Transmit power of non-orthogonal FDM

Question

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?

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Update

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.

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.