Digital Design Digital Sytems Number Systems

Digital design

1. 19Z204-Digital Design
Dr.S.Sivaranjani,AP-CSE

2. Course Topics(Unit –I)
•IntroductiontoDigitalSystems
•Numbersystems
•Number-Baseconversion
•ComplementsofNumbers(DiminishedRadix
complement,RadixComplement)
•SignedBinaryNumbers
•ArithmeticoperationwiththeBinaryNumbers
•BinaryCodes(BCD,8421code,Graycode,ASCII)
Dr.S.Sivaranjani,AP-CSE

3. •DigitalSystemsarepartofyourdailyactivities.
•EverythinginDigital(HighResolutionDisplay):Camera,
Laptop,Mobilephones,Tab-weareusingDigitalElectronics
Dr.S.Sivaranjani,AP-CSE
Data processing
Data transmission
Device control
Digital
Systems
Made or build using
Introduction to Digital Systems

4. •DigitalDeviceshaveGUI[GraphicalUser
Interface]
•Enablethedevicestoexecutecommandsthat
appeartotheusertobesimplebutwhich
involvespreciseexecutionofasequenceof
complexinternalinstructions
•DigitalSystemsarefast,accurateand
consumelesspower.
Dr.S.Sivaranjani,AP-CSE

5. Digital Systems
•Stemmingfromword“Digit”[means0,1,2,3…]
•Usedigitalsignalsorinformationtooperate
•DigitalSignals:arediscretesignalsthat
representinformationintheformofbinary
digits(bits)whichcantakeonlyoneoftwo
values(0&1)
Dr.S.Sivaranjani,AP-CSE

6. Processed
Stored
Transmitted
Dr.S.Sivaranjani,AP-CSE
Digital
Signals
Can be
Digital
Devices
Using
Smartphones,
Computers,Tablets&
other Electronic devices

7. Components of Digital Systems
•These components work together to process, store and transmit digital
information
Dr.S.Sivaranjani,AP-CSE
Digital
System
I/P Devices
[used to provide input to
the system in the form of
digital signals, such as
switches, sensors, and
keyboards]
Output Devices
[used to display or transmit
the processed information,
such as monitors, printers,
and speakers]
Processors
[The processing devices
perform the required logic
and arithmetic operations
on the input signals, such
as microprocessors and
microcontrollers. ]
Memory
Storage Devices
[storagedevicesareused
tostoredataand
instructions,suchas
memorychipsandhard
drives]

8. What is Digital System?
Dr.S.Sivaranjani,AP-CSE
Digital System
Digital Input Digital output
Discrete numbers
Process Digital information.
Processing will happen using Digital Logic

9. •Adigitalsystemisasystemthatprocesses
informationinadigitalform.
•Itconsistsofelectronicdevicesthat
manipulatedigitalsignals,suchasbinary
code,usingasetoflogicgatesandalgorithms.
•Thesedevicesincludemicroprocessors,
microcontrollers,digitalsignalprocessors,and
otherdigitalintegratedcircuits.
Dr.S.Sivaranjani,AP-CSE

10. Applications of digital system
Digitalsystemshaveawiderangeofapplicationsinvariousfields,including:
Communicationsystems:
•Digitalsystemsareusedincommunicationsystems,suchascellular
phones,satellites,andtheinternet.
•Theyenablereliabletransmissionandreceptionofdataandvoicesignals
overlongdistances.
Controlsystems:
•Digitalsystemsareusedincontrolsystems,suchasrobotics,process
control,andautomotivecontrolsystems.
•Theyallowforprecisecontrolofdevicesandprocesses,ensuringoptimal
performance.
Digitalsignalprocessing:
•Digitalsystemsareusedindigitalsignalprocessingapplications,suchas
audioandvideoprocessing,imageandspeechrecognition,andradarand
sonarsystems.
Dr.S.Sivaranjani,AP-CSE

11. Medical devices:
•Digitalsystemsareusedinmedicaldevices,such
aspacemakers,MRImachines,anddigitalX-ray
machines.
•Theyenableaccurateandreliablediagnosticand
treatmentprocedures.
Consumerelectronics:
•Digitalsystemsareusedinconsumerelectronics,
suchassmartphones,tablets,laptops,andsmart
TVs.
•Theyprovideawiderangeoffunctionsand
capabilities,suchasmultimediaplayback,
internetaccess,andgaming.
Dr.S.Sivaranjani,AP-CSE

12. Industrial automation:
•Digitalsystemsareusedinindustrialautomation
systems,suchasmanufacturingplants,assembly
lines,andlogisticssystems.
•Theyenableefficientandautomatedcontrolof
productionandlogisticsprocesses.
Defenseandaerospace:
•Digitalsystemsareusedindefenseand
aerospaceapplications,suchasmilitary
communicationsystems,guidanceandcontrol
systemsforaircraftandmissiles,andsatellite
communicationandnavigationsystems.
Dr.S.Sivaranjani,AP-CSE

13. Number System
Dr.S.Sivaranjani,AP-CSE

14. General Representation
•A number is represented as
a
5a
4a
3a
2a
1a
0.a
-1a
-2a
-3a
-4
•. -radix point
•a
j–coefficients(symbols used in a number representation)
•j –place value
•Example:
–Decimal number: 567.28
Dr.S.Sivaranjani,AP-CSE

15. Radix or Base
•Theradixofanumbersystemisalsoknownas
itsbaseoritsnumericalbase.
•Itreferstothenumberofuniquedigitsor
symbolsusedinthesystemtorepresent
numbers.
•Forexample,thedecimalsystemthatweuse
ineverydaylifehasaradixof10becauseit
uses10digits(0-9)torepresentnumbers.
Dr.S.Sivaranjani,AP-CSE

16. •Theradixofanumbersystemisanintegergreaterthan
1.
•Thevalueoftheradixdeterminestherangeofvalues
thatcanberepresentedusingthesystem,aswellas
thewayinwhichnumbersarewrittenand
manipulated.
•Forexample,inabinarysystemwitharadixof2,there
areonlytwopossibledigits(0and1),whichmeans
thatallnumbersarerepresentedusingonlythesetwo
digits.
•Differentnumbersystemsareusedindifferent
contexts,andeachsystemhasitsownradix.
Dr.S.Sivaranjani,AP-CSE

17. Common number systems and their
radices
•Binarysystem(radix2):Usestwodigits(0and1)to
representnumbers.
•Octalsystem(radix8):Useseightdigits(0-7)to
representnumbers.
•Decimalsystem(radix10):Usestendigits(0-9)to
representnumbers.
•Hexadecimalsystem(radix16):Usessixteendigits(0-9
andA-F)torepresentnumbers.
Understandingtheradixofanumbersystemis
importantforconvertingbetweendifferentsystems
andforunderstandinghowcomputersrepresentand
manipulatenumbers.
Dr.S.Sivaranjani,AP-CSE

18. Radix or Base
•In apositional numeral system(value of each
symbol depends on the position),
theradixorbaseis the number of
uniquedigits, including the digit zero, used to
represent numbers.
•Eg: Decimal number system:
–Uses 10 symbols (0,1,2,3,4,5,6,7,8,9)
–Hence radix or base = 10
Dr.S.Sivaranjani,AP-CSE

19. Positional Number System
•Another representation of a number:
a
nr
n
+a
n-1r
n-1
+…+a
1r
1
+a
0r
0
+a
-1r
-1
+a
-2r
-2
+…+a
-mr
-m
•a
j–coefficients
•j –place value/ positional value
•r –radix or base
•Eg: 567.28(decimal number)
–567.28 = 5x10
2
+6x10
1
+7x10
0
+2x10
-1
+8x10
-2
Dr.S.Sivaranjani,AP-CSE

20. Radix -2 (Binary Numbers)
•Symbols Used: 0,1
•Base: 2
Dr.S.Sivaranjani,AP-CSE

21. Radix -8 (Octal Numbers)
•Symbols Used: 0,1,2,3,4,5,6,7
•Base: 8
Dr.S.Sivaranjani,AP-CSE

22. Radix –10 (Decimal Numbers )
•Symbols Used: 0,1,2,3,4,5,6,7,8,9
•Base: 10
Dr.S.Sivaranjani,AP-CSE

23. Radix –16 (Hexadecimal Numbers)
•Symbols Used: 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F
•Base: 16
Dr.S.Sivaranjani,AP-CSE

24. Example
•The number “twenty-seven” can be represented in
different ways :
–IIIII IIIIIIIIIIIIIIIIIIIIII (sticks or unary code)
–27 (radix-10 or decimal code)
–11011 (radix-2 or binary code)
–XXVII (roman numerals)
•The use of radix-2 ( binary) numbers became popular with
the onset of electronic computers,
–binary digits or bits, having only two possible values 0 and 1, is
compatible with electronic signals
•Radix-8 (octal) and radix-16 (hexadecimal) numbers have
been used as shorthand notation for binary numbers.
Dr.S.Sivaranjani,AP-CSE

25. General representation
•Usually first 10 symbols in a number system of radix
r is represented by the symbols of decimal number
system and for the 11
th
symbol it starts with the
symbols of English alphabets.
–Eg:
•Radix-6 , Symbols: 0,1,2,3,4,5
•Radix-19, Symbols:0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F,G,H,I.
•To distinguish a number represented in a particular
number system, the number is usually written by
enclosing it in a parenthesis with a subscript of r.
–Eg:
•(56)
8 –octal
•(56)
10–decimal
•(1011)
2-binary
Dr.S.Sivaranjani,AP-CSE

26. Conversion between Number Systems
(Conversion from base-r to decimal)
•Conversion of any base-r number to decimal:
–Multiply each digit with its weight(radix raised to
its positional value) to get the resultant value of
each symbol.
–Add all the resultant symbol values.
Dr.S.Sivaranjani,AP-CSE

27. Conversion between Number Systems
(Binary to decimal number)
•100.111
(100.111)
2=1x2
2
+0x2
1
+0x2
0
+1x2
-1
+1x2
-2
+1x2
-3
= 4+0+0+0.5+0.25+0.125
= (4.875)
10
Dr.S.Sivaranjani,AP-CSE

28. To convert the binary number 11011.11 to decimal,
we can use the following method:
•Separate the integer and fractional parts of the binary number:
•Integer part: 11011
•Fractional part: 0.11
1.Convert the integer part to decimal by using the positional notation
of the binary system:11011 in binary equals
1 x 2^4 + 1 x 2^3 + 0 x 2^2 + 1 x 2^1 + 1 x 2^0= 16 + 8 + 0 + 2 + 1= 27
in decimal
2.Convert the fractional part to decimal by using the positional
notation of the binary system:
0.11 in binary equals 1 x 2^-1 + 1 x 2^-2= 0.5 + 0.25= 0.75 in decimal
3.Add the decimal values of the integer and fractional parts to get
the final decimal value:27 + 0.75= 27.75 in decimal
Therefore, the binary number 11011.11 is equal to the decimal
number 27.75.
Dr.S.Sivaranjani,AP-CSE

29. Conversion between Number Systems
(Octal to decimal number)
•517.35
(517.35)
8=5x8
2
+1x8
1
+7x8
0
+3x8
-1
+5x8
-2
= 320+8+7+0.375+0.078125
= (335.453125)
10
Dr.S.Sivaranjani,AP-CSE

30. Conversion between Number Systems
(Hexadecimal to decimal number)
•786.BC
(786.BC)
16=7x16
2
+8x16
1
+6x16
0
+Bx16
-1
+Cx16
-2
= 1792+128+6+0.6875+0.046875
= (1926.734375)
10
Dr.S.Sivaranjani,AP-CSE

31. Conversion between Number Systems
(Conversion from decimal to base-r)
•Conversion of a decimal integer to a number in
base r:
–Divide the integer and all its successive quotients by r
–accumulate the remainder in reverse order
–Range of remainder: 0 to r-1
•Conversion of a decimal fraction to a number in
base r:
–Multiply the fraction by r and its successive remainder
by r
–accumulate the quotients in the same order
–Range of coefficients: 0 to r-1
Dr.S.Sivaranjani,AP-CSE

32. Conversion between Number Systems
(decimal to binary number)
•343.392
(343)
10= (101010111)
2
0.392x2=0.784
0.784x2=1 .568
0.568x2=1.136
0.136x2=0 .272
0.272x2=0 .544
0.544x2=1.088
2343
2171-1
285-1
242-1
221-0
210-1
2 5-0
2 2-1
1-0
(0.392)
10= (0.011001…)
2
Ans: (343.392)
10= (101010111.011001)
2
Dr.S.Sivaranjani,AP-CSE

33. Conversion between Number Systems
(decimal to octal number)
•153.513
(153)
10= (231)
8
0.513x8=4.104
0.104x8=0 .832
0.832x8=6.656
0.656x8=5 .248
0.248x8=1 .984
0.984x8=7.872
8153
819-1
2-3
(0.513)
10= (0.406517…)
8
Ans: (153.513)
10= (231.406517)
8
Dr.S.Sivaranjani,AP-CSE

34. Conversion between Number Systems
(decimal to hexadecimal number)
•487.365
(487)
10= (1E7)
16
0.365x16=5.84
0.84x16=13 .44
0.44x16=7.04
0.04x16=0 .64
0.64x16=10 .24
0.24x16=3.84
16487
1630-7
1-E
(0.365)
10= (0.5D70A3…)
16
Ans: (487.365)
10= (1E7.5D70A3)
16
Dr.S.Sivaranjani,AP-CSE

35. Conversion from Binary to Octal and
Octal to Binary
•Conversion of a binary number to a octal number:
–Keep splitting the binary number into 3 bits from right to
left before radix point and left to right after radix point.
–If the leftmost part before radix point has lesser than 3
bits, add 0s to fill the places.
–If the rightmost part after radix point has lesser than 3 bits,
add 0s to fill the places.
–Write the corresponding octal symbol for each 3 bits and
accumulate them.
•Conversion of a octal number to a binary number:
–Write the 3 bit binary equivalent of each octal number and
accumulate them.
Dr.S.Sivaranjani,AP-CSE

36. Conversion from Binary to Octal
•10110001101011.111100000110
–Splitting the numbers into 3 bits:
010 110 001 101 011 . 111 100 000 110
2 6 1 5 3 . 7 4 0 6
–Answer:
(10110001101011.111100000110)
2=(26153.7406)
8
Dr.S.Sivaranjani,AP-CSE

37. Conversion from Octal to Binary
•673.124
–Writing 3-bit binary equivalent for each number.
6 7 3 . 1 2 4
110 111 011 001 010 100
–Answer: (673.124)
8=(110111011.001010100)
2
Dr.S.Sivaranjani,AP-CSE

38. Conversion from Binary to Hexadecimal
and Hexadecimal to Binary
•Conversion of a binary number to a hexadecimal number:
–Keep splitting the binary number into 4 bits from right to left
before radix point and left to right after radix point.
–If the leftmost part has lesser than 4 bits, add 0s to fill the
places.
–If the rightmost part after radix point has lesser than 4 bits, add
0s to fill the places.
–Write the corresponding hexadecimal symbol for each 4 bits and
accumulate them.
•Conversion of a hexadecimal number to a binary number:
–Write the 4 bit binary equivalent of each hexadecimal number
and accumulate them.
Dr.S.Sivaranjani,AP-CSE

39. Conversion from Binary to Hexadecimal
•10110001101011.111100000110
–Splitting the numbers into 4 bits:
0010 1100 0110 1011 . 1111 0000 0110
2 C 6 B . F 0 6
–Answer:
(10110001101011.111100000110)
2=(2C6B.F06)
16
Dr.S.Sivaranjani,AP-CSE

40. Conversion from Hexadecimal to Binary
•306.D
–Writing 4-bit binary equivalent for each number.
3 0 6 . D
0011 0000 0110 . 1101
–Answer: (306.D)
16=(001100000110.1101)
2
Dr.S.Sivaranjani,AP-CSE

41. Dr.S.Sivaranjani,AP-CSE

42. Dr.S.Sivaranjani,AP-CSE
Complements of Numbers
(Diminished Radix complement ,Radix
Complement)

43. Complement
•There are two types of complements for each
base-r number system:
–Radix complement (or) r’scomplement
–Diminished radix complement (or) (r-1)’s
complement
•Complements are used in digital computers
for simplifying the subtraction operation.
Dr.S.Sivaranjani,AP-CSE

44. r-1’s Complement/ Diminished Radix
Complement
•The (r-1)’s complement of N:
(r
n
-1)-N
N –Number
r –radix /base
n –number of digits in N
(r
n
-1) is the largest number with n digits in base r
Hence, subtraction is between the number N from the largest number
with n digits.
Dr.S.Sivaranjani,AP-CSE

45. r-1’s Complement of Binary Number/
1’s Complement
•N = 1011000; r=2
n = 7 ; r –1 = 1 (1’s complement)
1’s complement = (2
7
-1)
10-(1011000)
2
= ( 128 -1)
10–(1011000)
2
= ( 127)
10–(1011000)
2
= (1111111)
2–(1011000)
2
= 0100111
•N = 0101101 ; r=2
n =7
1’s complement = 1111111 –0101101
= 1010010
Largest 7 digit binary number
Dr.S.Sivaranjani,AP-CSE

46. 1’s complement –Another Method
•1’s complement of a binary number is formed
by
–changing 1’s to 0’s and 0’s to 1’s
•Eg: N = 1011000
–1’s complement is 0100111
Dr.S.Sivaranjani,AP-CSE

47. r-1’s Complement of Octal Number/
7’s Complement
•N = 563 ; r = 8
n= 3; r-1 = 7 (7’s complement)
7’s complement = (8
3
-1)
10-(563)
8
= (512-1)
10-(563)
8
= (511)
10-(563)
8
= (777)
8-(563)
8
= 214 Largest 3 digit octal number
Dr.S.Sivaranjani,AP-CSE

48. r-1’s Complement of Decimal Number/
9’s Complement
•N = 546700 ; r = 10
n= 6; r-1 = 9 (9’s complement)
9’s complement = (10
6
-1)
10–(546700)
10
= (1000000-1)
10–(546700)
10
= (999999)
10–(546700)
10
= 453299 Largest 6 digit decimal
number
Dr.S.Sivaranjani,AP-CSE

49. r-1’s Complement of HexdecimalNumber/
15’s Complement
•N = C3DF ; r= 16
n= 4; r-1 = 15 (15’s complement)
15’s complement = (16
4
-1)
10–(C3DF)
16
= (65536-1)
10–(C3DF)
16
= (65535)
10–(C3DF)
16
= (FFFF)
16–(C3DF)
16
= 3C20
Largest 4 digit hexadecimal
number
Dr.S.Sivaranjani,AP-CSE

50. r’sComplement/Radix Complement
•The r’scomplement of N:
r
n
-N for N ≠ 0
0 for N = 0
N –Number
r –radix /base
n –number of digits in N
•r’scomplement is obtained by adding 1 to (r-1)’s complement:
r
n
-N = [(r
n
-1)-N]+1
•Note: It is better to do (r-1)’s complement first for r’scomplement since it is
easy to do the subtraction in (r-1)’s complement(no borrow problem!).
Dr.S.Sivaranjani,AP-CSE

51. r’sComplement of Binary Number/ 2’s
Complement
•N = 1011000 ; r=2
n = 7 ; r = 2 (2’s complement)
1’s complement = (1111111)
2–(1011000)
2
= 0100111
2’s complement = 1’s complement + 1
= 0100111 + 1
= 0101000
Dr.S.Sivaranjani,AP-CSE

52. 2’s complement –Another Method
•2’s complement of a binary number is formed by
–Scan the numbers from right to left
–Till the first ‘1’ is found write the digits as such.
–After the first ‘1’, invert all the digits.
•Eg: N = 1011000
–Writing the digits till 1
st
‘1’ from right to left :
---1 0 0 0
–Inverting the rest of the numbers
0 1 0 1 0 0 0
–Hence 2’s complement is : 0101000
•Eg: N = 1011001
–Writing the digits till 1
st
‘1’ from right to left :
------1
–Inverting the rest of the numbers
0 1 0 0 1 1 1
–Hence 2’s complement is : 0100111
Dr.S.Sivaranjani,AP-CSE

53. r’sComplement of Octal Number/
8’s Complement
•N = 563 ; r = 8
n= 3; r = 8 (8’s complement)
7’s complement = (777)
8-(563)
8
= 214
8’s complement = 7’s complement + 1
= 214 +1
= 215
Dr.S.Sivaranjani,AP-CSE

54. r’sComplement of Decimal Number/
10’s Complement
•N = 546700 ; r = 10
n= 6; r = 10 (10’s complement)
9’s complement = (999999)
10–(546700)
10
= 453299
10’s complement = 9’s complement + 1
= 453299 + 1
= 453300
Dr.S.Sivaranjani,AP-CSE

55. r’sComplement of Hexadecimal Number/
16’s Complement
•N = C3DF ; r= 16
n= 4; r = 16 (16’s complement)
15’s complement = (FFFF)
16–(C3DF)
16
= 3C20
16’s complement = 15’s complement + 1
= 3C20 + 1
= 3C21
Dr.S.Sivaranjani,AP-CSE

56. Complement of a number with radix
point
•If the original number N contains a radix point,
–Temporarily remove the point to perform
complement.
–The radix point is then restored to the
complemented number in the same relative
position.
Dr.S.Sivaranjani,AP-CSE

57. Complement with radix point (Base-2)
•N = 1101.011
1101.011= 1101011 x 2
-3
1’s complement of 1101011=1111111-1101011
= 0010100
2’s complement of 1101011 = 0010100 + 1
= 0010101
1’s complement of 1101.011= 0010100 x 2
-3
= 0010.100
2’s complement of 1101.011 = 0010101 x2
-3
= 0010.101
Dr.S.Sivaranjani,AP-CSE

58. Complement with radix point (Base-8)
•N = 323.64
323.64= 32364x 8
-2
7’s complement of 32364 =77777-32364
= 45413
8’s complement of 32364 = 45413 + 1
= 45414
7’s complement of 323.64 = 45413 x 8
-2
= 454.13
8’s complement of 323.64 = 45414x8
-2
= 454.14
Dr.S.Sivaranjani,AP-CSE

59. Complement with radix point (Base-10)
•N = 325.93
325.93= 32593x 10
-2
9’s complement of 32593 =99999-32593
=67406
10’s complement of 32593 = 67406 + 1
= 67407
9’s complement of 325.93 = 67406 x 10
-2
= 674.06
10’s complement of 325.93 = 67407 x10
-2
= 674.07
Dr.S.Sivaranjani,AP-CSE

60. Complement with radix point (Base-16)
•N = ABC.3E2
ABC.3E2 = ABC3E2 x 16
-3
15’s complement of ABC3E2 =FFFFFF-ABC3E2
= 543C1D
16’s complement of ABC3E2 = 543C1D + 1
= 543C1E
15’s complement of ABC.3E2 = 543C1D x 16
-3
= 543.C1D
16’s complement of ABC.3E2 = 543C1E x16
-3
= 543.C1E
Dr.S.Sivaranjani,AP-CSE

61. Subtraction with Complements
r’sComplement Subtraction
•The subtraction of 2 n-digit unsigned numbers
M-N in base r can be done as follows:
–Add the minuend M to the r’scomplement of the
subtrahend N. This performs: M+(r
n
-N)=M-N+r
n
–If M≥ N, the sum will produce an end carry,r
n
,
which can be discarded. Hence the result is M-N
–If M< N, the sum does not produce an end carry
and is equal to r
n
-(N-M), which is the r’s
complement of N-M
Dr.S.Sivaranjani,AP-CSE

62. r’sComplement Subtraction(Base-2)
•1010100-1000011
2’s complement of 1000011=1111111-1000011+1 = 0111101
Answer =0010001
•1000011-1010100
2’s complement of 1010100=1111111-1010100+1 = 0101100
Answer =1101111 (or) -0010001
1 0 1 0 10 0
0 1 1 1 1 0 1
10 0 1 0 0 0 1
1 0 0 0 0 1 1
0 1 0 1 1 0 0
1 10 1 1 1 1
Dr.S.Sivaranjani,AP-CSE

63. r’sComplement Subtraction(Base-8)
•342-614
8’s complement of 614=777-614+1 = 164
Answer =526 (or) -252
•614-342
8’s complement of 342=777-342+1 = 436
Answer =252
3 4 2
1 6 4
5 2 6
6 1 4
4 3 6
12 5 2
Dr.S.Sivaranjani,AP-CSE

64. r’sComplement Subtraction(Base-10)
•72532-3250
10’s complement of 03250=99999-03250+1 = 96750
Answer =69282
•3250-72532
10’s complement of 72532=99999-72532+1 = 27468
Answer =30718 (or) -69282
7 2 5 3 2
9 6 7 5 0
16 9 2 8 2
0 3 2 5 0
2 7 4 68
3 0 7 1 8
Dr.S.Sivaranjani,AP-CSE

65. r’sComplement Subtraction(Base-16)
•CB2-672
16’s complement of 672=FFF-672+1 =98E
Answer =640
•672-CB2
16’s complement of CB2=FFF-CB2+1 = 34E
Answer =9C0 (or) -640
C B 2
9 8 E
16 4 0
6 7 2
3 4 E
9 C 0
Dr.S.Sivaranjani,AP-CSE

66. Subtraction with Complements
r-1’s Complement Subtraction
•The subtraction of 2 n-digit unsigned numbers M-N
in base r can be done as follows:
–Add the minuend M to the r’scomplement of the
subtrahend N.
–If M≥ N, the sum will produce an end carry, which is added
to the result since it produces a sum that is 1 less than the
correct difference(only if carry is generated).
•Removing the end carry and adding 1 to the sum is referred to as
an end-around carry.
–If M< N, the sum does not produce an end carry,whichis
the r-1’s complement of N-M
Dr.S.Sivaranjani,AP-CSE

67. r-1’s Complement Subtraction(Base-2)
•1010100-1000011
1’s complement of 1000011=1111111-1000011 = 0111100
Answer =0010001
•1000011-1010100
1’s complement of 1010100=1111111-1010100 = 0101011
Answer =1101110 (or) -0010001
1 0 0 0 0 1 1
0 1 0 1 0 1 1
1 10 1 1 1 0
1 0 1 0 10 0
0 1 1 1 1 0 0
10 0 1 0 0 0 0
1
0 0 1 0 0 0 1
Dr.S.Sivaranjani,AP-CSE

68. r-1’s Complement Subtraction(Base-8)
•402-314
7’s complement of 314=777-314 = 463
Answer =066
•314-402
7’s complement of 402=777-402 = 375
Answer =711 (or) -066
3 14
3 7 5
7 1 1
4 0 2
4 6 3
10 6 5
1
0 6 6
Dr.S.Sivaranjani,AP-CSE

69. r-1’s Complement Subtraction(Base-10)
•4567-1234
9’s complement of 1234=99999-1234= 8765
Answer =3333
•1234-4567
9’s complement of 4567=99999-4567= 5432
Answer =6666 (or) -3333
1 2 3 4
5 4 3 2
6 6 6 6
4 5 6 7
8 7 6 5
13 3 3 2
1
3 3 3 3
Dr.S.Sivaranjani,AP-CSE

70. r-1’s Complement Subtraction(Base-16)
•B06-C7C
15’s complement of C7C=FFF-C7C = 383
Answer =E89 (or) -176
•C7C-B06
15’s complement of B06=FFF-B06 = 4F9
Answer =176
B 0 6
3 8 3
E 8 9
C 7 C
4 F 9
11 7 5
1
1 7 6
Dr.S.Sivaranjani,AP-CSE

71. Signed Binary Numbers
Dr.S.Sivaranjani,AP-CSE

72. Signed Binary Number Representations
•Both signed and unsigned binary numbers
consists of a string of bits when represented in
computer.
•User determines whether a number is signed or
not.
•Representation Types:
–Signed-magnitude representation
–Signed-Complement representation
•1’s Complement
•2’s Complement
Dr.S.Sivaranjani,AP-CSE

73. Sign-Magnitude Representation
•The number consists of two parts:
Sign bit (leftmost bit)
Magnitude bits (other than leftmost bit)
•If the leftmost bit is
0 –positive number
1 –negative number
•The negative number has the same magnitude bits as the
corresponding positive number but the sign bit is 1 rather than 0.
•Eg: 8-bit representation of ‘fifteen’
[phone]
-15 –1 0001111
It is used in ordinary arithmetic but usually not in computer
arithmetic, since sign and magnitude bits must be handled
separately.
Dr.S.Sivaranjani,AP-CSE

74. 1’s Complement Representation
•The negative number is the 1’s complement of
the corresponding positive number.
•Has some difficulties while used for arithmetic
operations.
•It is used in logical operations.
•There are two different representations for
zero.(i.e) 0000 and 1111 (4 bit +0 and -0).
•Eg: 8-bit representation of ‘fifteen’
[phone]
-15 –11110000
Dr.S.Sivaranjani,AP-CSE

75. 2’s Complement Representation
•The negative number is the 2’s complement of
the corresponding positive number.
•This is the most common representation used
in computer arithmetic
•Eg: 8-bit representation of ‘fifteen’
[phone]
-15 –11110001
Dr.S.Sivaranjani,AP-CSE

76. 2’s Complement Representation
Note: leftmost bit of the
representation acts a the sign bit (0
for positive values, 1 for negative
ones)
Dr.S.Sivaranjani,AP-CSE

77. Conversion of decimal numbers to
signed binary numbers
•Express decimal number -39 as 8-bit number in (a)sign-
magnitude (b)1’s complement and (c)2’s complement
representations.
–8-bit representation for +39
00100111
–(a)8-bit Sign magnitude representation for -39:
10100111
–(b) 8-bit 1’s complement representation for -39:
11011000
–(c) 8-bit 2’s complement representation for -39:
11011001
First represent the corresponding
positive number in the given number
of bits. Else the minimum number of
bits required to represent that
particular number should be taken.
Then use that
number
represented in
the required
number of bits to
find the negative
representation
Dr.S.Sivaranjani,AP-CSE

78. Conversion of a signed binary number
to decimal number
Determine the decimal value of signed binary number expressed
in sign-magnitude representation.
•10010101
–Computing the weights of rightmost 7 bits:
0x2
6
+0x2
5
+1x2
4
+0x2
3
+1x2
2
+0x2
1
+1x2
0
= 16+4+1=21
–Sign bit(leftmost bit) is 1.Hence it’s a negative number
–Therefore, the decimal number is -21
Dr.S.Sivaranjani,AP-CSE

79. Conversion of a signed binary number
to decimal number
Determine the decimal value of signed binary number expressed
in 1’s complement representation.
•00010111
–Computing the weights of the bits with the weight of the leftmost bit
as negative:
-0x2
7
+0x2
6
+0x2
5
+1x2
4
+0x2
3
+1x2
2
+1x2
1
+1x2
0
= 16+4+2+1= +23
-Therefore, the decimal number is +23
•11101000 (complement of the previous question)
–Computing the weights of the bits with the weight of the leftmost bit
as negative:
-1x2
7
+1x2
6
+1x2
5
+0x2
4
+1x2
3
+0x2
2
+0x2
1
+0x2
0
= -128+64+32+8=-24
-Adding 1 to the result (i.e)-24+1=-23
-Therefore, the decimal number is -23
Negative numbers alone add
1 if 1’s complement
representationDr.S.Sivaranjani,AP-CSE

80. Conversion of a signed binary number
to decimal number
Determine the decimal value of signed binary number expressed
in 2’s complement representation.
•01010110
–Computing the weights of the bits with the weight of the leftmost bit
as negative:
-0x2
7
+1x2
6
+0x2
5
+1x2
4
+0x2
3
+1x2
2
+1x2
1
+0x2
0
= 64+16+4+2= +86
-Therefore, the decimal number is +86
•10101010 (complement of the previous question)
–Computing the weights of the bits with the weight of the leftmost bit
as negative:
-1x2
7
+0x2
6
+1x2
5
+0x2
4
+1x2
3
+0x2
2
+1x2
1
+0x2
0
= -128+32+8+2=-86
-Therefore, the decimal number is -86
Need not add 1 like 1’s
complement representation
Dr.S.Sivaranjani,AP-CSE

81. 3-bit representation of signed numbers
NoPossible
3-bit
represent
ations
Ifonly
positive
numbers
represented
Ifnegative
numbers also
should be
represented
(sign-
magnitude)
Ifnegative
numbers also
should be
represented
(1’s
complement)
Ifnegative
numbers also
should be
represented
(2’s
complement)
1000 0 +0 +0 0
2001 1 +1 +1 +1
3010 2 +2 +2 +2
4011 3 +3 +3 +3
5100 4 -0 -3 -4
6101 5 -1 -2 -3
7110 6 -2 -1 -2
8111 7 -3 -0 -1
With the available combinations of binary numbers for a given number of bits,
positive and negative numbers must be represented(FOR SIGNED NUMBERS)!Dr.S.Sivaranjani,AP-CSE

82. Arithmetic operation with the Binary Numbers
Dr.S.Sivaranjani,AP-CSE

83. Binary Addition
•Addition
–The two numbers in addition:
•Augend
•Addend
–Result is:
•Sum
–Eg:
–Note: Generally, negative numbers are considered in 2’s
complement representation
0 0 0 00 1 1 1 Augend
+0 0 0 0 0 1 0 0 Addend
0 0 0 01 0 1 1 Sum
Dr.S.Sivaranjani,AP-CSE

84. Binary Addition
•Both numbers are positive (consider 8 bit)
•Positive number with magnitude larger than
negative number
0 0 0 00 1 1 1 (+7)
+0 0 0 0 0 1 0 0+(+4)
0 0 0 01 0 1 1 11
0 0 0 01 1 1 1 (+15)
[phone]+(-6)
10 0 0 0 1 0 0 1 9
Discard CarryDr.S.Sivaranjani,AP-CSE

85. Binary Addition
•Negative number with magnitude larger than
positive number
•Both numbers are negative
0 0 0 10 0 0 0 (+16)
[phone]+(-24)
1 1 1 11 0 0 0 -8
11 1 11 0 1 1 (-5)
[phone]+(-9)
11 1 1 10 0 1 0 -14
Discard Carry
Dr.S.Sivaranjani,AP-CSE

86. Binary Addition
•Overflow condition
–When two numbers are added and the number of bits required to
represent that sum exceeds the number of bits in the two numbers, an
overflow condition occurs.
–It can occur only if both numbers are positive or both numbers are
negative.
–If the sign bit of the result is different than the sign bit of the numbers
that are added, overflow is indicated.
0 1 1 11 1 0 1 (+125)
+0 0 1 11 0 1 0+(+58)
1 0 1 10 1 1 1 183
Incorrect Sign bit
Dr.S.Sivaranjani,AP-CSE

87. Binary Subtraction
•Special case of addition is subtraction
•Subtraction
–The two numbers in subtraction:
•Minuend
•Subtrahend
–Result is:
•Difference
•Eg:
1 0 1 1 0 1Minuend
+1 0 0 1 1 1Subtrahend
0 0 0 1 1 0Difference
Dr.S.Sivaranjani,AP-CSE

88. Binary Subtraction
•The sign of the number is changed by taking
2’s complement
•To subtract 2 numbers
–take the 2’s complement of the subtrahend and
add.
–Discard any final carry
Dr.S.Sivaranjani,AP-CSE

89. Binary Subtraction
•00001000-00000011
–2’s complement of 00000011=11111101
0 0 0 01 0 0 0 (+8)
[phone]+(-3)
10 0 0 00 1 0 1 +5
Dr.S.Sivaranjani,AP-CSE

90. Binary Multiplication
•Multiplication
–The numbers in multiplication:
•Multiplicand
•Multiplier
–Result is:
•Product
•(Partial Product)
Dr.S.Sivaranjani,AP-CSE

91. Binary Multiplication
•1011 x 101
1 0 1 1Multiplicand(11)
x 1 0 1Multiplier(5)
1 0 1 1Partial Product
0 0 0 0 Partial Product
1 0 1 1 Partial Product
1 1 0 1 1 1Product(55)
Dr.S.Sivaranjani,AP-CSE

92. Binary Division
•Division
–The numbers in division:
•Dividend
•Divisor
–Result is:
•Quotient
•Remainder
Dr.S.Sivaranjani,AP-CSE

93. Binary Division
•11000101 ÷1010
1 0 0 1 1 Quotient(19)
10101 1 0 0 0 1 0 1Divident(197)
1 0 1 0
0 1 0 0
0 0 0 0
1 0 0 1
0 0 0 0
10 0 1 0
1 0 1 0
1 0 0 0 1
1 0 10
1 1 1 Remainder(7)
Divisor(10)
Dr.S.Sivaranjani,AP-CSE

94. Binary Codes
( BCD, 8421 code,Graycode,ASCII)
Dr.S.Sivaranjani,AP-CSE

95. Binary Codes
•Any discrete elements of information that is distinct
among a group of quantities can be represented with
binary codes
•Sample Binary Codes:
–Binary Coded Decimal(BCD)/8421
–Gray Code
–Excess-3 Code
–2421 Code
–ASCII Code
.
.
.
Dr.S.Sivaranjani,AP-CSE

96. Binary Coded Decimal(BCD)/8421
•Straight binary assignment of the decimal numbers.
•It is a weighted code (codeswhich obey the
positionalweightprinciple.)
•10 decimal digits requires 4 bits for representation. But 6
out of 16 4-bit possible combination remains unassigned.
–A number with k decimal digits will require 4k bits in BCD
•Eg:
–(185)
10= (0001 1000 0101)
BCD = (10111001)
2
•Applications: Digital clocks, digital meters, Seven segment
display etc…(simplify the display of decimal numbers)
•This code is not very efficient but useful if only limited
processing is required.
Dr.S.Sivaranjani,AP-CSE

97. Decimal & BCD
DECIMAL BCD
0 0 0 0 0
1 0 0 0 1
2 0 0 1 0
3 0 0 1 1
4 0 1 0 0
5 0 1 0 1
6 0 1 1 0
7 0 1 1 1
8 1 0 0 0
9 1 0 0 1
10 00 0 1 0 0 0 0
Dr.S.Sivaranjani,AP-CSE

98. BCD Addition
•Add 2 BCD numbers using the rules for binary
addition
•If a 4-bit sum is equal or less than 9, it’s a valid
BCD number
•If a 4-bit sum is greater than 9 or if a carry out of
the 4-bit group is generated, it is an invalid result.
–Add 6(0110) to the 4-bit sum in order to skip the
invalid states.
–If a carry results when 6 is added, simply add the carry
to the next 4-bit group
Dr.S.Sivaranjani,AP-CSE

99. BCD Addition
•0110 0111 + 0101 0011
0 1 1 00 1 1 1 67
+0 1 0 10 0 1 1+53
1 0 1 11 0 1 0
0 1 1 00 1 1 0
0 0 0 10 0 1 00 0 0 0 120
Invalid BCD
Invalid BCD
Dr.S.Sivaranjani,AP-CSE

100. BCD Addition
•The sign of the decimal number is represented
using 4 bits
–0000 represents positive
–1001 represents negative
•Sign-magnitude is seldom used in computers
•Sign-complement uses 9’s or 10’s complement
Dr.S.Sivaranjani,AP-CSE

101. BCD Addition (10’s complement)
•(+ 0011 0111 0101) + (-0010 0100 0000)
–0011 0111 0101 –> 375
–0010 0100 0000 –> 240
–10’s complement of 0240 = 9999-0240+1= 9760
–Answer:
(+0011 0111 0101) + (-0010 0100 0000) = (+0001 0011 0101)
0 3 7 5
9 7 6 0
10 1 3 5
Discard Carry
Sign
Dr.S.Sivaranjani,AP-CSE

102. BCD Subtraction
•At first the decimal equivalent of the given
Binary Coded Decimal (BCD) codes are found
out.
•Then the 10’s compliment of the subtrahend is
done and then that result is added to the
number from which the subtraction is to be
done.
•Discard Carry if generated
•Note: If 9’s complement is used, carry is added to the
result of subtraction!
Dr.S.Sivaranjani,AP-CSE

103. BCD Subtraction (9’s complement)
•0101 0001 − 0010 0001
–0101 0001 -> 51
–0010 0001 -> 21
–9’s complement of 21 = 99-21 = 78
–30 -> 0011 0000
–Answer: 0101 0001 − 0010 0001 = 0011 0000
5 1
+7 8
12 9
1
3 0
Carry
Carry generated added to
result
Dr.S.Sivaranjani,AP-CSE

104. Gray Code
•Un-weighted code
•Important Feature:
–It exhibits one a single
bit change from one
code word to the next
in sequence
DecimalBinary Gray
0 0 0 0 00 0 0 0
1 0 0 0 100 0 1
2 0 0 1 00 0 1 1
3 0 0 1 10 0 1 0
4 0 1 0 00 1 1 0
5 0 1 0 10 1 1 1
6 0 1 1 00 1 0 1
7 0 1 1 10 1 0 0
8 10 0 01 1 0 0
9 1 0 0 11 1 0 1
10 1 0 10 11 1 1
11 1 0 1 11 1 1 0
12 1 1 0 01 0 1 0
12 1 1 0 11 0 1 1
14 1 1 1 01 0 0 1
15 1 1 1 11 0 0 0
Dr.S.Sivaranjani,AP-CSE

105. Binary to Gray Code Conversion
•The MSB in the gray code is the same as the
corresponding MSB in the binary number
•Going from left to right, add each adjacent
pair of binary code bits to get the next gray
code bit.
•Discard Carries.
Dr.S.Sivaranjani,AP-CSE

106. Binary to Gray Code Conversion
•Binary number = 10110
•(10110)
2=(11101)
Gray
•Note: Can perform XOR operation instead of addition. Hence
need not think about carry!
1 0 1 1 0 Binary
+ + + +
1 1 1 0 1 Gray
1+0 0+1 1+1 1+0
Dr.S.Sivaranjani,AP-CSE

107. Gray to Binary Code Conversion
•MSB in the binary code is the same as the
corresponding bit in the gray code.
•Add each binary code bit generated to the
next gray code bit in the next adjacent
position.
•Discard carries.
Dr.S.Sivaranjani,AP-CSE

108. Gray to Binary Code Conversion
•Gray code = 11101
•(11101)
Gray= (10110)
2
•Note: Can perform XOR operation instead of addition. Hence
need not think about carry!
1 1 1 0 1 Gray
+ + + +
1 0 1 1 0 Binary
1+1 0+1 1+0 1+1
Dr.S.Sivaranjani,AP-CSE

109. Excess-3 Code
•Un-weighted Code
•Codes are obtained from
the corresponding decimal
value plus 3 in 4-bit binary
•Self complementing
code(9’s complement of a
number is directly obtained
by changing 1s to 0s and 0s
to 1s).
–9’s complement of 4 is 5.
–9’s complement of 3 is 6
Decimal BCD Excess-3
[(BCD+3) in
binary]
0 0 0 0 0 0 0 1 1
1 0 0 0 1 0 1 0 0
2 0 0 1 0 0 1 0 1
3 0 0 1 1 0 1 1 0
4 0 1 0 0 0 1 1 1
5 0 1 0 1 10 0 0
6 0 1 1 0 1 0 0 1
7 0 1 1 1 1 0 1 0
8 10 0 0 1 0 1 1
9 1 0 0 1 1 10 0
10 0001 000001000011
Dr.S.Sivaranjani,AP-CSE

110. Converting Decimal to Excess-3
•(23)
10
–Add both the digits separately by 3
•2+3 =5
•3+3 = 6
–Convert each corresponding decimal number to
equivalent 4 bit binary code
•5 –0101
•6 –0110
–Answer: (23)
10=(0101 0110)
XS-3
Dr.S.Sivaranjani,AP-CSE

111. Converting Decimal to Excess-3
•(359.8)
10
–Add both the digits separately by 3
•3+3 =6
•5+3 = 8
•9+3 =12
•8 +3=11
–Convert each corresponding decimal number to equivalent
4 bit binary code
•6 –0110
•8 –1000
•12 –1100
•11 –1011
–Answer: (359.8)
10=(0110 1000 1100.1011)
XS-3
Dr.S.Sivaranjani,AP-CSE

112. American Standard Code for
Information Interchange (ASCII)
•An alphanumeric character set is a set of
elements that includes
–10 decimal digits
–26 letters of alphabets
–a number of special characters
•Uses 7 bit code and 128 characters
–Eg: A -1000001
Dr.S.Sivaranjani,AP-CSE

113. ASCII
•It contains
–94 graphic characters that can be printed
•26 uppercase letters( A –Z)
•26 lowercase letters( a –z)
•10 numerals ( 0 –9 )
•32 special printable characters such as %,*,$ …
–34 non-printable characters used for various
control functions.
Totally 128
(94+34) characters
Dr.S.Sivaranjani,AP-CSE

114. ASCII
•Control Characters
–Used for routing data and arranging printed text into prescribed
format
–Three types of control characters
•Format Effectors
–Characters that affect the layout of printing
–They include word processor and type writer controls:
»Backspace (BS)
»Horizontal tabulation (HT)
»Carriage return (CR)
•Information Separators
–Separate data into divisions such as paragraphs and pages
–They include:
»Record separator(RS)
»File separator(FS)
Dr.S.Sivaranjani,AP-CSE

115. ASCII
•Communication Control Characters
–Useful during the transmission of text between remote
devices so that it can be distinguished from other messages
using the same communication channel before it and after it
»Start of Text (STX)
»End of Text (ETX)
•Mostly computers manipulate an 8-bit quantity as a
single quantity. Hence it is stored as one ASCII
character per byte
–Extra bit is used for other purposes depending on the
application
Dr.S.Sivaranjani,AP-CSE

116. ASCII TABLE
Dr.S.Sivaranjani,AP-CSE