Circuit design that outputs square of binary input - boolean-logic

Question

So for my digital logic course, we were asked to design a combinational circuit with 3 inputs, and an output that generates the square of the binary input. I assume she means the inputs are 3 bit binary numbers 0-7. In describing the solution, she mentioned 3 general steps of 1. Finding truth table, 2. deriving the function and 3. Simplifying
I understand perfectly how to do this when the output is a single function (combine the instances of a true function, simplify etc.) In our only relevant example, we input BCD numbers 0-9 and output their excess-3 equivalent, then simplified a k-map for each of the 4 output variables (a map for W, the leftmost bit, X the second-to-left bit, etc.). Not sure what to do from here tho. Thanks for the help

Answer 1

As you already mentioned, you need to fetch the inputs in binary, calculate the square in binary format (the binary length will increase from 3 to 6).
But, you don't need those BCD numbers and output their excess-3 equivalent. That would rather make the problem more complex. Simply do as directed below.
So, for the 3 different bit combinations, you need to generate a function in terms of combinational circuit that will yield a 6-bit square of the given number.
So, your input set = {b0,b1,b2} and your output set = {B1,B2,B3,B4,B5,B6}.
Then, you need to draw the truth table of 3-bit to 6-bit square converter as :-
Input B6 B5 B4 B3 B2 B1 B0
b0
b1
b2
Fill the entries, that's the homework for you.
And, then when you have output bits B6,B5,...,B0 in terms of b0,b1 and b2, just simplify all the functions.
If you still have any doubt, comment below. I will help you with your issue. Good luck.

Answer 2

I have used Logic Friday 1 to derive the following minimized expression for the six outputs of your multiplier:
ab5 = a2 a1 b2 b1 + a2 a1 a0 b2 b0 + a2 a0 b2 b1 b0;
ab4 = a2 a1' a0' b2 + a2 a1' b2 b1' + a2 a0' b2 b1'
+ a2 a1' b2 b0' + a2 b2 b1' b0' + a2' a1 a0 b2 b1
+ a2 a1 b2' b1 b0 + a1 a0 b2 b1 b0;
ab3 = a2' a1 a0' b2 + a2 a1' b2' b1 + a2' a1 b2 b1'
+ a2 b2' b1 b0' + a2' a1 a0 b2' b1 b0 + a2 a1' a0 b2 b1' b0
+ a1 a0' b2 b1' + a2 a1' b1 b0' + a2 a0' b2 b1 b0
+ a2 a1 a0 b2 b0';
ab2 = a2' a1 a0' b1 + a2 a1' a0' b0 + a2 a0' b1' b0
+ a1' a0 b2 b0' + a1 a0' b1 b0' + a1 b2' b1 b0'
+ a0 b2 b1' b0' + a2' a0 b2 b0 + a2 a0 b2' b0;
ab1 = a1' a0 b1 + a1 a0' b0 + a1 b1' b0 + a0 b1 b0';
ab0 = a0 b0;
As a truth table:
The original truth table has 64 rows.
The largest output number is 49 = 7 * 7 = 32 + 16 + 1.
Therefore, six (= 3 + 3) output bits are sufficient.