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![Executing an Instruction
Fetch the contents of the memory location pointed to by
the PC. The contents of this location are loaded into the
IR (fetch phase).
IR ← [[PC]]
Assuming that the memory is byte addressable,
increment the contents of the PC by 4 (fetch phase).
PC ← [PC] + 4
Carry out the actions specified by the instruction in the
IR (execution phase).](https://image.slidesharecdn.com/compuerorganizaion-processingunit-121107001331-phpapp01/85/Compuer-organizaion-processing-unit-3-320.jpg)
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Compuer organizaion processing unit
This document discusses the architecture and operation of processing units. It begins by outlining fundamental concepts like how the processor fetches and executes instructions sequentially using the program counter. It then describes the execution of a single instruction through fetch, execution, and incrementing the program counter phases. The document details the components involved in instruction execution like registers, arithmetic logic unit, and memory. It explains the sequencing of control signals to perform operations like register transfers, arithmetic/logic, and memory access. The document compares hardwired and microprogrammed approaches to control and their advantages/disadvantages. It concludes with details on arithmetic operations like addition, subtraction, overflow detection, and fast addition techniques.
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Compuer organizaion processing unit
- 1. Processing Unit Deepak John,Department Computer Applications, SJCET-Pala
- 2. Fundamental Concepts Processorfetches one instruction at a time and perform the operation specified. Instructions are fetched from successive memory locations until a branch or a jump instruction is encountered. Processor keeps track of the address of the memory location containing the next instruction to be fetched using Program Counter (PC). Instruction Register (IR)
- 3. Executing an Instruction Fetch the contents of the memory location pointed to by the PC. The contents of this location are loaded into the IR (fetch phase). IR ← [[PC]] Assuming that the memory is byte addressable, increment the contents of the PC by 4 (fetch phase). PC ← [PC] + 4 Carry out the actions specified by the instruction in the IR (execution phase).
- 4.
- 5. Executing an Instruction Transfer a word of data from one processor register to another or to the ALU. Perform an arithmetic or a logic operation and store the result in a processor register. Fetch the contents of a given memory location and load them into a processor register. Store a word of data from a processor register into a given memory location. Register Transfers For each register two control signals are used 1. To place the contents of that register on the bus 2. To load the data on the bus into register The input and output of register are connected to the bus via switches controlled by the signals Rin and Rout
- 7. Performing an Arithmeticor Logic Operation The ALU is a combinational circuit that has no internal storage. ALU gets the two operands from MUX and bus. The result is temporarily stored in register Z. What is the sequence of operations to add the contents of register R1 to those of R2 and store the result in R3? 1. R1out, Yin 2. R2out, SelectY, Add, Zin 3. Zout, R3in
- 8. Fetching a Wordfrom Memory Address into MAR; issue Read operation; data into MDR. Memory-Function-Completed (MFC)
- 10. Execution of BranchInstructions A branch instruction replaces the contents of PC with the branch target address, which is usually obtained by adding an offset X given in the branch instruction. The offset X is usually the difference between the branch target address and the address immediately following the branch instruction. Conditional branch Step Action 1 PCou t , MAR in , Read , Select4,Add, Zi n 2 Zou t , PCi n , Y in , WM F C 3 MDR out , IRi n 4 Offset-field-of-IR out , Add , Z in 5 Zou t , PCi n , En d Figure 7.7. Control sequence for an unconditional branch instruction.
- 11. Multiple-Bus Organization Register file:allgeneral purpose registers are combined Buses A and B:used to transfer the source operands to the ALU Step Action 1 PCout, R=B, MAR in, Read, IncPC 2 WMFC 3 MDRoutB, R=B, IRin 4 R4outA, R5outB, SelectA, Add, R6in, End Figure 7.9. Control sequence for the instruction. Add R4,R5,R6,for the three-bus organization
- 12.
- 13. To executeinstructions, the processor must have some means of generating the control signals needed in the proper sequence. Two categories: hardwired control and microprogrammed control Hardwired system can operate at high speed; but with little flexibility. Control Unit Organization CLK Clock Control step IR encoder Decoder/ Control signals codes counter inputs Condition External
- 14. Detailed Block Description StepDecoder :provides seperate signal line for each step or time slot in the ccontrol sequence. Instruction decoder :o/p consists of a seperate line for each machine instruction. For any instruction in IR, one of the output lines INS1 through INSm is set to 1,all other lines are set to 0 Input signals to the encoder block are combined to generate the individual control signals Yin,Pcout,ADD,End etc
- 15. Generating Zin Zin= T1 + T6 • ADD + T4 • BR + … T1 AddBranch T4 T6
- 16. ADVANTAGES Hardwired ControlUnit is fast because control signals are generated by combinational circuits. The delay in generation of control signals depends upon the number of gates. DISADVANTAGES more complex will be the design of control unit. Modifications in control signal are very difficult. That means it requires rearranging of wires in the hardware circuit. It is difficult to correct mistake in original design or adding new feature in existing design of control unit.
- 17.
- 18.
- 19. Control signalsare generated by a program similar to machine language programs. Control Word (CW):is a word whose individual bits represent the various control signals; Each step of the instruction execution is represented by a control word with all of the bits corresponding to the control signals needed for the step set to one. Microroutine:a sequence of CW’s corresponding to the control sequence of a machine instruction microinstruction:individual control words in microroutine, consists of: One or more micro-operations to be executed. Address of next microinstruction to be executed.
- 21. Control store: microroutines for all instructions in the instruction set of a computer are stored
- 22. The previousorganization cannot handle the situation when the control unit is required to check the status of the condition codes or external inputs to choose between alternative courses of action. Use conditional branch microinstruction. Address Microinstruction 0 PCout , MARin , Read,Select4,Add, Zin 1 Zout , PCin , Yin , WMFC 2 MDRout , IRin 3 Branchto startingaddress of appropriate microroutine . ... .. ... ... .. ... .. ... ... .. ... ... .. ... .. ... ... .. ... .. ... ... .. ... .. 25 If N=0, then branchto microinstruction 0 26 Offset-field-of-IRout , SelectY, Add, Zin 27 Zout , PCin , End Figure 7.17. Microroutine for the instruction Branch<0.
- 23. Figure 7.18. Organizationof the control unit to allow conditional branching in the microprogram. Control store Clock generator Starting and branch address Condition codes inputs External CW IR mPC
- 24. Microinstructions A straightforwardway to structure microinstructions is to assign one bit position to each control signal. However, this is very inefficient. The length can be reduced: most signals are not needed simultaneously, and many signals are mutually exclusive. All mutually exclusive signals are placed in the same group in binary coding.
- 25. Partial Format forthe Microinstructions
- 26. Further Improvement Verticalorganization: 1. each micro instruction specify only a small number of control functions. 2. Slower operating speed due to the need of more micro instructions to perform the desired function. 3. less hardware is needed to handle the execution of microinstructions. Horizontal organization: 1. each micro instruction specify many control signals. 2. its useful when higher operating speed is desired and the machine structure allows parallel use of resources
- 27. Microprogram sequencing Several operationsexecute with varying addressing modes • For example, consider ADD r1, r2, r3; ADD (r1), r2,r3; ADD x(r1), r2, r3; and ADD (r1)+, r2, r3 • two disadvantages: Ø Having a separate microroutine for each machine instruction results in a large total number of microinstructions and a large control store. Ø Longer execution time because it takes more time to carry out the required branches. § a separate microroutine for each combination of instruction would produce considerable duplication of common parts. § Organize the micro program so that the micro routines share as many common parts as possible.
- 28. Prefetching microinstructions Microprogrammedcontrol leads to a slower operating speed because of the time it take to fetch microinstructions from control store. Avoid this by prefetching the next microinstruction while the current one is executed. prefetching the microinstruction had some difficulties like the status flags and the result of the current microinstruction are needed to determine the address of next microinstruction. Thus a straight forward prefetching occasionally prefetches wrong microinstruction. In this case the fetch must be repeated with correct address.
- 29. emulation Allows usto replace obsolete equipment with more up to date machines. If the replacement computer fully emulates the original one then no software changes have have to be made to run existing programs. Facilitates transition to new computer system with minimal disruption.
- 30. ADVANTAGES The designof micro-program control unit is less complex because micro-programs are implemented using software routines. The micro-programmed control unit is more flexible because design modifications, correction and enhancement is easily possible. The new or modified instruction set of CPU can be easily implemented by simply rewriting or modifying the contents of control memory. The fault can be easily diagnosed in the micro-program control unit using diagnostics tools by maintaining the contents of flags, registers and counters.
- 31. DISADVANTAGES The micro-programcontrol unit is slower than hardwired control unit. That means to execute an instruction in micro-program control unit requires more time. The micro-program control unit is expensive than hardwired control unit in case of limited hardware resources. The design duration of micro-program control unit is more than hardwired control unit for smaller CPU.
- 33.
- 34. Signed Numbers Leftmost bit is sign bit 0 means positive 1 means negative +18 = 00010010 -18 = 10010010 •To compute representation of a negative number faster, find the representation of its absolute value in n bits. •invert all bits of this number and add 1 to it. •For example, −10 in 8 bits 10 in 8 bits is 00001010. Invert all bits to get: 11110101. Add 1 to it to get: 11110110. 2’s complement number system
- 35. Addition/subtraction of signednumbers si = ci +1 = 13 7 + Y 1 0 0 0 1 0 1 1 0 0 1 1 0 1 1 0 0 1 1 0 1 0 0 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 Example: 1 0= = 0 0 1 1 1 1 1 0 0 1 1 1 10 xi yi Carry-in ci Sumsi Carry-outci+1 X Z + 6 0+ xi yi si Carry-out ci+1 Carry-in ci xi yi ci xi yi ci xi yi ci xi yi ci xi yi ciÅ Å=+ + + yi ci xi ci xi yi+ + At the ith stage: Input: ci is the carry-in Output: si is the sum ci+1 carry-out to (i+1)st state
- 36. Addition logic fora single stage Full Adder (FA): Symbol for the complete circuit for a single stage of addition. Full adder (FA) ci ci 1+ s i Sum Carry yi xi c i yi xi c i yi x i xi ci yi si c i 1+
- 37. n-bit adder •Cascade nfull adder (FA) blocks to form a n-bit adder. •Carries propagate or ripple through this cascade, n-bit ripple carry adder. FA c0 y1 x1 s1 FA c 1 y0 x0 s0 FA cn 1- y n 1- xn 1- cn sn 1- Most significant bit (MSB) position Least significant bit (LSB) position Carry-in c0 into the LSB position provides a convenient way to perform subtraction. K n-bit adder K n-bit numbers can be added by cascading k n-bit adders. n-bit c 0 yn xn s n cn y0 xn 1- s 0 ckn s k 1-( )n x0 yn 1- y2n 1- x2n 1- ykn 1- s n 1- s 2n 1- s kn 1- xkn 1- adder n-bit adder n-bit adder Each n-bit adder forms a block, so this is cascading of blocks. Carries ripple or propagate through blocks, Blocked Ripple Carry Adder
- 38. n-bit subtractor FA 1 y 1 x 1 s 1 FA c 1 y 0 x 0 s 0 FA c n1- y n 1- x n 1- c n s n 1- Most significant bit (MSB) position Least significant bit (LSB) position •Recall X – Y is equivalent to adding 2’s complement of Y to X. •2’s complement is equivalent to 1’s complement + 1. •X – Y = X + Y + 1 •2’s complement of positive and negative numbers is computed similarly.
- 39. n-bit adder/subtractor •Add/sub control= 0, addition. •Add/sub control = 1, subtraction. Add/Sub control n-bit adder x n 1- x 1 x 0 c n s n 1- s 1 s 0 c 0 y n 1- y 1 y 0
- 40. Detecting overflows Overflowscan only occur when the sign of the two operands is the same. Overflow occurs if the sign of the result is different from the sign of the operands. Recall that the MSB represents the sign. xn-1, yn-1, sn-1 represent the sign of operand x, operand y and result s respectively. Circuit to detect overflow can be implemented by the following logic expressions: 111111 −−−−−− += nnnnnn syxsyxOverflow 1−⊕= nn ccOverflow
- 41. Computing the addtime Consider 0th stage: x0 y0 c0 c1 s0 FA •c1 is available after 2 gate delays. •s1 is available after 1 gate delay. c i yi xi c i yi x i xi ci yi si c i 1+ Sum Carry
- 42. Computing the addtime (contd..) x0 y0 s2 FA x0 y0x0 y0 s1 FA c2 s0 FA c1c3 c0 x0 y0 s3 FA c4 Cascade of 4 Full Adders, or a 4-bit adder •s0 available after 1 gate delays, c1 available after 2 gate delays. •s1 available after 3 gate delays, c2 available after 4 gate delays. •s2 available after 5 gate delays, c3 available after 6 gate delays. •s3 available after 7 gate delays, c4 available after 8 gate delays. For an n-bit adder, sn-1 is available after 2n-1 gate delays ,cn is available after 2n gate delays.
- 43. Fast addition Recall theequations: iiiiiii iiii cycxyxc cyxs ++= ⊕⊕= +1 Second equation can be written as: iiiiii cyxyxc )(1 ++=+ We can write: iiiiii iiii yxPandyxGwhere cPGc +== +=+1 •Gi is called generate function and Pi is called propagate function •Gi and Pi are computed only from xi and yi and not ci, thus they can be computed in one gate delay after X and Y are applied to the inputs of an n- bit adder.
- 44. Carry lookahead ci +1= Gi + Pi ci ci = Gi −1 + Pi −1ci −1 ⇒ ci+1 = Gi + Pi (Gi −1 + Pi −1ci −1) continuing ⇒ ci+1 = Gi + Pi (Gi −1 + Pi −1(Gi − 2 + Pi− 2ci −2 )) until ci+1 = Gi + PiGi −1 + Pi Pi−1Gi −2 + .. + Pi Pi −1..P1G0 + Pi Pi −1...P0c0 •All carries can be obtained 3 gate delays after X, Y and c0 are applied. -One gate delay for Pi and Gi -Two gate delays in the AND-OR circuit for ci+1 •All sums can be obtained 1 gate delay after the carries are computed. •Independent of n, n-bit addition requires only 4 gate delays. •This is called Carry Lookahead adder.
- 45. Carry-lookahead adder Carry-lookahead logic Bcell B cell B cell B cell s 3 P3 G3 c 3 P2 G2 c 2 s 2 G 1 c 1 P 1 s 1 G 0 c 0 P 0 s 0 .c4 x 1 y 1 x 3 y 3 x 2 y 2 x 0 y 0 G i c i .. . P i s i x i y i B cell 4-bit carry-lookahead adder B-cell for a single stage C1=G0+P0C0 C2=G1+P1G0+P1P0C0 C3=G2+P2G1+P2P1G0+P2P1P0C0 C4=G3+P3G2+P3P2G1+P3P2P1G0+P3P2P1P0C0
- 46. Blocked Carry-Lookahead adder c4= G3 + P3G2 + P3P2G1 + P3P2 P1G0 + P3P2 P1P0c0 Carry-out from a 4-bit block can be given as: Rewrite this as: P0 I = P3P2 P1P0 G0 I = G3 + P3G2 + P3P2G1 + P3P2P1G0 Subscript I denotes the blocked carry lookahead and identifies the block. Cascade 4 4-bit adders, c16 can be expressed as: c16 = G3 I + P3 I G2 I + P3 I P2 I G1 I + P3 I P2 I P1 0 G0 I + P3 I P2 I P1 0 P0 0 c0 In order to add operands longer than 4 bits, we can cascade 4- bit Carry-Lookahead adders. Cascade of Carry-Lookahead adders is called Blocked Carry-Lookahead adder.
- 47. Blocked Carry-Lookahead adder Carry-lookaheadlogic 4-bit adder 4-bit adder 4-bit adder 4-bit adder s15-12 P3 IG3 I c12 P2 IG2 I c8 s11-8 G1 I c4 P1 I s7-4 G0 I c0 P0 I s3-0 c16 x15-12 y15-12 x11-8 y11-8 x7-4 y7-4 x3-0 y3-0 . After xi, yi and c0 are applied as inputs: - Gi and Pi for each stage are available after 1 gate delay. - PI is available after 2 and GI after 3 gate delays. - All carries are available after 5 gate delays. - c16 is available after 5 gate delays. - s15 which depends on c12 is available after 8 (5+3)gate delays (Recall that for a 4-bit carry lookahead adder, the last sum bit is available 3 gate delays after all inputs are available)
- 48.
- 49. Multiplication of unsignednumbers Product of 2 n-bit numbers is at most a 2n-bit number. Unsigned multiplication can be viewed as addition of shifted versions of the multiplicand.
- 50. Multiplication of unsignednumbers (contd..) Rules to implement multiplication are: If the 0th bit of multiplier is 1, then add the multiplicand with 0 If the 0th bit of multiplier is 0 ,then current partial product is 0. If the ith bit (except 0th bit) of the multiplier is 1, shift the multiplicand and add the shifted multiplicand to the current value of the partial product. Hand over the partial product to the next stage If the ith bit of the multiplier is 0, shift the multiplicand and add the current partial product to the 0. Value of the partial product at the start stage is 0.
- 51. Multiplication of unsignednumbers ith multiplier bit carry incarry out jth multiplicand bit Bit of incoming partial product (PPi) Bit of outgoing partial product (PP(i+1)) FA Typical multiplication cell
- 52. Combinational array multiplier Multiplicand m3 m 2 m 1 m 00 0 0 0 q3 q 2 q 1 q0 0 p 2 p 1 p 0 0 0 0 p3 p 4 p 5 p 6 p 7 PP1 PP2 PP3 (PP0) , Product is: p7,p6,..p0 Multiplicand is shifted by displacing it through an array of adders.
- 53. Combinatorial arraymultipliers are: Extremely inefficient. Have a high gate count for multiplying numbers of practical size such as 32-bit or 64-bit numbers. Perform only one function, namely, unsigned integer product. Improve gate efficiency by using a mixture of combinatorial array techniques and sequential techniques requiring less combinational logic.
- 54. Sequential Circuit Multiplier q n1- m n 1- n-bit Adder Multiplicand M Control sequencer Multiplier Q 0 C Shift right Register A (initially 0) Add/Noadd control a n 1- a 0 q 0 m 0 0 MUX
- 55. 1 1 11 1 0 1 1 1 1 1 1 1 1 1 0 1 1 1 0 1 1 0 1 1 1 0 1 Initial configuration Add M 1 1 0 1 C First cycle Second cycle Third cycle Fourth cycle No add Shift Shift Add Shift Shift Add 1 1 1 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 1 1 0 1 1 0 1 0 0 1 1 1 0 0 1 0 1 0 0 0 0 0 1 1 0 0 0 1 0 0 1 1 0 1 1 QA Product
- 56.
- 57. Signed Multiplication Considering2’s-complement signed operands, Extend the sign bit value of the multiplicand to the left as far as the product will extend (-13)(+11) 1 0 11 11 1 1 0 0 1 1 110 110 1 0 1000111011 000000 1100111 00000000 110011111 13-( ) 143-( ) 11+( )
- 58. Signed Multiplication Fora negative multiplier, a straightforward solution is to form the 2’s-complement of both the multiplier and the multiplicand and proceed as in the case of a positive multiplier. This is possible because complementation of both operands does not change the value or the sign of the product. A technique that works equally well for both negative and positive multipliers – Booth algorithm.
- 59. Booth Algorithm Multiplier Bit iBit i 1- V ersion of multiplicand selected by bit 0 1 0 0 01 1 1 0 M 1+ M 1 M 0 M Booth multiplier recoding table. X X X X Booth multiplication with a negative multiplier. 01 0 1 1 1 1 0 1 1 0 0 0 0 0 0 0 0 0 00 0110 0 0 0 0 1 1 0 1100111 0 0 0 0 0 0 01000 11111 1 1 0 1 1 0 1 1 1 0 1 0 6-( ) 13+( ) X 78-( ) +11- 1- 2's complement of the multiplicand
- 60.
- 61. Bit-Pair Recoding ofMultipliers Bit-pair recoding halves the maximum number of summands (versions of the multiplicand). 1+1 (a) Example of bit-pair recoding derived from Booth recoding 0 000 1 1 0 1 0 Implied 0 to right of LSB 1 0 Sign extension 1 21
- 62. 1- 0000 1 1 11 1 0 0 0 0 0 11 1 1 1 1 10 0 0 0 0 0 0 0 0000 111111 0 1 1 0 1 0 1 010011111 1 1 1 1 0 0 1 1 0 0 0 0 0 0 1 1 1 0 1 1 0 0 1 0 0 1 0 0 1 0 1 0 0 0 0 1 0 0 1 10 0 010 0 1 1 0 1 11 1- 6-( ) 13+( ) 1+ 78-( ) 1- 2- ´ Figure 6.15. Multiplication requiring only n/2 summands. i 1+ i 1 (b) Table of multiplicand selection decisions selected at positioni MultiplicandMultiplier bit-pair i 0 0 1 1 1 0 1 0 1 1 1 1 0 0 0 1 1 0 0 1 0 0 1 Multiplier bit on the right 0 0 X M 1+ 1 1+ 0 1 2 2+ X M X M X M X M X M X M X M
- 63. Carry-Save Addition ofSummands CSA speeds up the addition process. P7 P6 P5 P4 P3 P2 P1 P0
- 64.
- 65. Consider theaddition of many summands, we can: Ø Group the summands in threes and perform carry-save addition on each of these groups in parallel to generate a set of S and C vectors in one full-adder delay Ø Group all of the S and C vectors into threes, and perform carry- save addition on them, generating a further set of S and C vectors in one more full-adder delay Ø Continue with this process until there are only two vectors remaining Ø They can be added in a RCA or CLA to produce the desired product
- 66. Figure 6.17. Amultiplication example used to illustrate carry-save addition as shown in Figure 6.18. 100 1 11 100 1 11 100 1 11 11111 1 100 1 11 M Q A B C D E F (2,835) X (45) (63) 100 1 11 100 1 11 100 1 11 000 1 11 111 0 00 Product
- 67. Figure 6.18. Themultiplication example from Figure 6.17 performed using carry-save addition. 00000101 0 10 10010000 1 11 1 + 1000011 1 10010111 0 10 1 0110 1 10 0 00011010 0 00 10001011 1 0 1 110001 1 0 00111100 00110 1 10 11001 0 01 100 1 11 100 1 11 100 1 11 00110 1 10 11001 0 01 100 1 11 100 1 11 100 1 11 11111 1 100 1 11 M Q A B C S 1 C 1 D E F S 2 C 2 S1 C 1 S2 S 3 C3 C2 S4 C 4 Product x
- 68.
- 69. Longhand Division Steps Position the divisor appropriately with respect to the dividend and performs a subtraction. If the remainder is zero or positive, a quotient bit of 1 is determined, the remainder is extended by another bit of the dividend, the divisor is repositioned, and another subtraction is performed. If the remainder is negative, a quotient bit of 0 is determined, the dividend is restored by adding back the divisor, and the divisor is repositioned for another subtraction.
- 70. 001111 Division of UnsignedBinary Integers 1011 00001101 10010011 1011 001110 1011 1011 100 Quotient Dividend Remainder Partial Remainders Divisor
- 71. Circuit Arrangement Figure 6.21.Circuit arrangement for binary division. qn-1 Divisor M Control Sequencer Dividend Q Shift left N+1 bit adder q0 Add/Subtract Quotient Setting A m00 mn-1 a0an an-1
- 72. Restoring Division ShiftA and Q left one binary position Subtract M from A, and place the answer back in A If the sign of A is 1, set q0 to 0 and add M back to A (restore A); otherwise, set q0 to 1 Repeat these steps n times
- 73. Examples 10111 Figure 6.22. Arestoring-division example. 1 1 1 1 1 01111 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 01 11 1 1 01 0001 Subtract Shift Restore 1 0000 1 0000 1 1 Initially Subtract Shift 10111 10000 11000 00000 Subtract Shift Restore 10111 01000 10000 1 1 QuotientRemainder Shift 10111 1 0000 Subtract Second cycle First cycle Third cycle Fourth cycle 0 0 0 0 0 0 1 0 1 10000 1 1 1 0000 11111 Restore q0Set q0Set q0Set q0Set
- 74. Nonrestoring Division Avoidthe need for restoring A after an unsuccessful subtraction. Step 1: (Repeat n times) Ø If the sign of A is 0, shift A and Q left one bit position and subtract M from A; otherwise, shift A and Q left and add M to A. Ø Now, if the sign of A is 0, set q0 to 1; otherwise, set q0 to 0. Step2: If the sign of A is 1, add M to A
- 75. Examples A nonrestoring-division example. Add Restore remainder Remainder 00 0 01 1 1 1 1 1 0 0 0 1 1 1 Quotient 0 0 1 01 1 1 1 1 0 0 0 01 1 1 1 1 Shift 0 0 0 11000 01111 Add 0 0 0 1 1 0 0 0 0 1 0 0 0 1 1 1 0 1 Shift Subtract Initially 0 0 0 0 0 1 0 0 0 1 1 1 0 0000 1 1 1 0 0 0 0 0 1 1 0 0 0Shift Add 0 0 10 0 0 01 1 1 1 0 1 Shift Subtract 0 0 0 110000 Fourth cycle Third cycle Second cycle First cycle q 0Set q 0 Set q 0 Set q 0 Set
- 76.
- 77. Fractions If b isa binary vector, then we have seen that it can be interpreted as an unsigned integer by: V(b) = b31.231 + b30.230 + bn-3.229 + .... + b1.21 + b0.20 This vector has an implicit binary point to its immediate right: b31b30b29....................b1b0. implicit binary point Suppose if the binary vector is interpreted with the implicit binary point is just left of the sign bit: implicit binary point .b31b30b29....................b1b0 The value of b is then given by: V(b) = b31.2-1 + b30.2-2 + b29.2-3 + .... + b1.2-31 + b0.2-32
- 78. Range of fractions Thevalue of the unsigned binary fraction is: V(b) = b31.2-1 + b30.2-2 + b29.2-3 + .... + b1.2-31 + b0.2-32 The range of the numbers represented in this format is: In general for a n-bit binary fraction (a number with an assumed binary point at the immediate left of the vector), then the range of values is: 9999999998.021)(0 32 ≈−≤≤ − bV n bV − −≤≤ 21)(0
- 79. Scientific notation •Previous representationshave a fixed point. Either the point is to the immediate right or it is to the immediate left. This is called Fixed point representation. •Fixed point representation suffers from a drawback that the representation can only represent a finite range (and quite small) range of numbers. A more convenient representation is the scientific representation, where the numbers are represented in the form: x = m1.m2m3m4 × b±e Components of these numbers are: Mantissa (m), implied base (b), and exponent (e)
- 80. Significant digits A numbersuch as the following is said to have 7 significant digits x = ±0.m1m2m3m4m5m6m7 × b±e Fractions in the range 0.0 to 0.9999999 need about 24 bits of precision (in binary). For example the binary fraction with 24 1’s: 111111111111111111111111 = 0.9999999404 Not every real number between 0 and 0.9999999404 can be represented by a 24-bit fractional number. The smallest non-zero number that can be represented is: 000000000000000000000001 = 5.96046 x 10-8 Every other non-zero number is constructed in increments of this value.
- 81. Sign and exponentdigits •In a 32-bit number, suppose we allocate 24 bits to represent a fractional mantissa. •Assume that the mantissa is represented in sign and magnitude format, and we have allocated one bit to represent the sign. •We allocate 7 bits to represent the exponent, and assume that the exponent is represented as a 2’s complement integer. •There are no bits allocated to represent the base, we assume that the base is implied for now, that is the base is 2. •Since a 7-bit 2’s complement number can represent values in the range 64 to 63, the range of numbers that can be represented is: 0.0000001 x 2-64 < = | x | <= 0.9999999 x 263 •In decimal representation this range is: 0.5421 x 10-20 < = | x | <= 9.2237 x 1018
- 82. A sample representation SignExponent Fractional mantissa bit 1 7 24 •24-bit mantissa with an implied binary point to the immediate left •7-bit exponent in 2’s complement form, and implied base is 2.
- 83. Normalization If the numberis to be represented using only 7 significant mantissa digits, the representation ignoring rounding is: Consider the number: x = 0.0004056781 x 1012 x = 0.0004056 x 1012 If the number is shifted so that as many significant digits are brought into 7 available slots: x = 0.4056781 x 109 = 0.0004056 x 1012 Exponent of x was decreased by 1 for every left shift of x. A number which is brought into a form so that all of the available mantissa digits are optimally used (this is different from all occupied which may not hold), is called a normalized number. Same methodology holds in the case of binary mantissas 0001101000(10110) x 28 = 1101000101(10) x 25
- 84. Normalization (contd..) •A floatingpoint number is in normalized form if the most significant 1 in the mantissa is in the most significant bit of the mantissa. •All normalized floating point numbers in this system will be of the form: 0.1xxxxx.......xx Range of numbers representable in this system, if every number must be normalized is: 0.5 x 2-64 <= | x | < 1 x 263
- 85. Normalization, overflow andunderflow The procedure for normalizing a floating point number is: Do (until MSB of mantissa = = 1) Shift the mantissa left (or right) Decrement (increment) the exponent by 1 end do Applying the normalization procedure to: .000111001110....0010 x 2-62 gives: .111001110........ x 2-65 But we cannot represent an exponent of –65, in trying to normalize the number we have underflowed our representation. Applying the normalization procedure to: 1.00111000............x 263 gives: 0.100111..............x 264 This overflows the representation.
- 86. Changing the impliedbase So far we have assumed an implied base of 2, that is our floating point numbers are of the form: x = m 2e If we choose an implied base of 16, then: x = m 16e Then: y = (m.16) .16e-1 (m.24) .16e-1 = m . 16e = x •Thus, every four left shifts of a binary mantissa results in a decrease of 1 in a base 16 exponent. •Normalization in this case means shifting the mantissa until there is a 1 in the first four bits of the mantissa.
- 87. Excess notation •Rather thanrepresenting an exponent in 2’s complement form, it turns out to be more beneficial to represent the exponent in excess notation. •If 7 bits are allocated to the exponent, exponents can be represented in the range of -64 to +63, that is: -64 <= e <= 63 Exponent can also be represented using the following coding called as excess-64: E’ = Etrue + 64 In general, excess-p coding is represented as: E’ = Etrue + p True exponent of -64 is represented as 0 0 is represented as 64 63 is represented as 127 This enables efficient comparison of the relative sizes of two floating point numbers.
- 88. IEEE notation IEEE FloatingPoint notation is the standard representation in use. There are two representations: - Single precision. - Double precision. Both have an implied base of 2. Single precision: - 32 bits (23-bit mantissa, 8-bit exponent in excess-127 representation) Double precision: - 64 bits (52-bit mantissa, 11-bit exponent in excess-1023 representation) Fractional mantissa, with an implied binary point at immediate left. Sign Exponent Mantissa 1 8 or 11 23 or 52
- 89. Peculiarities of IEEEnotation •Floating point numbers have to be represented in a normalized form to maximize the use of available mantissa digits. •In a base-2 representation, this implies that the MSB of the mantissa is always equal to 1. •If every number is normalized, then the MSB of the mantissa is always 1. We can do away without storing the MSB. •IEEE notation assumes that all numbers are normalized so that the MSB of the mantissa is a 1 and does not store this bit. •So the real MSB of a number in the IEEE notation is either a 0 or a 1. •The values of the numbers represented in the IEEE single precision notation are of the form: (+,-) 1.M x 2(E - 127) •The hidden 1 forms the integer part of the mantissa. •Note that excess-127 and excess-1023 (not excess-128 or excess-1024) are used to represent the exponent.
- 90. Exponent field In theIEEE representation, the exponent is in excess-127 (excess-1023) notation. The actual exponents represented are: -126 <= E <= 127 and -1022 <= E <= 1023 not -127 <= E <= 128 and -1023 <= E <= 1024 This is because the IEEE uses the exponents -127 and 128 (and -1023 and 1024), that is the actual values 0 and 255 to represent special conditions: - Exact zero - Infinity
- 91. Floating point arithmetic Addition: 3.1415x 108 + 1.19 x 106 = 3.1415 x 108 + 0.0119 x 108 = 3.1534 x 108 Multiplication: 3.1415 x 108 x 1.19 x 106 = (3.1415 x 1.19 ) x 10(8+6) Division: 3.1415 x 108 / 1.19 x 106 = (3.1415 / 1.19 ) x 10(8-6) Biased exponent problem: If a true exponent e is represented in excess-p notation, that is as e+p. Then consider what happens under multiplication: a. 10(x + p) * b. 10(y + p) = (a.b). 10(x + p + y +p) = (a.b). 10(x +y + 2p) Representing the result in excess-p notation implies that the exponent should be x+y+p. Instead it is x+y+2p. Biases should be handled in floating point arithmetic.
- 92. Floating point arithmetic:ADD/SUB rule Choose the number with the smaller exponent. Shift its mantissa right until the exponents of both the numbers are equal. Add or subtract the mantissas. Determine the sign of the result. Normalize the result if necessary and truncate/round to the number of mantissa bits. Note: This does not consider the possibility of overflow/underflow.
- 93. Floating point arithmetic:MUL rule Add the exponents. Subtract the bias. Multiply the mantissas and determine the sign of the result. Normalize the result (if necessary). Truncate/round the mantissa of the result.
- 94. Floating point arithmetic:DIV rule Subtract the exponents Add the bias. Divide the mantissas and determine the sign of the result. Normalize the result if necessary. Truncate/round the mantissa of the result. Note: Multiplication and division does not require alignment of the mantissas the way addition and subtraction does.
- 95. Guard bits While addingtwo floating point numbers with 24-bit mantissas, we shift the mantissa of the number with the smaller exponent to the right until the two exponents are equalized. This implies that mantissa bits may be lost during the right shift (that is, bits of precision may be shifted out of the mantissa being shifted). To prevent this, floating point operations are implemented by keeping guard bits, that is, extra bits of precision at the least significant end of the mantissa. The arithmetic on the mantissas is performed with these extra bits of precision. After an arithmetic operation, the guarded mantissas are: - Normalized (if necessary) - Converted back by a process called truncation/rounding to a 24-bit mantissa.
- 96. Truncation/rounding Straight chopping: The guard bits (excess bits of precision) are dropped. Von Neumann rounding: If the guard bits are all 0, they are dropped. However, if any bit of the guard bit is a 1, then the LSB of the retained bit is set to 1. Rounding: If there is a 1 in the MSB of the guard bit then a 1 is added to the LSB of the retained bits.
- 97. Rounding Rounding isevidently the most accurate truncation method. However, Rounding requires an addition operation. Rounding may require a renormalization, if the addition operation de- normalizes the truncated number. IEEE uses the rounding method. 0.111111100000 rounds to 0.111111 + 0.000001 =1.000000 which must be renormalized to 0.100000
- 98.