Need explanation-circuit design & Boolean

[math&science]

Please refer to attached. Can someone please explain to me how they got the Boolean expression. I understand how to use Karnaugh maps but how did they come up with the equation in order to make the Karnaugh map? Thanks!

 

[berkeman]

They have combined terms to get the first combination XOR thing. The initial term with the two XORs ANDed expands out into four minterms, so the first thing gives you four and the last two minterms gives you the total of six. The ones in the K-map are isolated, so the map doesn't help you to combine them. To expand the XORs, remember that A XOR B = AB'+A'B.

BTW, when you see an isolated K-map like that without any good way to combine 1's, look to see if you can do a more effective job by combining 0's instead, and use the inverted form of the expression...

 

[ranger]

Can someone please explain to me how they got the Boolean expression.

Using a Truth Table:

The circuit has 4 inputs, making the highest binary number 1111. Your first step is to make a truth table for the all possble states for the 4 inputs. Then the output x, would only be high when the input is divisible by three. The only case you care about is when the output is high, ignore when it is low. Your next step is to write the equation for the high output then put all of them together.

EDIT: Just saw berkeman's reply

 

[math&science]

Thanks Berkeman and Ranger. I guess I'm still a bit confused. When you make the truth table for all possible states, are you just multiplying ABCD to find the output. Aren't there a whole bunch of ways like A+B+C+D, etc. to get an output so how am I supposed to know which one to use to find the output to begin with. I mean usually when I make a truth table, I need to know the equation first or at least know the schematic diagram to find the equation. Then I can make the Karnaugh map and combine terms to find the equation. And is there a general rule for making circuits that tell you when an n-bit number is divisible by x? Thanks!

 

[berkeman]

Hi Math. In this case, the function that you are asked to implement with the logic is a divisible by 3 output indicator. The truth table is the 16 lines, starting with ABCD=0000 and ending with ABCD=1111. Read each biniary number, and the output of the truth table is 1 if the binary number is divisible by 3, and 0 otherwise. Like, ABCD=0000 and ABCD = 0011 and ABCD = 1001 are all evenly divisible by 3, but ABCD=0001 is not.

In the case of this problem, you can go straight from the problem statement to the K-map, as long as you are comfortable with working with K-maps. The truth table is needed in times where the function is complicated. The K-map is usually the best way to get a simplified output logic expression, especially if there are groups of 1's that you can combine visually on the map.

 

[math&science]

Thanks Berkeman. I get it now:)