3 ms·

He touches on it - but I’d love to see an intuitive explanation of why the response of each frequency to the input function is linearly independent. i.e the fac

He touches on it - but I’d love to see an intuitive explanation of why the response of each frequency to the input function is linearly independent. i.e the fact that Fourier transform of the sum is equal to the sum of the Fourier transforms. This is “why it works” - it’s what makes the frequency space an orthonormal basis - but it’s never been intuitively obvious to me. Otherwise, there would be more than one way of decomposing a function into a superposition. e.g. what would be useful is to give an example of a set of functions which are not linearly independent.

totalZero 9y ago
Here's how I think about it. You can play the individual notes of a chord on one piano or several, but they still come together to produce the same chorus of frequencies. The Fourier series representation of a waveform is itself a sum, since trigonometric functions are waveforms as well. Thus, a combination thereof should be commutative because a sum of two sums retains the properties of addition.
rocqua 9y ago
It actually follows from the 'centre of mass' explanation. If you take the centre of mass as described of f + g you get the centre of mass of f plus the centre of mass of g. One way to explain this is to just say the + can be moved out of the integral. Alternatively, consider the centre of mass only over the horizontal axis. Now say we only look at the 'contribution' of f + g at time t (ignoring the issues of that contribution being infinitesimal). That contribution is (f(x) + g(x)) * sin (theta) where theta is the angle of our point. Clearly this equals f(x) * sin (theta) + g(x) * sin (theta). These are the separate contributions of f(x) and g(x). The same argument holds for the centre of mass over the vertical axis (replacing sin with cos). If we were to make the alternative explanation formal, we get back to the + being able to move outside the integral. Note that our decomposition into the horizontal and vertical part is an alternative way to de the fourier transform without complex numbers. The vertical part here is essentially the imaginary part of the fourier transform.
kortex 9y ago
Take your wrapping function from t1 to t2, at a frequency ƒ (signal) != Fs (sampling freq), then take the limit as t1/t2 goes to -∞/+∞. As your window gets longer, more cycles of the oscillation "cancel out", moving the center of mass towards 0+0i. This means the peak around ƒ narrows and raises. At ∞, ƒ becomes infinitely narrow and high (Dirac delta * ). This is also why peaks on an fft are gaussian (finite window), and get sharper as the fft window is increased. * for cosine, technically there is a peak at -ƒ too. this is because a real cosine signal is ambiguous whether it is "moving forward or backwards in time". Hence it has a peak at +/-ƒ. A complex exponetial (helix through time) has chirality due to the real and imag components, so it has a single peak at ƒ. And if you take a +ƒ (lefthanded) and -ƒ helix (righthanded) and add them, the complex part cancels out, leaving only a real "up and down" wave.
chestervonwinch 9y ago
The orthogonality is essentially follows from (1) integer frequency complex sinusoids have an average value of zero over [0,2π], and (2) if you multiply two distinct integer frequency complex sinusoids, you get another integer frequency complex sinusoid. I'm not sure that this is any more intuitive. > what would be useful is to give an example of a set of functions which are not linearly independent. See [1,2] for example, which (I believe) has applications in compressed sensing and dictionary learning. [1]: https://en.wikipedia.org/wiki/Frame_(linear_algebra) https://en.wikipedia.org/wiki/Frame_(linear_algebra) [2]: https://en.wikipedia.org/wiki/Overcompleteness https://en.wikipedia.org/wiki/Overcompleteness
- mlevental 9y ago
  >The orthogonality is essentially follows from (1) integer frequency complex sinusoids have an average value of zero over [0,2π], and (2) if you multiply two distinct integer frequency complex sinusoids, you get another integer frequency complex sinusoid. I'm not sure that this is any more intuitive. i think these kinds of explanations are hilariously pointless. and i don't mean to disparage because you're just trying to answer op's question but all you've done is restated the proof in english - i.e. of course it follows from that because what you've just said is the inner product of basis functions is 0. well yes of course that's definition of orthogonal.

totalZero 9y ago

Here's how I think about it. You can play the individual notes of a chord on one piano or several, but they still come together to produce the same chorus of frequencies. The Fourier series representation of a waveform is itself a sum, since trigonometric functions are waveforms as well. Thus, a combination thereof should be commutative because a sum of two sums retains the properties of addition.

rocqua 9y ago

It actually follows from the 'centre of mass' explanation. If you take the centre of mass as described of f + g you get the centre of mass of f plus the centre of mass of g. One way to explain this is to just say the + can be moved out of the integral. Alternatively, consider the centre of mass only over the horizontal axis. Now say we only look at the 'contribution' of f + g at time t (ignoring the issues of that contribution being infinitesimal). That contribution is (f(x) + g(x)) * sin (theta) where theta is the angle of our point. Clearly this equals f(x) * sin (theta) + g(x) * sin (theta). These are the separate contributions of f(x) and g(x). The same argument holds for the centre of mass over the vertical axis (replacing sin with cos). If we were to make the alternative explanation formal, we get back to the + being able to move outside the integral. Note that our decomposition into the horizontal and vertical part is an alternative way to de the fourier transform without complex numbers. The vertical part here is essentially the imaginary part of the fourier transform.

kortex 9y ago

Take your wrapping function from t1 to t2, at a frequency ƒ (signal) != Fs (sampling freq), then take the limit as t1/t2 goes to -∞/+∞. As your window gets longer, more cycles of the oscillation "cancel out", moving the center of mass towards 0+0i. This means the peak around ƒ narrows and raises. At ∞, ƒ becomes infinitely narrow and high (Dirac delta * ). This is also why peaks on an fft are gaussian (finite window), and get sharper as the fft window is increased. * for cosine, technically there is a peak at -ƒ too. this is because a real cosine signal is ambiguous whether it is "moving forward or backwards in time". Hence it has a peak at +/-ƒ. A complex exponetial (helix through time) has chirality due to the real and imag components, so it has a single peak at ƒ. And if you take a +ƒ (lefthanded) and -ƒ helix (righthanded) and add them, the complex part cancels out, leaving only a real "up and down" wave.

chestervonwinch 9y ago

The orthogonality is essentially follows from (1) integer frequency complex sinusoids have an average value of zero over [0,2π], and (2) if you multiply two distinct integer frequency complex sinusoids, you get another integer frequency complex sinusoid. I'm not sure that this is any more intuitive. > what would be useful is to give an example of a set of functions which are not linearly independent. See [1,2] for example, which (I believe) has applications in compressed sensing and dictionary learning. [1]: https://en.wikipedia.org/wiki/Frame_(linear_algebra) https://en.wikipedia.org/wiki/Frame_(linear_algebra) [2]: https://en.wikipedia.org/wiki/Overcompleteness https://en.wikipedia.org/wiki/Overcompleteness

mlevental 9y ago
>The orthogonality is essentially follows from (1) integer frequency complex sinusoids have an average value of zero over [0,2π], and (2) if you multiply two distinct integer frequency complex sinusoids, you get another integer frequency complex sinusoid. I'm not sure that this is any more intuitive. i think these kinds of explanations are hilariously pointless. and i don't mean to disparage because you're just trying to answer op's question but all you've done is restated the proof in english - i.e. of course it follows from that because what you've just said is the inner product of basis functions is 0. well yes of course that's definition of orthogonal.