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Channel bandwidth
Channel bandwidth






channel bandwidth

We compare the performance of the 802.11ac OBSSs implementing the CWSS and the standard 802.11ac OBSSs by a computer simulation using C + +. 2b, if we assumed that all STAs are capable of 80 MHz operation. The BW max is 80 MHz for STAs in each BSS in Fig. The STA obtains the channel width supported by the BSS from a beacon frame sent out periodically by an access point (AP). The BW max of the STA equals the maximum channel width supported by both the STA and the BSS. The BW min is 20 MHz for STAs in each BSS in Fig. Thus, each STA knows which channels are in use by OBSSs and is able to set BW min. The 802.11ac STAs associated with the BSS are required to monitor the channels affected by the frequency range of the BSS . The STA does not decrease the channel width below BW min because either BW min is the narrowest channel in which the STA is allowed to transmit (BW min = 20 MHz) or it knows that no other STA is able to use the released bandwidth (BW min = 40 MHz). 2b), if the primary 40 MHz channel of this BSS overlaps with the primary 20 MHz channel of the other BSS, whereas the BW min equals to 40 MHz, if the primary 40 MHz channel of the given BSS does not overlap with the primary 20 MHz channel of the other BSS. The BW min of the STA in the 80 MHz BSS equals to 20 MHz (Fig. They specify the range of channel width in which the STA transmits data. Such method is proposed in the following Section.ī Different primary 20 MHz channels Channel width selection schemeĮach STA implementing the channel width selection scheme (CWSS) maintains two parameters: the minimum channel bandwidth BW min and the maximum channel bandwidth BW max. The key issue is the method by which the STA selects the width of the channel for data frame transmission. However, this change alone is not sufficient to improve the bandwidth utilisation, because the STA, which wins contention for the channel access, still occupies the entire 80 MHz bandwidth. Although the primary 20 MHz channels do not overlap, the virtual carrier sense in each BSS works correctly because the PHY packet duration can be obtained from the legacy portion of the preamble which is replicated over each 20 MHz channel of the 80 MHz bandwidth (Fig 1b). As a result, the transmission in the channel narrower than 80 MHz by the STA from one BSS does not prevent the use of other parts of the 80 MHz bandwidth by STAs belonging to other BSSs. In this Letter, we propose assigning a different primary 20 MHz channel to each BSS (Fig. The reduction in the channel width by a transmitting STA to 40 or 20 MHz makes the remaining part of the 80 MHz bandwidth (40 or 60 MHz) go to waste, because the primary 20 MHz channel of all BSSs is busy. In such a scenario, STAs from all OBSSs compete for access to the same 80 MHz channel. The 802.11ac recommends that OBSSs should have a common primary 20 MHz channel (Fig. Consider a scenario with four 80 MHz overlapping BSSs (OBSSs). Channel allocation for overlapping BSSsīecause of the wider channels in 802.11ac, there is a high probability that BSSs will overlap in frequency and space in areas with a high density of BSSs. In this Letter, we propose a method by which 802.11ac STA selects the channel bandwidth, taking into account the length of the transmitted data frame. Thus, from the point of view of frequency bandwidth utilisation, it makes sense to allow the STA to send short data frames in a narrower channel even when a wider channel is available, leaving the rest of the bandwidth to STAs in other BSSs. Further reduction in the channel width to 20 MHz increases the maximum channel utilisation to 0.46. As a result, the maximum channel utilisation increases to 0.29. A reduction in the channel width to 40 MHz roughly doubles T d, whereas T phy/mac remains unchanged. For example, if we assume the PHY data rate equal to 292.5 Mbit/s and other system parameters as listed in the Simulation results Section, then the maximum channel utilisation, defined as T d/( T mac/phy + T d), is only 0.17 for the data frame length equal to 1500 bytes and for the 80 MHz channel. All other components express the constant medium access control (MAC) layer and the PHY overhead, labelled T mac/phy. (1)Only T d in ( 1) depends on the width of the channel in which the transmission is carried out. IET Generation, Transmission & Distribution.IET Electrical Systems in Transportation.IET Cyber-Physical Systems: Theory & Applications.IET Collaborative Intelligent Manufacturing.CAAI Transactions on Intelligence Technology.








Channel bandwidth