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Solution to Exercise 8

Solution to Exercise 8. The algorithm. I == 0 : Op  Decode_Op(TerminalInput) I == 1 : Dest  Decode_Dest(TerminalInput) I == 2 : Dig1  Decode_Digit(TerminalInput) I == 3 : Dig2  Decode_Digit(TerminalInput) If Op == ‘e’ E  0 Tmp TerminalInput, I = I+1 (I == 3) && Tmp == 20h (space):

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Solution to Exercise 8

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  1. Solution to Exercise 8

  2. The algorithm • I == 0 : Op  Decode_Op(TerminalInput)I == 1 : Dest  Decode_Dest(TerminalInput)I == 2 : Dig1  Decode_Digit(TerminalInput)I == 3 : Dig2  Decode_Digit(TerminalInput) • If Op == ‘e’ E  0 • Tmp TerminalInput, I = I+1 • (I == 3) && Tmp == 20h (space): • Reg[Dest]  Op(Reg[Dest],Dig1) • I = 0 (I==4): • Reg[Dest]  Op(Dig1,Dig2) • I = 0

  3. The algorithm • I == 0 : Op  Decode_Op(TerminalInput)I == 1 : Dest  Decode_Dest(TerminalInput)I == 2 : Dig1  Decode_Digit(TerminalInput)I == 3 : Dig2  Decode_Digit(TerminalInput) • If Op == ‘e’ E  0 • Tmp TerminalInput, I = I+1 • (I == 3) && Tmp == 20h (space): • Reg[Dest]  Op(Reg[Dest],Dig1) • I = 0 (I==4): • Reg[Dest]  Op(Dig1,Dig2) • I = 0 Execute ifCondition

  4. The algorithm • I == 0 : Op  Decode_Op(TerminalInput)I == 1 : Dest  Decode_Dest(TerminalInput)I == 2 : Dig1  Decode_Digit(TerminalInput)I == 3 : Dig2  Decode_Digit(TerminalInput) • If Op == ‘e’ E  0 • Tmp TerminalInput, I = I+1 • (I == 3) && Tmp == 20h (space): • Reg[Dest]  Op(Reg[Dest],Dig1) • I = 0 (I==4): • Reg[Dest]  Op(Dig1,Dig2) • I = 0 Put resultin the reg Written in dest

  5. Register sizes: • Op – 2 bit • Dest – 2 bit • Dig1 – 8 bits • Dig2 – 8 bits • a to d – 8 bits • Tmp – 8 bits • E = 1

  6. Cond2 Cond1

  7. Control of the circuit • NextChar • Decode1Enable • Decode2Enable • Mux2 • Mux3 • Iwrite • Tmpwrite

  8. The control of the circuit Input Output

  9. Op_Decode

  10. Dest_Decode

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