## Reverse FFT Size calculation/estimation of a Video Down-link. Started by 3 years ago66 views

Interested to determine the size of the FFT used by a proprietary video link transmitting 720p@30fps video.

(This can usually be achieved by 5-10 Mbps bit rate).

Recorded the signal at 20Msps and analyzed the structure of the burst and the related sizes(preamble, training, data symbols), reveals that the FFT size is 914 samples.

Then using the following code for calculations:

fs_hz = 20e6    %Sampling Rare

fft_smp = 914   %FFT Size samples

bw_hz = 10e6   %Signal BW

fft_time_sec = fft_smp/fs_hz   %Symbol duration time in seconds

bin_size_hz = 1/fft_time_sec %Bin size in Hz or Carrier spacing

number_of_bins = bw_hz/bin_size_hz %Number of bins in the FFT

constellation = 2  %QPSK - Constellation used - determined after time and frequency compensation

tx_rate = number_of_bins*constellation*(1/(1.125*fft_time_sec)) %Number of bits per second transmitted

Symbol duration(without the cp – just the FFT size) was calculated as 4.5700e-05 [sec].

As a result, it can be assumed that the bin size is 1/( 4.5700e-05) = 2.1882e+04 [Hz].

Taking into account the fact that the BW of the signal is 10Mhz, the number of Bins is 10e6/2.1882e+04 = 462[ bins].

Let’s assume 512 bins.

Knowing that the constellation is QPSK, means that 2 bits are being transmitted every symbol.

The number of symbols can be estimated as 1/(1.125*fft_time_sec) (neglecting the training and the preamble), yielding 19451 symbols per second.

The number of transmitted bits will then be - 1.7778e+07 = 17.7 Mbits.

This seem to be aligned with the expected 5-10Mbps, however, the size of the FFT of 512 as opposed to the one used in WiFi for 720p@30 transition(FFT of 64) looks peculiar for me.

Would be glad to know whether there are anything I am not taking into account during my calculations.