802.11b data rates.

Answered Question
Aug 3rd, 2011

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

I have this problem too.
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Correct Answer by dperowne@cco about 2 years 7 months ago

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.

Correct Answer by Nicolas Darchis about 2 years 8 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.

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dazza_johnson Thu, 08/04/2011 - 01:33

Hey there, thanks for the reply. I will review the link, in the meantime any idea how you get to the 1 and 2mbps using barker coding?

Thanks

Dazzler

dazza_johnson Thu, 08/04/2011 - 01:36

Hang on, that link anyone could find with a simple google serach. The question I hope was clear, how are the data rates calculated to be 1, 2, 5.5 and 11mbps. Sorry, but wikipedia links do not cut the mustard for this question....

dazza_johnson Thu, 08/04/2011 - 18:47

Hi Nicolas, thanks again for taking the time to reply. I have concluded the following, which I hope someone/somewhere could validate.

So, 802.11b uses a 22Mhz wide channel. According the the 802.11b standard (clause 15), only 11Mhz is actually used by the chipping process (I cannot find what happens to the other 11Mhz sadly)......

"The chipping rate is 11 MHz, which is the same as the DSSS system described in Clause 15 of IEEE Std 802.11, 1999 Edition, thus providing the same occupied channel bandwidth"

Source; http://pdos.csail.mit.edu/decouto/papers/802.11b.pdf

So, if each oscillation/wave/cycles equates to a single bit, then 11Mhz = 11 million cycles per second = 11 million bits per second. Cool. Barker coding, (BPSK) uses 11 bits to represent 1 real data bit. Therefore, 11million/11 = 1Mbps :-)

Using QPSK, each 11 bits = 2 real bits, 11million/11 multipled by 2 equals 2Mbps :-)

In the absence that someone can prove otherwise, I'm going to settle on this as the correct formula....

Thanks

Dazzler

Correct Answer
Nicolas Darchis Thu, 08/04/2011 - 23:37

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.

dazza_johnson Thu, 08/04/2011 - 23:47

Hey Nicolas, thanks for the update. What you are saying validates my theory then, and makes a lot of sense. I can't believe there is hardly anything on the internet about this. Oh well.... Thanks again.

Dazzler

rdvorak Fri, 08/05/2011 - 02:14

Hi Dazzler,

as per the CWNA "bible" one bit uses 2MHz....

http://books.google.com/books?id=woefmaA_YfEC&pg=PA177&lpg=PA177&dq=cwnp+dbpsk+22mhz&source=bl&ots=Yo1DemvAm0&sig=QLThNFdGacphaY7-YX4FRYoxBIQ&hl=en&ei=-bI7TvflG8zasgasgbHjDw&sa=X&oi=book_result&ct=result&resnum=1&ved=0CBQQ6AEwAA#v=onepage&q&f=false

Both the CWNA and CWDP book are really great resources that provide a good unterstanding about 802.11

Kind regards,

Ron

dazza_johnson Fri, 08/05/2011 - 04:38

Hey Ron, thanks for the response. I have the CWNA 'bible' but its 4th edition so maybe there are far newer copies out there with better content. Regarding you point and the link you attached, if 1 bit uses 2Mhz and therefore 11 bits = 22Mhz, doesn't this equate to 1 bit per second!?!? (11 bits over 22Mhz (22million cycles per second) / 11 bits for barker coding)

I appreciate the reference you sent stipulates this, but the maths doesn't add up, 1 bit per second would be extremely ssssslllllooooowwwwww ;-)

Dazzler

Nicolas Darchis Fri, 08/05/2011 - 04:49

Don't confuse bandwitdh and frequency.

The fact that this happens at 2.4Ghz means there are 2.4 billions oscillations per second. There is then the fact that the information is spread over 22 Mhz of bandwitdh.

If one bit takes 2Mhz of bandwidth, doesn't mean that there is not billions of oscilliations per second and thus possible changes in the bit signification per second.

This said, it always gives me headaches when I try to find the maths to match the numbers so I'm just being the devil's advocates here :-)

Correct Answer
dperowne@cco Fri, 09/09/2011 - 20:25

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.

dazza_johnson Mon, 09/12/2011 - 23:33

Awesome Dan, you are a legend - but you know this anyway.

So, to break it all down.

Each wi-fi channel uses only 11Mhz of the 22Mhz that it can use (see earlier posting on this above).

That is 11Mhz = 11 million waves per second.

For 1Mbps the calculations are;

     11 chipping bits = 1 'real' bit.

     11 million waves per second / 11 chipping bit = 1 million - THIS IS THE SYMBOL RATE

     For each symbol, one 'real' bit is represented = hence 1Mbps

For 2Mbps the calculations are;

     11 chipping bits = 2 'real' bits.

     11 million waves per second / 11 chipping bit = 1 million - THIS IS THE SYMBOL RATE

     For each symbol, 2 'real' bits are represented = hence 2Mbps

For 5.5Mbps the calculations are;

     8 chipping bits = 4 'real' bits.

     11 million waves per second / 8 chipping bit = 1.375 million - THIS IS THE SYMBOL RATE

     For each symbol, 4 'real' bits are represented = hence 5.5Mbps (1.375 million X 4)

For 11Mbps the calculations are;

     8 chipping bits = 8 'real' bits.

     11 million waves per second / 8 chipping bit = 1.375 million - THIS IS THE SYMBOL RATE

     For each symbol, 8 'real' bits are represented = hence 11Mbps (1.375 million X 8)

This is useful reference.

802.11b Data Rate Specifications

Data Rate

Code Length

Modulation

Symbol Rate

Bits/Symbol

1 Mbps

11 (Barker Sequence)

BPSK

1 MSps

1

2 Mbps

11 (Barker Sequence)

QPSK

1 MSps

2

5.5 Mbps

8 (CCK)

QPSK

1.375 MSps

4

11 Mbps

8 (CCK)

QPSK

1.375 MSps

8

Thanks to all those who helped me udnerstand this :-)

Dazzler

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Posted August 3, 2011 at 10:54 PM
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