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1 Hough transform Some Fourier basics: –Nyquist frequency: 1/2 , with  the difference between time samples. If signal is bandwidth limited below Nyquist.

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Presentation on theme: "1 Hough transform Some Fourier basics: –Nyquist frequency: 1/2 , with  the difference between time samples. If signal is bandwidth limited below Nyquist."— Presentation transcript:

1 1 Hough transform Some Fourier basics: –Nyquist frequency: 1/2 , with  the difference between time samples. If signal is bandwidth limited below Nyquist frequency, the complete information of the continuous response f(t) is obtained in the discrete set of FFT samples –FFT: time is order N log(N) steps, with N number of samples. So 1 FFT of 1 million seconds takes about 20 times longer than 1million FFTs of 1 second. –distance between frequency bins: 1/T, with T the total observation time. If you sample more frequently, the maximum frequency goes up, but not the frequency resolution.

2 2 PSS suppose: –signal strength h~ 10 -26 –noise 10 -22 /sqrt(Hz) –In 1 second: average amplitudes of 10 -22 in every frequency bin Needed: coherent measurement of at least 10*signal/noise seconds. Optimal sensitivity: matched filter: measurement is compared to theoretical signal with exact same shape as signal. Needed: signal should line up in 1 frequency bin, or equivalently the matched filter has the same frequency dependence as the signal with resolution of 1/T.

3 3 orbital movements daily rotational velocity around 300m/s: Doppler shift 10 -6 yearly: orbital velocity 30km/s: Doppler shift 10 -4 frequency shifts: frequency of source times Doppler shift. in order to contain the signal in 1 frequency bin, the frequency shifts should be smaller than the bin width. T= 3 hour: Doppler shift < 10 -6, frequency shift ~  * 10 -6 < 1 bin if  < 100 Hz. T = 10 6 s : frequency shift ~  * 10 -4 =~ 10000 bins.

4 4 Computing requirements Brady paper: PRD (1997), gr-qc/9702050 –coherent search 10million seconds: young pulsars, f/(df/dt) > 40 years old : f/(df/dt) > 1000 years slow: f<200 Hz, fast f<1000 Hz parameters: spin-down direction in sky: parameters –introduce canonical time t, depending on these parameters. –for each set of parameters, a separate FFT is needed.

5 5 computing requirements number of parameters: e.g 1 day, young pulsars: 4Tflops. For old, slow pulsars one could do about 20 days in same computing time.

6 6 Hierarchical search divide observation time in small steps. Make a time-frequency database with on the time axis the average time of the FFT and on the z-axis the power in the FFT frequency bin. original FFT database: signal should be contained in 1 bin. Do Hough transform on this map (next slides) for parameters above threshold, repeat search in longer times and finer parameter binning.

7 7 Hough transform Hough transform: designed for pattern recognition map N coordinates onto M other coordinates, every point in first set maps to a line in the other set. basic example: mapping points (x,y) to lines y=ax+b

8 8 Hough transform many points on same line will line up in one point in the Hough space:

9 9 Hough example

10 10 Hierarchical search make short FFT base: –500-2000 Hz: T~1000 s –125-500 Hz: T~4000 s –31-125 Hz: T~8000 s remove “events” (short-duration signals) create peak map: all peaks above threshold, store beginning of FFT and frequency bin from this time-frequency map, one could do a Hough transform.

11 11 peak map Nautilus

12 12 Pulsars young old binary

13 13 Binaries orbital periods as small as 2 hours doppler shift: example: 2 hour time spectrum (thesis Aulbert, Max Plank Institute)

14

15 15 binaries in order to retain the power in one FFT bin, one needs to limit the length of the FFT to about 100 seconds for the case of 2-hour period binaries! This is 40 times shorter than the FFT base possible for stationary pulsars, implying 6 times worse S/N ratio.

16 16 binaries the neutron star can be in an elliptical orbit, further complicating the analysis. the plane in which the neutron star orbits will be tilted with respect to the line of sight from the earth, so the frequency shift cannot be calculated from the orbital period 5 extra parameters needed! However, spin- down may be neglected. suppose: frequency of binary source is

17 17 binaries for a 500 Hz neutron star with the maximal value of Delta, the strength is spread over 1.5 Hz. (1500 bins). (for long t_observed). for 1000 sec FFT, with period 2 hours, the strength may spread in about 500 1mHz bins. to get the strength into 1 bin, for a matched-filter analysis of 1000 seconds of data, one needs to try about a billion combinations of parameters for Delta, omega, and phi. This looks prohibitive. Alternatively: 40 seconds of data: all strength is contained in 1 25mHz bin. This could be start of peak map. Search over 10 7 s of data: daily rotational Doppler shift is smaller than 25 mHz, can be ignored. Yearly Doppler shift ~ 50 mHz: only about 9 patches of the sky need to be distinguished.

18 18 binaries, 25 mHz bins binary orbital movement: differences of 25 mHz over 10 7 s imply that the resolution on orbital period omega is smaller than about 1/40 times 10 -7 Hz. The maximal value of omega is about 10 -4 Hz, so this implies about 25000 values of omega to study (for the maximal value of Delta). For each omega, about 100 different starting phases are needed. 40 distinct values of Delta necessary(0 – 3 times 10-3). For smaller Delta, less values of omega and starting phases are needed. For a 10 Hz band, there are 400 values of f0. in peak map: one should try ~ 400*25000*40*20 ( *9, sky position) parameters; –Hough: each entry in the time-frequency domain fills a hypersurface in this parameter space ( several Gb 4-dimensional histograms)

19 19 Hough map alternatively, instead of making a Hough map between the time-frequency FFTs and the billion parameter Hough space, one may search over the billion parameters and sum the strength and just store results above a certain threshold (i.e. analysis without Hough) resolution in this first step of hierarchical search: 25 mHz bins, 10 7 s observation time: –noise is 0.025* 10 -22 * sqrt(10 7 ), –signal = 10 7 * h –sensitivity: signal at least 10* average noise, h> 10 -26


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