digital communication principle

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TEL 213/05 Telecommunication Principle

Tutorial 2: Digital Communication Principle

Semester January 2012



Basic Roadmap of Unit 2



Digital Versus Analog Communication

Digital Communication Analogue CommunicationMore immune to noise as signal can be regenerated if it is below the threshold.

Less immune to noise

Has error detection and correction techniques

No error detection and correction

Compatible with time division multiplexing

Compatible with frequency division multiplexing

Smaller ICs possible with greater processing capability

Smaller ICs not possible

Can be processed using digital signal processing techniques including signal manipulation

Signal manipulation not possible with DSP

More bandwidth required Less bandwidth requiredMore complex and requires more circuitry

Less complex with less required circuitry



Sampling

• The challenge is always to represent analog signals in digital form to ease transmission.

• To do this, sampling of analog data needs to be done.• Sampling is a process of approximation (estimation) of an analog quantity. • After sampling, mathematical modeling can be done to represent the signal

in digital form.

• Sampling must be done at regular intervals and must cover most of the data (at least twice the bandwidth frequency) to have an accurate depiction of the whole data.

• This concept is called the Nyquist-Shannon Sampling Theory• The phenomenon called “aliasing” (misrepresentation) will happen if not

enough samples are taken to represent the whole population.

Sampling Frequency = 2 * Bandwidth



Sampling an Analog Signal



Example

• If a 12 bits A/D converter were to be used, how many voltage increments are there?

• 12 bits would produce 212 or 4096 voltage increments.



Example

• Calculate the minimum voltage step increment for a 10 bit A/D converter assuming that the input voltage is from 0V to 6V.

• Number of voltage levels =2 to the power of 10=1024 voltage levels• Number of Increments = 1024-1=1023• Minimum voltage step increment or maximum amount of error=



Example

• An information signal to be transmitted digitally is a rectangular wave with a period of . It is given that the wave will be adequately passed if the bandwidth includes the fourth harmonic. Calculate the signal frequency, the frequency of the fourth harmonic and the minimum sampling frequency (Nyquist rate).



Noise and S/N

• Noise is an electronic signal that is a mixture of many random frequencies at different amplitudes that gets added to a radio or information signal as it is transmitted from one place to another as it is processed

• The signal-noise (S/N) ratio is also called SNR, and is an indication of the relative strengths of the signal and noise in a communication system. The stronger the signal and the weaker the noise, the higher the S/N ratio



Formula for S/N calculation



Example



Bit Error Ratio in digital Communication System

• Bit error ratio (BER) is defined as the possibility of a bit being received in error in a digital communication system



Example



Solution – ERF Table



Bit Error Rate (BER)



Example

• Find the BER of a 100kbits/s assuming unipolar transmission. The SNR is given as 1.2dB.



Data Transmission

• All data need to be converted to ASCII code first: http://www.ascii-code.com/

http://www.ascii-code.com/



Value “M” being transmitted serially

M = 010011101(first 0 is not taken into account, and total is 8 bits). t is the time between each bit, known as bit interval. Bit time is the total time taken and can be expressed in bps



Example 1

What is the bit time at 230.4kbps?

st µ34.410*34.4230400

1 6 === −



No PAM

• Compute the bit rate that it will take to transmit a decimal number '201' using a bit interval of 1 microseconds using no modulation (serial transmission)



With PAM

• Repeat the transmission with 2 bits of PAM transmission and calculate the bit rate of this transmission.



PCM

• Digitizing = converting analogue signals to digital signals.• Pulse code modulation (PCM) is commonly used

The resulting 4 bit PAM of a signal is found to be 10,9,8,11. Draw the resultant PCM.

Solution10=10109=10018=100011=1011

10 9



FSK

• Frequency-shift keying (FSK) uses two sine wave frequencies are used to represent binary 0s and 1s

Binary Signal

FSK Signal



Problem with FSK



Binary Phase Shift Keying



DPSK



Line Encoding



Example

• Transmit the word “Data Com” in the transmission line and calculate the LRC/BCC and parity of VRC (odd)



Solution – Step 1

• D = 01000100

• A = 01000001

• T = 01010100

• A = 01000001

• <space> = 00100000

• C = 01000011

• O = 01001111

• M = 01001101



Solution – Step 2

A (input) B (input) Q (even parity) (odd parity)0 0 0 10 1 1 01 0 1 01 1 0 1

Q

Character D A T A C O M LRC or

BCC

(LSB) 0 1 0 1 0 1 1 1 10 0 0 0 0 1 1 0 01 0 1 0 0 0 1 1 00 0 0 0 0 0 1 1 00 0 1 0 0 0 0 0 10 0 0 0 1 0 0 0 1

(MSB) 1 1 1 1 0 1 1 1 1Parity of VRC

(odd)1 1 0 1 0 0 0 1 0



XOR Addition

•When the number of 1's is ODD, the result of the XOR operation is '1'.

•When the number of 1's is EVEN (or none present), the result of the XOR operation is '0'.



Detecting Errors

Error Bit



Hamming Detection (FEC) Example

• The data word is 01101010. Use Hamming FEC Method to transmit this data across the transmission line



Step 1 – Hamming – find the value of n

8 bit data word to be transmitted

Experimented Value (2)

False

8 bit data word to be transmitted

Experimented Value (4)

True (therefore n=4)



Step 2: Initial Hamming Table

• Total bits required = 8+4=12 bits

• Insert the 4 required Hamming bits between the transmitted data:

12 11 10 9 8 7 6 5 4 3 2 1

H 0 1 H 1 0 H 1 0 H 1 0



Step 3:

• Determine which bits are already ‘1’12 11 10 9 8 7 6 5 4 3 2 1

H 0 1 H 1 0 H 1 0 H 1 0

Position 2 = 0010

Position 5 = 0101

Position 8 = 1000

Position 10 = 1010



Step 4: XOR Bits that are ‘1’



Step 5: Insert Final XOR into Hamming Table

12 11 10 9 8 7 6 5 4 3 2 1H 0 1 H 1 0 H 1 0 H 1 012 11 10 9 8 7 6 5 4 3 2 10 0 1 1 1 0 0 1 0 1 1 0

Final Transmitted Data



Hartley’s Law

C=2B

whereby C = channel capacity expressed in bits per second and B is the channel bandwidth

whereby the S/N is the signal to noise ratio in power.

)1)(2log( 10 N

SBC +=

With noise



Example – Shanon-Hartley’s Law

Calculate the maximum channel capacity of a voice-graded telephone line with a bandwidth of 3100hz and a S/N of 30dB.

10003log)10

30(log

)10/log(

log10

11 ===

==

−−P

dBantiP

PdB

bpsC

N

SBC

31000)10(3100

1097.9)3(32.31001log32.31001log

1001log3100)10001(log3100)1(log

102

222

==≈===

=+=+=

32

5log

56200

31000

)3100(2

31000

2log

2

2

==

====

N

antiN

B

CN

32 channels of multilayer encoding is required for a maximum channel capacityOf 31kbps with S/N of 30dB



Summary

• A signal needs to be sampled as the first process in converting an analogue signal to a digital signal

• To properly sample a signal and to ensure aliasing does not occur and the whole signal can be adequately reconstructed from its digital representation, the sampling frequency needs to be twice the value of the original signal.

• The process of quantization and mapping changes the digitally acquired signals to become numerical values

• Pulse Code Modulation (PCM) uses multiplexing to convert the signals into a stream of data.

• The encoding process is the last process of the A/D converter and serves to change the numerical values for optimization prior to transmission.

• There are 3 main types of encoding available namely source encoding, channel encoding and line encoding.



Thank you!

digital communication principle

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