Digital-electronics 简明教程

Canonical and Standard Form

通过组合两个变量 x 和 y 与逻辑 AND 操作，我们将获得四个布尔乘积项。这些布尔乘积项被称为 Min Terms 或 Standard Product Terms 。最小项为 x’y', x’y, xy' 和 xy。

We will get four Boolean product terms by combining two variables x and y with logical AND operation. These Boolean product terms are called as Min Terms or Standard Product Terms. The min terms are x’y', x’y, xy' and xy.

类似地，通过组合两个变量 x 和 y 与逻辑 OR 操作，我们将获得四个布尔和项。这些布尔和项被称为 Max terms 或 standard sum terms 。最大项为 x + y, x + y', x' + y 和 x' + y’。

Similarly, we will get four Boolean sum terms by combining two variables x and y with logical OR operation. These Boolean sum terms are called as Max terms or standard sum terms. The Max terms are x + y, x + y', x' + y and x' + y’.

下表显示了 2 个变量的最大项和最小项的表示。

The following table shows the representation of min terms and MAX terms for 2 variables.

x

y

Min Terms

Max Terms

0

0

m0 = x’y’

M0 = x + y

0

1

m1 = x’y

M1 = x + y’

1

0

m2 = xy’

M2 = x’ + y

1

1

m3 = xy

M3 = x’ + y’

如果二进制变量为“0”，则它被表示为最小项中变量的补码，在最大项中则表示为变量本身。同样，如果二进制变量为“1”，则它被表示为最大项中变量的补码，在最小项中则表示为变量本身。

If the binary variable is '0', then it is represented as complement of variable in min term and as the variable itself in Max term. Similarly, if the binary variable is '1', then it is represented as complement of variable in Max term and as the variable itself in min term.

从上表中，我们可以很容易地注意到最小项和最大项互为补码。如果有“n”个布尔变量，那么将有 2n 个最小项和 2n 个最大项。

From the above table, we can easily notice that min terms and Max terms are complement of each other. If there are 'n' Boolean variables, then there will be 2n min terms and 2n Max terms.

Canonical SoP and PoS Forms

真值表包含一组输入和输出。如果有“n”个输入变量，那么将有 2n 个由零和一组成的可能组合。因此，每个输出变量的值取决于输入变量的组合。因此，对于输入变量的某些组合，每个输出变量将具有“1”，对于输入变量的某些其他组合，则具有“0”。

A truth table consists of a set of inputs and output(s). If there are ‘n’ input variables, then there will be 2n possible combinations with zeros and ones. So the value of each output variable depends on the combination of input variables. So, each output variable will have ‘1’ for some combination of input variables and ‘0’ for some other combination of input variables.

因此，我们可以用以下两种方式表示每个输出变量。

Therefore, we can express each output variable in following two ways.

  1. Canonical SoP Form

  2. Canonical PoS Form

Canonical SoP Form

规范 SoP 形式表示规范乘积和形式。在此形式中，每个乘积项都包含所有字面量。因此，这些乘积项只是最小项。因此，规范 SoP 形式也被称为 sum of min terms 形式。

Canonical SoP form means Canonical Sum of Products form. In this form, each product term contains all literals. So, these product terms are nothing but the min terms. Hence, canonical SoP form is also called as sum of min terms form.

首先，确定输出变量为一的最小项，然后执行这些最小项的逻辑 OR 以获得对应于该输出变量的布尔表达式（函数）。此布尔函数将以最小项相加的形式。

First, identify the min terms for which, the output variable is one and then do the logical OR of those min terms in order to get the Boolean expression (function) corresponding to that output variable. This Boolean function will be in the form of sum of min terms.

如果有多个输出变量，请对其他输出变量也执行相同的过程。

Follow the same procedure for other output variables also, if there is more than one output variable.

请考虑以下 truth table 。

Consider the following truth table.

Inputs

Output

p

q

r

f

0

0

0

0

0

0

1

0

0

1

0

0

0

1

1

1

1

0

0

0

1

0

1

1

1

1

0

1

1

1

1

1

这里，输出 (f) 对四个输入组合为“1”。相应的最小项为 p’qr, pq’r, pqr', pqr。通过对这四个最小项执行逻辑 OR，我们将获得输出 (f) 的布尔函数。

Here, the output (f) is '1' for four combinations of inputs. The corresponding min terms are p’qr, pq’r, pqr', pqr. By doing logical OR of these four min terms, we will get the Boolean function of output (f).

因此，输出的布尔函数为 f = p’qr + pq’r + pqr' + pqr。这是输出 f 的 canonical SoP form 。我们还可以在以下两个符号中表示此函数。

Therefore, the Boolean function of output is, f = p’qr + pq’r + pqr' + pqr. This is the canonical SoP form of output, f. We can also represent this function in following two notations.

\mathrm{f \: = \: m_{3} \: + \: m_{5} \: + \: m_{6} \: + \: m_{7}}

\mathrm{f \: = \: \sum \: m\left ( 3, \:5, \:6, \:7 \right )}

在第一个等式中，我们将函数表示为相应极小项之和。在第二个等式中，我们使用了这些极小项求和的符号。

In one equation, we represented the function as sum of respective min terms. In other equation, we used the symbol for summation of those min terms.

Canonical PoS Form

典范化 PoS 形式表示典范化积和形式。在这种形式中，每个和项包含所有字面量。因此，这些和项无非是最大项。因此，典范化 PoS 形式也称为 product of Max terms 形式。

Canonical PoS form means Canonical Product of Sums form. In this form, each sum term contains all literals. So, these sum terms are nothing but the Max terms. Hence, canonical PoS form is also called as product of Max terms form.

首先，找出输出变量为零的最大项，然后取这些最大项的逻辑与，以得到对应于该输出变量的布尔表达式（函数）。该布尔函数将具有最大项乘积的形式。

First, identify the Max terms for which, the output variable is zero and then do the logical AND of those Max terms in order to get the Boolean expression (function) corresponding to that output variable. This Boolean function will be in the form of product of Max terms.

如果有多个输出变量，请对其他输出变量也执行相同的过程。

Follow the same procedure for other output variables also, if there is more than one output variable.

考虑前一例的真值表。在这里，对于四种输入组合，输出 (f) 为“0”。相应最大项为 p + q + r、p + q + r'、p + q' + r、p' + q + r。通过取这四个最大项的逻辑与，我们将得到输出 (f) 的布尔函数。

Consider the same truth table of previous example. Here, the output (f) is '0' for four combinations of inputs. The corresponding Max terms are p + q + r, p + q + r', p + q' + r, p' + q + r. By doing logical AND of these four Max terms, we will get the Boolean function of output (f).

因此，输出的布尔函数为 f = (p + q + r)·(p + q + r')·(p + q' + r)·(p' + q + r)。这是输出 f 的 canonical PoS form 。我们还可以在以下两个符号中表示此函数。

Therefore, the Boolean function of output is, f = (p + q + r)·(p + q + r')·(p + q' + r)·(p' + q + r). This is the canonical PoS form of output, f. We can also represent this function in following two notations.

\mathrm{f \: = \: M_{0}\cdot M_{1} \cdot M_{2} \cdot M_{4}}

\mathrm{f \: = \: \prod M\left ( 0, \: 1, \: 2, \: 4 \right )}

在第一个等式中，我们将函数表示为相应最大项的乘积。在第二个等式中，我们使用了这些最大项的乘法符号。

In one equation, we represented the function as product of respective Max terms. In other equation, we used the symbol for multiplication of those Max terms.

布尔函数 f = (p + q + r)·(p + q + r’)·(p + q’ + r)·(p’ + q + r) 是 f = p’qr + pq’r + pqr' + pqr 的对偶。

The Boolean function, f = (p + q + r)·(p + q + r’)·(p + q’ + r)·(p’ + q + r) is the dual of the Boolean function, f = p’qr + pq’r + pqr' + pqr.

因此，典范化 SoP 形式和典范化 PoS 形式 Dual 彼此相同。在功能上，这两种形式相同。根据要求，我们可以使用这两种形式之一。

Therefore, both canonical SoP and canonical PoS forms are Dual to each other. Functionally, these two forms are same. Based on the requirement, we can use one of these two forms.

Standard SoP and PoS Forms

我们讨论了布尔输出的两种典范形式。类似地，还有两种标准形式来表示布尔输出。这些是典范形式的简化版本。

We discussed two canonical forms of representing the Boolean output(s). Similarly, there are two standard forms of representing the Boolean output(s). These are the simplified version of canonical forms.

  1. Standard SoP Form

  2. Standard PoS Form

我们将在后面的章节中讨论逻辑门。标准形式的主要 advantage 是施加到逻辑门上的输入数量可以最小化。有时，所需的逻辑门总数将减少。

We will discuss about Logic gates in later chapters. The main advantage of standard forms is that the number of inputs applied to logic gates can be minimized. Sometimes, there will be reduction in the total number of logic gates required.

Standard SoP Form

标准 SoP 形式表示 Standard Sum of Products 形式。在这种形式中，每个积项不必包含所有字面量。因此，积项也可以不是极小项。因此，标准 SoP 形式是典范化 SoP 形式的简化形式。

Standard SoP form means Standard Sum of Products form. In this form, each product term need not contain all literals. So, the product terms may or may not be the min terms. Therefore, the Standard SoP form is the simplified form of canonical SoP form.

我们将通过两个步骤得到输出变量的标准 SoP 形式。

We will get Standard SoP form of output variable in two steps.

  1. Get the canonical SoP form of output variable

  2. Simplify the above Boolean function, which is in canonical SoP form.

如果有多个输出变量，请对其他输出变量也遵循相同过程。有时，简化典范化 SoP 形式可能并不总是可行的。在这种情况下，典范化 SoP 形式和标准 SoP 形式是相同的。

Follow the same procedure for other output variables also, if there is more than one output variable. Sometimes, it may not possible to simplify the canonical SoP form. In that case, both canonical and standard SoP forms are same.

将以下布尔函数转换为标准 SoP 形式。

Convert the following Boolean function into Standard SoP form.

\mathrm{f \: = \: p’qr \: + \: pq’r \: + \: pqr' \: + \: pqr}

\mathrm{f \: = \: p’qr \: + \: pq’r \: + \: pqr' \: + \: pqr}

给定的布尔函数为规范 SoP 形式。现在，我们必须简化此布尔函数才能获得标准 SoP 形式。

The given Boolean function is in canonical SoP form. Now, we have to simplify this Boolean function in order to get standard SoP form.

Step 1 - 使用 Boolean postulate ，x + x = x。这意味着，与任何布尔变量执行 n 次逻辑或操作将等于同一变量。因此，我们可以再写两遍最后一项 pqr。

Step 1 − Use the Boolean postulate, x + x = x. That means, the Logical OR operation with any Boolean variable ‘n' times will be equal to the same variable. So, we can write the last term pqr two more times.

\mathrm{\Rightarrow \: f \: = \: p’qr \: + \: pq’r \: + \: pqr' \: + \: pqr \: + \: pqr \: + \: pqr}

\mathrm{\Rightarrow \: f \: = \: p’qr \: + \: pq’r \: + \: pqr' \: + \: pqr \: + \: pqr \: + \: pqr}

Step 2 - 对第 1 和第 4 项、第 2 和第 5 项、第 3 和第 6 项使用 Distributive law 。

Step 2 − Use Distributive law for 1st and 4th terms, 2nd and 5th terms, 3rd and 6th terms.

\mathrm{\Rightarrow \: f \: = \: qr(p' + p) \: + \: pr(q' + q) \: + \: pq(r' + r)}

Step 3 - 对每个括号中存在的项使用 Boolean postulate ，x + x' = 1 进行简化。

Step 3 − Use Boolean postulate, x + x' = 1 for simplifying the terms present in each parenthesis.

\mathrm{\Rightarrow \: f \: = \: qr(1) \: + \: pr(1) \: + \: pq(1)}

Step 4 - 对以上三项使用 Boolean postulate ，x.1 = x 进行简化。

Step 4 − Use Boolean postulate, x.1 = x for simplifying above three terms.

\mathrm{\Rightarrow \: f \: = \: qr \: + \: pr \: + \: pq}

\mathrm{\Rightarrow \: f \: = \: pq \: + \: qr \: + \: pr}

这是简化的布尔函数。因此，对应于给定的规范 SoP 形式的 standard SoP form 是 f = pq + qr + pr

This is the simplified Boolean function. Therefore, the standard SoP form corresponding to given canonical SoP form is f = pq + qr + pr

Standard PoS Form

标准 PoS 形式表示 Standard Product of Sums 形式。在此形式中，每个总项不必包含所有字面量。因此，总项可能是最大项，也可能不是。因此，标准 PoS 形式是规范 PoS 形式的简化形式。

Standard PoS form means Standard Product of Sums form. In this form, each sum term need not contain all literals. So, the sum terms may or may not be the Max terms. Therefore, the Standard PoS form is the simplified form of canonical PoS form.

我们将通过两步获得输出变量的标准 PoS 形式。

We will get Standard PoS form of output variable in two steps.

  1. Get the canonical PoS form of output variable

  2. Simplify the above Boolean function, which is in canonical PoS form.

如果存在多个输出变量，也对其他输出变量遵循相同的过程。有时，可能无法简化规范 PoS 形式。在这种情况下，规范 PoS 形式和标准 PoS 形式相同。

Follow the same procedure for other output variables also, if there is more than one output variable. Sometimes, it may not possible to simplify the canonical PoS form. In that case, both canonical and standard PoS forms are same.

将以下布尔函数转换为标准 PoS 形式。

Convert the following Boolean function into Standard PoS form.

f:=（p + q + r）（p + q + r'）（p + q' + r）（p' + q + r）

\mathrm{f \: = \: (p + q + r)\cdot(p + q + r')\cdot(p + q' + r)\cdot(p' + q + r)}

给定的布尔函数采用规范型 PoS 形式。现在，我们必须简化此布尔函数，以获得标准 PoS 形式。

The given Boolean function is in canonical PoS form. Now, we have to simplify this Boolean function in order to get standard PoS form.

Step 1 − 使用 Boolean postulate ，x · x = x。这意味着，与布尔变量“n”多次执行的逻辑 AND 操作将等于相同的变量。因此，我们可以再写两个第一项 p+q+r。

Step 1 − Use the Boolean postulate, x · x = x. That means, the Logical AND operation with any Boolean variable ‘n' times will be equal to the same variable. So, we can write the first term p+q+r two more times.

⇒ f := （p + q + r）·（p + q + r）·（p + q + r）·（p + q + r'）·（p +q' + r）·（p' + q + r）

\mathrm{\Rightarrow \: f \: = \: (p + q + r)\cdot(p + q + r)\cdot(p + q + r)\cdot(p + q + r')\cdot(p +q' + r)\cdot(p' + q + r)}

Step 2 − 将 Distributive law, x + (y · z) = (x + y)·(x + z) 用于第 1 和第 4 个括号、第 2 和第 5 个括号、第 3 和第 6 个括号。

Step 2 − Use Distributive law, x + (y · z) = (x + y)·(x + z) for 1st and 4th parenthesis, 2nd and 5th parenthesis, 3rd and 6th parenthesis.

⇒ f := （p + q + rr'）·（p + r + qq'）·（q + r + pp'）

\mathrm{\Rightarrow \: f \: = \: (p + q + rr')\cdot(p + r + qq')\cdot(q + r + pp')}

Step 3 − 使用 Boolean postulate ，x.x'=0 来简化每个括号中存在的项。

Step 3 − Use Boolean postulate, x.x'=0 for simplifying the terms present in each parenthesis.

⇒ f := （p + q + 0）·（p + r + 0）·（q + r + 0）

\mathrm{\Rightarrow \: f \: = \: (p + q + 0)\cdot(p + r + 0)\cdot(q + r + 0)}

Step 4 − 使用 Boolean postulate ，x + 0 = x 来简化每个括号中存在的项

Step 4 − Use Boolean postulate, x + 0 = x for simplifying the terms present in each parenthesis

⇒ f := （p + q）·（p + r）·（q + r）

\mathrm{\Rightarrow \: f \: = \: (p + q)\cdot(p + r)\cdot(q + r)}

⇒ f := （p + q）·（q + r）·（p + r）

\mathrm{\Rightarrow \: f \: = \: (p + q)\cdot(q + r)\cdot(p + r)}

这是简化的布尔函数。因此，对应于给定的规范型 PoS 形式的 standard PoS form 是 f = (p + q)·(q + r)·(p + r) 。这是布尔函数 f = pq + qr + pr 的 dual 。

This is the simplified Boolean function. Therefore, the standard PoS form corresponding to given canonical PoS form is f = (p + q)·(q + r)·(p + r). This is the dual of the Boolean function, f = pq + qr + pr.

因此，标准 SoP 和标准 PoS 形式彼此对偶。

Therefore, both Standard SoP and Standard PoS forms are Dual to each other.