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
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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)
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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
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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
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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