Question

What is the amount of ROM needed to implement a 4-bit multiplier?

• A. 1 Kbits
• B. 2 Kbits
• C. 4 Kbits
• D. 64 bits

Hint:

ROM size for a combinational circuit = (Number of input combinations) $\times$ (Output bits).

Answer 1

The Correct Option is B

Step 1: A ROM used as a lookup table is sized as (number of address lines) giving \(2^n\) words, each word being \(m\) bits wide, so total size = \(2^n \times m\) bits.
Step 2: Here the address comes from two 4-bit inputs, so \(n = 4+4 = 8\), giving \(2^8 = 256\) words, and each word (the product) needs \(m = 8\) bits.
Step 3: Total ROM size \( = 256 \times 8 = 2048 \) bits \( = 2 \) Kbits.
\[ \boxed{2 \text{ Kbits}} \]

Answer 2

We can also sanity-check the answer purely by units and order of magnitude, without redoing the full multiplication, by asking what each option implies about the ROM's shape:

1. 1 Kbit: If this were correct, the ROM could only be organized as, at best, 128 words of 8 bits (\(128 \times 8 = 1024\)), but we need 256 distinct input combinations covered, so 128 words isn't enough rows.
2. 2 Kbits: This works out to 256 words of 8 bits (\(256\times8=2048\)), and 256 is precisely the number of distinct 4-bit-by-4-bit input pairs, so every combination gets its own row with the right width.
3. 4 Kbits: This would fit 256 words of 16 bits, or 512 words of 8 bits, neither of which matches a table that only needs 256 rows of 8-bit products.
4. 64 bits: This barely covers 8 words of 8 bits, nowhere near the 256 rows the table requires.

Only one option lines up with a 256-row, 8-bit-wide table.

Therefore, the correct answer is 2 Kbits.