4.1 Digital Electronic Principles Introduction | unit 4 digital electronic principles

BEX

Module 04

Digital Electronic Principles

4.1 Digital Electronic Principles: Introduction

4.2 Binary digits

A digital system understands positional number system where there are few symbols called digits and these symbol represents different values depending on their position in the number.

A value of the digit is determined by using

The digit

Its position

The base of the number system

Binary Number System

It uses two digits 0 and 1.

It is also called as base 2 number system.

Here, each position in any binary number represents a power of the base (2). Example: 23

The last position represents a y power of the base (2). Example: 2y where y represents the last position.

Example

Binary Number: 101112

Calculating the Decimal Equivalent of binary number −

Step	Binary Number	Decimal Number
Step 1	101012	((1 × 24) + (0 × 23) + (1 × 22) + (1 × 21) + (1 × 20))10
Step 2	101012	(16 + 0 + 4 + 2 + 1)10
Step 3	101012	2310

Note: 101112 is normally written as 10111.

4.3 Logic levels and Digital waveforms

Computers are digital in nature.

They process, store, and communicate information in binary form which is the combination of 1s and 0s.

A bit is an individual 1 or O.

Multiple bit streams are used in a computer network.

Here, signal varies in voltage to positive (+) voltage and a null or zero (0) voltage (unipolar) or a positive (+) or a negative (-) voltage (bipolar).

Waveforms Positive and Negative logic

In this a high-level H is represented for logic 1 and a low level L is represented for logic 0.

4.4 Introduction to basic Logic operation

Digital electronic circuits operate at two logic levels namely Logic Low and Logic High.

The Logic Low is represented with ‘0’and Logic High is represented with ‘1’.

The basic digital electronic circuit which has one or more inputs and single output is known as Logic gate.

4.5 Number system, Decimal numbers, Binary numbers

A digital system understands positional number system where there are few symbols called digits and these symbol represents different values depending on their position in the number.

A value of the digit is determined by using

The digit

Its position

The base of the number system

Decimal Number System

Its the system that we use in our daily life. It has a base 10 and uses 10 digits from 0 to 9.

Here, the successive positions towards the left of the decimal point represent units, tens, hundreds, thousands and so on.

Each and every position represents a specific power of the base (10). For example, the decimal number 4321 consists of the digit 1 in the units position, 2 in the tens position, 3 in the hundreds position, and 4 in the thousands position, and its value can be written as

(1×1000) + (2×100) + (3×10) + (4×l)

(1×103) + (2×102) + (3×101) + (4×l00)

1000 + 200 + 30 + 1

1234

As a computer programmer, we should understand the following number systems used in computers.

S.N.	Number System & Description
1	Binary Number System Base 2. Digits used: 0 and 1
2	Octal Number System Base 8. Digits used: 0 to 7
3	Hexa Decimal Number System Base 16. Digits used: 0 to 9, Letters used: A- F

Binary Number System

It uses two digits 0 and 1.

It is also called as base 2 number system.

Here, each position in any binary number represents a power of the base (2). Example: 23

The last position represents a y power of the base (2). Example: 2y where y represents the last position.

4.6 Decimal-to-Binary conversion

Example

Binary Number: 101112

Calculating the Decimal Equivalent of binary number −

Step	Binary Number	Decimal Number
Step 1	101012	((1 × 24) + (0 × 23) + (1 × 22) + (1 × 21) + (1 × 20))10
Step 2	101012	(16 + 0 + 4 + 2 + 1)10
Step 3	101012	2310

Note: 101112 is normally written as 10111.

2. Decimal Number: 2710

Calculating Binary Equivalent −

Step	Operation	Result	Remainder
Step 1	27 / 2	13	1
Step 2	13 / 2	6	1
Step 3	6 / 2	3	0
Step 4	3 / 2	1	1
Step 5	1 / 2	0	1

Hence, the remainders are arranged in the reverse order and we get:

Decimal Number − 2710 = Binary Number − 110112.

4.7 Simple binary arithmetic

It is an essential part of all the digital calculations.

Binary Addition

It is a basis for binary subtraction, multiplication, division.

There are four rules which are as follows:

Addition Table

Fig. Rules of Binary addition

In fourth step, a sum (1 + 1 = 10) i.e. 0 is written in the given column and a carry of 1 over to the next column is done.

For Example −

Addition Example

Fig. Binary addition

Binary Subtraction

Subtraction and Borrow, these are the two words that will be used very frequently for binary subtraction. There rules of binary subtraction are:

Subtraction Table

Fig. Rules of Binary Subtraction

For Example

Subtraction Example

Fig. Binary subtraction

Binary Multiplication

It is similar to decimal multiplication.

It is also simpler than decimal multiplication as only 0s and 1s are involved.

There are four rules of binary multiplication which are:

Multiplication Table

Fig. Rules of Binary Multiplication (Ref. 1)

For Example

Fig. Binary Multiplication

Binary Division

It is similar to decimal division.

It is also called as the long division procedure.

For Example

Division Example

Fig. Binary Division

4.8 Logic Gates

The basic gates are namely AND gate, OR gate & NOT gate.

AND gate

It is a digital circuit that consists of two or more inputs and a single output which is the logical AND of all those inputs. It is represented with the symbol ‘.’.

The following is the truth table of 2-input AND gate.

A	B	Y = A.B
0	0	0
0	1	0
1	0	0
1	1	1

Here A, B are the inputs and Y is the output of two input AND gate.

If both inputs are ‘1’, then only the output, Y is ‘1’. For remaining combinations of inputs, the output, Y is ‘0’.

The figure below shows the symbol of an AND gate, which is having two inputs A, B and one output, Y.

And Gate

Fig. : AND gate (ref. 1)

Timing Diagram:

OR gate

It is a digital circuit which has two or more inputs and a single output which is the logical OR of all those inputs. It is represented with the symbol ‘+’.

The truth table of 2-input OR gate is:

A	B	Y = A + B
0	0	0
0	1	1
1	0	1
1	1	1

Here A, B are the inputs and Y is the output of two input OR gate.

When both inputs are ‘0’, then only the output, Y is ‘0’. For remaining combinations of inputs, the output, Y is ‘1’.

The figure below shows the symbol of an OR gate, which is having two inputs A, B and one output, Y.

Fig. : OR gate (ref. 1)

Timing Diagram:

NOT gate

It is a digital circuit that has one input and one output. Here the output is the logical inversion of input. Hence, it is also called as an inverter.

The truth table of NOT gate is:

A	Y = A’
0	1
1	0

Here A and Y are the corresponding input and output of NOT gate. When A is ‘0’, then, Y is ‘1’. Similarly, when, A is ‘1’, then, Y is ‘0’.

The figure below shows the symbol of NOT gate, which has one input, A and one output, Y.

Fig. : NOT gate (ref. 1)

Timing Diagram:

4.9 The NAND NOR, Exclusive-OR and Exclusive-NOR gate

NAND gate

It is a digital circuit which has two or more inputs and single output and it is the inversion of logical AND gate.

The truth table of 2-input NAND gate is:

A	B	Y = (A.B)’
0	0	1
0	1	1
1	0	1
1	1	0

Here A, B are the inputs and Y is the output of two input NAND gate. When both inputs are ‘1’, then the output, Y is ‘0’. If at least one of the input is zero, then the output, Y is ‘1’. This is just the inverse of AND operation.

The image shows the symbol of NAND gate:

Fig.: NAND gate (ref. 1)

NAND gate works same as AND gate followed by an inverter.

Timing Diagram:

NOR gate

It is a digital circuit that has two or more inputs and a single output which is the inversion of logical OR of all inputs.

The truth table of 2-input NOR gate is:

A	B	Y = (A+B)’
0	0	1
0	1	0
1	0	0
1	1	0

Here A and B are the two inputs and Y is the output. If both inputs are ‘0’, then the output is ‘1’. If any one of the input is ‘1’, then the output is ‘0’. This is exactly opposite to two input OR gate operation.

The symbol of NOR gate is:

Fig.: NOR gate (ref. 1)

NOR gate works exactly same as that of OR gate followed by an inverter.

Timing Diagram:

Special Gates

Ex-OR gate

It stands for Exclusive-OR gate. Its function varies when the inputs have even number of ones.

The truth table of 2-input Ex-OR gate is:

A	B	Y = A⊕B
0	0	0
0	1	1
1	0	1
1	1	0

Here A, B are the inputs and Y is the output of two input Ex-OR gate. The output (Y) is zero instead of one when both the inputs are one.

Therefore, the output of Ex-OR gate is ‘1’, when only one of the two inputs is ‘1’. And it is zero, when both inputs are same.

The symbol of Ex-OR gate is as follows:

Fig.: XOR gate (ref. 1)

It is similar to that of OR gate with an exception for few combination(s) of inputs. Hence, the output is also known as an odd function.

Timing Diagram:

Ex-NOR gate

It stands for Exclusive-NOR gate. Its function is same as that of NOR gate except when the inputs having even number of ones.

The truth table of 2-input Ex-NOR gate is:

A	B	Y = A⊙B
0	0	1
0	1	0
1	0	0
1	1	1

Here A, B are the inputs and Y is the output. It is same as Ex-NOR gate with the only modification in the fourth row. The output is 1 instead of 0, when both the inputs are one.

Hence the output of Ex-NOR gate is ‘1’, when both inputs are same and 0, when both the inputs are different.

The symbol of Ex-NOR gate is:

Fig.: XNOR gate (ref. 1)

It is similar to NOR gate except for few combination(s) of inputs. Here the output is ‘1’, when even number of 1 is present at the inputs. Hence is also called as an even function.

Timing Diagram:

4.10 Boolean operations and expressions

It deals with binary numbers & variables.

Therefore, also known as Binary Algebra or logical Algebra.

A mathematician named as George Boole had developed this algebra in 1854.

The variables that are used in this algebra are known as Boolean variables.

Considering the range of voltages as logic ‘High’ is represented with ‘1’ and logic ‘Low’ is represented with ‘0’.

Postulates and Basic Laws of Boolean Algebra

Here, the Boolean postulates and basic laws that are used are given underneath.

Boolean Postulates

Considering the binary numbers 0 and 1, boolean variable (x) and its complement (x’).

They known as literal.

The possible logical OR operations are:

x + 0 = x

x + 1 = 1

x + x = x

x + x’ = 1

Similarly, the possible logical AND operations are:

x.1 = x

x.0 = 0

x.x = x

x.x’ = 0

These are the simple Boolean postulates and verification can be done by substituting the Boolean variable with ‘0’ or ‘1’.

4.11 Laws and Rules of Boolean algebra

Basic Laws of Boolean Algebra

The three basic laws of Boolean Algebra are:

Commutative law

Associative law

Distributive law

Commutative Law

The logical operation carried between two Boolean variables when gives the same result irrespective of the order the two variables, then that operation is said to be Commutative. The logical OR & logical AND operations between x & y are shown below

x + y = y + x

x.y = y.x

The symbol ‘+’ and ‘.’ indicates logical OR operation and logical AND operation.

Commutative law holds good for logical OR & logical AND operations.

Associative Law

If a logical OR operation of any two Boolean variables is performed first and then the same operation is performed with the remaining variable providing the same result, then that operation is said to be Associative. The logical OR & logical AND operations of x, y & z are:

x + (y + z) = (x + y) + z

x.(y.z) = (x.y).z

Associative law holds good for logical OR & logical AND operations.

Distributive Law

If a logical OR operation of any two Boolean variables is performed first and then AND operation is performed with the remaining variable, then that logical operation is said to be Distributive. The distribution of logical OR & logical AND operations between variables x, y & z are :

x.(y + z) = x.y + x.z

x + (y.z) = (x + y).(x + z)

Distributive law holds good for logical OR and logical AND operations.

These are the Basic laws of Boolean algebra and we can verify them by substituting the Boolean variables with ‘0’ or ‘1’.

4.12 De Morgan’s theorem

It is useful in finding the complement of Boolean function.

It states that “The complement of logical OR of at least two Boolean variables is equal to the logical AND of each complemented variable”.

It can be represented using 2 Boolean variables x and y as

(x + y)’ = x’.y’

The dual of the above Boolean function is

(x.y)’ = x’ + y’

Therefore, the complement of logical AND of the two Boolean variables is equivalent to the logical OR of each complemented variable.

Similarly, DeMorgan’s theorem can be applied for more than 2 Boolean variables also.

4.13 Boolean analysis of logic circuits

Numerical

Simplify the Boolean function,

f = p’qr + pq’r + pqr’ + pqr

Method 1

Given

f = p’qr + pq’r + pqr’ +pqr.

In first and second term r is common and in third and fourth terms pq is common.

So, taking out the common terms by using Distributive law we get,

⇒ f = (p’q + pq’)r + pq(r’ + r)

The terms present in first parenthesis can be simplified by using Ex-OR operation.

The terms present in second parenthesis is equal to ‘1’ using Boolean postulate we get

⇒ f = (p ⊕q)r + pq(1)

The first term can’t be simplified further.

But, the second term is equal to pq using Boolean postulate.

⇒ f = (p ⊕q)r + pq

Therefore, the simplified Boolean function is f = (p⊕q)r + pq

Method 2

Given f = p’qr + pq’r + pqr’ + pqr.

Using the Boolean postulate, x + x = x.

Hence we can write the last term pqr two more times.

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

Now using the Distributive law for 1st and 4th terms, 2nd and 5th terms, 3rdand 6th terms we get.

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

Using Boolean postulate, x + x’ = 1 and x.1 = x for further simplification .

⇒ f = qr(1) + pr(1) + pq(1)

⇒ f = qr + pr + pq

⇒ f = pq + qr + pr

Therefore, the simplified Boolean function is f = pq + qr + pr.

Hence we got two different Boolean functions after simplification of the given Boolean function. Functionally, these two functions are same. As per requirement, we can choose one of them.

Numerical

Find the complement of the Boolean function,

f = p’q + pq’.

Solution:

Using DeMorgan’s theorem, (x + y)’ = x’.y’ we get

⇒ f’ = (p’q)’.(pq’)’

Then by second law, (x.y)’ = x’ + y’ we get

⇒ f’ = {(p’)’ + q’}.{p’ + (q’)’}

Then by using, (x’)’=x we get

⇒ f’ = {p + q’}.{p’ + q}

⇒ f’ = pp’ + pq + p’q’ + qq’

Using x.x’=0 we get

⇒ f = 0 + pq + p’q’ + 0

⇒ f = pq + p’q’

Therefore, the complement of Boolean function, p’q + pq’ is pq + p’q’.

4.14 Standard forms of Boolean expressions

Four product combinations is obtained by combining two variables x and y with logical AND operation. They are called as min terms or standard product terms. The min terms are given as x’y’, x’y, xy’ and xy.

In the same way, four Boolean sum terms is obtained by combining two variables x and y with logical OR operation. They are called as Max terms or standard sum terms. The Max terms are given as x + y, x + y’, x’ + y and x’ + y’.

The following table represents the 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’

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 it 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 2n min terms and 2n Max terms.

Canonical SoP and PoS forms

A truth table comprises of a set of inputs and output(s).

If there are ‘n’ input variables, then there shall be 2n possible combinations comprising of zeros and ones.

So the value of every output variable depends on the combination of input variables.

Hence, each output variable have ‘1’ for some combination and ‘0’ for other combination of input variables.

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

Canonical SoP form

Canonical PoS form

Canonical SoP form

It means Canonical Sum of Products form.

In this, each product term contains all literals.

So that these product terms are nothing but the min terms.

Hence is also known as sum of min terms form.

Firstly, identification of the min terms is done and then the logical OR of those min terms is taken in order to get the Boolean expression (function) corresponding to that output variable.

This Boolean function will be in sum of min terms form.

Then following the same procedure for other output variables too.

Example

Considering 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 only four combinations of inputs.

The corresponding min terms are given as p’qr, pq’r, pqr’, pqr.

By doing logical OR, we get the Boolean function of output (f).

Hence, the Boolean function of output is,

f = p’qr + pq’r + pqr’ + pqr.

This is the desired canonical SoP form of output, f.

It can also be represented as:

f=m3+m5+m6+m7f=m3+m5+m6+m7

f=∑m(3,5,6,7)f=∑m(3,5,6,7)

First, we represented the function as sum of respective min terms and then, the symbol for summation of those min terms is used.

Canonical PoS form

It means Canonical Product of Sums form.

Here In this form, each sum term contains all literals.

These sum terms are the Max terms.

Hence, canonical PoS form is also known as product of Max terms form.

Identification of the Max terms for which the output variable is zero is done and then the logical AND of those Max terms is done in order to get the Boolean expression corresponding to that output variable.

This Boolean function is in the form of product of Max terms.

Following the same procedure for other output variables too.

Standard SoP and PoS forms

Standard SoP form

It stands for Standard Sum of Products form.

In this, each product term need not contain all literals.

So, the product terms can or cannot be the min terms.

Therefore, it is therefore the simplified form of canonical SoP form.

Standard SoP of output variable can be obtained by two steps.

Getting the canonical SoP form of output variable

Simplification the above Boolean function.

The same procedure is followed for other output variables too, if there is more than one output variable.

Numerical

Convert the Boolean function into Standard SoP form.

f = p’qr + pq’r + pqr’ + pqr

Solution:

Step 1 – By using the Boolean postulate, x + x = x and also writing the last term pqr two more times we get

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

Step 2 – By Using Distributive law for 1st and 4th terms, 2nd and 5th terms, 3rdand 6th terms.

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

Step 3 – Then Using Boolean postulate, x + x’ = 1 we get

⇒ f = qr(1) + pr(1) + pq(1)

Step 4 – hence using Boolean postulate, x.1 = x we get

⇒ f = qr + pr + pq

⇒ f = pq + qr + pr

This is the required Boolean function.

Standard PoS form

It stands for Standard Product of Sum form.

Here, each sum term need not contain all literals.

So, the sum terms can or cannot be the Max terms.

Therefore, it is the desired simplified form of canonical PoS form.

Standard PoS form of output variable is obtained by two steps.

Getting the canonical PoS form of output variable

Simplification of the above Boolean function.

The same procedure is followed for other output variables too.

Numerical

Convert the Boolean function into Standard PoS form.

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

Solution:

Step 1 – By using the Boolean postulate, x.x = x and writing the first term p+q+r two more times we get

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

Step 2 – Now by using Distributive law, x + (y.z) = (x + y).(x + z) for 1st and 4thparenthesis, 2nd and 5th parenthesis, 3rd and 6th parenthesis.

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

Step 3 − Applying Boolean postulate, x.x’=0 for simplifying of the terms present in each parenthesis.

⇒ f = (p + q + 0).(p + r + 0).(q + r + 0)

Step 4 − Using Boolean postulate, x + 0 = x we get

⇒ f = (p + q).(p + r).(q + r)

⇒ f = (p + q).(q + r).(p + r)

This is the simplified Boolean function.

Hence, both Standard SoP and Standard PoS forms are Dual to one another.

4.15 Basic adders

A Combinational circuit is a circuit in which we combine the different gates. For example, encoder, decoder, multiplexer, demultiplexer etc. Some of the characteristics are as follows −

The output of combinational circuit depends only on the levels present at input terminals at any instant of time.

It does not use any memory element. Therefore, the previous state of input does not have any effect on the present state of the circuit.

It can have n number of inputs and m number of outputs.

Block diagram

Block Diagram of combinational circuit