Digital Data Communications Techniques Updated: 2/9/2009

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

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!  rising edge

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.

Asynchronous Transmission - Implementation

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

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!

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

Synchronous Transmission  Assume Synchronous transmission  Overhead = 48 bits for every 1000 bytes (characters) of data  Calculate the overhead ratio in percentage: 48/( x8)  0.6 %

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 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! Higher data rate  more errors

Bit Error Rate (BER) F is the frame size With no error detection mechanism

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?

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

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

Error Detection

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 (P=0 if odd ones)  always odd number of ones in the frame  Asynchronous applications and Standard in PC memory Even (even parity)  Synchronous applications F( )  odd parity Parity Bit + Data Block

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

CRC generator and checker Refer to your notes for examples! transmits n bits which is exactly divisible by some number (predetermined divisor) receiver divides frame by that number n-k bit n-k+1 bit n-k bit k + (n-k) bits

Error Detection & Correction Common Techniques  Example : Division in CRC Encoder

Error Detection & Correction Common Techniques  Example : Division in CRC Decoder

Example Message: Pattern: T  Step through to calculate the remainder!

n-k bit n-k+1 bit n-k bit k + (n-k) bits Example

Block Code – Error Detection and Correction  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 Table1

 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 d min  We can call this (Table 2) coding scheme C(5,2) with d min =3  Note : the coding scheme for (Table 1) is C(3,2) with d min =2 Table2 Block Code – Error Detection and Correction

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 is sent and 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 d min = t + 1  t = d min -1  To guarantee the maximum t correctable errors in all cases

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 What will be the likely original dataword? Rotate to left: t = dmin -1=3-1=2 t = 1

Error Detection and Correction  The larger the d min 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

System Performance  Assume n=4, k=2  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 BER=1-^-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 Ebit=2xEdata  For very high BER, adding coding requires higher Eb/No due to overhead 10^-6 Channel bit error rate

Slide for later ….

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)

Line Configuration - Topology

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

Summary  asynchronous verses synchronous transmission  error detection and correction  line configuration issues