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 — difx:pcalimplicitalgorithm [2015/10/21 10:08] (current) 2011/07/05 17:57 johnmorgan created 2011/07/05 17:57 johnmorgan created Line 1: Line 1: + + 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 {{:​difx:​evn_doc_2.pdf|EVN Doc 2}} {{:​difx:​evn_doc_8.pdf|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 -- 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. 