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I suppose that OFDM can't modulate analog signal. The basic mathematical description for OFDM is $$ v(t)=\sum_{k=0}^{N-1}X_{k}\exp(j2\pi k t /T),\quad 0<t<T$$ In this equation, $X_k$ is a discrete value. So OFDM can not modulate an analog signal.

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You are correct. However, the coefficients $X_k$ could be samples taken from an analog signal. These could be used in the receiver to recover the analog signal using interpolation.

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Well, the equation you show is an inverse DFT with a positive exponent. Analog DFT or FFT processors have long been known to exist. A search for "analog FFT processor" plus OFDM plus transceiver yields about 119 hits.

Although many of the targeted applications are for receivers, an analog FFT processor can also be used in transmission, as, for example, Fig. 1 in the following:

http://www.ece.ubc.ca/~nimas/edu/TCAS-I_small.pdf

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  • $\begingroup$ If $t$ in the question is the continous time, the equation dose not represent an inverse DFT. $\endgroup$ – Deve Mar 18 '15 at 12:04
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Let's assume for a moment that the $X_k$ are time-continous, i.e. analog, signals $X_k(t)$. Let $x_k(t)$ be the summand signals of $v(t)$: $$ x_k(t) = X_k(t)\exp(j2\pi k t/T) \tag{1} $$ To form an orthogonal frequency division multiplex (OFDM) signal, the functions $x_k(t)$ must be pairwise orthogonal, i.e. $$ \int x_n(t)x^*_m(t) \mathrm d t = C \delta(n-m) \tag{2} $$ must hold, where $\delta(0) = 1$ and zero otherwise, and $C$ is an arbitrary constant. Plugging (1) into the l.h.s. of (2) we get $$ \int X_n(t)X^*_m(t)\exp(j2\pi(n-m)t/T) \mathrm dt = C \delta(n-m) \tag{3} $$ It's clear, that generally the condition in (3) is not fulfilled. For example for $n=m$, (3) becomes $$ \int X_n(t)X^*_m(t) \mathrm dt = C\tag{4} $$ which is generally not true. One possible set of functions $X_k(t)$ that fulfills (4) is: $X_k(t) = \text{const. }$ in the interval $[0,T]$ and zero otherwise - which is a time-discret signal. Assuming such a set of signals will also fulfill (3).

In conclusion, arbitrary analog signals cannot be used for OFDM modulation because they violate the orthogonality condition. In contrast, discrete-time signals do fulfill this condition.

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  • $\begingroup$ thanks for your answer! But what you are talking about is the reason why the signal can't be time-variant in a period. The question is why the $X_k$ should be a discrete value, instead of a continuous one. $\endgroup$ – user1297181 Mar 19 '15 at 11:13
  • $\begingroup$ @user1297181 The problem is, that the equation in your question mixes discrete ($X_k$) and continuous time ($t$) signals. To build a bridge between the digital and analog world I have assumed in my answer that $X_k$ is actually the ideally analog-converted version of a discrete signal. With that assumption, I have shown that the orthogonality condition is met and that, in contrast, it is violated for arbitrary continous-time signals. $\endgroup$ – Deve Mar 24 '15 at 12:25

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