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: Understanding inputs and outputs.
A 4-bit multiplier has two 4-bit inputs. Therefore, total input bits = $4 + 4 = 8$ bits.
Step 2: Calculating number of input combinations.
Number of possible input combinations = $2^8 = 256$.
Step 3: Determining output size.
Multiplying two 4-bit numbers produces an 8-bit result.
Step 4: Calculating ROM size.
Each of the 256 input combinations stores an 8-bit output.
Total ROM size = $256 \times 8 = 2048$ bits = $2$ Kbits.
Step 5: Final conclusion.
Therefore, the amount of ROM needed to implement a 4-bit multiplier is 2 Kbits.

Answer 2

A ROM-based multiplier works as a giant lookup table: every possible pair of inputs is pre-multiplied and stored, and the inputs themselves form the address into that table. Let's size it and check each option:

1. 1 Kbit: \(1\text{Kbit} = 1024\) bits. The table needs \(2^8 \times 8 = 2048\) bits, so 1 Kbit is only half of what's required, too small to hold every input/output pair.
2. 2 Kbits: With two 4-bit inputs, there are \(2^4 \times 2^4 = 2^8 = 256\) possible input combinations, and each produces an 8-bit product (since multiplying two 4-bit numbers can need up to 8 bits). Storing all 256 outputs of 8 bits each needs \(256 \times 8 = 2048\) bits, which is exactly 2 Kbits.
3. 4 Kbits: \(4\text{Kbits}=4096\) bits is double what's actually required, this would be the right size only if the output needed 16 bits per entry, which it doesn't for a 4-bit by 4-bit multiply.
4. 64 bits: This is far too small, it wouldn't even cover a single row of output for all 256 combinations, let alone the whole table.

Matching the ROM's required word count (256) and word width (8 bits) to the options singles out one size.

Therefore, the correct answer is 2 Kbits.