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 1Mbs10^-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

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

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/(48+1000x8) 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)

3

2

1

)/1_&0_(

_)/1_(

)/0_(

_)1_(

PframeerrordtctederrorundtctedP

ErrorFramePframeerrorundtctedP

PframeerrorbitP

ErrorBitPerrorbitP b

12

1

1

)1(

PP

PP Fb

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(1110001) odd parity 1 111 000 1Parity 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

http://www.macs.hw.ac.uk/~pjbk/nets/crc/

Message: 1010001101Pattern: 110101

T 1010001101 + 01110

Step through to calculate the remainder!

n-k bit

n-k+1 bit

n-k bit

k + (n-k) bits

Example

1010001101

110101

01110

1010001101 + 01110

1010001101 + 01110

110101

00000

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 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

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 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

2

1mindt

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?

],_([minmin jj wCodeRXdt

Rotate to left: 11110

t = dmin -1=3-1=2

t = 1

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

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