comparatordigital-logic
I have stuck on implementing magnitude comparator for 2-bit numbers (three functions – greater, equal, less).
It is pretty easy to implement it with AND and OR gates, but the point is the task is to implement it with AND and XOR gates only. A little bit of help will be appreciated.
Thanks in advance.
First let's see how to implement a function using single decoder. It is a straight-forward task having the truth table. Decoder will output 1 on the lines selected by the input, so if we just "or" the outputs corresponding to the truth table lines with X=1, we will get the exact required function:
Unfortunately there is no straightforward way to convert it to 2->4 decoders. But maybe we can divide the original function to smaller ones, suitable for 2->4 representation? Let's look at it. After some manipulations I've omitted here, the function can be represented as $$A_1B_1'+A_0B_0'(B_1'+A_1)$$
So we can see here three different 2-variable functions combined with ANDs and ORs:
1) \$A_1B_1'\$
2) \$A_0B_0'\$
3) \$B_1'+A_1\$
while (1) and (3) are functions of the same variables and can share decoder.
So the solution would be: Implement functions (1)-(3) using the method above, and interconnect the outputs using the gates:
Change the circuit above, replacing the circuit on the left with the circuit on the right. The RED/GREEN LED combined can be found as a standard bi-colour LED with two pins.
I think this is what is intended:
simulate this circuit – Schematic created using CircuitLab
Best Answer
It should be easier to implement "equals" with XOR than with AND and OR. Take a look at the Karnaugh map for a single bit of "equals" and then look at the Karnaugh map for XOR.
I assume you have NOT as well (at least for your variables? (otherwise I don't know how you did it with AND and OR alone.)) If you don't have NOT, then XOR can be used to fake one: A XOR 1 = NOT A.
For the "greater" and the "less": In the worst case you can take your answer in sum-of-products form and turn it into all NAND gates (by DeMorgan AB+CD = (A NAND B) NAND (C NAND D)). But you can probably do better. Again: look at the Karnaugh map for your problem. Then look at the Karnaugh map for (2-bit) XOR. Any time you have a checkerboard-like pattern in your Karnaugh map you can probably put an XOR to good use.