1. JNTUH UNIVERSITY COLLEGE OF ENGINEERING MANTHANI
DIGITALELECTRONICS
(Fundamentals)
By
Kumar Saliganti
DEPARTMENT OF
ELECTRICALAND ELECTRONICS ENGINEERING
2. DIGITALELECTRONICS
It’s a field of electronics involving the study of digital
signals and the engineering of the devices that use or
produce them.
What is signal?
A signal is a function that represents the variation of a
physical quantity with respect to any parameter.
(Independent quantity like Time, Distance).
3. • The system which process or works on the digital signals.
• Extensively used in computation of data processing,
control systems, communications & measurements .
DIGITAL SYSTEMS
LOGIC
DESIGN
CIRCUIT
DESIGN
SYSTEM
DESIGN
4. It has replaced many tasks of Analog systems.
Also digital systems are easy to design
Information storage is easy.
Accuracy & precision are greater.
It’s more versatile.
They are less affected by noise.
More digital circuitry can be fabricated on IC
chips.
Reliability is more.
ADVANTAGES OVER ANALOG SYSTEM
5. Many physical quantities are Analog in nature.
& outputs,
Which are often the inputs
continually monitored, operated and
controlled by a system.
When they are processed & expressed
digitally, we are really making a digital
approximation to an inherently analog
quantity.
LIMITATIONS OF DIGITALTECHNIQUES
The real world
isAnalog.
7. Introduction to Number Systems
Counting in Decimal and Binary
Decimal to Binary Conversion
Binary to Decimal Conversion
Decimal to Binary Conversion
Hexadecimal Numbers
Octal Numbers
NUMBER SYSTEMS
8. Information Representation
Elementary storage units inside computer are electronic switches.
Each switch holds one of two states: on (1) or off (0).
We use a bit (binary digit), 0 or 1, to represent the state.
ON OFF
In general, N bits can represent 2N different values.
bits are needed.
1 bit - represents up to 2 values (0 or 1)
2 bits - rep. up to 4 values (00, 01, 10 or 11)
3 bits - rep. up to 8 values (000, 001, 010. …, 110, 111)
4 bits - rep. up to 16 values (0000, 0001, 0010, …, 1111)
For M values, log2M
9. Positional Notations
Decimal number system, symbols = { 0, 1, 2, 3, …, 9 }
Position is important
Example:(7594)10 = (7x103) + (5x102) + (9x101) +(4x100)
In general, (anan-1… a0)10 = (an x 10n) + (an-1 x 10n-1) + … + (a0 x
100)
(2.75)10 = (2 x 100) + (7 x 10-1) + (5 x10-2)
In general, (anan-1… a0 . f1f2 … fm)10 = (an x 10n) + (an-1x10n-1) + …
+ (a0 x 100) + (f1 x 10-1) + (f2 x 10-2) + … + (fm x10-m)
10. Other Number Systems
Binary (base 2): weights in powers-of-2. Binary digits (bits): 0,1.
Decimal(base 10): weights in powers-of-10, Decimal digits: 0, 1,
2, 3, …, 9
Octal (base 8): weights in powers-of-8. Octal digits:
0,1,2,3,4,5,6,7
Hexadecimal (base 16): weights in powers-of-16. Hexadecimal
digits: 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F
18. • Groupinto3's startingat least significant symbol
• Write one octal digit for each group
Example: Convert
2
(10011111)2to( )8
010 011 111
3 7
Answer = (237)8
BINARY to OCTAL
if the numberof
bits is not evenly
divisible by 3, then
add0's at the most
significant end
20. BINARY to HEXADECIMAL
• Group into 4's starting at leastsignificantsymbol write1 hexa digit
for each group.
E.g Convert (1010111010)2 to( )16
10 1011 1010
A
2 B
Answer= (2BA)16
If the number of
bits is not evenly
divisible by 4,
then add 0's at
the most
significant end.
21. HEXADECIMAL to BINARY
0011
0101 1011
0001
Answer = (0011010110110001)2
• For each of the Hexa digit write its binary equivalent.
E.g: Convert (35B1)16 to( )2
3 5B1
use 4bits
to
represent
22. Steps
Convert octal number to its binaryequivalent.
Convert binary numberto its hexadecimal equivalent.
E.g.: Convert (736.35)8 to ( )16
7 3 6 . 3 5
000 111 011 110 . 011 101 00
1 D E . 7 4
OCTAL to HEXADECIMAL
ANSWER: (1DE.74 )16
23. Steps
1.Convert hexadecimalnumber to its binaryequivalent.
2.Convert binary number toits octal equivalent.
E.g. Convert (A4C.6 )16 = ( )8
HEXADECIMAL TO OCTAL
A 4 C . 6
1010 0100 1100 . 0110
00
5 1 1 4 . 3 0
ANSWER: (5114.30 )8
25. DECIMAL to ANYBASE
Decimal To any
Base
IntegerPart
Successive
Division Method
Fractional Part Successive
Multiplication
Method
26. Steps in Successive Division Method
1. Divide the integer part ofdecimal number by desired base number, store
Quotient (Q)and remainder (R).
2. Considerquotient asa new decimalnumber and repeat Step1until
quotient becomes 0.
3. List the remainders in the reverseorder.
Steps in Successive Multiplication Method
1. Multiply the fractional part of decimalnumber by desired base number.
2. Record the integer part of product as carry and fractionalpart as new fractional
part.
3. RepeatSteps1and 2 until fractional part of product becomes 0 or until you
have many digits as necessaryforyourapplication.
4. Read carriesdownwardsto get desired base number.
27. Decimal to Binary Conversion
Convert (108)10 into binary number
(108)10 = (1101100)2
Decimal to Octal Conversion
Convert (2477.64)10 into octal number
DECIMAL TO ANY BASE
(2477.64)10 = (4655.5075)8
28. Decimal to Hexadecimal Conversion
Convert (2479)10 into Hexadecimal number
(2479)10 = (9AF)16
29. 1’sCOMPLEMENT
The 1’s complement of a binary number is the number that
results when we change all 1’s to zeros and the zeros toones.
1 1 0 1 0 0 1 0
NOT OPEARATION
0 0 1 0 1 1 0 1
30. 2’s COMPLEMENT
The 2’s complement thebinary number that results when
add 1 to the 1’s complement. It is givenas,
2’s complement = 1’s complement + 1
Example: Express 35 in8-bit 2’s complement form.
35in8-bit form is 00100011
+
0 0 1 0 0 0 1 1
1 1 0 1 1 1 0 0
1
1 1 0 1 1 1 0 1
1’s complement
2’s complement
31. 9’s COMPLEMENT
The 9’s complement of a decimal digit isthe number that must be
addedto it to produce9. The complement of 3 is 6, the complement
of 7 is 2.
Example: Obtain9’s complementof 7493
Solution:
9 9 99
- 7 4 9 3
2 506 9’s complement
32. 10’s COMPLEMENT
The10’s complementof the given number is obtainedby adding1 to
the 9’s complement. It is given as,
10’s complement = 9’s complement + 1
Example: Obtain10’s complement of 6492
9999 3507
- 6 49 2 + 1
3 5 07 3 508 10’s complement
Solution:
34. BINARY SUBTRACTION
• The Subtraction Consists of Four Possible Elementary
Operations.
• In Case of Second Operation the Minu end bit is Smaller
than the Subtrahend bit, hence 1 is borrowed. 0-0=0
0-1=1(borrow 1)
1-0=1
1-1=0
0
- 0
1
0
1
1
0
1 1 1 1
1
1
1
Borro
w
EXAMPLE:
Operatio
ns
