Data processing and processor organisation
- 1. 1 Unit-2 Data Processing and Processor Organization By Siddhasen R Patil G H Raisoni College of Engineering & Management, Wagholi, Pune An Empowered Autonomous Institute Affiliated to Savitribai Phule Pune University
- 2. 2 Syllabus •Fixed and Floating Point arithmetic •Booths multiplication •Division algorithm: restoring and non-restoring division •Data representation method: – IEEE standard single and – double precision format and examples •Assembly language programming •Instruction: elements of machine instruction
- 3. 3 Syllabus •Instruction format •Instruction types •Addressing modes and formats •Processor Structure and Function - Processor Organization •Register Organization. •Single cycle processor •Multi-cycle processor •Instruction-Level Parallelism: Concepts and Challenges
- 4. Fixed point Addition, Subtraction, Multiplication and Division • Addition: – Binary Addition 0 0 1 1 + 0 1 0 1 0 1 1 1 0 – Four bit adder 1100 12 + 0101 + 05 1 0001 17
- 5. Fixed point Addition, Subtraction, Multiplication and Division • Subtraction: – Binary Subtraction 0 1 1 0 - 0 1 0 1 0 0 1 1 1 – Four bit Subtractor 1100 12 - 0101 - 05 0111 07
- 6. Fixed point Addition, Subtraction, Multiplication and Division • Multiplication: – Four bit Multiplication
- 7. IEEE Standard Single & Double Precision Format & Examples • The IEEE-754 floating-point formats is used to store a real number or floating point. • There are two formats. – One is a Single-precision format • It consumes 32 bits to store all this information such as Significand, Exponent, and the Sign. – Second is a Double-precision format • It consumes 64 bits to store all this information such as Significand, Exponent, and the Sign.
- 8. Single Precision Format • IEEE-754 Single Precision Format – 32-bit format – 23-bits are given to store the significand – 8-bits are used to store the Exponent – 1-bit is used to store the Sign
- 9. Single Precision Format • IEEE-754 Single Precision Format – Example: Given number 15.6875 – Convert 15 to Binary • 1111 – Convert 0.6875 to Binary • 0.1011 – Combine 15 and 0.6875 • 1111.1011 => (1111.1011 X 20 ) – Shift the point towards left by 3-bit (i.e. after first 1) • 1.1111011 X 23 – So the significant = 111101100………0 (total 23-bits), – Sign bit is 0 (Since the number is positive), and – Exponent is 130 (i.e. 3+127) => 1000 0010. – Final Answer: 0 10000010 11110110000……….0
- 10. Single Precision Format • IEEE-754 Single Precision Format – Example: Given number -2015.4375 – Convert 2015 to Binary • 11111011111 – Convert 0.4375 to Binary • 0.0111 – Combine 2015 and 0.4375 • 11111011111.0111 => (11111011111.0111 X 20 ) – Shift the point towards left by 10-bit (i.e. after first 1) • 1.11110111110111 X 210 – So the significant = 111101111101110……0(total 23-bits), – Sign bit is 1 (Since the number is negative), and – Exponent is 137 (i.e. 10+127) => 1000 1001. – Final Answer: 1 10001001 111101111101110……0
- 11. Double Precision Format • IEEE-754 Double Precision Format – 64-bit format – 52-bits are given to store the significand – 11-bits are used to store the Exponent – 1-bit is used to store the Sign
- 12. Double Precision Format • IEEE-754 Double Precision Format – Example: Given number 25.625 – Convert 25 to Binary • 11001 – Convert 0.625 to Binary • 0.101 – Combine 25 and 0.625 • 11001.101 => (11001.101 X 20 ) – Shift the point towards left by 4-bit • 1.1001101 X 24 – So the significant = 1001101000………0 (total 52-bits), – Sign bit is 0 (Since number is positive), and – Exponent is 1027 (i.e. 4+1023) => 10000000011. – Final Answer: 0 100000000011 1001101000……….0
- 13. Floating Point Arithmetic • Floating Numbers Addition – Step 1: Compare the exponents of two numbers and calculate the absolute value of difference between the two exponents. Take the larger exponent as the tentative exponent of the result. – Step 2: Shift the significand of the number with the smaller exponent, right through a number of bit positions that is equal to the exponent difference. – Step 3 Add the two signed-magnitude significands. – Step 4 Check SUM for carry out (Cout) from the MSB position during addition. Shift SUM right by one bit position if a carry out is detected and increment the tentative exponent by 1.
- 14. Floating Point Arithmetic • Floating Numbers Addition • Example: A + B = C 2 + 3 = 5 A = 0 1000 0000 0000 0000 0000 0000 0000 000 B = 0 1000 0000 1000 0000 0000 0000 0000 000 Sign Exponent Mantissa A 0 1000 0000 = 128 – 127 = 1 0000 0000 0000 0000 0000 000 = 1.0 B 0 1000 0000 = 128 – 127 = 1 1000 0000 0000 0000 0000 000 = 1.1 C 0 1+1 = 2 => 22 = 4 0.10 + 0.11 = 1.01 = 1.25 0 4 X 1.25 = 5
- 15. Floating Point Arithmetic • Floating Number Subtraction – Step 1: Compare the exponents of two numbers and calculate the absolute value of difference between the two exponents. Take the larger exponent as the tentative exponent of the result. – Step 2: Shift the significand of the number with the smaller exponent, right through a number of bit positions that is equal to the exponent difference. – Step 3 Subtract the two signed-magnitude significands. Let the result of this is Subtraction. – Step 4 : During subtraction, check SUM for leading zeros. Shift SUM left until the MSB of the shifted result is a 1. Subtract the leading zero count from tentative exponent.
- 16. Floating Point Arithmetic • Floating Number Subtraction • Example: A - B = C 3 - 3.5 = - 0.5 A = 0 1000 0000 1000 0000 0000 0000 0000 000 B = 1 1000 0000 1100 0000 0000 0000 0000 000 Sign Exponent Mantissa A 0 1000 0000 = 128 – 127 = 1 1000 0000 0000 0000 0000 000 = 1.1 B 1 1000 0000 = 128 – 127 = 1 1100 0000 0000 0000 0000 000 = 1.11 C 1 1 - 3 = -2 => 2-2 = 0.25 0.11 – 0.111 = 0.001 << 1.0 1 0.25 X 2 = - 0.5
- 17. Floating Point Arithmetic • Floating Numbers Multiplication – Step 1: Calculate the tentative exponent of the product by adding the biased exponents of the two numbers, subtracting the bias i.e. 127. – Step 2: If the sign of two floating point numbers are the same, set the sign of product to ‘+’, else set it to ‘-’. – Step 3: Multiply the two significands. For 24-bit significand the product is 48-bits wide (including the leading hidden bit (1)).
- 18. Floating Point Arithmetic • Floating Numbers Multiplication – Step 4: Normalize the product if MSB of the product is 1, by shifting the product right by 1 bit position and incrementing the tentative exponent.
- 19. Floating Point Arithmetic • Floating Numbers Multiplication • Example: A X B = C 3 X -3.5 = -10.5 A = 0 1000 0000 1000 0000 0000 0000 0000 000 B = 1 1000 0000 1100 0000 0000 0000 0000 000 Sign Exponent Mantissa A 0 1000 0000 = 128 – 127 = 1 1000 0000 0000 0000 0000 000 = 1.1 B 1 1000 0000 = 128 – 127 = 1 1100 0000 0000 0000 0000 000 = 1.11 C 1 1+1 = 2 => 3 => 23 = 8 1.1 X 1.11 = 10.101 = 1.0101 = 1.3125 1 8 X 1.3125 = 10.5 => - 10.5
- 20. Floating Point Arithmetic • Floating Numbers Multiplication • Example: A X B = C 2 X 3 = 6 A = 0 1000 0000 0000 0000 0000 0000 0000 000 B = 0 1000 0000 1000 0000 0000 0000 0000 000 Sign Exponent Mantissa A 0 1000 0000 = 128 – 127 = 1 0000 0000 0000 0000 0000 000 = 1.0 B 0 1000 0000 = 128 – 127 = 1 1000 0000 0000 0000 0000 000 = 1.1 C 0 1+1 = 2 => 22 = 4 1.0 X 1.1 = 1.10 = 1.5 0 4 X 1.5 = 6
- 21. Floating Point Arithmetic • Floating Numbers Division – Step 1: Calculate the tentative exponent of the product by adding the biased exponents of the two numbers, subtracting the bias i.e. 127. – Step 2: If the sign of two floating point numbers are the same, set the sign of product to ‘+’, else set it to ‘-’. – Step 3: Multiply the two significands. For 24-bit significand the product is 48-bits wide (including the leading hidden bit (1)).
- 22. Floating Point Arithmetic • Floating Numbers Division – Step 4: Normalize the product if MSB of the product is 1, by shifting the product right by 1 bit position and incrementing the tentative exponent.
- 23. Floating Point Arithmetic • Floating Numbers Division • Example: A / B = C 3 / 2 = 1.5 A = 0 1000 0000 1000 0000 0000 0000 0000 000 B = 0 1000 0000 0000 0000 0000 0000 0000 000 Sign Exponent Mantissa A 0 1000 0000 = 128 – 127 = 1 1000 0000 0000 0000 0000 000 = 1.1 B 0 1000 0000 = 128 – 127 = 1 0000 0000 0000 0000 0000 000 = 1.0 C 0 1-1 = 0 => 20 = 1 1.1 / 1.0 = 1.1 0 1 X 1.1 = 1.5
- 24. Floating Point Arithmetic • Floating Numbers Division • Example: A / B = C 3.5 / 2 = 1.5 A = 0 1000 0000 1100 0000 0000 0000 0000 000 B = 0 1000 0000 0000 0000 0000 0000 0000 000 Sign Exponent Mantissa A 0 1000 0000 = 128 – 127 = 1 1100 0000 0000 0000 0000 000 = 1.11 B 0 1000 0000 = 128 – 127 = 1 0000 0000 0000 0000 0000 000 = 1.0 C 0 1-1 = 0 => 20 = 1 1.11 / 1.0 = 1.11 0 1 X 1.11 = 1.75
- 25. Floating Point Arithmetic • Floating Numbers Division • Example: A / B = C 15 / 4 = 3.75 A = 0 1000 0010 1110 0000 0000 0000 0000 000 B = 0 1000 0001 0000 0000 0000 0000 0000 000 Sign Exponent Mantissa A 0 1000 0000 = 130 – 127 = 3 1110 0000 0000 0000 0000 000 = 1.111 B 0 1000 0000 = 129 – 127 = 2 0000 0000 0000 0000 0000 000 = 1.0 C 0 3-2 = 1 => 21 = 2 1.111 / 1.0 = 1.111 0 2 X 1.111 = 3.75
- 26. Booth’s Multiplication • The booth algorithm is a multiplication algorithm that allows us to multiply the two signed binary integers. • It is also used to speed up the performance of the multiplication process. • It is very efficient too. • Booth algorithm works equally well for both negative and positive multipliers. • In Booth algorithm, the Multiplier recoded before multiplication.
- 27. Booth’s Multiplication • For example: – If the Multiplier is 0110 (6), then add one zero at right side of multiplier: 01100 – Then scan the multiplier from right to left 01100 – During scanning of multiplier, we have to follow rule given in below table. Bit i Bit i+1 Recoded Value 0 0 0 0 1 -1 1 0 +1 1 1 0
- 28. Booth’s Multiplication • 0 1 1 0 0 +1 0 -1 0 • Now if our multiplicand is 0101 then the multiplication of 0101 (5) and 0110 (6) is : 0 1 0 1 X +1 0 -1 0 0 0 0 0 0 0 0 0 0101 X 0 1 1 1 1 0 1 1 x 0101 X -1 0 0 0 0 0 0 x x 0101 X 0 0 0 1 0 1 x x x 0101 X +1 0 0 0 1 1 1 1 0 30
- 29. Booth’s Multiplication • Flow-Chart of Multiplication using Booth’s algorithm 6 X 5 =30 6 => 0110, 5 => 0101 Count A Q Q-1 Operation 4 0000 0101 0 4 A = A - M 0000 - 1010 1010 1101 0101 0101 0010 0 0 1 Right Shift 3 A = A + M 1101 + 0110 0011 0001 0010 0010 1001 1 1 0 Right Shift
- 30. Booth’s Multiplication • Flow-Chart of Multiplication using Booth’s algorithm 6 X 5 =30 6 => 0110, 5 => 0101 Count A Q Q-1 Operation 2 A = A - M 0001 - 1010 1011 1101 1001 1001 1100 0 0 1 Right Shift 1 A = A + M 1101 + 0110 0011 0001 1100 1100 1110 1 1 0 Right Shift 0 11110 => 30
- 31. Booth’s Multiplication • Flow-Chart of Multiplication using Booth’s algorithm -5 X -4 = 20 -5 => 1011, -4 => 1100 Count A Q Q-1 Operation 4 0000 1100 0 4 0 to 0 means Right Shift 0000 0110 0 Right Shift 3 0 to 0 means Right Shift 0000 0011 0 Right Shift 2 1 to 0 means A = A - M 0000 0101 0101 0010 0011 0011 1001 0 0 1 Right Shift 1 1 to 1 means Shift Right 0001 0100 1 Right Shift 0 00010100 => +20
- 32. Division
- 34. 34 Assembly Language Programming • An assembly language is a type of programming language that translates high-level languages into machine language. • It is a necessary bridge between software programs and their underlying hardware platforms. • Assembly language relies on language syntax, labels, operators, and directives to convert code into usable machine instruction.
- 35. 35 Instruction • Elements of machine instruction – The typical elements of a machine instruction are • Operation code • Source Operand Reference • Result Operand reference • Next Instruction Reference • Example: – MOV A, B – JMP 2000H
- 36. 36 Instruction format • An assembly language statement consists of four possible fields: – Label field – Mnemonic field – Operand field – Comment file. • The lable and comment fields are always optional.
- 37. 37 Instruction Types • Data Transfer Instruction – MOV A,B • Arithmetic Instruction – ADD A,B • Logical Instruction – ANL A,B • Branch Instruction – JNZ UP
- 38. 38 Addressing Modes and Formats
- 39. 39 Processor Structure and Function
- 41. 41 Register Organization • General Register Organization is the processor architecture that stores and manipulates data for computations. • The main components of a register organization include registers, memory, and instructions. • The registers act as memory within the processor and are used to process instructions as they are executed.
- 43. 43 Single Cycle and Multi Cycle Processor • Single cycle processor is a processor in which the instructions are fetched from the memory, and then executed, and the results are further stored in a single cycle • Multicycle implementation, each i.e. fetch, decode, execute and write back, all are required one clock cycle to execute.
- 44. Instruction-Level Parallelism: Concepts and Challenges • Instruction-Level Parallelism – Multiple operations can be performed parallelly 44 Sequential execution of operations Instruction Level Parallelism