Show the following problems in a diagram(all 6 steps). The leftmost hexadecimal number or the leftmost four binary bits are the opcode 0001B or 1HX = Load AC from memory 0010B or 2HX = Store AC to memory 0101B or 5HX = Add to AC from memory 1. The PC(program counter) contains 350, the address of the first instruction. This instruction (the value 1940 in hexadecimal) is loaded into the IR (Instruction set register), and the PC is incremented. Note that this process involves the use of a memory address register and a memory buffer register. For simplicity, these intermediate registers are ignored. 2. The first 4 bits (first hexadecimal digit) in the IR indicate that the AC is to be loaded. The remaining 12 bits (three hexadecimal digits) specify the address (940) from which data are to be loaded. 3. The next instruction (5930) is fetched from location 351, and the PC is incremented. 4. The old contents of the AC and the contents of location 930 are added, and the result is stored in the AC. 5. The next instruction (2941) is fetched from location 352, and the PC is incremented. 6. The contents of the AC are stored in location 941
Show the following problems in a diagram(all 6 steps).
The leftmost hexadecimal number or the leftmost four binary bits are the opcode
0001B or 1HX = Load AC from memory
0010B or 2HX = Store AC to memory
0101B or 5HX = Add to AC from memory
1. The PC(program counter) contains 350, the address of the first instruction. This instruction (the value 1940 in hexadecimal) is loaded into the IR (Instruction set register), and the PC is incremented. Note that this process involves the use of a memory address register and a memory buffer register. For simplicity, these intermediate registers are ignored.
2. The first 4 bits (first hexadecimal digit) in the IR indicate that the AC is to be loaded. The remaining 12 bits (three hexadecimal digits) specify the address (940) from which data are to be loaded.
3. The next instruction (5930) is fetched from location 351, and the PC is incremented.
4. The old contents of the AC and the contents of location 930 are added, and the result is stored in the AC.
5. The next instruction (2941) is fetched from location 352, and the PC is incremented.
6. The contents of the AC are stored in location 941
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