Rotate image (square matrix) by 90 deg
Tue, 06 May 2025
If there are just 2 bits in the address, then there can only be 4 addresses, viz.
00, 01, 10, 11
If 3 bits are allowed in the address, then we can have maximum 8 addresses:
000, 001, 010, 011, 100, 101, 110, 111
So, if we have n bits to store an Address (n-bits in Address Register) or n-wires to transfer the address (Address bus of size 10), then we can have RAM of 2n addresses.
Now there are 2n addresses, and each address is of 1 byte (because its a byte-addressable memory, so every byte will have a unique address or every address will be of 1-byte long).
Hence, total memory will be equal to
size of each address * number of addresses = 8 * 2n
If n=12 (as given in the question)
Size of memory = 8 * 212 bits = 212 bytes = 22 Kilo Bytes (210 Bytes = 1 KB) = 4 KB
Tue, 06 May 2025
Tue, 06 May 2025
Tue, 06 May 2025
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