Organizational Remarks

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Organizational Remarks

• I don’t use the exercises in Mano/Kime; this way you can use them as review exercises (for many of them there are solutions on the web: www.prenhall.com/mano).

• On the 8th of october there will lab sessions instead of the lecture on the 10th of october

Digital Techniques Fall 2007 André Deutz, Leiden University

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Remarks• Two’s complement

• For the math minded, otherwise ignore:

• Boolean ring is not a Boolean algebra;

• given a Boolean ring you can equip it with a Boolean algebra structure (and conversely):

a b := ab (AND)∧ a b := a+b – ab (OR)∨

a’ := 1 + a (NOT)


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Remarks

• For the math minded, otherwise ignore:

• given a Boolean algebra you can equip it with a Boolean ring structure:

xy := x ∧ y,

x + y := (x ∨ y) ~(∧ x ∧ y).

( =AND, =OR, ~=NOT; the given ∧ ∨operations of the Boolean algebra)


4Logic diagrams, Boolean expressions, and Truth(1/0)

Tables• LDs BEs TTs


5Logic diagrams (LDs), Boolean Expressions (BEs), and Truth Tables

(TTs) (or 1/0-tables)• The object which is lurking behind the scenes

is of the course the notion of a Boolean logic function (or for short: a Boolean function): a Boolean function of n variables is a mapping from the set {0,1}n to the set {0,1}. (What do you mean by {0,1}n ? … )

• Of course, a Boolean function is completely determined by a Truth table (1/0-table)


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LDs, BEs, and TTs

• LDs BEs TTs

• Each Boolean function has a unique TT

• In general: a Boolean function or a TT can have more than one LD (or for that matter BE) associated to it.

• For any two of (LD, BE, TT) can readily go back and forth


7LDs, BEs, and TTs• For any two of (LD, BE, TT) can readily go back and forth• From TT to BE: for each 1 in the output introduce an

appropriate minterm (a product term which • contains each of the

variables or their negation), then S is the

» sum of these minterms:


A B Cin

0 0 0

0 0 1

0 1 0

0 1 1

1 0 0

1 0 1

1 1 0

1 1 1

S

0

1

1

0

1

0

0

1

A’B’CA’BC’

AB’C’

ABC

S = A’B’C +A’BC’ + AB’C’ + ABC

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LDs, BEs, and TTs; BE TT


S = A’B’C +A’BC’ + AB’C’ + ABC

• How do you get the Truth Table (1/0-table)? Etc.

• We have done also the transition from TT to LD. Given the LD it is not difficult to come up with the TT.

• It is also not difficult to imagine how you would go directly from LD to BE and vice versa.

• Make sure that you are conversant with all possible transitions

•

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Reduction (Simplification) of Boolean Expressions

• It is usually possible to simplify the canonical SOP (or POS) forms.

• A smaller Boolean equation generally translates to a lower gate count in the target circuit.

• We cover three methods: algebraic reduction, Karnaugh map reduction, and tabular (Quine-McCluskey) reduction.


10Reduced Majority Function Circuit• Compared with the AND-OR circuit for the unreduced majority function, the

inverter for C has been eliminated, one AND gate has been eliminated, and one AND gate has only two inputs instead of three inputs. Can the function by reduced further? How do we go about it?


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The Algebraic Method• Consider the majority function, F. We apply the algebraic

method to reduce F to its minimal two-level form:


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An Aside from Alfred Tarski• What did you mean in high school by the

following statement: (a+b)(a-b) = a2 – b2?

• As such it can be confusing; it is not an equation but an identity; the following is much better:

For all a in R, for all b in R [(a+b)(a-b) = a2 – b2 ]; R stands for the real numbers.

• And so it goes when we say in Boolean algebra context: aa=a. We mean: For all a in B={0, 1} [aa =a]


13An Aside from Alfred Tarski1. And so it goes when we say in Boolean algebra

context: aa=a. We mean: For all a in B={01,} [aa =a]

2. Or another example:

3. a’bc+ab’c+abc’+abc = bc+ac+ab (pertains to the majority function)

4. Let B={0,1}.

5. For all a in B, for all b in B, for all c in B: [a’bc+ab’c+abc’+abc = bc+ac+ab]

6. Or say: 3) is shorthand for 5)


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The Algebraic Method• This majority circuit is functionally equivalent to the previous

majority circuit, but this one is in its minimal two-level form:


15Karnaugh Maps: Venn Diagram Representation of Majority Function

• Each distinct region in the “Universe” represents a minterm.

• This diagram can be transformed into a Karnaugh Map.


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K-Map for Majority Function• Place a “1” in each cell that corresponds to that minterm.

• Cells on the outer edge of the map “wrap around”


A

B

C

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Adjacency Groupings for Majority Function

• F = BC + AC + AB


C

10 2

4

3

5 67

A

B

C C

A

B

C

BB C

10 2

4

3

5 67

A

0

1

00 01 11 10

A 1 11

1

Slightly differentbookkeeping (no conceptual change):

B

C

A

0 2 6 4

1 3 57

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Minimized AND-OR Majority Circuit

• F = BC + AC + AB

• The K-map approach yields the same minimal two-level form as the algebraic approach.


19K-Map Groupings• Minimal grouping is on the left, non-minimal (but logically equivalent) grouping

is on the right.

• To obtain minimal grouping, create smallest groups first.


D

B A

C

20K-Map Corners are Logically Adjacent


A

D

C

B

B

ABCD

00 01 11 10

00

01

11

10

C

A

D

1

1 1 1 1

1

1

1

1

Slightly differentbookkeeping (no conceptual change):

F = B C D + B D + A B

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K-Maps and Don’t Cares• There can be more than one minimal grouping, as a result of don’t

cares.


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Five-Variable K-Map• Visualize two 4-variable K-maps stacked one on top of the other;

groupings are made in three dimensional cubes.


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Six-Variable K-Map• Visualize four 4-variable K-maps stacked one on top of the other;

groupings are made in three dimensional cubes.

Organizational Remarks

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