Загрузка видео...

Не удалось загрузить видео

На главную

Sharp feather tips can be approximated by adding smooth sine and cosine waves. This hummingbird’s outline is described by two Fourier series, one for each coordinate. Their rotating chains position the pen where the horizontal and vertical guides meet. Low frequencies capture the broad shape; higher frequencies recover smaller...

38,515 просмотров • 21 дней назад •via X (Twitter)

Комментарии: 7

Фото профиля Peter MOUEZA🇲🇫
Peter MOUEZA🇲🇫21 дней назад

By Fourier

Фото профиля Klaudivan
Klaudivan21 дней назад

@threadreaderapp unroll please

Фото профиля Giovanni
Giovanni21 дней назад

love that. wow

Фото профиля Master Steward of New London
Master Steward of New London21 дней назад

This is how they program drawing automatons, by the way

Фото профиля Ariz Ashraf
Ariz Ashraf21 дней назад

Basically the first few waves just sketch the rough outline of the bird, then the higher ones fill in the sharp feathers and finer bits. Gets pretty accurate once you add enough of them.

Фото профиля TheRoloway
TheRoloway21 дней назад

Using Fourier series to redraw a hummingbird outline is a beautiful way to show how low frequencies hold the shape and the high ones chase the feathers.

Фото профиля Nícolas Abreu
Nícolas Abreu21 дней назад

@grok me explique o que tá acontecendo de forma leiga e resumida.

Похожие видео

Discrete Fourier Transform by hand ✍️ ~ 12 steps walkthrough below Here is a little-known secret about the DFT and the inverse DFT: it is just matrix multiplication in both directions, one the transpose of the other, exactly like the forward pass and backpropagation I drew in other examples. Goal: recover which cosine waves a signal is made of, using nothing but multiplication and addition. = 1. Given = Three signals written as sums of cosines, and a fourth, X, that we do not know yet. = 2. Frequency matrix F = Let us write the coefficients as a matrix. Each signal is a row, each frequency a column, so A = cos(w) + 2cos(2w) becomes [1, 2, 0, 0]. = 3. Sample the waves = We read the four cosine waves at ten discrete time points. That word "discrete" is the whole difference between this and the continuous transform. = 4. Cosine matrix W = Let us write those samples as a matrix: each frequency a row, each time point a column. = 5. Frequency to time = We multiply F by W. That combines the four cosine waves in the proportions F specifies, and the result T is the three signals as they would look in time. = 6. Transpose = Let us stand each signal up as a column. = 7. Time to frequency = We multiply W by that transpose. Every cell is the dot product of one signal with one cosine wave, which measures how much of that wave the signal contains. Zero means none of it. = 8. Scale = Let us multiply by 2/n, with n = 10. The projections come out five times too large, and this is the correction. = 9. Transpose back = We turn it back around, and it is F again, exactly. That is the check: the transform recovered the coefficients we started from. = 10. Now solve for X = Let us run the same multiplication on the one signal whose recipe we never knew. = 11. Scale = We divide by 5 again. = 12. Transpose back = And X reads [0, 0, 3, 2], which says X = 3cos(3w) + 2cos(4w). Note: I originally drew this to show that the DFT is a special case of a convolution layer, its filters fixed to sine and cosine waves rather than learned. No wonder, then, that a convolution layer free to learn its own filters can be trained to process signals. 💾 Save this post!

Tom Yeh

13,602 просмотров • 1 месяц назад

Discrete Fourier Transform by hand ✍️ ~ 12 steps walkthrough below Here is a little-known secret about the DFT and the inverse DFT: it is just matrix multiplication in both directions, one the transpose of the other, exactly like the forward pass and backpropagation I drew in other examples. Goal: recover which cosine waves a signal is made of, using nothing but multiplication and addition. = 1. Given = Three signals written as sums of cosines, and a fourth, X, that we do not know yet. = 2. Frequency matrix F = Let us write the coefficients as a matrix. Each signal is a row, each frequency a column, so A = cos(w) + 2cos(2w) becomes [1, 2, 0, 0]. = 3. Sample the waves = We read the four cosine waves at ten discrete time points. That word "discrete" is the whole difference between this and the continuous transform. = 4. Cosine matrix W = Let us write those samples as a matrix: each frequency a row, each time point a column. = 5. Frequency to time = We multiply F by W. That combines the four cosine waves in the proportions F specifies, and the result T is the three signals as they would look in time. = 6. Transpose = Let us stand each signal up as a column. = 7. Time to frequency = We multiply W by that transpose. Every cell is the dot product of one signal with one cosine wave, which measures how much of that wave the signal contains. Zero means none of it. = 8. Scale = Let us multiply by 2/n, with n = 10. The projections come out five times too large, and this is the correction. = 9. Transpose back = We turn it back around, and it is F again, exactly. That is the check: the transform recovered the coefficients we started from. = 10. Now solve for X = Let us run the same multiplication on the one signal whose recipe we never knew. = 11. Scale = We divide by 5 again. = 12. Transpose back = And X reads [0, 0, 3, 2], which says X = 3cos(3w) + 2cos(4w). Note: I originally drew this to show that the DFT is a special case of a convolution layer, its filters fixed to sine and cosine waves rather than learned. No wonder, then, that a convolution layer free to learn its own filters can be trained to process signals. 💾 Save this post!

Tom Yeh

25,684 просмотров • 3 месяцев назад