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1 answer:
CPU Special Registers
The CPU contains a number of specialpurpose registers:
Instruction Register (IR): The instruction register holds the instruction currently being executed.
Memory Data Register (MDR): The memory data register (also known as the memory buffer register or data buffer) holds the piece of data that has been fetched from memory.
Memory Address Register (MAR): The memory address register holds the address of the next piece of memory to be fetched.
Program Counter (PC): The program counter holds the location of the next instruction to be fetched from memory. It is automatically incremented between supplying the address of the next instruction and the instruction being executed.
Accumulator: The accumulator is an internal CPU register used as the default location to store any calculations performed by the arithmetic and logic unit.
Next: General Purpose Registers
2 answer:
3 answer:
We begin with truth tables for the sum and carryout bits produced when two binary numbers are added together with a carryin. First we will look at the sum expression. given two input bits and a carryin, we have three variables and need a truth table with eight rows to represent all possible combinations of three binary variables.
After adding all combinations we see that the sum should be 1 whenever the combination of inputs is odd. This is an odd function, which could easily be represented by exclusiveor gates. But or goal here is to practice with truth tables, disjunctive normal form, karnaugh maps and designing digital circuits. So we begin by developing the disjunctive normal form for the following truth table representing the sum:
Binary Onebit 

x 
y 
C_{in} 
Sum 

0 
0 
0 
0 

0 
0 
1 
1 
x’y’Cin 
0 
1 
0 
1 
x’yCin’ 
0 
1 
1 
0 

1 
0 
0 
1 
xy’Cin’ 
1 
0 
1 
0 

1 
1 
0 
0 

1 
1 
1 
1 
xyCin 
Sum = x’y’Cin + x’yCin’ + xy’Cin’ + xyCin.
Now that we have determined the sumofproducts (disjunctive normal form) for the sum expression, we attempt to reduce this to a simpler expression requiring fewer gates to build. Here we will use the karnaugh map:
x \ y Cin 
0 0 
0 1 
1 1 
1 0 
0 
0 
1 
0 
1 
1 
1 
0 
1 
0 
No simplification up to this point. So we move on to the carryout expression.
The carryout truth table represents a majority circuit, the output is high whenever the number of high inputs is greater than the number of low inputs. Again we determine the disjunctive normal form for the Cout expression:
Binary Onebit 

x 
y 
C_{in} 
C_{out} 

0 
0 
0 
0 

0 
0 
1 
0 

0 
1 
0 
0 

0 
1 
1 
1 
x’yCin 
1 
0 
0 
0 

1 
0 
1 
1 
xy’Cin 
1 
1 
0 
1 
xyCin’ 
1 
1 
1 
1 
xyCin 
Cout = x’yCin + xy’Cin + xyCin’ + xyCin.
Using the karnaugh map we get:
x \ y Cin 
0 0 
0 1 
1 1 
1 0 
0 
0 
0 
1 
0 
1 
0 
1 
1 
1 
We can make three blocks of two to get a reduced circuit:
Cout = xCin + xy + yCin.
We should take a moment to verify that this is indeed an equivalent
circuit by ensuring that the truth table is the same for both.
x 
y 
C_{in} 
xCin + xy + yCin 

0 
0 
0 
0 
0 
0 
0 
0 
0 
1 
0 
0 
0 
0 
0 
1 
0 
0 
0 
0 
0 
0 
1 
1 
0 
0 
1 
1 
1 
0 
0 
0 
0 
0 
0 
1 
0 
1 
1 
0 
0 
1 
1 
1 
0 
0 
1 
0 
1 
1 
1 
1 
1 
1 
1 
1 
So this reduced circuit is the same. We combine the results into a single circuit with three inputs and two outputs to get the following drawing:
4 answer:
Full adder
Full Subtractor
5 answer:
There are two approaches to bus arbitration: Centralized and distributed.
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