Документ взят из кэша поисковой машины. Адрес
оригинального документа
: http://www.atnf.csiro.au/vlbi/dokuwiki/doku.php/difx/pcalimplicitalgorithm
Дата изменения: Unknown Дата индексирования: Mon Apr 11 02:54:35 2016 Кодировка: IBM-866 Поисковые слова: m 13 |
Walter's original email about the pcal extraction algorithm is copied in below.
There are also a couple of old documents from Sergei Pogrebenko which detail it EVN Doc 2 EVN Doc 8
A key parameter is the greatest common factor of the sample frequency and the frequency offset. Ths number leads to a number I will call N that tells you how many baseband data samples are required before the phase cal time series repeats itself exactly yielding the same phase.
The algorithm I envision works as follows:
1. create a real-valued accumulator array, A, with dimension N.
2. for the duration of an accumulation period, fold the voltage data, V, into this array:
for(i = 0; i < nsamples; i++) A[i % N] += V[i]
3. Perform a real-to-complex Fourier transform on this array into an N/2 element complex array B
B = FFT[A]
4. Extract the tones wanted. The B array has N/2 points, but one will only be interested in a very small subset of those points тАУ the ones the correspond to the pulse cal tone frequencies. Note that it _might_ be faster to do a DFT only for the desired points rather than performing a full FFT to generate the entire spectrum.
As a concrete example, lets consider the VLBI case with a single baseband channel, bandwidth bw = 2^n MHz, with 1 tone per MHz, with the first tone nu = 10*d kHz offset from the DC edge of the channel for integer d (note this is a standard case for VLBI).
The condition for return-to-phase is:
delta phi = 2*pi*nu*deltaT = R*2*pi for integer R
deltaT must be an integral number of samples which we call N (the same N as above). Each sample is 2^(-n-1) microseconds in duration. Thus
d/100 MHz * N*2^(-n-1) mus = R
or
d * N = 200 * 2^n * R
This expression is where the greatest common divisor concept comes in. It is easy to see that N = 200 * 2^n will meet the required condition for all values of d, meaning for all possible tones in the baseband. For 8 MHz this implies N = 1600 points which forms a spectrum with 800 complex points. There will be 8 tones to recover, each separated by 1 MHz = 100 points in the spectrum. The position of the first tone is at spectrum bin 800*nu/bw . The remaining 792 points of the spectrum are to be discarded.
Note the key aspect of this is that all of the averaging is done in the time domain and a single transform to the frequency domain is needed. There might be some logistical reason for doing the FFT and tone selection once per sub-integration in order to minimize data transfer from core processes to the manager process.