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Chapter 3: Karnaugh MapsPowerPoint Presentation

Chapter 3: Karnaugh Maps

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Chapter 3: Karnaugh Maps. Uchechukwu Ofoegbu Temple University. Riddle.

Chapter 3: Karnaugh Maps

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Three people check into a hotel. They pay $30 to the manager and go to their room. The manager suddenly remembers that the room rate is $25 and gives $5 to the bellboy to return to the people. On the way to the room the bellboy reasons that $5 would be difficult to share among three people so he pockets $2 and gives $1 to each person. Now each person paid $10 and got back $1. So they paid $9 each, totalling $27. The bellboy has $2, totalling $29. Where is the missing $1?

- If implemented correctly, they almost always produce a minimum solution.
- They are more straightforward that algebraic manipulations
- They generally produce SOPs, but POS can be generated from their complements if required.

Product terms corresponding to groups of two.

- An implicant of a function is a product term.
- From the point of view of the map, an implicant is a rectangle of 1, 2, 4, 8, . . . (any power of 2) 1’s. That rectangle may not include any 0’s.
- Example:
- f = A’B’C’D’+A’B’CD+A’BCD+AB’CD+ABC’D’+ABC’D+ABCD

AB

00

01

11

10

CD

00

1

1

01

1

1

1

1

11

1

10

- The implicants of f are:

Minterms (1 implicant)

ABCD

ABCD

ABCD

ABCD

ABCD

ABCD

ABCD

Groups of 2

ACD

BCD

ACD

BCD

ABC’

ABD

Groups of 4

CD

Prime and Essential Prime Implicants

- Prime Implicant:
- an implicant that (from the point of view of the map) is not fully contained in any one other implicant.

- Essential Prime Implicant:
- a prime implicant that includes at least one 1 that is not included in any other prime implicant.

AB

00

01

11

10

CD

00

1

1

01

1

1

1

1

11

1

10

Minimum SOP Expressions From Karnaugh Maps

- Find all essential prime implicants.
- Circle them on the map and mark the minterm(s) that make them essential with an asterisk (*).

- 2.Find enough other prime implicants to cover the function. Do this using two criteria:
- a.Choose a prime implicant that covers as many new 1’s (that is, those not already covered by a chosen prime implicant).
- b.Avoid leaving isolated uncovered 1’s.

- The main idea is
- To Have all ones covered
- To Have as few terms as possible
- To have several rectangles with more 1’s and few rectangles with less 1’s

f = w’x’y’z’+w’xy’z’+ w’xy’z+ w’xyz+ wx’y’z’+ w’xyz+ wxy’z’+ wxyz

AB

00

01

11

10

CD

* * *

00

1

1

1

1

unnecessary

1

01

1

1

*

11

1

10

f = y’z’+wyz+w’xz

f = b + a' c

- Prime implicant
- A rectangle of 1, 2, 4, 8, . . . 1’s or X’s not included in any one larger rectangle.
- From the point of view of finding prime implicants, X’s (don’t cares) are treated as 1’s.

- Essential prime implicant
- A prime implicant that covers at least one 1 not covered by any other prime implicant (as always).
- Don’t cares (X’s) do not make a prime implicant essential.

f = Σm(1,7,10,11,13) + Σd(5,8,15)

AB

00

01

11

10

CD

00

x

Use don’t cares to get as many minterms in each tem as possible

1

x

01

1

1

x

1

11

10

1

F = BD + ACD + ABC

- For the following problem, find the minimum SOP expression within the options given
- h(a,b,c) = Σm(0,1,5) + d(3,4,6,7)

- h = a'b' + c + a
- h = a + c + b’
- h = c + b’
- h = b’
- h = c

F = A’B’C’+A’BC’+ABC’+ABC; G = A’B’C+A’BC+ABC’+ABC

AB

AB

00

01

11

10

00

01

11

10

C

C

0

0

1

1

1

1

1

1

1

1

1

1

F = A’C’+AB

G = A’C+AB

F = AC +ACD+ABCG = AC+ACD+ABC

- Many electronic systems automatically invert gates
- Easier to fabricate with electronic components
- Basic gates used in integrated circuits (IC) digital logic families.
- NAND gate
- universal gate
- Could be used to construct any logic gate

NAND Gate Implementation

When we have a circuit consisting of AND and OR gates such that

the output of the circuit comes from an OR,

the inputs to all OR gates come either from a system input or from the output of an AND, and

the inputs to all AND gates come either from a system input or from the output of an OR.

All gates are replaced by NAND gates, and any input coming directly into an OR is complemented.

Example

Try: g = wx(y+z)+x’y

NOR Gate Implementation

When we have a circuit consisting of AND and OR gates such that

the output of the circuit comes from an AND,

the inputs to all OR gates come either from a system input or from the output of an AND, and

the inputs to all AND gates come either from a system input or from the output of an OR.

All gates are replaced by NOR gates, and any input coming directly into an AND is complemented.

Example

Try: g = (x+y’)(x’+y)(x’+z)

A xor B is 1 if a = 1 or b is 1 and 0 if both are 1 or 0;

Develop a truth table for XOR

- 1-17
- 20
- 21
- 22
- 23