Phones OFF Please Layer 1: The Physical Layer Transmission Basics: Signals and Media Parminder Singh Kang Home: pkang Email: [email protected].

Phones OFF Please

Layer 1: The Physical Layer

Transmission Basics: Signals and Media

Parminder Singh Kang

Home: www.cse.dmu.ac.uk/~pkang

Email: [email protected]

Topics:

1. Understanding communication model.

2. Role of Physical layer.

3. Encoding and Decoding.

4. Signals.

5. Amplitude, Wavelength and Frequency.

6. Bandwidth, Effect of Noise, Baud Rate and data Rate.

7. Composition of Signal.

8. Digital Signal Transmission Problems.

9. Modulation.

10. Analogue Communication.

11. Media Technologies.

1. Communication model

1.1 Purpose of Communication System?

1. Communication model

• Sharing of Resources.

• Exchange of Data

1.2 Components of Communication model?

• Source.

• Transmitter.

• Transmission System.

• Receiver.

• Destination.

Source Transmitter Receiver Destination

Data Flow

Transmission system

What is the Difference between Source and Transmitter/ Destination and Receiver?

2. Physical Layer

2.1 Definition:

• First Layer of OSI model that controls Functional Interface.

• Physical layer is not only Hardware. It also defines data transmission related functions.

2.2 Role of Physical Layer:

• To transmit bits across a medium.

• Hardware Specifications: cables, connectors, wireless radio transceivers, network interface cards and other hardware devices.

Eg: RJ45, RJ11, RS-232, 802.11 b/g.

• Encoding and Signalling: define how 1s and 0s are to be represented in the medium. e.g. voltage, frequency, etc.

• To define how a physical connection is set up and closed down;

eg: wireless communication uses RTS/CTS .

• provide handshaking and flow control, i.e. so a fast machine does not overrun a slow one.

Eg: sender is sending at 100mbps and receiver is working at 10mbps.

• Topology and Physical Network Design:

eg: Bus topology, Physical Star topology, Ring and Mesh.

3. Encoding and Decoding

3.1 Need of Encoding and Decoding?

Problem: two stations communicating over the public communications system:

• Computer transmits digital data as binary bits (0’s and 1’s).

• Public telephone system was designed initially for analogue voice data (continuous signals)

Modem encodes digital signal to suitable transmission format.

Decoding back to Digital Signal.

• Encoding is the process of putting a sequence of characters into a specialized format for efficient transmission or storage.

• e.g: In communication system: digital to analogue, Manchester Encoding.

In Storage: ASCII to Unicode

• I.e. data is passed from source to destination by changing physical properties with respect to physical media used.

For example; Signal, Voltage, Frequency etc.

• More specifically in Digital System:

o A 0 volt level represents 0 and 5volt level represents 1.

o   A 1KHz signal represents 0 and a 2KHz signal represents a 1.

3.2 Encoding:

Encryption, Encoding and conversion are different Terms.

3.3 Decoding:

• Decoding is the opposite process -- the conversion of an encoded format back into the original sequence of characters.

• e.g: In communication system: back Analogue to Digital.

• The receiver recognises these different changes and decodes the data, i.e. determines whether a 1 or a 0 is being sent at a particular time.

For example; Signal, Voltage, Frequency etc.

• Need an interface at each end to perform encoding and decoding.

• Both Encoder and Decoder agrees on protocol or standard.

3.4 Points to Remember:

4. Signals

4.1 Definition:

• Signal is a physical quantity that can carry information.

• It can be either Analogue or Digital.

• Signal can be represented as change in media characteristics.

E.g. voltage.

4.2 Analogue:

• Continuous signal.

• Real life Signal.

4.2 Digital:

• Discrete level.

• e.g. 0 and 1. (Computer)

4.3 Baud Rate:

• In general terms, baud rate of a data communications system is the number of symbols per second generated.

OR

• number of changes per second that the hardware generates

5. Amplitude, Frequency and Wavelength

5.1 Amplitude: Amplitude is magnitude (size) of signal.

5.2 Frequency: Frequency is the number of cycles per second – measured in Hz (cycles per second).

5.3 Wavelength: Wavelength is the distance between identical points in the adjacent cycles of a waveform signal propagated in space or along a wire.

Frequency of Digital Signals:

in bits/second second signal is approximately twice the frequency of the first

6. Bandwidth, Effect of Noise, Baud Rate and data Rate

6.1 Bandwidth:

• The amount of data that can be carried from one point to another in a given time period (usually a second).

Or

Range of frequencies that medium can pass.

• bandwidth is normally expressed in terms of bits per sec. (bps)

• Hi-Fi audio amplifier bandwidth of 25Hz to 250000Hz to give good quality sound.

• Voice telephone bandwidth 300Hz to 3400Hz – does not need good quality.

Bandwidth = High frequency – Low frequency

6.2 Effect of noise on bandwidth:

• All transmission media are degraded by ‘noise’.

• Shannons Theorem:

Cmax = B log2 (1 + SNR)

SNR = 10 log10 (S/N)

Cmax is called the channel capacity of the medium

i.e. the max speed at which data can be transmitted down that channel

B is the bandwidthS is the power of the signalN is the power of the noiseSNR is the Signal To Noise ratio. The unit for SNR is decibels.

• The greater the value of SNR, the greater the value of Cmax, i.e. the more

noise the less the channel capacity.

• SNR always calculated at receiver end. Because signal processing takes place at receiver end to remove the unwanted noise.

Doubling the bandwidth doubles the data Rate. At the same noise level it also increases the error rate. HOW?

6.3 What is Difference Between Bit rate and Baud Rate?

• baud rate - number of changes per second that the hardware generates.

• data rate - bits per second - equal or higher than baud rate.

• The bandwidth of many systems is limited (e.g. POTS was 3KHz) so how can higher data rates be achieved?

• use more than 2 signaling levels. (I.e. increasing the bits per second and keeping baud rate same).

• Number of changes made per second is known as the baud rate

• Baud rate is same as data rate only for 2-level signals

• e.g. for 4-level signals, data rate in bps (bits/sec) is twice that for 2-level signal

POTS can only support up to 3000 baud (in practice up to about 10000 baud)

Using a mixture of modulation techniques (see below) data rates of more than 50,000 bps are possible

7. Composition of Signal

• Every wave-form, no matter how complex, can be considered to be a combination of simple sine-waves of different frequencies.

7.1 What about digital signals?

• Square wave = sum of base frequency + an infinite no. of odd harmonics.

• sin(Ø) + sin(3Ø) / 3 + sin(5Ø) / 5 + sin(7Ø) / 7 + sin(9Ø) / 9 + sin(11Ø) / 11

• All components must be present simultaneously for the wave to be truly square.

• If any are absent or removed, the wave will lose some of its ‘squareness’ and become distorted.

• up to 9th harmonic – contains over 95% of signal power – then still possible to receive/ interpret the signal

7.2 A square wave build up from a fundamental and odd harmonics:

• Fundamental + 3rd harmonic (showing fundamental, 3rd harmonic and result).

• Fundamental + 3rd + 5th harmonic (showing 5th harmonic and result).

• Fundamental + 3rd + 5 th + 7th + 9th + 11th harmonic (showing 11th harmonic and result).

8. Digital Signal Transmission Problems

8.1 Attenuation

• As signal travels, its amplitude decreases due to losses in the medium.

• conductors have resistance so losses appear as heat.

• Thus for a given medium there will be a maximum distance beyond which communication is unreliable.

• 5 volt signal applied at one end appears as 2 volts at the other end.

• Solution? Repeaters and Amplifiers.

8.2 Band-Limiting:

• If the medium is of poor quality, it will not be able to change value at the rate of the higher frequency components of the signal.

• i.e. high frequency components cannot pass.

• e.g. When a square wave arrives at the receiver, it appears distorted.

• Can the receiver still detect the difference between a ‘1’ bit and a ‘0’ bit ?

• If not , we say that digital signals are unable to pass through this medium reliably.

• Solution? using good quality cables and connectors.

8.3 Delay

• The time taken to send a block of data across a network link is mostly influenced by 2 factors.

• Transmission Delay Time taken to put the bits of the block onto the medium.

• Transmission Delay = Length of block in bits / Data Rate

• Propagation Delay Time taken for a signal to travel from the transmitter to the receiver.

• Propagation Delay = Length of the link / Velocity of Propagation.

• Overall Delay = Transmission Delay + Propagation Delay.

• Solution?

8.4 Delay distortion

• Higher frequency components of a wave travel faster than lower frequency components. • Therefore all components will not arrive at the receiver at the same time!

• The diagram below shows a square wave is distorted: (up to the 11th harmonic) where the 3rd harmonic is 10 degrees before the fundamental, the 5th is 20 degrees, etc. (the fundamental and 3rd harmonic are also shown so one can see the phase shift)

8.5 Noise:

• The background electrical environment will induce rogue signals in the medium.

• This may alter the shape of the signal and cause a ‘0’ bit to be read as a ‘1’ or vice versa.

• The following diagram shows a square wave (up to 11th harmonic) with a noise level of 0.5 signal level.

• Solution:

• Fibre Optic media instead of Copper Media.• STP instead of UTP.• using separate route for electricity and data cables.

9. Modulation

9.1 Definition:

• Modulation is technique used to Encode 1s and 0s in an analogue medium by altering the amplitude, frequency or phase of an analogue carrier wave.

9.2 Amplitude Modulation:

• Voltage applied to signal varied over time.

• In this example a 0 signal is twice the amplitude of a 1.

9.3 Frequency Modulation:

• varies frequency of signal with respect to time.

• In this example a 1 signal is twice the frequency of a 0.

9.3 Phase Modulation:

• Implemented by Phase change.

• Phase change can be 0, 90, 180 or 270 degree.

• In this example a 0 signal has no phase change a 1 signal has a 180degree phase change

10 Analogue Communication 10.1 Problem:

• The analogue medium would distort the digital signals.

• The old analogue telephone system had a bandwidth 300Hz to 3400Hz thus the higher frequency components of a digital signal would be lost.

10.2 Solution? Modems

• Modems modulates a carrier wave in response to the transmitted digital data.

 Modulator > MODEM < Demodulator

 • Modems allow data to be sent down ordinary analogue telephone line.

• Takes stream of binary data (0/1) and imposes it on an analogue signal (modulation).

• Receiving end takes analogue signal and demodulates it to recover digital data (0s and 1s).

• PCs connected via modems and the telephone system

10.2 Quadrature Amplitude Modulation

• Consider phase modulation using 0, 900, 1800 and 2700 shifts.

00 900

1800 2700

10.3 Importance of Modulation? 

• Combinations

• Can use two or more of these modulation effects to increase the data rate (bps) without altering the signalling rate (baud)

 • e.g. Combining phase and amplitude modulation

 • If we allow:

(a) 2 possible phase shifts

(b) 2 possible amplitude values 

therefore 1 of 4 possible signal values can be sent 

therefore each signal can be used to encode 2 bits (00 01 10 and 11) 

therefore Data Rate = 2 x Signalling Rate 

therefore can transmit at 4800bps over a 2400 baud line

10.4 Implementation: 

• 2 possible amplitudes and 4 possible phase shifts

Bit Values Amplitude of Generated Signal

Phase Shift of Generated Signal

000 Lower 00

001 higher 00

010 Lower 900

011 higher 900

100 Lower 1800

101 higher 1800

110 Lower 2700

111 higher 2700

therefore

Data Rate = 3 x Signalling Rate

11 Media technologies:

• Media technologies used include.

• copper cable.• optical fibre.• radio and microwave links. (The latter often used on large sites or to link separate sites which are in line-of-sight of each other)

11.1 Copper media:

Coaxial used to be very popular (had 57% of LANS in late 1980's)

• well-shielded, rugged, versatile.

• high-speed (10 to 1000 Mbits/sec) over medium distances .

• moderately priced.

• Example: 10base2 Ethernet and 10base5 (10 Mbits/sec)

Twisted Pair (UTP and STP)

• most popular for LANs.

• as used in telephone links - very cheap - easy to install.

• Speed: up to 10Mbits/sec typical, 100+ Mbits/sec with good shielding.

• Example: 10baseT Ethernet (10 Mbits/sec) and 100baseT Ethernet (100 Mbits/sec)

11.2 Optical media (Fibre Optic cables); can be multimode or single mode.

• used where noise immunity is critical and over long distances.

• very high speed (>G bits/sec possible) limited by transceiver complexity and price.

• immune to EMI (electromagnetic interference), e.g. from nearby cables and equipment.

• The problem with twisted pair and coaxial cable is that signal loss and noise over a long distance can cause problems.

• Fibre optic cable has low loss, high noise immunity and very large bandwidth (thus many separate signals can be sent down the same cable using Frequency Division Multiplexing).  

• Fibre optic cable is difficult to tap and tends to be used for long distances (1000's of metres) and in environments subject to large amounts of electromagnetic noise (e.g. a factory).

11.3 Wireless• Problem: with LANs is complexity and cost of physically wiring buildings.

• WLANs (wireless local-area network) can overcome this problem.

• 802.11b is the standard WLAN technology for both business and home.

• Provides 11 Mbps transmission (with a fallback to 5.5, 2 and 1 Mbps) in the 2.4 GHz band.

• Modes of Operation can be; Infrastructure or Adhoc.

WLAN problem: any nearby station can pick up the signal and tap it

Answer: use WEP (wireless encryption protocol)