Prześlij pliki do '09/do_sprawdzenia'
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09/do_sprawdzenia/9_6.ipynb
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09/do_sprawdzenia/9_6.ipynb
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09/do_sprawdzenia/9_7.ipynb
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09/do_sprawdzenia/9_7.ipynb
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"cells": [
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"### Exercise 7.1 \n",
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"#### The notebook for this chapter, chap07.ipynb, contains additional examples and explanations. Read through it and run the code."
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Done"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"### Exercise 1\n",
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"#### In this chapter, I showed how we can express the DFT and inverse DFT as matrix multiplications. These operations take time proportional to $N^2$, where $N$ is the length of the wave array. That is fast enough for many applications, but there is a faster algorithm, the Fast Fourier Transform (FFT), which takes time proportional to $N \\log N$.\n",
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"\n",
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"##### The key to the FFT is the Danielson-Lanczos lemma:\n",
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"\n",
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"$DFT(y)[n] = DFT(e)[n] + \\exp(-2 \\pi i n / N) DFT(o)[n]$\n",
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"\n",
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"##### Where $ DFT(y)[n]$ is the $n$th element of the DFT of $y$; $e$ is the even elements of $y$, and $o$ is the odd elements of $y$. This lemma suggests a recursive algorithm for the DFT:\n",
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"##### Given a wave array, $y$, split it into its even elements, $e$, and its odd elements, $o$.\n",
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"##### Compute the DFT of $e$ and $o$ by making recursive calls.\n",
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"##### Compute $DFT(y)$ for each value of $n$ using the Danielson-Lanczos lemma.\n",
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"##### For the base case of this recursion, you could wait until the length of $y$ is 1. In that case, $DFT(y) = y$. Or if the length of $y$ is sufficiently small, you could compute its DFT by matrix multiplication, possibly using a precomputed matrix.\n",
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"##### Hint: I suggest you implement this algorithm incrementally by starting with a version that is not truly recursive. In Step 2, instead of making a recursive call, use dft or np.fft.fft. Get Step 3 working, and confirm that the results are consistent with the other implementations. Then add a base case and confirm that it works. Finally, replace Step 2 with recursive calls.\n",
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"##### One more hint: Remember that the DFT is periodic; you might find np.tile useful.\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 15,
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"metadata": {},
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"outputs": [],
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"source": [
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"import numpy as np\n",
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"import sys\n",
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"sys.setrecursionlimit(10000)"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 51,
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"metadata": {},
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"outputs": [],
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"source": [
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"ys = [-0.2, 0.1, -0.1, -0.5]"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 53,
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"metadata": {},
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"outputs": [],
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"source": [
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"#not truly recursive version\n",
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"def FFT(ys):\n",
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" N = len(ys)\n",
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" if N == 1:\n",
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" return ys\n",
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" e = np.fft.fft(ys[::2]) #even\n",
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" o = np.fft.fft(ys[1::2]) #odd\n",
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" n = np.arange(N)\n",
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" exp = np.exp(-1j * 2 * np.pi * n / N)\n",
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" \n",
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" return np.tile(e, 2) + exp * np.tile(o, 2) #Danielson-Lanczos lemma"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 54,
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"metadata": {},
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"outputs": [],
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"source": [
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"#book implementation\n",
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"def dft(ys):\n",
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" N = len(ys)\n",
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" ts = np.arange(N) / N\n",
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" fs = np.arange(N)\n",
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" args = np.outer(ts, fs)\n",
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" M = np.exp(1j * PI2 * args)\n",
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" amps = M.conj().transpose().dot(ys)\n",
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" return amps"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 62,
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"metadata": {},
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"outputs": [],
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"source": [
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"#resursive\n",
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"def FFT_recursive(ys):\n",
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" N = len(ys)\n",
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" if N == 1:\n",
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" return ys \n",
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" e = fft(ys[::2])\n",
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" o = fft(ys[1::2]) \n",
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" n = np.arange(N)\n",
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" exp = np.exp(-1j * PI2 * n / N)\n",
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" \n",
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" return np.tile(e, 2) + exp * np.tile(o, 2)"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 59,
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"array([-0.7+0.0000000e+00j, -0.1-6.0000000e-01j, 0.1+4.8985872e-17j,\n",
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" -0.1+6.0000000e-01j])"
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]
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},
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"execution_count": 59,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"FFT(ys)"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 60,
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"array([-0.7+0.00000000e+00j, -0.1-6.00000000e-01j, 0.1+1.46957616e-16j,\n",
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" -0.1+6.00000000e-01j])"
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]
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},
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"execution_count": 60,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"dft(ys)"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 63,
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"array([-0.7+0.0000000e+00j, -0.1-6.0000000e-01j, 0.1+4.8985872e-17j,\n",
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" -0.1+6.0000000e-01j])"
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]
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},
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"execution_count": 63,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"FFT_recursive(ys)"
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]
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 3",
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"language": "python",
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"name": "python3"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.8.5"
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}
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},
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"nbformat": 4,
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"nbformat_minor": 4
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}
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