Changes between Version 23 and Version 24 of SpectraObject
- Timestamp:
- 01/13/15 15:56:07 (10 years ago)
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SpectraObject
v23 v24 96 96 The FFT performs a Discrete Fourier Transform (DFT) which is defined as 97 97 98 [[latex($F_k=\displaystyle \sum_{x=1}^{N} e^{ \frac{2 \pi i}{N} k x} f_{x}$)]] 98 [[latex($F_k=\displaystyle \sum_{x=1}^{N} e^{ \frac{2 \pi i}{N} k x} f_{x}$)]] and the inverse is 99 100 [[latex($F_k=\frac{1}{N^2} \displaystyle \sum_{x=1}^{N} e^{ \frac{-2 \pi i}{N} k x} f_{x}$)]] and the inverse is 99 101 100 102 where [[latex($k$)]] is any integer. 101 103 104 We always normalize the FFT by [[latex($1/N$)]] so that the transform has the same 'units' as the function. 105 102 106 We can see how this corresponds to a real fourier transform by making the substitutions [[latex($x'=x \Delta x$)]], [[latex($k'=k \Delta k $)]], [[latex($\Delta k = \frac{2 \pi}{L}$)]] and [[latex($N=\frac{L}{\Delta x}$)]] and take the limit as [[latex($\Delta x \rightarrow 0$)]] 103 107 104 [[latex($F(k')= \frac{ N}{L}\displaystyle \int_{x'=0}^{L} e^{ik' x'} f(x') dx$)]]108 [[latex($F(k')= \frac{1}{L} \mathcal{F}(k') = \frac{1}{L} \displaystyle \int_{x'=0}^{L} e^{ik' x'} f(x') dx$)]] 105 109 106 So that the DFT corresponds to a fourier transform where the physical wavelength is normalized by [[latex($\Delta k=\frac{2 \pi}{L}$)]] and the magnitude is normalized by [[latex($N/L$)]] 110 Note that [[latex($F(k')$)]] and [[latex($f(x')$)]] have the same units. 111 112 So the DFT corresponds to a Fourier transform where the physical wavelength is normalized by [[latex($\Delta k=\frac{2 \pi}{L}$)]] and the magnitude is normalized by [[latex($N/L$)]] 107 113 108 114 109 Now the FFT returns the array of transforms with [[latex($k=[0,1,2,...,N _x-1]$)]]115 Now the FFT returns the array of transforms with [[latex($k=[0,1,2,...,N-1]$)]] 110 116 111 117 However, [[latex($F_k$)]] is periodic in [[latex($k$)]] since 112 118 113 [[latex($F_{k+N _x}=\displaystyle \sum_{x=1}^{N_x} e^{ \frac{2 \pi i}{N_x} \left(k+N_x\right) x} f_{x} = \displaystyle \sum_{x=1}^{N_x} e^{2 \pi i x} \times e^{ \frac{2 \pi i}{N_x} k x} f_{x}=\displaystyle \sum_{x=1}^{N_x} e^{ \frac{2 \pi i}{N_x} kx} f_{x} =F_k$)]]119 [[latex($F_{k+N}=\frac{1}{N}\displaystyle \sum_{x=1}^{N} e^{ \frac{2 \pi i}{N} \left(k+N\right) x} f_{x} = \frac{1}{N} \displaystyle \sum_{x=1}^{N} e^{2 \pi i x} \times e^{ \frac{2 \pi i}{N} k x} f_{x}=\frac{1}{N}\displaystyle \sum_{x=1}^{N} e^{ \frac{2 \pi i}{N} kx} f_{x} =F_k$)]] 114 120 115 We can visually see this by plotting the discrete sampling of the real part of the continuous functions for [[latex($k=-1$)]] and [[latex($k=9$)]] for a grid where [[latex($N _x=10$)]]121 We can visually see this by plotting the discrete sampling of the real part of the continuous functions for [[latex($k=-1$)]] and [[latex($k=9$)]] for a grid where [[latex($N=10$)]] 116 122 117 123 [[Image(Screen Shot 2015-01-13 at 10.57.02 AM.png,width=600)]] … … 127 133 [[Image(Screen Shot 2015-01-13 at 12.05.47 PM.png, width=400)]] 128 134 129 === MultiDimensional FFTs ===135 === !MultiDimensional FFTs === 130 136 A multidimensional DFT can be thought of as a sequence of 1D transforms in the way that a multidimensional Fourier transform is a sequence of 1D integrals. That is 131 137
