Boolean Algebra

Basic Operations and Truth Table

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L, H: 高電位,低電位
F, T: false, true
complement(補數): 把零變一之類的
圈圈代表一個 inverter。之後寫得簡潔一點就會把前面的三角形拿掉

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And gate
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OR gate
沒有特別強調就是 inclusive or
inclusive or: 可以兩個都是1 或 0

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布林公式,可以試著推推看

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這是所謂的真值表,左邊是 input 右邊是最後的結果
加法:OR
乘法:AND

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左邊是 or 右邊是 and
下面那題,後面一個加一,前面都不用看,就是等於一。

Basic & Advanced Theorems

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上圖的下面定理,電路化簡會用到。

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這裡很重要
or=並聯。 and=串連

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or=並聯。 and=串連

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串連一個 x+x' = 1,不影響結果
最後並聯一個 xy,不影響結果

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分配律的寫法
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也可以由串連並聯的寫法來解釋

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重點:需要假設好幾個 group 來簡化電路

Multiplying Out and Factoring

Multiplying Out: 把式子乘開

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  • SOP

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    sum term = A+B, A
    product term = AB, A

    SOP: (AND) OR (AND) OR (AND)
    括號內不能有 OR

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  • POS

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    第三題要用 group 的概念會比較好理解

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不算 inverter 只算 and, or 只會有兩層

DeMorgan's Law and Duality

  • DeMorgan's Law

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    為什麼要學DeMorgan's Law : 這裡是使用正邏輯來判斷什麼時候 f = 0,對 f 取 complement(補數),原本 x+y = 0 會變成 1 ,再由正邏輯來判斷就可以得知何時 f = 0 。

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    DeMorgan's Law -> 如何快速得到函數的補數
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    法則就是,把補數的符號分配給函數內部的變數,並且轉換裡面的 operators(AND->OR)

  • Duality(對偶)

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    變數不需要補數,可以降低簡化難度

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    回過頭看,這兩個表格個別是彼此的對偶。

More Multiplying Out and Factoring

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x(y+z) = xy + xz (背)
x+ yz = (x+y)(x+z) (背)
(x+y)(x'+z) = xz + yx' (背)
example 第二題: ABC 可以再拿掉

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©證明 : and 一個 (x+x')=1
ab + abc -> 其中 abc 可以直接拿掉
因為 ab(1+c) = ab.
1+c = 1

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Exclusive-OR and Equivalence Operations

  • Exclusive-OR

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    Exclusive-OR: 排他,我有你就不能有。兩個一樣不行
    X (EOR) Y = xy' + x'y -> 下面板書有解釋如何得到
    要對 (EOR) DeMorgan's Law 時,要先轉換成 AND, OR

    • Exclusive-OR用越多,電路面積就越省
  • Equivalence Operations

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    式子裡面如果有 (EOR), Equivalence Operations 可以先用上面的式子做替換

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如何由真值表求得 F
把所有的1 OR 起來
然後對照個別的 x, y 。遇到零的給他補數
例如 x=0, y=1 -> x'y
x=1, y=0 -> xy'
x=1, y=1 -> xy
F = x'y + xy' + xy
-> xy' + y
-> x + y

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The Consensus Theorem

Consensus (共識)

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此處的證明之前講過了,練習一很重要需要運用 dual form

Simplification of Switching Expression - Techniques


第二個例子說明,可以加一個一樣的邏輯閘幫助化簡。


xy + x'z add yz
x add xy
(x+x')

Example: 可以發現 x, x' 使用 The Consensus Theorem
加 wz' 是為了刪掉 wy'z'

Simplification of Switching Expression - Combining All


這個題型用很多 The Consensus Theorem


(A+C)(A+B+C) = (A+C) (背)

Proving the Validity of Equation