hexadecimal numbers code
DESCRIPTION
WRITING one’s and zero’s can be error prone when dealing with large numbers. IBM came up with the following method of dealing with these numbers. BASE 16 with digits 0 - 15 in base 10 are represented by the following notation in Hexadecimal. 0 16 = 0000 2 = 0 10 8 16 = 1000 2 = 8 10 - PowerPoint PPT PresentationTRANSCRIPT
Introduction to MicroprocessorsNumber Systems and Conversions
No. 1-1 9/6/00
HEXADECIMAL NUMBERS
CodeWRITING one’s and zero’s can be error prone when dealing with large numbers. IBM came up with the following method of dealing with these numbers. BASE 16 with digits 0 - 15 in base 10 are represented by the following notation in Hexadecimal.
016 = 00002 = 010 816 = 10002 = 810
116 = 00012 = 110 916 = 10012 = 910
216 = 00102 = 210 A16 = 10102 = 1010
316 = 00112 = 310 B16 = 10112 = 1110
416 = 01002 = 410 C16 = 11002 = 1210
516 = 01012 = 510 D16 = 11012 = 1310
616 = 01102 = 610 E16 = 11102 = 1410
716 = 01112 = 710 F16 = 11112 = 1510
Introduction to MicroprocessorsNumber Systems and Conversions
No. 1-2 9/6/00
Base 16 Conversion
EXAMPLE
Convert A716 to binaryNote: This is easy because of the relationship between the two bases
A = 1010 7 = 0111
Therefore 0A7H or $A7 = 10100111
Note leading 0 before A.Since this is a number a different base than base 10, absence of the zero may confuse a compiler. Therefore it is customary to add the leading 0 to any hex character that begins with a letter of the alphabet(A,B,C,D,E,F) !! These are NUMBERS in base 16 !!
Introduction to MicroprocessorsNumber Systems and Conversions
No. 1-3 9/6/00
HEXADECIMAL TO DECIMAL
• Similar Rules as to those in binary to decimal– Convert to binary then weighted multiplication -OR-
– Leave in HEX and use HEX multiplication
• EXAMPLE (repeated multiplication)
$F8E6H = F(16)3 + 8 (16)2 + E(16)1 + 6(16)0
= 15(16)3 + 8 (16)2 + 14(16)1 + 6(16)0
= 61,440 + 2048 + 224 +6
= 63,718
Introduction to MicroprocessorsNumber Systems and Conversions
No. 1-4 9/6/00
DECIMAL TO HEXADECIMAL
• Similar Rules as to those in decimal to binary Repeated HEX division
• EXAMPLE (repeated division)
• Divisors are for 16 binary digits (4 HEX)
247910 =
2479/16 = 154 remainder 15 or F
154/16 = 9 remainder 10 or A
9/16 = 9 remainder 9 or 9
Answer = 9AFH
Introduction to MicroprocessorsNumber Systems and Conversions
No. 1-5 9/6/00
DECIMAL TO HEXADECIMAL
ALTERNATE
• Same EXAMPLE (but weighted division)
• Divisors are for 16 binary digits (4 HEX)
4096, 256, 16, 1
247910 =
2479/256 = 9 remainder .68359375 x256 =
175/16 = 10 or A remainder .9375 x 16 =
15/1 = 15 or F
Answer = 9AFH
Introduction to MicroprocessorsNumber Systems and Conversions
No. 1-6 9/6/00
BINARY CODES
DECIMAL 8421 BINARY0 0000 0000
1 0001 0001
2 0010 0010
3 0011 0011
4 0100 0100
5 0101 0101
6 0110 0110
7 0111 0111
8 1000 1000
9 1001 1001
10 0001 0001 1010
11 0001 0010 1011
....
98 1001 1000 1100010
99 1001 1001 1100011
7421 6311 5421 5311 5211
0000 0000 0000 0000 0000
0001 0001 0001 0000 0001
0010 0011 0010 0011 0011
0011 0100 0011 0100 0101
0100 0101 0100 0101 0111
0101 0111 1000 1000 1000
0110 1000 1001 1001 1001
1000 1001 1010 1011 1011
1001 1011 1011 1100 1101
1010 1100 1100 1101 1111
4 Bit BCD CODESBCD
Introduction to MicroprocessorsNumber Systems and Conversions
No. 1-7 9/6/00
MORE 4-BIT BCD CODES
Decimal 4221 3321 2421 84/2/1 74/2/1
0 0000 0000 0000 0000 0000
1 0001 0001 0001 0111 0111
2 0010 0010 0010 0110 0110
3 0011 0011 0011 0101 0101
4 1000 0101 0100 0100 0100
5 0111 1010 1011 1011 1010
6 1100 1100 1100 1010 1001
7 1101 1101 1101 1001 1000
8 1110 1110 1110 1000 1111
9 1111 1111 1111 1111 1110
The / is subtractweight
Introduction to MicroprocessorsNumber Systems and Conversions
No. 1-8 9/6/00
5-BIT CODES
Decimal 2-out of-5 63210 Shift-Counter 86421 51111
0 00011 00110 00000 00000 00000
1 00101 00011 00001 00001 00001
2 00110 00101 00011 00010 00011
3 01001 01001 00111 00011 00111
4 01010 01010 01111 00100 10000
5 01100 01100 11111 00101 100006 10001 10001 11110 01000 11000
7 10010 10010 11100 01001 11100
8 10100 10100 11000 10000 11110
9 11000 11000 10000 10001 11111
Introduction to MicroprocessorsNumber Systems and Conversions
No. 1-9 9/6/00
OTHER CODES
• Alphanumeric Codes– ASCII CODE 7 BITS
– EBCDIC CODE 8 BITS
– UNICODE 16 Bits
Introduction to MicroprocessorsNumber Systems and Conversions
No. 1-10 9/6/00
ASCII CODEPart 1
b6b5b4 (column)
b3b2b1b0
Row(hex)
000 0
001 1
010 2
011 3
100 4
101 5
110 6
111 7
0000 0 NUL DLE SP 0 @ P ` p0001 1 SOH DC1 ! 1 A Q a q0010 2 STX DC2 " 2 B R b r0011 3 ETX DC3 # 3 C S c s0100 4 EOT DC4 $ 4 D T d t0101 5 ENQ NAK % 5 E U e u0110 6 ACK SYN & 6 F V f v0111 7 BEL ETB ' 7 G W g w1000 8 BS CAN ( 8 H X h x1001 9 HT EM ) 9 I Y I y1010 A LF SUB * : J Z j z1011 B VT ESC + ; K [ k {1100 C FF FS , < L / l |1101 D CR GS - = M ] m }1110 E SO RS . N ^ n ~1111 F SI US / ? O _ o DEL
Introduction to MicroprocessorsNumber Systems and Conversions
No. 1-11 9/6/00
ASCII CODECONTROL CODES
NUL Null DLE Data link escapeSOH Start of heading DC1 Device control 1STX Start of text DC2 Device control 2ETX End of text DC3 Device control 3EOT End of transmission DC4 Device control 4ENQ Enquiry NAK Negative acknowledgeACK Acknowledge SYN SynchronizeBEL Bell ETC End transmitted blockBS Backspace CAN CancelHT Horizontal tab EM End of mediumLF Line feed SUB SubstituteVT Vertical tab ESC EscapeFF Form feed FS File separatorCR Carriage return GS Group separatorSO Shift out RS Record separatorSI Shift in US Unit separatorSP Space DEL Delete or rubout
Introduction to MicroprocessorsNumber Systems and Conversions
No. 1-12 9/6/00
EBCDICPart I
Most Significant Hexadecimal DigitLeast SignificantDigit 0 1 2 3 4 5 6 7
0 Null Data link escape Digit select Space
& -
1 Start of heading Device control 1Start ofsignificance
/
2 Start of text Device control 2 Field separatorSynchronizationcharacter
3 End of text Tape mark 4 Punch off Restore Bypass PN 5 Horizontal tab New line Line feed Record separator
6 Lowercase BackspaceEnd oftransmission block
Uppercase
7 Delete Idle EscapeEnd oftransmission
8 Cancel 9 RLF End of medium \
AStart of manualmessage
Cursor control Set mode
centsign
! | :
B Vertical tab Customer use 1 Customer use 2 Customer use 3 . $ , #
C Form feedInterchange fileseparator
Device control 4 < * % @
D Carriage returnInterchange groupseparator
EnquiryNegativeacknowledgment
( ) _ '
E Shift outInterchange recordseparator
Acknowledge + ; > =
F Shift inInterchange unitseparator
BellStart of specialsequence
| or[
]or]
? "
Introduction to MicroprocessorsNumber Systems and Conversions
No. 1-13 9/6/00
EBCDICPART II
LeastSignificant
Most Significant Hexadecimal Digit
Digit 8 9 A B C D E F0 { } ` 01 a j A J 12 b k s B K S 23 c l t C L T 34 d m u D M U 45 e n v E N V 56 f o w F O W 67 g p x G P X 78 h q y H Q Y 89 i r z I R Z 9A B C D E F
Introduction to MicroprocessorsNumber Systems and Conversions
No. 1-14 9/6/00
UNICODE
16 BIT CODE
First Page 0000H - 00FFH Same as ASCII
Has not been standardized the remaining
0FFFFH minus 00FFH available for all remaining
countries and languages!
Introduction to MicroprocessorsNumber Systems and Conversions
No. 1-15 9/6/00
Number Systems
Representation of Negative Numbers
Three major schemes:
sign and magnitude
ones complement
twos complement
nines complement
tens complement
Assumptions:
we'll assume a 4 bit machine word
16 different values can be represented
roughly half are positive, half are negative
Introduction to MicroprocessorsNumber Systems and Conversions
No. 1-16 9/6/00
Number SystemsSign and Magnitude Representation
0000
0111
0011
1011
11111110
1101
1100
1010
1001
1000
0110
0101
0100
0010
0001
+0+1
+2
+3
+4
+5
+6
+7-0
-1
-2
-3
-4
-5
-6
-7
0 100 = + 4 1 100 = - 4
+
-
High order bit is sign: 0 = positive (or zero), 1 = negative
Three low order bits is the magnitude: 0 (000) thru 7 (111)
Number range for n bits = +/-2 -1
Representations for 0 Two +/-
n-1
Introduction to MicroprocessorsNumber Systems and Conversions
No. 1-17 9/6/00
Number SystemsOnes Complement
Subtraction implemented by addition & 1's complement
Still two representations of 0! This causes some problems
Some complexities in addition
0000
0111
0011
1011
11111110
1101
1100
1010
1001
1000
0110
0101
0100
0010
0001
+0+1
+2
+3
+4
+5
+6
+7-7
-6
-5
-4
-3
-2
-1
-0
0 100 = + 4 1 011 = - 4
+
-
Introduction to MicroprocessorsNumber Systems and Conversions
No. 1-18 9/6/00
Number RepresentationsTwos Complement
0000
0111
0011
1011
11111110
1101
1100
1010
1001
1000
0110
0101
0100
0010
0001
+0+1
+2
+3
+4
+5
+6
+7-8
-7
-6
-5
-4
-3
-2
-1
0 100 = + 4 1 100 = - 4
+
-
Only one representation for 0
One more negative number than positive number
like 1's compexcept shiftedone positionclockwise
Introduction to MicroprocessorsNumber Systems and Conversions
No. 1-19 9/6/00
BINARY ARITHMETIC
RULES FOR ADDITION
0 + 0 = 0
0 + 1 = 1
1 + 0 = 1
1 + 1 = 0 + carry 1 to next column
RULES FOR SUBTRACTION
0 - 0 = 0
0 - 1 = 1 and borrow from next column
1 - 0 = 1
1 - 1 = 0
Borrowing 1 from next column is equivalent to subtracting 1 from that column
Introduction to MicroprocessorsNumber Systems and Conversions
No. 1-20 9/6/00
Addition
EXAMPLE
ADD 1010 to 1101
1010
+ 1101
10111
SUBTRACT 1010 from 1101
borrow 4 subtract 2
1101 nothing in 4 position
- 1010
0011
Introduction to MicroprocessorsNumber Systems and Conversions
No. 1-21 9/6/00
Number Representations
Addition and Subtraction of NumbersSign and Magnitude
4
+ 3
7
0100
0011
0111
-4
+ (-3)
-7
1100
1011
1111
When signs is same,result sign bit is thesame as the operands'sign. Just add magnitudes
4
- 3
1
0100
1011
0001
-4
+ 3
-1
1100
0011
1001
When signs differ,operation is subtract,sign of result dependson sign of number withthe larger magnitude. Thisis what we do in base 10.
Introduction to MicroprocessorsNumber Systems and Conversions
No. 1-22 9/6/00
Number Systems
Sign and Magnitude
Cumbersome addition/subtraction.
Must compare magnitudes to determine the sign of result.
Introduction to MicroprocessorsNumber Systems and Conversions
No. 1-23 9/6/00
Ones Complement
The general formula for finding
/N = (2n - 1) - N
To find 1's complement of 7
2 = 10000
-1 = 00001
1111
-7 = 0111
1000 = -7 in 1's comp.
4
The symbol N, with a bar over it, is used to represent the complement. Also used to indicate inversion are /, ‘, or ! (CUPL and ABEL)
N is positive number, then N is its negative 1's complement. N is the precision, as many 1’s as required. Binary can be any precision, but BCD requires 4 bits.
Precision 4 bitsTherefore,
Introduction to MicroprocessorsNumber Systems and Conversions
No. 1-24 9/6/00
Ones Complement
ZERO
Applying the general formula/N = (2n - 1) - N
To find 1's complement of 0
2 = 10000
-1 = 00001
1111
-0 = 0000
1111 = -0 in 1's comp.
4
NOTE:
The complement of 0 is 1111, which means, there are2 representations of 0.
0000 Positive 0
1111 Negative 0
Introduction to MicroprocessorsNumber Systems and Conversions
No. 1-25 9/6/00
SHORTCUT1’s Complement
Shortcut method:
Replace all 0’s with 1’s, and all 1’s with 0’s
simply compute bit wise complement
7 in 1’s complement 0111 -> 1000
Introduction to MicroprocessorsNumber Systems and Conversions
No. 1-26 9/6/00
Number SystemsAddition and Subtraction of Numbers
Ones Complement Calculations
4
+ 3
7
0100
0011
0111
-4
+ (-3)
-7
1011
1100
10111
1
1000
4
- 3
1
0100
1100
10000
1
0001
-4
+ 3
-1
1011
0011
1110
End around carry
End around carry
Introduction to MicroprocessorsNumber Systems and Conversions
No. 1-27 9/6/00
Number SystemsAddition and Subtraction of Binary Numbers
Ones Complement Calculations
Why does end-around carry work?
Its equivalent to subtracting 2 and adding 1n
M - N = M + / N = M + (2n - N -1) = (M - N) + 2n - 1
For (M > N)
-M + (-N) = M + N = (2n - M - 1) + (2n - N - 1)
= 2n + [2n - 1 - (M + N)] - 1
For M + N < 2n-1
after end around carry the -1 is canceled if a carry, else not:
= 2n - 1 - (M + N)
this is the correct form for representing -(M + N) in 1's complement!
Introduction to MicroprocessorsNumber Systems and Conversions
No. 1-28 9/6/00
Number SystemsTwo’s Complement Numbers
/N = 2n - N
Note: subtract number immediately.
Example: Twos complement of 7
Example: Twos complement of 0
Only 4bits can represent number.The most significant 1 is dropped!
42 = 10000
7 = 0111
1001 = represents -7
sub
2 = 10000
-0 = 0000
0000 = 0
4
sub
Shortcut method Number 1:
Twos complement =1’s complement + 1
0111 -> 1000 + 1 -> 1001 (representation of -7)
1001 -> 0110 + 1 -> 0111 (representation of 7)
Introduction to MicroprocessorsNumber Systems and Conversions
No. 1-29 9/6/00
SHORTCUT
Shortcut method Number 2:
Starting from the right, least significant bit, if 0 leave unchanged.Continue from right to left, if 0, no change. When the first 1 is encountered,we leave it unchanged, but complement each digit to the left.
710 = 01112 Apply method 10012 = -710 .
6810 = 011001002 yields 100111002 . The lead 0 is important so that the number to be converted is positive! In base 10, the - sign
does that for us. With only on and off (0,1) we need this extra bit
of information to describe the signed number.
-6810 = 10011100 yields 011001002 = + 6810
Introduction to MicroprocessorsNumber Systems and Conversions
No. 1-30 9/6/00
Number SystemsAddition and Subtraction of Binary Numbers
Twos Complement Calculations
4
+ 3
7
0100
0011
0111
-4
+ (-3)
-7
1100
1101
11001
4
- 3
1
0100
1101
10001
-4
+ 3
-1
1100
0011
1111
Simpler addition scheme makes twos complement the most commonchoice for integer number systems within digital systems
Introduction to MicroprocessorsNumber Systems and Conversions
No. 1-31 9/6/00
Number SystemsAddition and Subtraction of Binary Numbers
Twos Complement Calculations
Why can the carry-out be ignored?
-M + N when N > M:
M* + N = (2 - M) + N = 2 + (N - M)n n
Ignoring carry-out is just like subtracting 2n
-M + -N where N + M < or = 2 n-1
-M + (-N) = M* + N* = (2 - M) + (2 - N)
= 2 - (M + N) + 2n n
After ignoring the carry, this is just the right twos compl.representation for -(M + N)!
n n
Introduction to MicroprocessorsNumber Systems and Conversions
No. 1-32 9/6/00
Number SystemsOverflow Conditions
Add two positive numbers to get a negative number
or two negative numbers to get a positive number
5 0101+3 0011-8 1000
-7 - 2 = +7
0000
0001
0010
0011
1000
0101
0110
0100
1001
1010
1011
1100
1101
0111
1110
1111
+0
+1
+2
+3
+4
+5
+6
+7-8
-7
-6
-5
-4
-3
-2
-1
0000
0001
0010
0011
1000
0101
0110
0100
1001
1010
1011
1100
1101
0111
1110
1111
+0
+1
+2
+3
+4
+5
+6
+7-8
-7
-6
-5
-4
-3
-2
-1
4 bit representation3 bits + sign in msbrange -8 to +7
+1=6 (ok)
+2=7 (ok)+3=-8 (error)
Introduction to MicroprocessorsNumber Systems and Conversions
No. 1-33 9/6/00
OVERFLOW
If carry-in (penultimate) is added to sign bit and there is a carry-outthen no overflow.
if carry-in differs fromcarry-out then overflow
This is true for both 1’s and 2’s complement addition.
Introduction to MicroprocessorsNumber Systems and Conversions
No. 1-34 9/6/00
Number SystemsOverflow Conditions
Example is using 2’s Complement
5
3
-8
0 1 1 1 carry 0 1 0 1
0 0 1 1
1 0 0 0
-7
-2
7
1 0 0 0 carry 1 0 0 1
1 1 1 0
1 0 1 1 1
Note: the carry and penultimate carry are not the same!This can be very easily implemented in hardware. XOR
Overflow
5
2
7
0 0 0 0 carry 0 1 0 1
0 0 1 0
0 1 1 1
-3
-5
-8
1 1 1 1 carry 1 1 0 1
1 0 1 1
1 1 0 0 0
No overflow No overflow
Overflow when carry in to sign does not equal carry out
Overflow
Introduction to MicroprocessorsNumber Systems and Conversions
No. 1-35 9/6/00
BCD AdditionBCD Number Representation
Decimal digits 0 thru 9 represented as 0000 thru 1001 in binary
Addition:
5 = 0101
3 = 0011
1000 = 8
5 = 0101
8 = 1000
1101 = 13!
Problemwhen digit
sum exceeds 9
Solution: add 6 (0110) if sum exceeds 9.
5 = 0101
8 = 1000
1101
6 = 0110
1 0011 = 1 3 in BCD
9 = 1001
7 = 0111
1 0000 = 16 in binary
6 = 0110
1 0110 = 1 6 in BCD
Introduction to MicroprocessorsNumber Systems and Conversions
No. 1-36 9/6/00
BCD Subtractin
Must use 10’s complement
To find 10’s complement, first generate 9’s complement by subtractingThe number from 9.
If n = 9, 9-9 =0 9’s complement of 9 is 0 or 0000If n = 8, 9-8 =1 9’s complement of 8 is 1 or 0001If n = 7, 9-7 =2 9’s complement of 7 is 2 or 0010If n = 6, 9-6 =3 9’s complement of 6 is 3 or 0011If n = 5, 9-5 =4 9’s complement of 5 is 4 or 0100If n = 4, 9-4 =5 9’s complement of 4 is 5 or 0101If n = 3, 9-3 =6 9’s complement of 3 is 6 or 0110If n = 2, 9-2 =7 9’s complement of 2 is 7 or 0111If n = 1, 9-1 =8 9’s complement of 1 is 8 or 1000If n = 0, 9-0 =9 9’s complement of 0 is 9 or 1001
Add 1 to 9’s complement to find 10’s complement.
Introduction to MicroprocessorsNumber Systems and Conversions
No. 1-37 9/6/00
Example of BCD Subtract
To subtract 2 from 5, find 10’s complement of 2 which is 10002 4 BitRepresentation.
5 5 0101-2 8 1000
3 1 3 1101 The 1 carry, is normal form and indicates answer is 3. In binary, 1001 is exceeded! Must add 6 to correct.
Known as BCD adjust. 0110 1 0011 The 1 carry is normal form answer is 3 base 10.
Introduction to MicroprocessorsNumber Systems and Conversions
No. 1-38 9/6/00
Addition Examples
4 BITS signed andunsigned
Decimal BinaryUnsigned
1’sComplement
2’sComplement
9’sComplement
10’sComplement
9’sComplementBinaryCodedDecimal
10’sComplementBinaryCodedDecimal
No Sign No Sign 4 BitSigned
NoOverflow
4 BitSigned
NoOverflow
Base10
SignedNo
Overflow
Base10
SignedNo
Overflow
Base2
Signed NoOverflow4 Bit BCD
Base2
Signed NoOverflow4 Bit BCD
1+3
4
00010011
0 0 11
0100
00010011
0 0 1 1
0100
00010011
0 0 1 1
0100
1+3
4
1+3
4
00010011
0 0 1 1
0100
00010011
0 0 1 1
0100
4 BitNo Sign
NoOverflow
4 BitSigned
Overflow
4 BitSigned
Overflow
Base10
SignedOverflow
Base10
SignedOverflow
Base2 9’sSigned
Overflow4 Bit BCD
Base2 10’sSigned
Overflow4 Bit BCD
5+7
12
01010111
0 1 1 1
1100
01010111
0 1 1 1
1100
01010111
0 1 1 1
1100
05+07
12
5+7
12
01010111
0 1 1 1
1100
01010111
0 1 1 1
1100
Introduction to MicroprocessorsNumber Systems and Conversions
No. 1-39 9/6/00
Addition Examples
4 BITS signed andunsignedDecimal Binary
Unsigned1’sComplement
2’sComplement
9’sComplement
10’sComplement
9’sComplementBinaryCodedDecimal
10’sComplementBinaryCodedDecimal
No Sign No Sign 4 BitSigned
NoOverflow
4 BitSigned
NoOverflow
Base10
SignedNo
Overflow
Base10
SignedNo
Overflow
Base2
Signed NoOverflow4 Bit BCD
Base2
Signed NoOverflow4 Bit BCD
7-2
5
01111101
1 1 1 1
01001
0101
01111110
1 1 1 1
0101
797
1 1
041
05
798
1 1
05
01111101
1 1 1 1
01001
0101
01111110
1 1 1 1
0101
4 BitSigned
NoOverflow
4 BitSigned
NoOverflow
Base10
SignedNo
Overflow
Base10
SignedNo
Overflow
Base2 9’sSigned
NoOverflow4 Bit BCD
Base2 10’sSigned
NoOverflow4 Bit BCD
5-7
-2
01011000
0 0 0 0
11010
1101 (-2) 0010
01011001
0 0 0 1
1110
00011
(-2) 0010
0592
97
-2
593
98
011
(-2) 02
01011000
0 0 0 0
11010
1101 (-2) 0010
01011001
0 0 0 1
1110
00011
(-2) 0010
Introduction to MicroprocessorsNumber Systems and Conversions
No. 1-40 9/6/00
Excess 3 Code
Excess-3 code is another important BCD Code.
To encode a decimal number, add 3 to each digit before converting to binary.
EXAMPLE (Note: 2 digits represented by 2 4 bit numbers)
12 to excess-3 = 1+3=4 2+3=5
4 5
0100 0101
29 to excess-3 = 2+3=5 9+3=12
5 12
0101 1100
Introduction to MicroprocessorsNumber Systems and Conversions
No. 1-41 9/6/00
EXAMPLES OF EXCESS ADDITION
In excess-3 addition, whenever we add two numbers whose sum is 9 or less, an excess-6 number is formed. To return to excess-3 we must subtract 3.
EXAMPLE
2 +5 = 7
0101 2
1000 5
1101 excess-6
- 0011
1010 excess-3 equivalent of 7
CASE 1
Introduction to MicroprocessorsNumber Systems and Conversions
No. 1-42 9/6/00
EXAMPLES OF EXCESS ADDITION
In excess-3 addition, whenever we add two numbers whose sum is greater than 9, there will be a carry from one group to the next. When this happens, the group that produced the carry will revert to 8421 (BCD). To return to excess-3 we must subtract 3 from that group.
EXAMPLE 0 1 carry not carried to 10’s because of sum
29 0101 1100 excess-3 for 29
+ 39 0110 1100 excess-3 for 39
68 1100 1000 first result
- 0011 + 0011 subtract less than 9, add 3
1001 1011 excess-3 for 68
CASE 2