Cs151 introduction to digital design
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CS151 Introduction to Digital Design. Chapter 2-4-1 Map Simplification. Circuit Optimization. Goal: To obtain the simplest implementation for a given function Optimization is a more formal approach to simplification that is performed using a specific procedure or algorithm

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CS151 Introduction to Digital Design

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Cs151 introduction to digital design

CS151Introduction to Digital Design

Chapter 2-4-1

Map Simplification


Circuit optimization

CircuitOptimization

  • Goal: To obtain the simplest implementation for a given function

  • Optimization is a more formal approach to simplification that is performed using a specific procedure or algorithm

  • Optimization requires a cost criterion to measure the simplicity of a circuit

  • To distinct cost criteria we will use:

    • Literal cost (L)

    • Gate input cost (G)

    • Gate input cost with NOTs (GN)

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Literal cost

B

C

D

D

C

B

B

C

D

B

LiteralCost

  • Literal – a variable or its complement

  • Literal cost – the number of literal appearances in a Boolean expression corresponding to the logic circuit diagram

  • Examples:

    • F = BD + A C + A L = 8

    • F = BD + A C + A + AB L =

    • F = (A + B)(A + D)(B + C + )( + + D) L =

    • Which solution is best?

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Gate input cost

B

C

D

B

C

B

D

B

D

C

B

GateInputCost

  • Gate input costs - the number of inputs to the gates in the implementation corresponding exactly to the given equation or equations. (G - inverters not counted, GN - inverters counted)

  • For SOP and POS equations, it can be found from the equation(s) by finding the sum of:

    • all literal appearances

    • the number of terms excluding terms consisting only of a single literal,(G) and

    • optionally, the number of distinct complemented single literals (GN).

  • Example:

    • F = BD + A C + A G = 11, GN = 14

    • F = BD + A C + A + AB G = , GN =

    • F = (A + )(A + D)(B + C + )( + + D) G = , GN =

    • Which solution is best?

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Cost criteria continued

GN = G + 2 = 9

L = 5

G =L+ 2 = 7

B

C

A

F

  • L (literal count) counts the AND inputs and the single literal OR input.

  • G (gate input count) adds the remaining OR gate inputs

C

B

  • GN(gate input count with NOTs) adds the inverter inputs

CostCriteria(continued)

  • Example 1:

  • F = A + B C +

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Cost criteria continued1

B

A

A

B

C

F

A

B

C

F

A

C

C

B

CostCriteria(continued)

  • Example 2:

  • F = A B C +

  • L = 6 G = 8 GN = 11

  • F = (A + )( + C)( + B)

  • L = 6 G = 9 GN = 12

  • Same function and sameliteral cost

  • But first circuit has bettergate input count and bettergate input count with NOTs

  • Select it!

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Map simplification

Map Simplification

  • Simplifying Boolean Expressions

    • Algebraic manipulation

      • Lacks rules for predicting steps in the process

      • Difficult to determine if the simplest expression has been reached

    • Map simplification (Karnough map or K-map)

      • Straightforward procedure for function of up to 4 variables

      • Provides a pictorial form of the truth table

      • Maps for five and six variables can be drawn, but are more cumbersome to use

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Karnaugh maps k map

Karnaugh Maps (K-map)

  • A K-map is a collection of squares

    • Each square represents a minterm

    • Adjacent squares differ in the value of one variable

    • Visual diagram of all possible ways a function may be expressed in standard forms

      • Sum-of-Products

      • Product-of-Sums

    • Alternative algebraic expressions for the same function are derived by recognizing patterns of squares select simplest.

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Two variable map

Y’

Y

X’

X

Two-Variable Map

  • 2 variables  4 minterms  4 squares.

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K map function representation

K-Map Function Representation

  • Example: F(X,Y) = XY’ + XY

  • From the map, we see that F (X,Y) = X.

  • This can be justified using algebraic manipulations:

    F(X,Y) = XY’ + XY = X(Y’ +Y) = X.1 = X

11

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K map function representation1

1

1

1

K-Map Function Representation

  • Example:G(x,y) = m1 + m2 + m3

G(x,y) = m1 + m2 + m3

= X’Y + XY’ + XY

From the map, we can see that G = X + Y

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Three variable maps

Three-Variable Maps

3 variables 8 minterms (m0 – m7).

How can we locate a minterm square on the map?

 Use figure (a) OR  use column # and row # from figure (b)E.g. m5 is in row 1 column 01 (5 10 = 101 2)

Q. Show the area representing X’? Y’? Z’?

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Alternative map labeling

y

0

3

x

2

1

x

4

7

6

5

z

z

z

Alternative Map Labeling

  • Map use largely involves:

    • Entering values into the map, and

    • Reading off product terms from the map.

  • Alternate labelings are useful:

y

y z

y

x

00

01

11

10

0

3

2

1

0

4

7

6

5

1

x

z

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Example functions

1

1

1

1

Y

Y

YZ

YZ

00

00

01

01

11

11

10

10

X

X

0

0

1

1

X

X

Z

Z

Example Functions

  • By convention, we represent the minterms of F by a "1" in the map and leave the minterms of F’ blank

  • Example:

  • Example:

  • Learn the locations of the 8 indices based on the variable order shown (x, most significantand z, least significant) on themap boundaries

1

1

1

1

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