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I just read 4G LTE/LTE-Advanced for Mobile Broadband, and found the content as below,

If the DFT size M equals the IDFT size N, the cascaded DFT/IDFT processing would obviously completely cancel each other out. However, if M is smaller than N and the remaining inputs to the IDFT are set to zero, the output of the IDFT will be a signal with “single-carrier” properties – that is, a signal with low power variations, and with a bandwidth that depends on M.

My question is that why DFT-SPREAD OFDM has lower variations in comparison with OFDM? The picture is DFT-SPREAD OFDM modulation scheme.

DFT-SPREAD OFDM

Thank you.

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  • $\begingroup$ This scheme is more commonly called single-carrier frequency division multiple access (SC-FDMA) and there are some interesting answers here already. Basically, SC-FDMA is a way of shifting a single-carrier signal in frequency by using a pair of DFT/IDFT with different lengths. $\endgroup$ – Deve Jul 23 '17 at 9:17
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It is because the time domain symbols have lower PAPR than the frequency domain symbols.

The PAPR of OFDM is due to the 2 reasons. One is due to the IDFT(IFFT) and the other is due to the symbol type. The IDFT converts the time domain symbols to frequency domain symbols that have very high PAPR. The PAPR of 16QAM is also higher than that of QPSK. However, even when these symbol types are the same, the PAPR of DFT-SPREAD OFDM is still lower than that of regular OFDM. The reason is the (N point) IDFT effect is mostly cancelled out by the (M point) DFT. If the M=N (no zero), they are completely canceled out and the symbols would be just a time domain symbols. Even though N is much larger than M, the resulting symbol has more like time domain symbols, then the PAPR is lower. LTE up-link uses this type of OFDM that is called SC-FDMA. SC means Single Carrier because the symbols look like single carrier time domain symbols even though it has an OFDM structure. They use SC-FDMA for the up-link LTE for the lower PAPR seeking the longer battery life.

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  • $\begingroup$ thank you for your detailed explanation and now I understand the technique. In comparison with $M = N$, it to some extent decreased the PAPR with $M < N$. Many thanks for you~ $\endgroup$ – Charles Hou Jul 23 '17 at 4:59

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