In OFDM, how the increasing of sampling frequency didn't cause increase on the required channel bandwidth (fs>>BW), as i know sampling frequency means number of samples per second, if this number of samples increased by logic the the needed bandwidth increase.
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1 Answer
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You can say the null subcarriers have their own bandwidth if you define an alphabet including "zeros" and use null carriers to transmit these "zeros". As the comment of MBaz,
If we start with the discrete-time OFDM symbol s[n], and increase the sampling rate (defined by the time interval between successive samples of s[n]), then the OFDM symbol becomes "shorter" and therefore its bandwidth will increase.
But it is not the case. The OFDM symbol $T_u = N_{dft} T_s = 1/\Delta f$.
In this bandwidth interpretation, I think the term "virtual subcarrier" is more appropriate.
What you send over by electromagetic wave is data in $N$ active subcarriers $< N_{dft}$, or the frequency positions of null/virtual subcarriers can be used for other systems; because when you revert back the time domain signal to the frequency domain, what you need is the frequency positions of these active subcarriers.
It is easier to think OFDM as FDM. In this case you have $N$ narrowbands of bandwidth $B_{nb}$ and a sample rate $F_s > N B_{nb}$ is required for practical implementations, such as easily-realizable anti-aliasing filters. OFDM is simply one implementation of this system by using $\Delta f = B_{nb}$ and $F_s = N_{dft} \Delta f = N_{dft} B_{nb} > N B_{nb}$.
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$\begingroup$ Thanks alot @Alex TP, your words " the frequency positions of null/virtual subcarriers can be used for other systems" means that these subcarriers have its own frequency position. For example if i have 600 active subcarriers, 424 null subcarriers and these 424 null subcarriers are centerened at the middle, that mean the first active data subcarrier will be centered at (424/2*Δf). If that what happens, that means the bandwidth increased to be equal sampling frequency. correct for me please. $\endgroup$ Commented May 20, 2017 at 18:19
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$\begingroup$ @user24907 the subcarrier positions should be like this image i.sstatic.net/CxTXu.png $\endgroup$– AlexTPCommented May 20, 2017 at 21:59
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$\begingroup$ null subcarriers centered at middle is a delusion of DFT implementation. If you use Matlab/Octave, you should use fftshift function. $\endgroup$– AlexTPCommented May 20, 2017 at 22:00
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$\begingroup$ TP Could you please give more explaination or any link to understand it? $\endgroup$ Commented May 21, 2017 at 17:23
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$\begingroup$ @user24907 for fftshift ? fr.mathworks.com/help/matlab/ref/fftshift.html $\endgroup$– AlexTPCommented May 21, 2017 at 18:15
s[n], and increase the sampling rate (defined by the time interval between successive samples ofs[n], then the OFDM symbol becomes "shorter" and therefore its bandwidth will increase. $\endgroup$