来自 criwits 的 2022 春《计算方法》和 2021 秋《数字逻辑设计》笔记 (HITSZ-OpenCS#78)

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大二上/数字逻辑设计/note/2021_hanswan/Digital Logic Design 05bbc93ece5d4edd95c960bf21bada63/基本逻辑门 c5a80174c1ca46d7ae2c7190d0d77fa8.md

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+# 基本逻辑门
+
+Logic gates are the basic building blocks of any digital system. A logic gate is a electronic circuit having one or more than one *input port(s)* and only one *output port*. The relationship between its input(s) and output is determined by a certain logic. 
+
+It’s necessary to memorise the function, symbol as well as algebraic expression of some common logic gates introduced in this chapter.
+
+# Basic Operations
+
+## AND Gate
+
+The AND gate only outputs a positive signal when all of its inputs are getting positive signal.
+
+Expression: $F=A\cdot B=AB$. Intuitively the AND logic can be treated as multiplication.
+
+Symbol:
+
+![Untitled](%E5%9F%BA%E6%9C%AC%E9%80%BB%E8%BE%91%E9%97%A8%20c5a80174c1ca46d7ae2c7190d0d77fa8/Untitled.png)
+
+## OR Gate
+
+The OR gate only outputs a negative signal when all of its inputs are getting negative signal.
+
+Expression: $F=A+B$. Intuitively the OR logic can be treated as addition.
+
+Symbol:
+
+![Untitled](%E5%9F%BA%E6%9C%AC%E9%80%BB%E8%BE%91%E9%97%A8%20c5a80174c1ca46d7ae2c7190d0d77fa8/Untitled%201.png)
+
+## NOT Gate
+
+The NOT gate will produce a signal different from its input.
+
+Expression: $F=\bar{A}=A'$.
+
+Symbol:
+
+![Untitled](%E5%9F%BA%E6%9C%AC%E9%80%BB%E8%BE%91%E9%97%A8%20c5a80174c1ca46d7ae2c7190d0d77fa8/Untitled%202.png)
+
+# Other Operations
+
+## NAND Gate
+
+NAND = NOT + AND, i.e. $F=\overline{AB}$.
+
+Symbol:
+
+![Untitled](%E5%9F%BA%E6%9C%AC%E9%80%BB%E8%BE%91%E9%97%A8%20c5a80174c1ca46d7ae2c7190d0d77fa8/Untitled%203.png)
+
+## NOR Gate
+
+NOR = NOT + OR, i.e. $F=\overline{A+B}$.
+
+Symbol:
+
+![The symbol on the right side is so ugly that I’m wondering if my teacher drew it with line & curve tools in PowerPoint.](%E5%9F%BA%E6%9C%AC%E9%80%BB%E8%BE%91%E9%97%A8%20c5a80174c1ca46d7ae2c7190d0d77fa8/Untitled%204.png)
+
+The symbol on the right side is so ugly that I’m wondering if my teacher drew it with line & curve tools in PowerPoint.
+
+## AND-OR-NOT Gate
+
+**This is not a common logic gate**.
+
+This kind of logic gate has 4 input ports, wiring them with AND gate two by two and then connects the two outputs with a NOR gate.
+
+Expression: $F=\overline{AB+CD}$.
+
+Symbol:
+
+![Untitled](%E5%9F%BA%E6%9C%AC%E9%80%BB%E8%BE%91%E9%97%A8%20c5a80174c1ca46d7ae2c7190d0d77fa8/Untitled%205.png)
+
+The fact is, I hold the opinion that this is not a logic gate. It’s a kind of combinational logic block.
+
+## XOR Gate
+
+Exclusive OR gate, or XOR gate, outputs a positive signal only when its two input signals are different from each other.
+
+Expression: $F=A\oplus B=\bar{A}B+A\bar{B}$.
+
+Symbol:
+
+![Untitled](%E5%9F%BA%E6%9C%AC%E9%80%BB%E8%BE%91%E9%97%A8%20c5a80174c1ca46d7ae2c7190d0d77fa8/Untitled%206.png)
+
+## XNOR Gate
+
+XNOR = NOT + XOR, i.e. $F=AB+\bar{A}\bar{B}$.
+
+XNOR gate outputs a positive signal only when its two input signals are the same one.
+
+Symbol:
+
+![The bottom-left symbol is more or less... rough.](%E5%9F%BA%E6%9C%AC%E9%80%BB%E8%BE%91%E9%97%A8%20c5a80174c1ca46d7ae2c7190d0d77fa8/Untitled%207.png)
+
+The bottom-left symbol is more or less... rough.

大二上/数字逻辑设计/note/2021_hanswan/Digital Logic Design 05bbc93ece5d4edd95c960bf21bada63/布尔代数 68f91aab71134af48483ec3eda18e153.md

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+# 布尔代数
+
+The Boolean algebra is used to describe certain logic design made up with several logic gates and wires connecting them.
+
+# Laws and Theorems in Boolean Algebra
+
+## Basic Theorems
+
+- $A+B=B+A$, and $AB=BA$.
+- $(A+B)+C=A+(B+C)$, and $(AB)C=A(BC)$.
+- $A(B+C)=AB+AC$.
+- $A+AB=A$, and $A(A+B)=A$.
+
+    This theorem shows that $B$ will be ‘absorbed’ in this form of expression.
+
+- $AB+A\bar{B}=A$, and $(A+B)(A+\bar{B})=A$.
+- $A+\bar{A}B=A+B$.
+
+    $A+\bar{A}B=A+AB+\bar{A}B=A+(A+\bar{A})B=A+B$.
+
+- ${A+BC=(A+B)(A+C)}$
+
+    Why: $(A+B)(A+C)=A+AC+BA+BC$, and $A+AC+BA=A$.
+
+- $AB+\bar{A}C+BC=AB+\bar{A}C$, and $AB+\bar{A}C+BCD=AB+\bar{A}C$.
+
+    Use $BC$ to absorb $BCD$.
+
+- $\overline{A\bar{B}+\bar{A}B}=\bar{A}\bar{B}+AB$.
+
+    XNOR = XOR + NOT.
+
+
+## DeMorgan’s Law
+
+$$
+\begin{align*}\overline{X_1X_2\cdots X_n}=\bar{X}_1+\bar{X}_2+\cdots+\bar{X}_n\\\overline{X_1+X_2+\cdots+X_n}=\bar{X}_1\bar{X}_2\cdots\bar{X}_n\end{align*}
+$$
+
+## Dual Rule
+
+By changing operators in an expression in this way you get its dual expression:
+
+- $+\leftrightarrow\cdot$
+- $\oplus\leftrightarrow\odot$
+
+For a equation it’s dual expression is still valid. For example:
+
+$$
+A+BCD=(A+B)(A+C)(A+D)\Leftrightarrow A(B+C+D)=AB+AC+AD
+$$
+
+# Maxterm and Minterm
+
+It’s easy for us to realise that for a boolean expression containing some variables—say, $A$, $B$, $C$ and $D$—it can be written in this form: only several AND items added together, i.e. addiction of multiplications. An example:
+
+$$
+f(A, B, C, D)=AB+ACD+D
+$$
+
+And if we add invisible items (which is always 1) such as $(A+A')$ and $(D+D')$ to make each AND item has all four variables appear, then the formula can be expanded to:
+
+$$
+f(A,B,C,D)=AB(C+C')(D+D')+A(B+B')CD+(A+A')(B+B')(C+C')D
+$$
+
+Simplify this expression using the law $A(B+C)=AB+AC$, we will get 
+
+$$
+f(A,B,C,D)=ABCD+ABCD'+ABC'D+ABC'D'+AB'CD+A'BCD+A'B'CD+A'BC'D+A'B'C'D
+$$
+
+If we use $1011$ to represent $AB'CD$, use $0110$ to represent $A'BCD'$, then the items in the formula above are:
+
+$$
+1111, 1110, 1101, 1100, 1011, 0111, 0011, 0101, 0001
+
+$$
+
+All of them are 4-bit binary numbers so we just transform them into decimal numbers. They are
+
+$$
+15, 14, 13, 12, 11, 7, 3, 5, 1
+$$
+
+Intuitively, the original boolean expression can be expressed with a series of numbers. This is called the *minterm* expression. More detailed, we use minterm expression in this algebraic way:
+
+$$
+f(A,B,C,D)=\sum m(1, 3, 5, 7, 11, 12, 13, 14, 15)
+$$
+
+Dually, an expression can also be written in this way:
+
+$$
+g(A, B, C)=(A'+B+C)(A+B+C')(A'+B'+C)
+$$
+
+Treat $(A'+B+C)$ as binary number $100$, then we have the *maxterm* expression. It is expressed in this way:
+
+$$
+g(A,B,C)=\prod M(1, 5, 6)
+$$
+
+According to the dual rule it’s easy for us to change an expression into the other one. For instance:
+
+$$
+f(A,B,C)=\sum m(1, 4, 5, 7)=\prod M(0, 2, 3, 6)
+$$
+
+This figure shows some interesting relationships.
+
+![Untitled](%E5%B8%83%E5%B0%94%E4%BB%A3%E6%95%B0%2068f91aab71134af48483ec3eda18e153/Untitled.png)
+
+# Karnaugh Map
+
+Karnaugh Map (K. Map) is a powerful tool used in expression simplification and other scenarios. 
+
+K. Map is a table containing $2^n$ element cells. Each cell stands for a minterm. The cells are arranged in a Gray code way, which means the two cells next to each other have only one different bit. For 4 variables the K. Map usually likes this:
+
+![Untitled](%E5%B8%83%E5%B0%94%E4%BB%A3%E6%95%B0%2068f91aab71134af48483ec3eda18e153/Untitled%201.png)
+
+Steps when using K. Map to simplify logic expression:
+
+## Fulfil the K. Map
+
+To fulfil the K. Map you firstly transform your expression into the form ‘addiction of multiplications’. For example,
+
+$$
+F=\overline{A\oplus C\cdot\overline{\bar{B}(A\bar{C}\bar{D}+\bar{A}C\bar{D})}}=AC+\bar{A}\bar{C}+A\bar{B}\bar{C}\bar{D}+\bar{A}\bar{B}C\bar{D}
+$$
+
+Then for each multiplication fill its corresponding K. Map cells with 1. For example, for $\bar{A}\bar{C}$, we find the row where $A$ is 0, and mark 1 in those rows whose $C$ is 0. Finally we will have a K. Map like this:
+
+![Untitled](%E5%B8%83%E5%B0%94%E4%BB%A3%E6%95%B0%2068f91aab71134af48483ec3eda18e153/Untitled%202.png)
+
+*An interesting fact is that you can now read the minterm expression directly on this K. Map.*
+
+## Circle the Prime Implicants in K. Map
+
+Now, we make rectangular groups containing total terms in power of 2 and try to cover as many elements as you can in one group.
+
+![Untitled](%E5%B8%83%E5%B0%94%E4%BB%A3%E6%95%B0%2068f91aab71134af48483ec3eda18e153/Untitled%203.png)
+
+Note that the four red groups are the same group for the parallel edges and corners of such K. Map are connected together.
+
+This kind of rectangular groups, or circles, are called ‘prime implicants’. ‘Implicant’ means the rectangular group only consists 1 and ‘prime’ means it has been expanded as large as possible.
+
+## Get Simplified Expression from the Implicants
+
+Each group in the figure above stands for a multiplication item. Just write them and add them together and you get a simplified addiction of multiplications.
+
+Look at the figure above.
+
+- For four elements in the red circle, they have both $B$ and $D$ as 0, while their $A$ and $C$ values differ. So for this group it’s implicant is $B'D'$.
+- For four elements in the green circle, they have both $A$ and $C$ as 0, so this group represents $A'C'$.
+- For four elements in the yellow circle, they have both $A$ and $C$ as 1, so this group represents $AC$.
+
+So finally we can read out this expression:
+
+$$
+F=A'C'+AC+B'D'
+$$
+
+## Distinguished 1-Cell and Essential Prime Implicant
+
+- Distinguished 1-cell means the cell in a K. map who is only covered by one prime implicant.
+- Essential prime implicant means the prime implicant containing distinguished 1-cell.