# ROM logic design?

You are asked to design a comparator circuit that has an input of two 2-bit

numbers. The comparator has 3 outputs, representing the cases where the first number

(A1A2) is equal to the second number (B1B0) (in this case E should equal 1), smaller than

the second number (in this case L equals 1), and greater than the second number (G

equals 1), respectively. Complete the Truth Table below to define E, L and G. If the

comparator is realized with a ROM, what size of ROM is required? (i.e.: how many bits?)

Assuming the ROM below is used to generate all minterms, use dots to indicate

connections which realize each of the required functions.

### 2 Answers

- Robert JLv 78 months agoFavorite Answer
The ROM will have a total of four input (address) signals, the 2+2 bit numbers.

That's 2^4 combinations; 16.

Three output signals, so three bits used at each address = 48 bits used in total.

To work it out, make a 4x4 grid with one number across the top 0-3 and the other 0-3 down the side.

Write the e-l-g output bit state for each combination in the grid.

Then, if you use the top number for the lower two address bits and the side number for the higher two, just take the grid content working left to right and top to bottom.

That gives you 16 results for all the input combinations, in sequence.

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- D gLv 78 months ago
I DID THE truth table it involves 4 inputs and 3 outputs only one output active at any time..

16 inputs these can be address inputs

the outputs can be the data so you need 3 data bits fore each address

so 48 bit rom is fine

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