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Chapter 2 Review

• De Morgan’s Theorems• “SFA Theorems” for XOR and XNOR• Nothing but NAND gates

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Chapter 3Karnaugh Maps

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What are Karnaugh Maps?These are grids that all you to easily find the

simplest algebraic expression for a truth table.

K-Maps consists of one square for each possible minterm in a function.

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Karnaugh Mapping Steps1. Sketch a Karnaugh map grid for the problem.2. Fill in the 1’s and 0’s from the truth table.3. Circle groups of 1’s.

Circle the largest groups of 2, 4, 8, etc. first. Minimize the number of circles but make sure that every 1 is in a circle.

4. Write an equation using these circles.

Example: f(x,y,z) = m(2,3,4,5,7)

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Map 2.12 x yz + x yz + xy z + xy z + xyz. Map 2.13 A better solution.

Map 2.14 The minimum solutions.

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Karnaugh Mapping Steps1. Sketch a Karnaugh map grid for the problem.2. Fill in the 1’s and 0’s from the truth table.3. Circle groups of 1’s.

Circle the largest groups of 2, 4, 8, etc. first. Minimize the number of circles but make sure that every 1 is in a circle.

4. Write an equation using these circles.

Example: f(w,x,y,z) = m(1,3,5,6,7,12,13,14,15)

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= wx + wz + wxyz= wx + w(z + xyz)= wx + w(z + xy) [P10a]= wx + wz + wxy= wz + x(w + wy)= wz + x(w + y) [P10a]= wz + wx + xy

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Karnaugh Mapping Steps1. Sketch a Karnaugh map grid for the problem.2. Fill in the 1’s and 0’s from the truth table.3. Circle groups of 1’s.

Circle the largest groups of 2, 4, 8, etc. first. Minimize the number of circles but make sure that every 1 is in a circle.

4. Write an equation using these circles.

Example: Problem 1 (a-e)

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wx + wxy + wyz + wyz + wxyz

= (wx + wxy) + (wyz + wyz) + wxyz

= wx + wz + wxyz [P12a, P9a]

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Map 2.16 Consensus.

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Map 2.1 Two-variable Karnaugh maps.

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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.

The implicants of F areMinterms Groups of 2 Groups of 4ABCD ACD CDABCD BCDABCD ACDABCD BCDABCD ABCABCD ABDABCD

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A prime implicant is an implicant that (from the point of view of the map) is not fully contained in any one other implicant.

An essential prime implicant is a prime implicant that includes at least one 1 that is not included in any other prime implicant.

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minimum all prime implicants

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F = ABC + ABE + ABCE + ABCDE + BDE

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F = ACE + ABCD + CDE + BCE + BCDE + ACD

F = ACE + ABCD + CDE + BCE + BCDE + ADE

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F = ACD + ACD + ACD + ACD + BD + AB

F = ACD + ACD + ACD + ACD + BD + BC

F = ACD+ ACD + ACD + ACD+ AB+ BC

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f = acd + bcd + acd + bcd

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f = abd + abc + abd + abc

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

An essential prime implicant is 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.

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minimum other p.i.s

F = BD + ACD + ABC

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Map Method 3

1. Find all essential prime implicants (using either Map Method 1 or 2).

2. Replace all 1’s covered by the essential prime implicants with X’s. This highlights the 1’s that remain to be covered.

3. Then choose enough of the other prime implicants (as in Methods 1 and 2).

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Finding a minimum product of sums expression requires no new theory. The following approach is the simplest:

1. Map the complement of the function. (If there is already a map for the function, replace all 0’s by 1’s, all 1’s by 0’s and leave X’s unchanged.)

2. Find the minimum sum of products expression for the complement of the function (using the techniques of the last two sections).

3. Use DeMorgan’s theorem (P11) to complement that expression, producing a product of sums expression.

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f = ac + abc + acd f = ac + ac + abd

f = ac + ac + bcd

f = (a + c)(a + c)(a + b + d)

f = (a + c)(a + c)(b + c + d)

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F = AC + AB G = AC + AB

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F = AB + ABC G = AB + BC

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F = AC + ACD + ABC G = AC + ACD + ABC

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f = ab + bc g = ab + ac

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F = BC + ABC+ ABD + ABD

G = C + ABD

H = BC + ABC + ABD

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g = xz+ wz+ wyz+ wxy

g = xz+ wz+ wyz+ xyz

g = xz+ wz+ xyz + wxy

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not used minimum

G = ABC + ACD + ABC + ACD

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Page 109 – Chapter Test

• Problems starting at #5.

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Chapter 2

• 13• 17• Simplify the equation below using a

Karnaugh Map. Express your answer in algebraic form.

• f(A,B,C,D,E)=m(0,1,2,3,6,8,9,10,11,17,20,21,23,25,28,30,31)

Announcements• Homework #3 is due today.• Lab #3 will involve Electronic Workbench.• Exam #1 will be given on Wednesday.

– Open Book– The review sheet is online.

• Lecture– DeMorgan’s Theorems– Consensus Therorem– Questions over Homework #3?

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Example Problems

• 3.9(e)• Homework #3