1. Dr. Clincy Professor of CS
CS Chapter 2
Dr. Clincy
Professor of CS
Dr. Clincy
Lecture 3

2. Error Control Techniques
Trade off between detection probability and processing requirements
Approaches
Detection
Simple Parity (Double Parity) – Already Covered
Cyclic Redundancy Checksum
Error correction
Hamming Approach
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Lecture 9 and 10

3. Conceptually, How does CRC works ?
On the Tx side
An agreed-to divisor is used by both the Tx and Rx
The dataword is augmented (zeros added) based on the size of the divisor (setting up the dataword to be divided)
The “augmented dataword” is divided by the divisor
The quotient is discarded and the remainder is subtracted from the divisor to get the data that needs to be appended to the “augmented dataword”
Append the results in 4 to the “augmented dataword”
The result in 5 is sent to the receiver
On the Rx side
The augmented dataword with “divisor subtract remainder” appended is received
The Rx uses the same divisor as the Tx and divides the data in 1
If no errors occurred, the remainder will be zero
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4. Conceptually, How does CRC works ?
Example using decimal numbers
Data needing to be sent is 546 and let the common divisor be 8
On the Tx side
Shift 546 to the left (adding the appropriate amount of zeros)
Divide 5460 by 8 – get quotient 682 and remainder 4
Subtract remainder 4 from divisor 8 – get 4
Append or add the results in step 3 to
Send 5464 to the receiver
On the Rx side
Divide 5464 by the same divisor – 5464/8
Quotient equals 683 and the remainder equals 0
No errors occurred because the remainder is zero
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5. CRC using regular binary division
Appropriate # of zeros added to dividend 1001
Determine remainder
Rx performs the same calculation using the new dividend and gets a remainder of zero
Subtract remainder from divisor to determine what to append
Append result to dividend
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6. In making the CRC approach less complex, Modulo-2 division is used
Recall that “Black Boxes” do not subtract.
In making the CRC approach less complex, Modulo-2 division is used
Modulo-2 division uses XOR versus regular subtraction
The remainder result from Modulo-2 division is the final product (no need to subtract it from the divisor) – the Tx simply adds it to the augmented dataword
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7. Introducing Modulo-2 Arithmetic
Modulo-2 is used for error detection and correction
Modulo-2 has no carry over for addition and subtraction
Use the exclusive OR (XOR) for modulo-2
XORing of two single bits or two words
0 if two bits are the same
1 if two bits are different
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8. Division for CRC using Modulo-2Tx Rx
Looking at the # of bits to divide
Module-2 arithmetic here
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9. CRC using Modulo-2 Division
More Examples
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10. Cyclic Code Representation
Typically, a frame or packet of data is represented as a mathematical polynomial
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11. Error Detection: Cyclic Redundancy Check
Treat the frame or packet of data/bits as a mathematical polynomial
Right-most bit corresponds to x0
Next bit corresponds to x1
…Etc…
nth bit corresponds to xn
If the bit is 1, include x(i) in the polynomial
If the bit is 0, do not include x(i) in the polynomial
For example, would correspond to:
x8 + x6 + x3 + x2 + x0
The Tx would take this polynomial and divide it by a given generated polynomial
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12. CRC Approach using polynomials
Example
Using the generator polynomial of x5 + x4 + x2 + 1, send the message
1st: append some zeros to the end of the message you are sending. The number of zeros to append will depend on the degree of the polynomial (i.e. x4 + x3 +x0 has a degree of 4)
In our example, we have a degree of 5 – so add five 0’s
2nd: x5+x4+x2+1 translate to bit stream
3rd: Divide by using modulo-2
4th: Append remainder to original message and send to Rx
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13. Understanding Hamming Distance Concept
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14. Error Correction – Hamming Code Concept
Computers make errors occasionally due to voltage spikes and etc.
Recall Encoding Concept – codes representing characters
Hamming Distance of 1: change in 1 bit creates a new code
A-000
D-001
F-110
C-011
H-101
G-111
B-010
E-100
What happens with 1 bit in error ?
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15. Hamming Distance of 2 What happens with 1 bit in error ?
001
C-110
B-011
D-101
111
010
100
What happens with 1 bit in error ?
What happens with 2 bits in error ?
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16. Hamming Distance of 3 What happens with 1 bit in error ?
001
110
011
101
B-111
010
100
What happens with 1 bit in error ?
What happens with 2 bits in error ?
What happens with 3 bits in error ?
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17. Hamming Error Correcting Approach
1st: Determine the number of parity bits to add to the code word for checking
2nd: determine bit positions of each added parity bits
3rd: Determine what each parity bit checks
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18. Hamming Code Example (Understand the “how” vs “why”)
Example: Given a 4-bit code and even parity request – recall parity
Determine number of parity bits to add: 20=1, 21=2, 22=4: also determine bit positions of parity bits – 1, 2 and 4 (with labels C1, C2 and C4)
Let the 4-bit code have labels I3, I5, I6 and I7 due to the parity bit positions
Therefore the seven bits would be transmitted in the following order: C1 C2 I3 C4 I5 I6 I7
Determine what each parity bit checks: add the parity positions to determine this (must be less than or equal to 7 in this case): C1 case: 1+2=3, 1+4=5, 1+2+4=7; C2 case: 2+1=3, 2+1+4=7, 2+4=6; C4 case: 4+1=5, 4+2=6, 4+2+1=7
For example, transmitting the 4-bit code of 0101 (I3I5I6I7) would be (red bits are parity bits)
If I3 was corrupted during transmission, C1 and C2 would detect it (1+2=3)
If I5 was corrupted during transmission, C1 and C4 would detect it (1+4=5)
If I6 was corrupted during transmission, C2 and C4 would detect it (2+4=6)
If I7 was corrupted during transmission, C1 , C2 and C4 would detect it (1+2+4=7)
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19. Floating Point Numbers

20. Floating-Point Representation
The signed magnitude, one’s complement, and two’s complement representation that we have just presented deal with signed integer values only.
Without modification, these formats are not useful in scientific or business applications that deal with real number values.
Floating-point representation solves this problem.
For example, how would be stored compared to ?
How can we use the existing infrastructure of the computer to store real numbers ?

21. Scientific-Notation Vs Floating-Point Representation
Computers use a form of scientific notation for floating-point representation
Numbers written in scientific notation have three components:
Computer representation of a floating-point number consists of three fixed-size fields:
Size determines range
Size determines precision

22. Floating-Point Representation
Floating-point numbers allow an arbitrary number of decimal places to the right of the decimal point.
For example: 0.5  0.25 = 0.125
They are often expressed in scientific notation.
For example:
0.125 = 1.25  10-1
5,000,000 = 5.0  106

23. The Concept

24. The Simple Model
We introduce a hypothetical “Simple Model” to explain the concepts
In this model:
A floating-point number is 14 bits in length
The exponent field is 5 bits
The significand field is 8 bits
The significand is always preceded by an implied binary point.
Thus, the significand always contains a fractional binary value.
The exponent indicates the power of 2 by which the significand is multiplied.

25. Simple Model Illustrated
Example:
Express 3210 in the simplified 14-bit floating-point model.
We know that 32 is 25. So in (binary) scientific notation 32 = 1.0 x 25 = 0.1 x 26.
In a moment, we’ll explain why we prefer the second notation versus the first.
Using this information, we put 110 (= 610) in the exponent field and 1 in the significand as shown.

26. Normalization
To resolve the problem of synonymous forms, we establish a rule that the first digit of the significand must be 1, with no ones to the left of the radix point.
This process, called normalization, results in a unique pattern for each floating-point number.
In our simple model, all significands must have the form 0.1xxxxxxxx
For example, 4.5 = x 20 = x 22 = x 23. The last expression is correctly normalized.
In our simple instructional model, we use no implied bits.

27. Negative Exponents
To provide for negative exponents, we will use a biased exponent.
A bias is a number that is approximately midway in the range of values expressible by the exponent. We subtract the bias from the value in the exponent to determine its true value.
In our case, we have a 5-bit exponent. We will use 16 for our bias. This is called excess-16 representation.
In our model, exponent values greater than 16 are positive.
an exponent of 4, would be 16+4=20
an exponent of 10, would be 16+10=26
In our model, exponent values less than 16 are negative, representing fractional numbers.
an exponent of -4, would be 16-4=12
an exponent of -10, would be 16-10=6

28. Examples Example 1: Example 2:
Express 3210 in the revised 14-bit floating-point model.
We know that 32 = 1.0 x 25 = 0.1 x 26.
To use our excess 16 biased exponent, we add 16 to 6, giving 2210 (=101102).
So we have:
Example 2:
Express in the revised 14-bit floating-point model.
We know that is So in (binary) scientific notation = 1.0 x 2-4 = 0.1 x 2 -3.
To use our excess 16 biased exponent, we add 16 to -3, giving 1310 (=011012).