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Sorry for this question, I'm still new in signal processing area.

In OFDM system, let $x$ be the received signal, what do we call the process of multiplying the received signal $x$ by $e^{-j2\pi F_c t}$ , where $F_c$ is frequency carrier and $t$ is interval 0 : sampling rate: sampling rate * length of signal.

So the question, what is that function? and why do we use it.

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You can check the basic of OFDM and its MATLAB code from here http://read.pudn.com/downloads159/ebook/716250/OFDM_tutorial.pdf and check comments here and animations to imagine how ofdm works https://www.quora.com/Why-does-OFDM-involve-the-FFT-I-find-this-confusing-as-the-FFT-seems-like-a-coarse-tool-which-would-severely-limit-the-data-rate-because-of-the-large-number-of-samples-required-to-do-an-FFT# Good luck

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  • $\begingroup$ Ok, I will check those links and try to understand it well $\endgroup$ – Fatima_Ali Jul 30 '18 at 8:43
  • $\begingroup$ Generally, link-only answers are a bit frowned upon – explain what you're linking to, give a bit of insight into what is written there. $\endgroup$ – Marcus Müller Jul 30 '18 at 13:33
  • $\begingroup$ Hello Marcus, Yes I totally agree with you. But, I just gave links which explain basic things in OFDM ..that will simplify the idea for many students, as soon as they have the basic meaning of OFDM, then they can go further and read different books $\endgroup$ – Zeyad_Zeyad Jul 31 '18 at 7:20
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That is just a simulation of a direct conversion downmixer.

You're probably looking at a simulation of OFDM in e.g. Matlab. In a real OFDM system, the signal at carrier frequency is analog, and has to be mixed down to complex baseband with an analog mixer. Of course, in an analog mixer, there's no discrete time $t$, but just an oscillator of the right frequency with which the I and Q output branches of the mixer are generated.

Since complex baseband is so fundamentally important to how OFDM works, I strongly recommend refreshing your knowledge of it and of direct conversion mixers, if this sounds new to you.

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  • $\begingroup$ Thanks for your answer. OK, I will try to understand it more. $\endgroup$ – Fatima_Ali Jul 30 '18 at 8:42

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