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

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1 Wireless Physical Layer Q1. Is it possible to transmit a digital signal, e.g., coded as square wave as used inside a computer, using radio transmission without any loss? Why? It is not possible to transmit a square wave without loss because radio signal undergoes attenuation and hence the square wave will lose the information it carries (if it represents bit 1 it may be attenuated so much that the signal may be interpreted as bit 0 at the receiver!) Q2. What are the main problems of signal propagation? Why do radio waves not always follow a straight line? Why is reflection both useful and harmful? Radio signal propagation faces the following problems: Attenuation (amplitude of the wave loses strength thereby the signal power) Refraction Reflection Shadowing Scattering Diffraction Radio waves do not follow a straight line because of blocking objects in its path. Reflection is useful because in non-line-of-sight environments (where there is no direct path from the transmitter to receiver for example in offices, town and cities) it allows the radio signal to reach from the transmitter to the receiver. Reflection can be harmful because multiple copies of the same signal can reach the receiver at different times. Q3. Define transmission range, reception range, detection range, and interference range. Transmission range: Maximum distance from a transmitting node at which the signals of this node can be successfully decoded with very low error rates Reception range: Maximum distance at which signals can be received with low error rates (same as transmission range) Detection range: Distance from a transmitting node at which its signals can be detected but not successfully decoded Interference range: Distance from a transmitting node at which its signals contribute to the background noise Q4. Explain the term interference in the space, time, frequency, and code domain. What are countermeasures in SDMA, TDMA, FDMA, and CDMA systems?

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).

4 2. The nodes clocks can drift in time by some maximum known value. 3. Nodes can have three modes: Transmit, Listening/Receiving, Sleep. While transmitting nodes consume twice in comparison to when they are receiving. In sleep mode nodes consume 1% power in comparison with transmit mode. Only the mobile nodes are power limited (the access point is not power limited). 4. You have two types of traffic: emergency (delay constrained and requires high reliability: for example alarms), and best-effort traffic (with a time to live. Which means that if the frames are not transmitted within TTL they can be dropped). The emergency traffic is small in comparison with besteffort traffic. You may have different emergency-type connections with different delay bounds. 5. You goal is to design a MAC that satisfies both constraints while maximizing battery life. There is no single or simple solution. The goal of this exercise is to make you think about the main issues in Wireless MAC design. Propose one or several schemes. Discuss the limitations of your proposed scheme(s). You can also make additional assumption when comparing schemes. Design problem. Refer slides. No exact answer. Q2. Suppose you are the IT manager of a corporate firm and you are looking to provide seamless and fast connectivity to the enterprise network to the corporate employees. Since the current wireless infrastructure in place in the corporate office employs IEEE b based Wireless LANs the data rate supported is just 11Mbps in the best case. You need to make a decision to upgrade the infrastructure to either IEEE a or IEEE g both of which provides upto 54Mbps of data rate. Which one will you choose for upgrading the existing wireless infrastructure? Remember that as an IT manager your goal is to provide a cost-effective solution to meet the desired goals. Be specific in your answer g will be the preferred choice given that both a/g have the same physical link data rate of 54Mbps: a) g operates at 2.4 GHz and a operates at around 5 GHz. Thus, g has a larger range (range inversely proportional to frequency) compared to a. Deploying a would require more access points and hence more investments compared to g. b) Deploying a access points would create interoperability problem as existing clients in the enterprise may not have the interface card which also operates at 5 GHz. c) Since a operate at high frequency (signals have smaller wavelengths) the signals have problems penetrating office walls and cubicles thereby leading to poor connectivity in enterprise environments.

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?

7 It uses CSMA with Collision Avoidance which is based on carrier sense function in PHY called Clear Channel Assessment (CCA). It reduce collision probability where mostly needed. Efficient backoff algorithm stable at high loads. Please refer slides and readings for details. Q8. This problem will help you understand why you cannot achieve the physical layer rate at the medium access layer in and will give you an idea that medium access layer throughput is roughly half of the physical layer data rate physical layer transmission rate = 54Mbps MAC layer data payload = 1452 bytes MAC header = 28 bytes ACK Frame Size = 14 bytes RTS length = 20 bytes CTS length = 14 bytes Propagation Delay = 1μs Slot Time = 9 μs SIFS Time = 16μs DIFS Time = 34μs Physical layer overhead = 20μs Define MAC layer throughput as the number of bits sent by the MAC layer in a given period of time. a) Assuming that there are two stations exchanging data using DCF but without using RTS/CTS transaction, what is MAC layer throughput? Total data in 1 frame = 1452 byte MAC Header length = 28 bytes Total frame size = = 1480 byte Time required for transmission at 54 Mbps = 1480*8 /(54 Mbps) = μs Total time required for transmission of 1 Frame = DIFS (34μs ) + Data Time ( μs ) + propagation time(1μs ) + Physical overhead (20 μs ) + SIFS(16 μs ) + ACK time(2.07 μs ) + propagation time for Ack (1 μs )+ Physical overhead for Ack (20 μs ) = μs Hence we are using μs to transmit 1452 bytes Hence MAC layer throughput = 37 Mbps b) What is the throughput when RTS/CTS transaction is used? Total time required for the transmission of 1 Frame DIFS (34μs ) + RTS (2.96 μs )+ propagation time(1μs ) + Physical overhead (20 μs )+

8 SIFS(16 μs ) + CTS (2.07 μs )+ propagation time(1μs ) + Physical overhead (20 μs )+ SIFS(16 μs )+ Data Time ( μs )+ propagation time(1μs ) + Physical overhead (20 μs )+ SIFS (16 μs ) + ACK(20 μs ) + propagation time(1μs ) + Physical overhead (20 μs ) = 410 μs Hence we are using 410μs to transmit 1452 bytes Hence MAC layer throughput = Mbps Q9. How are fairness problems regarding channel access solved in IEEE ? implement a back-off mechanism that tries to offer fair access to the medium in the standard case (no polling from the access point). If all systems behave well this mechanism gives a fair share of the overall bandwidth to all stations. Q10. Suppose an 802.1lb station is configured to always reserve the channel with the RTS/CTS sequence. Suppose this station suddenly wants to transmit 1,000 bytes of data, and all other stations are idle at this time. As a function of SIFS and DIFS, and ignoring propagation delay and assuming no bit errors, calculate the time required to transmit the frame and receive the acknowledgment. A frame excluding data is 34 bytes long (272 bits). The time required to transmit the 1,000 bytes data frame is (8272 bits)/ (11 Mbps) = 752 usec. RTS frame is 14 bytes; the time required to transmit a RTS frame is (112 bits)/ (11 Mbps) = 10.2 usec. CTS frame is 12 bytes; the time required to transmit a CTS frame is (96 bits)/(11 Mbps) = 8.7 usec. ACK frame is 12 bytes; the time required to transmit a ACK frame is (96 bits)/(11 Mbps) = 8.7 usec. DIFS + RTS + SIFS + CTS + SIFS + FRAME + SIFS + ACK = DIFS + 3SIFS + ( ) usec = DIFS + 3SIFS usec Q11. Suppose there are two ISPs providing Wi-Fi access in a particular café, with each ISP operating its own AP and having its own IP address block. a. Further suppose that by accident, each ISP has configured its AP to operate over channel 11. a) Will the protocol completely break down in this situation? Discuss what happens when two stations, each associated with a different ISP, attempt to transmit at the same time. The two APs will typically have different SSIDs and MAC addresses. A wireless station arriving to the café will associate with one of the SSIDs (that is, one of the APs). After association, there is a virtual link between the new station and the AP. Label the APs AP1 and AP2. Suppose the new station associates with AP1. When the new station sends a frame, it will be addressed to AP1. Although AP2 will also receive the frame, it will not process the frame because the frame is not addressed to it. Thus, the two ISPs can work in parallel over the same channel. However, the two ISPs will be sharing the same wireless bandwidth. If wireless stations in different ISPs transmit at the same

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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