Answer:
For 32 bits Instruction Format:
OPCODE DR SR1 SR2 Unused bits
a) Minimum number of bits required to represent the OPCODE = 3 bits
There are 8 opcodes. Patterns required for these opcodes must be unique. For this purpose, take log base 2 of 8 and then ceil the result.
Ceil (log2 (8)) = 3
b) Minimum number of bits For Destination Register(DR) = 4 bits
There are 10 registers. For unique register values take log base 2 of 10 and then ceil the value. 4 bits are required for each register. Hence, DR, SR1 and SR2 all require 12 bits in all.
Ceil (log2 (10)) = 4
c) Maximum number of UNUSED bits in Instruction encoding = 17 bits
Total number of bits used = bits used for registers + bits used for OPCODE
= 12 + 3 = 15
Total number of bits for instruction format = 32
Maximum No. of Unused bits = 32 – 15 = 17 bits
OPCODE DR SR1 SR2 Unused bits
3 bits 4 bits 4 bits 4 bits 17 bits
Answer:
Total Memory= 4 KB = 4096 bytes = 32768 bits
Explanation:
<em><u>1. Data lines are 8 From D0 to D7</u></em>
so
Total memory at single address locations is 8 bits.
<em><u>2. Address lines are 12 (A0 to A11)</u></em>
There are 12 address lines but 3 out 12 are for selction of chip or memory bank.
so only 9 pins are there to address the locations on one chip.
Total No. of address locations on single chip = 2^9 = 512 locations
as 1 location is 1 byte so total memory of single chip is 512 bytes.
<u><em>3. Total Memory Bank </em></u>
There are total 3 selection pins for memory bank.
so
Total chips = 2^3 = 8.
<em><u>4. Total Memory </u></em>
Total size of 1 chip = 512 bytes
Total size of 8 chip = 8x512 bytes = 4096 bytes = 4096/1024 kb = 4 kb
<em>So total memory of system is 4 Kb = 4096 bytes = 32768 bits</em>
Answer:
EOF stands for End Of File
Google explains it well: EOF is a condition in a computer operating system where no more data can be read from a data source. The data source is usually called a file or stream.
What do you want to skip in brainly? There isn't anything to skip that I can think of.
