Homework Answers
a)
Given Instruction is :
Hexadecimal : C450
Instruction(binary): 1100 0100 0101 0000 contains
Instruction RTL : M[450] <- R1 ;(consider C450)
So, xor operation between these two locations..
0000 1001 1000 0000 = 0890
1100 0100 0101 0000 = C450
----------------------------------
1100 1101 1101 0000 = cDD0
------------------------------------
In Hexadecimal: CDD0
In binary: 1100 1101 1101 0000
------------------------------------------------------------------------------------------------------------------------------
b)
Given data:
Hexadecimal: 4450
Binary instruction: 0100 0100 0101 0000
Instruction: M[450] <- R1 ;consider R1 = 4450H
Performing XOR operation..
0000 1001 1001 0000 ----> 0990
0100 0100 0101 0000 ----> 4450
-----------------------------------
0100 1101 1100 0000 ----> 4DC0
Final answer in
Hexadecimal : 4DC0
Binary: 0100 1101 1100 0000
-----------------------------------------------------------------------------------------
(c)
Given data
Hexadecimal : 7040
Binary: 0111 0000 0100 0000
RTL instruction
Ac <- 7040
Accumulator operation, so addition performing
0101 0000 0000 0111 ---> 5007
0111 0000 0100 0000 ---> 7040
----------------------------------
1100 0000 0100 0111 ----> C047
-------------------------------------
Final answer is:
Hexadecimal : C047
Binary : 1100 0000 0100 0111
Add Answer to:
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