Übung 6: IEEE WLAN

Transkript

1 WCOM2, Rumc, 1/7 Übung 6: IEEE WLAN Problem 1: Rate versus radio range for g WLAN. If a WLAN-device moves away from the Access Point (AP), the received signal strength decreases and the device and/or the AP will automatically switch to a lower but more robust data rate. In this problem, the compromise between rate and radio range is investigated which, by the way, is typical for all mobile radio systems. a) Please estimate the receiver sensitivity, i.e. the minimum required Rx signal-power S min, for the different g modulation and FEC coding schemes given in the following table and then complement the table. Rate Modulation Rate FEC S min R indoor R outdoor 6 Mbps BPSK 1/2 24 Mbps 16-QAM 1/2 54 Mbps 64-QAM 3/4 Please proceed as follows: - Determine the noise floor N [dbm] within an g WLAN-channel of 25 MHz bandwidth. - Assume that the WLAN-receiver has a noise figure F of 2 db. - Remember that BPSK modulation requires a minimum SNR min of 10 db, 16-QAM a SNR min of 15 db and 64-QAM a SNR min of 20 db on an AWGN channel without FEC. - Assume that a coding gain of 7 db for R = 1/2 FEC and 3 db for R = 3/4 FEC can be realised. - Include a (Rayleigh-) fading margin of 20 db into the receiver sensitivity S min to account for a good connection. b) Please estimate the maximum radio range R indoor for indoor coverage assuming a propagation path loss of 40 db / decade and then complement the table above. Note that a simplified indoor propagation model is used in this example since the propagation loss is usually smaller, but there are walls etc. adding additional loss. Please also estimate the maximum radio range R outdoor for very optimistic outdoor coverage with strong LOS assuming a (free space) propagation path loss of 20 db / dec. Please proceed as follows: - Assume an EIRP of 20 dbm (100 mw) and no receiver antenna gain, i.e. assume that the maximum tolerable path loss PL path = EIRP - S min. - Assume further that the propagation model has the form PL path = A + B log 10 (R[m]) where A is the free space path 1m (and 2.45 GHz).

2 WCOM2, Rumc, 2/7 Problem 2: MAC-DCF. Consider an BSS with an Access Point (AP) and 3 stations (STAx) sending data packets to the AP using the asynchronous data service. a) Please complement the following timing diagram with all transmitted packets under the assumption that - no RTS- and CTS-packet are used - DIFS = 50 µs, SIFS = 10 µs and a slot time = 20 µs - the transmission period for each data packet is 500 µs - the transmission period for each -packet is 140 µs - STA1 has a data packet pending and a remaining backoff of 20 slots - STA2 has a data packet pending and a remaining backoff of 5 slots - STA3 gets a request to send a packet from a higher layer at t = 750 µs and chooses a random backoff of 10 slots AP Zeit [µs] 0 STA1 backoff timer STA2 backoff timer STA3 backoff timer b) Which multiple access method is used in a) and which mechanisms are used? c) What happened in a) if STA3 had randomly selected a backoff timer value of 15 slots instead of 10 slots? d) What happened in a) if STA1 and STA3 could not hear each other but would hear the AP only? Which options are provided by the MAC to deal with this hidden node problem? e) Which disadvantages has the DCF-procedure for applications like VoIP and video streaming?

3 WCOM2, Rumc, 3/7 Problem 3: Short questions. a) How does the maximum data rate R max = 54 Mbit/s come about in an IEEE g WLAN? b) What happens if the G-only mode is activated in an IEEE g access point (AP)? c) What is the maximum transfer rate in an IEEE g WLAN under ideal conditions? And why is the maximum transfer rate smaller than 54 Mbit/s? d) The OFDM-guard-period of IEEE g is t g = 800 ns. Please determine the maximum tolerable length-difference Δ max between a LOS- and a NLOS-path. e) Assume that an g AP of the WLAN network has 2 antennas. Please explain what they are good for? f) Sie konfigurieren das Wi-Fi-Modul ODIN-W1von ublox als AP und stellen eine passende SSID und den Kanal ein. Das dual-band Wi-Fi-Modul ODIN-W1 unterstützt die WLAN- Standards IEEE a/b/g/n auf diversen 2.4 GHz bzw. 5 GHz WLAN-Kanälen. f 1 ) Skizzieren und begründen Sie das Spektrum der IEEE g Beacon-Frame- Aussendung, wenn Sie den WLAN-Kanal 11 eingestellt haben. f 2 ) Skizzieren und begründen Sie das Spektrum der IEEE n Beacon-Frame- Aussendung, wenn Sie den WLAN-Kanal 36 eingestellt haben. f 3 ) Nehmen Sie an, dass das Wi-Fi-Modul als AP auf Kanal 1 konfiguriert und über ein SMA-Kabel direkt mit einem Spektrum-Analyzer verbunden worden ist. Nehmen Sie weiter an, dass am Spektrum-Analyzer die folgenden Einstellungen gemacht worden sind: center frequency = 2412 MHz, span = 0 Hz, sweep time = 1s Skizzieren und begründen Sie die resultierende Anzeige am Spektrum Analyzer. g) Erklären Sie die Wirkung der verschiedenen High-Throughput (HT-) Features von n.

4 WCOM2, Rumc, 4/7 Solution Problem 1 Noise level: N = *log 10 ( Hz WLAN-bandwidth) -100 dbm 54 Mbps SNR min = 20 db (64-QAM) - 3 db (R=3/4 FEC coding gain) = 17 db S min = N + F + SNR min + margin = = -61 dbm max. PL path = 20 - (-61) = 81 db PL path = 40 db + 40 log 10 (R indoor [m]) = 81 db => R indoor = 10 41/40 = 11 m PL path = 40 db + 20 log 10 (R outdoor [m]) = 81 db => R outdoor = 10 41/20 = 110 m 24 Mbps SNR min = 15-7 = 8 db S min = N + F + SNR min + margin = = -70 dbm max. PL path = 20 - (-70) = 90 db PL path = 40 db + 40 log 10 (R indoor [m]) = 90 db => R indoor = 10 50/40 = 18 m PL path = 40 db + 20 log 10 (R outdoor [m]) = 90 db => R outdoor = 10 50/20 = 316 m 6 Mbps SNR min = 10-7 = 3 db S min = N + F + SNR min + margin = = -75 dbm max. PL path = 20 - (-75) = 95 db PL path = 40 db + 40 log 10 (R indoor [m]) = 95 db => R indoor = 10 55/40 = 24 m PL path = 40 db + 20 log 10 (R outdoor [m]) = 95 db => R outdoor = 10 55/20 = 562 m Rate Modulation Rate FEC S min R indoor R outdoor 6 Mbps BPSK 1/2-75 dbm 24 m 562 m 24 Mbps 16-QAM 1/2-70 dbm 18 m 316 m 54 Mbps 64-QAM 3/4-61 dbm 11 m 110 m Please note that the figures above well compare with the figures reported in

5 DIFS CW (5 slots) SIFS DIFS CW (10 slots) SIFS DIFS CW (5 slots) SIFS WCOM2, Rumc, 5/7 Problem 2: MAC-DCF. a) Please also cf. J. Schiller, Mobile Communications, 2 nd edition, Addison Wesley, 2003, pages for a good description of the MAC-DCF. AP Zeit [µs] STA1 backoff timer Data STA2 backoff timer 5 0 Data STA3 backoff timer Data b) The mandatory basic multiple access method used on the MAC layer is Carrier Sense Multiple Access (CSMA) with Collision Avoidance (CA), because the nodes have to sense if the medium is idle for at least the duration of DIFS and collisions are avoided by a random backoff mechanism. c) If STA3 had randomly selected a backoff timer value of 15, there were a collision caused by STA1 and STA3. Then, STA1 and STA3 each had to select a random backoff within a 2-times larger backoff window and contend again for the channel to retransmit their packets without privileges. d) There were a collision between the packet sent by STA3 at t = 1050 μs and the packet sent by STA1 at t = 1150 μs. To deal with the hidden node problem, the MAC provides the option to send short RTS- and CTS-packets before a long data packet that virtually reserve the medium with the NAV timing information in the duration field in the packet headers. Then, collisions can only occur while the RTS is sent, a very short period at the beginning of a transmission period. e) The transmission delay of the data packets varies whereas applications like VoIP require a constant bandwidth or, equivalently, constant packet transmission delays. If a WLAN network is heavily loaded, the voice quality can suffer.

6 WCOM2, Rumc, 6/7 Problem 3 a) IEEE g uses OFDM with 48 data-subchannels having 4 μs symbol period each. The maximum data rate R max is achieved if 64 QAM-modulation and R=3/4 FEC are used. R max = 48 OFDM-subchannels (6 Bit / OFDM-subchannel) 0.75 / 4 μs = 54 Mbps b) IEEE b WLAN-devices are no longer supported in the corresponding BSS. On the other hand, there is no extra overhead for backward compatibility and, thus, the throughput increases (by 20-40%). c) about 20 Mbit/s if a WLAN-device and the access point communicate with each other and less than 10 Mbit/s if 2 WLAN-devices communicate with each other (using the air interface twice) The maximum transfer rate is smaller than 54 Mbit/s because - each data frame has to be acknowledged with an frame - the PHY-header has to be transmitted with 6 Mbit/s ( G-only -mode) such that WLANdevices at the border of the BSS can decode it d) Δ = t g c = 0.8 us m/s = 240 m => this is larger than the typical cell size => ISI-problems only in special outdoor situations e) The AP uses the 2 antennas for diversity, i.e. reducing the chance for a deep Rayleigh fade g does not support MIMO. f) f 1 ) IEEE g ist default-mässig rückwärts kompatibel zu IEEE b. Die Beacon Frames werden deshalb im DSSS-Format ausgesendet. Das 1 Mbps DBPSK-Signal wird mit dem Barker-Code der Länge 11 gespreizt. Das Spektrum ist sinc-förmig mit Nullstellen bei f c ±11 MHz, f c ±22 MHz,... Das pulse-shaping sorgt dafür, dass die Tx- Maske eingehalten wird. Spektrum (Quelle: Tektronix-Poster W-Fi Physical Layer..., f c = 2462 MHz):

7 WCOM2, Rumc, 7/7 f 2 ) IEEE n sendet wie alle anderen Standards im 5 GHz Bereich die Beacon Frames OFDM-moduliert aus. Wenn bei IEEE n nur 20 MHz Bandbreite verwendet wird, sind die OFDM-Spektren von IEEE n und g praktisch identisch. Die OFDM-Bandbreite von IEEE g beträgt 53 Subträger mal khz Trägerabstand 16.6 MHz, wobei seitlich ca. 20 db schwächer zwei Schultern erkennbar sind. Spektrum (Quelle: Tektronix-Poster W-Fi Physical Layer..., f c = 5180 MHz): f 3 ) Dank der zero-span Option ist das RSS-Zeitsignal auf dem WLAN-Kanal 1 sichtbar. Im 1s breiten Zeitfenster sind kurze Beacon-Frames im Abstand von 100 ms sichtbar. g) Erklären Sie die Wirkung der verschiedenen High-Throughput (HT-) Features von n. HT-Feature Wirkung 40 MHz channels höhere Datenrate wegen Verdoppelung Anzahl Subträger frame aggregation 400ns OFDM guard 1 für mehrere Frames, effizientere Nutzung des Mediums und weniger Tx-Rx-Wechsel 400ns mehr Tx-Zeit pro OFDM-Symbol 5/6 FEC weniger FEC-Redundanz wenn SNR hoch ist MIMO höhere Datenrate dank SDMA