Tutorial 3 : Reservation Schemes

Lund University ETSN01 Advanced Telecommunication Tutorial 3 : reservation schemes Author: Antonio Franco Tutor: Emma Fitzgerald Farnaz Moradi February 1, 2016 Contents I Before you start 3 II Exercises 3 1 Reservation schemes 3 1.1 .....................................3 1.2 .....................................3 1.3 .....................................3 1.4 .....................................4 1.5 .....................................4 1.6 .....................................6 1.7 .....................................6 III Solutions 7 2 Reservation schemes 7 2.1 .....................................7 2.2 .....................................7 2.3 .....................................8 2.4 .....................................9 2.5 .....................................9 2.6 ..................................... 11 2.7 ..................................... 11 Part I Before you start This tutorial is given to prepare you to the exam. Since time is limited, it is highly advised that you first try to solve the exercises (Part II) at home, then have a look at the solutions (Part III), and, finally, ask questions during the exercises sessions. Part II Exercises 1 Reservation schemes 1.1 For each of the following reservation schemes: • Briefly describe how the scheme separates multiple users. • List one advantage and one disadvantage of the scheme. • Give an example of a system in which the scheme is used. 1. Space Division Multiple Access 2. Time Divison Multiple Access 3. Frequency Division Multiple Access 4. Code Division Multiple Access 1.2 Consider a combined TDMA/FDMA scheme in which each slot lasts for 10ms, with a guard time of 1ms between slots. The data rate on each channel is 1 Mbps, and there are 3 channels, each with a bandwidth of 100 MHz and guard band of 5 MHz. In each slot, there is an 80% chance that the device allocated that slot transmits data, and a 20% chance that the device is silent. If a device transmits, it uses the entire slot. What is the rate of work for this scheme? 1.3 1. Explain the steps of the Reservation ALOHA algorithm and draw a diagram of its operation. 3 2. What are the advantages and disadvantages of using this algorithm over slotted ALOHA? 3. Consider a Reservation ALOHA system with N nodes. Assume a Poissonian packet generation process for each node with parameter λ. The round length is Ts seconds | that is, every Ts seconds contains a reservation phase and a data transmission phase. Each data transmission phase contains k timeslots. What is the probability that a given node will have at least one packet to send in the second round? You can assume that all nodes begin the first round with no packets queued to send. 4. What is the expected number of nodes with at least one packet to send in the second round? 5. Each node that wishes to transmit selects a slot to reserve at random (with a uniform distribution), and each node may only send one packet per round, even if it has more packets queued to send. For N = 3, what is the probability of a collision in the reservation phase of the second round? 1.4 Check whether those chip codes are mutually orthogonal: a) 1)1111 2) 1 -1 1 -1 3) 1 -1 -1 1 b) 1)11111111 2) 1 -1 -1 1 1 -1 -1 1 c) 1) 1 1 -1 -1 -1 -1 1 1 2) 1 -1 -1 -1 -1 -1 1 1 1.5 Below is a diagram of the signal transmitted from two nodes, A and B, which are sending data on a wireless channel at the same time using code division multiple acces (CDMA). Each node has two bits of data to send and its own CDMA code as pictured (with a spreading factor of 4). Draw the signals transmitted by nodes A and B. Draw the resulting (total) signal as it would appear to a receiving node that is in range of both transmitters. 4 Node A 1 data 0 -1 1 CDMA code 0 -1 2 1 transmitted 0 signal -1 -2 Node B 1 data 0 -1 1 CDMA code 0 -1 2 1 transmitted 0 signal -1 -2 Receiver node 2 1 received 0 signal -1 -2 Page 3 Below is a diagram of a signal received on the same network, along with the receiver's CDMA code. Draw the signal after decoding at the receiver, and the resulting data received. 1.6 In 802.11 PCF, why is there a sudden, large increase in delay once the number of users becomes too high? 1.7 We are using a technology that allows the following modulations with the corresponding data rates: Modulation type Coding rate Data rate (Mbps) SINR threshold (dB) BPSK 1/2 6.5 1.7609 QPSK 1/2 13 3.0103 QPSK 3/4 19.5 3.9794 16-QAM 1/2 26 4.7712 16-QAM 3/4 39 5.4407 64-QAM 2/3 52 6.0206 64-QAM 3/4 58.5 6.5321 64-QAM 5/6 65 6.9897 ; we have two transmitters and two receivers, that are positioned in a way they are both d=10m from the two transmitters; transmitter 2 needs a data rate of 50 Mbps while transmitter 1 just needs 15 Mbps; choose an appropriate modulation scheme and transmission power in order to ensure the simultaneous transmission from both the transmitters; assume there is no fast fading but only a pathloss such that, if Ω is the received power in W, d is the distance in meters from Tx to Rx and P the transmitted power in W, we have Ω = P d−4. 6 Part III Solutions 2 Reservation schemes 2.1 a) SDMA separates users in physical space, by segmenting space into different cells or sectors, for example by using directional antennas. Advantage: very simple to implement. Disadvantage: inflexible as the antennas are typically fixed, i.e. it is difficult to change the segmentation dynamically. Example system: GSM (cells) b) TDMA separates users by dividing time into timeslots, then each user has one or more slots when they are able to transmit and no other users may transmit. Advantage: fully digital so easy to adjust dynamically. Disad- vantage: synchronization can be difficult, especially considering multipath propagation. Example system: GSM c) FDMA separates users by allocating different frequency bands to different users. Advantage: simple and robust. Disadvantage: spectrum is a scarce resource so it may be difficult to have enough for all users. Example system: GSM d) CDMA uses orthoganal (or pseudo-orthoganal) codes to separate users. Users all transmit simultaneously but each user combines their data with their assigned code. A receiver can then decode a given transmission if they know the user's code. Advantage: flexible and less frequency planning needed since all users can use the entire spectrum all the time. Disadvantage: complex receivers are required. Example system: UMTS 2.2 The rate of work is given by n 1 X η = ρ n × ν i i=1 For each slot, we have 10 ms of time that can be used, plus 1 ms guard time, and a rate of 1 Mbps, giving 10 1 × 220 × = 953250:909 bits=s 11 for each channel. For each channel, we have 100 Mhz in usable bandwidth and a guard band of 5 MHz. The total resources available for each channel are then 1 × 220 ν = = 0:0099864381 bits=Hz=s 105 × 106 7 (i.e. the total bandwidth available at the given data rate). In each slot, we have an 80% chance to use the slot, so we have a utilised rate of ρ = 0:8×0:0099864381 = 0:00798915048, which is the same across all the three channels. So we have: 3 1 X η = 0:00798915048 3 × 0:0099864381 i=1 = 0:8 2.3 1. In Reservation ALOHA, there are two phases: a reservation phase, followed by a transmission phase. In the reservation phase, stations reserve slots for the transmission phase using ALOHA. If the transmission phase has k slots, then the reservation phase is divided into k mini-slots (of much shorter duration than the transmission slots), so that a transmission in mini-slot i reserves transmission slot i. These two phases are then repeated. See the MAC slides, pp12-13, for a diagram. 2. Advantage: efficiency can be significantly increased (up to 80% from only 36% for slotted ALOHA), as no collisions are possible in the transmission phase. Disadvantage: delay increases due to waiting for the reserved timeslot, especially if a collision occurs during the reservation, and syn- chronisation is required between all stations to keep the reservation list consistent. 3. P(at least one packet to send) = 1 - P(no packets to send) e−λTs · (λT )0 P = 1 − s 0! = 1 − e−λTs . −λT 4. N · 1 − e s 5. " 3 k − 1 k − 2 1 − 1 − e−λTs k k 2 k − 1 + 3 1 − e−λTs e−λTs k 2 + 3 1 − e−λTs e−λTs # 3 + e−λTs 8 that is, the opposite event of the sum of the probability of all three transmitting and choosing different slots, plus the probability of two of them transmitting and choosing different slots, plus the probability of only one of them tranmsitting plus the probability of no one transmitting. 2.4 In order to check whether a chip code is orthogonal to another we have to check wether the inner product is zero; for example if c1=1 1 1 1 and c2 =1 -1 -1 1 we have hc1; c2i = 1 · 1 + 1 · −1 + 1 · −1 + 1 · 1 = 0, so they are mutually orthogonal. a) they are all mutually orthoganal; b) they are mutually orthoganal; c) they are NOT mutually orthogonal. 2.5 For an example refer to slide 36 of "MAC 1:Reservation schemes\; 9 Node A 1 data 0 -1 1 CDMA code 0 -1 2 1 transmitted 0 signal -1 -2 Node B 1 data 0 -1 1 CDMA code 0 -1 2 1 transmitted 0 signal -1 -2 Receiver node 2 1 received 0 signal -1 -2 Page 3 2.6 The 802.11 PCF uses a fixed-length period for contention-free (i.e.

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