Computational Methods for Astrophysics: Fourier Transforms

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1 Computational Methods for Astrophysics: Fourier Transforms John T. Whelan (filling in for Joshua Faber) April 27, 2011 John T. Whelan April 27, 2011 Fourier Transforms 1/13

2 Fourier Analysis Outline: Fourier Transforms Continuous Fourier Transforms Discrete Fourier Transforms Sampling and Aliasing The Fast Fourier Transform John T. Whelan April 27, 2011 Fourier Transforms 2/13

3 Continuous Fourier Transforms Fourier transform h(f ) of time series h(t): h(f ) = h(t) e i2πf (t t0) dt h(t) = Alternate conventions: h(f ) = h(t) e i2πf (t t0) dt h(t) = h(f ) e i2πf (t t 0 ) df h(f ) e i2πf (t t 0 ) df John T. Whelan April 27, 2011 Fourier Transforms 3/13

4 Continuous Fourier Transforms Fourier transform h(f ) of time series h(t): h(f ) = h(t) e i2πf (t t0) dt h(t) = Alternate conventions: sign of i, h(f ) = h(t) e i2πf (t t0) dt h(t) = h(f ) e i2πf (t t 0 ) df h(f ) e i2πf (t t 0 ) df John T. Whelan April 27, 2011 Fourier Transforms 3/13

5 Continuous Fourier Transforms Fourier transform h(f ) of time series h(t): h(f ) = h(t) e i2πf (t t0) dt h(t) = Alternate conventions: sign of i, f vs ω, h ω = 1 h(t) e iω(t t0) dt 2π h(t) = h ω e iω(t t0) dω h(f ) e i2πf (t t 0 ) df John T. Whelan April 27, 2011 Fourier Transforms 3/13

6 Continuous Fourier Transforms Fourier transform h(f ) of time series h(t): h(f ) = h(t) e i2πf (t t0) dt h(t) = Alternate conventions: sign of i, f vs ω, normalization, h ω = h(t) e iω(t t0) dt h(t) = 1 h ω e iω(t t0) dω 2π h(f ) e i2πf (t t 0 ) df John T. Whelan April 27, 2011 Fourier Transforms 3/13

7 Continuous Fourier Transforms Fourier transform h(f ) of time series h(t): h(f ) = h(t) e i2πf (t t0) dt h(t) = Alternate conventions: sign of i, f vs ω, normalization, h ω = 1 2π h(t) = 1 2π h(t) e iω(t t 0) dt h ω e iω(t t 0) dω h(f ) e i2πf (t t 0 ) df John T. Whelan April 27, 2011 Fourier Transforms 3/13

8 Continuous Fourier Transforms Fourier transform h(f ) of time series h(t): h(f ) = h(t) e i2πf (t t0) dt h(t) = h(f ) e i2πf (t t 0 ) df Alternate conventions: sign of i, f vs ω, normalization, time origin h(f ) = h(t) e i2πf t dt h(t) = h(f ) e i2πf t df John T. Whelan April 27, 2011 Fourier Transforms 3/13

9 Continuous Fourier Transforms Fourier transform h(f ) of time series h(t): h(f ) = h(t) e i2πf (t t0) dt h(t) = h(f ) e i2πf (t t 0 ) df Alternate conventions: sign of i, f vs ω, normalization, time origin If h(t) is real, h( f ) = h (f ) Convolution theorem: g(t) = A(t t ) h(t ) dt g(f ) = Ã(f ) h(f ) Can Fourier transform in space as well as time; e.g., h(fx, f y ) = h(x, y)e i2π(fx x+fy y) dx dy John T. Whelan April 27, 2011 Fourier Transforms 3/13

10 A Few Words About Convolution g(t) = A(t t ) h(t ) dt Can be written in different-looking but equivalent ways, e.g. g(t) = A(τ) h(t τ) dτ Because A(τ) is a function of a time difference, its Fourier transform is always defined w/zero time origin: Ã(f ) = A(τ) e i2πf τ dτ Arises naturally in astrophysical science & technology: superposition of impulse responses, point-spread fcn of imaging device, linear transfer function,... John T. Whelan April 27, 2011 Fourier Transforms 4/13

11 Applications of the Fourier Transform Multiply Fourier transforms to calculate convolution Differential equations: Spectral analysis d n dt n h(t) (i2πf )n h(f ) Matched/optimal filtering when spectral properties of signal and/or noise are known John T. Whelan April 27, 2011 Fourier Transforms 5/13

12 Discrete Fourier Transforms DFT of N-point sequence {h j j = 0,..., N 1}: ĥ k = N 1 j=0 h j e i2πjk/n h j = 1 N Corresponds to CFT of discretized data: h j = h(t 0 + j δt) ĥ k δt h N 1 k=0 ĥ k e i2πjk/n ( k ) N δt Again, different conventions (mostly ±i & where to put 1 N ); always wise to check your FT package s documentation Note by construction ĥn+k = ĥk; means e.g., 8-pt FT packed ĥ 0 ĥ 1 ĥ 2 ĥ 3 ĥ 4 ĥ 3 ĥ 2 ĥ 1 Sometimes use fftshift fcn to swap halves of array & get ĥ 4 ĥ 3 ĥ 2 ĥ 1 ĥ 0 ĥ 1 ĥ 2 ĥ 3 John T. Whelan April 27, 2011 Fourier Transforms 6/13

13 DFTs of Real and Complex Data For complex data, N cmplx data points {h i i = 0,..., N 1} N independent complex Fourier components {ĥk k = 0,..., N 1} or {ĥk k = N 2,..., N 2 1} h 0 h 1 h 2 h 3 h 4 h 5 h 6 h 7 ĥ 0 ĥ 1 ĥ 2 ĥ 3 ĥ 4 ĥ 3 ĥ 2 ĥ 1 For real data, symmetry of FT means ĥk = ĥ k = ĥ N k N real data points {h i i = 0,..., N 1} 2 real Fourier cmpts ĥ0 & ĥn/2 N 2 1 indep cmplx Fourier cmpts {ĥk k = 1,..., N 2 1} h 0 h 1 h 2 h 3 h 4 h 5 h 6 h 7 ĥ 0 ĥ 1 ĥ 2 ĥ 3 ĥ 4 [ĥ 3 ] [ĥ 2 ] [ĥ 1 ] These assume N even; modification for odd N straightforward John T. Whelan April 27, 2011 Fourier Transforms 7/13

14 Time and Frequency Resolution Recall correspondence between continuous & discrete FT: h j = h(t 0 + j δt) ĥ k δt h (f k ) where f k = k δf δf = 1 N δt 1 T John T. Whelan April 27, 2011 Fourier Transforms 8/13

15 Time and Frequency Resolution Recall correspondence between continuous & discrete FT: where h j = h(t 0 + j δt) ĥ k δt h (f k ) f k = k δf δf = 1 N δt 1 T Aside: why not factor 1 N = δt δf & define DFT as h k ĥk δt hk = δt N 1 j=0 h j e i2πjk/n h j = δf N/2 1 k= N/2 hk e i2πjk/n? John T. Whelan April 27, 2011 Fourier Transforms 8/13

16 Time and Frequency Resolution Recall correspondence between continuous & discrete FT: where h j = h(t 0 + j δt) ĥ k δt h (f k ) f k = k δf δf = 1 N δt 1 T Aside: why not factor 1 N = δt δf & define DFT as h k ĥk δt hk = δt N 1 j=0 h j e i2πjk/n h j = δf N/2 1 k= N/2 hk e i2πjk/n? {ĥk} involves only data {h j }; { h k } mixes in additional metadata δt John T. Whelan April 27, 2011 Fourier Transforms 8/13

17 Time and Frequency Resolution Recall correspondence between continuous & discrete FT: h j = h(t 0 + j δt) ĥ k δt h (f k ) where f k = k δf δf = 1 N δt 1 T N time points represent duration of T = N δt If freq indices are N 2 k N 2 1 (complex) or 0 k N 2 (real), then f k N δf 2 = 1 2 δt f Ny. The Nyquist frequency f Ny is half the sampling rate 1 δt & is the largest independent frequency represented in the DFT John T. Whelan April 27, 2011 Fourier Transforms 8/13

18 Sampling and Aliasing ĥ N+k = ĥk, but we said ĥk δt h (f k ) and in general h (f N+k ) h (f k ). Actually ĥ k δt + h (f N+k ) + h (f k ) + h (f N+k ) + h (f 2N+k ) + If sampled time series had Fourier cmpts w/ f > f N/2 = f Ny, those will be aliased with data in the range f Ny f f Ny Generally low-pass filter time series to discard f > f Ny = 1 2 δt content before sampling at rate 1 δt Alternately, if data already band-limited ( h(f ) = 0 for f > B) 1 avoid aliasing by choosing δt so f Ny > B i.e., δt > 2B Confusingly, 2B is sometimes called the Nyquist rate John T. Whelan April 27, 2011 Fourier Transforms 9/13

19 Illustration of Aliasing John T. Whelan April 27, 2011 Fourier Transforms 10/13

20 Illustration of Aliasing John T. Whelan April 27, 2011 Fourier Transforms 10/13

21 Illustration of Aliasing John T. Whelan April 27, 2011 Fourier Transforms 10/13

22 The Fast Fourier Transform Naïve implementation of discrete FT as ĥ k = N 1 j=0 ( e i2π/n) kj hj would require O(N 2 ) ops & not be any faster than convolution g j = N 1 m=0 A j m h m Fast Fourier Transform (FFT) algorithms [Tukey & Cooley 1965] cut that to O(N log N) by writing N-pt FT two N/2-pt FTs four N/4-pt FTs Speedup is greatest for N =power of 2, but products of small prime factors are generally good Popular/fast GPLed implementation is FFTW John T. Whelan April 27, 2011 Fourier Transforms 11/13

23 Exercise For N = 16, generate the discrete time series h j = cos 3π(j 2) 8 Use an FFT routine to transform it & examine resulting ĥk Apply an inverse FFT routine to h k & compare results to h j Use the angle sum formula to write h j as a linear combination of cos 3πj 8 = ei3πj/8 +e i3πj/8 2 and sin 3πj 8 = ei3πj/8 e i3πj/8 2i and therefore as a linear combination of e ±i3πj/8 and compare the coëfficients to the components of the discrete Fourier transform. John T. Whelan April 27, 2011 Fourier Transforms 12/13

24 Grant Tremblay s PhD Defense Time 11:30 Location Carlson Auditorium, Title Feedback-Regulated Star Formation in Cool Core Clusters of Galaxies John T. Whelan April 27, 2011 Fourier Transforms 13/13

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