Question 3389 – Communication
December 11, 2023GATE 1998
December 11, 2023GATE 1996
| Question 21 |
A ROM is sued to store the table for multiplication of two 8-bit unsigned integers. The size of ROM required is
| 256 × 16 | |
| 64 K × 8 | |
| 4 K × 16 | |
| 64 K × 16 |
Question 21 Explanation:
When we multiply the two 8 bit numbers result will reach upto 16 bits. So we require 16 bits for each multiplication output.
No. of results possible = 28 × 28 = 216 = 64K
Then total size of ROM = 64K × 16
No. of results possible = 28 × 28 = 216 = 64K
Then total size of ROM = 64K × 16
Correct Answer: D
Question 21 Explanation:
When we multiply the two 8 bit numbers result will reach upto 16 bits. So we require 16 bits for each multiplication output.
No. of results possible = 28 × 28 = 216 = 64K
Then total size of ROM = 64K × 16
No. of results possible = 28 × 28 = 216 = 64K
Then total size of ROM = 64K × 16