Introduction to signals
The signals are broadly classified into
two categories:
1.
2.
Analog Signals.
Digital signals.

Analog Signals
These signals have infinite number of different magnitudes /
values.
They varies continuously with respect to time.
Ex= Sine wave , triangular wave .
+v
t
-v

Digital Signals
These signals have only finite number of predetermined distinct
magnitudes.
Ex= Binary signal.
1
t
Introduction to Analog & Digital system

Analog System
A system which process the analog signals
are called as Analog system.
Ex = Filter, Amplifier, Signal generator,
Motors speed controller.

Digital System
A system which work on digital signal is
called as digital system.
Ex = Register, Flip flop, Counters etc.
INTRODUCTION TO NUMBER SYSTEM
Def:= Number system defines a set of values used to represents
quantity.
Following are the different number systems.




Decimal Number system.
Binary Number system.
Octal Number system.
Hexadecimal Number system.
DECIMAL NUMBER SYSTEM




Contains 10 unique symbols 0,1,2,3,4,5,6,7,8 & 9.
The base / radix is 10.
Positional Weighted system & its positional weights are 100, 101
, 102 , 103 etc.
Ex. [1234]10
1) [1234]10 = 1*103 + 2*102 +3*101 +4*100
= 1234
2) [12.34]10 = 1*101 + 2*100 +3*10-1 +4*10-2
= 12.34
Decimal point
BINARY NUMBER SYSTEM





Contains only two symbols, 0 & 1.
The base / radix is 2.
Positional Weighted system & its positional weights are 23, 22 , 21
, 20 etc.
Binary digit is called as a BIT.
Ex. [1010]2
1) [1010]2 = 1*23 + 0*22 +1*21 +0*20
2) [10.10]2 = 1*21 + 0*20 +1*2-1 +0*2-2
OCTAL NUMBER SYSTEM





Contains eight symbols, 0,1,2,3,4,5,6 & 7.
The base / radix is 8.
Positional Weighted system & its positional weights are 83, 82 , 81
, 80 etc.
Octal numbers is 3 bit numbers.
Ex. [23]8
1) [23]8 = 2*81 +3*80
2) [23.16]8 = 2*81 +3*80 + 1*8-1 + 6*8-2
HEXADECIMAL NUMBER SYSTEM





Contains 16 symbols, 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E & F.
The base / radix is 16.
Positional Weighted system & its positional weights are 163, 162 ,
161 , 160 etc.
Hexadecimal numbers is 4 bit numbers.
Ex. [09AB]16
1) [09AB]16= 0*163 + 9* 162 + A*161 + B*160
2) [AB.12]16= A*161 + B* 160 + 1*16-1 + 2*16-2
NUMBER SYSTEM CONVERSION

1)
2)
3)
4)
5)
6)
7)
8)
9)
10)
11)
12)
There are 12 types of number system conversion.
Binary to Decimal number Conversion
Binary to Octal number Conversion
Binary to Hexadecimal number Conversion
Decimal to Binary number Conversion
Decimal to Octal number Conversion
Decimal to Hexadecimal number Conversion
Octal to Binary number Conversion
Octal to Decimal number Conversion
Octal to Hexadecimal number Conversion
Hexadecimal to Binary number Conversion
Hexadecimal to Decimal number Conversion
Hexadecimal to octal number Conversion
BINARY TO DECIMAL CONVERSION
Steps: 1) Write the Binary number.
2) Multiply each binary digit by its positional weight.
3) Add the product terms to obtain the decimal number.
Example:
1) [1011]2 = [ ]10
[1011]2 = 1*23 + 0*22 + 1*21 + 1*20
= 8 + 0 + 2 + 1 = 11
[1011]2 = [11]10
2) [10.10]2 = [ ]10
[10.10]2 = 1*21 + 0*20 + 1*2-1 + 0*2-2
= 2 + 0 + 0.5 + 0 = 2.5
[10.10]2 = [ 2.5 ]10
DECIMAL TO BINARY CONVERSION
The decimal integer is converted to binary integer by Successive
Division by 2 & the decimal fraction is converted into binary fraction
by Successive Multiplication by 2. This is known as Double-Dabble
Method.
1) Successive Division by 2
 The number is successively divided by 2 till the quotient is 0.
 Last Remainder is MSB.
 Remainders read from bottom to top, gives binary integer.
2) Successive Multiplication by 2
 The number is successively multiplied by 2 till the fraction part of the
product is 0 or till the desired accuracy.
 First Integer is MSB.
 Integers read from top to bottom, gives binary fraction.

DECIMAL TO BINARY CONVERSION
1) [10]10 = [ ]2
10/2
5/2
2/2
1/2
Q
5
2
1
0
2) [2.5]10 = [ ]2
a) For Integer part
Q
2/2
1
1/2
0
[2.5]10 =[10.1]2
R
0
1
0
1
[10]10 = [ 1010 ]2
b) For Fractional part
R
0
1
0.5*2 = 1. 0
1
OCTAL TO DECIMAL CONVERSION
Steps: 1) Write the Octal number.
2) Multiply each octal digit by its positional weight.
3) Add the product terms to obtain the decimal number.
Example:
1) [27]8 = [ ]10
[27]8 = 2×81 + 7×80
= 16 + 7 = 23
[27]8 = [ 23 ]10
2) [23.23]8 = [ ]10
= 2×81 + 3×80 + 2×8-1 + 3×8-2
= 19 + 0.29
[23.23]8 = [19.29] 10
DECIMAL TO OCTAL CONVERSION

The decimal integer is converted to octal integer by Successive
Division by 8 & the decimal fraction is converted into octal fraction by
Successive Multiplication by 8.
1) Successive Division by 8
 The number is successively divided by 8 till the quotient is 0.
 Last Remainder is MSB.
 Remainders read from bottom to top, gives octal integer.
2) Successive Multiplication by 8
 The number is successively multiplied by 8 till the fraction part of the
product is 0 or till the desired accuracy.
 First Integer is MSB.
 Integers read from top to bottom, gives octal fraction.
DECIMAL TO OCTAL CONVERSION
1) [19]10 = [ ]8
Q
19/8
2
2/8
0
[19]10 =[23]8
2) [0.296]10 =[ ]8
For Fractional part
0.296*8 = 2.368
2
0.368*8 = 2.994
2
0.944*8 = 7.552
7
[0.296]10 = [ 0.227 ]8
R
3
2
HEXADECIMAL TO DECIMAL CONVERSION
Steps: 1) Write the Hexadecimal number.
2) Multiply each hexa digit by its positional weight.
3) Add the product terms to obtain the decimal number.
Example:
1) [9.AB]16 = [ ]10
[9.AB]16= 9×160 + A×16-1 + B×16-2
= 9×1 + 10×16-1 + 11×16-2
= 9 + 0.625 + 0.0429
= [ 9.667 ] 10
DECIMAL TO HEXADECIMAL CONVERSION

The decimal integer is converted to hexadecimal integer by
Successive Division by 16 & decimal fraction is converted into hexa
fraction by Successive Multiplication by 16.
1) Successive Division by 16
 The number is successively divided by 16 till the quotient is 0.
 Last Remainder is MSB.
 Remainders read from bottom to top, gives hexadecimal integer.
2) Successive Multiplication by 16
 The number is successively multiplied by 16 till the fraction part of the
product is 0 or till the desired accuracy.
 First Integer is MSB.
 Integers read from top to bottom, gives hexadecimal fraction.
DECIMAL TO HEXADECIMAL CONVERSION
1)
2)
[95]10 =[ ]16
Q
95/16
5
5/16
0
[95]10 = [5F]16
R
15 (F)
5
[95.95]10 =[ ]16
a) For real part
[95]10 = [ 5F]16
[95.95]10 = [ 5F.F33 ]16
b) For Fractional part
0.95*16 = 15.2
15
0.2*16 = 3.2
3
0.2*16 = 3.2
3
[.95]10 = [ .F33 ]16
BINARY TO OCTAL CONVERSION
1)
2)
3)
Make a group of 3-bit.
Integer part: Start from LSB & moving towards MSB,
replace each group of 3-bit by its octal number.
Fractional part: Start from MSB & moving towards LSB,
replace each group of 3-bit by its octal number.
Example:
[001010101.101010100]2 = 001 010 101.101 010 100
= 1 2 5 . 5 2 4
= [125.524]8
OCTAL TO BINARY CONVERSION

Replace each octal digit by its 3 bit binary equivalent.
[23.36]8 = 2 3 . 3 6
= 010 011 . 011 110
= [010011.011110]2
BINARY TO HEXADECIMAL CONVERSION
1)
2)
3)
Make a group of 4-bit.
Integer part: Start from LSB & moving towards MSB, replace
each group of 4-bit by its hexadecimal number.
Fractional part: Start from MSB & moving towards LSB,
replace each group of 4-bit by its hexadecimal number.
[000111101010.101010101000]2= 0001 1110 1010 . 1010 1010 1000
= 1
E
A . A
A
8
= [1EA.AA8]16
HEXADECIMAL TO BINARY CONVERSION

Replace each hexadecimal digit by its 4 bit binary equivalent.
[ABC95.CDA]16 = A
B
C
9 . C
D
A
= 1010 1011 1100 1001 . 1100 1101 1010
= [1010101111001001.110011011010]2
HEXADECIMAL TO OCTAL CONVERSION
Step 1: Convert Hexadecimal number into Binary number.
Step 2: Convert binary number into Octal number.
Example:
1) [ABCD]16 = [ ]8
Step 1) Convert ABCD into binary equivalent.
[ABCD]16 = [1010 1011 1100 1101 ]2
Step 2) Convert binary number into octal equivalent.
[1010 1011 1100 1101 ]2 = [001 010 101 111 001 101 ]2
=[ 1 2
5 7 1 5]8
OCTAL TO HEXADECIMAL CONVERSION
Step 1: Convert Octal number into Binary number.
Step 2: Convert binary number into Hexadecimal number.
Example:
1) [235.12]8 = [ ]16
Step 1) Convert 235.12 into binary equivalent.
[235.12]8 = [010 011 101 . 001 010]2
Step 2) Convert binary number into Hexadecimal equivalent.
[010 011 101 . 001 010]2 = [1001 1101. 0010 1000]2
=[ 9
D . 2
8]16
BINARY CODED DECIMAL (BCD)

To express any decimal number in BCD, each decimal digit
should be replaced by the appropriate 4-bit code.

It is a weighted code.

There are 6 illegal combinations 1010, 1011, 1100, 1101, 1110
and 1111.

The rules of binary addition & subtraction do not apply to the
entire number but only to the individual 4-bit groups.
BINARY CODED DECIMAL (BCD)
Decimal No.
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Binary No.
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
Binary Coded Decimal (BCD)
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
0001 0000
0001 0001
0001 0010
0001 0011
0001 0100
0001 0101
BINARY CODED DECIMAL (BCD)
Ex: Give BCD Code for decimal number 874.
Decimal number: 874
BCD Code:
1000 0111 0100
Ex: Give BCD Code for decimal no. 96.42
Decimal number: 96.42
BCD Code:
1001 0110 . 0100 0010
EXCESS-3 CODE





It is also called as XS-3 Code.
It is Non-Weighted Code.
It is a Sequential code & can be used for arithmetic operation.
Excess-3 code word derived from the BCD code word by adding
(0011)2 or (3)10 to each word in BCD code.
Excess-3 code are obtained:
add (3)10

decimal no.
BCD code
Excess-3 code
It has six invalid states 0000, 0001, 0010, 1101, 1110 & 1111.
EXCESS-3 CODE
Eg. Obtain the Excess-3 code for [428]10
Given no.
4
2
8
Binary equivalent
0100
0010
1000
Excess-3 code
0111
0101
1011
GRAY CODE



It is a non-weighted code, not suitable for arithmetic operations.
Here only one bit in the code group changes, each time the
decimal no. is incremented.
As only one bit changes at a time, it is called as unit-distance
code.
Gray Code
Decimal
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Binary
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
Gray
0000
0001
0011
0010
0110
0111
0101
0100
1100
1101
1111
1110
1010
1011
1001
1000
ASCII CODE




ASCII stands for American Standard Code for Information
Interchange, is an alphanumeric code.
It is 7-bit code & the number of bit patterns are 2^7 = 128.
Used to encode both the uppercase & lowercase characters of
the alphabet, some special symbols & 10 decimal digits.
Extensively used for printers & terminals.
Binary to Gray Conversion

The MSB of gray code is same as first bit of binary no.
Moving from left to right, remaining bits are obtained by “EXORing” the adjacent pairs of binary no.
 Binary number= B4 B3 B2 B1 B0
Gray number= G4 G3 G2 G1 G0 then,
=> 0 + 0 = 0, 0 + 1 = 1, 1 + 0 = 1, 1 + 1 = 0

G4= B4
G3= B4
G2= B3
G1= B2
G0= B1
B4
+ B3
B3
B2
B1
B0
+ B2
+ B1
+ B0
G4
G3
G2
G1
G0
Binary to Gray Conversion
Ex: Convert [10101101]2 to its Gray code.
Solution:
1 + 0 + 1 + 0 + 1 + 1 +
0
1
1
1
1
1
1
0
+
1
Binary
1
Gray
Gray to Binary Conversion


The MSB of binary code is same as first bit of gray code.
Add by EX-OR method, each generated binary digit with gray
digit in the next adjacent position.
B4 = G4
B3 = B4
B2 = B3
B1 = B2
B0 = B1
+ G3
G4
G3
G2
G1
G0
+ G2
+ G1
+ G0
B4
B3
B2
B1
B0
Gray to Binary Conversion
Ex: Convert [10101101]G to its binary number.
Solution:
1
0
1
0
1
1
0
+
1
+
1
+
0
+
0
+
1
+
0
1
Gray
+
0
1 Binary
BINARY ARITHMETIC




Binary Addition
Binary Subtraction
Binary Multiplication
Binary division
Binary Addition
The rules for binary addition are:
0+0=0, 0+1=1, 1+0=1, 1+1=0 w/c 1
Example:
1)
10
+01
-------1 1 (no carry)
2) 1
11
+01
-------100
( 1 + 1 = 0 carry 1 to next column, then again 1 + 1
is 0 carry 1 to next column)
Binary Subtraction
The rules for binary subtraction are:
0-0=0, 0-1=1w/b 1 , 1-0=1, 1-1=0
Example:
100
- 001
------011
0 -1 = 1; with borrow of 1 from next column
0 -1(borrow) - 0 = 1, with borrow of 1
1 - 1(borrow) - 0 = 0.
Answer = 011.
Binary Multiplication
The rules for binary multiplication are:
0×0=0, 0×1=0 , 1×0=0, 1×1=1
Example: Multiply 100 by 101
100
×101
--------100
+000X
+100XX
----------------1010 0
Therefore, 100 × 101 = 10100
Binary Division
Long division procedures similar to those in decimals
are used.
Example: Divide 101101 by 110
1 1 0 ) 1 0 1 1 0 1 ( 1 1 1 . 1
1 1 0
--------------------1 0 1 0
1 1 0
----------------------1 0 0 1
1 1 0
----------------------1 1 0
1 1 0
-----------------------0 0 0
Therefore, 101101 ÷ 110 = 111.1
BCD ADDITION

1)
2)
3)
4)
Rules for addition of 2 BCD numbers are:
Add the 2 numbers using the rules for binary addition.
If a four bit number is equal to or less than 9, it is a valid BCD
number.
If a 4-bit sum is greater than 9, or if a carry-out of the group is
generated, it is an invalid result. Then add 6 (0110)2 to the four
bit sum in order to skip the six invalid states.
If a carry results, when 6 is added, add the carry to the next 4bit group.
BCD ADDITION
Example:
0001 1000 0110
+ 0110 1000 1000
-----------------------------------------1000 0000 1110
+
110 0110
-----------------------------------------1000 0111 0100
------------- (874)10
BCD SUBTRACTION

1)
2)
3)
4)
Rules for subtraction of 2 BCD numbers are:
Subtract the 2 numbers using the rules for binary subtraction.
If a four bit number is equal to or less than 9, it is a valid BCD
number.
if a borrow is taken from other four bit code, it is an invalid
result. Then subtract 6 (0110)2 to the four bit result in order to
skip the six invalid states.
If a borrow results, when 6 is subtracted, subtract the borrow to
the next 4-bit group.
BCD SUBTRACTION
Example:
1000 0110 0100
- 0010 1001 0011
-----------------------------------------0101 1101 0001
110
-----------------------------------------0101 0111 0001
------------- (571)10