Canonical expression and truth table

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SUMMARY

This discussion focuses on the process of creating a truth table and deriving a canonical expression from the Boolean expression A = B'.(C' + D). The steps involve first constructing a truth table for the three variables B, C, and D, which results in 8 rows. The canonical forms discussed are the Sum of Products (SOP) and Product of Sums (POS), with emphasis on using Boolean algebra or Karnaugh Maps for simplification.

PREREQUISITES
  • Understanding of Boolean algebra
  • Familiarity with truth tables
  • Knowledge of Karnaugh Maps
  • Basic concepts of canonical forms in logic
NEXT STEPS
  • Learn how to construct truth tables for multiple variables
  • Study the principles of Boolean algebra for simplification
  • Explore the use of Karnaugh Maps for minimizing logical expressions
  • Understand the differences between Sum of Products and Product of Sums canonical forms
USEFUL FOR

This discussion is beneficial for students and professionals in computer science, electrical engineering, and anyone involved in digital logic design or Boolean function optimization.

nelectrode
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Hey,

What are the steps to follow to draw a truth table and create canonical expression
from a given expression A?

Lets say I have A= B'.(C'+D)


1) first I have to draw the truth table for the expression above, but How could I do that since I only have one expression?


2)Then I have to create canonical expression for this A=B'.(C'+D)
( but isn't that already in canonical?)
 
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That one is not really in canonical form. There are two canonical forms: SUM OF PRODUCTS and PRODUCT of SUMS. You can use boolean algebra to make it into any of the canonical forms, or you can use a truth table, or you can use a Karnaugh Map.

A is the function here. So your variables are:
B, C, and D.

Since you have 3 variables your truth table will have 8 rows (2^3 = 8)

It's like saying A(B,C,D) = B'.(C'+D) [A as a function of B, C, and D]

You can make the truth table from there like this:
________________
B | C | D | A(B, C, D)|
0 | 0 | 0 | |
0 | 0 | 1 | |
0 | 1 | 0 | |
0 | 1 | 1 | |
1 | 0 | 0 | |
1 | 0 | 1 | |
1 | 1 | 0 | |
1 | 1 | 1 | |
--------------------------

Now just fill the value of A. That is, plug in the values of B, C, and D to get A. Don't forget to negate those that are suppose to be negated in your A function (B and C in this case).

2) To make the Canonical form of SUM OF PRODUCTS you just have to write down as product those BCD combinations that make A have a 1 and keep adding them. For example if the 000 make A(B,C,D) a 1 then you would have B'C'D'. Then if the 010 make the A(B,C,D) a 1 then you would have together with the last one: A = B'C'D' + B'CD'. And so on.

If you want to have it in its simplest form use boolean algebra or a Karnaugh map which I recommend here.
 

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