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# Summary - PowerPoint PPT Presentation

Summary. Associative Laws. Pp175 The associative laws are also applied to addition and multiplication. For addition, the associative law states. When ORing more than two variables, the result is the same regardless of the grouping of the variables. A + ( B +C ) = ( A + B ) + C.

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Summary

Associative Laws

Pp175 The associative laws are also applied to addition and multiplication. For addition, the associative law states

When ORing more than two variables, the result is the same regardless of the grouping of the variables.

A + (B +C) = (A + B) + C

For multiplication, the associative law states

When ANDing more than two variables, the result is the same regardless of the grouping of the variables.

A(BC) = (AB)C

Summary

Distributive Law

The distributive law is the factoring law. A common variable can be factored from an expression just as in ordinary algebra. That is

AB + AC = A(B+ C)

The distributive law can be illustrated with equivalent circuits: pp176 fig4-5

A(B+ C)

AB + AC

6. A + A = 1

11.A + AB = A + B

8. A.A = 0

=

9. A = A

Summary

Rules of Boolean Algebra

1. A + 0 = A

7. A.A = A

2. A + 1 = 1

3. A. 0 = 0

4. A. 1 = A

10.A + AB = A

5. A + A = A

12.(A + B)(A + C) = A + BC

AB = A + B

Summary

DeMorgan’s Theorem

DeMorgan’s 1st Theorem

The complement of a product of variables is equal to the sum of the complemented variables.

Applying DeMorgan’s first theorem to gates:

(A + B )

C (A + B )

= C (A + B )+ D

X = C (A B) + D = A B C + D

Summary

Boolean Analysis of Logic Circuits

Combinational logic circuits can be analyzed by writing the expression for each gate and combining the expressions according to the rules for Boolean algebra.

Apply Boolean algebra to derive the expression for X.

Example

Solution

Write the expression for each gate:

X

Applying DeMorgan’s theorem and the distribution law:

Convert X = A B + A B C to standard form.

The first term does not include the variable C. Therefore, multiply it by the (C + C), which = 1:

X = A B (C + C) + A B C

= A B C + A B C + A B C

Summary

SOP Standard form

In SOPstandard form, every variable in the domain must appear in each term. This form is useful for constructing truth tables or for implementing logic in PLDs.

You can expand a nonstandard term to standard form by multiplying the term by a term consisting of the sum of the missing variable and its complement.

Example

Solution

Convert X = (A + B)(A + B + C) to standard form.

The first sum term does not include the variable C. Therefore, add C C and expand the result by rule 12.

X = (A + B + C C)(A + B + C)

= (A +B + C )(A + B + C)(A + B + C)

Summary

POS Standard form

In POSstandard form, every variable in the domain must appear in each sum term of the expression.

You can expand a nonstandard POS expression to standard form by adding the product of the missing variable and its complement and applying rule 12, which states that (A + B)(A + C) = A + BC.

Example

Solution

求出使下列標準POS表示式等於0的二進位變數值。　　求出使下列標準POS表示式等於0的二進位變數值。

在A = 0, B = 0, C = 0, D = 0時，A + B + C + D等於0。

A + B + C + D = 0 + 0 + 0 + 0 = 0

在A = 0,, , D = 0時， 。

4-16

P201

上述三項中任一項等於0，就會使此POS表示式等於0。

(續)

P201

將下列SOP表示式轉換成等效的POS表示式：

000 + 010 + 011 + 101 + 111

此表示式的範圍集合含有三個變數，所以總共有8種

4-17

P202

建立標準SOP表示式 的真值表。

此表示式有三個變數，所以變數的二進位值共有8

: 111。如表所示，在這幾個二進位值組合的輸出欄位中

4-18

P203

(續)

P203

4-20

由表4-8的真值表求出標準SOP表示式，及其等效的

P205

輸出欄位中共有四個1，其對應的二進位數為011

，100，110，和111。將這些二進位數如下轉換成積項：

(續)

P206

至於POS表示式，則看輸出為0的二進位數000，001，　　至於POS表示式，則看輸出為0的二進位數000，001，

010，和101。將這些二進位數如下轉換成和項：

得到輸出X的標準POS表示式為：

(續)

P206

Gray code

Summary

Karnaugh maps

Cells are usually labeled using 0’s and 1’s to represent the variable and its complement.

The numbers are entered in gray code, to force adjacent cells to be different by only one variable.

Ones are read as the true variable and zeros are read as the complemented variable.

The cells are ABC and ABC.

Summary

Karnaugh maps

Alternatively, cells can be labeled with the variable letters. This makes it simple to read, but it takes more time preparing the map.

C C

Example

Read the terms for the yellow cells.

AB

AB

AB

AB

Solution

B changes across this boundary

C changes across this boundary

• The vertical group is read AC.
• The horizontal group is read AB.

X = AC +AB

Summary

Karnaugh maps

K-maps can simplify combinational logic by grouping cells and eliminating variables that change.

Example

Group the 1’s on the map and read the minimum logic.

Solution

• 1.Group the 1’s into two overlapping groups as indicated.
• Read each group by eliminating any variable that changes across a boundary.

C changes across outer boundary

B changes

• The upper (yellow) group is read as AD.

B changes

C changes

Summary

Karnaugh maps

Example

Group the 1’s on the map and read the minimum logic.

Solution

• 1.Group the 1’s into two separate groups as indicated.
• Read each group by eliminating any variable that changes across a boundary.
• The lower (green) group is read as AD.

X

將下列標準SOP表示式轉換成卡諾圖：

4-21

P210

(續)

P211

4-23

P212

(續)

對照所得的各個二進位數，在圖4-26之3變數卡諾圖

P213

求出圖4-31各卡諾圖的積項，並寫出所得的最小　　求出圖4-31各卡諾圖的積項，並寫出所得的最小

SOP表示式。

圖4-31中已顯示出各組的最小積項。此圖中各卡諾

(a)(b)

(c)(d)

SOP表示式的卡諾圖化簡法pp209 ex4-29

4-27

P216

SOP表示式的卡諾圖化簡法

(續)

P217

利用卡諾圖將下列SOP表示式最小化：

SOP表示式的卡諾圖化簡法

4-29

P218

注意所得兩組都有包捲式鄰接的情形。因為卡諾圖的最　　注意所得兩組都有包捲式鄰接的情形。因為卡諾圖的最

記住，最小表示式和原標準表示式是等效的。

SOP表示式的卡諾圖化簡法

(續)

P218

將下列標準POS表示式做成卡諾圖：

如下將表示式各和項的二進位數算出，再將0填入

4-30

P222

(續)

P222

利用卡諾圖將下列POS表示式最小化：

第一項必須先擴展成 和A + B + C + D形成

4-39所示。圖中已顯示出各組和項，最後得到的最小POS表

POS表示式的卡諾圖簡化法

4-32

P223

POS表示式的卡諾圖簡化法

(續)

P224

顯示段可用來表示不同的十進位數字，表4-9。

P230

P231

SOP表示式為：

e段的標準SOP表示式為：

P230

P232