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Computer Science Books/Sem7/Wt

Compiled By: Mr. Nilesh M. Patil [1] WIRELESS TECHNOLOGY SEMESTER-VII (CBGS) INFORMATION TECHNOLOGY BRANCH PREPARED BY MR. NILESH M. PATIL BATCH: 2016-17

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2 Notes Compiled By: Mr. Nilesh M. Patil IT Dept., RGIT CHAPTER 1 FUNDAMENTALS OF WIRELESS COMMUNICATION 1.1.Introduction  Wireless systems have a unique capability of maintaining the same contact number even if one moves from one location to another. This has made them increasingly popular.  The wireless telephones are not only convenient but also provide flexibility and versatility; there have been growing number of wireless phone subscribers as well as service providers.  A combination of wireless communication and computer technologies has revolutionized the world of telecommunications.  Wireless and mobile communications have found usefulness in areas such as commerce, education, defense, etc.  According to the nature of a particular application, they can be used in home- based and industrial systems or in commercial and military environment.  There can many novel applications of such a wireless system; for example, a bracelet worn can constantly monitor the body parameters and take needful actions (like informing the family physician about the problem).  In commercial system, the wireless communications can be employed for purchase or selling of goods and services, playing audio and video, payment of telephone bills, payment of electricity bills, airline/ railway/ bus reservations, etc.  The difference between wireless and mobile devices is not much and they are used interchangeably. However, mobile just means portable.  A laptop is a mobile device, as is a personal digital assistant (PDA). A desktop would be a mobile device if you had the inclination to carry it around with you.  A wireless device has some sort of network connectivity. A cell phone is wireless, and a laptop or a PDA would be wireless if they had a wireless modem.  Similarly, applications are wireless when they connect and exchange data with a network. 1.2.Wireless Communication System  The major function of the communication system is to convert information into a format appropriate for the transmission medium and to modulate analog signals or bits for transmission over channel.  The channel (either wired or wireless medium) propagates the electromagnetic waves (signals)and the intended recipient picks the signal. Wireless communication systems exchange electronic data among different users through a wireless media.  Analog communication systems convert (modulate) analog signals into modulated (analog) signals whereas digital communication systems convert information in

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3 Notes Compiled By: Mr. Nilesh M. Patil IT Dept., RGIT the form of bits into digital signals. Computers naturally generate information as bits.  Analog signals can be converted into bits by quantizing and digitizing for use in digital communication.  A typical wireless system communication consisting of sender and receiver is shown in figure 1.1 below. Figure 1.1 Wireless Communication System  The sender receives the information from the source and encodes it using a source encoder.  The source encoder encodes the information into a binary data sequence. Methods for source encoding are waveform coding, linear predictive coding, etc.  The channel encoder encodes the signal for error detection and correction by adding some redundant bits. Methods for channel encoding are hamming codes, cyclic codes, block codes, etc.  The encoded signal is modulated by using the digital modulation schemes such as binary phase shift keying (BPSK), quadrature phase shift keying (QPSK), minimum shift keying (MSK), Gaussian MSK (GMSK), etc.  The transmitter antenna, which is the interface between the transmitter hardware and the channel, converts the modulated signal into a suitable energy form that can be transmitted through the physical channel.  The transmitted waveforms propagate along the channel and presumably reach to the receiver antennas with additional noise whose extent mostly depends on the physical characteristics of the channel.  The waveforms detected at the receiver antennas are transmitted to the demodulator which converts the signal into a binary bit stream according to a predefined demodulation scheme.  The function of the channel decoder is the removal of the redundant bits.  The source decoder converts the binary stream into the symbols, and transmits them to the destination as the recovered information. The main design goals of the transmitter and the receiver are to mitigate distortion and noise from the channel. Performance metric for analog system is fidelity, whereas digital systems are analyzed based on data rate and bit error probability as performance metrics.

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4 Notes Compiled By: Mr. Nilesh M. Patil IT Dept., RGIT  Fidelity: Fidelity describes how close is the received signal to the original signal. Fidelity defines acceptability.  Data Rate: Data rates over channels with noise have a fundamental capacity limit. Data rate is limited by signal power, noise power, distortion, and bit error probability. Without distortion or noise, we can have maximum data rate with zero bit error probability.  Bit Error Probability: It is defined as the ratio of the number of bits, elements, characters, or blocks incorrectly received to the total number of bits, elements, characters, or blocks sent during a specified time interval. For example, if 10 bits are altered when 10,000 bits are transmitted, the bit error probability equals 10/10,000 = 0.001. An important parameter in communication channel is bandwidth. For digital communications, bandwidth of a channel is defined as the maximum number of bits transmitted in a second (bits per second or bps) whereas for analog systems bandwidth is defined in terms of hertz (Hz). Shannon capacity defines maximum possible data rate for systems with noise and distortion. In noisy channel, data rate C is defined as C = B log2 (1 + S/N) bps where B is the bandwidth in Hz and S/N is the signal-to-noise ratio. Problem 1.1 In a communication channel, the channel bandwidth is 3.4 kHz and output S/N power ratio is 20dB. Calculate the channel capacity. Solution: Given, channel bandwidth, B = 3.4 kHz Output S/N power ratio = 20 dB Therefore, 10 log10 S/N = 20 dB log10 S/N = 2 S/N = 102 = 100 Channel capacity is C = B log2 (1 + S/N) bps = 3.4 * 103log2(1 + 100) bps C = 22.638Kbps Problem 1.2 Calculate the minimum SNR required to support information transmission through the telephone channel of bandwidth 3.4 kHz at the data rate of 4800 bps. Solution: Given, channel data rate, C = 4800 bps; bandwidth, B = 3.4 kHz Channel capacity is C = B log2 (1 + S/N) bps

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5 Notes Compiled By: Mr. Nilesh M. Patil IT Dept., RGIT 4800 = 3.4 * 103log2(1 + S/N) 1.411 = log2(1 + S/N) 1 + S/N = 21.411 1 + S/N = 2.659 S/N = 1.659 (S/N) dB = 10 log101.659 = 2.2 dB S/N = 2.2 dB Problem 1.3 In a communication channel, the bandwidth is 10 MHz and SNR is 100. (a) Determine the channel capacity. (b) If SNR drops to 10, how much bandwidth is needed to achieve the same channel capacity as in (a). Solution: Given, channel bandwidth = 10MHz and S/N = 100 Channel capacity is C = B log2 (1 + S/N) bps C = 10 * 106 log2 (1 + 100) C = 66.6 Mbps If the SNR drops to 10, that is, S/N = 10, the bandwidth is B = C/ log2(1 + S/N) = C/log2(1 + 10) B = 66.6 * 106 / 3.47 B = 19.19 MHz 1.3.Advantages  Users can move around freely within the area of the network with their laptops, handheld devices, etc. and get an internet connection.  Users are also able to share files and other resources with other devices that are connected to the network without having to be cabled to a port.  Not having to lay lots of cables and put those through walls etc. can be a considerable advantage in terms of time and expense. It also makes it easier to add extra devices to the network, as no new cabling is needed.  If you are a business such as a cafe, having a wireless network that is accessible to customers can bring you extra business. Customers generally love wireless networks because they are convenient.  Wireless networks can sometimes handle a larger amount of users because they are not limited by a specific number of connection ports.  Instant transfer of information to social media is made much easier. For instance, taking a photograph and uploading it to Facebook can generally be done much quicker with wireless technology.

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6 Notes Compiled By: Mr. Nilesh M. Patil IT Dept., RGIT 1.4.Limitations (a) Bandwidth  Bandwidth is still a limited resource in wireless environments. When transmitting data, users must sometimes send smaller bits of data to accommodate within the available bandwidth so that the information moves very quickly.  The size of the device that is accessing the information is also still an issue. Most recent phones and PDAs have small screens, often only a couple of inches in diameter, and smaller memory, and it is hard to read large documents on them. These may require information of lesser bandwidth.  Larger computing devices connected in wireless environments may require more bandwidth information as there is no constraint on the screen and the memory.  The available wireless local networks in the market operate with a maximum of 55 Mbps whereas some of the user‘s applications aggregate demand is10 Gbps. This imbalance forces us to have clever wireless networking environments.  Many applications need to be reconfigured if they are going to be used through wireless connections. Most client/server applications rely on a persistent connection, which is not the case with wireless applications. Transactional systems require safeguards for dropped wireless connections (due to bandwidth limitations). (b) Frequency Spectrum  The frequency spectrum is limited and finite. The number of users who can be connected to a wireless network at a given time are limited. However, dynamic channel allocation schemes can be used to optimize the frequency usage of the given wireless communication area. (c) Power  The power density from a wireless antenna decreases rapidly with the square of the distance as one move away from the antenna. However, because radio frequency (RF) energy travels as waves, there are effects from reflections, interactions among waves from multiple antennas, and spikes of intensity due to each antenna pattern. This produces a pattern of peaks and valleys in field intensity as one move away from the source.  The intensity of RF energy depends onseveral factors, including design characteristics of the antenna, power transmitted to the antenna, height of the antenna, and distance from the antenna. (d) Interference  Radio transmission cannot be protected against interference using shielding as this is done in coaxial cable or shielded twisted pair.  For example, electrical engines and lightning cause severe interference and result in higher loss rates for transmitted data or higher bit error rates respectively.

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7 Notes Compiled By: Mr. Nilesh M. Patil IT Dept., RGIT (e) High delays, large delay variation  A serious problem for communication protocols used in today‘s Internet (TCP/IP) is the big variation in link characteristics.  In wireless systems, delays of several seconds can occur, and links can be very asymmetrical (i.e., the links offer different service quality depending on the direction to and from the wireless device).  Applications must be tolerant and use robust protocols. (f) Lower security, simpler to attack  Not only can portable devices be stolen more easily, but the radio interface is also prone to the dangers of eavesdropping.  Wireless access must always include encryption, authentication, and other security mechanisms that must be efficient and simple to use. 1.5.Applications The following applications describe the need of wireless communications. (a) Vehicles  Transmission of news, road conditions, weather, music via Digital Audio Broadcasting (DAB)  Current position of the vehicle can be known via the Global Positioning System (GPS)  Cars in the same area could build a local ad-hoc network to ensure a minimum safe distance from other cars. This network could also be used to alert other cars and hospitals in case of an accident. (b) Emergencies and Natural Disasters  The condition of the patient can be transmitted to the hospital from the ambulance itself. The hospital can then make the necessary arrangements to speed up the network.  In case of disasters like earthquakes, cyclones, heavy rains, etc. most of the wired networks and the infrastructure based networks completely fail. On demand, ad-hoc networks are the only way for communication in such cases.  Ad-hoc wireless networks are also useful on the battlefield (in wars) as the existent communication network might have already been destroyed (jammed) by the enemy. (c) Replacement of Fixed (Wired) Networks  Remote sensors used for weather forecasts, earthquake detection, etc. can be wireless, this allows freedom from miles of cabling.  Instead of fixed networks, wireless networks can be used for information display and enforcing security measures in historical monuments, as the cabling required for the fixed network may cause damage to the monument. (d) Businesses  A travelling salesman can have instant access to the company‘s database. Thus, he can easily provide the latest product information to the customers.  Also, the managers can keep track of the performance of the salesman.  The laptop can truly be turned into a mobile device.

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8 Notes Compiled By: Mr. Nilesh M. Patil IT Dept., RGIT (e) Infotainment and Entertainment  Wireless networks can provide instant outdoor internet access.  Travel guides can push historical information about a building or a place on to the user‘s mobile devices such as a PDA or a mobile phone.  Wireless networks can be used to set up ad-hoc gaming consoles during online gaming competitions. (f) Transport  Wireless communication could be used to create and co-ordinate car sharing schemes amongst villages where public transport modes are not easily available. (g) Micro-Commerce  Small businesses in rural areas often have to travel significant distances to markets or other places where they can distribute their goods, and hence they cannot make arrangements in advance with the buyers.  Mobile phones could significantly change the logistical issues faced by traders and home entrepreneurs, by providing affordable mobile-based ordering systems and the ability to make more reliable and advance arrangements with business partners or clients. (h) Healthcare  New mobile services in rural areas could allow for better connectivityamong rural communities.  Villages can now create networks to share and discuss health information and advice. (i) Education  Educational services could be provided to children in remote villages and communities, particularly where personal computers or connections to the Internet are not available.  Mobile phones could serve as an essential means for children to connect to one another for educational and peer-learning activities.  These are particularly important for communities that are either nomadic or transitional on account of displacements due to a natural disaster or for other reasons. 1.6.Wireless Media  ―Wireless‖ means transmitting signals over invisible radio waves instead of wires.  Garage door openers and television remote controls were the first wireless devices to become a part of everyday life.  Even though many mobile and wireless devices are available, there will be manymore in the future. There is no precise classification of such devices, by size,shape, weight, or computing power.

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9 Notes Compiled By: Mr. Nilesh M. Patil IT Dept., RGIT  Currently, laptops are considered the upper end of the mobile device range. The following list gives some examples of mobile and wireless devices graded by increasing performance (CPU, memory, display, input devices etc.).  However, there is no sharp line between the categories and companies tend to invent more and more new categories.

  1. Sensors: A very simple wireless device is represented by a sensor transmitting state information. One example could be a switch sensing the office door. If the door is closed, the switch transmits this to the mobile phone inside the office which will not accept incoming calls. Without user interaction, the semantics of a closed door is applied to phone calls.
  2. Embedded controllers: Many appliances already contain a simple or sometimes more complex controller. Keyboards, mice, headsets, washing machines, coffee machines, hair dryers and TV sets are just some examples. Why not have the hair dryer as a simple mobile and wireless device (from a communication point of view) that is able to communicate with the mobile phone? Then the dryer would switch off as soon as the phone starts ringing – that would be a nice application!
  3. Pager: As a very simple receiver, a pager can only display short text messages, has a tiny display, and cannot send any messages. Pagers can even be integrated into watches. The tremendous success of mobile phones has made the pager virtually redundant in many countries. Short messages have replaced paging. The situation is somewhat different for emergency services where it may be necessary to page a larger number of users reliably within short time.
  4. Mobile phones: The traditional mobile phone only had a simple black and white text display and could send/receive voice or short messages. Today,mobile phones migrate more and more toward PDAs. Mobile phones with fullcolor graphic display, touch screen, and Internet browser are easily available.
  5. Personal digital assistant: PDAs typically accompany a user and offersimple versions of office software (calendar, note-pad, mail). The typicalinput device is a pen, with built-in character recognition translating handwritinginto characters. Web browsers and many other software packagesare available for these devices.
  6. Pocket computer: The next steps toward full computers are pocket computersoffering tiny keyboards, color displays, and simple versions of programsfound on desktop computers (text processing, spreadsheets etc.).
  7. Notebook/laptop: Finally, laptops offer more or less the same performanceas standard desktop computers; they use the same software – the only technicaldifference being size, weight, and the ability to run on a battery. Ifoperated mainly via a sensitive display (touch sensitive or electromagnetic),the devices are also known as notepads or tablet PCs. 1.7.Frequency Spectrum • An ordered array of the components of an emission or wave is called spectrum. • A frequency spectrum is the range of frequencies of electromagnetic radiation from zero to infinity.

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10 Notes Compiled By: Mr. Nilesh M. Patil IT Dept., RGIT Figure 1.2 Frequency Spectrum  Radio transmission can take place using many different frequency bands. Each frequency band exhibits certain advantages and disadvantages. Figure above gives a rough overview of the frequency spectrum that can be used for data transmission.The figure shows frequencies starting at 300 Hz and going up to over 300 THz.  Directly coupled to the frequency is the wavelength λ via the equation: λ = c/f, where c ≅ 3·108 m/s (the speed of light in vacuum) and f the frequency.  For traditional wired networks, frequencies of up to several hundred kHz are used for distances up to some km with twisted pair copper wires, while frequencies of several hundred MHz are used with coaxial cable (new coding schemes work with several hundred MHz even with twisted pair copper wires over distances of some 100 m). Fiber optics are used for frequency ranges of several hundred THz, but here one typically refers to the wavelength which is, e.g., 1500 nm, 1350 nm etc. (infra red).  Radio transmission starts at several kHz, the very low frequency (VLF) range. These are very long waves. Waves in the low frequency (LF) range are used by submarines, because they can penetrate water and can follow the earth‘s surface. Some radio stations still use these frequencies, e.g., between 148.5 kHz and 283.5 kHz in Germany. The medium frequency (MF) and high frequency (HF) ranges are typical for transmission of hundreds of radio stations either as amplitude modulation (AM) between 520 kHz and 1605.5 kHz, as short wave (SW) between 5.9 MHz and 26.1 MHz, or as frequency modulation (FM) between 87.5 MHz and 108 MHz. The frequencies limiting these ranges are typically fixed by national regulation and, vary from country to country. Short waves are typically used for (amateur) radio transmission around the world, enabled by reflection at the ionosphere. Transmit power is up to 500 kW – which is quite high compared to the 1 W of a mobile phone.  As we move to higher frequencies, the TV stations follow. Conventional analog TV is transmitted in ranges of 174–230 MHz and 470–790 MHz using the very high frequency (VHF) and ultra-high frequency (UHF) bands. In this range, digital audio broadcasting (DAB) takes place as well (223–230 MHz and 1452–1472 MHz) and digital TV is planned or currently being installed (470– 862 MHz), reusing some of the old frequencies for analog TV. UHF is also used for mobile phones with analog technology (450–465 MHz), the digital GSM (890–960 MHz, 1710–1880 MHz), digital cordless telephones following the DECT standard (1880–1900 MHz), 3G cellular systems following the

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11 Notes Compiled By: Mr. Nilesh M. Patil IT Dept., RGIT UMTS standard (1900–1980 MHz, 2020–2025 MHz, 2110–2190 MHz) and many more. VHF and especially UHF allow for small antennas and relatively reliable connectionsfor mobile telephony.  Super high frequencies (SHF) are typically used for directed microwavelinks (approx. 2–40 GHz) and fixed satellite services in the C-band (4 and6 GHz), Ku-band (11 and 14 GHz), or Ka-band (19 and 29 GHz). Some systemsare planned in the extremely high frequency (EHF) range which comes closeto infrared. All radio frequencies are regulated to avoid interference, e.g., theGerman regulation covers 9 kHz–275 GHz.  The next step into higher frequencies involves optical transmission, whichis not only used for fiber optical links but also for wireless communications.Infrared (IR) transmission is used for directed links, e.g., to connect differentbuildings via laser links. The most widespread IR technology, infrared data association(IrDA), uses wavelengths of approximately 850–900 nm to connectlaptops, PDAs etc. Finally, visible light has been used for wireless transmissionfor thousands of years. While light is not very reliable due to interference, but itis nevertheless useful due to built-in human receivers. 1.8.Spread Spectrum  As the name implies, spread spectrum techniques involve spreading the bandwidthneeded to transmit data. Figure 1.3 Spreading and Despreading of the User Signal  The figure 1.3 above shows the basic steps involved in this technology. (a) It is the idealized narrowband signal that is to be transmitted by the sender. (b) This narrowband signal is first converted in to broadband signal i.e. the signal is spread. The energy needed to transmit the signal is still the same; however, the required power level reduces. (c) During transmission, narrowband and broadband interference gets added to the spread signal. The sum of interference and user signal is received. (d) The receiver now knows how to despread the signal, converting the spread user signal into a narrowband signal again, while spreading the narrowband interference and leaving the broadband interference. (e) The signal is now applied to a band pass filter that cuts off the frequencies to the left and right of the narrowband signal. The original user signal can now be recovered.  The above scenario can also be applied simultaneously to many channels.

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12 Notes Compiled By: Mr. Nilesh M. Patil IT Dept., RGIT  Figure 1.4 below shows a snapshot of the channel quality of six channels using FDM. The situation may be completely different at the next instant. Figure 1.4 Multiple Channels Making use of Narrowband Signals  Each channel has its own narrowband frequency of transmission. Guard spaces are required between the channels to avoid adjacent channel interference.  Using FDM requires careful planning of the frequencies. Also, we can see that the quality of channels 3 and 4 is too bad (due to narrowband interference) to recover the data.  In order to solve these problems, we can apply spread spectrum to all the six channels.  As shown in figure 1.5 below, the narrow band signals for each of the channels are converted to broadband signals. Figure 1.5 Spread Spectrum to avoid Narrowband Interference  Each channel now uses the same frequency band, CDM has to be used to separate each channel.  Each channel has to be assigned its own code for transmitting and recovering data. Advantages of Spread Spectrum  Provides resistance to narrowband interference.  It allows co-existence of several signals without the need of dynamic co-ordination among the signals.  Relatively high security allows use in military applications. In fact, one of the key applications of spread spectrum is for building anti-jam communication systems, which is a communication system designed to resist intentional jamming (by the enemy) in a hostile environment. Disadvantages of Spread Spectrum  Increased complexity of the senders and receivers.  The spread signal requires a larger frequency band.

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13 Notes Compiled By: Mr. Nilesh M. Patil IT Dept., RGIT  The spread signals increase the background noise level and may interfere with other transmission.  Precise power control is necessary. 1.9.Direct Sequence Spread Spectrum (DSSS)  In this scheme, the user signal is spread by performing an XOR with a fixed sequence called as a chipping sequence.  As shown in figure 1.6, a user signal 01, is XORed with the chipping sequence 0110101, the resulting signal is either 0110101 (if the user bit is 0) or its complement 1001010 (if the user bit is 1). Figure 1.6 Spreading of signal using DSSS  Each user bit has duration of tb. The chipping sequence consists of smaller pulses called chips each having duration of tc.  The spreading factor S = tb/tc, determines the bandwidth of the spread signal. Thus, if the original signal needed a bandwidth of B, the spread signal would require a bandwidth of S*B.  Barker codes are usually used for spreading the signals as these codes are insensitive to multipath propagation and also exhibit good robustness against interference. DSSS Transmitter  The figure 1.7 shows the typical DSSS transmitter.  The user data is first XORed with the chipping sequence to obtain the spread signal.  This signal is then converted to an analog baseband signal by digital modulation.  The baseband signal is then subject to analog modulation and transmitted.

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14 Notes Compiled By: Mr. Nilesh M. Patil IT Dept., RGIT Figure 1.7 DSSS Transmitter DSSS Receiver  The DSSS receiver is comparatively more complex than the transmitter (refer figure 1.8).  The received signal is first converted to the baseband signal by the demodulator. Additional mechanisms also need to be applied as data may be distorted due to noise and multipath propagation.  The low pass filtered signal is now applied to a correlator; the correlator performs two steps which need precise synchronization with the sender.  Firstly, the signal is once again XORed with a chipping sequence to generate products; this chipping sequence is the same as that used by the transmitter to transmit the data.  The integrator then adds all the products.  For each bit period, the output of the integrator is fed to the decision unit that decides whether the user data is a binary 0 or 1. Figure 1.8 DSSS Receiver Advantages of DSSS  Reduces frequency selective fading.  In case of cellular systems, several base stations can use the same frequency for transmission.  A soft handover is possible using DSSS. Disadvantages of DSSS  The overall system is complex to implement.  Precise power control required.  Synchronization required between the sender and the receiver.

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15 Notes Compiled By: Mr. Nilesh M. Patil IT Dept., RGIT 1.10. Frequency Hopping Spread Spectrum (FHSS)  For frequency hopping spread spectrum (FHSS) systems, the total available bandwidth is split into many channels of smaller bandwidth plus guard spaces between the channels.  Transmitter and receiver stay on one of these channels for a certain time and then hop to another channel. This system implements FDM and TDM.  The pattern of channel usage is called the hopping sequence, the time spend on a channel with a certain frequency is called the dwell time.  FHSS comes in two variants, slow and fast hopping (refer figure 1.9) Figure 1.9 Slow Hopping and Fast Hopping Sr. No. Slow Hopping Fast Hopping 1 Several user bits transmitted at the same frequency. Therefore, td>tb Several frequencies may be used to transmit a single user bit. 2 Provides lesser resistance to narrowband interference. Better resistance against narrowband interference and frequency selective fading as compared to slow hopping. 3 Lower security as compared to fast hopping. Better security as compared to slow hopping. 4 Slow hopping systems are cheaper and have relaxed tolerances. Comparatively costlier with smaller tolerances. 5 Very tight synchronization is not required. Very tight synchronization is required. 6 Can be used GSM. Used by Bluetooth.

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16 Notes Compiled By: Mr. Nilesh M. Patil IT Dept., RGIT FHSS Transmitter  The simplified block diagram of FHSS transmitter is shown in the figure 1.10.  The user data is first converted to a narrowband signal using digital modulation (FSK or BPSK).  Frequency hopping is then performed using the hopping sequence. The hopping sequence is applied to the frequency synthesizer that generates the corresponding carrier frequencies.  Analog modulation is then applied to shift the narrowband frequency by the carrier frequency.  Thus, the spread signal is generated.  The hopping sequences used by various transmitters should have low cross- correlation among them. Figure 1.10 FHSS Transmitter FHSS Receiver  As shown in figure 1.11, the reverse process needs to be applied at the receiver.  The received signal is first subject to a demodulation process to generate the narrowband signal. The same hopping sequence used to spread the data needs to be regenerated at the receiver and then applied to the frequency synthesizer.  The narrowband signal is then demodulated again to get the user data. Figure 1.11 FHSS Receiver

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17 Notes Compiled By: Mr. Nilesh M. Patil IT Dept., RGIT Advantages of FHSS  Spreading of signals is simpler.  Frequency selective fading and interference is limited to short period only.  Use only a small portion of the total bandwidth at any time. Disadvantages of FHSS  Not as robust as DSSS.  Signals are easier to detect, thus, lower security. Sr. No. DSSS FHSS 1 Implementation is complex. Simple to implement. 2 At any time, it uses all of the total available bandwidth. It uses only a small portion of the available bandwidth at any time. 3 It provides better security; without knowing the spreading code it is very hard to detect the signal. Signals are easier to detect. 4 More resistant to frequency selective fading. Less resistant to frequency selective fading. 5 More resistant to multipath propagation. Less resistant to multipath propagation. 1.11. Multiplexing  Multiplexing describes how several users can share a medium with minimum or no interference.  In wireless communication, multiplexing can be carried out in four dimensions viz. space, time, frequency, and code.  The goal of multiplexing is to assign space, time, frequency, and code to each communication channel for maximum medium utilization and minimum interference. (a) Space Division Multiplexing (SDM)  Consider six communication channels k1 to k6.  We use 3D co-ordinate system with frequency, time, and code as x, y and z- axis respectively.  Each of the channels are mapped on to the spaces (figure 1.12 shows for only three channels).

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18 Notes Compiled By: Mr. Nilesh M. Patil IT Dept., RGIT Figure 1.12 Space Division Multiplexing  The interference range of one space should not overlap with the interference range of the other space.  The space between interference ranges is called as the guard space.  Thus, SDM separates the senders via a space wide enough to avoid interference.  Advantages: Very simple and easy to implement.  Disadvantages: Clearly causes a waste of space. If multiple users want to use the space for communication, other multiplexing schemes like FDM, TDM or CDM need to be applied. (b) Frequency Division Multiplexing (FDM)  As shown in figure 1.13, this scheme divides the frequency dimension into several non-overlapping frequency bands.  Each channel is allocated its own frequency band.  Guard spaces are required between the bands to avoid adjacent channel interference.  To receive the signal, receiver has to tune in to a particular frequency.  Advantages:  No complex co-ordination required between the sender and the receiver.  Works for analog signals also.  Disadvantages:  Waste of bandwidth if the traffic is distributed unevenly.  Inflexible

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19 Notes Compiled By: Mr. Nilesh M. Patil IT Dept., RGIT Figure 1.13 Frequency Division Multiplexing (c) Time Division Multiplexing (TDM)  Each channel is given the entire bandwidth for a certain period of time.  As shown in figure 1.14, all the channels use same frequency but at different point of time.  To avoid co-channel interference, guard spaces are required.  In order to receive the signal, receive need to tune in at the right time.  Advantages:  Flexible: Time allocation can be done with respect to load.  Efficient channel utilization  High throughput  Disadvantages:  Tight synchronization required between the sender and the receiver.  Complex to implement. Figure 1.14 Time Division Multiplexing

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20 Notes Compiled By: Mr. Nilesh M. Patil IT Dept., RGIT (d) Code Division Multiplexing (CDM)  As shown in figure 1.15, in this scheme channel separation is done with the help of codes i.e. each channel is assigned a unique code.  All the senders now transmit at the same time with same frequency but different codes.  Guard spaces are required to avoid interference.  Advantages:  Efficient bandwidth utilization  No co-ordination required between sender and receiver  Provides good protection against tapping and interference  Disadvantages:  Lower data rates  Precise power control is necessary. Figure 1.15 Code Division Multiplexing 1.12. Modulation Techniques  Modulation is the process of converting analog or digital information to a waveform suitable for transmission over a given medium.  Consider a cosine function g(t) = At cos(2π ftt+ φt).  This function has three parameters: amplitude At, frequency ft, and phase φtwhich may be varied in accordance with data or another modulating signal.  For digital modulation, which is the main topic in this section, digital data (0 and 1) is translated into an analog signal (baseband signal).  Digital modulation is required if digital data has to be transmitted over a medium that only allows for analog transmission.

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21 Notes Compiled By: Mr. Nilesh M. Patil IT Dept., RGIT (a) Amplitude shift keying  Figure 1.16 illustrates amplitude shift keying (ASK), the most simple digital modulation scheme.  The two binary values, 1 and 0, are represented by two different amplitudes. In the example, one of the amplitudes is 0 (representing the binary 0).  This simple scheme only requires low bandwidth, but is very susceptible to interference.  Effects like multi-path propagation, noise, or path loss heavily influence the amplitude.  In a wireless environment, a constant amplitude cannot be guaranteed, so ASK is typically not used for wireless radio transmission.  However, the wired transmission scheme with the highest performance, namely optical transmission, uses ASK. Here, a light pulse may represent a 1, while the absence of light represents a 0.  ASK can also be applied to wireless infra red transmission, using a directed beam or diffuse light. Figure 1.16 Amplitude Shift Keying (b) Frequency shift keying  A modulation scheme often used for wireless transmission is frequency shiftkeying (FSK) (see figure 1.17).  The simplest form of FSK, also called binary FSK (BFSK), assigns one frequency f1 to the binary 1 and another frequency f2 to the binary 0.  A very simple way to implement FSK is to switch between two oscillators, one with the frequency f1 and the other with f2, depending on the input.  To avoid sudden changes in phase, special frequency modulators with continuous phase modulation, (CPM) can be used. Sudden changes in phase cause high frequencies, which is an undesired side-effect.

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22 Notes Compiled By: Mr. Nilesh M. Patil IT Dept., RGIT  A simple way to implement demodulation is by using two bandpass filters, one for f1 the other for f2.  A comparator can then compare the signal levels of the filter outputs to decide which of them is stronger.  FSK needs a larger bandwidth compared to ASK but is much less susceptible to errors. Digital information 1 0 1 1 0 0 1 Carrier 1 (frequency #1) FSK modulated signal Carrier 2 (frequency #2) Frequency varying- amplitude constant Figure 1.17 Frequency Shift Keying (c) Phase shift keying  Finally, phase shift keying (PSK) uses shifts in the phase of a signal to represent data.  Figure 1.18 shows a phase shift of 180° or π as the 0 follows the 1 (the same happens as the 1 follows the 0).  This simple scheme, shifting the phase by 180° each time the value of data changes, is also called binary PSK (BPSK).  A simple implementation of a BPSK modulator could multiply a frequency f with +1 if the binary data is 1 and with –1 if the binary data is 0.  To receive the signal correctly, the receiver must synchronize in frequencyand phase with the transmitter. This can be done using a phase lock loop (PLL).  Compared to FSK, PSK is more resistant to interference, but receiver and transmitterare also more complex.

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23 Notes Compiled By: Mr. Nilesh M. Patil IT Dept., RGIT Figure 1.18 Phase Shift Keying 1.13. Multiple Access Methods  Multiple access techniques are used to allow a large number of mobile users to share the allocated spectrum in the most efficient manner.  As the spectrum is limited, so the sharing is required to increase the capacity of cell or over a geographical area by allowing the available bandwidth to be used at the same time by different users.  And this must be done in a way such that the quality of service doesn‘t degrade within the existing users. (a) Space Division Multiple Access (SDMA)  It implements SDM.  It divides the available space into cells and further divides the cells into sectors. This allows for frequency reuse in different sectors by using directed antennas.  Typically, SDMA is never used in isolation. It is always combined with some other schemes like FDMA, TDMA, CDMA, etc.  A new application of SDMA comes up together with beam-forming antenna arrays.  Single users are separated in space by individual beams. This can improve the overall capacity of a cell (e.g., measured in bit/s/m2 or voice calls/m2) tremendously. (b) Frequency Division Multiple Access (FDMA)  It implements FDM.  In pure FDMA, non-overlapping frequency bands are assigned to users on a continuous time basis.

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24 Notes Compiled By: Mr. Nilesh M. Patil IT Dept., RGIT  FDMA is usually combined with TDMA (as in GSM); frequency allocation can now vary with time according to a certain pattern called as the hopping pattern.  In cellular networks, FDM is often used to provide the base station and the mobile station a simultaneous access to the medium. The two partners establish a duplex channel.  The scheme itself is called as Frequency Division Duplex (FDD). The mobile station communicates with the base station at one frequency whereas the base station communicates with the mobile host at another frequency.  Thus, the communication in the two directions is now separated via two different frequencies.  The frequency from the mobile station to the base station is called as uplink frequency and the frequency from the base station to the mobile station is called as downlink frequency. (c) Time Division Multiple Access (TDMA)  It implements TDM.  Different users can access the medium at different time.  This is much flexible scheme as compared to pure FDMA.  In this mode, the receiver just has to tune in to the frequency at the right time.  This technique works well with slow voice data signals, but it‘s also useful for compressed video and other high-speed data.  The basic GSM (Global System of Mobile Communications) cellular phone system is TDMA-based. (d) Code Division Multiple Access (CDMA)  CDMA (Code-Division Multiple Access) is a channel access method used by various radio communication technologies.  It is a form of multiplexing, which allows numerous signals to occupy a single transmission channel, optimizing the use of available bandwidth.  The technology is used in ultra-high-frequency (UHF) cellular telephone systems in the 800-MHz and 1.9-GHz bands.  CDMA employs analog-to-digital conversion (ADC) in combination with spread spectrum technology.  Audio input is first digitized into binary elements.  The frequency of the transmitted signal is then made to vary according to a defined pattern (code), so it can be intercepted only by a receiver whose frequency response is programmed with the same code, so it follows exactlyalong with the transmitter frequency.  There are trillions of possible frequency-sequencing codes, which enhance privacy and makes cloning difficult.  The original CDMA standard, also known as CDMA One and still common in cellular telephones in the U.S offers a transmission speed of only up to 14.4 Kbps in its single channel form and up to 115 Kbps in an eight-channel form. CDMA2000 and Wideband CDMA deliver data many times faster.

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25 Notes Compiled By: Mr. Nilesh M. Patil IT Dept., RGIT  CDMA devices use a rake receiver, which exploits multipath delay components to improve the performance of the system.  In a CDMA system, the same frequency can be used in every cell, because channelization is done using the pseudo-random codes.  Reusing the same frequency in every cell eliminates the need for frequency planning in a CDMA system.  CDMA systems use the soft hand off, which is undetectable and provides a more reliable and higher quality signal. (e) Carrier Sense Multiple Access (CSMA)  Carrier sense multiple access (CSMA) is a probabilistic media access control (MAC) protocol in which a node verifies the absence of other traffic before transmitting on a shared transmission medium, such as an electrical bus, or a band of the electromagnetic spectrum.  Carrier sense means that a transmitter uses feedback from a receiver to determine whether another transmission is in progress before initiating a transmission. That is, it tries to detect the presence of a carrier wave from another station before attempting to transmit.  If a carrier is sensed, the station waits for the transmission in progress to finish before initiating its own transmission. In other words, CSMA is based on the principle “sense before transmit” or “listen before talk”.  Multiple access means that multiple stations send and receive on the medium. Transmissions by one node are generally received by all other stations connected to the medium. CSMA Access Modes 1-persistent  1-persistent CSMA is an aggressive transmission algorithm.  When the sender (station) is ready to transmit data, it senses the transmission medium for idle or busy.  If idle, then it transmits immediately. If busy, then it senses the transmission medium continuously until it becomes idle, then transmits the message (a frame) unconditionally (i.e. with probability=1).  In case of a collision, the sender waits for a random period of time and attempts to transmit again unconditionally (i.e. with probability=1).  1-persistent CSMA is used in CSMA/CD systems including Ethernet. Non-persistent  Non persistent CSMA is a non aggressive transmission algorithm.  When the sender (station) is ready to transmit data, it senses the transmission medium for idle or busy.  If idle, then it transmits immediately. If busy, then it waits for a random period of time (during which it does not sense the transmission medium) before repeating the whole logic cycle (which started with sensing the transmission medium for idle or busy) again.

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26 Notes Compiled By: Mr. Nilesh M. Patil IT Dept., RGIT  This approach reduces collision, results in overall higher medium throughput but with a penalty of longer initial delay compared to 1–persistent. P-persistent  This is an approach between 1-persistent and non-persistent CSMA access modes.  When the sender (station) is ready to transmit data, it senses the transmission medium for idle or busy.  If idle, then it transmits immediately. If busy, then it senses the transmission medium continuously until it becomes idle, then transmits a frame with probabilityp.  If the sender chooses not to transmit (the probability of this event is 1-p), the sender waits until the next available time slot.  If the transmission medium is still not busy, it transmits again with the same probability p. This probabilistic hold-off repeats until the frame is finally transmitted or when the medium is found to become busy again (i.e. some other sender has already started transmitting their data).  In the latter case the sender repeats the whole logic cycle (which started with sensing the transmission medium for idle or busy) again.  p-persistent CSMA is used in CSMA/CA systems including Wi-Fi and other packet radio systems. (f) OFDMA  Orthogonal Frequency Division Multiplexing (OFDM) is a form of signal modulation that divides a high data rate modulating stream placing them onto many slowly modulated narrowband close-spaced subcarriers, and in this way is less sensitive to frequency selective fading.  Orthogonal Frequency Division Multiplexing or OFDM is a modulation format that is being used for many of the latest wireless and telecommunications standards.  OFDM has been adopted in the Wi-Fi arena where the standards like 802.11a, 802.11n, 802.11ac and more. It has also been chosen for the cellular telecommunications standard LTE / LTE-A, and in addition to this it has been adopted by other standards such as WiMAX and many more.  Orthogonal frequency division multiplexing has also been adopted for a number of broadcast standards from DAB Digital Radio to the Digital Video Broadcast standards, DVB.  Although OFDM, orthogonal frequency division multiplexing is more complicated than earlier forms of signal format, it provides some distinct advantages in terms of data transmission, especially where high data rates are needed along with relatively wide bandwidths.  OFDM is a form of multicarrier modulation. An OFDM signal consists of a number of closely spaced modulated carriers.

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27 Notes Compiled By: Mr. Nilesh M. Patil IT Dept., RGIT  When modulation of any form - voice, data, etc. is applied to a carrier, then sidebands spread out either side.  It is necessary for a receiver to be able to receive the whole signal to be able to successfully demodulate the data.  As a result when signals are transmitted close to one another they must be spaced so that the receiver can separate them using a filter and there must be a guard band between them. This is not the case with OFDM.  Although the sidebands from each carrier overlap, they can still be received without the interference that might be expected because they are orthogonal to each another. This is achieved by having the carrier spacing equal to the reciprocal of the symbol period. Traditional view of receiving signals carrying modulation  To see how OFDM works, it is necessary to look at the receiver. This acts as a bank of demodulators, translating each carrier down to DC. The resulting signal is integrated over the symbol period to regenerate the data from that carrier.  The same demodulator also demodulates the other carriers. As the carrier spacing is equal to the reciprocal of the symbol period means that they will have a whole number of cycles in the symbol period and their contribution will sum to zero - in other words there is no interference contribution. OFDM Spectrum

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28 Notes Compiled By: Mr. Nilesh M. Patil IT Dept., RGIT Comparison of SDMA, TDMA, FDMA and CDMA

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29 Notes Compiled By: Mr. Nilesh M. Patil IT Dept., RGIT CHAPTER 2 WIRELESS TECHNOLOGY 2.1 Cellular Concept  The cellular concept refers to a system-level concept that focuses on substituting a single, high-powered transmitter with several low-powered transmitters, each targeted at providing coverage to a small part of the service area.  The cellular systems for mobile communications implements space division multiplexing.  Base station covers a certain transmission area called cell wherein it provides its service.  Mobile stations communicate only via the base station.  We consider the cell shapes to be hexagonal.However, cells are never perfect hexagons or circles; they vary depending on the environment, weather conditions and even on the system load.  Cell radii sizes from some 100 m in cities to, e.g., 35 km on the country side (GSM). Cellular System with 3 Cell Clusters Advantages and Disadvantages of Cellular System Advantages:

  1. Higher capacity Implementing SDM allows frequency reuse.If one transmitter is far away from another,i.e. outside the interference range it can reuse the same frequency.As most of mobile phone systems assign frequencies to certain users,this frequency is blocked for other users .But as frequency is a scarce resource the number of concurrent users per cell is also very limited.
  2. Less transmission power While power aspects are not a big problem for base stations, they are indeed a problematic for mobile stations.A receiver far away from the base station would need much more transmit power than the current few watts.But energy is a serious problem for mobile handling devices.

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30 Notes Compiled By: Mr. Nilesh M. Patil IT Dept., RGIT 3. Local interference only Having long distances between sender and receiver results even more interference problems.With the small cell mobile stations and base stations only have to deal with local interference. 4. Robustness Cellular systems are decentralized and so,more robust again the failure of single components.If one antenna fails,this only influences communication within small area. Disadvantages:

  1. Infrastructure needed Cellular systems need a complex infrastructure to connect all base stations .This includes many antennas, switches for forwarding, location registers to find a mobile station etc,which make the whole system quite expensive.
  2. Handover needed The mobile station has to perform a handover when changing from one cell to another.Depending on the cell size and the speed of movement,this can happen quite soon
  3. Frequency planning To avoid interference, frequency spectrum should be distributed properly with a very less range of frequency spectrum. Some important cellular concepts are as follows:
  4. Frequency Reuse
  5. Channel Assignment Strategies
  6. Handoff strategies
  7. Interference and system capacity
  8. Frequency Reuse  Frequency reuse, or, frequency planning, is a technique of reusing frequencies and channels within a communication system to improve capacity and spectral efficiency.  Frequency reuse is one of the fundamental concepts on which commercial wireless systems are based that involve the partitioning of an RF radiating area into cells.  The increased capacity in a commercial wireless network, compared with a networkwith a single transmitter, comes from the fact that the same radio frequency can bereused in a different area for a completely different transmission.  Frequency reuse in mobile cellular systems means that frequencies allocated to the service are reused in a regular pattern of cells, each covered by one base station.  The repeating regular pattern of cells is called cluster.The radio coverage for a cell is popularly called a footprint.

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31 Notes Compiled By: Mr. Nilesh M. Patil IT Dept., RGIT  Since each cell is designed to use radio frequencies only within its boundaries, the same frequencies can be reused in other cells not far away without interference, in another cluster. Such cells are called ‘co-channel’ cells.  The reuse of frequencies enables a cellular system to handle a huge number of calls with a limited number of channels. Frequency reuse technique of a cellular system.  Figure showsa frequency planning with cluster size of 7, showing the co-channels cells in differentclusters by the same letter.  The closest distance between the co-channel cells (indifferent clusters) is determined by the choice of the cluster size and the layout ofthe cell cluster.  Consider a cellular system with S duplex channels available foruse and let N be the number of cells in a cluster. If each cell is allotted K duplexchannels with all being allotted unique and disjoint channel groups we have S = KNunder normal circumstances.  Now, if the cluster are repeated M times within thetotal area, the total number of duplex channels, or, the total number of users in thesystem would be T = MS = KMN.  Clearly, if K and N remain constant, then T ∝ M and, if T and K remain constant, then N ∝1/ M.  Hence the capacity gain achieved is directly proportional to the number of times a cluster is repeated, as well as, for a fixed cell size, small N decreases the size of the cluster which in turn results in the increase of the number of clusters and hence the capacity.  However for small N, co-channel cells arelocated much closer and hence more interference.  The value of N is determined by calculating the amount of interference that can be tolerated for a sufficient quality communication. Hence the smallest N having interference below the tolerated limit is used. However, the cluster size N cannot take on any value and is given only by the following equation K = i2+ij+j2 Where, K = number of cells per cluster or cluster size i = number of cells (centre to centre) along any chain of hexagon

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32 Notes Compiled By: Mr. Nilesh M. Patil IT Dept., RGIT j = number of cells (centre to centre) in 60 degree counterclockwise of i.

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  1. Prove that for a hexagonal geometry, the cell cluster size is given by the relation K = i2+ij+j2 Where, K = number of cells per cluster or cluster size i = number of cells (centre to centre) along any chain of hexagon j = number of cells (centre to centre) in 60 degree counterclockwise of i. Solution: Rules for determining the nearest co-channel cell using “Shift parameters” (i, j) to lay out a cellular system is: Step 1: Move i cells along any side of ahexagon. Step 2: Turn 60 degrees anticlockwise Step 3: Move j cells. where i and j are shift parameters and can haveinteger value 0, 1, 2, 3, and so on … Let R be the distance from the centre of a regular hexagon and any of its vertex. A regular hexagon is one whose sides are also equal to R. Let d be the distance between the centres of two adjacent regular hexagons. From the geometry of the figure, OA = R and AB = R/2, OQ = d Then, OB = OA + AB = R + R/2 = 3R/2 Then, in right-angled ∆OAP, OP = OA sin 60̊ = (√3/2)R Let the distance between the centres of two adjacent hexagonal cells, OQ, be denoted by d. Then, OQ = OP + PQ (where OP = PQ) Therefore, d = [(√3/2)R + (√3/2)R ] Hence, d= √3 R To establish relation between D, d and shift parameters.

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33 Notes Compiled By: Mr. Nilesh M. Patil IT Dept., RGIT Let d be the distance between two adjacent cells and D be the distance from the centre of the cell under consideration to the centre of nearest co-channel cell. Using Cosine formula for ∆XYZ in figure above, we have, XZ2 = XY2 + YZ2 – 2XYYZ cos 120̊ D2 = ( id )2 + ( jd )2 – 2*(id)(jd)cos 120̊ D2 = ( id )2 + ( jd )2 – 2(id)(jd) (-1/2) D2 = ( id )2 + ( jd )2 + (id)(jd) D2 = d2( i2 + j2 + ij) We have d= √3 R Therefore, D2 = 3R2 * ( i2 + j2 + i*j) Area of small hexagon, Asmallhexagon = (3√3/2) R2 By joining the centres of the six nearest neighboring cochannel cells, a large hexagon is formed with radius equal to D, which is also the cochannel cell separation. Area of large hexagon, Alargehexagon = (3√3/2) D2 = (3√3/2) * 3R2 * ( i2 + j2 + ij) Number of cells in largehexagon, L = D2∕R2 Number of cells enclosedby large hexagon,

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34 Notes Compiled By: Mr. Nilesh M. Patil IT Dept., RGIT L = K+6*[(1 ∕3)* K] = 3 * K K = D2∕ (3 * R2);;; K = i2 + j2 + i * j Note: Larger hexagon encloses the entire cluster of k cells plus (1/3)rd the number of cells associated with six other peripheral clusters in the first tier. 2. Prove that for a regular hexagonal geometry, the frequency reuse ratio is given by the relationship 𝑞 = √(3𝐾), where K = K = i2+ij+j2, i and j being the shift parameters. Solution: Frequency reuse ratio q = D/R Where, D= the distance from the centre of the cell under consideration to the centre of nearest co-channel cell. R = radius of the cell under consideration. From previous problem, we have, D2 = 3R2 * ( i2 + j2 + ij) Given, K = i2 + j2 + ij Therefore, D2 = 3R2 *K q= D/R = √(3𝐾)*R/R Hence, 𝑞 = √(3𝐾) 2. Channel Assignment Strategies  With the rapid increase in number of mobile users, the mobile service providers had to follow strategies which ensure the effective utilization of the limited radio spectrum.  With increased capacity and low interference being the prime objectives,a frequency reuse scheme was helpful in achieving these objectives.  A variety ofchannel assignment strategies have been followed to aid these objectives.  Channelassignment strategies are classified into two types: fixed and dynamic, as discussedbelow. a) Fixed Channel Assignment (FCA)  In fixed channel assignment strategy each cell is allocated a fixed number of voicechannels.  Any communication within the cell can only be made with the designated unused channels of that particular cell.  Suppose if all the channels are occupied, then the call is blocked and subscriber has to wait. This is simplest of the channel assignment strategies as it requires very simple circuitry but provides worst channel utilization.  Later there was another approach in which the channels were borrowed from adjacent cell if all of its own designated channels were occupied. This was named as borrowing strategy.  In such cases the MSC supervises the borrowing process and ensures that none of the calls in progress are interrupted. b) Dynamic Channel Assignment (DCA)  In dynamic channel assignment strategy channels are temporarily assigned for use in cells for the duration of the call.

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35 Notes Compiled By: Mr. Nilesh M. Patil IT Dept., RGIT  Each time a call attempt is made from a cell the corresponding BS requests a channel from MSC. The MSC then allocates a channel to the requesting the BS.  After the call is over the channel is returned and kept in a central pool.  To avoid co-channel interference any channel that in use in one cell can only be reassigned simultaneously to another cell in the system if the distance between the two cells is larger than minimum reuse distance.  When compared to the FCA, DCA has reduced the likelihood of blocking and even increased the trunking capacity of the network as all of the channels are available to all cells, i.e., good quality of service.  But this type of assignment strategy results in heavy load on switching center at heavy traffic condition. 3. Consider that a geographical service area of a cellular system is 4200 km2. A total of 1001 radio channels are available for handling traffic suppose the area of a cell is 12 km2. a) How many times would the cluster of size 7 have to be replicated in order to cover the entire service area? Calculate the number of channels per cell and the system capacity. b) If the cluster size is decreased from 7 to 4, then does it result into increase in system capacity? Comments on the results obtained. Solution: Service area of a cellular system, Asys = 4200 km2 Coverage area of a cell, Acell = 12 km2 Total number of channels available,N = 1001 a) To calculate number of clusters, cell capacity and system capacity Cluster size, K = 7 The coverage area of the cluster, Acluster = K * Acell = 7 * 12 km2 = 84 km2 The number of times that the cluster has to be replicated to cover the entire service area of cellular system, M = Asys/ Acluster = 4200/84 = 50 clusters Since total number of available channels are allocated to one cluster, therefore, the number of channels per cell, J = N/K = 1001/7 = 143 channels/cell The system capacity C = NM = 100150 = 50050 channels b) To calculate new system capacity for reduced K New cluster size, K = 4 The coverage area of the cluster, Acluster = K * Acell = 4 * 12 km2 = 48 km2 The number of times that the cluster has to be replicated to cover the entire service area of cellular system, M = Asys/ Acluster = 4200/48 = 87 clusters (approx.) Since total number of available channels are allocated to one cluster, therefore, the number of channels per cell, J = N/K = 1001/4 = 250 channels/cell The system capacity C = NM = 100187 = 87087 channels (87000 channels approx.) c) Comments on the result From (a) and (b), it is seen that, decreasing the cluster size, increases the number of clusters and hence, also increases the system capacity. 4. If a 20 MHz of total spectrum is allocated for a duplex wireless cellular system and each simplex channel has 25KHz RF bandwidth find

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36 Notes Compiled By: Mr. Nilesh M. Patil IT Dept., RGIT a) The no. of duplex channels. b) The no. of channels per cell site where N = 4, N = 7 Solution: Available bandwidth = 20 MHz Channel bandwidth = 25 KHz a) Total no. of duplex channels available = (20 * 106) / (225103) = 400 channels b) If, N=4, the number of channels per cell site = 400/4 = 100 If, N=7, the number of channels per cell site = 400/7 = 57 5. If a total of 33 MHz of bandwidth is allocated to a particular cellular system which uses two 25 KHz Simplex channels to provide full Duplex voice. Compute the number of channels available per cell if the system uses: (i) 4 cell reuse (ii) 7 cell reuse (iii) 12 cell reuse If 1 MHz of the allocated spectrum is dedicated to control channels, determine an equitable distribution of control channels and voice channels in each cell for each of the three systems Solution: Available bandwidth = 33 MHz Channel bandwidth = 25 KHz a) Total no. of duplex channels available = (33 * 106) / (225103) = 660 channels b) If, N=4, the number of channels per cell site = 660/4 = 165 If, N=7, the number of channels per cell site = 660/7 = 95 If, N=12, the number of channels per cell site = 660/12 = 55 A 1MHz spectrum of control channels means that (1 * 106) / (225103) = 20 control channels out of 660 channels available. If, N=4, the number of control channels = 20/4 = 5. Therefore, out of 165 channels, 5 will be assigned to control channels and 160 to voice channels. If, N=7, the number of control channels = 20/7 = 3 (approx.). Therefore, out of 95 channels, 3 will be assigned to control channels and 92 to voice channels. If, N=12, the number of control channels = 20/12 = 2 (approx.). Therefore, out of 55 channels, 2 will be assigned to control channels and 53 to voice channels. 6. Consider a single high-power transmitter that can support 40 voice channels over an area of 140km2 with the available spectrum. If the area is equally divided into seven smaller areas (cells), each supported by lower transmitters so that each cell supports 30% of the channels, then determine (a) Coverage area of the cell (b) Total number of voice channels available in cellular system (c) Comment on the results obtained. Solution: Total service area to be covered = 140km2 ……………..(given) Total number of channels available = 40 ……………….(given) Number of cells = 7………………………………………(given) (a) Coverage area of the cell = Total service area / Number of cells = 140km2/ 7 = 20km2 (b) Number of voice channels per cell = 30% of original channels = 0.3 * 40 = 12 channels/cell. Total number of voice channels available in cellular system = Number of channels per cell * Number of cells in the service area = 12 * 7 = 84 channels

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37 Notes Compiled By: Mr. Nilesh M. Patil IT Dept., RGIT (c) There is significant increase in the number of available channels in the cellular system. This means, the system capacity is increased. 7. Calculate the number of times the cluster of size 4 have to be replicated in order to approximately cover the entire service area of 1765km2 with the adequate number of uniform-sized cells of 7km2 each. Solution: Size of the cluster, K = 4…………………..(given) Area of the cell, Acell = 7km2……………….(given) Total service area, Asystem = 1765km2………(given) Area of cluster, Acluster = K * Acell = 4 * 7 = 28km2. Number of clusters in service area = Asystem / Acluster = 1765 / 28 = 63. Hence, the number of times the cluster of size 4 has to be replicated is 63. (a) Assume a cellular system of 32 cells with a cell radius of 1.6 km, a total frequency bandwidth that supports 336 traffic channels, and a reuse factor of N = 7. If there are 32 total cells, what geographic area is covered, how many channels are there per cell, and what is the total system capacity? Assume regular hexagonal cellular topology. (b) Let the cell size be reduced to the extent that the same area as covered in Part (a) with 128 cells. Find the radius of the new cell, and new system capacity. (c) Comment on the results obtained. Solution: (a) Total number of cells in service area = 32………………..(given) Radius of a cell, R = 1.6km……………………………….(given) The area of a hexagon of radius R = (3√3/2) R2. Acell= (3√3/2) * (1.6 km)2 = 6.65 km2 , Hence, the total service area covered = Acell * No. of cells in total area = 6.65 X 32 = 213km2. Total number of available traffic channels = 336 …………(given) Frequency reuse pattern (cluster size) = 7 ………………..(given) Hence, the number of channels per cell = 336/7 = 48 Total system capacity = Number of channels per cell * Number of cells = 4832 = 1536 channels. (b) Total number of available cells = 128 …………………..(given) Total service area = 213km2…………………………….(Part a) Area of a regular hexagonal area = Total service area / Number of cells = 213/128 = 1.66km2. The area of a hexagon of radius R = (3√3/2) *R2. Therefore, (3√3/2) *R2 = (3√3/2) *R2 i.e. R = 0.8km Hence, radius of new smaller cell, R = 0.8km. New system capacity = number of channels per cell * number of cells = 48 * 128 = 6144 channels Hence, new system capacity = 6144 channels. (c) It is observed that as the number of cells are increased from 32 to 128 to cover the same service area (213km2), the size of the cell (in terms of radius R) is decreased from 1.6km to 0.8km. Keeping the identical number of channels (48) per cell, the total system capacity is increased from 1536 channels to 6144 channels. 9. A mobile communication system is allocated RF spectrum of 25 MHz and uses RF channel bandwidth of 25 kHz so that a total number of 1000 voice channels can be supported in the system.

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38 Notes Compiled By: Mr. Nilesh M. Patil IT Dept., RGIT (a) If the service area is divided into 20 cells with a frequency reuse factor of 4, compute the system capacity. (b) The cell size is reduced to the extent that the service area is now covered with 100 cells. Compute the system capacity while keeping the frequency reuse factor as 4. (c) Consider the cell size is further reduced to the extent that the service area is now covered with 700 cells with the frequency reuse factor as 7. Compute the system capacity. (d) Comment on the results obtained. Solution: Number of available voice channels, N = 1000. Therefore, each cluster can serve 1000 active users simultaneously. In other words, the capacity of a cluster = 1000. (a) To compute the system capacity for given K Number of cells covering the area = 20 (given) Frequency reuse factor or cluster size = 4 (given) Number of clusters = number of cells / cluster size = 20 / 4 = 5 Thus, the number of channels in 5 clusters = 1000 * 5 = 5000 Hence, the system capacity = 5000 users. (b) To compute new system capacity for increased number of cells Number of cells covering the area = 100 (given) Frequency reuse factor or cluster size = 4 (given) Number of clusters = number of cells / cluster size = 100 / 4 = 25 Thus, the number of channels in 5 clusters = 1000 *25 = 25000 Hence, the system capacity = 25000 users. (c) To compute new system capacity for increased number of cells and cluster size Number of cells covering the area = 700 (given) Frequency reuse factor or cluster size = 7 (given) Number of clusters = number of cells / cluster size = 700 / 7 = 100 Thus, the number of channels in 5 clusters = 1000 *100 = 100,000 Hence, the system capacity = 100,000 users. (d) It is observed that as the number of cells covering a given service area is increased, the number of clusters having all available number of channels increases. This results into significant increase in the number of active users in the system or the system capacity. Hence, it is concluded that frequency reuse enhances system capacity. 10. Determine the distance from the nearest cochannel cell for a cell having a radius of 0.64km and a cochannel reuse factor of 12. Solution: The radius of a cell, R = 0.64km (given) The cochannel reuse factor, q = 12 (given) To determine the distance from the nearest cochannel cell, D We know that, q = D/R Therefore, D = q * R = 12 * 0.64 = 7.68km Hence, the distance from the nearest cochannel cell, D = 7.68km 11. Determine the frequency reuse ratio for a cell radius of 0.8km separated from the nearest cochannel cell by a distance of 6.4km. Solution: The radius of a cell, R = 0.8km (given) The distance between nearest cochannel cells, D = 6.4km (given) To determine the frequency reuse ratio, q

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39 Notes Compiled By: Mr. Nilesh M. Patil IT Dept., RGIT We know that, q = D/R Therefore, q = 6.4 / 0.8 = 8 Hence, the frequency reuse ratio, q = 8. 3. Handoff Process  When a user moves from one cell to the other, to keep the communication between the user pair, the user channel has to be shifted from one BS to the other without interrupting the call, i.e., when a MS moves into another cell, while the conversation is still in progress, the MSC automatically transfers the call to a new FDD channel without disturbing the conversation. This process is called as handoff.  A schematicdiagram of handoff is given in Figure below. Handoff scenario at two adjacent cell boundary.  Processing of handoff is an important task in any cellular system. Handoffs must be performed successfully and be imperceptible to the users. Factors Influencing Handoffs The following factors influence the entire handoff process: a) Transmitted power: As we know that the transmission power is different for different cells, the handoff threshold or the power margin varies from cell to cell. b) Received power: the received power mostly depends on the Line of Sight (LoS) path between the user and the BS. Especially when the user is on the boundary ofthe two cells, the LoS path plays a critical role in handoffs. c) Area and shape of the cell: Apart from the power levels, the cell structure also a plays an important role in the handoff process. d) Mobility of users: The number of mobile users entering or going out of a particularcell, also fixes the handoff strategy of a cell. Handoffs In Different Generations  In 1G analog cellular system, the signal strength measurements were made bythe BS and in turn supervised by the MSC. The handoffs in this generation canbe termed as Network Controlled Hand-Off (NCHO). The BS monitors the signal strengths of voice channels to determine the relative positions of the subscriber. The special receivers located on the BS are controlled by the MSC to monitor the signal strengths

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40 Notes Compiled By: Mr. Nilesh M. Patil IT Dept., RGIT of the users in the neighboring cells which appear to be in need of handoff. Based on the information received from the special receivers the MSC decides whether a handoff is required or not. The approximate time needed to make a handoff successful was about 5-10s.  In the 2G systems, the MSC was relieved from the entire operation. In thisgeneration, which started using the digital technology, handoff decisions were mobileassisted and therefore it is called Mobile Assisted Hand-Off (MAHO). In MAHO,the mobile center measures the power changes received from nearby base stationsand notifies the two BS. Accordingly the two BS communicate and channel transferoccurs. As compared to 1G, the circuit complexity was increased here whereas thedelay in handoff was reduced to 1-5s.  In the 3G systems, the MS measures the power from adjacent BS and automatically upgrades the channels to its nearer BS. Hence this can be termed as Mobile Controlled Hand-Off (MCHO). When compared to the other generations, delay during handoff is only 100ms. The Quality of Service (QoS) has improved a lot although the complexity of the circuitry has further increased which is inevitable. All these types of handoffs are usually termed as hard handoff as there is a shiftin the channels involved. There is also another kind of handoff, called soft handoff,as discussed below.  Handoff in CDMA: In spread spectrum cellular systems, the mobiles share the same channels in every cell. The MSC evaluates the signal strengths received from different BS for a single user and then shifts the user from one BS to the other without actually changing the channel. These types of handoffs are called as soft handoff as there is no change in the channel. Handoff Priority While assigning channels using either FCA or DCA strategy, a guard channel conceptmust be followed to facilitate the handoffs. This means, a fraction of total availablechannels must be kept for handoff requests. But this would reduce the carriedtraffic and only fewer channels can be assigned for the residual users of a cell. Agood solution to avoid such a dead-lock is to use DCA with handoff priority (demandbased allocation). A Few Practical Problems in Handoff Scenario a) Different speed of mobile users: With the increase of mobile users in urban areas, microcells are introduced in the cells to increase the capacity. The users with high speed frequently crossing the micro-cells become burdened to MSC as it has to take care of handoffs. Several schemes thus have been designed to handle the simultaneous traffic of high speed and low speed users while minimizing the handoff intervention from the MSC, one of them being the ‘Umbrella Cell’ approach. This technique provides large area coverage to high speed users while providing small area coverage to users traveling at low speed. By using different antenna heights and different power

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41 Notes Compiled By: Mr. Nilesh M. Patil IT Dept., RGIT levels, it is possible to provide larger and smaller cells at a same location. The umbrella cell is co-located with few other microcells. The BS can measure the speed of the user by its short term average signal strength and decides which cell to handle that call. If the speed is less, then the corresponding microcell handles the call so that there is good corner coverage. This approach assures that handoffs are minimized for high speed users and provides additional microcell channels for pedestrian users. b) Cell dragging problem: This is another practical problem in the urban area with additional microcells. For example, consider there is a LOS path between the MS and BS1 while the user is in the cell covered by BS2. Since there is a LOS with the BS1, the signal strength received from BS1 would be greater than that received from BS2. However, since the user is in cell covered by BS2, handoff cannot take place and as a result, it experiences a lot of interferences. This problem can be solved by judiciously choosing the handoff threshold along with adjusting the coverage area. c) Inter-system handoff: If one user is leaving the coverage area of one MSC and is entering the area of another MSC, then the call might be lost if there is no handoff in this case too. Such a handoff is called inter-system handoff and in order to facilitate this, mobiles usually have roaming facility. 4. Interference & System Capacity  Susceptibility and interference problems associated with mobile communicationsequipment are because of the problem of time congestion within the electromagneticspectrum. It is the limiting factor in the performance of cellular systems.  This interference can occur from clash with another mobile in the same cell or because of a call in the adjacent cell. There can be interference between the base stations operating at same frequency band or any other non-cellular system‘s energy leaking inadvertently into the frequency band of the cellular system.  If there is an interference in the voice channels, cross talk is heard will appear as noise between the users.  The interference in the control channels leads to missed and error calls because of digital signaling. Interference is more severe in urban areas because of the greater RF noise and greater density of mobiles and base stations.  The interference can be divided into 2 parts: co-channel interference and adjacent channel interference. Co-channel interference (CCI)  For the efficient use of available spectrum, it is necessary to reuse frequency bandwidth over relatively small geographical areas. However, increasing frequency reuse also increases interference, which decreases system capacity and service quality.  The cells where the same set of frequencies is used are call co-channel cells.  Co-channel interference is the cross talk between two different radio transmitters using the same radio frequency as is the case with the co-channel cells.

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42 Notes Compiled By: Mr. Nilesh M. Patil IT Dept., RGIT  The reasons of CCI can bebecause of either adverse weather conditions or poor frequency planning or overlycrowdedradio spectrum.  If the cell size and the power transmitted at the base stations are same then CCIwill become independent of the transmitted power and will depend on radius of thecell (R) and the distance between the interfering co-channel cells (D).  If D/R ratio is increased, then the effective distance between the co-channel cells will increaseand interference will decrease. The parameter Q is called the frequency reuse ratioand is related to the cluster size.  For hexagonal geometry Q = D/R = √(3K).  From the above equation, small of ‘Q’ means small value of cluster size ‘K’ andincrease in cellu

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