Wi-Fi: unobvious nuances (for example, home Wi-Fi) [part 1]
Now many people buy 802.11n access points, but not everyone can achieve good speeds. In this post we will talk about not very obvious small nuances that can significantly improve (or worsen) the work of Wi-Fi. Everything described below is applicable both to home Wi-Fi routers with standard and advanced (DD-WRT & Co.) firmware, as well as to corporate devices and networks. Therefore, as an example, take the "home" theme, as more familiar and close to the body. For even the most administrative of administrators and engineers of engineers live in apartment buildings (or villages with a sufficient density of neighbors), and everyone wants fast and reliable Wi-Fi.
[Attention!]: After the comments regarding the publication, the article is posted here . This article is left for an example of how to publish it is not necessary. Sorry for the mess :)
[1.1] It would seem - what is really there? He unscrewed the point at full power, got the maximum coverage possible - and rejoice. And now let's think: not only the signal of the access point must reach the client, but the signal of the client must reach the point. TD transmitter power is typically up to 100 mW (20 dBm). Now look in the datasheet to your laptop / phone / tablet and find the power of its Wi-Fi transmitter there. Found? You are very lucky! Often it does not indicate at all (you can search by FCC ID ). Nevertheless, we can confidently say that the power of typical mobile clients is in the range of 30-50 MW. Thus, if the TD is broadcasting at 100mW and the client is only at 50mW, there will be places in the coverage area where the client will hear the point well, and the client's TD will be bad (or not at all) - asymmetry. There is a signal - but there is no connection. Or downlink is fast and uplink is slow. This is true if you use Wi-Fi for online games or Skype, for normal Internet access is not so important (only if you are not on the edge of the cover). And we will complain about the poor provider, buggy point, driver curves, but not illiterate network planning.
Conclusion: it may be that in order to get a more stable connection, the power of the point will have to be reduced . That, you see, is not quite obvious :)
Rationale (for those who are interested in the details):
Our task is to provide the most symmetrical communication channel between the client (STA) and the point (AP) in order to equal the speeds of uplink and downlink. For this we will rely on SNR (signal-to-noise ratio).
SNR (STA) = Rx (AP) - RxSens (STA); SNR (AP) - Rx (STA) - RxSens (AP)
where Rx (AP / STA) is the received signal power from the point / client, RxSens (AP / STA) is the point / client reception sensitivity. For simplicity, let us assume that the background noise threshold is below the sensitivity threshold of the AP / STA receiver. Such a simplification is quite acceptable, since if the level of background noise for the AP and STA is the same - it does not affect the symmetry of the channel.
Further,
Rx (AP) = Tx (AP) [transmitter power point at the antenna port] + TxGain (AP) [transmit antenna gain point taking into account all losses, gains and directivity] - PathLoss [signal loss from point to client] + RxGain (STA) [client antenna gain gain with all losses, gains and directivity given].
Similarly, Rx (STA) = Tx (STA) + TxGain (STA) - PathLoss + RxGain (AP) .
It is worth noting the following:
Total we get:
SNR (AP) = Tx * (STA) [with antenna] - PathLoss + Gain (AP) - RxSens (AP)
SNR (STA) = Tx (AP) + Gain (AP) - PathLoss -RxSens * (STA) [including antenna]
The difference between the SNR at both ends will be the asymmetry of the channel, we use arithmetic: D = SNR (STA) -SNR (AP) = Tx * (STA) - Tx (AP) - (RxSens * (STA) - (RxSens (AP)) .
')
Thus, the asymmetry of the channel does not depend on the type of antenna on the point and on the client (again, it depends if you use MIMO, MRC and so on, but then it will be rather difficult to calculate something), but depends on the power difference and sensitivity of the receivers. When D <0, the dot will hear the client better than the dot client. Depending on the distance, this will mean either that the data flow from the client to the point will be slower than from the point to the client, or the client will not be able to reach the point at all.
For the points we have taken (100mW = 20dBm) and the client (30-50mW ~ = 15-17dBm), the power difference will be 3-5dB. As long as the receiver point is more sensitive than the client receiver for these same 3-5dB - there will be no problems. Unfortunately, this is not always the case. Let's carry out calculations for the HP 8440p laptop and D-Link points of the DIR-615 point for 802.11g@54Mbps:
[1.2] It is also far from the most well-known fact that adds to the asymmetry, that the majority of client devices have reduced transmitter power on the “extreme” channels (1 and 11/13 for 2.4 GHz). Here is an example for the iPhone from the FCC documentation (power at the antenna port).
As you can see, on the extreme channels, the transmitter power is ~ 2.3 times lower than on average. The reason is that Wi-Fi - broadband connection, to keep the signal clearly within the channel frame will not succeed. So we have to reduce the power in "borderline" cases, so as not to hurt the ranges adjacent to the ISM. Conclusion: if your tablet does not work well in the toilet - try to move to channel 6.
If we are talking about the channels - next time we will talk about them in more detail.
UPD : I have already stated that the note is too short. I already understood everything. But if I add the rest of the text here - many will not see it. The following post I will lay out all entirely (I will clean the first part under kat). Additional comments are welcome if they expand the topic of “literate” posting on Habré. For emotions there is this hub . Thank you for understanding.
[Attention!]: After the comments regarding the publication, the article is posted here . This article is left for an example of how to publish it is not necessary. Sorry for the mess :)
1. How to live well yourself and not interfere with the neighbors.
[1.1] It would seem - what is really there? He unscrewed the point at full power, got the maximum coverage possible - and rejoice. And now let's think: not only the signal of the access point must reach the client, but the signal of the client must reach the point. TD transmitter power is typically up to 100 mW (20 dBm). Now look in the datasheet to your laptop / phone / tablet and find the power of its Wi-Fi transmitter there. Found? You are very lucky! Often it does not indicate at all (you can search by FCC ID ). Nevertheless, we can confidently say that the power of typical mobile clients is in the range of 30-50 MW. Thus, if the TD is broadcasting at 100mW and the client is only at 50mW, there will be places in the coverage area where the client will hear the point well, and the client's TD will be bad (or not at all) - asymmetry. There is a signal - but there is no connection. Or downlink is fast and uplink is slow. This is true if you use Wi-Fi for online games or Skype, for normal Internet access is not so important (only if you are not on the edge of the cover). And we will complain about the poor provider, buggy point, driver curves, but not illiterate network planning.
Conclusion: it may be that in order to get a more stable connection, the power of the point will have to be reduced . That, you see, is not quite obvious :)
Rationale (for those who are interested in the details):
Our task is to provide the most symmetrical communication channel between the client (STA) and the point (AP) in order to equal the speeds of uplink and downlink. For this we will rely on SNR (signal-to-noise ratio).
SNR (STA) = Rx (AP) - RxSens (STA); SNR (AP) - Rx (STA) - RxSens (AP)
where Rx (AP / STA) is the received signal power from the point / client, RxSens (AP / STA) is the point / client reception sensitivity. For simplicity, let us assume that the background noise threshold is below the sensitivity threshold of the AP / STA receiver. Such a simplification is quite acceptable, since if the level of background noise for the AP and STA is the same - it does not affect the symmetry of the channel.
Further,
Rx (AP) = Tx (AP) [transmitter power point at the antenna port] + TxGain (AP) [transmit antenna gain point taking into account all losses, gains and directivity] - PathLoss [signal loss from point to client] + RxGain (STA) [client antenna gain gain with all losses, gains and directivity given].
Similarly, Rx (STA) = Tx (STA) + TxGain (STA) - PathLoss + RxGain (AP) .
It is worth noting the following:
- PathLoss is the same in both directions.
- TxGain and RxGain antennas are the same for conventional antennas (true for AP and for STA). It does not consider cases with MIMO, MRC, TxBF and other tricks. So you can accept: TxGain (AP) === RxGain (AP) = Gain (AP) , similarly for STA.
- The client's Rx / Tx Gain antenna is little known. Client devices are usually equipped with non-replaceable antennas, which allows you to specify the transmitter power and receiver sensitivity immediately taking into account the antenna. We note this in our calculations below.
Total we get:
SNR (AP) = Tx * (STA) [with antenna] - PathLoss + Gain (AP) - RxSens (AP)
SNR (STA) = Tx (AP) + Gain (AP) - PathLoss -RxSens * (STA) [including antenna]
The difference between the SNR at both ends will be the asymmetry of the channel, we use arithmetic: D = SNR (STA) -SNR (AP) = Tx * (STA) - Tx (AP) - (RxSens * (STA) - (RxSens (AP)) .
')
Thus, the asymmetry of the channel does not depend on the type of antenna on the point and on the client (again, it depends if you use MIMO, MRC and so on, but then it will be rather difficult to calculate something), but depends on the power difference and sensitivity of the receivers. When D <0, the dot will hear the client better than the dot client. Depending on the distance, this will mean either that the data flow from the client to the point will be slower than from the point to the client, or the client will not be able to reach the point at all.
For the points we have taken (100mW = 20dBm) and the client (30-50mW ~ = 15-17dBm), the power difference will be 3-5dB. As long as the receiver point is more sensitive than the client receiver for these same 3-5dB - there will be no problems. Unfortunately, this is not always the case. Let's carry out calculations for the HP 8440p laptop and D-Link points of the DIR-615 point for 802.11g@54Mbps:
- 8440p : Tx * (STA) = 17dBm, RxSens * (STA) = -76dBm @ 54Mbps
- DIR-615 : Tx (AP) = 20dBm, RxSens (AP) = -65dBm @ 54Mbps.
- D = (17 - 20) - (-76 +65) = 3 - 11 = -7dB.
[1.2] It is also far from the most well-known fact that adds to the asymmetry, that the majority of client devices have reduced transmitter power on the “extreme” channels (1 and 11/13 for 2.4 GHz). Here is an example for the iPhone from the FCC documentation (power at the antenna port).
As you can see, on the extreme channels, the transmitter power is ~ 2.3 times lower than on average. The reason is that Wi-Fi - broadband connection, to keep the signal clearly within the channel frame will not succeed. So we have to reduce the power in "borderline" cases, so as not to hurt the ranges adjacent to the ISM. Conclusion: if your tablet does not work well in the toilet - try to move to channel 6.
If we are talking about the channels - next time we will talk about them in more detail.
UPD : I have already stated that the note is too short. I already understood everything. But if I add the rest of the text here - many will not see it. The following post I will lay out all entirely (I will clean the first part under kat). Additional comments are welcome if they expand the topic of “literate” posting on Habré. For emotions there is this hub . Thank you for understanding.
Source: https://habr.com/ru/post/149418/
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