# Attenuation (amplitude of the wave loses strength thereby the signal power) Refraction Reflection Shadowing Scattering Diffraction

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2 SDMA: counteracts interference by separating two interfering signals in space. For example, a base station will use directional antennas to transmit to two spatially separated mobile stations so that the signal of one doesn t reach other (takes advantage of spatial separation of the mobile users. If the users are located at the same position then of course SDMA doesn t works!) TDMA: counteracts interference by separating two interfering signals in time. Two interfering signals would transmit at different times. FDMA: counteracts interference by separating two interfering signals in frequency. Two interfering signals would transmit at different frequency at the same time. CDMA: counteracts interference by separating two interfering signals in code domain. Two interfering signals would transmit using different codes (analogous to using different languages for communication in a room where every pair is using a different language for communication) at the same time at the same frequency. Q5. In an FDMA system, each user is allocated a duplex channel i.e. one for uplink and other for downlink voice communication. Also, these channels need to be guarded from each other for interference by providing sufficient separation in the spectrum through the use of guard bands. If a US cellular operator is allocated 25 MHz of spectrum, and guard band must be atleast 10 KHz and each simplex voice channel is 30 KHz, then a) Find the number of channels in an FDMA system. Each simplex channel must be separated from each other by a guard band for interference avoidance. Thus, if the number of simplex channels is x the number of guard bands needed will be x-1. Hence: (x-1)*10 + x*30 = 25*10 3 x = So the number of FDMA channels is 625. b) Find the number of users FDMA can support at the same time. Now suppose the cellular operator decides to use TDMA system and the TDMA system can support maximum of 8 time slots per TDMA frame, then what is the total number of users supported per TDMA frame? Each user requires duplex channel for voice communication. Hence number of users that can be accommodated at the same time: 625/2 = 312

3 Assuming each user will use time division duplex communication, each user requires 2 slots per TDMA frame for voice communication. One slot will be utilized for uplink and the other slot for downlink voice communication; Maximum no. of users supported is 4. If users use frequency division duplex communication (two different frequency channels for uplink and downlink), then each user requires 1 slot per TDMA frame; Maximum number of users supported is 8. Q6. If TDMA uses a frame structure where each frame consists of 8 time slots, and each time slot contains bits, and data is transmitted at Kbps in the channel, find a) the time duration of a bit 1 bit duration = 3.69 sec (1bit) / bits/sec = 3.69 μsec b) the time duration of slot slot time duration = (156.25bits) / bits/sec = msec c) the time duration of a frame frame duration = 8 slot time duration = 4.64msec d) How long a user must wait between two successive transmissions? Give the worst case and best case (assume each user is guaranteed transmission opportunity in each frame) solutions. In the worst case, a user may be allocated the first slot in a frame and the last slot in the next frame so it has to wait for 14 slots for transmitting again. In the best case, if there are not many users the user may be allocated successive slots in the same frame so it doesn t wait. Wireless LAN Q1. Assume that you are requested to design a Medium Access Control protocol for a set of low power wireless (mobile) nodes that have to communicate with a wireless access point. Your constraints are: 1. You can only use FDMA and TDMA like schemes (CDMA is not considered for cost and complexity reasons).

5 Q3. IEEE tries to solve the hidden node problem using RTS/CTS message exchange. Give a scenario when RTS/CTS message exchange cannot solve the hidden node problem. D is the distance between node B and node C. When node A starts to transmit, all the nodes within detection range (carrier sense) of node A defer their transmission until the end of the transmission. However, node C is outside detection range of node A but within detection range of node B. It can not sense the transmission from node A to node B. So it will not defer its transmission because of A s transmission. Because node C is outside transmit range of node B, it can t decode the CTS from node B. So it will not be blocked by the CTS. If node C tries to access channel before the end of transmission from node A, a collision will occur at node B. There is another hidden node problem. Node C sends RTS to some other node D, and at the same time, node B sends CTS to node A. Thus node C misses the CTS from node B. The collision happens when both node A and node C send data packet. Thus, the hidden node problem remains in IEEE based networks. Q4. While RTS/CTS addresses the hidden node problem it creates the exposed node problem. What are the cons of the exposed node problem? Suggest some optimization considerations to overcome exposed node problem without introducing any new signaling messages.

6 Exposed node problem prevents spatial reuse of the channel (even though two communication links can exist in the same space RTS/CTS doesn t allow). Thus, network throughput can degrade. Overcoming exposed node problem: A node overhearing an RTS (it is not the target of the RTS) but no CTS can assume that it is only the neighbor of the current transmitting node and not the receiver and hence can initiate transmission to another node. Note though, that this secondary transmission must be of shorter duration than the earlier transmission so as to prevent corruption of Acknowledgement packet at the neighboring transmitter. Q5. Explain the role of Distribution Set in IEEE Please refer readings and slides. Q6. What is the role of beacon signal and probe message in based WLANs? The SSID is normally carried in every BEACON sent by the APs (5 to 20 times per second). The BEACON also contains the time, capabilities, supported data rates, and physical layer parameter sets that regulate the smooth operation of a wireless network. The SSID in the BEACON is the network advertisement that appears to those not fully versed in the low-level workings of IEEE networks to be the risk exposure. A station preparing to roam in a WLAN whose BEACONs do not carry the SSID must actively scan to discover APs. The station sends out PROBE Requests sequentially on all channels with its SSID and listens for PROBE Responses. The station may do this channel scanning every 50 msec (20 times a second!) as it attempts to discover a stronger signal. In some office configurations, stations have been observed to bounce between APs, spending only minutes on one AP and then switching to another based on signal strength (Usually noticed when you try to connect UMASS access point). Thus a WLAN that has stations with weak signals from the APs will readily expose the SSID in all the PROBE Requests and ASSOCIATION frames. Q7. Why don t employ CSMA with collision detection rather than just try to avoid collision using CSMA/CA?

9 time, there will be a collision. For b, the maximum aggregate transmission rate for the two ISPs is 11 Mbps. b) Now suppose that one AP operates over channel I and the other over channel 11. Now if two wireless stations in different ISPs (and hence different channels) transmit at the same time, there will not be a collision. Thus, the maximum aggregate transmission rate for the two ISPs is 22 Mbps for b.

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