2
\$\begingroup\$

Shannon's channel capacity formula does not consider modulation techniques. How then does LTE achieve high data rate using advanced modulation techniques?

P.S: I am novice in RF and signal processing.

\$\endgroup\$
2
\$\begingroup\$

The Shannon capacity limit holds irrespective of modulation scheme used. It is the theoretical limit given an ideal choice of modulation and channel coding. The Shannon limit is as fundamental a rule in communications engineering as the first law of thermodynamics is in mechanical engineering.

For example, let's assume you have a 20 MHz wide AWGN channel with a 20 dB signal-to-noise ratio. The Shannon-Hartley theorem gives:

$$C = B\,\textrm{log}_2(1 + SNR) = 20E6 \times \textrm{log}_2(1 + 10^\frac{20}{10}) = 133.2 MBit/sec.$$

This is the upper limit. Provided you pick the optimum modulation scheme and forward error correction code, you can get 133.2 MBit/sec out of the channel, but no more. To get a higher error-free data rate, you need to either improve the SNR or increase your bandwidth.

Until the early 1990s, it was assumed that getting arbitrarily close to the Shannon limit was not feasible. That changed with the introduction of Turbo Codes, which underpin most 3G & LTE telecom networks. With a turbo code, you can get arbitrarily close to the Shannon limit provided that you can accept the processing required.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.