08.06.20 14:50

CSCI 150 Introduction to Digital and Computer

System Design Lecture 2: Combinational Logical Circuits IV

Jetic Gū2020 Summer Semester (S2)

Overview

• Focus: Boolean Algebra

• Architecture: Combinatory Logical Circuits

• Textbook v4: Ch2 2.4, 2.5; v5: Ch2 2.4, 2.5

• Core Ideas:

1. Boolean Algebra III: K-Map

Boolean Algebra I&II• AND, OR, NOT Operators and Gates

• Simple digital circuit implementation

• Algebraic manipulation using Binary Identities

• Standard Forms

• Minterm & Maxterm

• Sum of Products & Product of Sums

Review

P0 Review

Boolean Algebra III: K-Map

Summary

P1 Optimisation

Cost Criteria;Map and Map Manipulation

K-Map

• Karnaugh Map, or just K-Map

• For optimising 2-4 variable boolean expressions

• Skip: 5,6 variable K-Maps can also be drawn but are not very intuitive to use

Concep

t

P1 Optimisation

Two Variable Maps

Concep

t

P1 Optimisation

2-4 / Two-Level Circuit Optimization 63

than sum- of- products and product- of- sums, the gate- input count must be deter-mined directly from the implementation.

Regardless of the cost criteria used, we see later that the simplest expression is not necessarily unique. It is sometimes possible to !nd two or more expressions that satisfy the cost criterion applied. In that case, either solution is satisfactory from the cost standpoint.

Map Structures

We will consider maps for two, three, and four variables as shown in Figure"2-12. The number of squares in each map is equal to the number of minterms in the corre-sponding function. In our discussion of minterms, we de!ned a minterm mi to go with the row of the truth table with i in binary as the variable values. This use of i to

(a) (b)

0

1

0 1Y

X

X Y X Y

X YX Y

(c) (d)

ZX

ZX

0

6 4

13

57

2

X Z

02

8 10

1 3911

(f)(e)

00

01

00 01YZ

WX

Y

Z

W

11 10

11

10

X

X Z

0 1

2 3 X

Y

0

1

0 1Y

X

Y

X

Z

0 1 3 2

4 5 7 6

YZ

X

0

00 01 11 10

1

1 3 2

4 5 7 6

8 9 1011

12 13 1415

0

FIGURE!2-12Map Structures

M02_MANO0637_05_SE_C02.indd 63 23/01/15 1:47 PM

• Number of squares in each map is equal to the number of minterms for the same number of variables, light blue digit above is the index (of minterm)

• Two squares are adjacent if they only differ in one variable

• Binary value inside at each position indicates the truth table value for that term

Three Variable Maps

Concep

t

P1 Optimisation

• Number of squares in each map is equal to the number of minterms for the same number of variables, light blue digit above is the index (of minterm)

• Two squares are adjacent if they only differ in one variable

• Binary value inside at each position indicates the truth table value for that term

2-4 / Two-Level Circuit Optimization 63

than sum- of- products and product- of- sums, the gate- input count must be deter-mined directly from the implementation.

Regardless of the cost criteria used, we see later that the simplest expression is not necessarily unique. It is sometimes possible to !nd two or more expressions that satisfy the cost criterion applied. In that case, either solution is satisfactory from the cost standpoint.

Map Structures

We will consider maps for two, three, and four variables as shown in Figure"2-12. The number of squares in each map is equal to the number of minterms in the corre-sponding function. In our discussion of minterms, we de!ned a minterm mi to go with the row of the truth table with i in binary as the variable values. This use of i to

(a) (b)

0

1

0 1Y

X

X Y X Y

X YX Y

(c) (d)

ZX

ZX

0

6 4

13

57

2

X Z

02

8 10

1 3911

(f)(e)

00

01

00 01YZ

WX

Y

Z

W

11 10

11

10

X

X Z

0 1

2 3 X

Y

0

1

0 1Y

X

Y

X

Z

0 1 3 2

4 5 7 6

YZ

X

0

00 01 11 10

1

1 3 2

4 5 7 6

8 9 1011

12 13 1415

0

FIGURE!2-12Map Structures

M02_MANO0637_05_SE_C02.indd 63 23/01/15 1:47 PM

Four Variable Maps

Concep

t

P1 Optimisation

2-4 / Two-Level Circuit Optimization 63

than sum- of- products and product- of- sums, the gate- input count must be deter-mined directly from the implementation.

Regardless of the cost criteria used, we see later that the simplest expression is not necessarily unique. It is sometimes possible to !nd two or more expressions that satisfy the cost criterion applied. In that case, either solution is satisfactory from the cost standpoint.

Map Structures

We will consider maps for two, three, and four variables as shown in Figure"2-12. The number of squares in each map is equal to the number of minterms in the corre-sponding function. In our discussion of minterms, we de!ned a minterm mi to go with the row of the truth table with i in binary as the variable values. This use of i to

(a) (b)

0

1

0 1Y

X

X Y X Y

X YX Y

(c) (d)

ZX

ZX

0

6 4

13

57

2

X Z

02

8 10

1 3911

(f)(e)

00

01

00 01YZ

WX

Y

Z

W

11 10

11

10

X

X Z

0 1

2 3 X

Y

0

1

0 1Y

X

Y

X

Z

0 1 3 2

4 5 7 6

YZ

X

0

00 01 11 10

1

1 3 2

4 5 7 6

8 9 1011

12 13 1415

0

FIGURE!2-12Map Structures

M02_MANO0637_05_SE_C02.indd 63 23/01/15 1:47 PM

• Number of squares in each map is equal to the number of minterms for the same number of variables, light blue digit above is the index (of minterm)

• Two squares are adjacent if they only differ in one variable

• Binary value inside at each position indicates the truth table value for that term

Two Variable Maps Optimisation

Concep

t

P1 Optimisation

• Step 1: Enter the values

• Step 2: Identify the set of largest rectangles in which all values are 1, covering all 1s

• Step 3: Read off the selected rectangles. If rectangle has odd length edges (excluding 1), split

2-4 / Two-Level Circuit Optimization 63

than sum- of- products and product- of- sums, the gate- input count must be deter-mined directly from the implementation.

Regardless of the cost criteria used, we see later that the simplest expression is not necessarily unique. It is sometimes possible to !nd two or more expressions that satisfy the cost criterion applied. In that case, either solution is satisfactory from the cost standpoint.

Map Structures

We will consider maps for two, three, and four variables as shown in Figure"2-12. The number of squares in each map is equal to the number of minterms in the corre-sponding function. In our discussion of minterms, we de!ned a minterm mi to go with the row of the truth table with i in binary as the variable values. This use of i to

(a) (b)

0

1

0 1Y

X

X Y X Y

X YX Y

(c) (d)

ZX

ZX

0

6 4

13

57

2

X Z

02

8 10

1 3911

(f)(e)

00

01

00 01YZ

WX

Y

Z

W

11 10

11

10

X

X Z

0 1

2 3 X

Y

0

1

0 1Y

X

Y

X

Z

0 1 3 2

4 5 7 6

YZ

X

0

00 01 11 10

1

1 3 2

4 5 7 6

8 9 1011

12 13 1415

0

FIGURE!2-12Map Structures

M02_MANO0637_05_SE_C02.indd 63 23/01/15 1:47 PM

Truth Table

X Y F0 0 0

0 1 1

1 0 0

1 1 1

0

1

1

0

Two Variable Maps Optimisation

Demo

P1 Optimisation

2-4 / Two-Level Circuit Optimization 63

than sum- of- products and product- of- sums, the gate- input count must be deter-mined directly from the implementation.

Regardless of the cost criteria used, we see later that the simplest expression is not necessarily unique. It is sometimes possible to !nd two or more expressions that satisfy the cost criterion applied. In that case, either solution is satisfactory from the cost standpoint.

Map Structures

We will consider maps for two, three, and four variables as shown in Figure"2-12. The number of squares in each map is equal to the number of minterms in the corre-sponding function. In our discussion of minterms, we de!ned a minterm mi to go with the row of the truth table with i in binary as the variable values. This use of i to

(a) (b)

0

1

0 1Y

X

X Y X Y

X YX Y

(c) (d)

ZX

ZX

0

6 4

13

57

2

X Z

02

8 10

1 3911

(f)(e)

00

01

00 01YZ

WX

Y

Z

W

11 10

11

10

X

X Z

0 1

2 3 X

Y

0

1

0 1Y

X

Y

X

Z

0 1 3 2

4 5 7 6

YZ

X

0

00 01 11 10

1

1 3 2

4 5 7 6

8 9 1011

12 13 1415

0

FIGURE!2-12Map Structures

M02_MANO0637_05_SE_C02.indd 63 23/01/15 1:47 PM

Truth Table

X Y F0 0 1

0 1 1

1 0 0

1 1 1

1

1

1

0 X + Y

• Step 1: Enter the values

• Step 2: Identify the set of largest rectangles in which all values are 1, covering all 1s

• Step 3: Read off the selected rectangles. If rectangle has odd length edges (excluding 1), split

Three Variable Maps Optimisation

Concep

t

P1 Optimisation

• Step 1: Enter the values

• Step 2: Identify the set of largest rectangles in which all values are 1, covering all 1s

• Step 3: Read off the selected rectangles. If rectangle has odd length edges (excluding 1), split

F(X, Y, Z) = Σm(0,1,2,3,4,5)

1

1

1

1

1 1

= X + Y

Three Variable Maps Optimisation

• Step 1: Enter the values

• Step 2: Identify the set of largest rectangles in which all values are 1, covering all 1s

• Step 3: Read off the selected rectangles. If rectangle has odd length edges (excluding 1), split

Demo

P1 Optimisation

F(X, Y, Z) = Σm(0,2,4,5,6)

1

1 11

1

= XY + Z

Three Variable Maps Optimisation

• Step 1: Enter the values

• Step 2: Identify the set of largest rectangles in which all values are 1, covering all 1s

• Step 3: Read off the selected rectangles. If rectangle has odd length edges (excluding 1), split

Exerci

se

P1 Optimisation

F(X, Y, Z) = Σm(1,3,4,5,6)

1

1 11

1

= XZ + XY + XZ

Four Variable Maps Optimisation

Exerci

se

P1 Optimisation

F(W, X, Y, Z) = Σm(0,1,2,4,5,6,8,9,10,12,13)

• Step 1: Enter the values

• Step 2: Identify the set of largest rectangles in which all values are 1, covering all 1s

• Step 3: Read off the selected rectangles. If rectangle has odd length edges (excluding 1), split

2-4 / Two-Level Circuit Optimization 63

than sum- of- products and product- of- sums, the gate- input count must be deter-mined directly from the implementation.

Regardless of the cost criteria used, we see later that the simplest expression is not necessarily unique. It is sometimes possible to !nd two or more expressions that satisfy the cost criterion applied. In that case, either solution is satisfactory from the cost standpoint.

Map Structures

We will consider maps for two, three, and four variables as shown in Figure"2-12. The number of squares in each map is equal to the number of minterms in the corre-sponding function. In our discussion of minterms, we de!ned a minterm mi to go with the row of the truth table with i in binary as the variable values. This use of i to

(a) (b)

0

1

0 1Y

X

X Y X Y

X YX Y

(c) (d)

ZX

ZX

0

6 4

13

57

2

X Z

02

8 10

1 3911

(f)(e)

00

01

00 01YZ

WX

Y

Z

W

11 10

11

10

X

X Z

0 1

2 3 X

Y

0

1

0 1Y

X

Y

X

Z

0 1 3 2

4 5 7 6

YZ

X

0

00 01 11 10

1

1 3 2

4 5 7 6

8 9 1011

12 13 1415

0

FIGURE!2-12Map Structures

M02_MANO0637_05_SE_C02.indd 63 23/01/15 1:47 PM

1

1

1

1

11

1

1

1 1

1

= Y + XZ + WZ

Four Variable Maps Optimisation

Exerci

se

P1 Optimisation

F(W, X, Y, Z) = WYZ + WZ + XY + YZ + WXZ

• Step 1: Enter the values

• Step 2: Identify the set of largest rectangles in which all values are 1, covering all 1s

• Step 3: Read off the selected rectangles. If rectangle has odd length edges (excluding 1), split

2-4 / Two-Level Circuit Optimization 63

than sum- of- products and product- of- sums, the gate- input count must be deter-mined directly from the implementation.

Regardless of the cost criteria used, we see later that the simplest expression is not necessarily unique. It is sometimes possible to !nd two or more expressions that satisfy the cost criterion applied. In that case, either solution is satisfactory from the cost standpoint.

Map Structures

We will consider maps for two, three, and four variables as shown in Figure"2-12. The number of squares in each map is equal to the number of minterms in the corre-sponding function. In our discussion of minterms, we de!ned a minterm mi to go with the row of the truth table with i in binary as the variable values. This use of i to

(a) (b)

0

1

0 1Y

X

X Y X Y

X YX Y

(c) (d)

ZX

ZX

0

6 4

13

57

2

X Z

02

8 10

1 3911

(f)(e)

00

01

00 01YZ

WX

Y

Z

W

11 10

11

10

X

X Z

0 1

2 3 X

Y

0

1

0 1Y

X

Y

X

Z

0 1 3 2

4 5 7 6

YZ

X

0

00 01 11 10

1

1 3 2

4 5 7 6

8 9 1011

12 13 1415

0

FIGURE!2-12Map Structures

M02_MANO0637_05_SE_C02.indd 63 23/01/15 1:47 PM

1

1

1 1

11

1 1

1

1

1

Don’t Care Condition

• Sometimes we don’t care what the output is when the inputs are in certain combinations

Concep

t

P1 Optimisation

2-4 / Two-Level Circuit Optimization 63

than sum- of- products and product- of- sums, the gate- input count must be deter-mined directly from the implementation.

Regardless of the cost criteria used, we see later that the simplest expression is not necessarily unique. It is sometimes possible to !nd two or more expressions that satisfy the cost criterion applied. In that case, either solution is satisfactory from the cost standpoint.

Map Structures

We will consider maps for two, three, and four variables as shown in Figure"2-12. The number of squares in each map is equal to the number of minterms in the corre-sponding function. In our discussion of minterms, we de!ned a minterm mi to go with the row of the truth table with i in binary as the variable values. This use of i to

(a) (b)

0

1

0 1Y

X

X Y X Y

X YX Y

(c) (d)

ZX

ZX

0

6 4

13

57

2

X Z

02

8 10

1 3911

(f)(e)

00

01

00 01YZ

WX

Y

Z

W

11 10

11

10

X

X Z

0 1

2 3 X

Y

0

1

0 1Y

X

Y

X

Z

0 1 3 2

4 5 7 6

YZ

X

0

00 01 11 10

1

1 3 2

4 5 7 6

8 9 1011

12 13 1415

0

FIGURE!2-12Map Structures

M02_MANO0637_05_SE_C02.indd 63 23/01/15 1:47 PM

1 1

1 X

1

11

1

Don’t Care Condition

Demo

P1 Optimisation

2-4 / Two-Level Circuit Optimization 63

than sum- of- products and product- of- sums, the gate- input count must be deter-mined directly from the implementation.

Regardless of the cost criteria used, we see later that the simplest expression is not necessarily unique. It is sometimes possible to !nd two or more expressions that satisfy the cost criterion applied. In that case, either solution is satisfactory from the cost standpoint.

Map Structures

We will consider maps for two, three, and four variables as shown in Figure"2-12. The number of squares in each map is equal to the number of minterms in the corre-sponding function. In our discussion of minterms, we de!ned a minterm mi to go with the row of the truth table with i in binary as the variable values. This use of i to

(a) (b)

0

1

0 1Y

X

X Y X Y

X YX Y

(c) (d)

ZX

ZX

0

6 4

13

57

2

X Z

02

8 10

1 3911

(f)(e)

00

01

00 01YZ

WX

Y

Z

W

11 10

11

10

X

X Z

0 1

2 3 X

Y

0

1

0 1Y

X

Y

X

Z

0 1 3 2

4 5 7 6

YZ

X

0

00 01 11 10

1

1 3 2

4 5 7 6

8 9 1011

12 13 1415

0

FIGURE!2-12Map Structures

M02_MANO0637_05_SE_C02.indd 63 23/01/15 1:47 PM

1

1

• Sometimes we don’t care what the output is when the inputs are in certain combinations

X

1

1

1X X

F = YZ + WX

Don’t Care Condition

Demo

P1 Optimisation

2-4 / Two-Level Circuit Optimization 63

than sum- of- products and product- of- sums, the gate- input count must be deter-mined directly from the implementation.

Regardless of the cost criteria used, we see later that the simplest expression is not necessarily unique. It is sometimes possible to !nd two or more expressions that satisfy the cost criterion applied. In that case, either solution is satisfactory from the cost standpoint.

Map Structures

We will consider maps for two, three, and four variables as shown in Figure"2-12. The number of squares in each map is equal to the number of minterms in the corre-sponding function. In our discussion of minterms, we de!ned a minterm mi to go with the row of the truth table with i in binary as the variable values. This use of i to

(a) (b)

0

1

0 1Y

X

X Y X Y

X YX Y

(c) (d)

ZX

ZX

0

6 4

13

57

2

X Z

02

8 10

1 3911

(f)(e)

00

01

00 01YZ

WX

Y

Z

W

11 10

11

10

X

X Z

0 1

2 3 X

Y

0

1

0 1Y

X

Y

X

Z

0 1 3 2

4 5 7 6

YZ

X

0

00 01 11 10

1

1 3 2

4 5 7 6

8 9 1011

12 13 1415

0

FIGURE!2-12Map Structures

M02_MANO0637_05_SE_C02.indd 63 23/01/15 1:47 PM

1

1

• Sometimes we don’t care what the output is when the inputs are in certain combinations

X

1

1

1X X

F = YZ + WZ

Summary

• Boolean Algebra III: K-Map

• Two Variable K-Map

• Three Variable K-Map

• Four Variable K-Map

• Don’t care optimisation

Review

P1 Optimisation

Exercises

• Step 1: Enter the values

• Step 2: Identify the set of largest rectangles in which all values are 1, covering all 1s

• Step 3: Read off the selected rectangles. If rectangle has odd length edges (excluding 1), split

Exerci

se

P1 Optimisation

F(X, Y, Z) = Σm(0,2,6,7)

Exercises

• Step 1: Enter the values

• Step 2: Identify the set of largest rectangles in which all values are 1, covering all 1s

• Step 3: Read off the selected rectangles. If rectangle has odd length edges (excluding 1), split

Exerci

se

P1 Optimisation

F(X, Y, Z) = Σm(0,2,6,7)

1 1

1 1

Exercises

• Step 1: Enter the values

• Step 2: Identify the set of largest rectangles in which all values are 1, covering all 1s

• Step 3: Read off the selected rectangles. If rectangle has odd length edges (excluding 1), split

Exerci

se

P1 Optimisation

F(X, Y, Z) = Σm(0,2,6,7)

1 1

1 1

Exercises

• Step 1: Enter the values

• Step 2: Identify the set of largest rectangles in which all values are 1, covering all 1s

• Step 3: Read off the selected rectangles. If rectangle has odd length edges (excluding 1), split

Exerci

se

P1 Optimisation

F(X, Y, Z) = Σm(0,1,2,4)

Exercises

• Step 1: Enter the values

• Step 2: Identify the set of largest rectangles in which all values are 1, covering all 1s

• Step 3: Read off the selected rectangles. If rectangle has odd length edges (excluding 1), split

Exerci

se

P1 Optimisation

F(X, Y, Z) = Σm(0,1,2,4)

1 1 1

1

Exercises

• Step 1: Enter the values

• Step 2: Identify the set of largest rectangles in which all values are 1, covering all 1s

• Step 3: Read off the selected rectangles. If rectangle has odd length edges (excluding 1), split

Exerci

se

P1 Optimisation

F(X, Y, Z) = Σm(0,1,2,4)

1 1 1

1

Exercises

• Step 1: Enter the values

• Step 2: Identify the set of largest rectangles in which all values are 1, covering all 1s

• Step 3: Read off the selected rectangles. If rectangle has odd length edges (excluding 1), split

Exerci

se

P1 Optimisation

F(X, Y, Z) = Σm(0,2,3,4,6)

Exercises

• Step 1: Enter the values

• Step 2: Identify the set of largest rectangles in which all values are 1, covering all 1s

• Step 3: Read off the selected rectangles. If rectangle has odd length edges (excluding 1), split

Exerci

se

P1 Optimisation

F(X, Y, Z) = Σm(0,2,3,4,6)

1 1 1

1 1

Exercises

• Step 1: Enter the values

• Step 2: Identify the set of largest rectangles in which all values are 1, covering all 1s

• Step 3: Read off the selected rectangles. If rectangle has odd length edges (excluding 1), split

Exerci

se

P1 Optimisation

F(X, Y, Z) = Σm(0,2,3,4,6)

1 1 1

1 1

Exercises

• Step 1: Enter the values

• Step 2: Identify the set of largest rectangles in which all values are 1, covering all 1s

• Step 3: Read off the selected rectangles. If rectangle has odd length edges (excluding 1), split

Exerci

se

P1 Optimisation

F(X, Y, Z) = Σm(0,2,3,4,5,7)

Exercises

• Step 1: Enter the values

• Step 2: Identify the set of largest rectangles in which all values are 1, covering all 1s

• Step 3: Read off the selected rectangles. If rectangle has odd length edges (excluding 1), split

Exerci

se

P1 Optimisation

F(X, Y, Z) = Σm(0,2,3,4,5,7)

1 1 1

1 1 1

Exercises

• Step 1: Enter the values

• Step 2: Identify the set of largest rectangles in which all values are 1, covering all 1s

• Step 3: Read off the selected rectangles. If rectangle has odd length edges (excluding 1), split

Exerci

se

P1 Optimisation

F(X, Y, Z) = Σm(0,2,3,4,5,7)

1 1 1

1 1 1