sampling frequency in LTE&OFDM downlink.

I want to determine the system clock frequency under which my Fpga program is working.

I assume the bandwidth = subcarrier number * subcarrier space, and the subcarrier numuber equals the ifft number
so the bandwidth = ifft number * subcarrier space, 7.68MHZ = 512 * 15KHZ. bandwidth = 7.68MHZ.

And in the picture, the sampling frequency in time domain is also 7.68MHZ. For other different ifft number modulation in the picture, the bandwidth is also the same with the sampling frequency.

Could anyone please explain how the sampling frequency is calculated? I use this frequency as my system clock in FPGA design, but couldn`t figure out the math behind it.
thanks in advance!

Re: sampling frequency in LTE&OFDM downlink.

The sampling frequency is the minimum baseband digital symbol rate of data at the transmitter (before DAC and RF upconversion). Bandwidth here is not a precise number. Remember that all time-limited signals have infinite bandwidth. Here bandwidth refers to the width of the spectrum where the non-negligible signal energy lies. This is so it can fit within a channel and not interfere significantly with users of adjacent channels.

This effective bandwidth is determined by the guard (null) subcarriers which are on the edges (edge in terms of the frequency domain representation) of each OFDM symbol. In your table you can see that for 5 MHz channels, only 301 of 512 subcarriers are not guard subcarriers. Hence, we have 301 subcarriers before the edges, representing the fraction 301/512=0.59 of spectrum.

Note that 0.59 * 7.68 MHz = 4.53 MHz. The fraction of subcarriers not allocated to the guard subcarriers is conservative here since the transition region is not steep. This is the result of OFDM's orthogonal subcarrier components which are represented in the frequency domain as sinc functions, leading to significant spectral spread.

Re: sampling frequency in LTE&OFDM downlink.

rcdaniels, thanks for your help.

In the 5 Mhz case, you point out that the effective bandwidth is now 0.59*7.68 MHz = 4.53 MHz. The 7.68 Mhz = (512 subcarrier) * (15 KHz subcarrier space). This 7.68Mhz is the bandwidth but it coincides with the sampling rate, could you explain that? thanks.

As I understand, then the sampling rate seems to me to be two times than the effective bandwidth, 2 * 4.53 Mhz = 9.06 Mhz. So as long as my system clock works larger than 9.06 Mhz, it would be ok, right? I get confusion determining the clock frequency...

Thanks in advance.

Just to be clear, the effective bandwidth is 5 MHz, but the spectrum width before the edge transition is around 4.5 MHz = (301 non-edge subcarriers) * (15 kHz subcarrier width).

I think your confusion is very common.

The sampling theorem tells you that you need a sampling rate at least 2x the signal bandwidth to prevent signal information loss (perfect reconstruction possible). Remember, however, that isn't necessarily important for communication.

With matched filtering, it is pretty straightforward to show that the maximum SNR can be obtained for communication while sampling at only the symbol rate (7.68 MHz), which is less than 2x the signal bandwidth (~10 MHz).

Typical receivers still oversample anyway so that functionality can be pushed into the digital domain (AGC, bandlimiting front-end filters) and/or so that practical digital operations such as time and frequency synchronization have maximum performance.

Long story short, for simplicity I recommend sampling at an integer (>=2) multiple of 7.68 MHz if you have the computational resources. What that multiple is will depend on how the other functions of your receiver behave (especially synchronization). After synchronization you can easily decimate down to the symbol rate for equalization. You can probably get away with sampling at 7.68 MHz, but your performance will suffer.

yes, sample rate, that's what baffles me so long!
I come to know that the when the subcarrier space of OFDM is 15Khz, the OFDM signal lasts for 1/15K second. Now I'm going to to represent the OFDM signal by using 2048 subcarrier within the "1/15K second", then the interval between each subcarrier is (1/15K)/2048 second, whose inverse is the 30.72Mhz.

rcdaniels, thanks again for your help!