’z’+ x’y’+ y’z’+ x’y’z
[x+ x= x]
= x’+ x’z’+ x’y’+ y’ [z’+ x’z]
= x’+ x’z’+ x’y’+ y’ [z’+ x’]
[ x’+ xy = x’+ y]
= x’+ x’y’+ y’ [z’+ x’]
[x+ xy = x]
= x’+ x’y’+ y’z’+ x’y’
= x’+ y’z’+ x’y’
= x’+ y’z’.
[x+ xy = x]
[x+ xy = x]
23. xy+ xy’( x’z’)’
= xy+ xy’ (x’’+ z’’)
= xy+ xy’ (x+ z)
[x’’ = x]
= xy+ xy’x+ xy’z
= xy+ xy’+ xy’z
[x. x= x]
= xy+ xy’ [1+ z]
= xy+ xy’ [1]
[ 1+ x = 1 ]
= xy+ xy’
= x( y+ y’)
= x [1]
[ x+ x’= 1]
= x.
SMVEC/CSE/U23ITIC01/DDSA/UNIT 1
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24. [( xy’+ xyz)’+ x (y+ xy’)]’
= [ x( y’+yz)’+ x (y+ xy’)]’
= [ x( y’+z)’+ x (y+ x)]’
[ x’+ xy = x’+ y]; [ x+ x’y = x+ y]
= [ x( y’+z)’+ xy+ x.x)]’
= [ (xy’+xz)’+ xy+ x)]’
[x. x= x]
= [ ( xy’+xz)’+ x)]’
[x+ xy = x]
= [ (xy’)’. (xz)’+ x]’
= [ (x’+y’’). (x’+z’)+ x]’
= [ (x’+y). (x’+z’)+ x]’
[x’’ = x]
= [ (x’+ yz’)+ x]’
[ (x+ y) (x+ z)= x+ yz]
= [ x’+ yz’+ x]’
= [ 1+ yz’]’ = [1]’ =0
[ x+ x’= 1]
25. [(xy)’+ x’+ xy]’
= [ x’+ y’+ x’+ xy]’
= [ x’+ y’+ xy]’
[x+ x= x]
= [x’+ y’+ x]’
[ x’+ xy = x’+ y]
= [y’+ 1]’
[ x+ x’= 1]
= [ 1 ]’
[ 1+ x = 1 ]
= 0.
26. [ (xy+ z’) ((x+ y)’+z) ]’
= [ (xy+ z’) ((x’. y’)+z) ]’
= [ xy. x’y’+ xy. z+ z’. x’y’+ z’. z]’
= [ 0+ xyz+ x’y’z’+ 0]’
[x. x’= 0]
= [ xyz+ x’y’z’ ]’
= (xyz)’. ( x’y’z’)’
= ( x’+ y’+ z’). (x’’+ y’’+ z’’)
= ( x’+ y’+ z’). (x+ y+ z).
[x’’ = x]
27. (x+ y) (x’z’+ z) (y’+ xz)’
= (x+ y) (x’z’+ z) (y’’. (xz)’)
= (x+ y) (x’+ z) (y. (xz)’)
[ x+ x’y = x+ y]; [x’’ = x]
= (x+ y) (x’+ z) (y. (x’+z’))
= ( x.x’+ xz+ x’y+ yz) (x’y+ yz’)
= ( 0+ xz+ x’y+ yz) (x’y+ yz’)
= (xz+ x’y+ yz) (x’y+ yz’)
= xz. x’y+ xz. yz’+ x’y. x’y+ x’y. yz’+ yz. x’y+ yz. yz’
= 0+ 0+ x’y+ x’yz’+ x’yz+ 0
[x. x’= 0]; [x. x= x]
= x’y+ x’yz’+ x’yz
= x’y (1+ z’+ z)
= x’y (1)
[ 1+ x = 1 ]
= x’y.
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28. Y= ∑m (1, 3, 5, 7)
= x’y’z+ x’yz+ xy’z+ xyz
= x’z( y’+y) + xz( y’+y)
= x’z (1)+ xz (1)
[ x+ x’= 1]
= x’z+ xz
= z ( x’+ x)
= z (1)
[ x+ x’= 1]
= z.
7.5 COMPLEMENT OF A FUNCTION:
The complement of a function F is F’ and is obtained from an interchange of 0’s for 1’s and 1’s for
0’s in the value of F. The complement of a function may be derived algebraically through DeMorgan’s
theorem.
DeMorgan’s theorem for any number of variables resemble in form the two-variables and can be
derived by successive substitutions similar to the method used in the preceding derivation. These theorems
can be generalized as –
(A+ B+ C+ D+ … + F)’ = A’ B’ C’ D’ … F’
(A B C D … F)’ = A’+B’+ C’+ D’+ … +F’.
Find the complement of the following functions,
1. F= x’yz’+ x’y’z
F’= (x’yz’+ x’y’z)’
= (x”+ y’+ z”) . (x”+ y”+z’)
= (x+ y’+ z). (x+ y+ z’).
2. F= x (y’z’+ yz)
F’= [x (y’z’+ yz)]’
= x’+ (y’z’+ yz)’
= x’+ (y’z’)’. (yz)’
= x’+ (y”+ z”) . (y’+ z’)
= x’+ (y+ z) . (y’+ z’).
3. F= (xy + y’z + xz) x.
F’ = [(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]
SMVEC/CSE/U23ITIC01/DDSA/UNIT 1
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= 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]
= x’+ y’z’
4. F= xy’+ x’y
F’= (xy’+ x’y)’
= (xy’)’. (x’y)’
= (x’+y) (x+y’)
= x’x+ x’y’+ yx+ yy’
= x’y’+ xy.
5. F = wx’y + xy’+ wxz
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’
7.6
CANONICAL FORM AND STANDARD FORM:
7.6.1
Minterm and Maxterm
A binary variable may appear either in its normal form (x) or in its complement form (x’). When two
binary variables x and y combined with an AND operation, there are four possible combinations:
x’y’, x’y, xy’ and xy
Each of these four AND terms is called as ‘minterm’.
In a similar fashion, when two binary variables x and y combined with an OR operation, there are
four possible combinations:
x’+ y’, x’+ y, x+ y’ and x+ y
Each of these four OR terms is called as ‘maxterm’.
The minterms and maxterms of a 3- variable function can be represented as in table below.
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Variables
7.6.2
Minterms
Maxterms
X
Y
Z
mi
Mi
0
0
0
x’y’z’ = m0
x+ y+ z= M0
0
0
1
x’y’z = m1
x+ y+ z’= M1
0
1
0
x’yz’ = m2
x+ y’+ z= M2
0
1
1
x’yz = m3
x+ y’+ z’= M3
1
0
0
xy’z’ = m4
x’+ y+ z= M4
1
0
1
xy’z = m5
x’+ y+ z’= M5
1
1
0
xyz’ = m6
x’+ y’+ z= M6
1
1
1
xyz = m7
x’+ y’+ z’= M7
Sum of Products (SOP):
The logical sum of two or more logical product terms is called sum of products expression. It is
logically an OR operation of AND operated variables such as
7.6.3
Product of Sums (POS):
A product of sums expression is a logical product of two or more logical sum terms. It is basically an
AND operation of OR operated variables such as,
7.6.4
Canonical Sum of Product Expression:
If each term in SOP form contains all the literals then the SOP is known as Standard (or)
Canonical SOP form. Each individual term in standard SOP form is called minterm canonical form.
F (A, B, C) = AB’C+ ABC+ ABC’
Steps to convert general SOP to standard SOP form:
1. Find the missing literals in each product term if any.
2. AND each product term having missing literals by ORing the literal and its complement.
3. Expand the term by applying distributive law and reorder the literals in the product term.
4. Reduce the expression by omitting repeated product terms if any.
Obtain the canonical SOP form of the function:
1. Y(A, B) = A+ B
SMVEC/CSE/U23ITIC01/DDSA/UNIT 1
40
= A. (B+ B’)+ B (A+ A’)
= AB+ AB’+ AB+ A’B
= AB+ AB’+ A’B.
= m3+ m2+ m1
= ∑m (1, 2, 3).
2. Y (A, B, C) = A+ ABC
= A. (B+ B’). (C+ C’)+ ABC
= (AB+ AB’). (C+ C’)+ ABC
= ABC+ ABC’+ AB’C+ AB’C’+ ABC
= ABC+ ABC’+ AB’C+ AB’C’
= m7+ m6+ m5+ m4
= ∑m (4, 5, 6, 7).
3. Y (A, B, C) = A+ BC
= A. (B+ B’). (C+ C’)+(A+ A’). BC
= (AB+ AB’). (C+ C’)+ ABC+ A’BC
= ABC+ ABC’+ AB’C+ AB’C’+ ABC+ A’BC
= ABC+ ABC’+ AB’C+ AB’C’+ A’BC
= m7+ m6+ m5+ m4+ m3
= ∑m (3, 4, 5, 6, 7).
4. Y (A, B, C) = AC+ AB+ BC
= AC (B+ B’)+ AB (C+ C’)+ BC (A+ A’)
= ABC+ AB’C+ ABC+ ABC’+ ABC+ A’BC
= ABC+ AB’C+ ABC’+ A’BC
= ∑m (3, 5, 6, 7).
5. Y (A, B, C, D) = AB+ ACD
= AB (C+ C’) (D+ D’) + ACD (B+ B’)
= (ABC+ ABC’) (D+ D’) + ABCD+ AB’CD
= ABCD+ ABCD’+ ABC’D+ ABC’D’+ ABCD+ AB’CD
= ABCD+ ABCD’+ ABC’D+ ABC’D’+ AB’CD.
= m15+ m14+ m13+ m12+ m11
= ∑m (11, 12, 13, 14, 15).
7.6.5
Canonical Product of Sum Expression:
If each term in POS form contains all literals then the POS is known as standard (or) Canonical
POS form. Each individual term in standard POS form is called Maxterm canonical form.
•
F (A, B, C) = (A+ B+ C). (A+ B’+ C). (A+ B+ C’)
•
F (x, y, z) = (x+ y’+ z’). (x’+ y+ z). (x+ y+ z)
SMVEC/CSE/U23ITIC01/DDSA/UNIT 1
41
Steps to convert general POS to standard POS form:
1. Find the missing literals in each sum term if any.
2. OR each sum term having missing literals by ANDing the literal and its complement.
3. Expand the term by applying distributive law and reorder the literals in the sum term.
4. Reduce the expression by omitting repeated sum terms if any.
Obtain the canonical POS expression of the functions:
1. Y= A+ B’C
= (A+ B’) (A+ C)
[ A+ BC = (A+B) (A+C)]
= (A+ B’+ C.C’) (A+ C+ B.B’)
= (A+ B’+C) (A+ B’+C’) (A+ B+ C) (A+ B’+ C)
= (A+ B’+C). (A+ B’+C’). (A+ B+ C)
= M2. M3. M0
= ∏M (0, 2, 3)
2. Y= (A+B) (B+C) (A+C)
= (A+B+ C.C’) (B+ C+ A.A’) (A+C+B.B’)
= (A+B+C) (A+B+C’) (A+B+C) (A’+B+C) (A+B+C) (A+B’+C)
= (A+B+C) (A+B+C’) (A’+B+C) (A+B’+C)
= M0. M1. M4. M2
= ∏M (0, 1, 2, 4)
3. Y= A. (B+ C+ A)
= (A+ B.B’+ C.C’). (A+ B+ C)
= (A+B+C) (A+B+C’) (A+B’+C) (A+ B’+C’) (A+B+C)
= (A+B+C) (A+B+C’) (A+B’+C) (A+ B’+C’)
= M0. M1. M2. M3
= ∏M (0, 1, 2, 3)
4. Y= (A+B’) (B+C) (A+C’)
= (A+B’+C.C’) (B+C+ A.A’) (A+C’+ B.B’)
= (A+B’+C) (A+B’+C’) (A+B+C) (A’+B+C) (A+B+C’) (A+B’+C’)
= (A+B’+C) (A+B’+C’) (A+B+C) (A’+B+C) (A+B+C’)
= M2. M3. M0. M4. M1
= ∏M (0, 1, 2, 3, 4)
5. Y= xy+ x’z
= (xy+ x’) (xy+ z)
= (x+x’) (y+x’) (x+z) (y+z)
Using distributive law, convert the function into OR terms.
[x+ x’=1]
= (x’+y) (x+z) (y+z)
= (x’+y+ z.z’) (x+z+y.y’) (y+z+ x.x’)
= (x’+ y+ z) (x’+ y+ z’) (x+ y+ z) (x+ y’+ z) (x+ y+ z) (x’+ y+ z)
SMVEC/CSE/U23ITIC01/DDSA/UNIT 1
42
= (x’+ y+ z) (x’+ y+ z’) (x+ y+ z) (x+ y’+ z)
= M4. M5. M0. M2
= ∏M (0, 2, 4, 5).
Practice problems:
1. Show that 𝑨 + 𝑨𝑩 = 𝑨 + 𝑩
A + AB = A. 1 + AB
= A (1 + B) + AB
∵ [( 𝐴. 1) = 𝐴]
∵ [( 𝐴 + 1) = 1]
= A + AB + AB
= A + B (A + A)
= A + (B. 1)
𝐀 + 𝐀𝐁 = 𝐀 + 𝐁
Using the truth Table,
A
B
A+B
0
0
1
1
2.
0
1
0
1
0
1
1
1
∵ [( 𝐴 + 1) = 1]
𝑨𝑩
0
1
0
0
𝑨+ 𝑨𝑩
0
1
1
1
Prove that 𝑭 = 𝑨. 𝑩 + 𝑨. 𝑩 is exclusive OR operation and it equals (𝑨. 𝑩). 𝑨. (𝑨. 𝑩). 𝑩 [May’13]
(𝐴. 𝐵). 𝐴. (𝐴. 𝐵). 𝐵 = (𝐴. 𝐵). 𝐴 + (𝐴. 𝐵). 𝐵
∵ 𝐵𝑦 𝐷𝑒 𝑚𝑜𝑟𝑔𝑎𝑛′ 𝑠 𝐿𝑎𝑤
= (𝐴. 𝐵). 𝐴 + (𝐴. 𝐵). 𝐵
∵ [ 𝐴 = A]
= (𝐴 + 𝐵). 𝐴 + (𝐴 + 𝐵). 𝐵
∵ 𝐵𝑦 𝐷𝑒 𝑚𝑜𝑟𝑔𝑎𝑛′ 𝑠 𝐿𝑎𝑤
= 𝐴. 𝐴 + 𝐴. 𝐵 + 𝐴. 𝐵 + 𝐵. 𝐵
= 0 + 𝐴. 𝐵 + 𝐴. 𝐵 + 0
∵ [( 𝐴. 𝐴) = 0]
= 𝐴. 𝐵 + 𝐴. 𝐵
∵ [( 𝐴 + 0) = 𝐴]
(𝑨. 𝑩). 𝑨. (𝑨. 𝑩). 𝑩 = 𝑨. 𝑩 + 𝑨. 𝑩
3. Reduce the given expressions using Boolean algebra:
i) x’y’z’+x’y’z+x’yz+xy’z+xyz
ii) abc’+ab’c+a’bc+abc
iii) p’q’r+p’qr’+p’qr+pqr’+pq’r’ [May’11]
𝑖).
𝑥̅ 𝑦̅𝑧̅ + 𝑥̅ 𝑦̅𝑧 + 𝑥̅ 𝑦𝑧 + 𝑥𝑦̅𝑧 + 𝑥𝑦𝑧 = 𝑥̅ 𝑦̅[𝑧̅ + 𝑧] + 𝑦𝑧[ 𝑥 + 𝑥̅ ] + 𝑥𝑦̅𝑧
= 𝑥̅ 𝑦̅ + 𝑦𝑧 + 𝑥𝑦̅𝑧
= 𝑦̅[𝑥̅ + 𝑥𝑧] + 𝑦𝑧
= 𝑦̅[𝑥̅ + 𝑧] + 𝑦𝑧
= 𝑥̅ 𝑦̅ + 𝑦̅𝑧 + 𝑦𝑧
= 𝑥̅ 𝑦̅ + 𝑧[𝑦̅ + 𝑦]
̅𝒚
̅𝒛̅ + 𝒙
̅𝒚
̅𝒛 + 𝒙
̅𝒚𝒛 + 𝒙𝒚
̅𝒛 + 𝒙𝒚𝒛 = 𝒙
̅𝒚
̅+𝒛
𝒙
𝑖𝑖).
̅ + 𝑐] + 𝑎𝑏̅ 𝑐 + 𝑎̅ 𝑏𝑐
𝑎𝑏𝑐̅ + 𝑎𝑏̅ 𝑐 + 𝑎̅ 𝑏𝑐 + 𝑎𝑏𝑐 = 𝑎𝑏[𝑐
= 𝑎𝑏 + 𝑎𝑏̅ 𝑐 + 𝑎̅ 𝑏𝑐
= 𝑎[𝑏 + 𝑏̅ 𝑐] + 𝑎̅ 𝑏𝑐
SMVEC/CSE/U23ITIC01/DDSA/UNIT 1
∵ [( 𝐴 + 𝐴) = 1]
∵ [( 𝐴 + 𝐴𝐵) = [𝐴 + 𝐵]
∵ [( 𝐴 + 𝐴) = 1]
43
= 𝑎[𝑏 + 𝑐] + 𝑎̅ 𝑏𝑐
= 𝑎𝑏 + 𝑎𝑐 + 𝑎̅ 𝑏𝑐
= 𝑎𝑏 + 𝑐[𝑎 + 𝑎̅ 𝑏]
= 𝑎𝑏 + 𝑐[𝑎 + 𝑏]
𝒂𝒃𝒄̅ + 𝒂𝒃̅ 𝒄 + 𝒂̅ 𝒃𝒄 + 𝒂𝒃𝒄 = 𝒂𝒃 + 𝒂𝒄 + 𝒃𝒄
∵ [( 𝐴 + 𝐴𝐵) = [𝐴 + 𝐵]
𝑖𝑖𝑖). 𝑝̅ 𝑞̅ 𝑟 + 𝑝̅ 𝑞𝑟̅ + 𝑝̅ 𝑞𝑟 + 𝑝𝑞𝑟̅ + 𝑝𝑞̅ 𝑟̅ = 𝑝̅ 𝑞̅ 𝑟 + 𝑞𝑟̅ [𝑝̅ + 𝑝] + 𝑝̅ 𝑞𝑟 + 𝑝𝑞̅ 𝑟̅
= 𝑝̅ 𝑞̅ 𝑟 + 𝑞𝑟̅ + 𝑝̅ 𝑞𝑟 + 𝑝𝑞̅ 𝑟̅
= 𝑝̅ 𝑞̅ 𝑟 + 𝑞[ 𝑟̅ + 𝑝̅ 𝑟] + 𝑝𝑞̅ 𝑟̅
= 𝑝̅ 𝑞̅ 𝑟 + 𝑞[ 𝑟̅ + 𝑝̅ ] + 𝑝𝑞̅ 𝑟̅
= 𝑝̅ 𝑞̅ 𝑟 + 𝑞 𝑟̅ + 𝑝̅ 𝑞 + 𝑝𝑞̅ 𝑟̅
= 𝑝̅ [𝑞̅ 𝑟 + 𝑞] + 𝑟̅ [𝑞 + 𝑝𝑞̅ ]
̅𝒒
̅𝒓 + 𝒑
̅ 𝒒𝒓̅ + 𝒑
̅ 𝒒𝒓 + 𝒑𝒒𝒓̅ + 𝒑𝒒
̅𝒓̅ = 𝒑
̅ [𝒒 + 𝒓] + 𝒓̅ [𝒑 + 𝒒]
𝒑
4. Simplify the following expression
i) Y = (A + B)(A + C’) (B' + C’)
ii) (X + Y' + XY)(X + Y') (X'Y) = 0
𝑖). 𝑌 = (𝐴 + 𝐵)(𝐴 + 𝐶̅ )(𝐵̅ + 𝐶̅ )
= (𝐴𝐴 + 𝐴𝐶̅ + 𝐴𝐵 + 𝐵𝐶̅ )(𝐵̅ + 𝐶̅ )
= (𝐴 + 𝐴𝐶̅ + 𝐴𝐵 + 𝐵𝐶̅ )(𝐵̅ + 𝐶̅ )
= (𝐴[1 + 𝐶̅ + 𝐵] + 𝐵𝐶̅ )(𝐵̅ + 𝐶̅ )
= (𝐴[1] + 𝐵𝐶̅ )(𝐵̅ + 𝐶̅ )
= (𝐴 + 𝐵𝐶̅ )(𝐵̅ + 𝐶̅ )
= 𝐴𝐵̅ + 𝐵𝐵̅𝐶̅ + 𝐴𝐶̅ + 𝐵𝐶̅ 𝐶̅
= 𝐴𝐵̅ + 0 + 𝐴𝐶̅ + 𝐵𝐶̅
̅ + 𝑩𝑪
̅
̅ + 𝑨𝑪
𝒀 = 𝑨𝑩
∵ [( 𝐴 + 𝐴) = 1]
∵ [( 𝐴 + 𝐴𝐵) = [𝐴 + 𝐵]
∵ [( 𝐴. 𝐴) = 𝐴]
∵ [( 𝐴 + 1) = 1]
∵ [( 𝐴. 1) = 𝐴]
∵ [( 𝐴. 𝐴̅) = 0]
∵ [( 𝐴 + 0) = 𝐴]
𝑖𝑖). (𝑋 + 𝑌̅ + 𝑋𝑌)(𝑋 + 𝑌̅)(𝑋̅𝑌) = 0
= (𝑋 + 𝑌̅ + 𝑋𝑌)(𝑋𝑋̅𝑌 + 𝑋̅𝑌𝑌̅)
= (𝑋 + 𝑌̅ + 𝑋𝑌)(0 + 0)
∵ [( 𝐴. 𝐴̅) = 0]
̅ + 𝑿𝒀)(𝟎)
= (𝑿 + 𝒀
∵ [( 𝟎 + 𝟎) = 𝟎]
̅ + 𝑿𝒀)(𝑿 + 𝒀
̅ )(𝑿
̅ 𝒀) = 𝟎
(𝑿 + 𝒀
∵ [( 𝐴. 0) = 0]
5. Prove that the expression ABC + ABC' + AB'C + A'BC = AB + AC + BC. (Apr’18)
𝐴𝐵𝐶 + 𝐴𝐵𝐶̅ + 𝐴𝐵̅𝐶 + 𝐴̅𝐵𝐶 = 𝐴𝐵[𝐶 + 𝐶̅ ] + 𝐴𝐵̅𝐶 + 𝐴̅𝐵𝐶
= 𝐴𝐵 + 𝐴𝐵̅𝐶 + 𝐴̅𝐵𝐶
∵ [( 𝐴 + 𝐴) = 1]
̅
̅
= 𝐴[𝐵 + 𝐵𝐶] + 𝐴𝐵𝐶
= 𝐴[𝐵 + 𝐶] + 𝐴̅𝐵𝐶
= 𝐴𝐵 + 𝐴𝐶 + 𝐴̅𝐵𝐶
= 𝐵[𝐴 + 𝐴̅𝐶 ] + 𝐴𝐶
= 𝐵[𝐴 + 𝐶] + 𝐴𝐶
̅ + 𝑨𝑩
̅ 𝑩𝑪 = 𝑨𝑩 + 𝑩𝑪 + 𝑨𝑪
̅𝑪 + 𝑨
𝑨𝑩𝑪 + 𝑨𝑩𝑪
∵ [( 𝐴 + 𝐴𝐵) = [𝐴 + 𝐵]
9. Convert the given expression in canonical SOP form Y=AC+AB+BC. (Apr’18)
𝑌 = 𝐴𝐵 + 𝐵𝐶 + 𝐴𝐶
= 𝐴𝐵[𝐶 + 𝐶̅ ] + [𝐴 + 𝐴̅] 𝐵𝐶 + 𝐴[𝐵 + 𝐵̅]𝐶
= 𝐴𝐵𝐶 + 𝐴𝐵 𝐶̅ + 𝐴 𝐵𝐶 + 𝐴̅ 𝐵𝐶 + 𝐴𝐵𝐶 + 𝐴𝐵̅𝐶
̅+𝑨
̅ 𝑩𝑪 + 𝑨𝑩
̅𝑪
𝒀 = 𝑨𝑩𝑪 + 𝑨𝑩 𝑪
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8.
8.1
KARNAUGH MAPS (10 Marks or 5 Marks)
Introduction
The simplification of the functions using Boolean laws and theorems becomes complex with the
increase in the number of variables and terms. The map method, first proposed by Veitch and slightly
improvised by Karnaugh, provides a simple, straightforward procedure for the simplification of Boolean
functions. The method is called Veitch diagram or Karnaugh map, which may be regarded as a pictorial
representation of a truth table.
The Karnaugh map technique provides a systematic method for simplifying and manipulation of
Boolean expressions. A K-map is a diagram made up of squares, with each square representing one
minterm of the function that is to be minimized. For n variables on a Karnaugh map there are 2n numbers of
squares. Each square or cell represents one of the minterms. It can be drawn directly from either minterm
(sum-of-products) or maxterm (product-of-sums) Boolean expressions.
Two- Variable, Three Variable and Four Variable Maps
Karnaugh maps can be used for expressions with two, three, four and five variables. The number of
cells in a Karnaugh map is equal to the total number of possible input variable combinations as is the
number of rows in a truth table. For three variables, the number of cells is 23 = 8. For four variables, the
number of cells is 24 = 16.
Product terms are assigned to the cells of a K-map by labeling each row and each column of a map
with a variable, with its complement or with a combination of variables & complements. The below figure
shows the way to label the rows & columns of a 1, 2, 3 and 4- variable maps and the product terms
corresponding to each cell.
It is important to note that when we move from one cell to the next along any row or from one cell to
the next along any column, one and only one variable in the product term changes (to a complement or to
SMVEC/CSE/U23ITIC01/DDSA/UNIT 1
45
an uncomplemented form). Irrespective of number of variables the labels along each row and column must
conform to a single change.
Hence gray code is used to label the rows and columns of K-map as shown below.
8.2
Grouping cells for Simplification
The grouping is nothing but combining terms in adjacent cells. The simplification is achieved by
grouping adjacent 1’s or 0’s in groups of 2i, where i = 1, 2, …, n and n is the number of variables. When
adjacent 1’s are grouped then we get result in the sum of product form; otherwise we get result in the
product of sum form.
Grouping Two Adjacent 1’s: (Pair)
In a Karnaugh map we can group two adjacent 1’s. The resultant group is called Pair.
SMVEC/CSE/U23ITIC01/DDSA/UNIT 1
46
Examples of Pairs
Grouping Four Adjacent 1’s: (Quad)
In a Karnaugh map we can group four adjacent 1’s. The resultant group is called Quad. Fig (a)
shows the four 1’s are horizontally adjacent and Fig (b) shows they are vertically adjacent. Fig (c) contains
four 1’s in a square, and they are considered adjacent to each other.
Examples of Quads
The four 1’s in fig (d) and fig (e) are also adjacent, as are those in fig (f) because, the top and
bottom rows are considered to be adjacent to each other and the leftmost and rightmost columns are also
adjacent to each other.
Grouping Eight Adjacent 1’s: (Octet)
In a Karnaugh map we can group eight adjacent 1’s. The resultant group is called Octet.
SMVEC/CSE/U23ITIC01/DDSA/UNIT 1
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8.3
Simplification of Sum of Products Expressions: (Minimal Sums)
The generalized procedure to simplify Boolean expressions as follows:
1. Plot the K-map and place 1’s in those cells corresponding to the 1’s in the sum of product
expression. Place 0’s in the other cells.
2. Check the K-map for adjacent 1’s and encircle those 1’s which are not adjacent to any other 1’s.
These are called isolated 1’s.
3. Check for those 1’s which are adjacent to only one other 1 and encircle such pairs.
4. Check for quads and octets of adjacent 1’s even if it contains some 1’s that have already been
encircled. While doing this make sure that there are minimum number of groups.
5. Combine any pairs necessary to include any 1’s that have not yet been grouped.
6. Form the simplified expression by summing product terms of all the groups.
8.4
Three- Variable Map
1. Simplify the Boolean expression,
F(x, y, z) = ∑m (3, 4, 6, 7).
Soln:
F = yz+ xz’
2. F(x, y, z) = ∑m (0, 2, 4, 5, 6).
Soln:
F = z’+ xy’
3. F = A’C + A’B + AB’C + BC
= A’C (B+ B’) + A’B (C+ C’) + AB’C + BC (A+ A’)
= A’BC+ A’B’C + A’BC + A’BC’ + AB’C + ABC + A’BC
SMVEC/CSE/U23ITIC01/DDSA/UNIT 1
48
= A’BC+ A’B’C + A’BC’ + AB’C + ABC
= m3+ m1+ m2+ m5+ m7
= ∑ m (1, 2, 3, 5, 7)
F = C + A’B
4. AB’C + A’B’C + A’BC + AB’C’ + A’B’C’
Soln:
= m5 + m1 + m3 + m4 + m0
= ∑ m (0, 1, 3, 4, 5)
F = A’C + B’
1.5.5
Four - Variable Map
1. Simplify the Boolean expression,
Y = A’BC’D’ + A’BC’D + ABC’D’ + ABC’D + AB’C’D + A’B’CD’
Soln:
Therefore,
Y= A’B’CD’+ AC’D+ BC’
2. F (w, x, y, z) = ∑ m(0, 1, 2, 4, 5, 6, 8, 9, 12, 13, 14)
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49
Therefore,
F= y’+ w’z’+ xz’
3. F= A’B’C’+ B’CD’+ A’BCD’+ AB’C’
= A’B’C’ (D+ D’) + B’CD’ (A+ A’) + A’BCD’+ AB’C’ (D+ D’)
= A’B’C’D+ A’B’C’D’+ AB’CD’+ A’B’CD’+ A’BCD’+ AB’C’D+ AB’C’D’
= m1+ m0+ m10+ m2+ m6+ m9+ m8
= ∑ m (0, 1, 2, 6, 8, 9, 10)
Therefore,
F= B’D’+ B’C’+ A’CD’.
4. Y= ABCD+ AB’C’D’+ AB’C+ AB
= ABCD+ AB’C’D’+ AB’C (D+D’)+ AB (C+C’) (D+D’)
= ABCD+ AB’C’D’+ AB’CD+ AB’CD’+ (ABC+ ABC’) (D+ D’)
= ABCD+ AB’C’D’+ AB’CD+ AB’CD’+ ABCD+ ABCD’+ ABC’D+ ABC’D’
= ABCD+ AB’C’D’+ AB’CD+ AB’CD’+ ABCD’+ ABC’D+ ABC’D’
= m15+ m8+ m11+ m10+ m14+ m13+ m12
= ∑ m (8, 10, 11, 12, 13, 14, 15)
Therefore,
SMVEC/CSE/U23ITIC01/DDSA/UNIT 1
Y= AB+ AC+ AD’
50
5. Y (A, B, C, D)= ∑ m (7, 9, 10, 11, 12, 13, 14, 15)
Therefore,
Y= AB+ AC+ AD+BCD.
6. Y= A’B’C’D+ A’BC’D+ A’BCD+ A’BCD’+ ABC’D+ ABCD+ AB’CD
= m1+ m5+ m7+ m6+ m13+ m15+ m11
= ∑ m (1, 5, 6, 7, 11, 13, 15)
In the above K-map, the cells 5, 7, 13 and 15 can be grouped to form a quad as indicated by the
dotted lines. In order to group the remaining 1’s, four pairs have to be formed. However, all the four 1’s
covered by the quad are also covered by the pairs. So, the quad in the above k-map is redundant.
Therefore, the simplified expression will be,
Y = A’C’D+ A’BC+ ABD+ ACD.
7. Y= ∑ m (1, 5, 10, 11, 12, 13, 15)
Therefore, Y= A’C’D+ ABC’+ ACD+ AB’C.
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8. Y= A’B’CD’+ ABCD’+ AB’CD’+ AB’CD+ AB’C’D’+ ABC’D’+ A’B’CD+ A’B’C’D’
Therefore, Y= AD’+ B’C+ B’D’
9. F (A, B, C, D) = ∑ m (0, 1, 4, 8, 9, 10)
Therefore, F= A’C’D’+ AB’D’+ B’C’.
Simplification of Sum of Products Expressions: (Minimal Sums)
1. Y= (A+ B+ C’) (A+ B’+ C’) (A’+ B’+ C’) (A’+ B+ C) (A+ B+ C)
= M1. M3. M7. M4. M0
=∏ M (0, 1, 3, 4, 7)
= ∑ m (2, 5, 6)
Y’ = B’C’+ A’C+ BC.
Y= Y” = (B’C’+ A’C+ BC)’
= (B’C’)’. (A’C)’. (BC)’
= (B”+ C”). (A”+C’). (B’+ C’)
Therefore, Y = (B+ C). (A+C’). (B’+ C’)
2. Y= (A’+ B’+ C+ D) (A’+ B’+ C’+ D) (A’+ B’+ C’+ D’) (A’+ B+ C+ D)
=(A+ B’+ C’+ D) (A+ B’+ C’+ D’) (A+ B+ C+ D) (A’+ B’+ C+ D’)
= M12. M14. M15. M8. M6. M7. M0. M13
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= ∏M (0, 6, 7, 8, 12, 13, 14, 15)
Y’ = B’C’D’+ AB+ BC
Y= Y” = (B’C’D’+ AB+ BC)’
= (B’C’D’)’. (AB)’. (BC)’
= (B”+ C”+D”). (A’+B’). (B’+ C’)
= (B+ C+ D). (A’+ B’). (B’+ C’)
Therefore, Y= (B+ C+ D). (A’+ B’). (B’+ C’)
3. F(A, B, C, D)= ∏M (0, 2, 3, 8, 9, 12, 13, 14, 15) (aug 21)
Soln:
Y’ = A’B’D’+ A’B’C+ ABD+ AC’
Y= Y” = (A’B’D’+ A’B’C+ ABD+ AC’)’
= (A’B’D’)’. (A’B’C)’. (ABD)’. (AC’)’
= (A”+ B”+ D”). (A”+ B”+C’). (A’+ B’+ D’). (A’+ C”)
= (A+ B+ D). (A+ B+ C’). (A’+ B’+ D’). (A’+ C)
Therefore, Y= (A+ B+ D). (A+ B+ C’). (A’+ B’+ D’). (A’+ C)
4. F(A, B, C, D)= ∑m (0, 1, 2, 5, 8, 9, 10) (aug 21)
= ∏M (3, 4, 6, 7, 11, 12, 13, 14, 15)
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Y’ = BD’+ CD+ AB
Y= Y” = (BD’+ CD+ AB)’
= (BD’)’. (CD)’. (AB)’
= (B’+ D”). (C’+ D’). (A’+ B’)
= (B’+ D). (C’+ D’). (A’+ B’)
Therefore, Y= (B’+ D). (C’+ D’). (A’+ B’)
8.5 Don’t care Conditions:
A don’t care minterm is a combination of variables whose logical value is not specified. When
choosing adjacent squares to simplify the function in a map, the don’t care minterms may be assumed to
be either 0 or 1. When simplifying the function, we can choose to include each don’t care minterm with
either the 1’s or the 0’s, depending on which combination gives the simplest expression.
1. F (x, y, z) = ∑m (0, 1, 2, 4, 5)+ ∑d (3, 6, 7)
Soln:
F (x, y, z) = 1
2. F (w, x, y, z) = ∑m (1, 3, 7, 11, 15)+ ∑d (0, 2, 5)
F (w, x, y, z) = w’x’+ yz
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3. F (w, x, y, z) = ∑m (0, 7, 8, 9, 10, 12)+ ∑d (2, 5, 13) (Dec 2021)
Soln:
F (w, x, y, z) = w’xz+ wy’+ x’z’.
4. F (w, x, y, z) = ∑m (0, 1, 4, 8, 9, 10)+ ∑d (2, 11)
Soln:
F (w, x, y, z) = wx’+ x’y’+ w’y’z’.
5. F( A, B, C, D) = ∑m (0, 6, 8, 13, 14)+ ∑d (2, 4, 10)
F( A, B, C, D) = CD’+ B’D’+ A’B’C’D’
Practice problems
1. Reduce the following expression using K-map f=x’y’z+w’xz+wxyz’+wxz+w’xyz. [May’11]
𝑓 = 𝑥̅ 𝑦̅ 𝑧 + 𝑤
̅ 𝑥𝑧 + 𝑤𝑥𝑦𝑧̅ + wxz + 𝑤
̅ 𝑥𝑦𝑧
= [𝑤 + 𝑤
̅] 𝑥̅ 𝑦̅ 𝑧 + 𝑤
̅ 𝑥[ 𝑦 + 𝑦̅]𝑧 + 𝑤𝑥𝑦𝑧̅ + wx[y + 𝑦̅]z + 𝑤
̅ 𝑥𝑦𝑧
= 𝑤𝑥̅ 𝑦̅ 𝑧 + 𝑤
̅𝑥̅ 𝑦̅ 𝑧 + 𝑤
̅ 𝑥𝑦𝑧 + 𝑤
̅ 𝑥𝑦̅𝑧 + 𝑤𝑥𝑦𝑧̅ + wxyz + wx 𝑦̅z + 𝑤
̅ 𝑥𝑦𝑧
= 𝑤𝑥̅ 𝑦̅ 𝑧 + 𝑤
̅𝑥̅ 𝑦̅ 𝑧 + 𝑤
̅ 𝑥𝑦𝑧 + 𝑤
̅ 𝑥𝑦̅𝑧 + 𝑤𝑥𝑦𝑧̅ + wxyz + wx 𝑦̅z
= 1001 + 0001 + 0111 + 0101 + 1110 + 1111 + 1101
= 𝑚9 + 𝑚1 + 𝑚7 + 𝑚5 + 𝑚14 + m15 + m13
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𝑓 = Σ𝑚( 1, 5, 7, 9, 13, 14, 15)
K-map:
𝒇 = 𝒘𝒛𝒚 + 𝒙𝒛 + 𝒚𝒛
2. Given the following Boolean function F = A’C + A’B + AB’C + BC.
(i) Express it in sum of minterms.
(ii) Find the minimal sum of products expression. [Nov’18]
̅C + A
̅ B + A B̅ C + BC
F= A
̅ [B + B
̅ B[C + C̅] + A B̅ C + [A + A
̅ ]BC
̅]C + A
= A
̅ BC + A
̅B
̅BC+ A
̅ BC̅ + A B̅ C + ABC + A
̅ BC
̅C + A
= A
̅
̅
̅
̅
̅
̅
= A BC + A BC + A BC + A B C + ABC
= 011 + 001 + 010 + 101 + 111
= m3 + m1 + m2 + m5 + m7
F(A, B, C) = Σm(1, 2, 3, 5, 7)
K- Map:
̅𝐁
𝐅(𝐀, 𝐁, 𝐂) = 𝐂 + 𝐀
3. Plot the logical expression ABCD+A𝑩𝑪𝑫 +A𝑩C+AB on a 4-variable k-map,obtain the
simplified expression from the map.(Nov’17)
F = ABCD + A𝐵 𝐶 𝐷 + A𝐵C + AB
= ABCD + A𝐵 𝐶 𝐷 + A𝐵C[D + 𝐷] + AB[C + 𝐶][D + 𝐷]
= ABCD + A𝐵 𝐶 𝐷 + A𝐵CD + A𝐵C𝐷 + ABC[D + 𝐷] + AB𝐶[D + 𝐷]
= ABCD + A𝐵 𝐶 𝐷 + A𝐵CD + A𝐵C𝐷 + ABCD + ABC𝐷 + AB𝐶D + AB𝐶 𝐷
= 1111 + 1000 + 1011 + 1010 + 1111 + 1110 + 1101 + 1100
= m15, m8, m1, m15, m14, m13, m12
𝐹 = Σ𝑚( 8,11,12,13,14,15)
̅𝑫
𝑭 = 𝑨𝑩 + 𝑨𝑪
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4. Simplify the following using K map. X=A’B+A’B’C+ABC’+AB’C’
𝑋(𝐴, 𝐵, 𝐶) = 𝐴𝐵 + 𝐴 𝐵𝐶 + 𝐴𝐵 𝐶 + 𝐴 𝐵 𝐶
= 𝐴𝐵[𝐶 + 𝐶] + 𝐴 𝐵𝐶 + 𝐴𝐵 𝐶 + 𝐴 𝐵 𝐶
= 𝐴𝐵𝐶 + 𝐴𝐵𝐶 + 𝐴 𝐵𝐶 + 𝐴𝐵 𝐶 + 𝐴 𝐵 𝐶
= 011 + 010 + 001 + 110 + 100
= 𝑚3 + 𝑚2 + 𝑚1 + 𝑚6 + 𝑚4
𝑋(𝐴, 𝐵, 𝐶) = Σ𝑚(1, 2, 3, 4, 6 )
𝑿(𝑨, 𝑩, 𝑪) = 𝑨𝑪 + 𝑩 𝑪 + 𝑨 𝑪
5. Obtain the canonical POS for F (A, B, C) = (A+B’) (B+C) (A+C’) ii) Apply De Morgan theorem
for the function [(A+B+C) D]’ iii) Find the compliment of A+BC+AB.
[𝐴 + 𝐵̅][ 𝐵 + 𝐶][𝐴 + 𝐶̅ ]
𝑖).
= [𝐴 + 𝐵̅ + (𝐶. 𝐶̅ )][ (𝐴. 𝐴̅) + 𝐵 + 𝐶][𝐴 + (𝐵. 𝐵̅) + 𝐶̅ ]
= [(𝐴 + 𝐵̅ + 𝐶)(𝐴 + 𝐵̅ + 𝐶̅ )][ (𝐴 + 𝐵 + 𝐶). (𝐴̅ + 𝐵 + 𝐶)][(𝐴 + 𝐵 + 𝐶̅ )(𝐴 + 𝐵̅ + 𝐶̅ )]
̅ )(𝑨 + 𝑩 + 𝑪)(𝑨 + 𝑩 + 𝑪
̅ ). (𝑨
̅ + 𝑩 + 𝑪)
̅ + 𝑪)(𝑨 + 𝑩
̅ +𝑪
𝑭(𝑨, 𝑩, 𝑪) = (𝑨 + 𝑩
̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅
[(𝐴 + 𝐵 + 𝐶)𝐷]
̅
[(𝐴 + 𝐵 + 𝐶)] + 𝐷
= ̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅
̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅
̅𝑩
̅+ 𝑫
̅𝑪
̅
[(𝑨 + 𝑩 + 𝑪)𝑫] = 𝑨
̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅
[𝐴 + 𝐵𝐶 + 𝐴𝐵]
̅̅̅̅̅̅̅. (𝐴𝐵)
̅̅̅̅̅̅̅
= 𝐴̅ . (𝐵𝐶)
= 𝐴̅ . (𝐵̅ + 𝐶̅ ) (𝐴̅ + 𝐵̅)
= (𝐴̅𝐵̅ + 𝐴̅𝐶̅ ) (𝐴̅ + 𝐵̅)
= 𝐴̅ 𝐴̅𝐵̅ + 𝐴̅𝐴̅𝐶̅ + 𝐴̅𝐵̅𝐵̅ + 𝐴̅𝐵̅𝐶̅
= 𝐴̅𝐵̅ + 𝐴̅𝐶̅ + 𝐴̅𝐵̅ + 𝐴̅𝐵̅𝐶̅
̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅
̅𝑩
̅𝑪
̅+𝑨
̅𝑩
̅
̅ +𝑨
̅𝑪
[𝑨
+ 𝑩𝑪 + 𝑨𝑩] = 𝑨
𝑖𝑖).
𝑖𝑖𝑖).
6. Minimize the four variable logic function using k-map. F(A,B,C,D) = Σm (0, 1, 2, 3, 5, 7, 8, 9,
11, 14)
𝐹(𝐴, 𝐵, 𝐶, 𝐷) = Σ𝑚(0,1, 2,3, 5, 7, 8, 9, 11, 15 )
𝑭(𝑨, 𝑩, 𝑪, 𝑫) = 𝑨 𝑩 + 𝑩 𝑪 + 𝑩𝑫 + 𝑨 𝑩𝑪𝑫
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7. Reduce the Following Boolean Function Using 4 Variable K Map. (oct 22)
F(W,X,Y,Z) =Σm (2,3,10,11,12,13,14,15)
𝑭(𝑾, 𝑿, 𝒀, 𝒁) = 𝑾𝑿 + 𝑿 𝒀
8. Simplify using K map F (W, X, Y, Z) = Σm ( 0, 1, 2, 4, 5, 6, 8, 9, 12, 13, 14) [ Apr’10]
𝐅( 𝐰, 𝐱, 𝐲, 𝐳) = 𝐲 + 𝐱𝐳 + 𝐰𝐳
9. Reduce the following function using Karnaugh map technique. F(A,B,C,D) = Σm (5, 6, 7,
12,13) +Σd (4,9,14,15).
F=B
10. Simplify the following function using karnaugh map. F (W, X, Y, Z) = Σm (0, 1, 3, 9, 10, 12, 13,
14) + Σd (2, 5, 6, 11) [May’16]
𝑭(𝑾, 𝑿, 𝒀, 𝒁) = 𝑾 𝑿 + 𝒀𝒁 + 𝑾𝑿𝒁 + 𝒀𝒁
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11. Reduce the following minterms using k-map. F (w, x, y, z) =Σm (0, 1, 3, 5, 6, 7, 8, 12, 14) + Σd
(9, 15). [May’17]
𝑭(𝑿, 𝒀, 𝒁) = 𝑾𝒁 + 𝑿𝒀 + 𝑿 𝒀 + 𝑾𝒀 𝒁
12. Simplify the Boolean function using K-map F (W,X,Y,Z) =Σ (1, 3, 7, 11, 15) which has the don’t
care conditions d (W,X,Y,Z)= Σ (0, 2, 5). [May’13]
𝑭(𝑾, 𝑿, 𝒀, 𝒁) = 𝒘 𝒙 + 𝐲𝐳
13. Simplify using k-map F (A,B,C,D) = Σm (7,8,9) + d(10,11,12,13,14,15) [Nov’13]
𝑭(𝑨, 𝑩, 𝑪, 𝑫) = 𝑨 + 𝑩𝑪𝑫
14. Use K-map to obtain the POS forms for F=A’B’D’+A’CD+A’BC’D+AB’D’.
̅ + 𝐴̅ 𝐶𝐷 + 𝐴̅ 𝐵𝐶̅ 𝐷 + 𝐴𝐵̅𝐷
̅
𝐹 = 𝐴̅𝐵̅𝐷
̅ + 𝐴̅ [𝐵 + 𝐵̅]𝐶𝐷 + 𝐴̅ 𝐵𝐶̅ 𝐷 + 𝐴𝐵̅[𝐶 + 𝐶̅ ]𝐷
̅
= 𝐴̅𝐵̅[𝐶 + 𝐶̅ ]𝐷
̅ + 𝐴̅𝐵̅𝐶̅ 𝐷
̅ + 𝐴̅𝐵𝐶𝐷 + 𝐴̅𝐵̅𝐶𝐷 + 𝐴̅ 𝐵𝐶̅ 𝐷 + 𝐴𝐵̅𝐶𝐷
̅ + 𝐴𝐵̅𝐶̅ 𝐷
̅
= 𝐴̅𝐵̅𝐶𝐷
= 0010 + 0000 + 0111 + 0011 + 0101 + 1010 + 1000
= 𝑚2 + 𝑚0 + 𝑚7 + 𝑚3 + 𝑚5 + 𝑚10 + 𝑚8
𝐹(𝐴, 𝐵, 𝐶, 𝐷) = Σ𝑚( 0, 2, 3, 5, 7, 8, 10)
𝐹𝑜𝑟 𝑃𝑂𝑆 𝑓𝑜𝑟𝑚
𝐹(𝐴, 𝐵, 𝐶, 𝐷) = 𝜋𝑀(1, 4, 6, 9, 11, 12, 13,14, 15)
SMVEC/CSE/U23ITIC01/DDSA/UNIT 1
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̅+ 𝑫
̅+ 𝑩
̅ ][ 𝑨
̅ ][𝑨
̅ ][ 𝑩
̅ + 𝑫]
𝑭 = [𝑩 + 𝑪 + 𝑫
15. Reduce the following using k map F (A, B, C, D) = П(0, 3, 4, 7, 8, 10, 12, 14) + d (2, 6)
𝑭(𝑨, 𝑩, 𝑪, 𝑫) = 𝑫[ 𝑨 + 𝑪]
16. Reduce the following function using K-map, f(A,B,C,D) = π M(0,2,3,8,9,12,13,15) [ Apr’15]
F(A,B,C,D) = π M(0,2,3,8,9,12,13,15)
̅ + 𝑪][𝑨
̅+ 𝑩
̅ ][𝑨
̅+ 𝑩
̅ +𝑪
̅+ 𝑫
̅ ][ 𝑩 + 𝑪 + 𝑫]
𝑭(𝑨, 𝑩, 𝑪, 𝑫) = [ 𝑨
i. A’B’C+A’B’C’+A’BC+AB’C+ABC.
ii.
(A+B) (A+C) (B+C’)
iii. F=XY+X’Z
𝑖).
𝐴̅ 𝐵̅ 𝐶 + 𝐴̅ 𝐵̅ 𝐶̅ + 𝐴̅ 𝐵 𝐶 + 𝐴 𝐵̅ 𝐶 + 𝐴𝐵𝐶
= 001 + 000 + 011 + 101 + 111
= 𝑚1 + 𝑚0 + 𝑚3 + 𝑚5 + 𝑚7
𝐹(𝐴, 𝐵, 𝐶) = Σ𝑚(0, 1, 3, 5, 7)
𝐹𝑜𝑟 𝑃𝑂𝑆 𝑓𝑜𝑟𝑚
𝐹(𝐴, 𝐵, 𝐶) = 𝜋𝑀(2, 4, 6)
̅ + 𝑪]
̅ + 𝑪] [𝑨
𝑭(𝑨, 𝑩, 𝑪, 𝑫) = [𝑩
SMVEC/CSE/U23ITIC01/DDSA/UNIT 1
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𝑖𝑖).
𝑖𝑖𝑖).
[𝐴 + 𝐵][ 𝐴 + 𝐶][𝐵 + 𝐶̅ ]
= [𝐴 + 𝐵 + (𝐶. 𝐶̅ ) ][ 𝐴 + (𝐵. 𝐵̅) + 𝐶][(𝐴. 𝐴̅) + 𝐵 + 𝐶̅ ]
= [(𝐴 + 𝐵 + 𝐶)(𝐴 + 𝐵 + 𝐶̅ ) ][ (𝐴 + 𝐵 + 𝐶)(𝐴 + 𝐵̅ + 𝐶)][(𝐴 + 𝐵 + 𝐶̅ ) (𝐴̅ + 𝐵 + 𝐶̅ )]
̅ )(𝑨 + 𝑩
̅+𝑩+𝑪
̅)
̅ + 𝑪)( 𝑨
𝑭(𝑨, 𝑩, 𝑪) = (𝑨 + 𝑩 + 𝑪)(𝑨 + 𝑩 + 𝑪
𝐹 = 𝑋𝑌 + 𝑋̅𝑍
= 𝑋𝑌[𝑍 + 𝑍̅] + 𝑋̅ [𝑌 + 𝑌̅]𝑍
= 𝑋𝑌𝑍 + 𝑋𝑌𝑍̅ + 𝑋̅𝑌𝑍 + 𝑋̅𝑌̅𝑍
= 111 + 110 + 011 + 001
= 𝑚7 + 𝑚6 + 𝑚3 + 𝑚1
𝐹 = Σ𝑚( 1, 3, 6, 7)
𝐹𝑜𝑟 𝑃𝑂𝑆 𝑓𝑜𝑟𝑚
𝐹 = 𝜋𝑀(0, 2, 4, 5)
𝐹 = [𝑋 + 𝑌] [𝑋̅ + 𝑌]
17. Express the function F=A+B’C in Canonical SOP form and Canonical POS form [Nov’13,
Nov’17]
𝐹 = 𝐴 + 𝐵̅ 𝐶
= [𝐴 [ 𝐵 + 𝐵̅][ 𝐶 + 𝐶̅ ]] + [ 𝐴 + 𝐴̅]𝐵̅ 𝐶
= [[𝐴 𝐵 + 𝐴 𝐵̅][ 𝐶 + 𝐶̅ ]] + [ 𝐴𝐵̅ 𝐶 + 𝐴̅𝐵̅ 𝐶]
= 𝐴 𝐵 𝐶 + 𝐴 𝐵̅𝐶 + 𝐴 𝐵𝐶̅ + 𝐴 𝐵̅𝐶̅ + 𝐴𝐵̅ 𝐶 + 𝐴̅𝐵̅ 𝐶
= 𝐴 𝐵 𝐶 + 𝐴 𝐵̅𝐶 + 𝐴 𝐵𝐶̅ + 𝐴 𝐵̅𝐶̅ + 𝐴̅𝐵̅ 𝐶
= 111 + 101 + 110 + 100 + 001
= 𝑚7 + 𝑚5 + 𝑚6 + 𝑚4 + 𝑚1
= Σ𝑚( 1, 4, 5, 6, 7)
𝐹𝑜𝑟 𝑃𝑂𝑆 𝑓𝑜𝑟𝑚 = 𝜋𝑀(0, 2,3)
̅]
𝑭 = [𝑨 + 𝑪] [𝑨 + 𝑩
18. Simplify the following expression in (1) Sum of Products and (2) Products of Sums.
a). x’z’ + y’z’ + yz’ + xy
b). AC’ + B’D + A’CD + ABCD
c). (A’ + B’ + D’) (A + B’ + C’) (A’ + B +D’) (B + C’ + D’) [Nov’19]
a). x’z’ + y’z’ + yz’ + xy
𝐹(𝑥, 𝑦, 𝑧) = 𝑥̅ 𝑧̅ + 𝑦̅𝑧̅ + 𝑦𝑧̅ + 𝑥𝑦
= 𝑥̅ [𝑦 + 𝑦̅] 𝑧̅ + [𝑥 + 𝑥̅ ]𝑦̅𝑧̅ + [𝑥 + 𝑥̅ ]𝑦𝑧̅ + 𝑥𝑦[𝑧 + 𝑧̅]
= 𝑥̅ 𝑦𝑧̅ + 𝑥̅ 𝑦̅𝑧̅ + 𝑥𝑦̅𝑧̅ + 𝑥̅ 𝑦̅𝑧̅ + 𝑥𝑦𝑧̅ + 𝑥̅ 𝑦𝑧̅ + 𝑥𝑦𝑧 + 𝑥𝑦𝑧̅
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= 𝑥̅ 𝑦𝑧̅ + 𝑥̅ 𝑦̅𝑧̅ + 𝑥𝑦̅𝑧̅ + 𝑥𝑦𝑧̅ + 𝑥𝑦𝑧
= 010 + 000 + 100 + 110 + 111
= 𝑚2 + 𝑚0 + 𝑚4 + 𝑚6 + 𝑚7
𝐹(𝑥, 𝑦, 𝑧) = Σ𝑚( 0, 2, 4, 6, 7)
𝐹𝑜𝑟 𝑃𝑂𝑆 𝑓𝑜𝑟𝑚 𝐹(𝑥, 𝑦, 𝑧) = 𝜋𝑀(1, 3, 5)
𝑭(𝒙, 𝒚, 𝒛) = [𝒙 + 𝒛̅] [𝒚 + 𝒛̅]
b). AC’ + B’D + A’CD + ABCD
𝐹(𝐴, 𝐵, 𝐶, 𝐷) = 𝐴𝐶̅ + 𝐵̅𝐷 + 𝐴̅𝐶𝐷 + 𝐴𝐵𝐶𝐷
̅ ] + [𝐴 + 𝐴̅]𝐵̅[𝐶 + 𝐶̅ ]𝐷 + 𝐴̅[𝐵 + 𝐵̅]𝐶𝐷 + 𝐴𝐵𝐶𝐷
= 𝐴[𝐵 + 𝐵̅ ] 𝐶̅ [𝐷 + 𝐷
̅ ] + [𝐴𝐵̅𝐷 + 𝐴̅𝐵̅𝐷][𝐶 + 𝐶̅ ] + 𝐴̅𝐵𝐶𝐷 + 𝐴̅𝐵̅𝐶𝐷 + 𝐴𝐵𝐶𝐷
= [𝐴𝐵𝐶̅ + 𝐴𝐵̅𝐶̅ ] [𝐷 + 𝐷
̅
̅
̅
̅ + 𝐴𝐵̅𝐶̅ 𝐷
̅ + 𝐴𝐵̅𝐶𝐷 + 𝐴̅𝐵̅𝐶𝐷 + 𝐴𝐵̅𝐶̅ 𝐷 + 𝐴̅𝐵̅𝐶̅ 𝐷 + 𝐴̅𝐵𝐶𝐷
= 𝐴𝐵𝐶 𝐷 + 𝐴𝐵𝐶 𝐷 + 𝐴𝐵𝐶̅ 𝐷
+ 𝐴̅𝐵̅𝐶𝐷 + 𝐴𝐵𝐶𝐷
̅ + 𝐴𝐵̅𝐶̅ 𝐷
̅ + 𝐴𝐵̅𝐶𝐷 + 𝐴̅𝐵̅𝐶𝐷 + 𝐴̅𝐵̅𝐶̅ 𝐷 + 𝐴̅𝐵𝐶𝐷 + 𝐴𝐵𝐶𝐷
= 𝐴𝐵𝐶̅ 𝐷 + 𝐴𝐵̅𝐶̅ 𝐷 + 𝐴𝐵𝐶̅ 𝐷
= 1101 + 1001 + 1100 + 1000 + 1011 + 0011 + 0001 + 0111 + 1111
= 𝑚13 + 𝑚9 + 𝑚2 + 𝑚8 + 𝑚11 + 𝑚3 + 𝑚1 + 𝑚7 + 𝑚15
𝐹(𝐴, 𝐵, 𝐶, 𝐷) = Σ𝑚( 1, 3, 7, 8, 9, 11, 12, 13,15)
𝐹𝑜𝑟 𝑃𝑂𝑆 𝑓𝑜𝑟𝑚
𝐹(𝐴, 𝐵, 𝐶, 𝐷) = 𝜋𝑀(0, 2, 4, 5, 6, 10, 14)
̅ + 𝑫]
̅ + 𝑪][𝑪
𝑭(𝑨, 𝑩, 𝑪, 𝑫) = [𝑨 + 𝑫][𝑨 + 𝑩
c). (A’ + B’ + D’) (A + B’ + C’) (A’ + B +D’) (B + C’ + D’)
̅ ][𝐴 + 𝐵̅ + 𝐶̅ ][𝐴̅ + 𝐵 + 𝐷
̅ ][𝐵 + 𝐶̅ + 𝐷
̅]
𝐹(𝐴, 𝐵, 𝐶, 𝐷) = [𝐴̅ + 𝐵̅ + 𝐷
̅
̅
̅
̅
̅
̅
̅
̅
̅ ][(𝐴. 𝐴̅) + 𝐵 + 𝐶̅ + 𝐷
̅]
= [𝐴 + 𝐵 + (𝐶. 𝐶 ) + 𝐷 ][𝐴 + 𝐵 + 𝐶 + (𝐷. 𝐷 )][𝐴 + 𝐵 + (𝐶. 𝐶̅ ) + 𝐷
̅ ][𝐴̅ + 𝐵̅ + 𝐶̅ + 𝐷
̅ ][𝐴 + 𝐵̅ + 𝐶̅ + 𝐷][𝐴 + 𝐵̅ + 𝐶̅ + 𝐷
̅ ][𝐴̅ + 𝐵 + 𝐶 + 𝐷
̅ ][𝐴̅
= [𝐴̅ + 𝐵̅ + 𝐶 + 𝐷
̅ ][𝐴 + 𝐵 + 𝐶̅ + 𝐷
̅ ][𝐴̅ + 𝐵 + 𝐶̅ + 𝐷
̅]
+ 𝐵 + 𝐶̅ + 𝐷
̅ ][𝐴̅ + 𝐵̅ + 𝐶̅ + 𝐷
̅ ][𝐴 + 𝐵̅ + 𝐶̅ + 𝐷][𝐴 + 𝐵̅ + 𝐶̅ + 𝐷
̅ ][𝐴̅ + 𝐵 + 𝐶 + 𝐷
̅ ][𝐴̅
= [𝐴̅ + 𝐵̅ + 𝐶 + 𝐷
̅ ][𝐴 + 𝐵 + 𝐶̅ + 𝐷
̅]
+ 𝐵 + 𝐶̅ + 𝐷
= [0 + 0 + 1 + 0][0 + 0 + 0 + 0][1 + 0 + 0 + 1][1 + 0 + 0 + 0][0 + 1 + 1 + 0][0 + 1 + 0
+ 0][1 + 1 + 0 + 0]
= [𝑀13][𝑀15][𝑀6][𝑀7][𝑀9][𝑀11][𝑀3]
𝐹(𝐴, 𝐵, 𝐶, 𝐷) = 𝜋𝑀(3, 6, 7, 9, 11, 13, 15)
𝐹𝑜𝑟 𝑆𝑂𝑃 𝑓𝑜𝑟𝑚
𝐹(𝐴, 𝐵, 𝐶, 𝐷) = Σ𝑚(0, 1, 2, 4, 5, 8, 10, 12,14)
SMVEC/CSE/U23ITIC01/DDSA/UNIT 1
62
̅𝑪
̅+𝑩
̅𝑫
̅𝑫
̅ + 𝑨𝑫
̅ +𝑪
̅
𝑭(𝑨, 𝑩, 𝑪, 𝑫) = 𝑨
19. Using K map method obtain the minimal SOP & POS expression for the function.
F(W, X,Y,Z) =Σm(1,3,4,5,6,7,9,12,13)
(i). minimal SOP: F(X,Y,Z,W) =∑ m(1,3,4,5,6,7,9,12,13)
𝑭(𝑾, 𝑿, 𝒀, 𝒁) = 𝑾𝒁 + 𝑾 𝑿 + 𝑿 𝒀 + 𝒀𝒁
(ii). minimal POS:
𝑭(𝑾, 𝑿, 𝒀, 𝒁) = [𝑾 + 𝒀][𝑿 + 𝒁]
20. Simplify T( x,y,z)=(x+y)[x’’+(y’+z’)’]+x’y’+x’z’
[May 2015]
=(x+y)+[x+yz]+x’y’+x’z’
=xx+xyz+xy+yyz+ x’y’+x’z’
=x+xyz+xy+yz+ x’y’+x’z’
=x+xyz+yz+ x’y’+x’z’
=x+yz+ x’y’+x’z’
=x+y’+yz+x’z’
=x+z’+y’+yz
= x+z’+y’+z
=1+X+Y’
=1+Y’=1
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21.Convert the following function into product of max terms. F(a,b,c)=(a+b’)(b+c)(a+c’)[Nov 14]
=[(a+b’)+(c*c’)][(b+c)+(a*a’)][ (a+c’)+(b*b’)
= (a+b+c) (a+b’+c’) (a+b+c) (a’+b+c) (a+b+c’) (a+b’+c’)
= (a+b+c) (a+b’+c’) (a’+b+c) (a+b+c’)
=∏M(0,2,4,1)
22. Simplify the given Boolean expression using K-map
23. Simplify the given Boolean expression using K-map
24. Simplify the given Boolean expression using K-map
SMVEC/CSE/U23ITIC01/DDSA/UNIT 1
64
25. Simplify the logic function specified by the truth table using the –map method. Y is the output
variable, and A, B and C are the input variables.
8.6
Five- Variable Map
A 5-Variable K- map requires 25= 32 cells, but adjacent cells are difficult to identify on a single 32cell map. Therefore, two 16 cell K-maps are used. If the variables are A, B, C, D and E, two identical 16cell maps containing B, C, D and E can be constructed. One map is used for A and other for A’.
In order to identify the adjacent grouping in the 5- variable map, we must imagine the two maps
superimposed on one another ie., every cell in one map is adjacent to the corresponding cell in the other
map, because only one variable changes between such corresponding cells.
SMVEC/CSE/U23ITIC01/DDSA/UNIT 1
65
z
Five- Variable Karnaugh map (Layer Structure)
Thus, every row on one map is adjacent to the corresponding row (the one occupying the same
position) on the other map, as are corresponding columns. Also, the rightmost and leftmost columns within
each 16- cell map are adjacent, just as they are in any 16- cell map, as are the top and bottom rows.
Typical subcubes on a five-variable map
However, the rightmost column of the map is not adjacent to the leftmost column of the other map.
1. Simplify the Boolean function
F (A, B, C, D, E) = ∑m (0, 2, 4, 6, 9, 11, 13, 15, 17, 21, 25, 27, 29, 31)
Therefore, F (A, B, C, D, E) = A’B’E’+ BE+ AD’E
2. F (A, B, C, D, E) = ∑m (0, 5, 6, 8, 9, 10, 11, 16, 20, 24, 25, 26, 27, 29, 31)
SMVEC/CSE/U23ITIC01/DDSA/UNIT 1
66
Soln:
Therefore, F (A, B, C, D, E) = C’D’E’+ A’B’CD’E+ A’B’CDE’+ AB’D’E’+ ABE+ BC’
3. F (A, B, C, D, E) = ∑m ( 1, 4, 8, 10, 11, 20, 22, 24, 25, 26)+∑d (0, 12, 16, 17)
Soln:
Therefore, F (A, B, C, D, E) = B’C’D’+ A’D’E’+ BC’E’+ A’BC’D+ AC’D’+ AB’CE’
4. F (A, B, C, D, E) = ∑m (0, 1, 2, 6, 7, 9, 12, 28, 29, 31)
Soln:
Therefore, F (A, B, C, D, E) = BCD’E’+ ABCE+ A’B’C’E’+ A’C’D’E+ A’B’CD
5. F (x1, x2, x3, x4, x5) = ∑m (2, 3, 6, 7, 11, 12, 13, 14, 15, 23, 28, 29, 30, 31 )
Soln:
SMVEC/CSE/U23ITIC01/DDSA/UNIT 1
67
Therefore, F (x1, x2, x3, x4, x5) = x2x3+ x3x4x5+ x1’x2’x4+ x1’x3’x4x5
6. F (x1, x2, x3, x4, x5) = ∑m (1, 2, 3, 6, 8, 9, 14, 17, 24, 25, 26, 27, 30, 31 )+ ∑d (4, 5)
Soln:
Therefore, F (x1, x2, x3, x4, x5) = x2x3’x4’+ x2x3x4x5’+ x3’x4’x5+ x1x2x4+ x1’x2’x3x5’+ x1’x2’x3’x4
9.
QUINE-MCCLUSKEY (OR) TABULAR METHOD (10 MARKS)
The K-map is a very effective tool for minimization of logic functions with 4 or less variables. For
logic expressions with more than 4 variables, the visualization of adjacent cells and drawing of the K-map
becomes more difficult.
The Quine-McCluskey method, also known as the tabular method, can be employed in such cases
to minimize switching functions. This method employs a systematic, step-by-step procedure to produce a
simplified standard form of expression for a function with any number of variables.
The steps to be followed in Quine-McCluskey method are –
1. A set of all prime implicants of the function must be obtained.
2. From the set of all prime implicants, a set of essential implicants must be determined by preparing a
prime implicants chart.
3. The minterms which are not covered by the essential implicants are taken into consideration and a
minimum cover is obtained from the remaining prime implicants.
Selecting prime implicants:
1. List all minterms in the binary form.
2. Arrange the minterms according to the number of 1’s.
3. Compare each binary number with every term in the adjacent next higher category and if they differ
only by one position, put a check mark (
position that they differed.
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) and copy the term in the next column with ‘—‘ in the
68
4. Apply the same process described in step 3 for the resultant column and continue these cycles until
a single pass through cycle yields no further elimination of literals.
5. List all prime implicants.
6. Select the minimum number of prime implicants which must cover all the minterms.
1. Find the minimal SOP expression for the following function
F (a, b, c, d) = ∑ m (0, 3, 4, 6, 11) + ∑ d (8, 10, 12, 13) using Quine-McCluskey method.
Step 1:
These minterms are represented in the binary form.
Minterm
a
b
c
d
0
0
0
0
0
3
0
0
1
1
4
0
1
0
0
6
0
1
1
0
11
1
0
1
1
8
1
0
0
0
10
1
0
1
0
12
1
1
0
0
13
1
1
0
1
Binary representation of minterms
Step 2:
The above binary representations are grouped into a number of sections in terms of the number of 1’s.
No of 1’s
0
1
2
Minterm
a
b
c
d
0
0
0
0
0
4
0
1
0
0
8
1
0
0
0
3
0
0
1
1
6
0
1
1
0
10
1
0
1
0
1
1
0
0
11
1
0
1
1
13
1
1
0
1
12
3
Group of minterms for different number of 1’s
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Step 3:
Minterm Combination
a
b
c
d
(0, 4)
0
-
0
0
(0, 8)
-
0
0
0
(4, 6)
0
1
-
0
(4, 12)
-
1
0
0
(8, 10)
1
0
-
0
(8, 12)
1
-
0
0
(3, 11)
-
0
1
1
(10, 11)
1
0
1
-
(12, 13)
1
1
0
-
2-cell combinations
Step 4:
From the two-cell combinations, one variable and a dash (-) in the same position can be combined
to form 4-cell combination.
Minterm Combination
a
b
c
d
(0, 4, 8, 12)
-
-
0
0
(0,8,4,12)
-
-
0
0
4-cell combination
Step 5:
The column having only one cross mark corresponds to the essential prime implicants. A tick mark
is put against every column which has only one cross mark. A star mark is put against every essential
prime implicants.
Prime Implicants
a’bd’
*
ab’d’
b’cd
0
(4, 6)
3
4
6
11
(3, 11)
x
ab’c
(10, 11)
x
abc’
(12, 13)
c’d’
*
10
x
x
12
13
x
x
x
(8, 10)
*
8
(0, 4, 8, 12)
x
x
x
x
F (a, b, c, d) = a’bd’+ b’cd+ c’d’
2. Find the minimal SOP expression for the following function
F (w, x, y, z) = ∑ m (0, 1, 2, 3, 10, 11, 12, 13, 14, 15) using Quine-McCluskey method.
Solution:
Step 1:
These minterms are represented in the binary form.
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70
Minterm
w
x
y
z
0
0
0
0
0
1
0
0
0
1
2
0
0
1
0
3
0
0
1
1
10
1
0
1
0
11
1
0
1
1
12
1
1
0
0
13
1
1
0
1
14
1
1
1
0
15
1
1
1
1
Binary representation of minterms
Step 2:
No of 1’s
0
1
2
3
4
Minterm
w
x
y
z
0
0
0
0
0
1
0
0
0
1
2
0
0
1
0
3
0
0
1
1
10
1
0
1
0
12
1
1
0
0
11
1
0
1
1
13
1
1
0
1
14
1
1
1
0
15
1
1
1
1
Group of minterms for different number of 1’s
Step 3:
Minterm Combination
w
X
y
z
(0, 1)
0
0
0
-
(0, 2)
0
0
-
0
(1, 3)
0
0
-
1
(2, 3)
0
0
1
-
(2, 10)
-
0
1
0
-
0
1
1
1
0
1
-
(3, 11)
SMVEC/CSE/U23ITIC01/DDSA/UNIT 1
71
1
-
1
0
1
1
0
-
1
1
-
0
(11, 15)
1
-
1
1
(13, 15)
1
1
-
1
(14, 15)
1
1
1
-
(10, 11)
(10, 14)
(12, 13)
(12, 14)
2-cell combinations
Step 4:
Minterm Combination
w
x
y
z
(0, 1, 2, 3)
0
0
-
-
(2, 3, 10, 11)
-
0
1
-
(10, 11, 14, 15)
1
-
1
-
(12, 13, 14, 15)
1
1
-
-
4-cell combinations
Step 5:
Prime Implicants
0
w’x’
*
(0, 1, 2, 3)
x’y
*
(2, 3, 10, 11)
wy
wx
1
2
3
x
x
x
x
x
(10, 11, 14, 15)
*
10
11
x
X
x
X
12
(12, 13, 14, 15)
13
x
14
15
x
x
x
x
F (w, x, y, z) = w’x’+ wx+ (x’y or wy).
3. Find the minimal SOP expression for the following function
F (A, B, C, D) = ∑ m (1, 3, 4, 7, 8, 10, 11, 13, 15) using Quine-McCluskey method.
Soln:
Step 1:
These minterms are represented in the binary form.
Minterm
A
B
C
D
1
0
0
0
1
3
0
0
1
1
4
0
1
0
0
7
0
1
1
1
8
1
0
0
0
10
1
0
1
0
11
1
0
1
1
13
1
1
0
1
15
1
1
1
1
SMVEC/CSE/U23ITIC01/DDSA/UNIT 1
72
Binary representation of minterms
Step 2:
The above binary representations are grouped into a number of sections in terms of the number of 1’s.
No of 1’s
1
2
3
4
Minterm
A
B
C
D
1
0
0
0
1
4
0
1
0
0
8
1
0
0
0
3
0
0
1
1
10
1
0
1
0
7
0
1
1
1
11
1
0
1
1
13
1
1
0
1
15
1
1
1
1
Group of minterms for different number of 1’s
Step 3:
Minterm Combination
A
B
C
D
(1, 3)
0
0
-
1
(8, 10)
1
0
-
0
(3, 7)
0
-
1
1
(3, 11)
-
0
1
1
(10, 11)
1
0
1
-
(7, 15)
-
1
1
1
(11, 15)
1
-
1
1
(13, 15)
1
1
-
1
2-cell combinations
Step 4:
From the two-cell combinations, one variable and a dash (-) in the same position can be combined
to form 4-cell combination.
Minterm Combination
A
B
C
D
(3, 7, 11, 15)
-
-
1
1
4-cell combination
Step 5:
The column having only one cross mark corresponds to the essential prime implicants. A tick mark
is put against every column which has only one cross mark. A star mark is put against every essential
prime implicants.
Prime Implicants
SMVEC/CSE/U23ITIC01/DDSA/UNIT 1
1
3
4
7
8
10
11
13
15
73
A’BC’D’
*
4
A’B’D
*
(1, 3)
AB’D’
*
(8, 10)
x
(10, 11)
x
AB’C
x
ABD
*
(13, 15)
CD
*
(3, 7, 11, 15)
x
x
x
x
x
F (A, B, C, D) = A’BC’D’+ A’B’D+ AB’D’+ ABD+ CD.
4. Find the minimal SOP expression for the following function
F (A, B, C, D) = ∑ m (0, 1, 2, 8, 10, 11, 14, 15) using Quine-McCluskey method.
Soln:
Step 1:
These minterms are represented in the binary form.
Minterm
A
B
C
D
0
0
0
0
0
1
0
0
0
1
2
0
0
1
0
8
1
0
0
0
10
1
0
1
0
11
1
0
1
1
14
1
1
1
0
15
1
1
1
1
Binary representation of minterms
Step 2:
The above binary representations are grouped into a number of sections in terms of number of 1’s.
No of 1’s
0
1
2
3
4
Minterm
A
B
C
D
0
0
0
0
0
1
0
0
0
1
2
0
0
1
0
8
1
0
0
0
10
1
0
1
0
11
1
0
1
1
14
1
1
1
0
15
1
1
1
1
Group of minterms for different number of 1’s
SMVEC/CSE/U23ITIC01/DDSA/UNIT 1
74
Step 3:
Minterm
A
B
C
D
(0, 1)
0
0
0
-
(0, 2)
0
0
-
0
(0, 8)
-
0
0
0
(2, 10)
-
0
1
0
(8, 10)
1
0
-
0
(10, 11)
1
0
1
-
(10, 14)
1
-
1
0
(11, 15)
1
-
1
1
(14, 15)
1
1
1
-
Combination
2-cell combinations
Step 4:
From the two-cell combinations, one variable and a dash (-) in the same position can be combined to form
4-cell combination.
Minterm Combination
A
B
C
D
(0, 2, 8, 10)
-
0
-
0
(10, 11, 14, 15)
1
-
1
-
4-cell combinations
Step 5:
The column having only one cross mark corresponds to the essential prime implicants. A tick mark is put
against every column which has only one cross mark. A star mark is put against every essential prime
implicants.
Prime Implicants
A’B’C’ *
0
(0, 1)
x
x
B’D’
*
(0, 2, 8, 10)
AC
*
(10, 11, 14, 15)
1
2
8
10
x
x
11
x
14
15
x
x
F (A, B, C, D) = A’B’C’+ B’D’+ AC.
5. Find the minimal SOP expression for the following function using Quine-McCluskey method.
F (v, w, x, y, z) = ∑ m (0, 1, 3, 5, 6, 9, 11, 14, 21, 23, 24, 31) + ∑ d (25, 30)
Soln:
Step 1:
These minterms are represented in the binary form.
Minterm
SMVEC/CSE/U23ITIC01/DDSA/UNIT 1
v
w
x
y
z
75
0
0
0
0
0
0
1
0
0
0
0
1
3
0
0
0
1
1
5
0
0
1
0
1
6
0
0
1
1
0
9
0
1
0
0
1
11
0
1
0
0
0
14
0
1
1
1
1
21
1
0
1
1
0
23
1
0
1
0
1
24
1
1
0
0
1
31
1
1
1
1
1
25
1
1
0
1
0
30
1
1
1
1
1
Binary representation of minterms
Step 2:
The above binary representations are grouped into a number of sections in terms of the
number of 1’s.
No of 1’s
Minterm
v
w
X
y
z
0
0
0
0
0
0
0
1
1
0
0
0
0
1
3
0
0
0
1
1
5
0
0
1
0
1
6
0
0
1
1
0
9
0
1
0
0
1
1
1
0
0
0
11
0
1
0
1
1
14
0
1
1
1
0
21
1
0
1
0
1
1
1
0
0
1
23
1
0
1
1
1
30
1
1
1
1
0
31
1
1
1
1
1
2
24
3
25
4
5
Group of minterms for different number of 1’s
SMVEC/CSE/U23ITIC01/DDSA/UNIT 1
76
Step 3:
Minterm Combination
v
w
x
y
z
(0, 1)
0
0
0
0
-
(1, 3)
0
0
0
-
1
(1, 5)
0
0
-
0
1
(1, 9)
0
-
0
0
1
(3, 11)
0
-
0
1
1
(5, 21)
-
0
1
0
1
(6, 14)
0
-
1
1
0
(9, 11)
0
1
0
-
1
(9, 25)
-
1
0
0
1
(24, 25)
1
1
0
0
-
(14, 30)
-
1
1
1
0
(21, 23)
1
0
1
-
1
(23, 31)
1
-
1
1
1
(30, 31)
1
1
1
1
-
2-cell combinations
Step 4:
From the two-cell combinations, one variable and a dash (-) in the same position can be combined to form
4-cell combination.
Minterm Combination
v
w
x
y
z
(1, 3, 9, 11)
0
-
0
-
1
4-cell combination
Step 5:
The column having only one cross mark corresponds to the essential prime implicants. A tick mark is put
against every column which has only one cross mark. A star mark is put against every essential prime
implicants.
Prime Implicants
0
1
v’w’x’y’ *
(0, 1)
x
v’w’y’z
(1, 5)
x
w’xy’z
*
(5, 21)
v’xyz’
*
(6, 14)
wx’y’z
(9, 25)
SMVEC/CSE/U23ITIC01/DDSA/UNIT 1
3
5
6
9
11
14 21 23
24
31 25 30
x
x
x
X
x
x
77
vwx’y’
*
(24, 25)
wxyz’
(14, 30)
vw’xz
(21, 23)
vxyz
*
vwxy
v’x’z
x
X
x
x
x
(23, 31)
x
x
(30, 31)
*
x
(1, 3, 9,
11)
x
x
x
F (v, w, x, y, z) = v’w’x’y’+ w’xy’z+ v’xyz’+ vwx’y’+ vxyz+ v’x’z
6. Find the minimal SOP expression for the following function
F (A, B, C, D) = ∑ m (0, 2, 3, 6, 7, 8, 10, 12, 13) using Tabulation method.
Soln:
Step 1:
List all minterms in the binary form.
Minterm
A
B
C
D
0
0
0
0
0
2
0
0
1
0
3
0
0
1
1
6
0
1
1
0
7
0
1
1
1
8
1
0
0
0
10
1
0
1
0
12
1
1
0
0
13
1
1
0
1
Binary representation of minterms
Step 2:
The above binary representations are grouped into a number of sections in terms of the number of 1’s.
No of 1’s
0
1
2
Minterm
A
B
C
D
0
0
0
0
0
2
0
0
1
0
8
1
0
0
0
3
0
0
1
1
6
0
1
1
0
10
1
0
1
0
1
1
0
0
12
SMVEC/CSE/U23ITIC01/DDSA/UNIT 1
78
3
7
0
1
1
1
13
1
1
0
1
Group of minterms for different number of 1’s
Step 3:
Minterm Combination
A
B
C
D
(0, 2)
0
0
-
0
(0, 8)
-
0
0
0
(2, 3)
0
0
1
-
(2, 6)
0
-
1
0
(2, 10)
-
0
1
0
1
0
-
0
1
-
0
0
(3, 7)
0
-
1
1
(6, 7)
0
1
1
-
(12, 13)
1
1
0
-
(8, 10)
(8, 12)
2-cell combinations
Step 4:
From the two-cell combinations, one variable and a dash (-) in the same position can be combined to form
4-cell combination.
Minterm Combination
A
B
C
D
(0, 2, 8, 10)
-
0
-
0
(2, 3, 6, 7)
0
-
1
-
4-cell combinations
Step 5:
Prime Implicants
AC’D’
0
2
3
6
7
(8, 12)
8
10
x
ABC’
*
(12,13)
B’D’
*
(0, 2, 8, 10)
x
A’C
*
(2, 3, 6, 7)
x
12
13
x
x
x
x
F (A, B, C, D) = ABC’+ B’D’+ A’C
7. Find the minimal SOP expression for the following function
F (A, B, C, D) = ∑ m (1, 4, 6, 7, 8, 9, 10, 11, 15) using Quine-McCluskey method.
SMVEC/CSE/U23ITIC01/DDSA/UNIT 1
79
Soln:
Step 1:
List all minterms in the binary form.
Minterm
A
B
C
D
1
0
0
0
1
4
0
1
0
0
6
0
1
1
0
7
0
1
1
1
8
1
0
0
0
9
1
0
0
1
10
1
0
1
0
11
1
0
1
1
15
1
1
1
1
Binary representation of minterms
Step 2:
The above binary representations are grouped into a number of sections in terms of the number of 1’s.
No of 1’s
1
2
3
4
Minterm
A
B
C
D
1
0
0
0
1
4
0
1
0
0
8
1
0
0
0
6
0
1
1
0
9
1
0
0
1
10
1
0
1
0
7
0
1
1
1
11
1
0
1
1
15
1
1
1
1
Group of minterms for different number of 1’s
Step 3:
Minterm Combination
A
B
C
D
(1, 9)
-
0
0
1
(4, 6)
0
1
-
0
(8, 9)
1
0
0
-
(8, 10)
1
0
-
0
SMVEC/CSE/U23ITIC01/DDSA/UNIT 1
80
(6, 7)
0
1
1
-
(9, 11)
1
0
-
1
(10, 11)
1
0
1
-
(7, 15)
-
1
1
1
(11, 15)
1
-
1
1
2-cell combinations
Step 4:
From the two-cell combinations, one variable and a dash (-) in the same position can be combined to form
4-cell combination.
Minterm Combination
A
B
C
D
(8, 9, 10,11)
1
0
-
-
4-cell combination
Step 5:
The column having only one cross mark corresponds to the essential prime implicants. A tick mark is put
against every column which has only one cross mark. A star mark is put against every essential prime
implicants.
Prime Implicants
1
4
6
B’C’D
*
(1, 9)
A’BD’
*
(4, 6)
x
(6, 7)
x
A’BC
BCD
*
ACD
AB’
7
8
9
11
15
x
x
(7, 15)
x
x
(11, 15)
*
10
x
(8, 9, 10, 11)
x
x
x
F (A, B, C, D) = B’C’D+ A’BD’+ BCD+ AB’
8.
Using Tabulation method, simplify the Boolean function,
F (w, x, y, z) = ∑ m (1, 2, 3, 5, 9, 12, 14, 15) + ∑ d (4, 8, 11).
Soln:
Step 1:
List all minterms in the binary form.
Minterm
w
x
Y
z
1
0
0
0
1
2
0
0
1
0
3
0
0
1
1
5
0
1
0
1
9
1
0
0
1
12
1
1
0
0
SMVEC/CSE/U23ITIC01/DDSA/UNIT 1
81
14
1
1
1
0
15
1
1
1
1
4
0
1
0
0
8
1
0
0
0
11
1
0
1
1
Binary representation of minterms
Step 2: The above binary representations are grouped into a number of sections in terms of number of 1’s.
No of 1’s
1
Minterm
w
x
y
z
1
0
0
0
1
2
0
0
1
0
4
0
1
0
0
1
0
0
0
3
0
0
1
1
5
0
1
0
1
9
1
0
0
1
1
1
0
0
14
1
1
1
0
11
1
0
1
1
15
1
1
1
1
8
2
12
3
4
Group of minterms for different number of 1’s
Step 3:
Minterm Combination
w
X
y
z
(1, 3)
0
0
-
1
(1, 5)
0
-
0
1
(1, 9)
-
0
0
1
(2, 3)
0
0
1
-
(4, 5)
0
1
0
-
(4, 12)
-
1
0
0
(8, 9)
1
0
0
-
(8, 12)
1
-
0
0
(3, 11)
-
0
1
1
(9, 11)
1
0
-
1
(12, 14)
1
1
-
0
(11, 15)
1
-
1
1
(14, 15)
1
1
1
-
SMVEC/CSE/U23ITIC01/DDSA/UNIT 1
82
2-cell combinations
Step 4:
Minterm Combination
w
X
y
z
(1, 3, 9, 11)
-
0
-
1
4-cell combination
Step 5:
Prime Implicants
w’y’z
*
(1, 5)
w’x’y
*
(2, 3)
1
2
3
5
x
9
12
xy’z’
(4, 12)
wx’y’
(8, 9)
wy’z’
(8, 12)
x
(12, 14)
x
wyz
4
8
11
x
(4, 5)
*
15
x
w’xy’
wxz’
14
x
x
x
x
x
x
x
x
(11, 15)
x
wxy
*
(14, 15)
x
x’z
*
(1,3,9,11)
x
x
x
x
x
x
F (w, x, y, z) = w’y’z+ w’x’y+ wxz’+ wxy+ x’z
9. Using Tabulation method, simplify the Boolean function,
F (w, x, y, z) = ∑ m (2, 3, 4, 6, 7, 11, 12, 13, 15) + ∑ d (1, 5, 15).
Step 1:
List all minterms in the binary form.
Minterm
w
x
Y
z
2
0
0
1
0
3
0
0
1
1
4
0
1
0
0
6
0
1
1
0
7
0
1
1
1
11
1
0
1
1
12
1
1
0
0
13
1
1
0
1
14
1
1
1
0
1
0
0
0
1
5
0
1
0
1
SMVEC/CSE/U23ITIC01/DDSA/UNIT 1
83
15
1
1
1
1
Step 2:
No of 1’s
1
2
Minterm
w
x
y
z
2
0
0
1
0
4
0
1
0
0
1
0
0
0
1
3
0
0
1
1
6
0
1
1
0
12
1
1
0
0
0
1
0
1
7
0
1
1
1
11
1
0
1
1
13
1
1
0
1
1
1
1
0
1
1
1
1
5
3
14
4
15
Group of minterms for different number of 1’s
Step 3:
Minterm Combination
(2, 3)
w
X
y
z
0
0
1
-
(2, 6)
0
-
1
0
(4, 6)
0
1
-
0
(4, 12)
-
1
0
0
(4, 5)
0
1
0
-
0
0
-
1
0
-
0
1
0
-
1
1
-
0
1
1
(6, 7)
0
1
1
-
(6, 14)
-
1
1
0
(12, 13)
1
1
0
-
1
1
-
0
0
1
-
1
-
1
0
1
-
1
1
1
(1, 3)
(1, 5)
(3, 7)
(3, 11)
(12, 14)
(5, 7)
(5, 13)
(7, 15)
SMVEC/CSE/U23ITIC01/DDSA/UNIT 1
84
(11, 15)
(13, 15)
(14, 15)
1
-
1
1
1
1
-
1
1
1
1
-
2-cell combinations
Step 4:
From the two-cell combinations, one variable and a dash (-) in the same position can be combined to form
4-cell combination.
Minterm Combination
w
X
y
z
(2, 3, 6, 7)
0
-
1
-
(4, 6, 12, 14)
-
1
-
0
(4, 6, 5, 7)
0
1
-
-
(4, 12, 5, 13)
-
1
0
-
(1, 3, 5, 7)
0
-
-
1
(3, 7, 11, 15)
-
-
1
1
(6, 7, 14, 15)
-
1
1
-
(12, 13, 14, 15)
1
1
-
-
(5, 7, 13, 15)
-
1
-
1
4-cell combinations
From the two-cell combinations, one variable and a dash (-) in the same position can be combined to form
8-cell combination.
Minterm Combination
w
x
y
z
(4, 5, 6, 7, 12, 13, 14, 15)
-
1
-
-
8-cell combination
Step 5:
Prime Implicants
2
3
4
6
7
X
x
w’y *
(2,3,6,7)
x
w’z
(1,3,5,7)
x
x
yz *
(3,7,11,15)
x
x
x
(4, 5, 6, 7, 12,
13, 14, 15)
*
X
11
x
12
13
14
1
5
x
x
15
x
x
x
F (w, x, y, z) = w’y+ yz+ x.
10. Find the minimal SOP expression for the following function using Quine-McCluskey method.
F (v, w, x, y, z) = ∑ m (0, 1, 8, 11, 12, 15, 20, 21, 22, 24, 29, 31) + ∑ d (9, 18, 30)
Step 1:
These minterms are represented in the binary form.
SMVEC/CSE/U23ITIC01/DDSA/UNIT 1
85
Minterm
v
w
x
y
z
0
0
0
0
0
0
1
0
0
0
0
1
8
0
1
0
0
0
11
0
1
0
1
1
12
0
1
1
0
0
15
0
1
1
1
1
20
1
0
1
0
0
21
1
0
1
0
1
22
1
0
1
1
0
24
1
1
0
0
0
29
1
1
1
0
1
31
1
1
1
1
1
9
0
1
0
0
1
18
1
0
0
1
0
30
1
1
1
1
0
Binary representation of minterms
Step 2:
The above binary representations are grouped into a number of sections in terms of the number of 1’s.
No of 1’s
0
1
2
Minterm
V
w
X
y
z
0
0
0
0
0
0
1
0
0
0
0
1
8
0
1
0
0
0
12
0
1
1
0
0
20
1
0
1
0
0
24
1
1
0
0
0
9
0
1
0
0
1
1
0
0
1
0
11
0
1
0
1
1
21
1
0
1
0
1
22
1
0
1
1
0
15
0
1
1
1
1
29
1
1
1
0
1
30
1
1
1
1
0
31
1
1
1
1
1
18
3
4
5
Group of minterms for different number of 1’s
SMVEC/CSE/U23ITIC01/DDSA/UNIT 1
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Step 3:
Minterm Combination
v
w
x
y
z
(0, 1)
0
0
0
0
-
(0, 8)
0
-
0
0
0
(1, 9)
0
-
0
0
1
(8, 12)
0
1
-
0
0
(8, 24)
-
1
0
0
0
(8, 9)
0
1
0
0
-
(20, 21)
1
0
1
0
-
(20, 22)
1
0
1
-
0
(9, 11)
0
1
0
-
1
(18, 22)
1
0
-
1
0
(11, 15)
0
1
-
1
1
(21, 29)
1
-
1
0
1
(22, 30)
1
-
1
1
0
(15, 31)
-
1
1
1
1
(29, 31)
1
1
1
-
1
(30, 31)
1
1
1
1
-
2-cell combinations
Step 4:
From the two-cell combinations, one variable and a dash (-) in the same position can be
combined to form 4-cell combination.
Minterm Combination
v
w
X
y
z
(0, 1, 8, 9)
0
-
0
0
-
4-cell combination
Step 5:
Using 2-cell and 4-cell combination, prime implicants table can be drawn as,
Prime Implicants
0
1
8
11
12
15 20 21 22
v’wy’z’ *
(8, 12)
x
wx’y’z’ *
(8, 24)
x
v’wx’z’
(9, 11)
vw’yz’
(18, 22)
vw’xy’ *
(20, 21)
x
vw’xz’ *
(20, 22)
x
v’wyz
(11, 15)
*
vxy’z
(21, 29)
vxyz’
(22, 30)
SMVEC/CSE/U23ITIC01/DDSA/UNIT 1
24
29 31
x
18 30
x
X
x
9
x
x
X
x
x
x
X
x
87
wxyz
vwxz
(15, 31)
*
vwxy
v’x’y’
x
(29, 31)
x
(30, 31)
*
(0,1,8,9)
x
x
x
x
x
x
F (v, w, x, y, z) = v’wy’z’+ wx’y’z’+ vw’xy’+ vw’xz’+ v’wyz+ vwxz+ v’x’y’.
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