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# Chapter 3: Karnaugh Maps - PowerPoint PPT Presentation

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

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### Chapter 3: Karnaugh Maps

Uchechukwu Ofoegbu

Temple University

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.

A

0

1

B

B’

A’

A

0

A’B’

AB’

B

1

A’B

AB

A

0

1

B

0

m0

m2

1

m1

m3

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)

ABCD

ABCD

ABCD

ABCD

ABCD

ABCD

ABCD

Groups of 2

ACD

BCD

ACD

BCD

ABC’

ABD

Groups of 4

CD

• 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

• 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 + ACD + ABC

• 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 = AB+ABC

G = AB +BC

F = AB+ABC

G = AB +ABC

f = ab + bc g = ab + ac

f = ab + abc g = ab + abc

F = AC +ACD+ABCG = AC+ACD+ABC

• 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

Alternate symbol for NAND.

Symbols for NOR gate.

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.

Try: g = wx(y+z)+x’y

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.

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