It is known that the total power of an OFDM signal equals to the sum of the powers on each individual subcarriers. I am a bit puzzled by the total transmit power of a generalized FDM system if subcarriers are not orthogonal and if it should remain the same or changed due to the non-orthogonality.

Let us consider the following OFDM transmit signal: $$s(t) = \sum_{k = 0}^{N-1} a_k e^{j2\pi f_k t} = \sum_{k = 0}^{N-1} a_k e^{j2\pi \frac{k \: \phi}{T} t}, \quad 0\leq t\leq T,$$

where $k$ is the subcarrier index, $a_k$ are i.i.d. data symbols with zero mean, $N$ is the total number of subcarriers, $T$ is the FDM symbol duration, and $\phi < 1$ is introduced to break the orthogonality.

The transmit power is represented as

$$ P = \frac{1}{T} \int_{0}^{T} s(t) \: s^*(t) dt,$$

which can be further simplified as

$$P = \frac{1}{T} \sum_{k = 0}^{N-1} \sum_{k' = 0}^{N-1} a_k a^*_{k'} \int_{0}^{T} e^{j2\pi \phi (k - k')t/T} dt$$

Now, the value of the integral $\int_{0}^{T} e^{j2\pi \phi (k - k')t/T} dt$ does not equal to $0$ if $k \neq k'$ (due to the non-orthogonality introduced by $\phi$). This means that the transmit power in case of non-orthogonal subcarriers is no longer equals the sum of the power on each subcarrier? However, it is higher than the sum!!!

I feel there is something wrong in this simple derivation, as by intuition, if a power source spends a certain amount of power on each subcarrier, then how come the total power will be greater than the sum of these individual powers?



The following update to make the modeling of non-orthogonality more clear. Let us consider the FDM signal written as

$$s(t) = \sum_{k = 0}^{N-1} a_k e^{j2\pi (f_k + \delta_k) t} = \sum_{k = 0}^{N-1} a_k e^{j2\pi (\frac{k \:}{T} + \delta_k) t}, \quad 0\leq t\leq T,$$

where $\delta_k$ is the frequency offset on subcarrier $k$. Following similar approach, the power $P$ can be found as

$$P = \frac{1}{T} \sum_{k = 0}^{N-1} \sum_{k' = 0}^{N-1} a_k a^*_{k'} \int_{0}^{T} e^{j2\pi (k/T + \delta_k - k'/T - \delta_{k'})t} dt$$

which does not equal to the sum of power on individual subcarriers.

  • 1
    $\begingroup$ As a matter of terminology, I think that "non-orthogonal OFDM" sounds a bit self-contradictory. I'd maybe call it "generalized FDM" or "non-orthogonal FDM." $\endgroup$
    – Jason R
    May 4, 2016 at 1:13
  • $\begingroup$ @JasonR That makes sense. I will update the question. Thanks. $\endgroup$
    – Noor
    May 4, 2016 at 1:35

1 Answer 1


In your example, the sinusoids are still orthogonal to one another; the key is that you must integrate over the correct period when calculating $P$. All of the sinusoids have frequencies that are multiples of $\frac{\phi}{T}$, so their resulting sum will have period $\frac{T}{\phi}$.

You should therefore take this fact into account when calculating the average power over a period of $s(t)$:

$$ P = \frac{\phi}{T} \sum_{k = 0}^{N-1} \sum_{k' = 0}^{N-1} a_k a^*_{k'} \int_{0}^{\frac{T}{\phi}} e^{j2\pi \phi (k - k')t/T} dt $$

With this formulation, then you do get the useful property that the integral is zero for all $k \ne k'$, which allows you to simplify the expression to be simply the sum of the powers of each of the sinusoids.

  • $\begingroup$ But the OFDM/FDM symbol duration should not change, i.e. it should remain as is $T$ to model/reflect the non-orthogonality. $\endgroup$
    – Noor
    May 4, 2016 at 2:50
  • $\begingroup$ When you calculate the average power of a periodic signal, you must use the correct period. Otherwise, what you're calculating is some other thing that isn't the average power, so it's not really interesting. The sinusoids in your example are not periodic in $T$, so the equation you have for $P$ does not reflect the average power of $s(t)$. $\endgroup$
    – Jason R
    May 4, 2016 at 3:39
  • $\begingroup$ You mean that symbol duration changes such that subcarriers are still orthogonal? In other words, this means that if frequency offset happens, the FDM symbol duration changes? I updated the question with another model of offset that is different for each subcarrier. Still you conclusion holds? $\endgroup$
    – Noor
    May 4, 2016 at 4:20
  • $\begingroup$ Yes, my conclusion holds. In your updated example, the sinusoids are at frequencies $\big\{\frac{1}{T} + \delta_k\big\}$. If $\delta_k \ne \frac{n}{T}$ for some integer $n$ (i.e. the sinusoids are not orthogonal to one another), then the sinusoids are not periodic in $T$. Therefore, you cannot calculate the signal's average power using the equation for $P$ that you gave. This has nothing to do with FDM at all, and is simply a byproduct of how you calculate the average power of a signal. $\endgroup$
    – Jason R
    May 4, 2016 at 9:25
  • 1
    $\begingroup$ There is no consistent "power" definition in this case. The amount of energy delivered by the transmitter will vary from (FDM) symbol to symbol, as the extent to which the subcarriers cancel/reinforce one another will vary depending upon the combinations of modulation factors applied for each symbol. The average power is actually defined as $$P = \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} s(t) \: s^*(t) dt. $$ It just so happens that if $s(t)$ is periodic, it can be reduced to the form you gave above. However, in the general FDM case you're talking about, $s(t)$ is not periodic. $\endgroup$
    – Jason R
    May 4, 2016 at 15:49

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