802 11b data rates.

Hmmmm, I have question what a quick Internet search has failed.

We know 802. 11 b data rates are 1 Mbps, 2 Mbps, 5.5 Mbps and 11 Mbps. My question is, how can we get these numbers?

The 802. 11 b frequency spectrum signal is sent via, is 22 Mhz wide. In other words, the 22 million oscillations per second.

Lets take 1 MB/s first. It uses barker coding, with 11-bit sent (or chip) equal to the actual data bit 1. As a result, 22 million (oscillations) each result in a forest (a 1 or a 0). Of these 22 million bits, we know that a chip (11 bits) equals the actual data bit 1. As a result, 22 million/11 = 2 million or 2Mbps. If this shouldn't be 1 MB/s? The 2mbps is a unique swing is equivalent to two bits, doubling the 1mbps to 2 Mbps. Im trying to work on how these 802. 11 b rates are calculated. Math below is probably correlated to somehow how rates are calculated?

22 000 000 (22 Mhz) / 22 = 1 MB/s

22 000 000 (22 Mhz) / 11 = 2 Mbps

22 000 000 (22 Mhz) / 4 = 5.5 Mbps

22 000 000 (22 Mhz) / 2 = 11 Mbps

Anyone?

Dazzler

Watching a trace of Spectrum Analyzer also lets you understand.

If you look at a 802. 11B trace, you will see that it resembles a hill, a bump.

Which means that the centre 11 Mhz are at max power. The other 11 Mhz border are not useful signal, but they are still unusable by others.

22 Mhz is so the connection you "kill" when you will pass, but only the 11 mhz Center contain a useful signal.

To remedy this, the 802.11 g brings OFDM and there are several substrings and material. This allows the signal look like "bart simpsons hair", which is very steep on the border, you use not all 20 Mhz then.

Tags: Cisco Wireless

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