Digital Electronics

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Chapter 1 :

1.1 Introduction

Electronic systems usually deal with information. Representation of information is called a signal. Signal in electronics is generally in form of voltage or current. Value of a signal is proportional to some physical quantity and it gives information about it. For example, temperature represented in terms of voltage signal.

There are two types of signals which are different in terms of their characteristics with respect to time and value. 

1. Analog Signals                                                                                                                                                  

2. Digital Signals.

A signal whose value is defined at all instances of time is called continuous time signal. On the other hand signal whose values are defined only at discrete instances of time is called discrete time signal. Most of the signals that occur in nature are analog in form. A discrete time signal can be obtained from continuous time signal by process called sampling. This has been illustrated in Fi/g.1.1. 

(Pending – diagram)

Fig. 1.1: (a) Continuous time signal x(t) sampled at every T interval, (b) Resulting discrete time signal x(n)

Similarly if a signal can take any value in a given range between some minimum and maximum value then the signal is called continuous value signal. On the other hand if a signal takes only certain fixed values in a given range then it is called discrete value signal. The process of converting a continuous value signal to a discrete value signal is called quantization. This is illustrated in Fig.1.2.

(Pending – diagram)

Fig. 1.2: Continuous value signal (solid line) and discrete value signal (dotted line) 

Analog signal : Signals that are continuous in time and continuous in value are called analog signal. 

Digital signal : Signals that are discrete in time and discrete in values are called digital signals. Digital signals are generally processed by digital systems like computers and hence their values are represented in terms of binary as shown in
Fig.1.2 

  • Analog signal being continuous in time will have infinite values in any given period of time. Practically a digital system like computer cannot handle infinite values due to limited physical resources and processing power. This is the reason why a continuous time signal has to be sampled and converted to discrete time signal.  
  • Again analog signals are continuous in value and hence can take any value in a given range. Now ideally number of values in any given range will be infinite which cannot be represented by finite number of bits on a computer. For example, as shown in Fig.1.2, with three bits used for representing values only eight different values can be represented. Thus a continuous value signal has to be quantized and converted to discrete value signal.  

1.1.1 Levels of Integration

Digital electronic circuits have become increasingly popular and successful due to integrated circuit (IC) technology. Advancement in IC technology has made it possible to construct large number of devices (eg. transistor, diode, resistors, capacitors, etc) on a very small chip. Classification of IC technology based on number of components per chip is as follows

1. Small-scale integration (SSI), containing fewer than 100components
2. Medium-scale integration (MSI), containing 100 to 1000components
3. Large-scale integration (LSI), containing 1000 to 10,000components
4. Very large-scale integration (VLSI), containing more than 10,000components

1.1.2 Comparison of Analog and Digital Systems

(Pending – Comparision table)

1.2 Introduction Digital System

  • A digital system uses a building blocks approach. Many small operational units are interconnected to make up the overall system. 
  • Themostbasiclogicalunitsystemisgatecircuit.Thereareseveraldifferenttypesofgateswitheach perform differently from other logic gates. 
  • Digital signal consist of only two values, ‘0’ and ‘1’. These two values have logical meaning i.e. ‘1’ represents the existence of particular condition and ‘0’ represents the absence of condition.  
  • There are two types of tables are used in digital system:                                                                        Truth Table: Truth table plots inputs and outputs in terms of 1s and 0s.                                            Function Table: Function table plots inputs and outputs in term of HIGH and LOW voltage levels. 
  • The design of digital system may be roughly divided into three stages;                                            System Design: It involves breaking the overall system into subsystem and specifying the characteristics of each subsystem. For example, the system design of a digital computers involves specifying the number and type of memory, ALU and i/p – o/p devices.                              Logic Design: It involves how to interconnect basic logic building blocks to perform specific function. For example, to make a flip flop different logic gates are needs to be connected in specific manner.                                                                                                                                  Circuit Design: It involves specifying the interconnection of specific components like resistors, transistors, diodes, CMOS etc. to create a logic gates

1.2.1 Advantages of Digital Systems

  1. Digital systems are easier to design 
  2. Information storage is easy 
  3. Accuracy and precision are greater 
  4. Digital systems are more versatile 
  5. Digital circuits are less affected by noise 
  6. More digital circuitry can be fabricated on IC chips  

1.2.2 Logic Levels and Different types of Logics

  • Digital system use the binary number system. Therefore, two-state devices are used to represent the two binary digits 1s & 0s by two different voltage levels, called HIGH and LOW. 
  • Normally, the binary 0 and 1 are represented by the logic voltage levels 0 V and +5V. 
  • Usually any voltage between 0 V to 0.8 V represents the logic 0 and any voltage between 2 V to 5 V represents the logic 1. This voltage levels can be varies according to the different logical systems. 
  • There are three types of logics available in digital systems.    
  1. Positive Logic
  2. Negative Logic             
  3. Mixed Logic
      1. Positive Logic: In positive logic high voltage level is represent as logic 1 and low voltage level is represent as logic 0.

(Pending – diagram)

Fig. 1.3: Illustration of positive logic  

      2. Negative Logic: In positive logic high voltage level is represent as logic 0 and low voltage level is represent as logic 1.  
(Pending – diagram)

Fig. 1.4: Illustration of negative logic 

       3. Mixed Logic: This scheme uses positive logic in some portions (e.g inputs) of the system while applying negative logic (e.g. outputs) in other portion of the system.

Suppose some function X=AB’+A’B for this function the representation of all the logics areas follow;(Pending – diagram)

Fig. 1.5: Representation of function X = AB’ + A’B  
 

Truth table of the given function for all the logics is shown as follow ;
(Pending – table)

Table 1.1: Truth table of Positive logic, Negative logic, Mixed logic for X = AB’ + A’B  

1.3 NUMBERSYSTEMS:

1.3.1 Introduction to Number System:

  • Definition &Importance
  • Number system is the basis for counting various items. On hearing the word ‘number’, we immediately think of the familiar decimal number system with 10 digits 0 to 9. But modern computers communicate and operate with binary numbers which use only 2 digits 0 & 1. Also different types of number systems like octal and hexadecimal are also used widely. Depending upon the type of number system, we use different digits to represent various numbers.
  • Few Common Aspects to All Numbering Systems                                                                               

(i) Base or Radix 

The number of symbols used for the representation of numbers in a number system is known as its Base or Radix and is generally denoted by r.                                                    

(ii) Digit 

Each symbol in the number system is called a Digit.                                                               

(iii) The largest value of a digit is always one less than the base                                                  

For ex, in decimal system, the largest digit is 9 (since base is 10)                                               

(iv) Each digit position (i.e. place) represents a different multiple of base                                       

This means that the numbers have positional importance. Hence the number systems are known as Positional Weighted Number System. It means that the value attached to a symbol depends on its location with respect to the decimal point. 

  • For example decimal number 123.4 (base 10) can actually be represented as;                                  (123.4)10 = 1×102 + 2×101 + 3×100 +4×10-1 (<–Pending – edit eqn)
  • In general, a number of any radix can be expressed as,                                                                       Nr = … + D3 x r3 + D2 x r2 + D1 x r1 + D0 x r0 + D-1 x r -1 + D-2 x r -2 + D-3 x r -3 +…  (<–Pending – edit eqn)                                                                                                                                                   Where;                                                                                                                                                               r is the base and Di is any valid digit in the number system of base r. 
  • The digits on the left side of the decimal point form the integer part of a number and those on the right side form the fractional part. 
  • The left most digit in any number representation, which has the greatest positional weight out of all the digits present in that number is called the most significant digit(MSD). 
  • The right most digit in any number representation, which has the least positional weight out of all the digits present in that number is called the least significant digit(LSD).
  • Various Numbering Systems 
  • Different number systems are used in various applications. The commonly used number systems along with their base, 1st digit, last digit and available digits are as shown below:             (Pending – table)                                                                                                                                                                                                
Note : In hexadecimal number system, meaning of A≈10, B≈11, C≈12, D≈13, E≈14 & F≈15.  

1. DIGITAL NUMBER SYSTEM

  •  Decimal number system is the most familiar no. system used in day consists of 10 unique symbols. Hence the this system, any number(integer, fraction or mixed) of any magnitude can be represented by the use of these  ten symbols only.
  • The digits on the left side of the decimal point form the integer part of a decimal number while those on right side from the fractional part. The digits on the right of the decimal point have weights which are negative powers of 10 and the digits to the left of the decimal point have weights which are positive powers of 10. The sum of all the digits multiplied by their weights gives the total number being represented.
  • In general, the value of any mixed decimal number                                                                                dn dn-1 dn-2 . . . d1 d0 . d      (<–Pending – edit eqn)                                                                                  is given by                                                                                                                                                        (dn x 10n) + (dn-1 x 10n-1) + . . . + (d1 x 10     (<–Pending – edit eqn)
  • Consider a decimal no. 9256.26. We represent it as:                                                                                9256.26 = 9 x 1000 + 2 x 100 + 5 x 10 + 6 x 1 + 2 x (1/10) + 6 x = 9 x 103 + 2 x 102 + 5 x 10  (<–Pending – edit eqn)

MSD . … 103 102 101(<–Pending – edit eqn)

(Pending – diagram)

2. BINARY NUMBER SYSTEM 

  • The binary number system is a positional weighted system. The base or radix of this number system is 2. Hence, it has two independent symbols. The base itself cannot be a symbol. The symbols used are 0 & 1. A binary digit is called a bit either a 0 or a 1. The binary point separates the integer and fraction part. The weight of each bit position is one power of 2 greater than the weight of the position to its immediate right. The place values left on the binary point in binary are 64, 32, 16, 8, 4, 2 and 1. 
  • In general, the value of any mixed binary number                                                                                  bn bn-1 bn-2 . . . b1 b0 . b      (Pending – edit)                                                                                           is given by                                                                                                                                                   (bn x 2n) + (bn-1 x 2n-1) + . . . +     (Pending – eqn)                                                                            MSB . … 23 22 21 (Pending – edit)
(Pending – diagram)

Fig. 1.7: Binary position values as power of 2

  • Counting in Binary
  • Counting in binary is very similar to decimal counting. Start counting with 0, the next count is 1.
    Moving ahead, we put 1 in the column to the left and continue the count using two bits. Similarly, we can continue counting with 5, 6, …bits.

Table 1.3: Counting in Binary

(Pending – table)

  • Applications
  • The binary number system is used in digital computers because the switching circuits  used in these computers use two-state devices such as transistors, diodes, etc. These devices have to exist in one of the two possible states: ON of OFF, OPEN or CLOSED. So, these two states can be represented by the symbols 0 and 1,respectively.
3.  OCTAL NUMBERSYSTEM
  • The octal number system was extensively used by early minicomputers. It is also a positional weighted system. Its base or radix is 8. It has 8 independent symbols 0 to 7.
  • Sinceitsbase8=23(Edit as 2 power 3) ,every 3-bit group of binary can be represented by an octal 1/3 rd. the length of the corresponding binary number.
(Pending – diagram and table)
  • Usefulness of the Octal System 
  • In computer work, binary numbers up to 64 bits are not uncommon. always represent a numerical quantity; they often represent some type of code. While dealing with large binary numbers, it is convenient and more efficient for us to write the numbers in octal rather
    than binary. The ease with which conversions can be made between octal and binary makes the octal system more attractive as a shorthand means of expressing large binary numbers.

4. HEXADECIMAL NUMBERSYSTEM

  •  Binary numbers are too long. These numbers are fine format machines but are too lengthy to be handled by human beings. So, there is a need (check the word here ) to represent the binary numbers concisely. One number system developed with this objective is the hexadecimal number system (or Hex). Although it is somewhat difficult to interpret than the octal number system, it has become the most popular and retrieval in digital systems.
  • The hexadecimal number system is positional weighted system. The it has16 independent symbols. The symbols used are digit combination can be represented by one hexadecimal digit. So, a hexadecimal number is ¼ length of the corresponding binary number.
  • A 4-bit group is called a nibble. Since computer words come in 8, 16, 32 bits and so on, they
    easily represented in hexadecimal. The hexadecimal number system is particularly used for human communications with computers. It is used in both large and small computers.
    MSD . ..

Fig. 1.9: Hexadecimal position values as power of 16  

Table 1.5: Counting in Hexadecimal  

(Pending – diagram and table)

1.3.2 Number Base Conversions

  • Definition and Importance
  • The human beings use decimal number system while computer uses binary number system.
    Therefore, it is essential to convert decimal number into its equivalent binary while feeding number into computer and to convert binary number into its decimal equivalent while displaying result of operation to the human beings.
  • However, dealing with a large quantity of binary numbers of many bits is inconvenient for human beings. Therefore, octal and hexadecimal numbers are used as a short hand means of expressing large binary numbers. Hence inter conversion among different number systems is required.
  • Conversions between Decimal, Binary, Octal and Hexadecimal
  • The below table shows the decimal, binary, octal and hexadecimal numbers.
(Pending – table)
 
 1. Binary to Decimal Conversion
  • Binary numbers can be converted to their decimal equivalents by the positional weights method. In this method, each binary digit of the number is multiplied by its position weight and the product terms are added to obtain the decimal number.
    Ex 1: Convert (10101)2 to decimal
    Solution: (Pending – edit below solution )
    Positional weights are: 24 2

    Hence, (11011.101)2 = (27.625)10

          Ex 2: Convert (11011.101)2 to decimal 
         Solution: Positional weights are:  (Pending – edit below solution )
         Hence, (11011.101)2 = (27.625)10

 2. Octal to Decimal Conversion
  • To convert an octal number to a decimal number, multiply each digit in the octal number by the
    weight of its position and add all the product terms.
    Ex 3: Convert (4057.06)8 to decimal
    Solution: (Pending – edit below solution )
    Positional weights are: 83 8

    Hence, (4057.06)8 = (2095.0937)10

3. Hexadecimal to Decimal Conversion

  • Multiplyeachdigitinthehexnumberbyitspositionweightandaddallthoseproductterms.Inthis way, we get the decimal equivalent of the hexadecimal number.
    Ex 4: Convert (5C7)16 to decimal
    Solution: (Pending – edit below solution )
    Positional weights are: 162 161 160
    Hence, (5C7)16 = (1479)10
    Ex 5: Convert A0F9.0EB16 to decimal
    Solution: (Pending – edit below solution )
    Positional weights are: 163 162 161 160
    Hence, (A0F9.0EB)16 = (41209.0572)10

4. Decimal to Binary Conversion

  • The conversion of decimal number to binary is carried out in 2 steps. In step 1, we have to convert integer part and in step 2, we have to convert fractional part.
  • For integer part conversion, we use successive division-by-2 method. In this method we repeatedly divide the integer part of the decimal number by 2 until the quotient is zero. The remainder of each division becomes the numeral in the new radix. The remainders are taken in the reverse order to form a new radix number. This means that the first remainder is the LSD and the last remainder is the MSD in the new radix number. Thus the integers read from bottom to top give the equivalent binary fraction.
  • Similarly, for fractional part, we use successive multiplication-by-2 method. In this method, the number to be converted is multiplied by the radix of new number, producing a product that has an integer part and a fractional part. The integer part (carry) of the product becomes a numeral in the new radix number. The fractional part is again multiplied by the radix and this process is repeated until fractional part reaches 0 or until the new radix number is carried out to significant digits. The integer part (carry) of each product is read from top to bottom to represent the new radix number.

Ex 6: Convert 5210 to binary

Here the number is integer number so we need to divide the given decimal number by 2 and read the remainders from bottom to top to get the equivalent binary number.

(Pending – edit solution )

Hence, (52)10 = (110100)

Ex 7: Convert (163.875)10 to binary
Solution:
Step 1: Separate the integer and fractional parts of the decimal number. Now for integer part, we carry successive division-by-2
method as follows:
(Pending – edit solution )

So, 16310 = (10100011)2

Step 2: Now the fraction part is 0.87510. Carrying out successive multiplication-by-2 as follows: (Pending – edit solution )
So, 0.87510 = 0.1112
Hence. (163.875)10 = (10100011.111)2

5. Decimal to Octal Conversion
  • Decimal to Octal conversion can be done in similar way as decimal to binary conversion. The integer and fractional parts are to be separated and the same procedure is carried out. But the division and multiplication are carried out by 8 as the base of octal number is 8. Following the same steps, we can get the equivalent octal number of the given decimal number.
    Ex 8: Convert 378.9310 to octal
    Solution: (Pending – edit solution )
    Step1: Conversion of integer part by successive division-by-8 method.
    So, (378)10 = (572)8
    Step 2:  Conversion of fractional part by successive multiplication-by-8 method. (Pending – edit solution )
    So, 0.9310 = 0.73418
    Hence, (378.93)10 = (572.7341)8

6. Decimal to Hexadecimal Conversion

  • Decimal to hexadecimal conversion is carried out by 2 steps. In the first step, the integer part of the decimal number is divided by 16 successively and the remainder is noted. The remainders read from bottom to top gives the equivalent hexadecimal integer. In the second step, the successive multiplication of fractional part by 16 is done and the integers are noted down. Reading the integers from bottom to top gives the hexadecimal fraction.
    Ex 9: Convert (3509.75)10 to hexadecimal
    Solution: (Pending – edit solution )
    Step1: Conversion of integer part by successive division-by-16 method
    Remainder
    5
    11 =B
    13 =D
    So, (3509)10 =(DB5)16
    Step 2: Conversion of fractional part by successive multiplication-by-16 method.
    0.75 x 16 =12.0 12 =C
    So, (0.75) 10 = (0.C)16
    Hence, (3509.75)10 =(DB5.C)16

7. Octal to Binary Conversion

  • Toconvertagivenoctalnumbertobinary,justreplaceeachoctaldigitbyits3-bitbinaryequivalent.
    Ex 9: Convert 367.528 to binary.
    Solution: (Pending – edit solution )
    Given octal number is 3 6 7 . 5 2
    Convert each octal digit to binary 011 110 111 . 101 010
    Hence, (367.52)8 = (11110111.101010)

8. Hexadecimal to Binary Conversion

  • To convert a given hexadecimal number to binary, just replace each hexadecimal digit by its 4-bit binary equivalent.
    Ex 10: Convert 4BAC16 to binary.
    Solution: (Pending – edit solution )
    Given hexadecimal number is 4 B A C
    Convert each digit to 4-bit binary 0100 1011 1010 1100
    Hence, (4BAC)16 = (100101110101100)2
    Ex 11: Convert 3A9.B0D16 to binary.
    Solution: (Pending – edit solution )
    Given hexadecimal number is 3 A 9 . B 0 D
    Convert each digit to 4-bit binary 0011 1010 1001 . 1011 0000 110
    Hence, (3A9.B0D)16 = (1110101001.101100001101)2

9. Binary to Octal Conversion

  • To convert a binary number to an octal number, starting from the binary point make groups of 3 bits each, on either side of the binary point and replace each 3-bit binary group by the equivalent octal digit.
    Ex 12: Convert 110101.1010102 to octal.
    Solution: (Pending – edit solution )
    Group of 3 bits are 110 101 . 101 010
    Convert each group to octal
    Hence, (110101.101010)2 = (65.52)8
    6 5 . 5 2
    Ex 13: Convert 10101111001.01112 to octal.
    Solution: (Pending – edit solution )
    Group of 3bitsare 10 101 111 001 . 011 1
    = 010 101 111 001 . 011 100
    Convert each group to octal 2 5 7 1 . 3 4
    Hence, (10101111001.0111)2 = (2571.34)8

10. Binary to Hexadecimal Conversion

  • To convert a binary number to an octal number, starting from the binary point make groups of 4 bits each, on either side of the binary point and replace each 4-bit binary group by the equivalent
    hexadecimal digit.
    Ex 14: Convert 1011111011.0111112 to hexadecimal.
    Solution: (Pending – edit solution )
    Group of 4 bits are 10 1111 1011 . 0111 11
    0010 1111 1011 . 0111 1100
    Convert each group to hex 2 F B . 7 C
    Hence, (1011111011.011111)2 = (2FB.7C)16

11. Octal to Hexadecimal Conversion

  • To convert an octal number to hexadecimal, the simplest way is to first convert the given octal
    number to binary and then the binary number to hexadecimal.
    Ex 15: Convert 1245 to hex.
    Solution: (Pending – edit solution )
    Given octal number is 1 2 4 5
    Convert each octal digit to binary 001 010 100 110
    Group of 4 bits are 0010 1010 0110
    Convert each 4–bit group to hex 2 A 6
    Hence, (1245)8 = (2A6)16
    Ex 16: Convert 756.6038 to hex.
    Solution: (Pending – edit solution )
    Given octal number is 7 5 6 . 6 0 3
    Convert each octal digit to binary 111 101 110 . 110 000 011
    Group of 4bitsare 0001 1110 1110 . 1100 0001 1000
    Convert each 4–bit group to hex 1 E E . C 1 8
    Hence, (756.603)8 = (1EE.C18)16

12. Hexadecimal to Octal Conversion

  • To convert hexadecimal number to octal, the simplest way is to first convert the given hexadecimal number to binary and then the binary number to octal.

Ex 17: Convert B9F.AE16 to octal.
Solution: (Pending – edit solution )
Given hex number is B 9 F . A E
Convert each hex digit to binary 1011 1001 1111 . 1010 1110
Group of 3 bits are 101 110 011 111 . 101 011 100
Convert each 3–bit group to octal
Hence, (B9F.AE)16 = (5637.534)8
5 6 3 7 . 5 3 4

13. Any radix r number to Decimal Conversion

  • We can convert a given number in radix r to decimal by multiplying each digit by its positional weights and taking sum of all the products.

Ex 18: Convert 12213 to decimal.
Solution: (Pending – edit solution )
Here, the given number is in base 3. Its positional weights are: 33 3
Hence, (1221)3 = (52)10
0.35 x12 = 4.2 4
0.2 x12 = 2.4 2
0.4 x12 = 4.8 4
Ex 19: Convert 234.025 to decimal.
Solution: (Pending – edit solution )
Here, the given number is in base 5. Its positional weights are: 52 5

Hence, (234.02)5 = (69.08)10

14. Decimal to any radix r Conversion

  • Decimal number can be converted in any radix by 2 steps. In step1, the integer part of the decimal number is divided successively by the radix r and the remainders are noted down. Taking the remainders from bottom to top gives the radix r equivalent of the integer part. Similarly, the fractional part is successively multiplied by the radix r and the integer part of the result is noted down. Noting the carry from top to bottom gives the fractional part equivalent in radix r.
    Ex 20: Convert 1989.3510 to base 12.
    Solution: (Pending – edit solution )
    Step 1: Separate the integer and fractional parts of the decimal number. Now for integer part, we carry
    successive division-by-12 method as follows:

    Hence. (1989.35)10 = (1199.424)12

15. Find the value of unknown base

Ex 21: Determine b if (33)10 =
(201)b Solution: (Pending – edit solution )
We have, (33)10 = (201)b
33 = 2 x b2 + 0 x b1 + 1 x b0
= 2b2 +
1 2b2 =32
b
2 =16
b =±4
But base of any number cannot be negative. Hence value of b = 4.
Ex 22: Determine b if (193)b = (623)8
Solution: (Pending – edit solution )
We have, (193)b = (623)8
As base is always positive, value of b=16.

16. Binary Coded Decimal(BCD)

  • It is a numeric code that is used to represent decimal using binary bits i.e.1’s and 0’s. It is different from representation of a decimal number in binary system i.e. base 2 system.
  • In BCD representation each digit of a decimal number is represented by a group of four bits. These bits are given with weights of 8-4-2-1 and hence many a times BCD code is also called 8421 code. 

Code for each digit of decimal is as follows.

Table 1.7: Conversions between Decimal to BCD code

(Pending – table)

Ex 23: (58)10 = ( )BCD
Solution: (Pending – edit solution )
Decimal: 5 8
0101 1000
Thus (58)10 = (01011000)BCD.

  • It should be noted that binary representation of(58)10 will be (111010)2 which is quite different from the BCD representation. (Pending – edit numbers in the sentence)
    Ex 23: (001001011001)BCD = ()10
    Solution: (Pending – edit solution )
    BCD: 0010 01011001
    2 5 9
    (001001011001)BCD = (259)1
  • It can be observed that BCD codes are less efficient for representation compared to binary as it
    requires more number of bits then required in binary representation. However, it is popular because of its ease of conversion to and from decimal.

BOOLEAN ALGEBRA

  • (Pending MCQs)

THEORY: 

LOGIC GATES:

  • Logic gates are the fundamental building blocks of digital systems. They are the physical devices that performs the basic Boolean operations of AND, OR and NOT.
  • Input and outputs of logic gates (that is basically a voltage signal) can occur only in two levels. These two levels are termed as High and Low or True and False or ON and OFF or simply 1 and 0. In representation of higher of the two voltage levels is symbolized as 1 and lower symbolized as 0 the
    gate is said to be positive logic gate. However, if higher of the two voltage levels is symbolized as 0
    and lower as 1 then it is said to be negative logic gate.
  • Input output behavior of a gate is generally represented using truth table. It is a  table that lists output for all
    possible combinations of inputs.
  • There are total seven logic gates in which three are basic logic gates(AND,OR,NOT) and two are
    universal logic gates (NAND, NOR).
  • Various basic gates are discussed as follows;
NOTGate: 
  • NOT gate has one inputs and one output. The output becomes logic 1 when input is at logic 0 and output becomes logic 0 when the input is at logic 1. Thus it inverts or complements the logic available at input and hence called and inverter or complement. It is represented by a bar over the variable “̅”or with a symbol“ ’ ”. Thus, for example, X =A΄orX= A read as “X is equal to Not A or A bar or A complement”. NOT gate and its truth table are shown in fig.  
(Pending – diagram)
 
  • AND gate means all or nothing logic 
  • AND gate has two or more inputs and one output. The output becomes logic 1 only when each one of its input is at logic1. For all other input combinations it gives output logic 0. It is represented by a symbol •. 
  • Thus, for example, X = A · B (also written simply as X = AB) is read as “X is equal to A AND B”. Two input AND gate and its truth table is shown in fig.1.11. 
OR Gate: 
  • OR gate means any or all logic
  • OR gate has two or more inputs and one output. The output becomes logic 1 when at least
    (minimum) one of the inputs is at logic 1. It is represented by a symbol +. Thus, for example, X = A + B is read as “X is equal to A OR B”. Two input AND gate and its truth table is shown in fig.1.12.  
(Pending – diagram)
 
Exclusive OR Gate (EX-OR): 
  • It also means Inequality detector because it gives output high when both inputs are different. 
  • Exclusive OR gate give output equal to 1 when the two inputs are exclusively different. This is the reason why it is also known as inequality gate. The schematic symbol and truth table of the gate is shown in fig. 1.13. It is represented by a symbol.
  • Thus, for example, X  A Bis read as “X is  equal to A XOR B.” The logic expression this gate in terms of AND, OR and NOT operation is X  A  B  AB  AB . (Pending – edit boolean exp in sentence)
(Pending – diagram)
 
Exclusive NOR Gate (EX-NOR): 
  • It also means equality detector because it gives output high when both inputs are same. 
  • Exclusive NOR gate is XOR gate followed by inverter. Thus it is complement of XOR gate. This is the reason why it is also known as equality gate. The logic symbol, logic expression, schematic symbol, truth table of the gate is shown in fig.  X = AB +A’B’ 
(Pending – diagram)

Universal Gates: 

  • NAND and NOR gates are known as a universal gates because from this two gates all other gates can be constructed. 

NAND Gate:

  • NAND gate represents combination of AND gate followed by NOT gate. It represents complement of AND operation. Schematic symbol of NAND gate and its truth table are shown in fig. 1.15. The logic  Expression is given as X = 𝐴തതത.ത𝐵ത or X = (A·B) (Pending – edit boolean exp in sentence)
(Pending – diagram)
 

NAND gate as Universal gate

1. Implementing NOT gate
    All NAND input pins connect to the input signal A gives an output A’.
(Pending – diagram)
2. Implementing AND gate
    The AND is replaced by a NAND gate with its output complemented by a NAND gate inverter.  
(Pending – diagram)
3. Implementing OR gate
The OR gate is replaced by a NAND gate with all its inputs complemented by NAND gate inverters

(Pending – diagram)

NOR Gate: 

  • NOR gate represents combination of OR gate followed by NOT gate. It represents complement of OR operation. Schematic symbol of NOR gate and its truth table are shown in fig.4.16.The logic expression is given as X  ( A  B) or X = (A+B)’.
(Pending – diagram)

 

NOR gate as Universal gate.

1. Implementing NOT gate                                                                                                                                                                                             All NOR input pins connect to the input signal A gives an output A`. 

(Pending – diagram)

2. Implementing AND gate                                                                                                                                                                                            The AND gate is replaced by a NOR gate with all its inputs complemented by NOR gate inverters. 

(Pending – diagram)

3. Implementing OR gate                                                                                                                                                                                              The OR is replaced by a NOR gate with its output complemented by a NOR gate inverter. 

(Pending – diagram)

 

Gates with More Inputs 

  • Any logic gates can be designed to accept three or more input values. As an example, three-input AND gate produces an output of 1 only if all input values are1. Constructing Gates 
  • A transistor is a device that acts, depending on the voltage level of an input signal, either as a wire that conducts electricity or as a resistor that blocks the flow of electricity 
  • A transistor has no moving parts, yet acts like a switch. 
  • It is made of a semiconductor material, which is neither a particularly good conductor of electricity, such as copper, nor a particularly good insulator, such as rubber. 
  • A transistor is shown in fig.1.18. Fig. 1.18: 

(Pending – diagram)

  • A transistor has three terminals                                                                                                                                                                  1. A source                                                                                                                                                                                                      2. Abase                                                                                                                                                                                                            3. An emitter, typically connected to a ground wire 
  • If the electrical signal is grounded, it is allowed to flow through an alternative route to the ground (literally) where it can do no harm. 
  • It turns out that, because the way a transistor works, the easiest gates to create are the NOT,NAND, and NOR gates shown in fig.1.19. 

(Pending – diagram)

2.1 BOOLEAN ALGEBRA

  • Inventor of Boolean algebra was George Boole (1815 -1864).
  • Designing of any digital system there are three main objectives;
    1) Build a system which operates within given specifications
    2) Build a reliable system
    3) Minimize resources
  • Boolean algebra is a system of mathematical logic.
  • Any complex logic can be expressed by Boolean function.
  • Boolean algebra is governed by certain rules and laws.
  • Boolean algebra is different from ordinary algebra & binary number system.                                                                                    In ordinary algebra; A + A = 2A and AA = A2, here A is numeric value.
  • In Boolean algebra;
    A + A = A and AA = A, here A has logical significance, but no numeric significance.
    Table 2.1: Difference between Binary, Ordinary and Boolean system                                      (Pending – table)
  • In Boolean algebra there is nothing like subtracting or division, no negative or fractional numbers.
  • Boolean algebra represent logical operation only. Logical multiplication is same as AND operation and logical addition is same as OR operation.
  • Boolean algebra has only two values 0 &1. 
  • In Boolean algebra;
    If A = 0 then A ≠ 1. & If A = 1 then A ≠ 0.
  • Let’s introduce Boolean algebra by considering a practical problem:
  • Suppose a system which transmitting 2 bit binary information over 2 line to another system.
  • So with 2 line, 4 unique code could be represented.
  • The receiving system may need to identify the presence of certain transmitted codes.
  • As an example suppose system is to identify the occurrence of the code representing the decimal number 1 and 2. Each one of these code appear on the 2 lines, a circuit is to generate an output1.
  • When any other code present the output should be 0.

(Pending – table)

  • The method of implementing the system is shown in following fig. 
  • We want output 1 if decimal 1 & 2 otherwise 0 as defined above. 
  • Below circuit generate output 1 when A = 0 & B = 1 or A = 1 & B =0. 

(Pending – table)

 

2.1.1 Advantages of Boolean Algebra

 

  1. Minimize the no. of gates used in circuit.                                                                                              
  2. Decrease the cost of circuit. 
  3. Minimize there sources. 
  4. Less fabrication area is required to design a circuit. 
  5. Minimize the designer’s time. 
  6. Reducing to a simple form. Simpler the expression more simple will be hardware. 
  7. Reduce the complexity. 

2.1.2 Axioms of Boolean Algebra

 

  • Axioms or postulate of Boolean algebra are a set of logical expression that we accept without proof & upon which we can build a set of useful theorems.                                                                                                                                                                      Axioms 1: 0 · 0 =0                                                                                                                                                                                              Axioms 2: 0 · 1 =0                                                                                                                                                                                    Axioms 3: 1 · 0 =0                                                                                                                                                                                        Axioms 4: 1 · 1 =1                                                                                                                                                                                      Axioms 5: 0 + 0 =0                                                                                                                                                                                Axioms 6: 0 + 1 =1                                                                                                                                                                                Axioms 7: 1 + 0 =1                                                                                                                                                                                          Axioms 8: 1 · 1 = 1                                                                                                                                                                                    Axioms 9: 1’ = 0                                                                                                                                                                                    Axioms 10: 0’ = 1

2.1.3. Types of logic circuits  

  • There are two types of logic circuit; 
  1. Sequential circuit 
  2. Combinational circuit 
2.1.3.1 Difference between combinational & sequential circuit  
 (Pending – table)

2.2 LAWS OF BOOLEAN ALGEBRA

1. Complementation Laws:
    The term complement simply means to invert, i.e. to change 0’s to 1’s and 1’s to0’s.
    Law 1: 0’ =1
    Law 2: 1’ = 0
    Law 3: If A = 0 then A’ = 1
    Law 4: If A = 1 then A’ = 0
    Law 5: A’’ = A

2. AND Laws:
    Law 1: A · 0 = 0
    Law 2: A · 1 = A
    Law 3: A · A = A
    Law 4: A · A’ =0

3. OR Laws:
    Law 1: A + 0 = A
    Law 2: A + 1 = 1
    Law 3: A + A = A
    Law 4: A + A’ = 1

4. Commutative Laws:

  • Commutative laws allow change in position of AND or OR variables.
    Law 1: A + B = B + A
    Proof: (Pending – diagram)
  • This law can be extended to any numbers of variables for e.g.
    A + B + C = B + A + C = C + B + A = C + A +B
    Law 2: A · B = B · A
    Proof: (Pending – diagram)
  • This law can be extended to any numbers of variables for e.g.
    A · B · C = B · A · C = C · B · A = C · A · B

5. Associative Laws:

  • The associative laws allow grouping of variables.
    Law 1: (A + B) + C = A + (B + C)
    Proof: (Pending – diagram)
  • This law can be extended to any no. of variables for e.g.
    A + (B + C + D) = (A + B + C) + D = (A + B) + (C + D)
    Law 2: (A · B) · C = A · (B · C)
    Proof: (Pending – diagram)
  • This law can be extended to any no. of variables for e.g.
    A · (B · C · D) = (A · B · C) · D = (A · B) · (C · D)

6. DistributiveLaws:

  • The distributive laws allow factoring or multiplying out ofexpressions.
    Law 1: A (B + C) = AB + AC
    Proof: (Pending – diagram)
    Law 2: A + BC = (A + B) (A + C)
    Proof: R.H.S. = (A + B) (A + C)
    = AA + AC + BA + BC
    = A + AC + BA + BC
    = A+BC (B’cz 1 + C + B = 1 + B =1)
    =L.H.S.
    Law 3: A + A’B = A + B
    Proof: L.H.S. = A +A’B
    = (A + A’) (A + B)
    = A + B
    = R.H.S.

7. Idempotence Laws:

  • Idempotence means the same value.
    Law 1: A · A = A Proof: Case 1: If A = 0  A · A = 0 · 0 = 0 = A Case 2: If A = 1  A · A = 1 · 1 = 1 = A (Pending – diagram) Law 2: A + A = A Proof: Case 1: If A = 0  A + A = 0 + 0 = 0 = A Case 2: If A = 1  A + A = 1 + 1 = 1 = A (Pending – diagram)

8. Complementation Law / Negation Law:

    Law 1: A · A’ = 0
    Proof:
    Case 1: If A = 0  A’ = 1 So, A · A’ = 0 · 1 =0
    Case 2: If A = 1  A’ = 0 So, A · A’ = 1 · 0 =0 (Pending – diagram)
    Law 2: A + A’ = 1
    Proof:
    Case 1: If A = 0  A’ = 1 So, A + A’ = 0 + 1 = 1 Case 2: If A = 1  A’ = 0 So, A + A’ = 1 + 0 = 1 (Pending – diagram)

9. Double Negation Law:

  • This law states that double negation of a variables is equal to the variable itself.
    Law 1: A’’ = A
    Proof:
    Case 1: If A = 0  A’’ = 0’’ = 1’ = A
    Case 2: If A = 1  A’’ = 1’’ = 0’ = A (Pending – diagram)
  • Any odd no. of inversion is equivalent to single in version.
  • Any even no. of inversion is equivalent to no inversion at all.

10. Identity Law:
       Law 1: A · 1 = A
       Proof:
       Case 1: If A= 1  A · 1 = 1 · 1 = 1 =A
       Case 2: If A= 0  A · 0 = 0 · 0 = 0 =A (Pending – diagram)
       Law 2: A + 1 = 1
       Proof:
       Case 1: If A= 1  A + 1 = 1 + 1 = 1 =A
       Case 2: If A= 0  A + 0 = 0 + 0 = 0 =A (Pending – diagram)

11. Null Law:
       Law 1: A · 0 = 0
       Proof:
       Case 1: If A= 1  A · 0 = 1 · 0 = 0 =0
       Case 2: If A= 0  A · 0 = 0 · 0 = 0 =0 (Pending – diagram)
       Law 2: A + 0 = A
       Proof:
       Case 1: If A= 1  A + 0 = 1 + 0 = 1 =A
       Case 2: If A= 0  A + 0 = 0 + 0 = 0 =A (Pending – diagram)

12. Absorption Law:
       Law 1: A + AB = A
       Proof: L.H.S. = A +AB
                              = A (1 +B)
                              = A (1)
                              = A
                              = R.H.S.
       Law 2: A (A + B) = A Proof:
       L.H.S. = A (A + B)
                  = A · A + AB
                  = A + AB
                  = A (1 + B)
                  = A (1)
                  = A
                  = L.H.S.

13. Consensus Theorem:

Theorem 1: A · B + A’C + BC = AB + A’C
Proof: L.H.S. = AB + A’C +BC
= AB + A’C + BC (A +A’)
= AB + A’C + BCA + BCA’
= AB (1 + C) + A’C (1 + B)
= AB + A’C
= R.H.S.

  • This theorem can be extended as,
    AB + A’C + BCD = AB + A’C
    Theorem 2: (A + B) (A’ + C) (B + C) = (A + B) (A’ + C)
    Proof: L.H.S. = (A + B) (A’ + C) (B + C)
    = (AA’ + AC + A’B + BC) (B + C)
    = (0 + AC + A’B + BC) (B + C)
    = ACB +ACC +A’BB + A’BC + BCB + BCC
    = ABC + AC + A’B + A’BC + BC + BC
    = ABC + AC + A’B + A’BC + BC
    = AC (1 + B) + A’B (1 + C) + BC
    = AC + A’B+BC …………………………….. (1)
    R.H.S. = (A + B) (A’ + C)
    = AA’ + AC + BA’ + BC
    = 0 + AC + BA’ + BC
    = AC + A’B+BC …………………………….. (2)
    Equation (1) = Equation (2) So.
    L.H.S = R.H.S.
  • This theorem can be extended to any no. of variables.
    (A + B) (A’ + C) (B + C + D) = (A + B) (A’ + C)

14. Transposition theorem:

Theorem: AB + A’C = (A + C) (A’ +B)
Proof: R.H.S. = (A + C) (A’ +B)
= AA’ + AB + CA’ + CB
= 0 + AB + CA’ + CB
= AB + CA’ + CB
= AB+A’C (B’cz of AB + A’C + BC = AB +A’C)
=L.H.S.

15. De Morgan’sTheorem:
       Law 1: (A + B)’ = A’ · B’
       Proof: (Pending – diagram)
       Law 2: (A· B)’ = A’ + B’

16. Duality Theorem:

  • Duality theorem arises as a result of presence of two logic system i.e. positive & negative logic
    system.
  • This theorem helps to convert from one logic system to another.
  • From changing one logic system to another following steps are taken:
    1) 0 becomes 1, 1 becomes0.
    2) AND becomes OR, OR becomes AND.
    3) ‘+’ becomes ‘·’, ‘·’ becomes ‘+’.
    4) Variables are not complemented in the process. (Pending – expressions)

2.3 REDUCTION OF BOOLEAN EXPRESSIONS :

2.3.1 Demorganize the following functions

Example 1: [(A + B’) (C + D’)]’
Answer: [(A + B’) (C + D’)]’
               = (A + B’)’ + (C + D’)’
               = A’ B’’ + C’D’’
               = A’B + C’D

Example 2: [(AB)’ (CD + E’F) ((AB)’ + (CD)’)]’
Answer: [(AB)’ (CD + E’F) ((AB)’ + (CD)’)]’
               = (AB)’’ + (CD + E’F)’ + ((AB)’ + (CD)’)’
               = AB + [(CD)’ (E’F)’] + [(AB)’’ (CD)’’]
               = AB + (C’ + D’) (E + F’) + ABCD

Example 3: [(AB)’ + A’ + AB]’
Answer: [(AB)’ + A’ + AB]’
               = AB’’ · A’’ · AB’
               = ABA (A’ + B’)
               = AB (A’ + B’)
               = ABA’ + ABB’
               = 0

Example 4: [P (Q + R)]’
Answer: [P (Q + R)]’
               = P’ + (Q + R)’
               = P’ + Q’ R’

Example 5: [(P + Q’) (R’ + S)]’
Answer: [(P + Q’) (R’ + S)]’
               = (P + Q’)’ + (R’ + S)’
               = P’ Q’’ + R’’ S’
               = P’Q + RS’

Example 6: [[(A + B)’ (C + D)’]’ [(E + F)’ (G + H)’]’]’
Answer: [[(A + B)’ (C + D)’]’ [(E + F)’ (G + H)’]’]’
               = [(A + B)’ (C + D)’]’’ + [(E + F)’ (G + H)’]’’
               = [(A + B)’ (C + D)’] + [(E + F)’ (G + H)’]
               = A’B’C’D’ + E’F’G’H’

2.3.2 Reducing the following functions

Example 1: A [ B + C’ (AB + AC’)’ ]
Answer: A [ B + C’ (AB + AC’)’ ]
               = A [ B + C’ (AB)’ (AC’)’ ]
               = A [ B + C’ (A’ + B’) (A’ + C) ]
               = A [ B + (A’ C’ + B’ C’) (A’ + C) ]
               = A [ B + (A’ C’A’ + B’ C’A’) (A’ C’ C + B’ C’ C) ]
               = A [ B + (A’C’ + B’ C’A’) (0 + 0) ]
               =  A [ B + A’C’ ( 1 + B’) ]
               = A [ B +A’C’]
               = AB + A’AC’
               = AB + 0
               = AB

Example 2: A + B [ AC + (B + C’)D ]
Answer: A + B [ AC + (B + C’)D ]
               = A + B [ AC + (BD + C’D) ]
               = A + ABC + BBD + BC’D
               = A + ABC + BD + BC’D
               = A (1 + BC) + BD (1 + C’)
               = A (1) + BD (1)
               = A + BD

Example 3: (A + (BC)’)’ (AB’ + ABC)
Answer: (A + (BC)’)’ (AB’ + ABC)
               = (A’ (BC)’’) (AB’ + ABC)
               = (A’BC) (AB’ + ABC)
               = A’BCAB’ + A’BCABC
               = 0 + 0
               = 0

Example 4: (B + BC) (B + B’C) (B + D)
Answer: (B + BC) (B + B’C) (B + D)
               = (BB + BB’C + BBC + BCB’C) (B + D)
               = (B + 0 + BC + 0) (B + D)
               = B (1 + C) (B + D)
               = B (B + D)
               = BB + BD
               = B + BD
               = B (1 + D)
               = B

Example 5: AB + AB’C +BC’
Answer: AB + AB’C +BC’
               = A (B + B’C) + BC’
               = A (B + B’) (B + C) + BC’
               = A (1) (B + C) + BC’
               = AB + AC +BC’
               = CA + C’B +AB
               = CA+C’B (B’cz of Consensus theorem1)

Example 6: AB’C + B + BD’ + ABD’ + A’C
Answer: AB’C + B + BD’ + ABD’ + A’C
               = AB’C + B (1 + D’ + AD’) + A’C
               = AB’C + B + A’C
               = C (A’ + AB’) + B
               = C (A’ + A) (A’ + B’) + B
               = C (1) (A’ + B’) + B
               = C (A’ + B’) + B
               = A’C + CB’ + B
               = A’C + (C + B) (B’ + B)
               = A’C + (B + C) (1)
               = A’C + B + C
               = C (1 + A’) + B
               = B + C

Example 7: A’B’ + A’B
Answer: A’B’ + A’B = A’ (B’ + B) = A’

Example 8: A’B’ + AB’
Answer: A’B’ + AB’
               = B’ (A’ + A)
               = B’

Example 9: A’B + AB
Answer: A’B + AB
               = B (A’ + A)
               = B (1)
               = B 

Example 10: A’B’ + A’B + AB’ + AB
Answer: A’B’ + A’B + AB’ + AB
               = A’ (B’ + B) + A (B’ + B)
               = A’ + A
               = 1

Example 11: [(A + B) (A’ + B)] + [(A + B) (A + B’)]
Answer: [(A + B) (A’ + B)] + [(A + B) (A + B’)]
               = [AA’ + AB + BA’ + BB] + [AA + AB’ + BA + BB”]
               = [0 + AB + A’B + B] + [A + AB’ + AB + 0]
               = [B (A + A’ + 1)] + [A (1 + B’ + B)]
               = B +A
               = A +B

Example 12: [(A + B’) (A’ + B’)] + [(A’ + B’) (A’ + B’)]
Answer: [(A + B’) (A’ + B’)] + [(A’ + B’) (A’ + B’)]
               = [AA’ + AB’ + B’A’ + B’B’] + [A’A’ + A’B’ + B’A’ + B’B’]
               = [0 + AB’ + A’B’ + B’] + [A’ + A’B’ + B’]
               = [B’ (A + A’ + 1)] + [A’ + B’(1)]
               = B’ + A’ + B’
               = A’ + B’

Example 13: (A + B) (A + B’) (A’ + B) (A’ + B’)
Answer: (A + B) (A + B’) (A’ + B) (A’ + B’)
               = (AA + AB’ + BA + BB’) (A’A’ + A’B’ + BA’ + BB’)
               = [A (1+ B’ + B)] [A’ (1 + B’ + B)]
               = AA’
               = 0

2.4 DIFFERENT FORMS OF BOOLEAN ALGEBRA

There are two types of boolean form
1) Standard form
2) Canonical form

2.4.1 Standard Forms

  • In this configuration, the terms that form the function may contain one, two, or any number of
    literals.
  • There are two types of standard forms:                                                                                                  (i) sum of product (SOP)                                                                                                                           (ii) product of sum(POS). 

Sum of Product(SOP)

SOP is a Boolean expression containing AND terms, called product terms, of one or more literals each. The sum denote the ORing of these terms.
An example of a function expressed in sum of product is:
F = Y’ + XY + X’YZ’

Product of Sum(POS)

The OPS is a Boolean expression containing OR terms, called sum terms. Each terms may have any no. of literals. The
product denotes ANDing of these terms.
An example of a function expressed in product of sum is:
F = X (Y’ + Z) (X’ + Y + Z’ + W)

  • A Boolean expression function may be expressed in a nonstandard form. For example the function:
    F = (AB + CD) (A’B’ + C’D’)
    Above function is neither sum of product nor in product sums. It can be changed to a standard form by using
    distributive law as below;
    F = ABC’D’ + A’B’CD

2.4.2 Canonical Forms
          Any boolean expression can be expressed in Sum of Product (SOP) form or Product of Sum (POS) form, they are called canonical form.

SOP – Sum of Product

  • A standard SOP form is one in which a no. of product terms, each one of which contains all the
    variables of the function either in complemented or non-complemented form, summed together.
  • Each of the product term is called MINTERM.
  • For min terms,
    Each non-complemented variable  1 Each
    complemented variable  0
  • Decimal equivalent is expressed in terms of lower case‘m’.
     For example,
    1. XYZ = 111 = m7
    2. A’BC = 011 = m3
    3. P’Q’R’ = 000 = m0
    4. T’S’ = 00 =m0
    5. B’C = 01 =m1
     Example 1: F1 = X’Y’Z + XY’Z’ +XYZ
    = 001 + 100 + 111
    = m1 + m4 + m7
    = Σm(1,4,7)
     Example 2: F2 = P’Q’ +PQ
    = 00 + 11
    = m0 +m3
    =Σm(0,3)
     Example 3: F3 = XY’ZW + XYZ’W’ +X’Y’Z’W’
    = 1011 + 1100 + 0000
    = m11 + m12 + m0
    = Σm(0,11,12)

POS – Product ofSum

  • A standard POS form is one in which a no. of sum terms, each one of which contains all the variables of the function either in complemented or non-complemented form, are multiplied together.
  • Each of the product term is called MAXTERM.
  • For maxterms,
    Each non-complemented variable  0 Each
    complemented variable  1
  • Decimal equivalent is expressed in terms of upper case ‘M’.
     For example,
    1. X+Y+Z = 000 = M0
    2. P’+Q’+R’ = 111 = M7
    3. A’+B+C’+D = 1010 = M10
     Example 1: F1 =(P’+Q)(P+Q’)
    = (10)(01) = M2·M1
    = ΠM(1,2)
     Example 2: F2=(X’+Y’+Z’+W)(X’+Y+Z+W’)(X+Y’+Z+W’)
    = (1110)(1001)(0101)
    =M14·M9·M5
    =ΠM(5,9,14)
     Example 3: F3 =(A’+B+C)(A+B’+C)(A+B+C’)
    = (100) (010) (001)
    = M4 M2M1
    =ΠM(1,2,4).

Min terms & Max terms for 3variables

(Pending – table)

2.4.3 Conversion between Canonical forms 

  • The complement of a function expressed as the sum of min terms equals the sum of min terms missing from original function
  • This is because the original function is expressed by those min terms that make the function equal to 1, while its complement is 1 for those min terms that the function is0. 
  • Example 1: F(A,B,C)= Σ(1,4,5,6,7)  F’(A,B,C) = ΠM(0,2,3) 

STEP 1: Take complement of the given function; F’(A,B,C) = Σ(0,2,3) = (m0 + m2 +m3)’                    STEP 2: Put value of MINTERM in form of variables; F’= (A’B’C’ + A’BC’ +A’BC)’ =(A+B+C)(A+B’+C)(A+B’+C’) =M0·M2·M3 =ΠM(0,2,3) In general, mj’ = Mj 

  • Example 2: F(A,B,C,D)= ΠM(0,3,7,10,14,15) 

STEP 1: Take complement of the given function; F’(A,B,C,D)= ΠM(1,2,4,5,6,8,9,11,12,13) = (M1 M2 M4 M5 M6 M8 M9 M11 M12 M13)’                                                                                                                      STEP 2: Put value of MAXTERM in form of variables; F’= [(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’)(A’+B’+C+D)(A’+B’+C+D’)]’ = (A’B’C’D) + (A’B’CD’) + (A’BC’D’) + (A’BC’D) +(A’BCD’) (AB’C’D’) + (AB’C’D) + (AB’CD) + (ABC’D’) +(ABC’D) = m1 + m2 + m4 + m5 + m6 + m8 + m9 + m11 + m12 + m13 = Σm(1,2,4,5,6,8,9,11,12,13) 

Convert to Min terms 

  • Example1: F = A + B’C

Answer: A  B & C is missing. So multiply with (B + B’) & (C + C’).                                                              B’C  A is missing. So multiply with (A + A’).                                                                                                      A = A (B + B’) (C + C’)                                                                                                                                                    = (AB + AB’) (C +C’)                                                                                                                                                  = ABC + AB’C + ABC’ + AB’C’                                                                                                                              B’C = B’C (A + A’)                                                                                                                                                              = AB’C + A’B’C

 So, A + B’C = ABC + AB’C + ABC’ + AB’C’ + AB’C + A’B’C                                                                                                         = ABC + AB’C + ABC’ + AB’C’ + A’B’C                                                                                                                     = 111 + 101 + 110 + 100 + 001                                                                                                                               = m7 + m6 + m5 + m4 + m1                                                                                                                                   = Σm(1,4,5,6,7) 

Convert to Maxterms 

  • Example1: F = A (B + C’)

Answer: A  B & C is missing. So add BB’ & CC’.                                                                                                B + C’  A is missing. So add AA’.                                                                                                                          A = A + BB’ + CC’                                                                                                                                                           = (A + B) (A +B’) + CC = (A + B + CC’) (A + B’ + CC’) = (A + B + C) (A + B + C’) (A + B’ + C) (A + B’ + C’) B + C’ = B + C’ + AA’ = (A + B + C’) (A’ + B + C’) So, 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’) (A’ + B + C’) = (000) (001) (010) (011) (101) = ΠM(0,1,2,3,5) 

  • Example 2:F = XY +X’Y 

Answer: F = XY + X’Y = (XY + X’) (XY + Z) = (X +X’) (Y + X’) (X + Z) (Y + Z) = (Y + X’) (X + Z) (Y + Z) X’ + Y = X’ + Y + ZZ’ = (X’ + Y + Z) (X’ + Y + Z’) = (100) (101) X + Z = X’ + Z + YY’ = (X + Y + Z) (X + Y’ + Z) = (000) (010) Y + Z = Y + Z + XX’ = (X + Y + Z) (X’ + Y + Z) = (000) (100) So, F = XY + X’Z = (100) (101) (000) (010) = M4 M5 M0 M2 F = ΠM(0,2,4,5)  

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