# Determining sampling rate in OFDM demodulation

I have seen that in OFDM, the sampling rate is much lesser than twice the bandwidth of signal. why is it so? How is sampling rate calculated in OFDM? Does it depend on the modulation type used inside like QAM or BPSK. Does it relate to FFT length by any chance.

• Where have you seen this? It doesn't sound accurate to me. – Jason R Apr 19 '13 at 12:43
• 130.203.133.150/viewdoc/summary?doi=10.1.1.59.1295 – Karan Talasila Apr 19 '13 at 13:14
• The above paper has a table of fft size , bandwidth,sampling rate,cyclic prefix length used etc in wimax. The sampling frequency given there is just above the bandwidth. – Karan Talasila Apr 19 '13 at 13:15

It is incorrect to say that the sample rate is "much less" than the bandwidth of the signal. It is true to say that it is a "little less". Even that statement is only true for part of the demodulator.

Demodulators almost always use a sample rate that is higher than the bandwidth for two reasons: to ensure that all of the signal is captured rather than aliased due to Nyquist frequency limitations, and to do some version of early/late gating when syncing with the symbols. It is very common to sample at twice the symbol rate because that is very convenient for capturing the symbols and doing early/late gating.

OFDM is different because of the FFTs/IFFTs built into the signal. There is no advantage for the demodulator to do the FFT at higher than the symbol rate, so they don't. The roll-off in OFDM is very minimal (an example of typical OFDM roll-off is shown below), though, so the sample rate is not much less than the bandwidth. In some schemes, like 802.11a, the outer frequency bins are not used which reduces the bandwidth of the signal relative to the sample rate, meaning that the sample rate will be higher than the bandwidth even when taking into consideration the signal roll-off.

Also, before the demodulator does the FFT it has to eliminate as much carrier offset as it can. This is typically done at a sample rate that is higher than the bandwidth for the same Nyquist reasons as a typical demodulator.

EDIT: I see that I forgot to answer some of your questions. The sample rate for OFDM is exactly equal to the sample rate that the OFDM transmitter used when it inverse FFT'ed the data. No, the sample rate has nothing to do with the modulation types used inside the OFDM symbols (e.g. BPSK or QPSK). No, the sample rate has nothing to do with the FFT length.

• The paper i have attached says bandwidth is 1.25MHz and sampling frequency is 1.48MHZ.That's not slightly lesser than twice the bandwidth. Is the data wrong? secondly, you say that it should sample at twice the symbol rate. Isn't it twice the bandwidth of signal.Is there a equivalence relation between bandwidth and symbol rate here. – Karan Talasila Apr 19 '13 at 13:21
• @talasila: Is the sample rate that you're referring to complex or real? If it is complex and you're analyzing a complex baseband signal, then you haven't violated the Nyquist criterion. – Jason R Apr 19 '13 at 13:39
• @talasila Jason is correct, twice the symbol rate far exceeds what is needed in terms of the Nyquist criterion if the samples are complex. – Jim Clay Apr 19 '13 at 14:09
• @talasila Your own example says that the sample frequency is higher than the bandwidth. I said that traditional demodulators (e.g. PSK) often do twice the symbol rate. I said that OFDM demodulators do NOT do twice the symbol rate because they don't need to. Yes, there is an equivalence relationship between bandwidth and symbol rate. – Jim Clay Apr 19 '13 at 14:11
• @JimClay: In one way of looking at it, each of the subcarriers is oversampled by a large factor (the FFT size $N$). OFDM is really just the transmission of many low-rate modulated signals at carefully-chosen carrier frequencies, such that the modulation can be recovered easily using a DFT. So, the sample rate is usually many times larger than the "symbol rate" for OFDM. – Jason R Apr 19 '13 at 14:18