Here's a very quick outline of what we'll cover in this tutorial. The tutorial document at the end will cover these subtopics in greater detail, with examples and truth tables.
DEMORGAN’S LAWIt states that
(X+Y)’ = X’.Y’
(X.Y)’ = X’ + Y’
Steps to perform De Morgan’s law
LOGIC GATESA GATE is an electronic circuit that operates on one or more signals to give an output. Gates are two state circuits i.e. the input or output can be either low or high (0 or 1).
The three basic gates are
NOT gateA NOT gate performs the not logic that is, it gives the inverted value of input signal as output. It gives a high output for a low input and a low output for a high input. If the input is X then output is X’ hence the output expression for not gate is F(x) = x’.
AND gateAn AND gate operates on two or more inputs and gives one
output. It gives a low output if any of the inputs is low, and a high XOR gateAn XOR gate operates on two or more inputs and gives one
output. It gives a high output for odd number of high inputs and a low XNOR gateAn XNOR gate also operates on two or more inputs and gives one output. It gives a low output for odd number of high inputs and a high output for even number of high inputs. For a two input XNOR gate, output expression is F(x,y) = (x . y) = x’y’ + xy NAND gateA NAND gate also operates on two or more inputs and gives one output. It is an AND gate preceding a NOT gate. It gives a high output if any of the inputs is low, and a low output only when all the inputs are
high. For a two input NAND gate, output expression is F(x,y) = (x . y)’. NOR gateA NOR gate also operates on two or more inputs and gives one
output. It is a NOR gate preceding a NOT gate. It gives a low output Universality of NAND and NOR gatesBoth NAND and NOR gate are very special gates as any Boolean expression can be realized by using only NAND or NOR gate. The tutorial document will show you how. Realization of Boolean expressions using NAND and NOR gateConsider the expression
• XY+Z
Now,
X.Y+Z = ((X.Y)’)’ + Z = ((X.Y)’ . Z’)’
X.Y+Z = ((X’+Y’)’+Z) = (((X’+Y’)’+Z)’)’
The above conversions can be proved by using DeMorgan’s law. But the conversions seem to be hard and nonintuitive. A good way to realize a circuit using NAND or NOR gate is the following—
(XY+Z using NAND gate)
(XY+Z using NOR gate)
You might understand this better once you read the tutorial. Complete Tutorial with Truth Tables and Examples (You can also try out the companion quizzes at the end of this page):MCQ Quiz #1 Boolean AlgebraCompanion MCQ Quiz for Boolean Algebra test how much you know about the topic. Your score will be emailed to you at the address you provide. MCQ Quiz #1 Digital Electronics: Boolean AlgebraGoogle Spreadsheet Form MCQ Quiz #2 Logic GatesCompanion MCQ Quiz for Logic Gates test how much you know about the topic. Your score will be emailed to you at the address you provide.
MCQ Quiz #2  Logic GatesGoogle Spreadsheet Form Here's a list of all the tutorials we currently have in this area  Introductory Digital Electronic Circuits and Boolean logic
 Tutorials in Basic Digital Electronics: At a glance
