FIT3165 / FIT4165 COMPUTER NETWORKS
WEEK 9 – Physical Layer Transmission – Part 1
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9.1 DATA AND SIGNALS
9.2 DIGITAL TRANSMISSION
❑ Discuss the relationship between data and signals. We then show how data and signals can be both analog and digital.
❑ Then concentrate on digital transmission. We show how to convert digital and analog data to digital signals.
❑ Next, concentrate on analog transmission. We show how to convert digital and analog data to analog signals.
❑ Discuss about multiplexing techniques and how they can combine several channels.
❑ Finally, consider hardware level below the physical layer and discuss the transmission media.
DATA AND SIGNAL
DATA AND SIGNALS
• At the physical layer, the communication is node-to-node, but the nodes exchange electromagnetic signals.
Communication at the physical layer
Analog and Digital
Data can be analog or digital.
● Analog data is continuous.
● Digital data take on discrete values.
signals can be either analog or digital.
● An analog signal has infinitely many levels of intensity over a period of time.
● A digital signal, on the other hand, can have only a limited number of defined values. Although each value
can be any number, it is often as simple as 1 and 0.
Analog Signals
❖ Time and Frequency Domains ❖ Composite Signals
❖ Bandwidth
Digital Signals
❖ Bit Rate
❖ Bit Length
❖ Digital Signal as a Composite
Analog Signal
❖ Transmission of Digital Signals
❖ Baseband Transmission
❖ Broadband Transmission
Analog Signals
A sine wave
The time-domain and frequency-domain plots of a sine wave
Analog Signals
The frequency domain is more compact and useful when we are dealing with more than one sine wave. For example, the figure (top left) shows three sine waves, each with different amplitude and frequency. All can be represented by three spikes in the frequency domain as shown in the bottom left figure.
The time domain and frequency domain of three sine waves
The bandwidth of periodic and nonperiodic composite signals
Digital Signals
Two digital signals: one with two signal levels and the other with four signal levels
Baseband transmission
An example of a dedicated channel where the entire bandwidth of the medium is used as one single channel is a LAN. Almost every wired LAN today uses a dedicated channel for two stations communicating with each other.
Transmission Impairment
• Signals travel through transmission media, which are not perfect.
• The imperfection causes signal impairment.
• This means that the signal at the beginning of the medium may not be the same we see at the destination.
• What is sent is not what is received.
• Three causes of impairment are,
❑ Attenuation
❑ Distortion
❖ Signal-to-Noise Ratio (SNR)
Attenuation and amplification
Signal to Noise Ratio (SNR)
Signal Strength
■ During propagation of signal, there is attenuation. Compensation is possible using amplifiers or repeaters at regular intervals.
■ We use deciBELs (dB) to measure gains, losses and relative levels
■ Logarithmic representation is best when losses in transmission media are exponential ■ Gains and losses can be cascaded using simple addition of units in dB
P1 P2 = 1⁄2 P1 Gdb = 10 log10 —— P in
Example : Suppose a signal travels through a transmission medium and its power is reduced to half. This means that P2 = 0.5 P1. In this case, the attenuation (loss of power) can be calculated as
A loss of 3 dB (−3 dB) is equivalent to losing one-half the power.
Signal Strength and Noise
Distortion
Two cases of SNR: a high SNR and a low SNR
Data Rate Limits
A very important consideration in data communications is how fast we can send data called capacity in bits per second, over a channel. Data rate depends on three factors:
• The bandwidth available
• The levels of the signals we use
• The quality of the channel (the level of noise)
Two theoretical formulas were developed to calculate the data Capacity:
• one by Nyquist for a noiseless channel(Nyquist Equation)
C = 2 B Log2 L
• another by Shannon for a noisy channel (Shannon’s Equation)
C = B Log2 (1 + SNR)
• C = Capacity of the channel in bps
• B = Bandwidth of the channel in Hz
• L = Number of voltage levels in digital signal
• SNR = Signal to noise ratio (as a ratio NOT in dB)
A few useful Logarithm formulas and rules
A = Log b N N = bA
Logb (A B) = Logb(A) + Logb(B)
Logb (A /B) = Logb(A) – Logb(B)
Logb 1 = 0 for any base b
Logb (A ) = Logn(A) Logn(b)
Logb b = 1 for any base b
Logb (An) = n Logb(A)
We need to send 265 kbps over a noiseless (ideal) channel with a bandwidth of 20 kHz. How many signal levels do we need? We can use the Nyquist formula as shown:
Nyquist for a noiseless channel(Nyquist Equation) C = 2 B Log2 L
Since this result is not a power of 2, we need to either increase the number of levels or reduce the bit rate. If we have 128 levels, the bit rate is 280 kbps. If we have 64 levels, the bit rate is 240 kbps.
Consider an extremely noisy channel in which the value of the signal-to-noise ratio is almost zero. In other words, the noise is so strong that the signal is faint. For this channel the capacity C is calculated as shown below.
Shannon for a noisy channel (Shannon’s Equation) C = B Log2 (1 + SNR)
This means that the capacity of this channel is zero regardless of the bandwidth. In other words, the data is so
corrupted in this channel that it is useless when received.
We can calculate the theoretical highest bit rate of a regular telephone line. A telephone line normally has a bandwidth of 3000 Hz (300 to 3300 Hz) assigned for data communications. The signal-to-noise ratio is usually 3162. For this channel the capacity is calculated as shown below.
This means that the highest bit rate for a telephone line is 34.881 kbps. If we want to send data faster than this, we can either increase the bandwidth of the line or improve the signal-to noise ratio.
We have a channel with a 1-MHz bandwidth. The SNR for this channel is 63. What are the appropriate bit rate and signal level?
Performance
• Up to now, we have discussed the tools of transmitting data (signals) over a network and how the data behave.
• One important issue in networking is the performance of the network—how good is it?
❑ Bandwidth
❖ Bandwidth in Hertz (Hz)
❖ Bandwidth in Bits per Seconds (bps) ❖ Relationship
❑ Throughput
❑ Latency (Delay)
❑ Bandwidth-Delay Product
The bandwidth of a subscriber line is 4 kHz for voice or data. The bandwidth of this line for data transmission can be up to 56 kbps, using a sophisticated modem to change the digital signal to analog. If the telephone company improves the quality of the line and increases the bandwidth to 8 kHz, we can send 112 kbps.
B = 4000 Hz
C = 56,000 bps
Shannon for a noisy channel (Shannon’s Equation) C = B Log2 (1 + SNR) 56000 = 4000 x Log2 (1 + SNR)
SNR = 16383
112000 = 8000 x Log2 (1 + SNR) SNR = 16383
DIGITAL TRANSMISSION
Digital-to-Digital Conversion
In this section:
• We see how we can represent digital data by using digital signals. • The conversion involves three techniques:
• line coding, block coding, and scrambling. Line coding is always needed; block coding and scrambling may or may not be needed.
Line Coding
❖ Polar Schemes
❖ Bipolar Schemes
❖ Multilevel Schemes
Block Coding
❖ 4B/5B Coding
❖ 8B/10B Coding
❑ Scrambling
❖ B8ZS Coding
❖ HDB3 Coding
Line Coding
❖ Polar Schemes
❖ Bipolar Schemes
❖ Multilevel Schemes
Line coding and decoding
Polar schemes-NRZ
Polar Schemes – Manchesters
Line Coding
❖ Polar Schemes
❖ Bipolar Schemes
❖ Multilevel Schemes
Bipolar schemes: AMI and pseudoternary
Multilevel: 2B1Q
Block Coding
❖ 4B/5B Coding
❖ 8B/10B Coding
Block coding concept
• We need redundancy to ensure synchronization and to provide some kind of inherent error detecting.
• Block coding can give us this redundancy and improve the performance of line coding.
• In general, block coding changes a block of m bits into a block of n bits, where n is larger than m. Block coding is referred to as an mB/nB encoding technique.
❑ Scrambling
B8ZS Coding HDB3 Coding
Two cases of B8ZS scrambling technique
When you have data with long sequences of 00000…We are looking for a technique that does not increase the number of bits and yet provide synchronization.
Solution is to substitutes long zero-level pulses with a combination of other levels to provide synchronization.
This is called scrambling.
Analog-to-Digital Conversion
• Sometimes, we have an analog signal such as one created by a microphone or camera. The tendency today is to change an analog signal to digital data because the digital signal is less susceptible to noise.
• In this section we describe two techniques,
• Pulse code modulation
○ Sampling
○ Quantization
○ Encoding
○ Original Signal Recovery
○ PCM Bandwidth
• Delta modulation.
• After the digital data are created (digitization)
Components of PCM encoder and sampling
SNR dB = 6.02 n b + 1.76 ;
nb = Quantization encoded words – levels
Sampling Theorem:
Sampling Rate: important consideration is the sampling rate or frequency. According to the Nyquist theorem, to reproduce the original analog signal, one necessary condition is that the sampling rate be at least twice the highest frequency in the original signal.
Three different sampling methods for PCM
Quantization and encoding of a sampled signal
We want to digitize the human voice. What is the bit rate, assuming 8 bits per sample?
The human voice normally contains frequencies from 0 to 4000 Hz. So the sampling rate and bit rate are calculated as follows.
Components of a PCM decoder
SNRdB = 6.02 nb + 1.76
nb = Quantization encoded word – levels
The process of delta modulation
■ Analog input is approximated by a staircase function
■ can move up or down one level (δ) at each sample interval
■ Has binary behavior
■ since function only moves up or down at each sample interval
■ hence can encode each sample as single bit
■ 1 for up or 0 for down
Delta Modulation Example
● So far we have discussed
○ Analog signals
○ Digital Signals
○ Analog to Digital Conversion
■ Pulse-Code Modulation
■ Delta Modulation.
● Next week
○ Physical Layer Transmission – Part 2
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