Working on a college assignment and I am not sure about a certain part of the question.

In a previous question that required calculations done on an ALU, the next question looks for similar functions, but with ROM.
There is an:
A input, which is 4-bit input number
B input, which is also a 4-bit input number
S input, which is a 3-bit select input
and F, which is a 4-bit output number

The question asks how many addressable locations are required and what is the minimum memory cell size at each location.

In working out the first part I have worked out that there are 2^11 = 2048 addressable locations.

But for this second part I am stumped. Is it a case of working out how many bits can be stored at each location?

Thanks for the help

 

But for this second part I am stumped. Is it a case of working out how many bits can be stored at each location?

As many as you need. Memories are usually organized as 1, 4, 8, and 16 bits wide, but other sizes are possible in custom designs.

And thanks for clearly describing the problem, showing your attempt to solve it, and stating that it's homework.

 

MOD NOTE: You already have a thread on this. Please, one thread per topic. Your other thread has more activity, so let's stick with that one.

http://forum.allaboutcircuits.com/threads/rom-based-solution-needed-to-mirror-alu.130762/