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This tutorial points out that the peak power of an OFDM system is N times the average power - where N is the number of subcarriers. Thus, the PAPR of an OFDM system is N.

$PAPR=\frac{|{s}_{peak}{|}^{2}}{{S}_{S}}$ |

$PAPR=\frac{|{s}_{peak}{|}^{2}}{{S}_{S}}=\frac{N{S}_{S}}{{S}_{S}}=N$ |

Let's compare single carrier and OFDM systems. Initially both shall use QPSK modulation and transmit with a power of
$2\text{\hspace{0.17em}}{V}^{2}$. The difference is that the

OFDM transmission signal is the output of an IFFT that is power invariant normalized. The point is that the IFFT transforms some sets of QPSK symbols, i.e.

OFDM frequency domain symbols, to OFDM time symbols where all signals disappear but one, see examples in table. To keep the average power the peak

power of the remaining signal is N times the average power.

Single Carrier QPSK symbols | Signals of OFDM symbol |
---|---|

This applies to OFDM systems in general:

- OFDM adds a $PAPR=10\text{}log\text{}\left(N\right)[dB]$
on top of any preexisting PAPR.
- M-QAM: For any constellation size OFDM adds a $PAPR=10\text{}log\text{}\left(N\right)[dB]$ on top of the PAPR of M-QAM. (N peak power QAM symbols may result in an OFDM signal with N times the peak power.)
- Quadrature modulation contributes 3 dB PAPR.

- The guard interval doesn't change the average and peak power and PAPR.
- Unused (virtual) subcarriers don't count. N is the number of used subcarriers. See experiment
- Variation of the transmit power doesn't change the PAPR.