My Histogram

Lab 7 Task 3

For Task 3, we must prove through Logic Gates that DeMorgan's Law is true. For this I was able to create the following circuit in the Logic Gate Simulator:



After running different values of A and B on the circuit I was able to form the following truth table, proving DeMorgan's Law true:

A	B	NOT A	NOT B	A AND B	NOT (A AND B)	NOT A OR NOT B
0	0	1	1	0	1	1
0	1	1	0	0	1	1
1	0	0	1	0	1	1
1	1	0	0	1	0	0

As you can see from the red numbers, NOT (A AND B) is the equivalent of NOT (A OR B), therefore proving that DeMorgan's Law is true.

Lab 7 Task 2

For this task the final result for the logic gate is NOT (A XOR B), for this expression I came up with the following image in the Logic Gate Simulator:



After viewing the result of this expression at different values of A and B, I came up with the following Truth Table:

A	B	A XOR B	NOT (A XOR B)
0	0	0	1
0	1	1	0
1	0	1	0
1	1	0	1

This expression is only true when both A and B are the same value, if they differ in value then the expression is false. It is the opposite of the Exclusive-OR or XOR expression, and this is the result of the NOT being placed in front of the XOR expression. So the function of this circuit is to only allow for a "True" result when the values are equivalent, if they aren't then there is a "False" result.