I am reading articles about error correcting codes, facing to a term "spectral efficiency". Due to the context, I guessed that it means "bit rate", but I wonder are these two terms equivalent or not? I will be thankful if you guide me
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I had done some research some time ago on space communications, here is my memory test: spectral efficiency is the ratio of the datarate achieved by the signal bandwidth occupied, or how the initial bandwidth is altered - before it is actually transmitted. This is a property of the modulator, which modulates a high frequency carrier wave in function of the data to be transmitted in order to make it easier to propagate through the medium (air generally). As said below, it can also be used to increase the datarate in physical communication lines where the bandwidth is limited.
There are various modulation techniques, and some of them are more efficient than others; as I said in that thread, a simple multilevel amplitude modulation enables you to transmit several bits per clock period. For 4 levels for example, 2 bits are transmitted per clock period, so compared to the initial 1 bit per clock cycle that's a 2 bits/(s.Hz) spectral efficiency. Combining phase and amplitude can lead you to much higher values, but there are many techniques and here you'll find a comparison table (along with the definition of spectral efficiency...). 4G LTE achieves 30 bits/(s.Hz) for example, which is amazing. 64-QAM (combined phase and amplitude information) which I believe was used at the beginning of ADSL, is 6bits/(s.Hz) for comparison.
However, spectral efficiency must not be mixed up with encoding efficiency, which happens further upstream just before the modulation process and relates to how many bits are appended to the initial data frame in order to be able to detect on the receiving end if some of the data has been corrupted. I'm not sure whether the table I've linked above includes the encoding efficiency; probably since some efficiencies are below 1bit/(s.Hz). Some encoding schemes are very heavy, sometimes based on state machines that enable you to recover corrupted data in order to decrease drastically the bit error rate (which is a constraining requirement on your datarate).
To sum up, link data rate gets, with respect to the feed data rate, decreased by encoding, then increased by modulation - and it all depends on the techniques used (more or less complex). That's only when there is no encoding, and no modulation, that datarate = bandwidth in value (still different units) and the spectral efficiency is 1bit/(s.Hz). That's why you'll usually see "symbol rate" out of the modulator (and encoder I think) instead of "data rate".
answered Oct 29, 2014 at 15:54
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\$\begingroup\$ @Mystere I accept that encoding affects the bit rate, but how can modulation affect it as well? \$\endgroup\$– CLAUDECommented Oct 29, 2014 at 16:01
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1\$\begingroup\$ Suppose your datarate is 9600 bits/s. 1 bit is added by encoding to your 8bits frame, which means the datarate at the output of the encoder is now 8533 bits/s. If you're modulating using a standard AM or FM modulation, even though the bit error rate is going to change (because they have different sensitivities to blocking walls etc.), the datarate will remain the same at the receiver end. Now try to change the modulator and combine 2 bits every clock cycle (4 possible amplitude levels for example). You've effectively doubled the datarate. \$\endgroup\$ Commented Oct 29, 2014 at 16:07
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\$\begingroup\$ In your AM example, the data is transmitted with the rate 2 bit/second (Since with each clock we are sending 2 bits), and as \$\frac{1}{second}=Hz\$, I think the unit of spectral efficiency should be \$bit. Hz\$. Where am I wrong? \$\endgroup\$– CLAUDECommented Oct 29, 2014 at 16:20
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\$\begingroup\$ Woops sorry, typo. The spectral efficiency is the ratio of the datarate by the bandwidth occupied, so it should be bits/s/Hz. Updating my post. \$\endgroup\$ Commented Oct 29, 2014 at 16:29
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1\$\begingroup\$ I'm pretty busy usually and others on SE are certainly more experienced than me so I would advise you to continue posting your questions here, however I'm regularly checking the questions so I should answer your next ones if I can. \$\endgroup\$ Commented Oct 29, 2014 at 16:44