SWITCHING THEORY AND LOGIC CIRCUITS, Study notes of Logic

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SWITCHING THEORY AND

LOGIC CIRCUITS

COURSE OBJECTIVES

1. To understand the concepts and techniques associated with the number systems and codes
2. To understand the simplification methods (Boolean algebra & postulates, k-map method and tabular method) to simplify the given Boolean function.
3. To understand the fundamentals of digital logic and to design various combinational and sequential circuits.
4. To understand the concepts of programmable logic devices(PLDs)
5. To understand formal procedure for the analysis and design of synchronous and asynchronous sequential logic

COURSE OUTCOMES

After completion of the course the student will be able to

4. Design and analyze various combinational circuits like decoders, encoders, multiplexers, and de-multiplexers, arithmetic circuits (half adder, full adder, multiplier etc).
5. Design and analyze various sequential circuits like flip-flops, registers, counters etc.
6. Analyze and Design synchronous and asynchronous sequential circuits.

UNIT-I

Introductory Concepts

(Number systems, Base conversions)

Understanding Decimal Numbers

● Decimal numbers are made of decimal digits: (0,1,2,3,4,5,6,7,8,9)  Base = 10 ● How many items does decimal number 8653 represents? ● 8653 = 8 x10^3 + 6 x10^2 + 5 x10^1 + 3 x10^0 ● Number = d 3 x B^3 + d 2 x B^2 + d 1 x B^1 + d 0 x B^0 = Value ● What about fractions? ● 97654.35 = 9x10^4 + 7x10^3 + 6x10^2 + 5x10^1 + 4x10^0 + 3x10-1^ + 5x10- ● In formal notation → (97654.35) 10

1000 100 10 1 Weight

Understanding Octal Numbers

● Octal numbers are made of octal digits: (0,1,2,3,4,5,6,7) ● How many items does an octal number represent? ● 512 64 8 1 = Weights ● (4536) 8 = 4x8^3 + 5x8^2 + 3x8^1 + 6x8^0 = (2398) 10 ● What about fractions? ● (465.27) 8 = 4x8^2 + 6x8^1 + 5x8^0 + 2x8-1^ + 7x8- ● Octal numbers don’t use digits 8 or 9

Understanding Binary Numbers

● Binary numbers are made of binary digits (bits): ● 0 and 1 ● How many items does a binary number represent? ● 8 4 2 1 = Weights ● (1011) 2 = 1x2^3 + 0x2^2 + 1x2^1 + 1x2^0 = (11) 10 ● What about fractions? ● (110.10) 2 = 1x2^2 + 1x2^1 + 0x2^0 + 1x2-1^ + 0x2- ● Groups of eight bits are called a byte ● (11001001) 2 ● Groups of four bits are called a nibble ● (1101) 2

Putting It All Together

● Binary, octal, and hexadecimal are similar ● Easy to build circuits to operate on these representations ● Possible to convert between the three formats

Conversion Between Number Bases

Decimal (base 10)

Octal (base 8) Binary (base 2)

Hexadecimal (base 16)

● Learn to convert between bases ● Already demonstrated how to convert from binary to decimal

Convert an Integer from Decimal to Another Base

1. Divide decimal number by the base (e.g. 2)
2. The remainder is the lowest-order digit
3. Repeat first two steps until no divisor remains

For each digit position:

Example for (13) 10 : Quotient 13/2 =6/2 = 63 ++ 10 aa 0 = 1 3/2 = 1 + 1 a^12 = 0= 1 1/2 = 0 + 1 a 3 = 1

Remainder Coefficient

Answer (13) 10 = (a 3 a 2 a 1 a 0 ) 2 = (1101) 2 MSB LSB

The Growth of Binary Numbers

n 2 n 0 20 = 1 21 = 2 22 = 3 23 = 4 24 = 5 25 = 6 26 = 7 27 =

n 2 n 8 28 = 9 29 = 10 210 = 11 211 = 12 212 = 20 220 =1M 30 230 =1G 40 240 =1T

Mega Giga Tera

Kilo

Convert an Integer from Decimal to Octal

1. Divide decimal number by the base (8)
2. The remainder is the lowest-order digit
3. Repeat first two steps until no divisor remains

For each digit position:

Example for (175) 10 : Quotient 175/8 =21/8 = 212 ++ 75 aa 0 = 7 2/8 = 0 + 2 a^12 = 5= 2

Remainder Coefficient

Answer (175) 10 = (a 2 a 1 a 0 ) 8 = (257) 8

Conversion Between Base 16 and Base 2

● Conversion is easy! Determine the 4-bit binary value for each hex digit ● Note that there are 16 different values of four bits ● Easier to read and write in hexadecimal ● Representations are equivalent!

3A9F 16 = 0011 1010 1001 1111 2

3 A 9 F

Conversion Between Base 16 and Base 8

1. Convert from Base 16 to Base 2
2. Regroup bits into groups of three starting from right
3. Ignore leading zeros
4. Each group of three bits forms an octal digit

3 5 2 3 7

3A9F 16 = 0011 1010 1001 1111 2

3 A 9 F