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The "textbook" OFDM block diagram is as the below figure. But I wonder if it satisfies the sampling theorem. Here, we have N points $X[k]$ before IFFT, and

$$x[n]=\frac{1}{\sqrt{N}}\sum_{k=0}^{N-1} X[k] \mathrm{e}^{\displaystyle -j\frac{2\pi k n}{N}}$$

This derives from the analog signal

$$x(t)=\sum_{k=0}^{N-1}\mathrm{e}^{\displaystyle -j 2\pi f_k t}=\sum_{k=0}^{N-1}\mathrm{e}^{\displaystyle -j 2\pi k\Delta f t} $$, where $\Delta f$ is the subcarrier space.

We then sample this signal with a sampling frequency $f_s>2f_{max}=2(N-1)\Delta f$, which can be taken as $f_s=2N\Delta f$. The time domain sample points are at

$$nt_s=n/f_s=\frac{n}{2N\Delta f}$$

Then we define

$$x(nt_s):=x[n]=\sum_{k=0}^{N-1}\mathrm{e}^{\displaystyle -j \frac{2\pi k n}{2N}} $$

and then we define a expanded

$$\tilde{x}[n]:=\begin{cases} x[n], 0\le n\le N-1\\ 0, 2N-1\ge n\ge N\\\end{cases}$$ $$

That means we should do a length $2N$ IFFT rather than a length $N$ IFFT

$$ \tilde{x}[n]=\sum_{k=0}^{2N-1}\mathrm{e}^{\displaystyle -j \frac{2\pi k n}{2N}}$$

Is the textbook wrong?

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This may sound trivial but the sampling theorem can only be violated in the process of sampling an analog signal. In your question you would like to sample the signal $$ x(t) = \sum_{k=0}^{N-1}\mathrm{e}^{\displaystyle -j 2\pi k\Delta f t} $$ and you state correclty that, in order to not violate the sampling theorem, $f_\mathrm{s}>2(N-1)\Delta f$ must hold. In contrast, OFDM systems usually work with a sampling frequency $f_\mathrm{s}=N\Delta f$ and you may now ask where this discrepancy originates.

If we look at $x(t)$ we notice that it has a one-sided spectrum with discrete frequencies at $0, \Delta f, \ldots, (N-1)\Delta f$. That's an odd choice because the signal bandwidth is $2(N-1)\Delta f$ but only $N$ subcarriers are used. With other words: this system has potentially $2N$ subcarriers but the subcarriers $-N, -N+1, \ldots, -1$ are unused and the corresponding spectrum is wasted. We should rather define $x(t)$ like this: $$ x(t) = \sum_{k=-N/2}^{N/2-1}\mathrm{e}^{\displaystyle -j 2\pi k\Delta f t} $$ The absolute maximum frequency is now $\frac{N}{2}\Delta f$ and accordingly $f_\mathrm{s}>N\Delta f$.

For creating a discrete-time OFDM signal, however, $x(t)$ is quite irrelevant. You cannot violate the sampling theorem when generating a discrete-time signal, because it can be considered as already sampled. Furthermore, due to the cyclic property of the DFT, the limits of the sum in the definition of $x[n]$ are correct. Whether they are $0 \ldots N-1$, $-N/2\ldots N/2-1$ or $N+3\ldots 2N+2$ doesn't matter.

So I wouldn't say that the textbook is wrong. But if OFDM is motivated by sampling an analog signal, then $x(t)$ as defined in your answer is a poor choice.

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  • $\begingroup$ $x(t)$ is time-limited, so at the frequencies $k \Delta f$, instead of impulse, you'll be getting a sinc pulse, and the absolute maximum frequency won't be $\dfrac{N}{2} \Delta f$. I think it will be $\dfrac{N}{2} \Delta f + \dfrac{\Delta f}{2}$. If you use this to find Sampling Frequency, you won't end up with neat IDFT relation. Can you explain what needs to be done to get the IDFT relation? $\endgroup$ – Narendra Deconda Dec 4 '18 at 6:26
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The diagram from the textbook in the opening post appears to have some mistakes. For this IFFT method, the box that says 'QAM modulator' should just be a complex number generator (such as a digital QAM generator that produces a complex number from 'x' binary bits at a time - that is meant to represent a vector on a QAM grid).

And there should be two D/A blocks on the ouput --- one for handling the real part of a complex number, and the other for handling the imaginary part of a complex number. One D/A output goes to 1 carrier (eg. the in-phase carrier), while the other D/A output goes to the other carrier (the 90 degrees out of phase carrier).

Also, it seems that this IFFT method of this so-called 'OFDM' technique is pretty much based on using two carriers to transmit complex numbers. The main work is to properly be able to recover the complex numbers (or rather ..... not just the complex numbers, but the complex-number-sequences), in order to carry out an FFT ..... which provides the original complex number sequence that we started with (ie. the original data-containing QAM vectors).

This IFFT method (having digital aspects) is very much unlike (ie. different from) the analog OFDM method.

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No, you just made some random stuff up - you're not doing sampling, you're modulating things onto a carrier, so the sampling theorem is irrelevant.

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  • $\begingroup$ But if we don't do as sampling theorem, how can we derive the discrete time signal $x[n]$ as above? $\endgroup$ – Robert Fan Oct 20 '14 at 9:11

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