Homework Answers
Hexadecimal instruction: 0x00026140
000000 00000 00010
01100 00101 000000
SLL $12, $2, 5
$2 = 0x12345678
shift left $2 by 5 bits
(0001 0010 0011 0100 0101 0110 0111 1000) << 5
(0100 0110 1000 1010 1100 1111 0000 0000)
$12 = 0x468ACF00
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Assume that the instruction code is 0x00026140 and the content of some of the registers are...
Assume that the instruction code is 0x00225820 and the content of some of the registers are...
Assume that the instruction code is 0x00225820 and the content of some of the registers are as follows: Oxabbbdff to $0 = $ 1 = 0x88779955 $ 2 = 0x8456ffdc In Oxabcdbc 53 = ALU operation Instruction Zero ALU ALU Read register 1 Read Read data 1 register 2 Registers Write register Read Write data 2 data RegWrite result ALU operation 000 001 010 011 100 101 110 111 OPERATION ADD SUB AND OR NOR XOR SHIFT LEFT SHIFT RIGHT...
6. Consider a datapath similar to the one in figure below, but for a processor that only has one type of instruction: unconditional PC-relative branch. What would the cycle time be for this dat...
6. Consider a datapath similar to the one in figure below, but for a processor that only has one type of instruction: unconditional PC-relative branch. What would the cycle time be for this datapath? PCSrc Add ALU Add result Shift +( left 2 Read register 1 ALUSrc, 4 ALU operation PCRead PC-address Read data 1 Registers Read data 2 MemWrite Zero ALU ALU-I Address MemtoReg Instruction register 2 Instruction | Write Read data-M register Write Lu memory Write Data data...
Assume that ‘slt $1, $2, $3’ is executed with the implementation in the picture. Identify the value of the 9-bit control...
Assume that ‘slt $1, $2, $3’ is executed with the implementation
in the picture. Identify the value of the 9-bit control
signals.
Add u X ALU result 4 Add Shift left 2 RegDst Branch MemRead MemtoReg Control ALUOP Instruction [31-26 MemWrite ALUSRC RegWrite Instruction [25-21] Read register 1 Read Read PC address Instruction [20-16] data 1 Read Zero register 2 Instruction ALU ALU 31-0] Instruction memory Read data M Read Address Write result u M Instruction [15-11] register data 2...
Add 9 MUX 4 4 Addresult ALU Shift left 2 RegDst Branch MemRead Instruction (31-26) Control...
Add 9 MUX 4 4 Addresult ALU Shift left 2 RegDst Branch MemRead Instruction (31-26) Control Memto Reg ALUOD MemWrite ALUSC RegWrite Instruction [25-21) Read PC Read address register 1 Read Instruction (20-16] MUX1 MUX Zero ALU ALU MUX3 M Instruction (31-0) Instruction memory Road Address data Read data 1 register 2 Write Read register data 2 Write data Registers result Instruction (15-11] Fox SX) Data Write data memory 16 32 Instruction (150) Sign- extend ALU control Instruction (5-0)
ISA & Addressing Mode The instruction opcodes and formats for a computer system are as follows Format AD AD OP AD SA OP SA SA LDdir LDindir LDrel LDindex ACC ← 씨씨ADn ACC ← OP ACC ← MPC-AD]...
ISA & Addressing Mode The instruction opcodes and formats for a computer system are as follows Format AD AD OP AD SA OP SA SA LDdir LDindir LDrel LDindex ACC ← 씨씨ADn ACC ← OP ACC ← MPC-AD] ACC ← MRtSA].OP] ACC -RISA] 001 010 011 101 110 ·ISA Suppose the Instruction format ts as follows: AD: Address write the Operation for LDimm and LDreg (for immediate and register direct addressing) OP: Constant Operand SA : Register A ACC is...
MCS) Add Addresult ALU Shift left 2 RegDst Branch MemRead MemtoReg Instruction (31-26] Control ALUOP MemWrite...
MCS) Add Addresult ALU Shift left 2 RegDst Branch MemRead MemtoReg Instruction (31-26] Control ALUOP MemWrite ALUS RegWrite PC instruction (25-21] Instruction (20-16) Read address Instruction (31-0) Instruction memory Read register 1 Read Read data 1 register 2 Write Read Zoro ALU ALU result Address Read data instruction (15-11] register data 2 x3) Write data Registers Write Data data memory Instruction 15-01 16 Sign- extend ALU control Instruction (5-0) With regards to the single cycle implementation (as shown in the...
Add EX ALU Add dresult Shift left 2 Regst Branch MomRoad Instruction (31-26) MemtoReg Control ALUOO...
Add EX ALU Add dresult Shift left 2 Regst Branch MomRoad Instruction (31-26) MemtoReg Control ALUOO MemWrite ALUST RegWrite instruction [25-21] Read register 1 Read instruction (20-16) Read data 1 register 2 Write Read data 2 instruction (15-11) register Write data Registers Read address Zero ALU ALU Instruction (31-0) Instruction memory result Address Read data Write Data data memory Instruction (15-01 16 32 Sign- extend ALU control Instruction (5-0) With regards to the single cycle implementation (as shown in the...
Q4: Answer the following questions. [7 Marks] The single cycle implementation of MIPS is as shown...
Q4: Answer the following questions. [7 Marks] The single cycle implementation of MIPS is as shown below. Answer the following questions with reference to "beq $S1, $S2, 8H” instruction. Assume that the contents of the registers S1 = 10 H, S2 = 10H, and PC = 16H, pointing to the instruction under consideration. 1. What is the addressing mode of the instruction? [1] ii. Which part of the instruction format, address of S1 and S2 are stored? [1] 111. What...
Modify the circuit to support a MFCC instruction. MFCC Rd instruction: Move From Condition Codes MFCC copies into the fo...
Modify the circuit to support a MFCC
instruction.
MFCC Rd instruction: Move From Condition Codes
MFCC copies into the four rightmost bits of Rd the values of the
ALU signals Carry (C), Overflow (O), Zero (Z) and Negative (N) as
they were set by the previous R- type instruction. The remaining 28
bits of Rd are set to zero.
Describe the changes and additions needed for the
single-cycle MIPS processor datapath and control to support this
instruction.
Hints:
1) MFCC...
(o x Add Addresult ALU Shift left 2 Regst Branch MemRead Instruction (31-26) MemtoReg Controll ALUOP...
(o x Add Addresult ALU Shift left 2 Regst Branch MemRead Instruction (31-26) MemtoReg Controll ALUOP MemWrite ALUSC RogWrite Instruction [25-21] Read register 1 Read Instruction (20-16) Read data 1 register 2 Write Read Instruction (15-11) Write data Registers PC Read address Zoro ALU ALU Instruction (31-0) Instruction memory result Address Read data register data 2 **039 -25 Write Data data memory Instruction (15-01 16 Sign- extend ALU control Instruction 15-01 With regards to the single cycle implementation (as shown...
Assume that the instruction code is 0x00225820 and the content of some of the registers are as follows: Oxabbbdff to $0 = $ 1 = 0x88779955 $ 2 = 0x8456ffdc In Oxabcdbc 53 = ALU operation Instruction Zero ALU ALU Read register 1 Read Read data 1 register 2 Registers Write register Read Write data 2 data RegWrite result ALU operation 000 001 010 011 100 101 110 111 OPERATION ADD SUB AND OR NOR XOR SHIFT LEFT SHIFT RIGHT...
6. Consider a datapath similar to the one in figure below, but for a processor that only has one type of instruction: unconditional PC-relative branch. What would the cycle time be for this datapath? PCSrc Add ALU Add result Shift +( left 2 Read register 1 ALUSrc, 4 ALU operation PCRead PC-address Read data 1 Registers Read data 2 MemWrite Zero ALU ALU-I Address MemtoReg Instruction register 2 Instruction | Write Read data-M register Write Lu memory Write Data data...
Assume that ‘slt $1, $2, $3’ is executed with the implementation
in the picture. Identify the value of the 9-bit control
signals.
Add u X ALU result 4 Add Shift left 2 RegDst Branch MemRead MemtoReg Control ALUOP Instruction [31-26 MemWrite ALUSRC RegWrite Instruction [25-21] Read register 1 Read Read PC address Instruction [20-16] data 1 Read Zero register 2 Instruction ALU ALU 31-0] Instruction memory Read data M Read Address Write result u M Instruction [15-11] register data 2...
Add 9 MUX 4 4 Addresult ALU Shift left 2 RegDst Branch MemRead Instruction (31-26) Control Memto Reg ALUOD MemWrite ALUSC RegWrite Instruction [25-21) Read PC Read address register 1 Read Instruction (20-16] MUX1 MUX Zero ALU ALU MUX3 M Instruction (31-0) Instruction memory Road Address data Read data 1 register 2 Write Read register data 2 Write data Registers result Instruction (15-11] Fox SX) Data Write data memory 16 32 Instruction (150) Sign- extend ALU control Instruction (5-0)
ISA & Addressing Mode The instruction opcodes and formats for a computer system are as follows Format AD AD OP AD SA OP SA SA LDdir LDindir LDrel LDindex ACC ← 씨씨ADn ACC ← OP ACC ← MPC-AD] ACC ← MRtSA].OP] ACC -RISA] 001 010 011 101 110 ·ISA Suppose the Instruction format ts as follows: AD: Address write the Operation for LDimm and LDreg (for immediate and register direct addressing) OP: Constant Operand SA : Register A ACC is...
MCS) Add Addresult ALU Shift left 2 RegDst Branch MemRead MemtoReg Instruction (31-26] Control ALUOP MemWrite ALUS RegWrite PC instruction (25-21] Instruction (20-16) Read address Instruction (31-0) Instruction memory Read register 1 Read Read data 1 register 2 Write Read Zoro ALU ALU result Address Read data instruction (15-11] register data 2 x3) Write data Registers Write Data data memory Instruction 15-01 16 Sign- extend ALU control Instruction (5-0) With regards to the single cycle implementation (as shown in the...
Add EX ALU Add dresult Shift left 2 Regst Branch MomRoad Instruction (31-26) MemtoReg Control ALUOO MemWrite ALUST RegWrite instruction [25-21] Read register 1 Read instruction (20-16) Read data 1 register 2 Write Read data 2 instruction (15-11) register Write data Registers Read address Zero ALU ALU Instruction (31-0) Instruction memory result Address Read data Write Data data memory Instruction (15-01 16 32 Sign- extend ALU control Instruction (5-0) With regards to the single cycle implementation (as shown in the...
Q4: Answer the following questions. [7 Marks] The single cycle implementation of MIPS is as shown below. Answer the following questions with reference to "beq $S1, $S2, 8H” instruction. Assume that the contents of the registers S1 = 10 H, S2 = 10H, and PC = 16H, pointing to the instruction under consideration. 1. What is the addressing mode of the instruction? [1] ii. Which part of the instruction format, address of S1 and S2 are stored? [1] 111. What...
Modify the circuit to support a MFCC
instruction.
MFCC Rd instruction: Move From Condition Codes
MFCC copies into the four rightmost bits of Rd the values of the
ALU signals Carry (C), Overflow (O), Zero (Z) and Negative (N) as
they were set by the previous R- type instruction. The remaining 28
bits of Rd are set to zero.
Describe the changes and additions needed for the
single-cycle MIPS processor datapath and control to support this
instruction.
Hints:
1) MFCC...
(o x Add Addresult ALU Shift left 2 Regst Branch MemRead Instruction (31-26) MemtoReg Controll ALUOP MemWrite ALUSC RogWrite Instruction [25-21] Read register 1 Read Instruction (20-16) Read data 1 register 2 Write Read Instruction (15-11) Write data Registers PC Read address Zoro ALU ALU Instruction (31-0) Instruction memory result Address Read data register data 2 **039 -25 Write Data data memory Instruction (15-01 16 Sign- extend ALU control Instruction 15-01 With regards to the single cycle implementation (as shown...
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