Transcript
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EE2000 Logic Circuit Design
Number Systems & Arithmetic
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OutlineNumber systems
Conversion between different radix systemsBinary arithmetic
Complement of binary numbersSigned number representations
Codes in Digital WorldBCDASCIIGray codes
Parity and error correction
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Number Systems
Decimal number system employed in everyday arithmeticRepresent numbers by strings of digits called positional notation
Digits – the Latin word for fingersan-1 an-2 … a2 a1 a0 (0 ≤ ai < 10)Each digit is associated to a value depends on its position in the string
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Polynomial Form of IntegerThe no. can be expressed in power seriesN = an-1rn-1 + an-2rn-2 + … + a2r2 + a1r + a0
N: the decimal value of the integern: number of digitsr: radix (base)ai: coefficients (digit), 0 ≤ ai < r
e.g. (7672)104 digits, so n = 4decimal number, i.e. r = 10ai: 0 ≤ ai < 10, i.e. one of the ten digits (0, 1, 2, …, 9)N = 7 x 103 + 6 x 102 + 7 x 10 + 2
These two 7 symbols have different meaning
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Floating Point Numberan-1 an-2 … a2 a1 a0 . a-1 a-2 … a-(m-1) a-m (0 ≤ ai < 10)
“.” is the radix pointThe left most digit an-1 is the most significant digit (MSD)The right most digit a-m is the least significant digit (LSD)
General polynomial formN = an-1rn-1 + … + a1r1 + a0r0 + a-1r -1 + … + a-mr -m
Number of digits = n + me.g. (767.2)10
n = 3, m = 1N = 7 x 102 + 6 x 101 + 7 + 2 x 10-1
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Alternative Conversion Method
(…((an-1r + an-2)r + an-3)r + … + a1)r + a0
+(a-1+ (a-2 + (… + (a-m+1+ a-mr -1)r -1)…r -1)r -1
e.g. (767.295)10
= ((7 x 10 + 6)10 + 7)+ (2 + (9 + 5 x 10-1)10-1)10-1
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Other Number Systems
Decimal (base 10) are for human
Three number systems are commonly used in computer work
Binary (base 2)Octal (base 8)Hexadecimal (base 16)
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Binary Number SystemA base 2 system with two digits: 0 and 1Expressed with a string of 1s and 0sThe digits are called bits(binary digits).The left most bit called most significant bit (MSB)The right most bit called least significant bit (LSB)
Decimal Binary01234567
01
1011
100101110111
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Radix Conversion
Convert from binary to decimale.g. convert (1011.11)2 to decimal
= 1 x 23 + 0 x 22 + 1 x 2 + 1 + 1 x 2-1 + 1 x 2-2
= 8 + 0 + 2 + 1 + 0.5 + 0.25= (11.75)10
i.e. the summation of the power series only
Powers of 2n 0 1 2 3 4 … 202n 1 2 4 8 16 … 1,048,576
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From Decimal to Binary
Convert from decimal to binarye.g. convert (746)10 to binary
= 7 x 102 + 4 x 101 + 6 x 100
= 111 x 101010 + 100 x 10101 + 110 x 10100
Involve binary multiplication!Time consuming and not easy to compute
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Two MethodsMethod 1: Subtraction
Subtract from the largest power of 2 that less than or equal to that numberPut a 1 in the corresponding position of the binary equivalentRepeat the subtract procedure with the remainder until the reminder becomes 0
Method 2: DivisionDivide the number by 2The remainder (either 0 or 1) gives the least significant bitDivide the quotient by 2 repeatedly until the quotient becomes 0
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Method 1: SubtractionN = 746The largest power of 2 that less than or equal to 746 is 512 (i.e. 29)Put a “1” in the 29 positionCompute remainder: 746 - 512 = 234The next smaller power of 2 is 256 (28), but that is larger than 234.So put a “0” in the 28 positionThe next smaller power of 2 is 128 (27), which is smaller than 234.So put a “1” in the 27 positionThe new remainder is 234 - 128 = 106
Answer: - - - - - - - - - -
Answer: 1 - - - - - - - - -
Answer: 10 - - - - - - - -
Answer: 101 - - - - - - -
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Method 1: Subtraction
And so on until the remainder becomes zeroFinally we will have (746)10
= 1 x 29 + 0 x 28 + 1 x 27 + 1 x 26 + 1 x 25 + 0x 24 + 1 x 23 + 0 x 22 + 1 x 21 + 0= (1011101010)2
Do you think it is easy to compute?
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Method 2: DivisionN = 746
746 / 2 = 373 + 0/2373 / 2 = 186 + 1/2186 / 2 = 93 + 0/293 / 2 = 46 + 1/246 / 2 = 23 + 0/223 / 2 = 11 + 1/211 / 2 = 5 + 1/25 / 2 = 2 + 1/22 / 2 = 1 + 0/21 / 2 = 0 + 1/2
?????????0????????10???????010??????1010?????01010????101010???1101010??11101010?0111010101011101010
quotientremainder
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Method 2 (Another Format)
010
0101010
01010101010
110101011101010
0111010101011101010
7 4 623 7 32 … 01 8 62 … 1
9 32 … 04 62 … 12 32 … 01 12 … 1
52 … 122 … 112 … 00 … 1
quotient remainder
MSB
LSB
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How About Decimal Factions?Multiply by 2 to give an integer and a fractionRepeat multiplying the fraction until it becomes 0e.g. convert decimal 0.6875 to binary
0.6875 x 2 = 1.37500.3750 x 2 = 0.75000.7500 x 2 = 1.50000.5000 x 2 = 1.0000(0.6875)10 = (0.1011)2
Integer and fractional parts is done separately(746.6875)10 = (1011101010.1011)2
MSB
LSB
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Method 1 vs. Method 2
Subtraction produces binary digits from the MSB to the LSB (from left to right)Division produces binary digits reversely, from the LSB to the MSB (from right to left)
Both methods work slowly, do we have faster methods?
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Octal NumbersOctal digits are: 0, 1, 2, … 7e.g. (1352)8
= 1 x 83 + 3 x 82 + 5 x 81 + 2 x 80
= 512 + 192 + 40 + 2= (746)10
To convert decimal to octalDivide the number by 8 repeatedlye.g. (746)10
746 / 8 = 93 … 293 / 8 = 11 … 511 / 8 = 1 … 31 / 8 = 0 … 1
Answer: (1352)8
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Octal FractionMultiply by 8 to give an integer and a fractionRepeat multiplying the fraction until it becomes 0Convert decimal 0.513 to a 3-digit octal fraction
0.513 x 8 = 4.1040.104 x 8 = 0.8320.832 x 8 = 6.6560.656 x 8 = 5.248…(0.513)10 = (0.4065…)2 = (0.407)2 (rounded)
MSB
LSB
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Octal NumbersBase 8, a power of 2 (23 = 8)Each octal digit corresponds to 3 binary digitsTo convert binary to octal
Partition the binary digits into groups of 3 bits eachNote the way of groupingAdd 0s to the MSB and LSB to make the no. of bits a multiple of 3e.g. (1 011 101 010 . 001 01)2 = (001 011 101 010 . 001 010)2= (1 3 5 2.1 2)8
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Octal Numbers
To convert octal to binaryReplace each octal digit by 3-bit binary equivalentFinally remove the extra 0s from MSB & LSBe.g. (1352.12)8
= (001 011 101 010 . 001 010)2
= (001 011 101 010 . 001 010)2
= (1 011 101 010 . 001 01)2
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Hex Conversion
Hex digits are: 0, 1, … 9, A, B, C, D, E, FA is 10B is 11C is 12D is 13E is 14F is 15
Each hex digit can be in the range 0 to 15
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Hex ConversionTo convert hex to decimal
e.g. (2EA)16= 2 x 162 + 14 x 161 + 10 x 160
= 512 + 224 + 10= (746)10
To convert decimal to hexDivide the number by 16 repeatedlye.g. (746)10
746 / 16 = 46 … 10 (A)46 / 16 = 2 … 14 (E)2 / 16 = 0 … 2 Answer: (2EA)16
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Hex ConversionTo convert binary to hex
Partition the binary digits into groups of 4 bits eachAdd necessary 0s to left and righte.g. (10 1110 1010)2 = (2 E A)16
To convert hex to binaryReplace each hex digit by 4-bit binary equivalent and remove the extra 0se.g. (2EA)16 = (0010 1110 1010)2
2 E A
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Method 3: Convert via Octal
Convert decimal to octal firstDivide by 8 repeatedly, and we have(746)10 = (1352)8
Now convert octal to binaryReplace each octal digit by 3-bit binary equivalent and remove the extra 0s(1352)8 = (1 011 101 010)2
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Or Via Hexadecimal
Convert decimal to hex firstDivide by 16 repeatedly, and we have(746)10 = (2EA)16
Now convert hex to binaryReplace each hex digit by 4-bit binary equivalent and remove the extra 0s(2EA)16 = (10 1110 1010)2
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Octal and HexadecimalCompact representation of binary numbersMore convenient for people than using bit strings (3 or 4 times longer)Computer manuals usually use octal or hexadecimal numbers to specify binary quantities
e.g. memory address of computer usually written in hex. FormatWhich one more easy to read? 2EA or 001011101010
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Binary Arithmetic
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Binary Addition
Four cases only0 + 0 = 00 + 1 = 11 + 0 = 11 + 1 = 10 (a carry of 1 to the next bit)
An XOR relation0 1
0 0 1
1 1 0The carry bit is omitted here
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Binary Addition: Example
Compute the sum of (0110)2 and (0111)2(0110)2 = (6)10
(0111)2 = (7)10
0 1 1 00 1 1 1+)
1
00 1 1 00 1 1 1+)
0 1
1 0Carries
Sum the 2nd LSBs and the carry bit: 0 + (1 + 1) = 0 + 10 = 10
0 1 1 00 1 1 1+)
1 0 1
1 1 00 1 1 00 1 1 1+)
1 1 0 1
1 1 0
The final result is 11012 (1310)
Sum the LSBsfirst: 0 + 1 = 1
Augend
Addend
Sum
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Binary Addition: 2nd Example
Compute the sum of (1101)2 and (0101)2(1101)2 = (13)10 ,(0101)2 = (5)10
Result should be (18)10
But…
1 1 0 10 1 0 1+)
1 0 0 1 0
1 1 0 1
Overflow (result out of range)
Addition of n-bit positive integers may produces (n+1)-bit result
0 0 1 02 = 210
1 0 0 1 02 = 1810
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One-bit adder
A complete table defining the addition process
a b cin cout s00001111
00011001
101010
1 1
0 00 10 11 00 11 01 01 1
0 1 1 00 1 1 1+)
1 1 0 1
1 1 0ab
s
Cout cin
The resulting carry from the addition
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Signed NumberUnsigned numbers
Either zero or positive integers, e.g. 5, 3, 0Signed numbers
An integer has a sign to indicate positive or negative numbers
Signed-magnitude, e.g. +5, -5, -3How about binary system?
The MSB be a sign bit (0 means +, 1 means -)i.e. +5 = 0101, -5 = 1101 (4-bit system, 3 bits for magnitude)
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Negative Binary NumberHow to compute the sum if they involve negative numbers?
e.g. assume 4-bit binary system is used, compute (+5) + (-3)+5 = 01012-3 = 10112Result is 100002 (overflow and incorrect!)
Another problem0000 (positive zero)1000 (negative zero)Any difference between them? Cause confusion!
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How to represent negative binary?
The MSB is still a sign bit Positive number, +a, is still stored as usual
e.g. +5 = 01012, 0 = 00002 (0 means positive)Negative number, -a, is stored as 2n - a
e.g. -3= 24 – 3= 16 – 3= 13= 11012
This is called two’s complement format
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Two’s ComplementThe largest number can be stored is 2n-1 - 1for n-bit binary system
e.g. 7 (i.e. 01112) for 4-bit binary
The smallest number can be stored is -2n-1
e.g. -8 (i.e. 10002) for 4-bit binary
Binary Unsigned Signed0000 0 0
1010 10 -6
1011 11 -5
… … …
0001 1 +1
0010 2 +2
0011 3 +3
0100 4 +4
0101 5 +5
0110 6 +6
0111 7 +7
1000 8 -8
1001 9 -7
1110 14 -2
15 -11111
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Find the Two’s Complement
Negative number, -a, is stored as 2n - aNot convenient to compute 2n – aAn easier way to find the 2’s complement
Step 1) Find the binary equivalent of the magnitudeStep 2) Complement each bit (i.e. 0’s to 1’s, 1’s to 0’s) (this is 1’s complement)Step 3) Add 1 (this is 2’s complement)
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3 Examples
0 1 0 1
1 0 1 0
+)
1 0 1 1
1
Complement the bits
Magnitude
= 5 =
Add 1
0 0 0 1
1 1 1 0
+)
1 1 1 1
1
a) -5 b) -1
0 0 0 0
1 1 1 1
+)
0 0 0 0
1
c) -0
The carry bit is ignored here
Find the 2’s complement of -5, -1 and 0
Magnitude
= 1 =
Magnitude
= 0 =
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Find the magnitude
Now giving a negative number that stored in 2’s complement format, how can you tell the magnitude?Two methods
1) reverse the process of 2’s complementi.e. Subtract 1, then complement
2) complement first, then add 1Addition is easier than subtraction
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Example: Find the magnitude
1 0 1 1
1-)
1 0 1 0
0 1 0 1
subtract 1
= (5)10
complement
-5 1 0 1 1
0 1 0 0
+)
0 1 0 1
1
complement
add 1
Method 1: Method 2:
Compute the magnitude of binary no. (1011)2 which is stored in 2’s complement format
-5
= (5)10
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1’s & 2’s ComplementFor a n-digit binary number,
The sum of the number and its 1’s complement will be 1…1 (n 1s) = 2n - 1The sum of the number and its 2’s complement will be 10…0 (n 0s) = 2n
e.g. -a is represented as 2n - a in 2’s complement format
a + -a = a + (2n - a)= 2n
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Back to the Addition Problem
Now we know how to represent negative binary numbersHow to do addition?
1 0 1 1
1 1 1 0
0 0 1 1
- 5
- 2
+ 3
Example 1:
1 0 1 1
1 0 0 0 0
0 1 0 1
- 5
0
+ 5
Example 2:
The carry outs are ignored
1 0 1 1
1 0 0 1 0
0 1 1 1
- 5
2
+ 7
Example 3:
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Overflow
There is a limitation in additionWhen the sum is out of range, we call it overflowe.g. For 4-bit system, sum = a + b
The range of sum: -8 ≤ sum ≤ +7Overflow when sum < -8, or sum > 7Note: overflow will occur even if a and b are in the range!
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Overflow Example0 1 0 1
1 0 0 1
0 1 0 0
+ 5
+ 9
+ 4-7?
Both +5, +4 are in the range
1 1 0 1
(1) 0 1 1 1
1 1 0 0
- 5
- 9
- 4+7?
Both -5, -4 are in the range
Overflow may occur if the two no. are of the same signBut addition of opposite signs will never produces overflow (why?)
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Binary Subtraction
Subtraction can be accomplished by 1) taking the 2’s complement of the 2nd
operand2) then perform addition
Work for both signed and unsigned numbersi.e. a – b is computed as a + (-b)
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Binary Subtraction Example
0 1 1 1
(1) 0 0 1 0
1 0 1 1
+ 7
+ 2
– 52’s complement
of +5
Compute 7 – 5:
Ignore the carry out
0 1 1 1
1 1 0 0
0 1 0 1
+ 7
+12
– – 52’s complement
of -5
Compute 7 – (–5):
-4? overflow
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Radix ComplementThere are 2 types of complement for any radix-rnumber system
r’s complement & (r-1)’s complementBinary system: 2’s & 1’s complementDecimal system: 10’s & 9’s complement
For a n-digit radix-r number systemThe (r-1)’s complement of number a is
r n - a - 1
The r’s complement of number a isr n - a
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ExampleGiven a decimal number 35, compute its 10’s and 9’s complement
a = 35Radix r = 10No. of digits n = 29’s complement = 102 - 35 - 1 = 6410’s complement = 102 - 35 = 65
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Subtraction with ComplementFor radix-r number system, the general form of subtraction is
a – b= a + (r’s complement of b)= a + (rn – b)= a – b + rn
If a > b, the sum is a positive numbera – b + rn = a – b if we discard the carry out (i.e. rn)
If a < b, the sum is a negative numbera – b + rn = rn – (b – a) = r’s complement of (b – a)
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Codes in Digital World
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Decimal Codes
Computer systems operate on binary no.Human use decimalWhen communicate with computer systems
Input: convert decimal no. to binaryProcess: perform arithmetic calculations in binaryOutput : convert binary result back to decimal
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Binary Coded Decimal (BCD)
Computer can accept binary values onlyMust represent the decimal digits by a code that contains 1s and 0sCalled binary coded decimal (BCD)Possible to perform arithmetic operations directly with BCD
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BCD Example739 would be stored as
011100111001 (12 bits in series)
There are several BCD codes8421 code5421 code2121 codeExcess 3 code2 of 5 codes
7 3 9Compare with the binary representation of 73910
= 10111000112
10 bits only!
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8421 Code8421 code is a kind of weighted codeUse 4-bit to represent a decimal digit8, 4, 2, 1 are the weight of the bits
Code: a3a2a1a0Value: (8 x a3) + (4 x a2) + (2 x a1)+ (1 x a0)
ExampleCode: 0110Value: (8 x 0) + (4 x 1) + (2 x 1)+ (1 x 0)= 4 + 2 = 6
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Three Weighted CodesValue
(decimal digit) 8421 code 5421 code 2421 code
0 0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
0000
10000
0001
0010
0011
0100
1000
1001
1010
1011
1100
0101
0110
0111
1101
1110
0001
2 0010
3 0011
4 0100
5 1011
6 1100
7 1101
8 1110
9 1111
0101
0110
0111
1000
1001
1111 1010
unused
The code are the same for the first 5 digits
(5 x 1) + (4 x 0) + (2 + 1) + (1 x 1)= 5 + 2 + 1= 8
(2 x 1) + (4 x 1) + (2 + 1) + (1 x 0)= 2 + 4 + 2= 8
4-bit codes have 16 combinations, but we just use 10 of them. The remaining 6 codes are unused (redundancy)
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2421 CodeA self-complementing code (9’s complement)
The complement of 0 is 9 (0000 ↔ 1111)The complement of 1 is 8 (0001 ↔ 1110)The complement of 2 is 7 (0010 ↔ 1101)The complement of 3 is 6 (0011 ↔ 1100)The complement of 4 is 5 (0100 ↔ 1011)
Evenly distribution of 1s and 0sEach bit has exactly five 1s and five 0s for the ten decimal values
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Excess 3 Code (XS3)Value
(decimal digit) 8421code Excess 3 code
0 0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
10011
0100
0101
0110
0111
1000
1001
1010
1011
1100
0000
0001
0010
1101
1110
23456789
unused
1111
Excess 3 code is a shifted 8421 code
XS3’s 0 = 8421 code’s 3
XS3’s 1 = 8421 code’s 4
…
Evenly distribution of 1s and 0s
8421 code: the MSB has eight 0s, only two 1s
XS3: each bit has exactly five 1s and five 0s
9’s complement code:Complement of 0 is 9 (0011 ↔ 1100)Complement of 1 is 8 (0100 ↔ 1011)…Complement of 4 is 5 (0111 ↔ 1000)
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2 of 5 CodeValue
(decimal digit) 2 of 5 Code
0 11000
10100
10010
10001
01100
01010
01001
00110
00101
00011
the rest of the 22
patterns with 0, 1, 3, 4 or 5 1’s
123456789
unused
Use 5 bits to represent each digit
exactly two 1s and three 0s for all code
Advantage: capable for error detection
If an error is made in just one of the bits during storage or transmission (the result will contain either one or three 1’s), can be detected as an error
Any limitation in error detection?
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BCD SummaryWeighted code
8421, 5421, 2124 codesNon-weighted code
XS3, 2 of 5 codesCode that self-complementing
2124, XS3 codesCode that able to detect error
2 of 5 codeCode that easy to determine negative number
5421, 2421, XS3 codes (why?)
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Determine Negative NumbersFor a signed 10’s complement (decimal) number, the first digit of that number would be in the range of 5 to 9 for negative numbersIn 5421, 2421 and XS3 codes, correspond to the first bit of the no. being 1
For these 3 codes, the first bit of no. 0 to 4 is “0”, while the first bit of no. 5 to 9 is “1”
Therefore, only check 1 bit can determine if it is a negative number
More complex logic is required for 8421, 2 of 5 codes
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BCD drawbacks
Inefficiency of storageUsed 10 out of 16 combinations (weighted codes and XS3 code)Used 10 out of 32 combinations (2 of 5 code)
Complex arithmetic calculation
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Example: 8421 Code Addition
0 1 1 1
1 1 0 1
0 1 1 0
7
1 3
6
The 8421 code representation of decimal 7 and 6
Compute 7 + 6:
1 1 0 1
1 0 0 1 1
0 1 1 0If the result is an unused code, add 110 to it
1
The resultant value is unused in 8421 code!
3Result is correct now
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Example: 8421 Code Addition
1 0 0 1
1 0 0 1 0
1 0 0 1
9
1 8
9
The 8421 code representation of decimal 9
Compute 9 + 9:
1 0 0 1 0
1 1 0 0 0
0 1 1 0Also add 110 to it
1
The resultant value has a carry out!
8Result is correct now
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ASCII CodeAmerican Standard Code for Information InterchangeTo store alphanumeric information
AlphaLetters: A, B, C, …, Z, a, b, c, …, zSymbols: ?, !, #, @, (, space, tab …
Numeric: 0, 1, 2, …, 9Control signals: carriage return, delete, Esc, break…
7-bit, to code various characters on standard keyboard
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100 1100 110 1111
110 0111 110 1001 110 0011
ASCII Code TableASCII Code Value
000 0000 NULL
… …
010 0000 Space
010 0001 ! (exclamation mark)
010 0010 “ (double quote)
… …
011 0000 0
011 0001 1
… …
011 1010 : (colon)
… …
100 0001 A
… …
101 1010 Z
… …
110 0001 a
… …
111 1010 z
… …
control signals
symbols
numeric characters
capital letters
symbols
small letters
symbols
symbols
The word Logic would be coded as:
L o
g i c
(Please refer to the complete ASCII table in your book)
011 0111 011 0011 011 1001
739 would be coded as:
7 3 9
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Bit Changes in CodeLook at the BCD code table, the no. of bits that change from one binary value to the next variesMultiple bit changes may not be good for some applicationsIn 1953, Frank Gray invented a code that has one-bit fixed difference between adjacent code word
Value 8421 code0 0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
123456789
1-bit changes
2-bit changes
3-bit changes
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Gray CodeNumber Gray code
0123456789101112131415
0000
0001
0011
0010
0110
0111
0101
0100
1100
1101
1111
1110
1010
1011
1001
1000
00000001
01110101
11101010
Consecutive numbers differ in one bit only, but can you memorize the pattern?
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Gray Code
A class of unit distance codeNo need to memorize the Gray code tableWe can compute the Gray code easily
Step 1) Convert to binary number firstStep 2) Add a leading 0 to the MSBStep 3) XOR each two adjacent bits
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Example: Gray Code
Find the Gray code of decimal value 13(1 3)10 = (1 1 0 1)2
(0 1 1 0 1)2
1 0 1 1
Convert to binary number
Add a leading 0 to the MSB
XOR each two adjacent bits
The Gray code of 13♁ ♁ ♁ ♁
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Example: Convert to Binary
Given a Gray code 10001011, find its representing decimal value
1 0 0 0 1 0 1 1
1 1 1 10 0
10
= (1 1 1 1 0 0 1 0)2 = (242)10
If the code has leading 0s, remove them
Set the output bit as 1
Starting from MSB, replace each bit in the Gray code to the output bit
If meet an one in the Gray code, complement the output bit
And continue to replace process
The representing decimal value
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Parity and Error Correction
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Data Communication
The transmission of data over a channel may result in error (e.g. interference, noise)Some schemes are developed to detect and/or correct the error(s)We have to insert extra bits in the data
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Parity Bit
The simplest method for error detectionAn additional bit is added, called parity bitThe value of parity bit is set to make the total no. of 1s in the resulting code word either even or odd
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Even / Odd Parity Bit
1000 10001 10000
1010 10100 10101
Original code word Even parity Odd parity
no. of 1s = 2 (even) no. of 1s = 3 (odd)no. of 1s = 2
Original Code: a1 a2 a3 a4 (4-bit)Even Parity bit a5 = a1 ♁ a2 ♁ a3 ♁ a4
Odd Parity bit a5 = a1 ♁ a2 ♁ a3 ♁ a4 ♁ 1Parity: a1 a2 a3 a4 a5 (5-bit)
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Error DetectionAssume even parity is usedIf the received code word is 01011Total no. of 1s in code word is 3 (odd no.)Conclusion: errors occurred during transmission (at least 1 bit has been changed)Limitation
Able to detect 1, 3 or any odd no. of error bitsBut an even no. of errors is undetected!
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Vertical Longitudinal Parity Bits
The previous method can detect error, but can’t correct errorBecause it cannot locate the error bitHow about if we have a two-dimension parity bits?
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Example: Error Correction1 0 0 01 0 1 00 1 1 01 0 1 1
1 0 0 0 11 0 1 0 00 1 1 0 01 0 1 1 1
The original block of code
Even parity
1 1 1 1 0
Parity bit for its row
Parity bit for its column
1 0 0 0 11 0 1 0 00 1 0 0 01 0 1 1 11 1 1 1 0
Received block of code
Error detected in this row (one 1s)
Error detected in this column (three 1s)
Error bit has been located!
1 0 0 0 11 0 1 0 00 1 1 0 01 0 1 1 11 1 1 1 0
Received block of code
Error corrected
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Limitation
Consume a lot of bits for error detection and correctionOnly work for single error in a row (or column)Able to detect double error in the same row (or columns), but not able to correct them (why?)
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Further Reading
Introduction to Logic DesignChapter 1.1 A brief review of number systems
Digital Logic Circuit Analysis & DesignChapter 1 Number Systems and Codes
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OBTL: upon completion of this lecture, you should able to …
express a radix-r number in the polynomial formperform number conversion between different radix systemsperform binary arithmeticunderstand the concept of complement numbers and signed numbers representationsexpress signed number in signed magnitude, 1's complement, and 2's complement formatsperform 2’s complement and 1’s complement number arithmeticexpress numbers in different BCD formatperform BCD addition and subtractionrecognize Gray codeexpress number in the ASCII formatunderstand the concept of error detection and error correctionderive and perform single error detection and single error correction using parity bits
Self evaluation: Please tick the items that you are able to complete
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Level of achievementExcellent
Able to apply learned knowledge to solve unseen problems
WellAble to perform arithmetic operations on different number systemsAble to perform error detection and correction using parity bits
BasicUnderstand the concept of number systemsAble to apply correct formulae to convert numbers
Self evaluation: Please tick the appropriate box and mark the date of evaluation
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