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Comprehensive Guide to Number Systems and Boolean Algebra in Digital Logic Design

Explore decimal, binary, hexadecimal, and octal number systems, including conversions and representations, essential for understanding digital logic design fundamentals.

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Digital Logic Design Sanjivani Rural Education Society’s Sanjivani College of Engineering, Kopargaon-423603 (An Autonomous Institute Affiliated to Savitribai Phule Pune University, Pune) NAAC ‘A’ Grade Accredited, ISO 9001:2015 Certified Department of Information Technology (NBAAccredited) Dr.R.D.Chintamani Assistant Professor
Digital Logic Design Sanjivani Rural Education Society’s Sanjivani College of Engineering, Kopargaon-423603 (An Autonomous Institute Affiliated to Savitribai Phule Pune University, Pune) NAAC ‘A’ Grade Accredited, ISO 9001:2015 Certified Department of Information Technology (NBAAccredited) Dr.R.D.Chintamani Assistant Professor UNIT – I : NUMBER SYSTEM AND BOOLEAN ALGEBRA Topic 1.1-1.2 NUMBER SYSTEMS NUMBER CONVERSION
Number Systems Decimal Number System Binary Number System Hexadecimal Number System Octal Number System Conversion of numbers between Number Systems Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Decimal Number Systems Decimal Number System is the most popular number system used across the globe to represent numbers. This number system uses 10 different symbols to represent any number. It is said to be Base 10 or Radix 10 number system. Every digit in the decimal number system has weightage of some power of 10, e.g. a number 7,825 can be expressed as: 7 x 103 + 8 x 102 + 2 x 101 + 5 x 100 A fractional decimal number 12.34 can be expressed as: 1 x 101 + 2 x 100 + 3 x 10-1 + 4 x 10-2 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Binary Number Systems Binary Number System is used in digital devices and systems e.g. computers. This number system uses only 2 symbols 0 and 1 to represent any number. It is said to be Base 2 or Radix 2 number system. Every digit in the binary number system has weightage of some power of 2, e.g. a binary number 1011 can be expressed as: 1 x 23 + 0 x 22 + 1 x 21 + 1 x 20 A fractional binary number 11.01 can be expressed as: 1 x 21 + 1 x 20 + 0 x 2-1 + 1 x 2-2 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Hexademial and Octal Number Systems Binary Number System uses only 2 symbols to represent any number hence numbers represented in binary are usually very long numbers consisting of many binary bits. To quickly convert and represent numbers in number system with compact representation Hexadecimal or Octal Numbers systems are used. Hexadecimal number system has 16 symbols 0 to 9 and A to F. Octal numbers system has 8 symbols 0 to 7. Numbers can be quickly converted from Binary to Hexadecimal/Octal or vice versa. Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Hexadecimal to Binary and Vice versa Ex:- Convert (1011110001010011)2 to hex • Step 1: Make group of 4 binary bits from left to right 1011 1100 0101 0011 • Step 2: Replace each 4-bit binary group with its hex equivalent. 1011 1100 0101 0011 (B C 5 3)16 Binary Hex 0000 0 0001 1 0010 2 0011 3 0100 4 0101 5 0110 6 0111 7 1000 8 1001 9 1010 A 1011 B 1100 C 1101 D 1110 E 1111 F Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Hexadecimal to Binary and Vice versa Ex:- Convert (1011110001010011)2 to hex • Step 1: Make group of 4 binary bits from right to left 1011 1100 0101 0011 • Step 2: Replace each 4-bit binary group with its hex equivalent. 1011 1100 0101 0011 (B C 5 3)16 Binary Hex 0000 0 0001 1 0010 2 0011 3 0100 4 0101 5 0110 6 0111 7 1000 8 1001 9 1010 A 1011 B 1100 C 1101 D 1110 E 1111 F Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Hexadecimal to Binary and Vice versa Ex:- Convert (1011110001010011)2 to hex • Step 1: Make group of 4 binary bits from right to left 1011 1100 0101 0011 • Step 2: Replace each 4-bit binary group with its hex equivalent. 1011 1100 0101 0011 (B C 5 3)16 Binary Hex 0000 0 0001 1 0010 2 0011 3 0100 4 0101 5 0110 6 0111 7 1000 8 1001 9 1010 A 1011 B 1100 C 1101 D 1110 E 1111 F Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Hexadecimal to Binary and Vice versa Ex:- Convert (1011110001010011)2 to hex • Step 1: Make group of 4 binary bits from right to left 1011 1100 0101 0011 • Step 2: Replace each 4-bit binary group with its hex equivalent. 1011 1100 0101 0011 (B C 5 3)16 Binary Hex 0000 0 0001 1 0010 2 0011 3 0100 4 0101 5 0110 6 0111 7 1000 8 1001 9 1010 A 1011 B 1100 C 1101 D 1110 E 1111 F Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Hexadecimal to Binary and Vice versa Ex:- Convert (1011110001010011)2 to hex • Step 1: Make group of 4 binary bits from right to left 1011 1100 0101 0011 • Step 2: Replace each 4-bit binary group with its hex equivalent. 1011 1100 0101 0011 (B C 5 3)16 Binary Hex 0000 0 0001 1 0010 2 0011 3 0100 4 0101 5 0110 6 0111 7 1000 8 1001 9 1010 A 1011 B 1100 C 1101 D 1110 E 1111 F Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Hexadecimal to Binary and Vice versa Ex:- Convert (1011110001010011)2 to hex • Step 1: Make group of 4 binary bits from right to left 1011 1100 0101 0011 • Step 2: Replace each 4-bit binary group with its hex equivalent. 1011 1100 0101 0011 (B C 5 3)16 Binary Hex 0000 0 0001 1 0010 2 0011 3 0100 4 0101 5 0110 6 0111 7 1000 8 1001 9 1010 A 1011 B 1100 C 1101 D 1110 E 1111 F Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Hexadecimal to Binary and Vice versa Ex:- Convert (1011110001010011)2 to hex • Step 1: Make group of 4 binary bits from right to left 1011 1100 0101 0011 • Step 2: Replace each 4-bit binary group with its hex equivalent. 1011 1100 0101 0011 (B C 5 3)16 Binary Hex 0000 0 0001 1 0010 2 0011 3 0100 4 0101 5 0110 6 0111 7 1000 8 1001 9 1010 A 1011 B 1100 C 1101 D 1110 E 1111 F Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Hexadecimal to Binary and Vice versa Ex:- Convert (1011110001010011)2 to hex • Step 1: Make group of 4 binary bits from right to left 1011 1100 0101 0011 • Step 2: Replace each 4-bit binary group with its hex equivalent. 1011 1100 0101 0011 (B C 5 3)16 Binary Hex 0000 0 0001 1 0010 2 0011 3 0100 4 0101 5 0110 6 0111 7 1000 8 1001 9 1010 A 1011 B 1100 C 1101 D 1110 E 1111 F Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Hexadecimal to Binary and Vice versa Ex:- Convert (1011110001010011)2 to hex • Step 1: Make group of 4 binary bits from right to left 1011 1100 0101 0011 • Step 2: Replace each 4-bit binary group with its hex equivalent. 1011 1100 0101 0011 (B C 5 3)16 Binary Hex 0000 0 0001 1 0010 2 0011 3 0100 4 0101 5 0110 6 0111 7 1000 8 1001 9 1010 A 1011 B 1100 C 1101 D 1110 E 1111 F Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Hexadecimal to Binary and Vice versa Ex:- Convert (1011110001010011)2 to hex • Step 1: Make group of 4 binary bits from right to left 1011 1100 0101 0011 • Step 2: Replace each 4-bit binary group with its hex equivalent. 1011 1100 0101 0011 (B C 5 3)16 Binary Hex 0000 0 0001 1 0010 2 0011 3 0100 4 0101 5 0110 6 0111 7 1000 8 1001 9 1010 A 1011 B 1100 C 1101 D 1110 E 1111 F Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Hexadecimal to Binary and Vice versa Ex:- Convert (1011110001010011)2 to hex • Step 1: Make group of 4 binary bits from right to left 1011 1100 0101 0011 • Step 2: Replace each 4-bit binary group with its hex equivalent. 1011 1100 0101 0011 (B C 5 3)16 Binary Hex 0000 0 0001 1 0010 2 0011 3 0100 4 0101 5 0110 6 0111 7 1000 8 1001 9 1010 A 1011 B 1100 C 1101 D 1110 E 1111 F Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Hexadecimal to Binary and Vice versa Ex:- Convert (1011110001010011)2 to hex • Step 1: Make group of 4 binary bits from right to left 1011 1100 0101 0011 • Step 2: Replace each 4-bit binary group with its hex equivalent. 1011 1100 0101 0011 (B C 5 3)16 Binary Hex 0000 0 0001 1 0010 2 0011 3 0100 4 0101 5 0110 6 0111 7 1000 8 1001 9 1010 A 1011 B 1100 C 1101 D 1110 E 1111 F Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Hexadecimal to Binary and Vice versa Ex:- Convert (1011110001010011)2 to hex • Step 1: Make group of 4 binary bits from right to left 1011 1100 0101 0011 • Step 2: Replace each 4-bit binary group with its hex equivalent. 1011 1100 0101 0011 (B C 5 3)16 Thus (1011110001010011)2 = (BC53)16 Binary Hex 0000 0 0001 1 0010 2 0011 3 0100 4 0101 5 0110 6 0111 7 1000 8 1001 9 1010 A 1011 B 1100 C 1101 D 1110 E 1111 F Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Hexadecimal to Binary and Vice versa Ex:- Convert (1011110001010011)2 to hex • Step 1: Make group of 4 binary bits from right to left 1011 1100 0101 0011 • Step 2: Replace each 4-bit binary group with its hex equivalent. 1011 1100 0101 0011 (B C 5 3)16 Thus (1011110001010011)2 = (BC53)16 Note that there are 16 binary bits in the binary representation of the number where as its hex representation has only 4 digits. Binary Hex 0000 0 0001 1 0010 2 0011 3 0100 4 0101 5 0110 6 0111 7 1000 8 1001 9 1010 A 1011 B 1100 C 1101 D 1110 E 1111 F Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Hexadecimal to Binary and Vice versa Ex:- Convert hex (4A59)16 to binary • Step 1: Replace every hex digit with its 4-bit binary equivalent 4 A 5 9 0100 1010 0101 1001 Step 2: Arrange the bits in sequence. 0100101001011001 Thus (4A59)16 = (0100101001011001)2 Binary Hex 0000 0 0001 1 0010 2 0011 3 0100 4 0101 5 0110 6 0111 7 1000 8 1001 9 1010 A 1011 B 1100 C 1101 D 1110 E 1111 F Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Hexadecimal to Binary and Vice versa Ex:- Convert hex (4A59)16 to binary • Step 1: Replace every hex digit with its 4-bit binary equivalent 4 A 5 9 0100 1010 0101 1001 Step 2: Arrange the bits in sequence. 0100101001011001 Thus (4A59)16 = (0100101001011001)2 Binary Hex 0000 0 0001 1 0010 2 0011 3 0100 4 0101 5 0110 6 0111 7 1000 8 1001 9 1010 A 1011 B 1100 C 1101 D 1110 E 1111 F Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Hexadecimal to Binary and Vice versa Ex:- Convert hex (4A59)16 to binary • Step 1: Replace every hex digit with its 4-bit binary equivalent 4 A 5 9 0100 1010 0101 1001 Step 2: Arrange the bits in sequence. 0100101001011001 Thus (4A59)16 = (0100101001011001)2 Binary Hex 0000 0 0001 1 0010 2 0011 3 0100 4 0101 5 0110 6 0111 7 1000 8 1001 9 1010 A 1011 B 1100 C 1101 D 1110 E 1111 F Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Hexadecimal to Binary and Vice versa Ex:- Convert hex (4A59)16 to binary • Step 1: Replace every hex digit with its 4-bit binary equivalent 4 A 5 9 0100 1010 0101 1001 Step 2: Arrange the bits in sequence. 0100101001011001 Thus (4A59)16 = (0100101001011001)2 Binary Hex 0000 0 0001 1 0010 2 0011 3 0100 4 0101 5 0110 6 0111 7 1000 8 1001 9 1010 A 1011 B 1100 C 1101 D 1110 E 1111 F Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Hexadecimal to Binary and Vice versa Ex:- Convert hex (4A59)16 to binary • Step 1: Replace every hex digit with its 4-bit binary equivalent 4 A 5 9 0100 1010 0101 1001 Step 2: Arrange the bits in sequence. 0100101001011001 Thus (4A59)16 = (0100101001011001)2 Binary Hex 0000 0 0001 1 0010 2 0011 3 0100 4 0101 5 0110 6 0111 7 1000 8 1001 9 1010 A 1011 B 1100 C 1101 D 1110 E 1111 F Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Hexadecimal to Binary and Vice versa Ex:- Convert hex (4A59)16 to binary • Step 1: Replace every hex digit with its 4-bit binary equivalent 4 A 5 9 0100 1010 0101 1001 Step 2: Arrange the bits in sequence. 0100101001011001 Thus (4A59)16 = (0100101001011001)2 Binary Hex 0000 0 0001 1 0010 2 0011 3 0100 4 0101 5 0110 6 0111 7 1000 8 1001 9 1010 A 1011 B 1100 C 1101 D 1110 E 1111 F Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Hexadecimal to Binary and Vice versa Ex:- Convert hex (4A59)16 to binary • Step 1: Replace every hex digit with its 4-bit binary equivalent 4 A 5 9 0100 1010 0101 1001 Step 2: Arrange the bits in sequence. 0100101001011001 Thus (4A59)16 = (0100101001011001)2 Binary Hex 0000 0 0001 1 0010 2 0011 3 0100 4 0101 5 0110 6 0111 7 1000 8 1001 9 1010 A 1011 B 1100 C 1101 D 1110 E 1111 F Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Hexadecimal to Binary and Vice versa Ex:- Convert hex (4A59)16 to binary • Step 1: Replace every hex digit with its 4-bit binary equivalent 4 A 5 9 0100 1010 0101 1001 Step 2: Arrange the bits in sequence. 0100101001011001 Thus (4A59)16 = (0100101001011001)2 Binary Hex 0000 0 0001 1 0010 2 0011 3 0100 4 0101 5 0110 6 0111 7 1000 8 1001 9 1010 A 1011 B 1100 C 1101 D 1110 E 1111 F Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Hexadecimal to Binary and Vice versa Ex:- Convert hex (4A59)16 to binary • Step 1: Replace every hex digit with its 4-bit binary equivalent 4 A 5 9 0100 1010 0101 1001 Step 2: Arrange the bits in sequence. 0100101001011001 Thus (4A59)16 = (0100101001011001)2 Binary Hex 0000 0 0001 1 0010 2 0011 3 0100 4 0101 5 0110 6 0111 7 1000 8 1001 9 1010 A 1011 B 1100 C 1101 D 1110 E 1111 F Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Hexadecimal to Binary and Vice versa Ex:- Convert hex (4A59)16 to binary • Step 1: Replace every hex digit with its 4-bit binary equivalent 4 A 5 9 0100 1010 0101 1001 Step 2: Arrange the bits in sequence. 0100101001011001 Thus (4A59)16 = (0100101001011001)2 Binary Hex 0000 0 0001 1 0010 2 0011 3 0100 4 0101 5 0110 6 0111 7 1000 8 1001 9 1010 A 1011 B 1100 C 1101 D 1110 E 1111 F Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Octal to Binary and Vice versa Ex:- Convert (1011110001010011)2 to octal • Step 1: Make group of 3 binary bits from left to right 1 011 110 001 010 011 • Step 2: Replace each 3-bit binary group with its octal equivalent. 1 011 110 001 010 011 (1 3 6 1 2 3)8 Thus (1011110001010011)2 = (136123)8 Note that there are 16 binary bits in the binary representation of the number where as its octal representation has only 6 digits. Binary Octal 000 0 001 1 010 2 011 3 100 4 101 5 110 6 111 7 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Octal to Binary and Vice versa Ex:- Convert (1011110001010011)2 to octal • Step 1: Make group of 3 binary bits from right to left 1 011 110 001 010 011 • Step 2: Replace each 3-bit binary group with its octal equivalent. 1 011 110 001 010 011 (1 3 6 1 2 3)8 Thus (1011110001010011)2 = (136123)8 Note that there are 16 binary bits in the binary representation of the number where as its octal representation has only 6 digits. Binary Octal 000 0 001 1 010 2 011 3 100 4 101 5 110 6 111 7 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Octal to Binary and Vice versa Ex:- Convert (1011110001010011)2 to octal • Step 1: Make group of 3 binary bits from right to left 1 011 110 001 010 011 • Step 2: Replace each 3-bit binary group with its octal equivalent. 1 011 110 001 010 011 (1 3 6 1 2 3)8 Thus (1011110001010011)2 = (136123)8 Note that there are 16 binary bits in the binary representation of the number where as its octal representation has only 6 digits. Binary Octal 000 0 001 1 010 2 011 3 100 4 101 5 110 6 111 7 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Octal to Binary and Vice versa Ex:- Convert (1011110001010011)2 to octal • Step 1: Make group of 3 binary bits from right to left 1 011 110 001 010 011 • Step 2: Replace each 3-bit binary group with its octal equivalent. 1 011 110 001 010 011 (1 3 6 1 2 3)8 Thus (1011110001010011)2 = (136123)8 Note that there are 16 binary bits in the binary representation of the number where as its octal representation has only 6 digits. Binary Octal 000 0 001 1 010 2 011 3 100 4 101 5 110 6 111 7 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Octal to Binary and Vice versa Ex:- Convert (1011110001010011)2 to octal • Step 1: Make group of 3 binary bits from right to left 1 011 110 001 010 011 • Step 2: Replace each 3-bit binary group with its octal equivalent. 1 011 110 001 010 011 (1 3 6 1 2 3)8 Thus (1011110001010011)2 = (136123)8 Note that there are 16 binary bits in the binary representation of the number where as its octal representation has only 6 digits. Binary Octal 000 0 001 1 010 2 011 3 100 4 101 5 110 6 111 7 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Octal to Binary and Vice versa Ex:- Convert (1011110001010011)2 to octal • Step 1: Make group of 3 binary bits from right to left 1 011 110 001 010 011 • Step 2: Replace each 3-bit binary group with its octal equivalent. 1 011 110 001 010 011 (1 3 6 1 2 3)8 Thus (1011110001010011)2 = (136123)8 Note that there are 16 binary bits in the binary representation of the number where as its octal representation has only 6 digits. Binary Octal 000 0 001 1 010 2 011 3 100 4 101 5 110 6 111 7 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Octal to Binary and Vice versa Ex:- Convert (1011110001010011)2 to octal • Step 1: Make group of 3 binary bits from right to left 1 011 110 001 010 011 • Step 2: Replace each 3-bit binary group with its octal equivalent. 1 011 110 001 010 011 (1 3 6 1 2 3)8 Thus (1011110001010011)2 = (136123)8 Note that there are 16 binary bits in the binary representation of the number where as its octal representation has only 6 digits. Binary Octal 000 0 001 1 010 2 011 3 100 4 101 5 110 6 111 7 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Octal to Binary and Vice versa Ex:- Convert (1011110001010011)2 to octal • Step 1: Make group of 3 binary bits from right to left 1 011 110 001 010 011 • Step 2: Replace each 3-bit binary group with its octal equivalent. 1 011 110 001 010 011 (1 3 6 1 2 3)8 Thus (1011110001010011)2 = (136123)8 Note that there are 16 binary bits in the binary representation of the number where as its octal representation has only 6 digits. Binary Octal 000 0 001 1 010 2 011 3 100 4 101 5 110 6 111 7 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Octal to Binary and Vice versa Ex:- Convert (1011110001010011)2 to octal • Step 1: Make group of 3 binary bits from right to left 1 011 110 001 010 011 • Step 2: Replace each 3-bit binary group with its octal equivalent. 1 011 110 001 010 011 (1 3 6 1 2 3)8 Thus (1011110001010011)2 = (136123)8 Note that there are 16 binary bits in the binary representation of the number where as its octal representation has only 6 digits. Binary Octal 000 0 001 1 010 2 011 3 100 4 101 5 110 6 111 7 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Octal to Binary and Vice versa Ex:- Convert (1011110001010011)2 to octal • Step 1: Make group of 3 binary bits from right to left 1 011 110 001 010 011 • Step 2: Replace each 3-bit binary group with its octal equivalent. 1 011 110 001 010 011 (1 3 6 1 2 3)8 Thus (1011110001010011)2 = (136123)8 Note that there are 16 binary bits in the binary representation of the number where as its octal representation has only 6 digits. Binary Octal 000 0 001 1 010 2 011 3 100 4 101 5 110 6 111 7 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Octal to Binary and Vice versa Ex:- Convert (1011110001010011)2 to octal • Step 1: Make group of 3 binary bits from right to left 1 011 110 001 010 011 • Step 2: Replace each 3-bit binary group with its octal equivalent. 1 011 110 001 010 011 (1 3 6 1 2 3)8 Thus (1011110001010011)2 = (136123)8 Note that there are 16 binary bits in the binary representation of the number where as its octal representation has only 6 digits. Binary Octal 000 0 001 1 010 2 011 3 100 4 101 5 110 6 111 7 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Octal to Binary and Vice versa Ex:- Convert (1011110001010011)2 to octal • Step 1: Make group of 3 binary bits from right to left 1 011 110 001 010 011 • Step 2: Replace each 3-bit binary group with its octal equivalent. 1 011 110 001 010 011 (1 3 6 1 2 3)8 Thus (1011110001010011)2 = (136123)8 Note that there are 16 binary bits in the binary representation of the number where as its octal representation has only 6 digits. Binary Octal 000 0 001 1 010 2 011 3 100 4 101 5 110 6 111 7 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Octal to Binary and Vice versa Ex:- Convert (1011110001010011)2 to octal • Step 1: Make group of 3 binary bits from right to left 1 011 110 001 010 011 • Step 2: Replace each 3-bit binary group with its octal equivalent. 1 011 110 001 010 011 (1 3 6 1 2 3)8 Thus (1011110001010011)2 = (136123)8 Note that there are 16 binary bits in the binary representation of the number where as its octal representation has only 6 digits. Binary Octal 000 0 001 1 010 2 011 3 100 4 101 5 110 6 111 7 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Octal to Binary and Vice versa Ex:- Convert (1011110001010011)2 to octal • Step 1: Make group of 3 binary bits from right to left 1 011 110 001 010 011 • Step 2: Replace each 3-bit binary group with its octal equivalent. 1 011 110 001 010 011 (1 3 6 1 2 3)8 Thus (1011110001010011)2 = (136123)8 Note that there are 16 binary bits in the binary representation of the number where as its octal representation has only 6 digits. Binary Octal 000 0 001 1 010 2 011 3 100 4 101 5 110 6 111 7 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Octal to Binary and Vice versa Ex:- Convert (1011110001010011)2 to octal • Step 1: Make group of 3 binary bits from right to left 1 011 110 001 010 011 • Step 2: Replace each 3-bit binary group with its octal equivalent. 1 011 110 001 010 011 (1 3 6 1 2 3)8 Thus (1011110001010011)2 = (136123)8 Note that there are 16 binary bits in the binary representation of the number where as its octal representation has only 6 digits. Binary Octal 000 0 001 1 010 2 011 3 100 4 101 5 110 6 111 7 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Octal to Binary and Vice versa Ex:- Convert (1011110001010011)2 to octal • Step 1: Make group of 3 binary bits from right to left 1 011 110 001 010 011 • Step 2: Replace each 3-bit binary group with its octal equivalent. 1 011 110 001 010 011 (1 3 6 1 2 3)8 Thus (1011110001010011)2 = (136123)8 Note that there are 16 binary bits in the binary representation of the number where as its octal representation has only 6 digits. Binary Octal 000 0 001 1 010 2 011 3 100 4 101 5 110 6 111 7 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Octal to Binary and Vice versa Ex:- Convert (1011110001010011)2 to octal • Step 1: Make group of 3 binary bits from right to left 1 011 110 001 010 011 • Step 2: Replace each 3-bit binary group with its octal equivalent. 1 011 110 001 010 011 (1 3 6 1 2 3)8 Thus (1011110001010011)2 = (136123)8 Note that there are 16 binary bits in the binary representation of the number where as its octal representation has only 6 digits. Binary Octal 000 0 001 1 010 2 011 3 100 4 101 5 110 6 111 7 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Octal to Binary and Vice versa Ex:- Convert (1011110001010011)2 to octal • Step 1: Make group of 3 binary bits from right to left 1 011 110 001 010 011 • Step 2: Replace each 3-bit binary group with its octal equivalent. 1 011 110 001 010 011 (1 3 6 1 2 3)8 Thus (1011110001010011)2 = (136123)8 Note that there are 16 binary bits in the binary representation of the number where as its octal representation has only 6 digits. Binary Octal 000 0 001 1 010 2 011 3 100 4 101 5 110 6 111 7 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Binary/Octal/Hex to Decimal Ex:- Convert (10011)2, (456)8, (2B6)16 to decimal • Step 1: Multiply each digit value in the number by its weight represented as power 2, 8 or 16 • Step 2: Sum the product terms to get decimal number. a) (10011)2 10011 = 1 x 24 + 0 x 23 + 0 x 22 + 1 x 21 + 1 x 20 = 16 + 0 + 0 + 2 + 1 = (17)10 b) (456)8 = 4 x 82 + 5 x + 6 x 160 = 2 x 256 + 11 x 16 + 6 x 1 = 512 + 176 + 6 = (694)10 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Binary/Octal/Hex to Decimal Ex:- Convert (10011)2, (456)8, (2B6)16 to decimal • Step 1: Multiply each digit value in the number by its weight represented as power 2, 8 or 16 • Step 2: Sum the product terms to get decimal number. a) (10011)2 10011 = 1 x 24 + 0 x 23 + 0 x 22 + 1 x 21 + 1 x 20 = 16 + 0 + 0 + 2 + 512 + 176 + 6 = (694)10 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Binary/Octal/Hex to Decimal Ex:- Convert (10011)2, (456)8, (2B6)16 to decimal • Step 1: Multiply each digit value in the number by its weight represented as power 2, 8 or 16 • Step 2: Sum the product terms to get decimal number. a) (10011)2 10011 = 1 x 24 + 0 x 23 + 0 x 22 + 1 x 21 + 1 x 20 = 16 + 0 + 0 + 2 + 1 = (17)10 b) (456)8 = 4 x 82 + 5 x 81 + 6 x 80 = 4 x 16 + 5 x 8 + 6 x 1 = 64 + 40 + 6 = (110)10 c) (2B6)16 = 2 x 162 + 11 x 161 + 6 x 160 = 2 x 256 + 11 x 16 + 6 x 1 = 512 + 176 + 6 = (694)10 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Binary/Octal/Hex to Decimal Ex:- Convert (10011)2, (456)8, (2B6)16 to decimal • Step 1: Multiply each digit value in the number by its weight represented as power 2, 8 or 16 • Step 2: Sum the product terms to get decimal number. a) (10011)2 10011 = 1 x 24 + 0 x 23 + 0 x 22 + 1 x 21 + 1 x 20 = 16 + 0 + 0 + 2 + 1 = (17)10 b) (456)8 = 4 x 82 + 5 x 81 + 6 x 80 = 4 x 16 + 5 x 8 + 6 x 1 = 64 + 40 + 6 = (110)10 c) (2B6)16 = 2 x 162 + 11 x 161 + 6 x 160 = 2 x 256 + 11 x 16 + 6 x 1 = 512 + 176 + 6 = (694)10 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Binary/Octal/Hex to Decimal Ex:- Convert (10011)2, (456)8, (2B6)16 to decimal • Step 1: Multiply each digit value in the number by its weight represented as power 2, 8 or 16 • Step 2: Sum the product terms to get decimal number. a) (10011)2 10011 = 1 x 24 + 0 x 23 + 0 x 22 + 1 x 21 + 1 x 20 = 16 + 0 + 0 + 2 + 1 = (17)10 b) (456)8 = 4 x 82 + 5 x 81 + 6 x 80 = 4 x 16 + 5 x 8 + 6 x 1 = 64 + 40 + 6 = (110)10 c) (2B6)16 = 2 x 162 + 11 x 161 + 6 x 160 = 2 x 256 + 11 x 16 + 6 x 1 = 512 + 176 + 6 = (694)10 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Binary/Octal/Hex to Decimal Ex:- Convert (10011)2, (456)8, (2B6)16 to decimal • Step 1: Multiply each digit value in the number by its weight represented as power 2, 8 or 16 • Step 2: Sum the product terms to get decimal number. a) (10011)2 10011 = 1 x 24 + 0 x 23 + 0 x 22 + 1 x 21 + 1 x 20 = 16 + 0 + 0 + 2 + 1 = (19)10 b) (456)8 = 4 x 82 + 5 x 81 + 6 x 80 = 4 x 16 + 5 x 8 + 6 x 1 = 64 + 40 + 6 = (110)10 c) (2B6)16 = 2 x 162 + 11 x 161 + 6 x 160 = 2 x 256 + 11 x 16 + 6 x 1 = 512 + 176 + 6 = (694)10 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Binary/Octal/Hex to Decimal Ex:- Convert (10011)2, (456)8, (2B6)16 to decimal • Step 1: Multiply each digit value in the number by its weight represented as power 2, 8 or 16 • Step 2: Sum the product terms to get decimal number. a) (10011)2 10011 = 1 x 24 + 0 x 23 + 0 x 22 + 1 x 21 + 1 x 20 = 16 + 0 + 0 + 2 + 1 = (19)10 b) (456)8 = 4 x 82 + 5 x 81 + 6 x 80 = 4 x 16 + 5 x 8 + 6 x 1 = 64 + 40 + 6 = (110)10 c) (2B6)16 = 2 x 162 + 11 x 161 + 6 x 160 = 2 x 256 + 11 x 16 + 6 x 1 = 512 + 176 + 6 = (694)10 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Binary/Octal/Hex to Decimal Ex:- Convert (10011)2, (456)8, (2B6)16 to decimal • Step 1: Multiply each digit value in the number by its weight represented as power 2, 8 or 16 • Step 2: Sum the product terms to get decimal number. a) (10011)2 10011 = 1 x 24 + 0 x 23 + 0 x 22 + 1 x 21 + 1 x 20 = 16 + 0 + 0 + 2 + 1 = (19)10 b) (456)8 = 4 x 82 + 5 x 81 + 6 x 80 = 4 x 16 + 5 x 8 + 6 x 1 = 64 + 40 + 6 = (110)10 c) (2B6)16 = 2 x 162 + 11 x 161 + 6 x 160 = 2 x 256 + 11 x 16 + 6 x 1 = 512 + 176 + 6 = (694)10 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Binary/Octal/Hex to Decimal Ex:- Convert (10011)2, (456)8, (2B6)16 to decimal • Step 1: Multiply each digit value in the number by its weight represented as power 2, 8 or 16 • Step 2: Sum the product terms to get decimal number. a) (10011)2 10011 = 1 x 24 + 0 x 23 + 0 x 22 + 1 x 21 + 1 x 20 = 16 + 0 + 0 + 2 + 1 = (19)10 b) (456)8 = 4 x 82 + 5 x 81 + 6 x 80 = 4 x 64 + 5 x 8 + 6 x 1 = 64 + 40 + 6 = (110)10 c) (2B6)16 = 2 x 162 + 11 x 161 + 6 x 160 = 2 x 256 + 11 x 16 + 6 x 1 = 512 + 176 + 6 = (694)10 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Binary/Octal/Hex to Decimal Ex:- Convert (10011)2, (456)8, (2B6)16 to decimal • Step 1: Multiply each digit value in the number by its weight represented as power 2, 8 or 16 • Step 2: Sum the product terms to get decimal number. a) (10011)2 10011 = 1 x 24 + 0 x 23 + 0 x 22 + 1 x 21 + 1 x 20 = 16 + 0 + 0 + 2 + 1 = (19)10 b) (456)8 = 4 x 82 + 5 x 81 + 6 x 80 = 4 x 64 + 5 x 8 + 6 x 1 = 256 + 40 + 6 = (302)10 c) (2B6)16 = 2 x 162 + 11 x 161 + 6 x 160 = 2 x 256 + 11 x 16 + 6 x 1 = 512 + 176 + 6 = (694)10 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Binary/Octal/Hex to Decimal Ex:- Convert (10011)2, (456)8, (2B6)16 to decimal • Step 1: Multiply each digit value in the number by its weight represented as power 2, 8 or 16 • Step 2: Sum the product terms to get decimal number. a) (10011)2 10011 = 1 x 24 + 0 x 23 + 0 x 22 + 1 x 21 + 1 x 20 = 16 + 0 + 0 + 2 + 1 = (19)10 b) (456)8 = 4 x 82 + 5 x 81 + 6 x 80 = 4 x 64 + 5 x 8 + 6 x 1 = 256 + 40 + 6 = (302)10 c) (2B6)16 = 2 x 162 + 11 x 161 + 6 x 160 = 2 x 256 + 11 x 16 + 6 x 1 = 512 + 176 + 6 = (694)10 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Binary/Octal/Hex to Decimal Ex:- Convert (10011)2, (456)8, (2B6)16 to decimal • Step 1: Multiply each digit value in the number by its weight represented as power 2, 8 or 16 • Step 2: Sum the product terms to get decimal number. a) (10011)2 10011 = 1 x 24 + 0 x 23 + 0 x 22 + 1 x 21 + 1 x 20 = 16 + 0 + 0 + 2 + 1 = (19)10 b) (456)8 = 4 x 82 + 5 x 81 + 6 x 80 = 4 x 64 + 5 x 8 + 6 x 1 = 256 + 40 + 6 = (302)10 c) (2B6)16 = 2 x 162 + 11 x 161 + 6 x 160 = 2 x 256 + 11 x 16 + 6 x 1 = 512 + 176 + 6 = (694)10 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Binary/Octal/Hex to Decimal Ex:- Convert (10011)2, (456)8, (2B6)16 to decimal • Step 1: Multiply each digit value in the number by its weight represented as power 2, 8 or 16 • Step 2: Sum the product terms to get decimal number. a) (10011)2 10011 = 1 x 24 + 0 x 23 + 0 x 22 + 1 x 21 + 1 x 20 = 16 + 0 + 0 + 2 + 1 = (19)10 b) (456)8 = 4 x 82 + 5 x 81 + 6 x 80 = 4 x 64 + 5 x 8 + 6 x 1 = 256 + 40 + 6 = (302)10 c) (2B6)16 = 2 x 162 + 11 x 161 + 6 x 160 = 2 x 256 + 11 x 16 + 6 x 1 = 512 + 176 + 6 = (694)10 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Binary/Octal/Hex to Decimal Ex:- Convert (10011)2, (456)8, (2B6)16 to decimal • Step 1: Multiply each digit value in the number by its weight represented as power 2, 8 or 16 • Step 2: Sum the product terms to get decimal number. a) (10011)2 10011 = 1 x 24 + 0 x 23 + 0 x 22 + 1 x 21 + 1 x 20 = 16 + 0 + 0 + 2 + 1 = (19)10 b) (456)8 = 4 x 82 + 5 x 81 + 6 x 80 = 4 x 64 + 5 x 8 + 6 x 1 = 256 + 40 + 6 = (302)10 c) (2B6)16 = 2 x 162 + 11 x 161 + 6 x 160 = 2 x 256 + 11 x 16 + 6 x 1 = 512 + 176 + 6 = (694)10 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Decimal to Binary/Octal/Hexadecimal Ex:- Convert (861)10 to binary/octal/hex • Step 1: Divide the number by base value of destination number system i.e. 2, 8 or 16 and note the remainder and quotient. • Step 2: Divide quotient again and note remainder and new quotient repeatedly until quotient is less than base i.e. 2, 8 or 16 • Step 3: Arranging remainders in the reverse order of they obtained gives number represented in binary/octal/hex number system. Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Decimal to Binary/Octal/Hexadecimal Ex:- (861)10 to binary Quotient Remainder 861 ÷ 2 = 430 1 430 ÷ 2 = 215 0 215 ÷ 2 = 107 1 107 ÷ 2 = 58 0 58 ÷ 2 = 29 1 29 ÷ 2 = 14 1 14 ÷ 2 = 7 0 7 ÷ 2 = 3 1 3 ÷ 2 = 1 1 1 ÷ 2 = 0 1 (1110110101) 2 Ex:- (861)10 to octal Quotient Remainder 861 ÷ 8 = 107 6 107 ÷ 8 = 13 3 13 ÷ 8 = 1 5 1 ÷ 8 = 0 1 (1536) 8 Ex:- (861)10 to binary Quotient Remainder 861 ÷ 16 = 53 13 53 ÷ 16 = 3 5 3 ÷ 16 = 0 3 (35D) 16 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Decimal to Binary/Octal/Hexadecimal Ex:- (861)10 to binary Quotient Remainder 861 ÷ 2 = 430 1 430 ÷ 2 = 215 0 215 ÷ 2 = 107 1 107 ÷ 2 = 58 0 58 ÷ 2 = 29 1 29 ÷ 2 = 14 1 14 ÷ 2 = 7 0 7 ÷ 2 = 3 1 3 ÷ 2 = 1 1 1 ÷ 2 = 0 1 (1110110101) 2 Ex:- (861)10 to octal Quotient Remainder 861 ÷ 8 = 107 6 107 ÷ 8 = 13 3 13 ÷ 8 = 1 5 1 ÷ 8 = 0 1 (1536) 8 Ex:- (861)10 to binary Quotient Remainder 861 ÷ 16 = 53 13 53 ÷ 16 = 3 5 3 ÷ 16 = 0 3 (35D) 16 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Decimal to Binary/Octal/Hexadecimal Ex:- (861)10 to binary Quotient Remainder 861 ÷ 2 = 430 1 430 ÷ 2 = 215 0 215 ÷ 2 = 107 1 107 ÷ 2 = 58 0 58 ÷ 2 = 29 1 29 ÷ 2 = 14 1 14 ÷ 2 = 7 0 7 ÷ 2 = 3 1 3 ÷ 2 = 1 1 1 ÷ 2 = 0 1 (1110110101) 2 Ex:- (861)10 to octal Quotient Remainder 861 ÷ 8 = 107 6 107 ÷ 8 = 13 3 13 ÷ 8 = 1 5 1 ÷ 8 = 0 1 (1536) 8 Ex:- (861)10 to binary Quotient Remainder 861 ÷ 16 = 53 13 53 ÷ 16 = 3 5 3 ÷ 16 = 0 3 (35D) 16 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Decimal to Binary/Octal/Hexadecimal Ex:- (861)10 to binary Quotient Remainder 861 ÷ 2 = 430 1 430 ÷ 2 = 215 0 215 ÷ 2 = 107 1 107 ÷ 2 = 58 0 58 ÷ 2 = 29 1 29 ÷ 2 = 14 1 14 ÷ 2 = 7 0 7 ÷ 2 = 3 1 3 ÷ 2 = 1 1 1 ÷ 2 = 0 1 (1110110101) 2 Ex:- (861)10 to octal Quotient Remainder 861 ÷ 8 = 107 6 107 ÷ 8 = 13 3 13 ÷ 8 = 1 5 1 ÷ 8 = 0 1 (1536) 8 Ex:- (861)10 to binary Quotient Remainder 861 ÷ 16 = 53 13 53 ÷ 16 = 3 5 3 ÷ 16 = 0 3 (35D) 16 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Decimal to Binary/Octal/Hexadecimal Ex:- (861)10 to binary Quotient Remainder 861 ÷ 2 = 430 1 430 ÷ 2 = 215 0 215 ÷ 2 = 107 1 107 ÷ 2 = 58 0 58 ÷ 2 = 29 1 29 ÷ 2 = 14 1 14 ÷ 2 = 7 0 7 ÷ 2 = 3 1 3 ÷ 2 = 1 1 1 ÷ 2 = 0 1 (1110110101) 2 Ex:- (861)10 to octal Quotient Remainder 861 ÷ 8 = 107 6 107 ÷ 8 = 13 3 13 ÷ 8 = 1 5 1 ÷ 8 = 0 1 (1536) 8 Ex:- (861)10 to binary Quotient Remainder 861 ÷ 16 = 53 13 53 ÷ 16 = 3 5 3 ÷ 16 = 0 3 (35D) 16 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Decimal to Binary/Octal/Hexadecimal Ex:- (861)10 to binary Quotient Remainder 861 ÷ 2 = 430 1 430 ÷ 2 = 215 0 215 ÷ 2 = 107 1 107 ÷ 2 = 58 1 58 ÷ 2 = 29 1 29 ÷ 2 = 14 1 14 ÷ 2 = 7 0 7 ÷ 2 = 3 1 3 ÷ 2 = 1 1 1 ÷ 2 = 0 1 (1110110101) 2 Ex:- (861)10 to octal Quotient Remainder 861 ÷ 8 = 107 6 107 ÷ 8 = 13 3 13 ÷ 8 = 1 5 1 ÷ 8 = 0 1 (1536) 8 Ex:- (861)10 to binary Quotient Remainder 861 ÷ 16 = 53 13 53 ÷ 16 = 3 5 3 ÷ 16 = 0 3 (35D) 16 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Decimal to Binary/Octal/Hexadecimal Ex:- (861)10 to binary Quotient Remainder 861 ÷ 2 = 430 1 430 ÷ 2 = 215 0 215 ÷ 2 = 107 1 107 ÷ 2 = 58 1 58 ÷ 2 = 29 0 29 ÷ 2 = 14 1 14 ÷ 2 = 7 0 7 ÷ 2 = 3 1 3 ÷ 2 = 1 1 1 ÷ 2 = 0 1 (1110110101) 2 Ex:- (861)10 to octal Quotient Remainder 861 ÷ 8 = 107 6 107 ÷ 8 = 13 3 13 ÷ 8 = 1 5 1 ÷ 8 = 0 1 (1536) 8 Ex:- (861)10 to binary Quotient Remainder 861 ÷ 16 = 53 13 53 ÷ 16 = 3 5 3 ÷ 16 = 0 3 (35D) 16 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Decimal to Binary/Octal/Hexadecimal Ex:- (861)10 to binary Quotient Remainder 861 ÷ 2 = 430 1 430 ÷ 2 = 215 0 215 ÷ 2 = 107 1 107 ÷ 2 = 58 1 58 ÷ 2 = 29 0 29 ÷ 2 = 14 1 14 ÷ 2 = 7 0 7 ÷ 2 = 3 1 3 ÷ 2 = 1 1 1 ÷ 2 = 0 1 (1110110101) 2 Ex:- (861)10 to octal Quotient Remainder 861 ÷ 8 = 107 6 107 ÷ 8 = 13 3 13 ÷ 8 = 1 5 1 ÷ 8 = 0 1 (1536) 8 Ex:- (861)10 to binary Quotient Remainder 861 ÷ 16 = 53 13 53 ÷ 16 = 3 5 3 ÷ 16 = 0 3 (35D) 16 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Decimal to Binary/Octal/Hexadecimal Ex:- (861)10 to binary Quotient Remainder 861 ÷ 2 = 430 1 430 ÷ 2 = 215 0 215 ÷ 2 = 107 1 107 ÷ 2 = 58 1 58 ÷ 2 = 29 0 29 ÷ 2 = 14 1 14 ÷ 2 = 7 0 7 ÷ 2 = 3 1 3 ÷ 2 = 1 1 1 ÷ 2 = 0 1 (1110110101) 2 Ex:- (861)10 to octal Quotient Remainder 861 ÷ 8 = 107 6 107 ÷ 8 = 13 3 13 ÷ 8 = 1 5 1 ÷ 8 = 0 1 (1536) 8 Ex:- (861)10 to binary Quotient Remainder 861 ÷ 16 = 53 13 53 ÷ 16 = 3 5 3 ÷ 16 = 0 3 (35D) 16 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Decimal to Binary/Octal/Hexadecimal Ex:- (861)10 to binary Quotient Remainder 861 ÷ 2 = 430 1 430 ÷ 2 = 215 0 215 ÷ 2 = 107 1 107 ÷ 2 = 58 1 58 ÷ 2 = 29 0 29 ÷ 2 = 14 1 14 ÷ 2 = 7 0 7 ÷ 2 = 3 1 3 ÷ 2 = 1 1 1 ÷ 2 = 0 1 (1110110101) 2 Ex:- (861)10 to octal Quotient Remainder 861 ÷ 8 = 107 6 107 ÷ 8 = 13 3 13 ÷ 8 = 1 5 1 ÷ 8 = 0 1 (1536) 8 Ex:- (861)10 to binary Quotient Remainder 861 ÷ 16 = 53 13 53 ÷ 16 = 3 5 3 ÷ 16 = 0 3 (35D) 16 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Decimal to Binary/Octal/Hexadecimal Ex:- (861)10 to binary Quotient Remainder 861 ÷ 2 = 430 1 430 ÷ 2 = 215 0 215 ÷ 2 = 107 1 107 ÷ 2 = 58 1 58 ÷ 2 = 29 0 29 ÷ 2 = 14 1 14 ÷ 2 = 7 0 7 ÷ 2 = 3 1 3 ÷ 2 = 1 1 1 ÷ 2 = 0 1 (1110110101) 2 Ex:- (861)10 to octal Quotient Remainder 861 ÷ 8 = 107 6 107 ÷ 8 = 13 3 13 ÷ 8 = 1 5 1 ÷ 8 = 0 1 (1536) 8 Ex:- (861)10 to binary Quotient Remainder 861 ÷ 16 = 53 13 53 ÷ 16 = 3 5 3 ÷ 16 = 0 3 (35D) 16 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Decimal to Binary/Octal/Hexadecimal Ex:- (861)10 to binary Quotient Remainder 861 ÷ 2 = 430 1 430 ÷ 2 = 215 0 215 ÷ 2 = 107 1 107 ÷ 2 = 58 1 58 ÷ 2 = 29 0 29 ÷ 2 = 14 1 14 ÷ 2 = 7 0 7 ÷ 2 = 3 1 3 ÷ 2 = 1 1 1 ÷ 2 = 0 1 (1110110101) 2 Ex:- (861)10 to octal Quotient Remainder 861 ÷ 8 = 107 6 107 ÷ 8 = 13 3 13 ÷ 8 = 1 5 1 ÷ 8 = 0 1 (1536) 8 Ex:- (861)10 to binary Quotient Remainder 861 ÷ 16 = 53 13 53 ÷ 16 = 3 5 3 ÷ 16 = 0 3 (35D) 16 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Decimal to Binary/Octal/Hexadecimal Ex:- (861)10 to binary Quotient Remainder 861 ÷ 2 = 430 1 430 ÷ 2 = 215 0 215 ÷ 2 = 107 1 107 ÷ 2 = 58 1 58 ÷ 2 = 29 0 29 ÷ 2 = 14 1 14 ÷ 2 = 7 0 7 ÷ 2 = 3 1 3 ÷ 2 = 1 1 1 ÷ 2 = 0 1 (1110110101) 2 Ex:- (861)10 to octal Quotient Remainder 861 ÷ 8 = 107 6 107 ÷ 8 = 13 3 13 ÷ 8 = 1 5 1 ÷ 8 = 0 1 (1536) 8 Ex:- (861)10 to binary Quotient Remainder 861 ÷ 16 = 53 13 53 ÷ 16 = 3 5 3 ÷ 16 = 0 3 (35D) 16 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Decimal to Binary/Octal/Hexadecimal Ex:- (861)10 to binary Quotient Remainder 861 ÷ 2 = 430 1 430 ÷ 2 = 215 0 215 ÷ 2 = 107 1 107 ÷ 2 = 58 1 58 ÷ 2 = 29 0 29 ÷ 2 = 14 1 14 ÷ 2 = 7 0 7 ÷ 2 = 3 1 3 ÷ 2 = 1 1 1 ÷ 2 = 0 1 (1101011101) 2 Ex:- (861)10 to octal Quotient Remainder 861 ÷ 8 = 107 6 107 ÷ 8 = 13 3 13 ÷ 8 = 1 5 1 ÷ 8 = 0 1 (1536) 8 Ex:- (861)10 to binary Quotient Remainder 861 ÷ 16 = 53 13 53 ÷ 16 = 3 5 3 ÷ 16 = 0 3 (35D) 16 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Decimal to Binary/Octal/Hexadecimal Ex:- (861)10 to binary Quotient Remainder 861 ÷ 2 = 430 1 430 ÷ 2 = 215 0 215 ÷ 2 = 107 1 107 ÷ 2 = 58 1 58 ÷ 2 = 29 0 29 ÷ 2 = 14 1 14 ÷ 2 = 7 0 7 ÷ 2 = 3 1 3 ÷ 2 = 1 1 1 ÷ 2 = 0 1 (1110101101) 2 Ex:- (861)10 to octal Quotient Remainder 861 ÷ 8 = 107 6 107 ÷ 8 = 13 3 13 ÷ 8 = 1 5 1 ÷ 8 = 0 1 (1536) 8 Ex:- (861)10 to binary Quotient Remainder 861 ÷ 16 = 53 13 53 ÷ 16 = 3 5 3 ÷ 16 = 0 3 (35D) 16 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Decimal to Binary/Octal/Hexadecimal Ex:- (861)10 to binary Quotient Remainder 861 ÷ 2 = 430 1 430 ÷ 2 = 215 0 215 ÷ 2 = 107 1 107 ÷ 2 = 58 1 58 ÷ 2 = 29 0 29 ÷ 2 = 14 1 14 ÷ 2 = 7 0 7 ÷ 2 = 3 1 3 ÷ 2 = 1 1 1 ÷ 2 = 0 1 (1110101101) 2 Ex:- (861)10 to octal Quotient Remainder 861 ÷ 8 = 107 6 107 ÷ 8 = 13 3 13 ÷ 8 = 1 5 1 ÷ 8 = 0 1 (1536) 8 Ex:- (861)10 to binary Quotient Remainder 861 ÷ 16 = 53 13 53 ÷ 16 = 3 5 3 ÷ 16 = 0 3 (35D) 16 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Decimal to Binary/Octal/Hexadecimal Ex:- (861)10 to binary Quotient Remainder 861 ÷ 2 = 430 1 430 ÷ 2 = 215 0 215 ÷ 2 = 107 1 107 ÷ 2 = 58 1 58 ÷ 2 = 29 0 29 ÷ 2 = 14 1 14 ÷ 2 = 7 0 7 ÷ 2 = 3 1 3 ÷ 2 = 1 1 1 ÷ 2 = 0 1 (1110101101) 2 Ex:- (861)10 to octal Quotient Remainder 861 ÷ 8 = 107 6 107 ÷ 8 = 13 3 13 ÷ 8 = 1 5 1 ÷ 8 = 0 1 (1536) 8 Ex:- (861)10 to binary Quotient Remainder 861 ÷ 16 = 53 13 53 ÷ 16 = 3 5 3 ÷ 16 = 0 3 (35D) 16 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Decimal to Binary/Octal/Hexadecimal Ex:- (861)10 to binary Quotient Remainder 861 ÷ 2 = 430 1 430 ÷ 2 = 215 0 215 ÷ 2 = 107 1 107 ÷ 2 = 58 1 58 ÷ 2 = 29 0 29 ÷ 2 = 14 1 14 ÷ 2 = 7 0 7 ÷ 2 = 3 1 3 ÷ 2 = 1 1 1 ÷ 2 = 0 1 (1110101101) 2 Ex:- (861)10 to octal Quotient Remainder 861 ÷ 8 = 107 6 107 ÷ 8 = 13 3 13 ÷ 8 = 1 5 1 ÷ 8 = 0 1 (1536) 8 Ex:- (861)10 to binary Quotient Remainder 861 ÷ 16 = 53 13 53 ÷ 16 = 3 5 3 ÷ 16 = 0 3 (35D) 16 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Decimal to Binary/Octal/Hexadecimal Ex:- (861)10 to binary Quotient Remainder 861 ÷ 2 = 430 1 430 ÷ 2 = 215 0 215 ÷ 2 = 107 1 107 ÷ 2 = 58 1 58 ÷ 2 = 29 0 29 ÷ 2 = 14 1 14 ÷ 2 = 7 0 7 ÷ 2 = 3 1 3 ÷ 2 = 1 1 1 ÷ 2 = 0 1 (1110101101) 2 Ex:- (861)10 to octal Quotient Remainder 861 ÷ 8 = 107 6 107 ÷ 8 = 13 3 13 ÷ 8 = 1 5 1 ÷ 8 = 0 1 (1536) 8 Ex:- (861)10 to binary Quotient Remainder 861 ÷ 16 = 53 13 53 ÷ 16 = 3 5 3 ÷ 16 = 0 3 (35D) 16 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Decimal to Binary/Octal/Hexadecimal Ex:- (861)10 to binary Quotient Remainder 861 ÷ 2 = 430 1 430 ÷ 2 = 215 0 215 ÷ 2 = 107 1 107 ÷ 2 = 58 1 58 ÷ 2 = 29 0 29 ÷ 2 = 14 1 14 ÷ 2 = 7 0 7 ÷ 2 = 3 1 3 ÷ 2 = 1 1 1 ÷ 2 = 0 1 (1110101101) 2 Ex:- (861)10 to octal Quotient Remainder 861 ÷ 8 = 107 6 107 ÷ 8 = 13 3 13 ÷ 8 = 1 5 1 ÷ 8 = 0 1 (1536) 8 Ex:- (861)10 to binary Quotient Remainder 861 ÷ 16 = 53 13 53 ÷ 16 = 3 5 3 ÷ 16 = 0 3 (35D) 16 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Decimal to Binary/Octal/Hexadecimal Ex:- (861)10 to binary Quotient Remainder 861 ÷ 2 = 430 1 430 ÷ 2 = 215 0 215 ÷ 2 = 107 1 107 ÷ 2 = 58 1 58 ÷ 2 = 29 0 29 ÷ 2 = 14 1 14 ÷ 2 = 7 0 7 ÷ 2 = 3 1 3 ÷ 2 = 1 1 1 ÷ 2 = 0 1 (1110101101) 2 Ex:- (861)10 to octal Quotient Remainder 861 ÷ 8 = 107 6 107 ÷ 8 = 13 3 13 ÷ 8 = 1 5 1 ÷ 8 = 0 1 (1536) 8 Ex:- (861)10 to binary Quotient Remainder 861 ÷ 16 = 53 13 53 ÷ 16 = 3 5 3 ÷ 16 = 0 3 (35D) 16 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Decimal to Binary/Octal/Hexadecimal Ex:- (861)10 to binary Quotient Remainder 861 ÷ 2 = 430 1 430 ÷ 2 = 215 0 215 ÷ 2 = 107 1 107 ÷ 2 = 58 0 58 ÷ 2 = 29 1 29 ÷ 2 = 14 1 14 ÷ 2 = 7 0 7 ÷ 2 = 3 1 3 ÷ 2 = 1 1 1 ÷ 2 = 0 1 (1110110101) 2 Ex:- (861)10 to octal Quotient Remainder 861 ÷ 8 = 107 6 107 ÷ 8 = 13 3 13 ÷ 8 = 1 5 1 ÷ 8 = 0 1 (1536) 8 Ex:- (861)10 to Hex Quotient Remainder 861 ÷ 16 = 53 13 53 ÷ 16 = 3 5 3 ÷ 16 = 0 3 (35D) 16 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Decimal to Binary/Octal/Hexadecimal Ex:- (861)10 to binary Quotient Remainder 861 ÷ 2 = 430 1 430 ÷ 2 = 215 0 215 ÷ 2 = 107 1 107 ÷ 2 = 58 1 58 ÷ 2 = 29 0 29 ÷ 2 = 14 1 14 ÷ 2 = 7 0 7 ÷ 2 = 3 1 3 ÷ 2 = 1 1 1 ÷ 2 = 0 1 (1110101101) 2 Ex:- (861)10 to octal Quotient Remainder 861 ÷ 8 = 107 6 107 ÷ 8 = 13 3 13 ÷ 8 = 1 5 1 ÷ 8 = 0 1 (1536) 8 Ex:- (861)10 to Hex Quotient Remainder 861 ÷ 16 = 53 13 53 ÷ 16 = 3 5 3 ÷ 16 = 0 3 (35D) 16 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Decimal to Binary/Octal/Hexadecimal Ex:- (861)10 to binary Quotient Remainder 861 ÷ 2 = 430 1 430 ÷ 2 = 215 0 215 ÷ 2 = 107 1 107 ÷ 2 = 58 1 58 ÷ 2 = 29 0 29 ÷ 2 = 14 1 14 ÷ 2 = 7 0 7 ÷ 2 = 3 1 3 ÷ 2 = 1 1 1 ÷ 2 = 0 1 (1110101101) 2 Ex:- (861)10 to octal Quotient Remainder 861 ÷ 8 = 107 6 107 ÷ 8 = 13 3 13 ÷ 8 = 1 5 1 ÷ 8 = 0 1 (1536) 8 Ex:- (861)10 to Hex Quotient Remainder 861 ÷ 16 = 53 13 53 ÷ 16 = 3 5 3 ÷ 16 = 0 3 (35D) 16 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Decimal to Binary/Octal/Hexadecimal Ex:- (861)10 to binary Quotient Remainder 861 ÷ 2 = 430 1 430 ÷ 2 = 215 0 215 ÷ 2 = 107 1 107 ÷ 2 = 58 1 58 ÷ 2 = 29 0 29 ÷ 2 = 14 1 14 ÷ 2 = 7 0 7 ÷ 2 = 3 1 3 ÷ 2 = 1 1 1 ÷ 2 = 0 1 (1110101101) 2 Ex:- (861)10 to octal Quotient Remainder 861 ÷ 8 = 107 6 107 ÷ 8 = 13 3 13 ÷ 8 = 1 5 1 ÷ 8 = 0 1 (1536) 8 Ex:- (861)10 to Hex Quotient Remainder 861 ÷ 16 = 53 13 53 ÷ 16 = 3 5 3 ÷ 16 = 0 3 (35D) 16 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Decimal to Binary/Octal/Hexadecimal Ex:- (861)10 to binary Quotient Remainder 861 ÷ 2 = 430 1 430 ÷ 2 = 215 0 215 ÷ 2 = 107 1 107 ÷ 2 = 58 1 58 ÷ 2 = 29 0 29 ÷ 2 = 14 1 14 ÷ 2 = 7 0 7 ÷ 2 = 3 1 3 ÷ 2 = 1 1 1 ÷ 2 = 0 1 (1110101101) 2 Ex:- (861)10 to octal Quotient Remainder 861 ÷ 8 = 107 6 107 ÷ 8 = 13 3 13 ÷ 8 = 1 5 1 ÷ 8 = 0 1 (1536) 8 Ex:- (861)10 to Hex Quotient Remainder 861 ÷ 16 = 53 13 53 ÷ 16 = 3 5 3 ÷ 16 = 0 3 (35D) 16 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Decimal to Binary/Octal/Hexadecimal Ex:- (861)10 to binary Quotient Remainder 861 ÷ 2 = 430 1 430 ÷ 2 = 215 0 215 ÷ 2 = 107 1 107 ÷ 2 = 58 1 58 ÷ 2 = 29 0 29 ÷ 2 = 14 1 14 ÷ 2 = 7 0 7 ÷ 2 = 3 1 3 ÷ 2 = 1 1 1 ÷ 2 = 0 1 (1110101101) 2 Ex:- (861)10 to octal Quotient Remainder 861 ÷ 8 = 107 6 107 ÷ 8 = 13 3 13 ÷ 8 = 1 5 1 ÷ 8 = 0 1 (1536) 8 Ex:- (861)10 to Hex Quotient Remainder 861 ÷ 16 = 53 13 53 ÷ 16 = 3 5 3 ÷ 16 = 0 3 (35D) 16 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Decimal to Binary Fractional Numbers Ex:- (0.625)10 to binary Product Carry 0.625 x 2 = 1.25 1 0.25 x 2 = 0.5 0 0.5 x 2 = 1.0 1 (0.101) 2 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Decimal to Binary Fractional Numbers Ex:- (0.625)10 to binary Product Carry 0.625 x 2 = 1.25 1 0.25 x 2 = 0.5 0 0.5 x 2 = 1.0 1 (0.101) 2 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Decimal to Binary Fractional Numbers Ex:- (0.625)10 to binary Product Carry 0.625 x 2 = 1.25 1 0.25 x 2 = 0.5 0 0.5 x 2 = 1.0 1 (0.101) 2 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Decimal to Binary Fractional Numbers Ex:- (0.625)10 to binary Product Carry 0.625 x 2 = 1.25 1 0.25 x 2 = 0.5 0 0.5 x 2 = 1.0 1 (0.101) 2 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Decimal to Binary Fractional Numbers Ex:- (0.625)10 to binary Product Carry 0.625 x 2 = 1.25 1 0.25 x 2 = 0.5 0 0.5 x 2 = 1.0 1 (0.101) 2 Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Topic Summary – Number Systems Overview of Number Systems Decimal Number System Binary Number System Hexadecimal Number System Octal Number System Conversion of numbers between Number Systems Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Next Topic – Boolean Algebra Binary Operators Basic Rules of Binary Operations Laws of Boolean Algebra Demorgon’s Theorem Digital Logic Design – Unit-I: Topic 1.1 -1.2 Dr.R.D.Chintamani Department of Information Technology
Ask Questions & Share Responses You can ask your questions and share your responses at:  Sanjivani LMS  Email: chintamanirameshwarit@sanjivani.org.in Lecture Presentation and Literature To refer presentation of this lecture and literature visit:  Sanjivani LMS Course: Digital Electronics & Logic Design References: [1] M Morris Mano, “Digital Design”, Prentice Hall, 3rd Edition, ISBN: 0130621218. [2] R. P. Jain, “Modern Digital Electronics “, 3rd Edition, Tata McGraw Hill, ISBN: 0 07 0494924 [3] Flyod, “Digital Principles”, Pearson Education, ISBN:978 81 7758 643 6. Digital Electronics & Computer Organization – Unit-I: Topic 1.1 Dr.R.D.Chintamani Department of Information Technology