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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.

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Summary

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

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


Summary

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


Summary

布林代數的運算法則pp178 t4-2

11.

表4-3

P181


Summary

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:


Summary

(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:


Summary

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


Summary

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


Summary

  求出使下列標準POS表示式等於0的二進位變數值。

解:

  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, 。

標準POS形式 pp195 4-18

4-16

P201


Summary

  在A = 1, B = 1, C = 1, D = 1時, 。

  上述三項中任一項等於0,就會使此POS表示式等於0。

相關問題 求出使下列POS表示式等於0的各個二進位變數

值:

這是個標準POS表示式嗎?

標準POS形式

例4-16

(續)

P201


Summary

  將下列SOP表示式轉換成等效的POS表示式:

解:估算方式如下:

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 

可能的組合。這個SOP表示式含有其中的5種,所以POS必定

含有其它3種,即001,100,和110。記住,這些就是使和項

為0的二進位值。其等效POS表示式為:

相關問題 利用二進位數代入各表示式的方法,證明此例題

中的SOP和POS表示式為等效。

將標準SOP轉換成標準POS pp195 4-19

4-17

P202


Summary

  建立標準SOP表示式 的真值表。

解:

  此表示式有三個變數,所以變數的二進位值共有8

種可能組合,見表4-6左邊三個欄位。使表示式各積項等

於1的二進位值分別為:   : 001; : 100;ABC

: 111。如表所示,在這幾個二進位值組合的輸出欄位中

填入1。其餘二進位組合的輸出欄位中則填入0。

將SOP表示式轉換成真值表格式pp196 4-20

4-18

P203


Summary

相關問題 建立標準SOP表示式 的真值表。

將SOP表示式轉換成真值表格式

例4-18

(續)

表4-6

P203


Summary

由真值表求出標準表示式pp198 4-22

4-20

  由表4-8的真值表求出標準SOP表示式,及其等效的

標準POS表示式。

表4-8

P205


Summary

解:

  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:

得到輸出X的標準SOP表示式為:

由真值表求出標準表示式

例4-20

(續)

P206


Summary

  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:

相關問題 利用二進位數代入法,驗證此例題推導出的SOP

和POS表示式兩者等效;即同一組二進位數代入表示式時,

兩表示式應同時為1或同時為0。

由真值表求出標準表示式

例4-20

(續)

P206


Summary

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.


Summary

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


Summary

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.


Summary

C changes across outer boundary

B changes

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

B changes

C changes

X = AD +AD

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


Summary

  將下列標準SOP表示式轉換成卡諾圖:

解:

將表示式如下換成二進位值。在圖4-24的3變數卡

諾圖中,代表表示式中各標準積項的方格內填入1。

標準SOP表示式轉換至卡諾圖pp203 EX4-23

4-21

P210


Summary

相關問題 將標準SOP表示式          轉換

成卡諾圖。

標準SOP表示式轉換至卡諾圖

例4-21

(續)

圖4-24

P211


Summary

  將下列SOP表示式做成卡諾圖: 。

解:

很明顯,此SOP表示式並非標準形式,因為並非各

積項都有三個變數。第一項中少了兩個變數,第二項中

少了一個變數,第三項則是標準形式。先將各積項展開

成數字:

將非標準SOP表示式轉換成卡諾圖

4-23

P212


Summary

將非標準SOP表示式轉換成卡諾圖

例4-23

(續)

  對照所得的各個二進位數,在圖4-26之3變數卡諾圖

的適當方格內填入1。

相關問題 將SOP表示式     做成卡諾圖。

圖4-26

P213


Summary

  求出圖4-31各卡諾圖的積項,並寫出所得的最小

SOP表示式。

解:

  圖4-31中已顯示出各組的最小積項。此圖中各卡諾

圖的最小SOP表示式分別為:

(a)(b)

(c)(d)

SOP表示式的卡諾圖化簡法pp209 ex4-29

4-27

P216


Summary

SOP表示式的卡諾圖化簡法

例4-23

(續)

相關問題 在圖4-31(d) 卡諾圖的0111方格內填入1,再

求出其SOP表示式。

P217


Summary

  利用卡諾圖將下列SOP表示式最小化:

解:第一項   必須擴展成 和 ,形成標

準SOP表示式,才能做成卡諾圖,將圖中含1的方格分組,

如圖4-33所示。

SOP表示式的卡諾圖化簡法

4-29

圖4-33

P218


Summary

  注意所得兩組都有包捲式鄰接的情形。因為卡諾圖的最

外兩行中的方格相鄰接而形成8方格的一組。因為最上和最下

層方格相鄰,將其餘的兩個1拼湊起來便形成4方格的一組。

各組的積項都標示在圖上了,所得的最小SOP表示式為:

  記住,最小表示式和原標準表示式是等效的。

相關問題 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

例4-29

(續)

P218


Summary

  將下列標準POS表示式做成卡諾圖:

解:

  如下將表示式各和項的二進位數算出,再將0填入

圖4-37的4變數卡諾圖中。

將標準POS表示式做成卡諾圖

4-30

P222


Summary

將標準POS表示式做成卡諾圖

例4-30

(續)

圖4-37

相關問題 將下列標準POS表示式做成卡諾圖:

P222


Summary

  利用卡諾圖將下列POS表示式最小化:

解:

  第一項必須先擴展成 和A + B + C + D形成

標準POS表示式。接著便做出卡諾圖,並對方格分組,如圖

4-39所示。圖中已顯示出各組和項,最後得到的最小POS表

示式為

記住,最小POS表示式與原標準POS表示式是等效的。

POS表示式的卡諾圖簡化法

4-32

P223


Summary

POS表示式的卡諾圖簡化法

例4-32

(續)

圖4-39

相關問題 利用卡諾圖來簡化下列POS表示式:

P224


Summary

顯示段解碼邏輯 pp212 Fig4-39

  顯示段可用來表示不同的十進位數字,表4-9。

表4-9 各個十進位數字所要 運作的顯示段

P230


Summary

顯示段解碼邏輯

表4-10 七段邏輯電路的真值表

P231


Summary

顯示段邏輯電路的布林表示式 利用真值表,便能寫出

各顯示段的標準SOP或POS表示式。例如,a段的標準

SOP表示式為:

e段的標準SOP表示式為:

顯示段解碼邏輯

P230


Summary

顯示段解碼邏輯

多輸出邏輯閘的進一步簡化

P232


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