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LTE: Number of sub carrier vs FFT size

–1 vote
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I have asked this at LTEUNIVERSITY but no luck -

I'm confused with FFT size and number of occupied sub carrier in LTE,
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I can understand relation between sampling rate and FFT. But In my understanding, the number of frequency that we can get from FFT/DFT operation is only half of FFT size, for example NFFT 1024 can only analyze 512 frequency, SO, why in these LTE standard, for example 10 MHz bandwidth LTE which is have 600 sub carrier can use 1024 FFT since 1024 FFT only able to analyze 512 frequency.

Appreciate your help .

posted Jul 28, 2014 by anonymous

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Although the subcarriers are 600, there are only 300 unique frequencies. Subcarriers are on both side of null subcarrier. For ex 15 kHz and -15 kHz, the real part remains same for both of them while the imaginary part is at 180*.

2 Answers

0 votes

Hi

In LTE for occupied sub carrier calculation we do like below:

e.g: 10MHz BW

Guard Bands=2*492.5=985KHz.

BW(with DC Carrier) without Guard Band=10000-985=9015KHz.

The size of sub carrier in OFDM is 15KHz Then:

9015KHz/15KHz= 601 SubCarriers

Without DC=600 SubCarriers.

600/12=50 RB

answer Jul 28, 2014 by Ali Mohseni
0 votes

Not sure if I understood your question but here is my approach:

1 symbol in time domain is 66.7 micro seconds and 15KHz in time domain.
Ts = 32.552ns = is the sampling period (should be high but less then 2 x maximum baseband frequency accordingly to the sampling theorem) --> Fs=1/Ts=1/32.55ns=30.721MHz. This baseband frequency is higher than symbol(subcarrier) bandwidth = 15KHz.

"n" =symbol sampling rate=2048 --> 1 symbol [66.7us]is sampled with n=2048 samples (2048 s x 32.55ns=66.7us). 1 symbol can be sampled w/ up to 2048 samples but the chosen value is set depending of the EARFCN size.

FFT size for 10MHz EARFCN = 1024 (1 symbol sampled w/ 1024 samples)
"m" = # of Subcarriers per EARFCN = 600 for 10MHz EARFCN = (10000KHz-2x492.5)/15-1sc)

answer Jul 28, 2014 by anonymous
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