Hello, readers welcome to the new post. In this post, we will discuss Introduction to Karnaugh Maps. Karnaugh Maps is known as K maps and important tool for digital circuit designing that is used for making Boolean algebra expressions and solving them easily. In this post, we will discuss all the details of K-Maps and solve the practical example to have a detailed understanding
Introduction to Karnaugh Maps
- A Karnaugh map (K-map) is a method used for simplification of algebraic expressions for Boolean functions without resorting to complicated equations. The K Map can used as a special type of truth table that helps to make maps easily with different values and Boolean expressions.
- The K Map is best for functions with 2 or 4 variables. Though K amps are used for functions having 5 or 6 variables and processing is not easy. With use of a K map for functions through 7 or more variables is not easy to make
Steps to Solve Expression using K-map
- First of all, choose K amp according to the variables.
- Â Find minterm or max terms according to the question
- For SOP place ones in a block of the map concerning minterim (0’s elsewhere).
- For POS add o’s in a block of the map and max terms (1’s elsewhere).
- Make rectangle shape groups having terms in the power of 2 such as 2,4,8, .. and cover the number of elements as can in a single group.
- The group made in step 5 find the product terms and add them for SOp
Karnaugh Maps Truth Table
- The truth table is Boolean algebra for a combination of inputs and output for logical function. The table has columns for each input and one column for each output.
- Â rows in the table are 2 raised to the power inputs.
- Each row in the table shows a different combination of inputs with 1 being true and 0 is false. The output column has results according to the input.
- For example, let’s make logical functions having two input variables A, B, and output C.So the truth table will have 4 rows making possible combinations for inputs A and B
| A | B | C |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
- Here logic function is XOR gates that has output one or true if any input is true or one.
- Truth tables are used for verification of accurate results of logical functions and used for the base of amplification of Boolen expressions that work on K Masp
Steps to solve expression using K-map-
- Let us take an example of a Boolean function:
F(X, Y, Z) = Σ(0, 1, 3, 4, 6)
- These steps involved to solve this fucjtionsTo solve this function using a K-map, we can follow these steps:
- First of all, write the truth table for the function:
| X | Y | Z | F |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 1 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 0 | 1 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 0 |
- Utilize ones to adjacent group cells. We can divide it into two groups here.: First having cells (0, 1, 4, 5) and the other with cells (2, 3, 6, 7).
| 0 | 1 | 1 |
|---|
| 1 |
|---|
| 0 | |||
|---|---|---|---|
| 1 |
- Use the corresponding Boolean expression to label each group. The first group is related to the expression X’YZ, and the 2nd group corresponds to XY’Z.
| 0 | 1 | 1 |
|---|
| 1 |
|---|
| 2 | |||
|---|---|---|---|
| 3 |
- Write the simple Boolean expression through a combination of expressions for every group with the use of the OR operator
- F(X, Y, Z) = X’YZ + XY’Z.
- Through making comparisons verify the truth tables of the original and make simplification of an expression. Here truth tables are same defining that implication results are accurate
5 variable K-Map in Digital Logic
- Now we desing 5 variables K-Map to solve it
- The steps involved in the 5-variable K-Map are explained here
- Mak table has 32 cells (2^5) and then labels rows and columns with binary values with inputs variables X, Y, Z, G, and H.
- Now write the truth table of function
- Utilize ones to adjacent group cells. The groups should be rectangular and as large as possible
- Use the corresponding Boolean expression to label each group. All of the group’s cells should have the same expression.
- Utilizing the OR operator, combine the expressions for each group to create the simplified Boolean expression.
- Compare the truth tables of the original and simplified expressions to confirm the simplification.
For example, let us consider the Boolean function
F(X, Y, Z, G, H) = Σ(0, 1, 2, 3, 5, 6, 7, 11, 12, 13, 14, 15, 20, 21, 22, 23, 28, 29, 30, 31)
- For a solution of this function, we use a 5-variable K-Map, we can follow these steps:
- Make table 32 cells and label the rows and columns with binary values for the input variables X, Y, Z,G, and H.
- Now write the table for a function
| X | Y | Z | G | H | F |
|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 0 | 0 | 1 | 1 |
| 0 | 0 | 0 | 1 | 0 | 1 |
| 0 | 0 | 0 | 1 | 1 | 0 |
| 0 | 0 | 1 | 0 | 0 | 1 |
| 0 | 0 | 1 | 0 | 1 | 1 |
| 0 | 0 | 1 | 1 | 0 | 1 |
| 0 | 0 | 1 | 1 | 1 | 0 |
| 0 | 1 | 0 | 0 | 0 | 1 |
| 0 | 1 | 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 1 | 0 | 0 |
| 0 | 1 | 0 | 1 | 1 | 1 |
| 0 | 1 | 1 | 0 | 0 | 1 |
| 0 | 1 | 1 | 0 | 1 | 0 |
| 0 | 1 | 1 | 1 | 0 | 1 |
| 0 | 1 | 1 | 1 | 1 | 0 |
| 1 | 0 | 0 | 0 | 0 | 1 |
| 1 | 0 | 0 | 0 | 1 | 0 |
| 1 | 0 | 0 | 1 | 0 | 0 |
| 1 | 0 | 0 | 1 | 1 | 1 |
| 1 | 0 | 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 | 1 | 1 |
| 1 | 0 | 1 | 1 | 0 | 0 |
| 1 | 0 | 1 | 1 | 1 | 1 |
| 1 | 1 | 0 | 0 | 0 | 1 |
| 1 | 1 | 0 | 0 | 1 | 0 |
| 1 | 1 | 0 | 1 | 0 | 0 |
| 1 | 1 | 0 | 1 | 1 | 1 |
| 1 | 1 | 1 | 0 | 0 | 0 |
| 1 | 1 | 1 | 0 | 1 | 1 |
| 1 | 1 | 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 | 1 | 0 |
- Utilize ones to adjacent group cells. The cells can be grouped into the following rectangular shapes in this instance:
| 2 | 3 | ||||
|---|---|---|---|---|---|
| 1 | 1 | 0 | |||
| 5 | 1 | 1 | 1 | 0 | |
| 4 | 1 | 1 | 1 | 1 | |
| 1 | 1 | 0 | |||
| 1 | 1 |
- Label every group through the use of the Boolean expression. We can write expressions for every group as:
Group 1: Y’Z’G’H’ Group 2: XZ’GH’ Group 3: XY’GH’ Group 4: XY’Z’H’ Group 5: XY’ZG
- Now write the simplified Boolean expression through a combination of expressions for every group with sue of OR operator
F(X, Y, Z, G, H) = B’C’D’E’ + AC’DE’ + AB’DE’ + AB’C’E + AB’CD
- We Now do verification of simplification through a comparison of the truth tables of real and simplified expressions.
Faqs
What is the basic of Karnaugh map?
- (K-map is visual technique used for simplification of algebraic expression in Boolean functions without resort complex theorems
What is a K-map and its types?
- There are two types of K Moas two dimensional mal and 3 dimensional map. Two dimensional K map, is commonly used and that is a 2-dimensional grid of cells configured at columns and rows
What are the rules of K-map?
- The rule for the K map is that make sure each value is related to a single AND gate, Groups must have only 1s. Groups are in rectangles though rectangles can wrap about the upper or lower of the table. The sides of the rectangle of are power of two: 1, 2, or 4.
Why are Karnaugh maps used?
- K maps are used for simplifying different logic needed so they can apply with use of fewer logic gates. The sum of product expression used through AND gates feeding in OR gate and product of sum expression the apply to OR gates feeding an AND gate.
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