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MALWA INSTITUTE OF SCIENCE
AND
TECHNOLOGY, INDORE
COMPUTER ARCHITECTURE
SUBMITTED TO : GAYATRI LODHI MAM SUBMITTED BY :
SURYANSH MALVIYA
DIPLOMA CS 2ND YEAR
8602090865
•Instruction cycle
INSTRUCTION CYCLE
• The instruction cycle (also known as the fetch–decode–execute
cycle, or simply the fetch-execute cycle) is the cycle that the
central processing unit (CPU) follows from boot-up until the
computer has shut down in order to process instructions.
INSTRUCTION CYCLE
• ´A program consisting of sequence of instructions is executed
in the computer by going through a cycle for each instruction.
• ´ Each instruction cycle is subdivided in to sub cycles or
phases.They are
• ´Fetch an instruction from memory
• ´Decode instruction
• ´Read effective address from memory if instruction has an
indirect address
• ´Execute instruction
• This cycle repeats indefinitely unless a HALT
instruction is encountered.
• The basic computer has three instruction code formats.
• Each format has 16 bits.
• The operation code (opcode) part of the instruction contains
three bits and the meaning of the remaining 13 bits depends
on the operation code encountered.
• A memoryreference instruction uses 12 bits to specify an
address and one bit to specify the addressing mode I.
• I is equal to 0 for direct address and to 1 for indirect address
MICROOPERATIONS
•The microoperations for fetch and decode
phases are
• T0:AR←PC
• T1:IR←M[AR] , PC<- PC+1
• T2:D0,D1..D7←Decode IR(12-14), AR←IR(011), I←IR(15)
FETCH & DECORD
• At T0 Transfers the address from PC to AR.
• At T1 Instruction read from memory is placed in IR and PC is
incremented by 1 to get the address of next instruction.
• At T2 opcode in IR is decoded , Indirect bit is transferred to
flipflop I & address part is transferred to AR.
• After decoding next step is to determine the type of
instruction.
COMPONENT OF INSTRUCTION CYCLE
• Program counter (PC)
• Memory address register (MAR)
• Memory data register (MDR)
• Instruction register (IR)
• Control unit (CU)
• Arithmatic logic unit (ALU)
•flow chart of instruction cycle
• Note that the sequence counter SC is either incremented or
cleared to 0 with every positive clock transition.
• We will adopt the convention that if SC is incremented, we will
not write the statement SC ← SC + 1, but it will be implied that
the control goes to the next timing signal in sequence.
• When SC is to be cleared, we will include the statement Sc ← 0
DETERMINING TYPE OF INSTRUCTION
• Afterdecoding timing signal active is T3 during which instruction
type is identified.
´Memory Reference
If D7=0 opcode will be 000 through 110.
If D7=0 and I=1,indirect and D7=0 and I=0 direct.
Microoperations for indirect address should be initially AR←M[AR]
´Register reference/ io
If D7=1 and I=0 – Register.
If D7=1 and I=1 – i/o.
EXPLANATION
• Decoder output D7 is equal to 1 if the operation code is equal
to binary 111.
• We determine that if D1 = 1, the instruction must be a
register-reference or input-output type.
• If D7 = 0, the operation code must be one of the other seven
values 000 through 110, specifying memory reference
instruction.
• Control then inspects the value of the first bit of the
instruction, which is now available in flip-flop I.
Instruction cycle.pptx

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Instruction cycle.pptx

  • 1. MALWA INSTITUTE OF SCIENCE AND TECHNOLOGY, INDORE COMPUTER ARCHITECTURE SUBMITTED TO : GAYATRI LODHI MAM SUBMITTED BY : SURYANSH MALVIYA DIPLOMA CS 2ND YEAR 8602090865
  • 3. INSTRUCTION CYCLE • The instruction cycle (also known as the fetch–decode–execute cycle, or simply the fetch-execute cycle) is the cycle that the central processing unit (CPU) follows from boot-up until the computer has shut down in order to process instructions.
  • 4.
  • 5. INSTRUCTION CYCLE • ´A program consisting of sequence of instructions is executed in the computer by going through a cycle for each instruction. • ´ Each instruction cycle is subdivided in to sub cycles or phases.They are • ´Fetch an instruction from memory • ´Decode instruction • ´Read effective address from memory if instruction has an indirect address • ´Execute instruction
  • 6. • This cycle repeats indefinitely unless a HALT instruction is encountered. • The basic computer has three instruction code formats. • Each format has 16 bits. • The operation code (opcode) part of the instruction contains three bits and the meaning of the remaining 13 bits depends on the operation code encountered. • A memoryreference instruction uses 12 bits to specify an address and one bit to specify the addressing mode I. • I is equal to 0 for direct address and to 1 for indirect address
  • 7. MICROOPERATIONS •The microoperations for fetch and decode phases are • T0:AR←PC • T1:IR←M[AR] , PC<- PC+1 • T2:D0,D1..D7←Decode IR(12-14), AR←IR(011), I←IR(15)
  • 8. FETCH & DECORD • At T0 Transfers the address from PC to AR. • At T1 Instruction read from memory is placed in IR and PC is incremented by 1 to get the address of next instruction. • At T2 opcode in IR is decoded , Indirect bit is transferred to flipflop I & address part is transferred to AR. • After decoding next step is to determine the type of instruction.
  • 9. COMPONENT OF INSTRUCTION CYCLE • Program counter (PC) • Memory address register (MAR) • Memory data register (MDR) • Instruction register (IR) • Control unit (CU) • Arithmatic logic unit (ALU)
  • 10. •flow chart of instruction cycle
  • 11.
  • 12. • Note that the sequence counter SC is either incremented or cleared to 0 with every positive clock transition. • We will adopt the convention that if SC is incremented, we will not write the statement SC ← SC + 1, but it will be implied that the control goes to the next timing signal in sequence. • When SC is to be cleared, we will include the statement Sc ← 0
  • 13. DETERMINING TYPE OF INSTRUCTION • Afterdecoding timing signal active is T3 during which instruction type is identified. ´Memory Reference If D7=0 opcode will be 000 through 110. If D7=0 and I=1,indirect and D7=0 and I=0 direct. Microoperations for indirect address should be initially AR←M[AR] ´Register reference/ io If D7=1 and I=0 – Register. If D7=1 and I=1 – i/o.
  • 14. EXPLANATION • Decoder output D7 is equal to 1 if the operation code is equal to binary 111. • We determine that if D1 = 1, the instruction must be a register-reference or input-output type. • If D7 = 0, the operation code must be one of the other seven values 000 through 110, specifying memory reference instruction. • Control then inspects the value of the first bit of the instruction, which is now available in flip-flop I.