Answer:
a. LOAD register 3 with the hex value 56
2356
b. ROTATE register 5 three bits to the right.
A503
c. JUMP to the instruction at location F3 if the contents of register 7 are equal to the contents of register 0
B7F3
d. AND the contents of register A with the contents of register 5 and leave the result in register 0
80A5
Explanation:
To translate the English instructions to Machine language we have to consider Appendix C of A Simple Machine Language sheet.
a. LOAD register 3 with the hex value 56
The OP code for LOAD is:
2
Operand is:
RXY
Description is:
LOAD the register R with the bit pattern XY.
In a.
3 represents R
56 represents XY
Now following this we can join OP and Operand together to translate a.
2356
b. ROTATE register 5 three bits to the right
The OP code for ROTATE is:
A
Operand is:
R0X
Description is:
ROTATE the bit pattern in register R one bit to the right X times. Each time place the bit that started at the low-order end at the high-order end
In b.
5 represents R
three represents X
Now following this we can join OP and Operand together to translate b.
A503
c. JUMP to the instruction at location F3 if the contents of register 7 are equal to the contents of register 0.
The OP code for JUMP is:
B
Operand is:
RXY
Description is:
JUMP to the instruction located in the memory cell at address XY if the bit pattern in register R is equal to the bit pattern in register number 0.
In c.
7 represents R register
F3 represents XY
Now following this we can join OP and Operand together to translate c.
B7F3
d. AND the contents of register A with the contents of register 5 and leave the result in register 0.
The OP code for AND is:
8
Operand is:
RST
Description is:
AND the bit patterns in registers S and T and place the result in register R
In d.
A represents register S
5 represents register T
0 represents register R
Now following this we can join OP and Operand together to translate d.
80A5
