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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.

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’.

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

x | y | Min terms | Max terms |
---|---|---|---|

0 | 0 | m_{0}=x’y’ |
M_{0}=x + y |

0 | 1 | m_{1}=x’y |
M_{1}=x + y’ |

1 | 0 | m_{2}=xy’ |
M_{2}=x’ + y |

1 | 1 | m_{3}=xy |
M_{3}=x’ + y’ |

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.

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 2^{n} min terms and 2^{n} Max terms.

A truth table consists of a set of inputs and output(s). If there are ‘n’ input variables, then there will be 2^{n} 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.

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

- Canonical SoP form
- Canonical PoS form

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.

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.

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

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 |

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).

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.

$$f = m_{3}+m_{5}+m_{6}+m_{7}$$

$$f = \sum m\left ( 3,5,6,7 \right )$$

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 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.

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.

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

**Example**

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).

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.

$$f=M_{0}.M_{1}.M_{2}.M_{4}$$

$$f=\prod M\left ( 0,1,2,4 \right )$$

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.

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.

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.

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.

- Standard SoP form
- Standard PoS form

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 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.

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

- Get the canonical SoP form of output variable
- Simplify the above Boolean function, which is in canonical SoP form.

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.

**Example**

Convert the following Boolean function into Standard SoP form.

f = p’qr + pq’r + pqr’ + pqr

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** − 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.

⇒ f = p’qr + pq’r + pqr’ + pqr + pqr + pqr

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

⇒ f = qr(p’ + p) + pr(q’ + q) + pq(r’ + r)

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

⇒ f = qr(1) + pr(1) + pq(1)

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

⇒ f = qr + pr + pq

⇒ 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 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.

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

- Get the canonical PoS form of output variable
- Simplify the above Boolean function, which is in canonical PoS form.

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.

**Example**

Convert the following Boolean function into Standard PoS form.

f = (p + q + r).(p + q + r’).(p + q’ + r).(p’ + q + r)

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** − 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)

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

⇒ f = (p + q + rr’).(p + r + qq’).(q + r + pp’)

**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)

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

⇒ f = (p + q).(p + r).(q + r)

⇒ f = (p + q).(q + r).(p + r)

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.

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

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