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3 years ago

The a0 is the part of the central processing unit that performs arithmetic calculations for the computer.

Computers and Technology

1 answer:

Serga [27]3 years ago

6 0

Answer:

It is the ALU or the Arithmetic Logic Unit.

Explanation:

It is the ALU. However, keep in mind that registers and buses do a very important task. The number of registers we have, faster is the processing, and the opposite is true as well. And there is a reason behind this if we have different channels for sending and receiving the data from the memory, and several registers for storing the data, and we can formulate the requirement seeing the requirements for full adder and half adders. Remember we need to store several variables in case of the full adder, and which is the carry, and if we have separate registers for each of them, our task becomes easier. Remember its the CU that tells the ALU what operation is required to be performed. Also remember we have the same channel for input and output in the case of Van Neumann architecture, as we have a single bus. and we also have a single shared memory. And Harvard architecture is an advanced version of it.

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3 years ago

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3 years ago

The instruction format of a computer is given by one indirect bit, opcode bits, register address bits, immediate operand bits an

oksian1 [2.3K]

Answer:

Following are the answer to this question:

Explanation:

The memory size is 1 Giga Bytes which is equal to $2^{30}$

$\texttt{Number of address bits \ \ \ \ \ \ \ \ \ \ \ \ \ \ Number of addresses}\\\\ \ \ \ \ 30 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2^{30}= 1073741824$

B) $\texttt{I \ \ \ Opcode \ \ \ Register address \ \ \ Immediate operand \ \ \ Memory address}\\\\\\\textt{1 \ bit \ \ \ \ \ \ 128 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2 \ bits \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 24 \ bits \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 30 \ bits}\\\\\\$ $= 2^7 \\\\ = 7 \ bits$

calculating the register Bits:

$= 64-(1+7+24+30)\\\\=64 -62\ \ bits\\\\= 2\ \ bits\\$

Immediate value size while merging the additional benefit with the address field:

$= 2^{24} + 2^{30}\\\\= 2^{54}\\\\$ $\texttt{Range before combining(-,+) 24 bits \ \ \ \ \ \ Range after combining( -,+)54bits}\\\\$ $\textt{-2^{12} from + (2^{12}-1) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -2^{27} from + (2^{27}- 1)}$

$= \frac{24}{2} = 12\\\\= \frac{54}{2} = 27$

The range is accomplished by dividing the bits by 2 into the two sides of the o and the number is one short to 0.

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Ray Of Light [21]

One of 34 possible Values can be assigned to each axis of classification in the seven-character code.

<h3>Understanding Coding Guidelines</h3>

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Read more about Coding at; brainly.com/question/16397886

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