digital communication principle
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Digital Communication PrincipleTRANSCRIPT
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TEL 213/05 Telecommunication Principle
Tutorial 2: Digital Communication Principle
Semester January 2012
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Basic Roadmap of Unit 2
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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
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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
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Sampling an Analog Signal
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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.
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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=
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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).
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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
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Formula for S/N calculation
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Example
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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
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Example
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Solution – ERF Table
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Bit Error Rate (BER)
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Example
• Find the BER of a 100kbits/s assuming unipolar transmission. The SNR is given as 1.2dB.
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Data Transmission
• All data need to be converted to ASCII code first: http://www.ascii-code.com/
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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
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Example 1
What is the bit time at 230.4kbps?
st µ34.410*34.4230400
1 6 === −
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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)
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With PAM
• Repeat the transmission with 2 bits of PAM transmission and calculate the bit rate of this transmission.
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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
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FSK
• Frequency-shift keying (FSK) uses two sine wave frequencies are used to represent binary 0s and 1s
Binary Signal
FSK Signal
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Problem with FSK
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Binary Phase Shift Keying
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DPSK
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Line Encoding
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Example
• Transmit the word “Data Com” in the transmission line and calculate the LRC/BCC and parity of VRC (odd)
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Solution – Step 1
• D = 01000100
• A = 01000001
• T = 01010100
• A = 01000001
• <space> = 00100000
• C = 01000011
• O = 01001111
• M = 01001101
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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
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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'.
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Detecting Errors
Error Bit
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Hamming Detection (FEC) Example
• The data word is 01101010. Use Hamming FEC Method to transmit this data across the transmission line
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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)
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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
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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
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Step 4: XOR Bits that are ‘1’
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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
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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
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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
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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.
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Thank you!