WT Question Paper Solution (DEC 2023) – IT MU
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Q1. Solve any four (20 marks)
a) Explain the difference between an Infrastructures based Network andanAdhoc Network of WLAN. (5 marks WT )
Ans-
| What is it? | In infrastructure mode, thecommunication occurs onlybetween the wirelessand access points (AP)not directly betweennodes | In ad-hoc mode, eachcommunicates directly other nodes, so no access point nodes control is needed. , but wireless |
| ExternalCommunication | Access points acts as,abridge to other wireless/wired network | Nodes in Ad-hoc cancommunicate if they the same range. |
| Physical needs | Physical infrastructureneeded | No physical infrastructure is needed. |
| Complexity | Designing is simpleof the network functionalitylies within AP and client isjust a simple machine. | As no central co-ordination as most exists, we need to use decentralized MAC such as CSMA/CA, withnodes having samefunctionality. This shootsthe complexity and cost. |
| Whenused: | It can’t be used in critical it can’t be situations like disaster where no infrastructureleft. | It is not always fully connectedas two mobile nodes may relief temporarily be out of range. is |
| Applications | IEEE 802.11 &HIPERLAN2 are basedinfrastructure mode. | Bluetooth is a typical ad-hocnetwork. on |
| Channel Access | Most infrastructure basedWLAN uses TDMA-basedprotocols | Most Ad-hoc based WLANuses contention MAC (e.g. CSMA) |
| Topology | Based on topology, oneadvantage is the abilityinfrastructure WLANsprovide wired networkapplications and services | Ad-hoc WLANs are easier to main set-up and require no infrastructure |
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b) Explain Direct Sequence Spread Spectrum. (5 marks)
Ans-
1. In this scheme, the user signal is spread by performing an XOR with a fixedsequence called as a chipping sequence.
2. As shown in figure, a user signal 01, is XORed with the chipping sequence0110101, the resulting signal is either 0110101(if the user bit is 0) or its complement 1001010 (if the user bit is 1).
3. XOR of the signal with pseudo-random number (chipping sequence) – many chips per bit (e.g., 128) result in higher bandwidth DSSS Transmitter & Receiver
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Advantages
1. Reduces frequency selective fading
2. In cellular networks,
⮚ Base stations can use the same frequency range
⮚ Several base stations candetect and recover the signal
3. Soft handover
Disadvantages
1. Precise power control necessary
2. The overall system is complex to implement.
c) Compare FDMA, TDMA and CDMA. (5 marks)
Ans-
| Features FDMA TDMA CDMA | |||
| Full Forms | FDMAabbreviationFrequencyMultiple Access. | GPRS is an is an abbreviation for Time Division DivisionMultiple Access. | CDMA is an abbreviation for Code for Division Multiple Access. |
| ModeOperation | It distributes of single bandwidth among multiplestations byit into sub-channels | It only shares the a time of transmission through the dividing satellite, not thechannel. | It shares both time bandwidth among multiple stations allocating a unique to each slot. |
| Flexibility | It has a littleflexible. | It has moderateflexibility. | It has high flexibility. |
| Codeword | It doesn’t requirecodeword. | It also doesn’t a require a codeword. | It needs a codeword. |
| Rate of Data | It has a lowrate. | It has a medium data data rate. | It has a high data |
| Mode of Datatransfer | It uses continuoussignals for datatransmission. | It uses signals bursts for data transmission. | It uses digital signals in data transmission. |
| Synchronization | It doesn’t needsynchronization. | It requires any synchronization. | It also doesn’t requireany synchronization. |
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| Terminals | Every terminal hasits own constantfrequency. | Every terminal onthe same frequency is active for just a short periodtime. | Every terminal mayremain operational at same time and same location of interruption. |
| Cells Capacity | It has a limitedcapacity. | It also has a cell cell capacity. | It has no capacity limited restriction for a although it isinterference-limited. |
| Cost | It has a high | cost. It has a low | cost. Its installationhigh, but the operationalcost is low. |
| Guardand Bands | It needed times bands. | It needed guard times. | It needed both guard times and guard |
| FadingMitigation | It doesn’t requireequalizer. | It needed anequalizer | RAKE receiver maybe an possible in CDMA. |
| Advantages | It is a verywell-established,and straightforwardprotocol. | It is highly reliable, flexible, entirely digital, andestablished. | It is more flexible, needs less frequency and offers a softer signal well- handover. |
| Disadvantages | It is very flexible, and the frequenciesare limited. | It requires guardspace. | It works with extremelycomplicated receivers, and senders/transmittersneed a more complexpower control method. |
d) Outline the method that supports mobility in CISCO Unified Wireless Network. (5 marks)
Ans
In Cisco Unified Wireless Network (CUWN), mobility is supported through a combination of protocols, mechanisms, and features that enable seamless roaminganduninterrupted connectivity for wireless clients as they move within the network. Here’s an outline of the key components and methods supporting mobility in CUWN:
1. Wireless LAN Controller (WLC) Integration: Cisco WLCs play a central role inmanaging the wireless network infrastructure. They control and coordinate the operations of access points (APs) and provide centralized management for mobility related functions.
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2. Layer 2 Roaming: In CUWN, Layer 2 roaming allows wireless clients to roamseamlessly between APs within the same subnet or VLAN without changing their IPaddress. As clients move between APs, the WLC maintains session information andforwards traffic to the appropriate AP.
3. Inter-Controller Roaming: CUWN supports roaming across multiple controllers. When a client roams between APs managed by different controllers, the controllers exchange client session information to ensure continuity of service. This enables clients to roam between different parts of the network without losing connectivity.
4. Fast Roaming Protocols:
⮚ IEEE 802.11r (Fast BSS Transition): Accelerates the roaming process by pre- authenticating the client with neighboring APs, reducing the time required for reconnection.
⮚ IEEE 802.11k (Radio Resource Measurement): Enables APs to provide information about neighboring APs and network conditions, facilitating better roaming decisions.
⮚ IEEE 802.11v (BSS Transition Management): Allows APs to assist clients inmaking roaming decisions by providing information about available networks andcapabilities.
5. Client Stateful Roaming: Cisco WLCs maintain client session state information, including authentication and encryption context. This allows for smooth handoff as clients roam between APs, ensuring that session continuity is maintained without requiring reauthentication.
6. Optimized Roaming Policies: Administrators can configure roaming policies onWLCs to control client roaming behavior based on factors such as signal strength, load balancing, and AP utilization. These policies help optimize roaming decisions and enhance overall network performance.
7. Wireless LAN Controller Redundancy: CUWN supports controller redundancyto ensure high availability. Redundant controllers synchronize configuration and client session information, allowing for seamless failover in case of controller failure.
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8. Integration with Location Services: CUWN integrates with Cisco’s location- based services, such as Cisco Connected Mobile Experiences (CMX), to enhance mobility and enable applications such as asset tracking, wayfinding, and location- based analytics.
e) State the features of Wi-MAX. (5 marks)
Ans-
There are many features of WiMAX in terms of services compared with some other broadband services. Some features WiMAX features are: 1. Speed: The speed of WiMAX has ability to handle up to 70 megabits per second
for the transmission of data. The Downlink speed of WiMAX is 46 Mbps andUplink speed of WiMAX is 4Mbps
2. Bandwidth: The bandwidth of WiMAX which gives 3.5 MHz to 10 MHz. 3. Range: Range of a single WiMAX network connectivity is expected up to 40miles with 70 Mbps speed or more, but it optimized to 1.5km to 5 kmby reducingthe cell size.
4. Data Transfer: In WiMAX technology the data transfer from source to destination is that up to 120 kmph.
5. Cell Radius: Cell radius which is seen in the coach that are D and R parameters, cell are up to 27 km/h.
6. Cell Capacity: The cell capacity of WiMAX is 100-200 users. 7. Duplexing Mode: TDD, FDD
8. Legacy: IEEE 802.16a, 802.16b, 802.16c, 802.16d.
9. Handover: Optimized hard handover.
Q2. a. Draw and explain 4G architecture with its specifications. (10 marks) Ans-
4G Architecture
1. 4G stands for fourth generation cellular system.
2. 4G is evaluation of 3G to meet the forecasted rising demand. 3. It is an integration of various technologies including GSM,CDMA,GPRS,IMT-2000 ,Wireless LAN.
4. Data rate in 4G system will range from 20 to 100 Mbps.
Features:
1. Fully IP based Mobile System.
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2. It supports interactive multimedia, voice, streaming video, internet and other broadband service.
3. It has better spectrum efficiency.
4. It supports Ad-hoc and multi hop network.
4G Architecture
1. Figure shows Generic Mobile Communication architecture.
2. 4 G network is an integration of all heterogeneous wireless access networks such as Ad-hoc, cellular, hotspot and satellite radio component.
3. Technologies used in 4 G are smart antennas for multiple input and multiple output (MIMO), IPv6, VoIP, OFDM and Software defined radio (SDR) System.
Smart Antennas:
1. Smart Antennas are Transmitting and receiving antennas.
2. It does not require increase power and additional frequency.
IPV6 Technology:
1. 4G uses IPV6 Technology in order to support a large number of wireless enable devices.
2. It enables a number of application with better multicast, security and route optimization capabilities.
VoIP:
1. It stands for Voice over IP.
2. It allows only packet to be transferred eliminating complexity of 2 protocols over the same circuit.
OFDM:
1. OFDM stands for Orthogonal Frequency Division Multiplexing. 2. It is currently used as WiMax and WiFi.
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SDR:
1. SDR stands for Software Defined Radio.
2. It is the form of open wireless architecture.
Advantages:
1. It provides better spectral efficiency.
2. It has high speed, high capacity and low cost per bit.
Disadvantage:
1. Battery usage is more.
2. Hard to implement.
1. MME- Mobility Management Entity
It is used for Paging ,Authentication, Handover and Selection of Serving Gateway2. SGW- Serving gateway
It is used to Routing and Forwarding user data packet.
3. PDN-GW Packet Data Network Gateway
It is used for user equipment (UE) IP allocation
4. HSS -Home Subscriber Server
It is a user Database used for service subscriber, user identification and addressing5. PCRF -Policy and Charging Rule Function
It provide quality of service and charging
6. eNode B-evolved Node B
It is used as radio resources management and radio bearer control
