Computer instruction
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1. The document discusses the topics of computer instructions, timing and control, and the instruction cycle for a basic computer. 2. It describes the three instruction code formats used - memory reference, register reference, and input/output. Memory reference instructions use bits to specify an address and addressing mode. Register reference instructions specify an operation on the accumulator register. 3. The instruction cycle consists of four phases - fetch an instruction, decode the instruction, read the effective address if needed, and execute the instruction.
![INSTRUCTION CYCLEIn basic computer each instruction cycle consist of following phases:-1. Fetch an instruction from the memory.2. Decode the instruction. 3. Read the effective address from the memory if the instruction has an indirect address. 4. Execute the instruction.Fetch & decode ->Initially, the PC is loaded with the address of the first incrementer in the program. The SC(sequence counter) is cleared to 0,providing a decoded time signal T0.After each clock pulse ,SC is incremented by1 so that the timing go through a sequenceT0,T1,T3 and so on . The microperations for fetch & decode phases are:- T0: AR<-PC T1: IR<-M[AR],PC<-PC+1 T2: D0…..,D7<-Decode IR(12-14),AR<-IR(0-11),I<-IR(15)](https://image.slidesharecdn.com/computerinstruction-110818112455-phpapp01/85/Computer-instruction-10-320.jpg)
![FETCH and DECODE• Fetch and DecodeT0: AR PC (S0S1S2=010, T0=1)T1: IR M [AR], PC PC + 1 (S0S1S2=111, T1=1)T2: D0, . . . , D7 Decode IR(12-14), AR IR(0-11), I IR(15)T1S2BusT0S1S0Memory7unitAddressReadAR1LDPC2INRIR5LDClockCommon bus](https://image.slidesharecdn.com/computerinstruction-110818112455-phpapp01/85/Computer-instruction-11-320.jpg)
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Computer instruction
- 1. PRESENTATIONTOPICS :1. COMPUTER INSTRUCTIONS.2.TIMING & CONTROL.3. INSTRUCTION CYCLE.BY :- PAAWAN GUPTA109-38126B.SC.(C.SCIENCE)2ND YEAR
- 2. Computer InstructionsThe basic computer has three instruction code formats which are :-MEMORY- REFERENCE INSTRUCTION A memory – reference instruction uses 12 bits to specify an address and 1 bit to specify the addressing mode I. I is equal to 0 for direct address and to 1 for indirect addressREGISTER- REFERENCE INSTRUCTION A register reference instruction specifies an operation on or a test of the AC register .An operand from the memory is not needed therefore the other 12 bits are used to specify the operation to be executed.INPUT-OUTPUT INSTRUCTION An input-output instruction does not need a reference to memory and is recognized by operational code 111 with a 1 in the left most bit of the instruction the remaining 12 bits are used to specify the type of input output operation or test performed.
- 3. 15 12 110Register operation0 1 1 115 12 110I/O operation1 1 1 1 Basic Computer Instruction FormatMemory-Reference Instructions (OP-code = 000 ~ 110)15 1412 110OpcodeAddressIRegister-Reference Instructions (OP-code = 111, I = 0) Input-Output Instructions (OP-code =111, I = 1)
- 4. Basic Computer Instruction 15 14 12 11 0 15 14 12 11 0 15 14 12 11 0I Opcode Address0 1 1 1 Register Operation1 1 1 1 I/O Operation3 Instruction Code Formats :Memory-reference instruction
- 6. Input-Output instructionI=0 : Direct, I=1 : Indirect
- 7. INSTRUCTION SET COMPLETENESSA computer should have a set of instructions so that the user can construct machine language programs to evaluate any function that is known to be computable.Instruction Types :-
- 8. Functional Instructions - Arithmetic, logic, and shift instructions - ADD, CMA, INC, CIR, CIL, AND, CLATransfer Instructions - Data transfers between the main memory and the processor registers - LDA, STAControl Instructions - Program sequencing and control - BUN, BSA, ISZInput/OutputInstructions - Input and output - INP, OUT
- 9. The timing for all register in basic computer is controlled by a master clock generator.Clock pulses – The clock pulses are applied to all flip-flops and registers in the system , including the flip-flops in the control unit . The clock pulses do not change the state of a register unless the register is unable by a control signal. There are two types of control organisations :-Hardwired control – In hardwired organization, the control logic is implemented with gates , flip-flops , decoders , and other digital circuits. It has the advantage that it can be optimized to produce a fast mode of operation.2. Microprogramme control– The control information is stored in the control memory . The control memory is programmed to initiate the required sequence of microperations. TIMING AND CONTROLTIMING AND CONTROLControl unit of Basic ComputerInstruction register (IR)14 13 121511 - 0Other inputs3 x 8decoder 7 6 5 4 3 2 1 0D0CombinationalControllogicIDControlsignals7T15T015 14 . . . . 2 1 04 x 16decoderIncrement (INR)4-bitsequenceClear (CLR)counterClock(SC)
- 10. INSTRUCTION CYCLEIn basic computer each instruction cycle consist of following phases:-1. Fetch an instruction from the memory.2. Decode the instruction. 3. Read the effective address from the memory if the instruction has an indirect address. 4. Execute the instruction.Fetch & decode ->Initially, the PC is loaded with the address of the first incrementer in the program. The SC(sequence counter) is cleared to 0,providing a decoded time signal T0.After each clock pulse ,SC is incremented by1 so that the timing go through a sequenceT0,T1,T3 and so on . The microperations for fetch & decode phases are:- T0: AR<-PC T1: IR<-M[AR],PC<-PC+1 T2: D0…..,D7<-Decode IR(12-14),AR<-IR(0-11),I<-IR(15)
- 11. FETCH and DECODE• Fetch and DecodeT0: AR PC (S0S1S2=010, T0=1)T1: IR M [AR], PC PC + 1 (S0S1S2=111, T1=1)T2: D0, . . . , D7 Decode IR(12-14), AR IR(0-11), I IR(15)T1S2BusT0S1S0Memory7unitAddressReadAR1LDPC2INRIR5LDClockCommon bus
- 12. REGISTER REFERENCE INSTRUCTIONSRegister Reference Instructions are identified when- D7 = 1, I = 0- Register Ref. Instr. is specified in b0 ~ b11 of IR- Execution starts with timing signal T3r = D7 IT3 => Register Reference InstructionBi = IR(i) , i=0,1,2,...,11 r: SC 0CLA rB11: AC 0CLE rB10: E 0CMA rB9: AC AC’CME rB8: E E’CIR rB7: AC shr AC, AC(15) E, E AC(0)CIL rB6: AC shl AC, AC(0) E, E AC(15)INC rB5: AC AC + 1SPA rB4: if (AC(15) = 0) then (PC PC+1)SNA rB3: if (AC(15) = 1) then (PC PC+1)SZA rB2: if (AC = 0) then (PC PC+1)SZE rB1: if (E = 0) then (PC PC+1)HLT rB0: S 0 (S is a start-stop flip-flop)
- 13. THaNk you…….