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For a high bit rate in data communication we need a high bandwidth. Suppose there are hundreds of carriers in a particular region and they are all allocated microwave frequencies. Minimum spacing rules are also applied during frequency allocation. Due to this the resultant bandwidth, that each carrier gets, reduces. So because of this the resultant bit rate would also decrease if we consider the shannon-hartley theorem. Then how do carriers claim to have high-bit rates? Is there a different way through which frequencies are allocated?

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  • \$\begingroup\$ What counts as a "high bit rate" and are there any particular examples you have seen? \$\endgroup\$
    – Andy aka
    Jul 7, 2014 at 17:50
  • \$\begingroup\$ If you are talking about wireless phone service, typically there aren't "hundreds" of carriers but only 3 or 4. Some of these may sell service wholesale to other companies who then market it retail, so while there are many phone companies to choose from, they're all selling service on just 3 or 4 networks. \$\endgroup\$
    – The Photon
    Jul 7, 2014 at 17:52
  • \$\begingroup\$ If you're actually talking about microwave communications, this is typically sent point-to-point in a narrow beam, so that several links can use the same frequencies as long as they aren't geometrically overlapping. \$\endgroup\$
    – The Photon
    Jul 7, 2014 at 17:54
  • \$\begingroup\$ I'm talking about high bit rates with respect to bandwidths over here. More the bandwidth, more is the bit rate. But as stated above if it reduces, the bit rate decreases. The carriers in my country still claim to provide bit rates that are practically impossible for their bandwidth. How? \$\endgroup\$ Jul 7, 2014 at 17:57
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    \$\begingroup\$ Again, can you cite a specific example? \$\endgroup\$
    – Dave Tweed
    Jul 7, 2014 at 18:00

1 Answer 1

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According to Wikipedia, 4G networks (IMT-Advanced)

Have peak link spectral efficiency of 15 bit/s/Hz in the downlink, and 6.75 bit/s/Hz in the uplink (meaning that 1 Gbit/s in the downlink should be possible over less than 67 MHz bandwidth).

The possibility of transferring more than 1 bps/Hz is a direct consequence of the Shannon-Hartley theorem

\$C = B \log_2\left(1+\mathrm{SNR}\right)\$

where C is capacity and B is bandwidth.

Achieving 15 bps/Hz thus requires an SNR of at least 215-1 or about 33,000.

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  • \$\begingroup\$ In addition to that, there is MIMO and Beamforming. \$\endgroup\$ Jul 7, 2014 at 19:16
  • \$\begingroup\$ @SimonRichter, good points to add. Feel free edit my answer or write your own to point out how those technologies improve bandwidth efficiency. \$\endgroup\$
    – The Photon
    Jul 7, 2014 at 20:24

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