OFDM for Terahertz Wireless Communication Systems
155
PAPR [s] =
max
0≤n≤N J−1
|x[n/J ]|
2
N −1
n=0 {|x[n/J ]|
2
}
(27)
It is proved in [20] that an oversampling factor of at least 4 (J ≥ 4) is necessary for
a good approximation of the PAPR. We will show this property later in this chapter
(Fig. 9).
When the digital symbols sound from an M-state phase modulation (M-PSK),
the value of the peak power is identical to the mean value. The maximum PRPR is
worth:
PAPR max = N
(28)
When the numerical symbols come from an MAQ to M-states constellation
(MAQ-M), the maximum PRPR is worth:
PAPR = N .
3
√
M − 1
√
M + 1
(29)
It should be noted that to reach this theoretical PAPR, a number of realizations
of the order of M
2 is required, and the probability for such an achievement to occur
is 1/M
N −2 [21]. Thus, if one places oneself in the standard IEEE 802: 11a, the
probability that the PR is worth N (in the case where the modulation used is a
QPSK) is 4.62. It is therefore not realistic to consider this value of the PAPR.
We must therefore consider a probabilistic approach by realizing that the N
samples of the discrete OFDM signal x [n] follow a Gaussian law by virtue of the
central limit theorem. By taking critical sampling (J = 1), that is to say by calculating
the PAPR on the N time samples, we obtain the following distribution function:
Pr[PAPR{x} > γ ] J=1 ∼ 1 −
1 − e
−γ
N
(30)
The PAPR should then be considered as a random variable.
The average PAPR value which has been shown in [22] is given by,
E[P AP R] = 0.57721 + ln[N ]
(31)
In Fig. 8, we plot the average PAPR value as a function of the number of carriers.
This curve shows that the average PAPR value of the OFDM increases with the
number of subcarriers. For example for N = 200, the average value of PAPR is 6 dB
while it is 6.7 dB for N = 1200 subcarriers.
In [23], the authors give an upper limit of the distribution function for an
oversampling factor greater than 2,
Pr[PAPR{x} > γ ] J>2 < J N e
γ
1−
π 2
2 J 2
(32)
155
PAPR [s] =
max
0≤n≤N J−1
|x[n/J ]|
2
N −1
n=0 {|x[n/J ]|
2
}
(27)
It is proved in [20] that an oversampling factor of at least 4 (J ≥ 4) is necessary for
a good approximation of the PAPR. We will show this property later in this chapter
(Fig. 9).
When the digital symbols sound from an M-state phase modulation (M-PSK),
the value of the peak power is identical to the mean value. The maximum PRPR is
worth:
PAPR max = N
(28)
When the numerical symbols come from an MAQ to M-states constellation
(MAQ-M), the maximum PRPR is worth:
PAPR = N .
3
√
M − 1
√
M + 1
(29)
It should be noted that to reach this theoretical PAPR, a number of realizations
of the order of M
2 is required, and the probability for such an achievement to occur
is 1/M
N −2 [21]. Thus, if one places oneself in the standard IEEE 802: 11a, the
probability that the PR is worth N (in the case where the modulation used is a
QPSK) is 4.62. It is therefore not realistic to consider this value of the PAPR.
We must therefore consider a probabilistic approach by realizing that the N
samples of the discrete OFDM signal x [n] follow a Gaussian law by virtue of the
central limit theorem. By taking critical sampling (J = 1), that is to say by calculating
the PAPR on the N time samples, we obtain the following distribution function:
Pr[PAPR{x} > γ ] J=1 ∼ 1 −
1 − e
−γ
N
(30)
The PAPR should then be considered as a random variable.
The average PAPR value which has been shown in [22] is given by,
E[P AP R] = 0.57721 + ln[N ]
(31)
In Fig. 8, we plot the average PAPR value as a function of the number of carriers.
This curve shows that the average PAPR value of the OFDM increases with the
number of subcarriers. For example for N = 200, the average value of PAPR is 6 dB
while it is 6.7 dB for N = 1200 subcarriers.
In [23], the authors give an upper limit of the distribution function for an
oversampling factor greater than 2,
Pr[PAPR{x} > γ ] J>2 < J N e
γ
1−
π 2
2 J 2
(32)