Shannon formula for channel capacity states that $$C=\log\left(1+\frac{S}{N}\right)$$ If this formula applied for baseband transmission only? Is it applied with passband transmission? Are there any changes in the formula with different types of modulation, for example, OFDM??
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2$\begingroup$ As a side note: it seems like you could really benefit from getting a few good textbooks and reading them. This site is not really a good replacement for taking courses or focused self-study. Let me know if you need book recommendations. $\endgroup$– MBazCommented Dec 3, 2020 at 19:23
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1$\begingroup$ Forgot the channel bandwidth? $\endgroup$– GillesCommented Dec 3, 2020 at 20:06
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$\begingroup$ @Gilles in this version the units are bits per second per Hertz. Sometimes it is called spectral efficiency, multiplication by the bandwidth gives you bits per second $\endgroup$– EngineerCommented Dec 3, 2020 at 20:19
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1$\begingroup$ @Engineer I'm more used to this $C = B\log_2 \left(1 + \frac SN\right)$, or when it is the spectral efficiency then $\eta$ is used. $\endgroup$– GillesCommented Dec 3, 2020 at 20:30
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1$\begingroup$ i think it's missing the $B$. and it should be base 2 logarithm. an even more general form is when $S$ and $N$ are not constant with frequency: $$ C = \int\limits_0^B \log_2\left(1 + \frac{S(f)}{N(f)} \right) \ \mathrm{d}f $$ $\endgroup$– robert bristow-johnsonCommented Dec 5, 2020 at 4:52
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1 Answer
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This formula is for baseband with real symbols in an AWGN channel.
However, it is trivially easy to adjust it for quadrature modulation or complex symbols: just double the capacity.
OFDM and other wideband modulations operate over frequency-selective channels, which have different capacity. There are capacity formulas for many channels in the literature.
