Digital Data Communications Techniques

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Digital Data Communications Techniques

• Updated 2/9/2009

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Asynchronous and Synchronous Transmission

• Timing problems require a mechanism to
  synchronize the transmitter and receiver
• receiver samples stream at bit intervals
• if clocks not aligned and drifting will sample at
  wrong time after sufficient bits are sent
• Two solutions to synchronizing clocks
• asynchronous transmission
• synchronous transmission

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Timing Problems

• Assume transmitting at 1Mbs?10-6 second per bit
  (1 usec)
• If the clock is off by 1 percent
• After 50 cycles ? off by 0.5 usec
• Missing one bit!

How can we achieve the desired Synchronization?
Figure! 100010 ? rising edge
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Example of Timing Error

• Assume data rate 10Kpbs
• Bit period 0.1 msec or 100 usec
• Receiver is fast by 6? bit period is 94usec.

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Asynchronous Transmission- Implementation
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Asynchronous Transmission- Implementation

• The frame is often made up of
• Start bit
• Stop elements
• Parity bit
• 5-8 data bits

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Asynchronous - Behavior

• simple
• cheap
• overhead of 2 or 3 bits per char (20)
• Overhead ratio () Overhead bits/Total number
  of bits x 100
• good for data with large gaps (keyboard)

Longer blocks can generate less overhead ratio
but results in more accumulative timing error!
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Synchronous Transmission

• block of data transmitted sent as a frame
• clocks must be synchronized
• can use separate clock line over distance
  timing error may still occur
• or embed clock signal in data
• need to indicate start and end of block
• use preamble and postamble
• more efficient (lower overhead) than async

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Synchronous Transmission

• Assume Synchronous transmission
• Overhead 48 bits for every 1000 bytes
  (characters) of data
• Calculate the overhead ratio in percentage

48/(481000x8) ? 0.6
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Types of Error

• An error occurs when a bit is altered between
  transmission and reception
• Single bit errors
• only one bit altered
• caused by white noise
• Burst errors
• contiguous sequence of B bits in which first last
  and any number of intermediate bits in error
• caused by impulse noise or by fading in wireless
• effect greater at higher data rates

Higher data rate ? more errors
If Data rate is 10 Mbps and the signal is lost
for only 1usec, how many bit errors occur?
1usec / 0.1 usec 10 bits!
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Bit Error Rate (BER)
With no error detection mechanism
F is the frame size
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Example

• Assume bit rate is 64Kbps
• There are 1000 bits per frame
• Assume BER is 10-6
• Calculate the Frame Error Rate (in one day how
  many errors)
• If we get one frame error per day, calculate the
  frame error rate
• Which is larger?

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Error Correction

• correction of detected errors usually requires
  data block to be retransmitted
• not appropriate for wireless applications
• bit error rate is high causing lots of
  retransmissions
• when propagation delay long (satellite) compared
  with frame transmission time, resulting in
  retransmission of frame in error plus many
  subsequent frames
• instead need to correct errors on basis of bits
  received

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Error Correction Basic Idea

• Adds redundancy to transmitted message
• Can deduce original despite some errors
• Errors are detected using error-detecting code
• Error-detecting code added by transmitter
• Error-detecting code are recalculated and checked
  by receiver
• map k bit input onto an n bit codeword
• each distinctly different
• if get error assume codeword sent was closest to
  that received

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Error Detection
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Error Detection Parity Check

• Basic idea
• Errors are detected using error-detecting code
• Error-detecting code added by transmitter
• error-detecting code are recalculated and checked
  by receiver
• Parity bit
• Odd (odd parity)
• If it had an even number of ones, the parity bit
  is set to a one, otherwise it is set to a zero
  (P0 if odd ones)? always odd number of ones in
  the frame
• Asynchronous applications and Standard in PC
  memory
• Even (even parity)
• Synchronous applications

F(1110001)? odd parity 1 111 000 1 Parity Bit
Data Block
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Cyclic Redundancy Check

• one of most common and powerful checks
• for block of k bits transmitter generates an n
  bit frame check sequence (FCS)

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transmits n bits which is exactly divisible by
some number (predetermined divisor) receiver
divides frame by that number
CRC generator and checker
Refer to your notes for examples!
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Error Detection Correction Common Techniques

• Example Division in CRC Encoder

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Error Detection Correction Common Techniques

• Example Division in CRC Decoder

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Example
http//www.macs.hw.ac.uk/pjbk/nets/crc/
Message 1010001101 Pattern 110101 T?
1010001101 01110
Step through to calculate the remainder!
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Example
1010001101 01110
1010001101
110101
110101
1010001101 01110
00000
01110
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Block Code Error Detection and Correction
Table1

• Hamming Distance
• The Hamming distance between two words is the
  number of differences between corresponding bits.
• Easily found by applying the XOR operation on the
  two words and count the number of 1s in the
  result.
• Ex Hamming distance d (000,011) 2
• Ex Hamming distance d (10101, 11110) 3
• Minimum Hamming Distance
• The minimum Hamming distance is the smallest
  Hamming distance between all possible pairs in a
  set of words
• Ex Find the minimum Hamming distance of the
  coding scheme in table 1

Sol 2
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Block Code Error Detection and Correction

• Minimum Hamming Distance
• Ex Find the minimum Hamming distance of the
  coding scheme in the table.
• 3 parameters to define the coding scheme
• Codeword size n
• Dataword size k
• The minimum Hamming distance dmin
• We can call this (Table 2) coding scheme C(5,2)
  with dmin 3
• Note the coding scheme for (Table 1) is C(3,2)
  with dmin 2

Table2
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Error Detection and Correction

• Relation between Hamming Distance and Error
• When a codeword is corrupted during transmission,
  the Hamming distance between the sent and
  received codewords is the number of bits affected
  by the error
• Ex if the codeword 00000 is sent and 01101 is
  received, 3 bits are in error and the Hamming
  distance between the two is d (00000, 01101) 3

• To guarantee the detection of up to t errors in
  all cases, the minimum Hamming distance in a
  block code must be
• dmin t 1 ? t dmin -1
• To guarantee the maximum t correctable errors in
  all cases

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Error Detection and Correction

• Example Give the above coding scheme we
  experience 1000 bit errors when transmitting
  1Gigbit bits
• Calculate the bit error rate.
• Calculate the ratio of data to coding rate.
• Using the given encoding technique, what is the
  codeword for 11?
• What is the coding scheme C(n,k),What is dmin?
• How many errors can be detected using the given
  encoding technique?
• How many errors can be corrected using the given
  encoding technique?
• Calculate the code gain.
• Assume the received codeword is 00101. What will
  be the likely original dataword?

Rotate to left 11110
t dmin -13-12
t 1
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Error Detection and Correction

• The larger the dmin the better
• The code should be relatively easy to
  encode/decode
• We like n-k to be small ? reduce bandwidth
• We like n-k to be large ? reduce error rate

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System Performance

• Assume n4, k2
• Given BER, coding can improve Eb/No
• Lower Eb/No is required)
• Code gain is the reduction in dB in Eb/No for a
  given BER
• E.g., for BER1--6 ? code gain is 2.77 dB
• Energy per coded bit (Eb) ½ data bit (Ed)
• If BET 10-6 ? required energy is 11 dB
• Hence, bit error rate will be 3dB les
• This is because Ebit2xEdata
• For very high BER, adding coding requires higher
  Eb/No due to overhead

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Slide for later .
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Line Configuration - Topology

• physical arrangement of stations on medium
• point to point - two stations
• such as between two routers / computers
• multi point - multiple stations
• traditionally mainframe computer and terminals
• now typically a local area network (LAN)

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Line Configuration - Topology
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Line Configuration - Duplex

• classify data exchange as half or full duplex
• half duplex (two-way alternate)
• only one station may transmit at a time
• requires one data path
• full duplex (two-way simultaneous)
• simultaneous transmission and reception between
  two stations
• requires two data paths
• separate media or frequencies used for each
  direction
• or echo canceling

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Summary

• asynchronous verses synchronous transmission
• error detection and correction
• line configuration issues

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