Shannon formula C= B log (1+S/N) Is this formula applied for baseband transmission only? or for both baseband and passband modulation? Can we apply it without any changes to any modulated signal? What about QAM, CDMA and OFDM, can this formula be applied with these techniques without any changes?
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\$\begingroup\$ Huh. I never knew there was a dignal processing SE: dsp.stackexchange.com/questions/82111/… \$\endgroup\$– DKNguyenCommented Mar 22, 2022 at 14:21
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\$\begingroup\$ The Shannon formula doesn't tell you what but rate is achieved by your modulation scheme. It tells you what's the maximum rate you can achieve with any scheme. \$\endgroup\$– The PhotonCommented Mar 22, 2022 at 14:36
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The formula is an upper bound, indicating the best performance that any modulation scheme could hope to achieve in a given bandwidth at a given SNR. QAM on its own will likely not achieve this theoretical limit, but a suitable modulation scheme with a suitable FEC code can approach the theoretical limit.
In an example of this, Turbo codes are described as the first practical codes that approached the Shannon limit (note that this is a property of the FEC, without referencing any particular modulation scheme).
The actual formula does not care whether communication is done at baseband or at passband. As a thought experiment, insert an ideal, noise-free I-Q mixer driven by an ideal local oscillator into the signal path to shift the signal between the baseband and a passband. The SNR is unchanged, the bandwidth is unchanged, and the information content is also unchanged.
answered Mar 22, 2022 at 14:48