Digital Data Communications Techniques

1. Digital Data Communications Techniques Updated: 2/9/2009

2. 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

3. 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! Figure! 100010  rising edge How can we achieve the desired Synchronization?

4. 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.

5. Asynchronous Transmission- Implementation

6. Asynchronous Transmission- Implementation • The frame is often made up of • Start bit • Stop elements • Parity bit • 5-8 data bits

7. 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!

8. 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

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

10. 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!

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

12. 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?

13. 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

14. 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

15. Error Detection

16. 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(1110001) odd parity 1111 000 1 Parity Bit + Data Block

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

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

19. Error Detection & Correction Common Techniques • Example : Division in CRC Encoder

20. Error Detection & Correction Common Techniques • Example : Division in CRC Decoder

21. http://www.macs.hw.ac.uk/~pjbk/nets/crc/ Example Message: 1010001101 Pattern: 110101 T 1010001101 + 01110 Step through to calculate the remainder!

22. n-k bit k + (n-k) bits n-k+1 bit n-k bit Example 1010001101 + 01110 1010001101 110101 110101 1010001101 + 01110 00000 01110

23. 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

24. 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

25. 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 tcorrectable errors in all cases Error Detection and Correction

26. 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 -1=3-1=2 t = 1

27. 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

28. Channel bit error rate 10^-6 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