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4

If you look at the questions dealing with the other side (Convolution on time domain and element wise multiplication of frequency domain) you'd see that in order to have proper equality one must pay attention to the borders and type of convolution. For example, have a look at Applying a 2D Convolution Using 2D FFT which aggregates many other answers. ...


2

Matt's answer is excellent. For anyone wanting to do the same in Python, I thought I'd add the code (with a slight modification to use a sine wave input, which I prefer for debugging): import numpy as np N=1536 N1=512 N2=3 # 512 x 3 = 1536 nCycles = 2 # Number of cycles in source frame time=np.arange(0,N ) # some input ...


2

Please recognise $N$-th root of unity (on the complex-plane): $$ 1^{1/N} = e^{j \frac{2\pi}{N} k} ~~,~~ k = 0,1,...,N-1 $$ where $j$ is the imaginary unit in DSP notation. Hence $1^{1/N} = 1$ is true but incomplete, for you ignore the complex roots. So in your derivation last two lines should follow as: $$ 1^{m / N} = (1^{1/N})^m $$ $$ e^{j\frac{2\pi}{N} k m ...


1

$$e^{ (\pi \cdot i) \cdot (-2 \cdot j / n)} \color{red}{\ne} \left(e^{\pi \cdot i}\right)^{-2 \cdot j / n}$$ You can't just $e^{a\cdot b}= \left(e^a\right)^b$ for arbitrary complex $a,b$, because $e^a$ is not necessarily a positive real number .


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One way to interpret the DFT, that I personally find most useful, is by comparing it to the DTFT (Discrete Tome Fourier Transform, which has an infinite input domain) of a repeated function. That is, imagine tiling your image out to infinity, and applying the DTFT, to obtain a periodic, continuous Fourier domain that you then sample. Because of the ...


1

Perceval's theorem is satisfied with you apply scale factor of $1/\sqrt(N)$ for both the forward and inverse transform. I.e. If $$X(k) = \frac{1}{\sqrt{N}}\sum_{n=0}^{N-1} x(n) e^{-j2\pi\frac{kn}{N}}$$ then $$x(n) = \frac{1}{\sqrt{N}}\sum_{k=0}^{N-1} x(n) e^{j2\pi\frac{kn}{N}}$$ and $$\sum_{n=0}^{N-1}|x(n)|^2 = \sum_{k=0}^{N-1}|X(k)|^2 $$


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