The first article on Boolean logic covered the basics. Here is a link to the first article:
http://www.associatedcontent.com/article/2261239/electronics_introduction_to_boolean.html?cat=59
This article covers the advanced topics.
Review Of Boolean Logic:
A* B means A AND B
Both A and B must be at a logic level 1 for the output to be at a logic level 1.
Truth table
A = 0, B = 0, A*B = 0
A = 0, B = 1, A*B = 0
A = 1, B = 0, A*B = 0
A = 1, B = 1, A*B = 1
In electronics, the logic levels are voltages. For purposes of our discussion, a logic level of 0 is 0 volts and a logic level of 1 is 5 volts. Implementing this in the truth table yields
Truth table
A = 0 volts, B = 0 volts, A*B = 0 volts
A = 0 volts, B = 5 volts, A*B = 0 volts
A = 5 volts, B = 0 volts, A*B = 0 volts
A = 5 volts, B = 5 volts, A*B = 5 volts
A + B means A OR B
If either A or B is at a logic one level, then the output is at a logic one level.
Truth table
A = 0, B = 0, A + B = 0
A = 0, B = 1, A + B = 1
A = 1, B = 0, A + B = 1
A = 1, B = 1, A + B = 1
In electronics, the logic levels are voltages. For purposes of our discussion, a logic level of 0 is 0 volts and a logic level of 1 is 5 volts. Implementing this in the truth table yields
Truth table
A = 0 volts, B = 0 volts, A + B = 0 volts
A = 0 volts, B = 5 volts, A + B = 5 volts
A = 5 volts, B = 0 volts, A + B = 5 volts
A = 5 volts, B = 5 volts, A + B = 5 volts
A’ means NOT A
A = 1, A’ = 0
A = 0, A’ = 1
In electronics:
A = 5 volts, A’ = 0 volts
A = 0 volts, A’ = 5 volts
Problem 1:
Write the truth table in terms of inputs A and B for
(A*B)’
Solution:
A = 0, B = 0, A*B = 0, (A*B)’ = 1
A = 0, B = 1, A*B = 0, (A*B)’ = 1
A = 1, B = 0, A*B = 0, (A*B)’ = 1
A = 1, B = 1, A*B = 1, (A*B)’ = 0
The truth table only contains inputs and outputs. Hence the term A*B is not in the truth table:
A = 0, B = 0, (A*B)’ = 1
A = 0, B = 1, (A*B)’ = 1
A = 1, B = 0, (A*B)’ = 1
A = 1, B = 1, (A*B)’ = 0
Note that if any input is at logic level zero, the output is at logic level one.
In electronics, the gate that performs this function is a NAND gate. The symbol for the NAND gate is shown in figure one.
The truth table for the NAND gate is:
A = 0 volts, B = 0 volts, (A*B)’ = 5 volts
A = 0 volts, B = 5 volts, (A*B)’ = 5 volts
A = 5 volts, B = 0 volts, (A*B)’ = 5 volts
A = 5 volts, B = 5 volts, (A*B)’ = 0 volts
Problem 2:
Write the truth table in terms of inputs A and B for
(A + B)’
Solution:
A = 0, B = 0, A + B = 0, (A + B)’ = 1
A = 0, B = 1, A + B = 1, (A + B)’ = 0
A = 1, B = 0, A + B = 1, (A + B)’ = 0
A = 1, B = 1, A + B = 1, (A + B)’ = 0
The truth table is:
A = 0, B = 0, (A + B)’ = 1
A = 0, B = 1, (A + B)’ = 0
A = 1, B = 0, (A + B)’ = 0
A = 1, B = 1, (A + B)’ = 0
In electonics, the gate performs this function is a NOR gate. The symbol for a NOR gate is shown in figure one.
The truth table for the NOR gate is:
A = 0 volts, B = 0 volts, (A + B)’ = 5 volts
A = 0 volts, B = 5 volts, (A + B)’ = 0 volts
A = 5 volts, B = 0 volts, (A + B)’ = 0 volts
A = 5 volts, B = 5 volts, (A + B)’ = 0 volts
Problem 3:
Write the truth table in terms of inputs A and B for
A*B’ + B*A’
Solution:
First Set of Inputs: A = 0, B = 0
Lets proceed step by step.
A = 0, A’ = 1
B = 0, B’ = 1
The first line has five entries.
The entries are
A = 0, B = 0, A*B’ = 0, B*A’ = 0,
A*B’ + B*A’ = 0 + 0 = 0
And the first line of the truth table in terms of A and B is
A = 0, B = 0, A*B’ + B*A’ = 0
Second Set of Inputs: A = 0, B = 1
A = 0, A’ = 1
B = 1, B’ = 0
A*B’ = 0
B*A’ = 1
A*B’ + B*A’ = 1
Third Set of Inputs: A=1, B = 0
If A = 1, A’ = 0
B = 0, B’ = 1
A*B’ = 1
B*A’ = 0
A*B’ + B*A’ = 1
Fourth Set of Inputs: A = 1, B = 1
If A = 1, A’ = 0
B = 1, B’ = 0
A*B’ = 0
B*A’ = 0
A*B’ + B*A’ = 0
So our truth table is as follows:
A = 0, B = 0, A*B’ + B*A’ = 0
A = 0, B = 1, A*B’ + B*A’ = 1
A = 1, B = 0, A*B’ + B*A’ = 1
A = 1, B = 1, A*B’ + B*A’ = 0
Note that the output is one when the difference is one.
In digital electronics the gate that performs this function is the Exclusive Or gate. The symbol for the Exclusive Or gate is shown in figure one.
The Truth Table for the EXCLUSIVE OR gate is:
A = 0 volts, B = 0 volts, A*B’ + B*A’ = 0 volts
A = 0 volts, B = 5 volts, A*B’ + B*A’ = 5 volts
A = 5 volts, B = 0 volts, A*B’ + B*A’ = 5 volts
A = 5 volts, B = 5 volts, A*B’ + B*A’ = 0 volts
References:
I have a Bachelor of Science in Electrical Engineering along with experience in digital logic design,testing and troubleshooting.
Micro-Electronics
Digital and Analog Circuits and Systems
ISBN 0-07-042327-X
Digital Circuits and Microprocessors
ISBN 0-07-062945-5
