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EE-2623 Mikroprosesor & Antarmuka

EE-2623 Mikroprosesor & Antarmuka. Materi 4 Tracing Instruksi Team Dosen 2006. Programming (JHR). Tahap Decode. Dari siklus FDX, proses yang paling rumit adalah decode Beberapa istilah : Program  susunan instruksi (user) Instruksi  bagian terkecil dari program (user)

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EE-2623 Mikroprosesor & Antarmuka

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  1. EE-2623Mikroprosesor & Antarmuka Materi 4 Tracing Instruksi Team Dosen 2006

  2. Programming (JHR)

  3. Tahap Decode • Dari siklus FDX, proses yang paling rumit adalah decode • Beberapa istilah : • Program  susunan instruksi (user) • Instruksi  bagian terkecil dari program (user) • Mikroprogram  susunan pekerjaan yang diperintahkan Control System • Mikroinstruksi  pekerjaan yang diperintahkan Control System • Instruksi –fetch Control System –decode mikroprogram

  4. Control System /Control Unit /Instruction Decoder ADD A,B • Opcode ADD dibawake CU • Didecodekanmenjadi • B (dimemori) ALU [2] • B (di ALU)  ACC [3] • A (dimemori) ALU [4] • B (di ACC)  ALU [5] • + A & B [6] • Hasil ACC [7] 1 ADD 4 2,5 A B 6 + 3,7 B,Hasil 5 B ACC Memory

  5. Control System /Control Unit /Instruction Decoder MOV A,B • Opcode ADD dibawake CU • Didecodekanmenjadi • B (dimemori) ALU [2] • B (di ALU)  ACC [3] • ACC  A (dimemori) [4] 1 MOV 2 B 3 B 4 B ACC Memory

  6. Misalkansebuah program sbb: • 136B:0100 BB0F00 MOV BX,000F • 136B:0103 89D8 MOV AX,BX • 136B:0105 8B07 MOV AX,[BX] • 136B:0107 8B47FF MOV AX,[BX-01] • 136B:010A A10D00 MOV AX,[000D] • Perintah MOV BX,000F  BX = 000F • Perintah MOV AX,BX  AX = BX • Perintah MOV AX,[BX]  AL = [DS:BX] AH = [DS:BX+1]

  7. Perhitungan Timing • Misal • NOP : 2 clock • IN : 10 clock • JMP : 15 clock • Clock : 1 MHz • Ingin input data ke AL setiap 75 mikrodetik = (75 us : 1 MHz) = 75 clock LOOP: IN COM1 10 NOP x 25 50 JMP LOOP 15

  8. ContohSoal: • Berapa clock yang dibutuhkan program ini: MOV BX,00FC 4 clock LOOP: MOV AX,00FF 4 clock INC BX 3 clock XOR AX,BX 3 clock JNZ LOOP 4/16 clock MOV AX,BX 2 clock

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