Karnaugh map
The Karnaugh map
provides a simple and straight-forward method of minimising boolean
expressions. With the Karnaugh map Boolean expressions having up to four and
even six variables can be simplified.
So what is a Karnaugh map?
A Karnaugh map provides a pictorial method of grouping together expressions with common factors and therefore eliminating unwanted variables. The Karnaugh map can also be described as a special arrangement of a truth table.
So what is a Karnaugh map?
A Karnaugh map provides a pictorial method of grouping together expressions with common factors and therefore eliminating unwanted variables. The Karnaugh map can also be described as a special arrangement of a truth table.
The diagram below
illustrates the correspondence between the Karnaugh map and the truth table for
the general case of a two variable problem.
diagram
The values inside the squares are copied from the output column of the truth table, therefore there is one square in the map for every row in the truth table. Around the edge of the Karnaugh map are the values of the two input variable. A is along the top and B is down the left hand side. The diagram below explains this:
The values inside the squares are copied from the output column of the truth table, therefore there is one square in the map for every row in the truth table. Around the edge of the Karnaugh map are the values of the two input variable. A is along the top and B is down the left hand side. The diagram below explains this:
The values around the edge of the map can be thought of as coordinates. So as an example, the square on the top right hand corner of the map in the above diagram has coordinates A=1 and B=0. This square corresponds to the row in the truth table where A=1 and B=0 and F=1. Note that the value in the F column represents a particular function to which the Karnaugh map corresponds.
Example of 2 Variable K-Map
Function F (A, B)
F = ∑ (m0, m1, m2) = A̅B̅ +A̅B +AB̅
F = ∑ (m0, m1, m2) = A̅B̅ +A̅B +AB̅
K-map from Truth table
We made 2 groups of 1’s. each group contains 2 minterms.
In the first group, variable A is changing & B remains unchanged. So the first term of the output expression will be B̅ (because B = 0 in this group).
In the 2nd group, Variable B is changing and variable A remains unchanged. So the second term will be of the output expression will be A̅ (because A=0 in this group).
Now the simplifies expression will be the sum of these two terms as given below,
F = A̅ + B̅
Compare this expression with the original expression of the function, this expression only uses one gate during its implementation
Example 1:
Consider the following
map. The function plotted is: Z = f(A,B) = AB' + AB
- Note that values of the input variables form the rows
and columns. That is the logic values of the variables A and B (with one
denoting true form and zero denoting false form) form the head of the rows
and columns respectively.
- Bear in mind that the above map is a one dimensional
type which can be used to simplify an expression in two variables.
- There is a two-dimensional map that can be used for up
to four variables, and a three-dimensional map for up to six variables.
Using algebraic
simplification,
z=AB'+AB
z=A(B'+B)
Z=A
z=AB'+AB
z=A(B'+B)
Z=A
Variable
B becomes redundant due to bollean theorm.
Referring
to the map above, the two adjust 1's are grouped together. Through
inspection it can be seen that variable B has its true and false form within
the group. This eliminates variable B leaving only variable A which only has
its true form. The minimised answer therefore is Z = A.
Pairs of 1's are grouped as shown above, and the simplified answer is obtained by using the following steps:
Note that two groups can be formed for the example given above, bearing in mind that the largest rectangular clusters that can be made consist of two 1s. Notice that a 1 can belong to more than one group.
The first group labelled I, consists of two 1s which correspond to A = 0, B = 0 and A = 1, B = 0. Put in another way, all squares in this example that correspond to the area of the map where B = 0 contains 1s, independent of the value of A. So when B = 0 the output is 1. The expression of the output will contain the term B'
For group labelled II corresponds to the area of the map where A = 0. The group can therefore be defined as A' This implies that when A = 0 the output is 1. The output is therefore 1 whenever B = 0 and A = 0
Hence the simplified answer is Z =A'+B'
example
Note that two groups can be formed for the example given above, bearing in mind that the largest rectangular clusters that can be made consist of two 1s. Notice that a 1 can belong to more than one group.
The first group labelled I, consists of two 1s which correspond to A = 0, B = 0 and A = 1, B = 0. Put in another way, all squares in this example that correspond to the area of the map where B = 0 contains 1s, independent of the value of A. So when B = 0 the output is 1. The expression of the output will contain the term B'
For group labelled II corresponds to the area of the map where A = 0. The group can therefore be defined as A' This implies that when A = 0 the output is 1. The output is therefore 1 whenever B = 0 and A = 0
Hence the simplified answer is Z =A'+B'
example:
f=A'BC'+A'BC+A'B'C+ABC
f=A'B+A'C+BC
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