## 802.11b data rates.

Hmmmm, I have question that a quick internet search didn't answer.

We know 802.11b data rates are 1mbps, 2mbps, 5.5mbps and 11mbps. My question is, how do we get these numbers?

The 802.11b frequency spectrum that a signal is sent over, is 22Mhz wide. That is, 22 million oscillations per second.

Lets take 1mbps first. It uses barker coding, with 11 sent bits (or a chip) equal to 1 real data bit. Therefore, 22 million (oscillations) each translate to a bit (a 1 or a 0). From these 22 million bits, we know that a chip (11 bits) equates to 1 real data bit. Therefore, 22 million/11 = 2 million or 2Mbps. Shouldnt this be 1mbps? The 2mbps is when a single oscillation equals two bits, effectively doubling the 1mbps to 2mbps. Im trying to work out how these 802.11b data rates are calculated. The maths below surely correlate somehow to how the data rates are calculated?

22,000,000 (22Mhz) / 22 = 1Mbps

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

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

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

Anyone?

Dazzler

**Correct Answer**by dperowne@cco about

*3 years 8 months*ago

**Correct Answer**by Nicolas Darchis about

*3 years 9 months*ago

Looking at a spectrum analyzer trace also helps to understand.

If you look at a 802.11b trace, you will see that it looks like a hill, a bump.

Meaning the 11 center Mhz are at max signal strength. The other border 11Mhz are not useful signal but are still unusable by other.

So 22Mhz is the bandwitdh you "kill" when you transmit but only the center 11mhz contain a useful signal.

To tackle this, the 802.11g brings OFDM and there are several subchannels and subcarriers. This allows the signal to look like "bart simpsons hair", i.e. very steep on the border, so you do use all the 20Mhz then.

The different data rates have different chipping methods, with 1 and 2 Mbps using the Barker code and 5.5 and 11 using CCK. To calculate the total throughput, you have relate chipping method to the bits each data rate sends along with the symbol rate (modulation symbol changes per second) they use. Both run at 11 Mchips/s so the symbol rates look like this:

Barker Code - 11 chip code

Symbol Rate = 11,000,000 / 11 = 1 Msps

CCK - 8 chip code

Symbol Rate = 11,000,000 / 8 = 1.375 Msps

Then you take the Number of Bits each rate transmits and multiply is by the symbol rate:

1 Mbps = 1 bit x 1 Msps

2 Mbps = 2 bits x 1 Msps

5.5 Mbps = 4 bits x 1.375 Msps

11 Mbps = 8 bits x 1.375 Msps

The key here is that the symbol rate is only different between the code types, not the data rates themselves. It's those extra bits being sent over the same rate.