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Let's say that I have two different wireless technologies, A and B.

A has a bandwidth of BA MHz, B has a bandwidth of BB MHz. I can compute the capacity of (for instance A) with Shannon:

CA = BA * log(1 + SNR)

but what if A is on frequency band FA, and B on FB? If, for instance, B is on a higher bandwidth, the total amount of information that B can transfer would be higher compared to A. How could be possible to compute this? I'd like to compute this regardless of the technology, whether FEC is implied or not etc..

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The center frequency of the band does not matter. A 1 MHz band around 5 MHz carries as much information as a 1 MHz band around 1 GHz.

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  • \$\begingroup\$ Ok, but on the same unit time, at 10 GHz i can send much more information than, for instance, 1 GHz. Am I right? \$\endgroup\$ – lbedogni May 12 '14 at 14:51
  • \$\begingroup\$ Only if you increase the bandwidth B in Shannon's formula. \$\endgroup\$ – The Photon May 12 '14 at 14:53
  • \$\begingroup\$ Yes, but let's take an example: Wifi on 5 GHz have a much higher throughput compared to standard 2.4 GHz. This is because on the same unit time, you can transmit more information due to the higher frequency. \$\endgroup\$ – lbedogni May 12 '14 at 14:57
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    \$\begingroup\$ You're missing the point. Information capacity depends only on bandwidth, not center frequency (note the absence of center frequency in Shannon's formula). The only advantage of using higher center frequencies is that there is more available bandwidth. For example, if you have an information bandwidth of 100 MHz, then you could barely transmit it with a center frequency of 100 MHz. With a center frequency of 200 MHz, it would be much easier. \$\endgroup\$ – Barry May 12 '14 at 15:05
  • \$\begingroup\$ @lbedogni, I don't know the wifi standards, but it sounds like the 5 GHz standard uses either a wider bandwidth, or has better SNR, or uses an ecoding scheme that allows it to achieve closer to the Shannon limit. In your original question, you specified that the B term is a given of the problem. \$\endgroup\$ – The Photon May 12 '14 at 15:52

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