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JNTUH UNIVERSITY COLLEGE OF ENGINEERING MANTHANI
DIGITALELECTRONICS
(Fundamentals)
By
Kumar Saliganti
DEPARTMENT OF
ELECTRICALAND ELECTRONICS ENGINEERING
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).
• 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
 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
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.
APPLICATIONS
BUSINESS
TRANSACTIONS
INDUSTRIAL
APPLICA
TIONS
MEDICAL
TREA
TMENT
DIGITAL TV,
DISKS,
TELEPHONES,
CAMERAS &
COMPUTERS
WEATHER
MONITORING
TRAFFIC
CONTOL &
COMMUNICA
TION
SYSTEM
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
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 
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)
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
BINARY NUMBER SYSTEM
DIGITS
BASE
0,1
2
1*21
1*20
1011
1*23
0*22
DECIMALNUMBER SYSTEM
0,1,2,3,4,5,6,7,8,9
DIGITS
10
BASE
5*10
7245
4*10
7*10
2*10
3
0
OCTAL NUMBER SYSTEM
DIGITS
8
BASE
0,1,2,3,4,5,6,7
7*80
4*83
4637
6*82
3*81
HEXADECIMAL NUMBER
SYSTEM
DIGITS
BASE
• 0,1,2,3,4,5,6,7,
• 8,9,A,B,C,D,E,F
16
6*161
A*160
B*16
2
2*16
3
2B6A
CONVERSION AMONG BASES
Hexadecimal
Decimal Octal
Binary
• 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
Answer = (011111110)2
OCTAL to BINARY
Example: Convert (376)8
376
011 111
110
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.
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
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
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
ANY BASE to DECIMAL
Converting fromany base to decimal is done by
multiplying each digit byits weight and summing.
(1101.101)2 = 123 + 122 + 120 + 12-1 + 12-3
= 8 + 4 + 1 + 0.5 + 0.125 = (13.625)10
(572.6)8 = 582 + 781 + 280 + 68-1
= 320 + 56 + 2 + 0.75 = (378.75)10
(2A.8)16 = 2161 + 10160 + 816-1
= 32 + 10 + 0.5 = (42.5)10
(341.24)5 = 352 + 451 + 150 + 25-1 + 45-2
= 75 + 20 + 1 + 0.4 + 0.16 = (96.56)10
DECIMAL to ANYBASE
Decimal To any
Base
IntegerPart
Successive
Division Method
Fractional Part Successive
Multiplication
Method
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.
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
Decimal to Hexadecimal Conversion
Convert (2479)10 into Hexadecimal number
(2479)10 = (9AF)16
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
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
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
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:
BINARY ADDITION
EXAMPLE:
+
1
0
1
1
0
1
0
0
1 0 0 1 0
Carry
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
THANK YOU!

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ADE UNIT-III (Digital Fundamentals).pptx

  • 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
  • 11.
  • 15. HEXADECIMAL NUMBER SYSTEM DIGITS BASE • 0,1,2,3,4,5,6,7, • 8,9,A,B,C,D,E,F 16 6*161 A*160 B*16 2 2*16 3 2B6A
  • 16.
  • 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
  • 19. Answer = (011111110)2 OCTAL to BINARY Example: Convert (376)8 376 011 111 110
  • 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
  • 24. ANY BASE to DECIMAL Converting fromany base to decimal is done by multiplying each digit byits weight and summing. (1101.101)2 = 123 + 122 + 120 + 12-1 + 12-3 = 8 + 4 + 1 + 0.5 + 0.125 = (13.625)10 (572.6)8 = 582 + 781 + 280 + 68-1 = 320 + 56 + 2 + 0.75 = (378.75)10 (2A.8)16 = 2161 + 10160 + 816-1 = 32 + 10 + 0.5 = (42.5)10 (341.24)5 = 352 + 451 + 150 + 25-1 + 45-2 = 75 + 20 + 1 + 0.4 + 0.16 = (96.56)10
  • 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