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ECE 15A
Fundamentals of Logic Design
Lecture 8
Malgorzata MarekSadowska
Electrical and Computer Engineering Department
UCSB
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Last time: Karnaugh Map Method to find
minimum SOP
Step 1:
Choose an element of ONset not already covered by an implicant
Step 2:
Find "maximal" groupings of 1's and X's adjacent to that element.
Remember to consider top/bottom row, left/right column, and
corner adjacencies.
This forms
prime implicants
(always a power
of 2 number of elements).
Repeat Steps 1 and 2 to find all prime implicants
Step 3:
Revisit the 1's elements in the Kmap.
If covered by single prime
implicant, it is
essential
, and participates in final cover.
The 1's it
covers do not need to be revisited
Step 4:
If there remain 1's not covered by essential prime implicants, then
select the smallest number of prime implicants that cover the
remaining 1's
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Discussion
Karnaugh maps are:
Very effective for functions up to 6 variables
Not useful for functions dependent on >6
variables
Kmaps depend on human’s visual ability to
identify prime implicants
Kmaps are not helpful in developing CAD
tools
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2level simplification approaches
Algebraic simplification
Not a systematic procedure
Difficult to tell when the minimum expression has been found
Hand methods
Kmaps
QuineMcCluskey method (aka Tabular Method)
Computeraided tools
High quality solutions for large functions ( more than 10
variables)
Heuristic methods applied
Espresso
Exact solutions for unate function
Shannon expansion
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QuineMcCluskey method
Tabular method to systematically find all
prime implicants
Can be programmed
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Outline of the QuineMcCluskey
Method
1. Produce a minterm expansion (canonical
sumofproducts form) for a function F
2. Eliminate as many literals as possible by
systematically
applying
XY + XY’ = X
.
3. Use a prime implicant chart to select a
minimum set of prime implicants that
when ORed together produce F, and that
contains a minimum number of literals.
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Determining Prime Implicants
AB’CD’ + AB’CD = AB’C
1 0 1 0
+ 1 0 1 1 = 1 0 1 
(The dash indicates a missing variable)
A’BC’D + A’BCD’
0 1 0 1
+ 0 1 1 0
We can combine the minterms above because they
differ by a single bit.
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 Winter '08
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