StormNalinHW13
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Nalin Storm EECE 2322 Digital Design & Computer Organization Professor Kimani Homework 13 19 April, 2023 Question 1: The output provided by Hwk_Main_datapath_MSD.sv provided the expected output for the decoded instruction jal that replaced beq. The encoded instruction successfully jumped the PC register -40 registers to the very first instruction, add. I verified my implementation was successful because the PC register that followed the jal instruction was 0x00010000 which is the very first instruction, exactly 40 registers previous. Below are a summary of the changes made to the file:
Question 2a: a.
Express the following base 10 number in IEEE 754 single-precision floating-point format. Express your answer in hexadecimal
. (show your steps)
−13.5625
Step 1. Convert 13 to binary:
13/2 = 6, Remainder = 1 6/2 = 3, Remainder = 0 3/2 = 1, Remainder = 1 1/2 = 0, Remainder = 1 13 in decimal is 1101 in binary Step 2. Convert 0.5625 to binary
0.5625 * 2 = 1.125
1.125
1 0.125 * 2 = 0.25
0.25
0 0.25 * 2 = 0.5
0.5
0 0.5 * 2 = 1
1
1 0.5625 in binary is 1001
13.5625
1101.1001 Step 3. Convert to
base 2 scientific notation
1101.1001
1.1011001 * 10
3
Step 4. Determine sign bit
& exponent bits
Sign bit is 1 (negative) Exponent bits are 127 + 3 = 130 130 converted to binary is 10000010 Step 5. Combine all three parts (sign, exponent, mantissa)
to form binary number
Sign bit = 1 Exponent bits = 10000010 Mantissa = 1011001 -13.5625 in binary is 1 10000010 10110010000000000000000 Step 6. Convert to hexadecimal using the table below
1100 0001 0101 1001 0000 0000 0000 0000 C 1 5 9 0 0 0 0
Hexadecimal Binary 0 0000 1 0001 2 0010 3 0011 4 0100 5 0101 6 0110 7 0111 8 1000 9 1001 A 1010 B 1011 C 1100 D 1101 E 1110 F 1111
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After execution the far jump instruction: JMP A3000127h; the new value of Physical Address=A3127h. (when updated Physical Address=10005h)
False
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Add
Add Sum
1
Shift
left 1
Branch
MemRead
Instruction [6-0]
MemtoReg
ALUOP
Control
MemWrite
ALUSCC
RegWrite
L3
Instruction [19-15]
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address
Read
register 1
PC
Instruction (24-20]
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Instruction
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Instruction [11-7)
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result
data
Instruction
register
memory
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data Registers
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Instruction (31-0)
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Imm
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ALU
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x8=72. x18=56, DM(64) =48
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51868 Id x8.8(x18)
Write below the value of the bus L1 in decimal:
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Write below the value of the bus L2 in decimal:
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Write below the value of the bus L3 in decimal:
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ALU
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Sign
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16
32
ALUOP
RegDst
Figure 3: For EACH of the following registers, indicate how many bits are in the reg-
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are: regWrite, regDest, aluOP, ALUsrc, Branch, MemRead, MemWrite, MemToReg.
Register
Number of bits
Control Signals
PC
32
None
IF/ID
64*
None
ID/EX
EX/MEM
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inst ...
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inst ... 4
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always zero
imm[12]
0 000000 01010 | 10011
imm[11]
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