master > master: notebook - wk10
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notebooks/week-10.ipynb
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notebooks/week-10.ipynb
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{
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"cells": [
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"# Woche 10 #\n",
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"\n",
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"Diese Wochen beschäftigen wir uns mit Matrizen und Vektoren.\n",
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"In diesem Notebook rechnen wir ein paar (Teil)aufgaben aus dem ÜB."
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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": null,
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"metadata": {},
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"outputs": [],
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"source": [
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"'''IMPORTS'''\n",
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"import os;\n",
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"import sys; \n",
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"\n",
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"# NOTE: need this to force jupyter to reload imports:\n",
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"for key in list(sys.modules.keys()):\n",
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" if key.startswith('src.'):\n",
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" del sys.modules[key];\n",
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"\n",
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"os.chdir(os.path.dirname(_dh[0]));\n",
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"sys.path.insert(0, os.getcwd());\n",
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"\n",
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"from src.thirdparty.maths import *;\n",
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"from src.thirdparty.misc import *;\n",
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"from src.thirdparty.render import *;\n",
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"from src.maths.diagrams import *;\n",
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"\n",
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"np.random.seed(8007253); # zur Wiederholbarkeit"
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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": null,
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"metadata": {},
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"outputs": [],
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"source": [
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"'''BEISPIEL-SIGNAL'''\n",
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"n = 100;\n",
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"u = np.zeros((n,), dtype=complex);\n",
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"u[8] = -0.7\n",
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"u[58] = 1j * sqrt(2);\n",
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"\n",
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"display_signal(u, 'Beispiel eines Vektors als Signal dargestellt');"
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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": null,
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"metadata": {},
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"outputs": [],
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"source": [
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"'''SEMINARAUFGABE 10.4 a)'''\n",
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"n = 3;\n",
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"t = linspace(0, n-1, n);\n",
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"i, j = np.meshgrid(t, t);\n",
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"theta = 2*pi * i * j / n;\n",
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"\n",
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"# definiere die Matrix F_n + die Vektoren u_nk\n",
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"F_n = (1/sqrt(n)) * exp(-1j * theta);\n",
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"u_n = [ (1/sqrt(n)) * exp(1j * theta[:, k]) for k in range(n) ];\n",
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"\n",
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"print(dedent(\n",
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" f'''\n",
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" Matrix:\n",
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"\n",
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" F_{n} = \\n{np.round(F_n, 3)}\n",
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" '''\n",
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"));\n",
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"print('');\n",
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"\n",
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"# NOTE: Programmiersprachen können nicht exakt berechnen.\n",
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"# Darum entstehen winzige Fehler, die sich durch Runden entfernen lassen.\n",
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"print('Matrix-Vektor-Produkte:');\n",
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"for k in range(n):\n",
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" result = F_n @ u_n[k];\n",
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" print(dedent(\n",
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" f'''\n",
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"\n",
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" u_{{n {k}}} = {np.round(u_n[k], 3)}\n",
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" F_n · u_{{n {k}}} = {np.round(result, 3)}\n",
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" '''\n",
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" ));"
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]
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},
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Seminaraufgabe 10.4 b) ##\n",
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"\n",
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"Für $i\\in\\{0,1,\\ldots,n-1\\}$ ist die $i$-te von $\\mathcal{F}_{n}\\,u_{n,k}$\n",
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"\n",
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"$$\n",
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" (\\mathcal{F}_{n}\\,u_{n,k})_{i}\n",
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" = \\sum_{j=0}^{n-1}\n",
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" (\\mathcal{F}_{n})_{ij}\\,(u_{n,k})_{j}\n",
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" = \\sum_{j=0}^{n-1}\n",
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" \\tfrac{1}{\\sqrt{n}}\n",
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" \\exp(-\\imath\\tfrac{2\\pi\\,ij}{n})\n",
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" \\cdot\n",
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" \\tfrac{1}{\\sqrt{n}}\n",
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" \\exp(\\imath\\tfrac{2\\pi\\,kj}{n})\n",
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" = \\frac{1}{n}\\sum_{j=0}^{n-1}\n",
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" \\underbrace{\n",
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" \\exp(\\imath\\tfrac{2\\pi\\,(k-i)j}{n})\n",
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" }_{= r^{j}}\n",
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"$$\n",
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"\n",
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"wobei $r \\colonequals \\exp(\\imath\\tfrac{2\\pi\\,(k-i)}{n})$.\n",
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"Wir haben, dass\n",
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" $r = 1$\n",
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" $\\Leftrightarrow$\n",
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" $\\tfrac{k-i}{n} \\in \\Z$\n",
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" $\\Leftrightarrow$\n",
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" $i = k$,\n",
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"da $0\\leq i,k \\leq n-1$.\n",
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"<br>\n",
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"Es gilt außerdem $r^{n} = \\exp(\\imath 2\\pi\\,(k-i)) = 1$.\n",
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"<br>\n",
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"Aus der o.s. Berechnung erhält man also\n",
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"\n",
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"$$\n",
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" (\\mathcal{F}_{n}\\,u_{n,k})_{i}\n",
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" = \\tfrac{1}{n}\\displaystyle\\sum_{j=0}^{n-1}r^{j}\n",
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" = \\tfrac{1}{n}\n",
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" \\begin{cases}\n",
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" \\displaystyle\\sum_{j=0}^{n-1}1 &: &i = k\\\\\n",
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" \\frac{1 - r^{n}}{1 - r} &: &\\text{sonst}\\\\\n",
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" \\end{cases}\n",
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" = \\tfrac{1}{n}\n",
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" \\begin{cases}\n",
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" n &: &i = k\\\\\n",
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" \\frac{1 - 1}{1 - r} &: &\\text{sonst}\\\\\n",
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" \\end{cases}\n",
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" = \\delta_{ik}.\n",
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"$$\n",
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"\n",
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"Darum $\n",
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" \\mathcal{F}_{n}\\,u_{n,k}\n",
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" = \\left(\\begin{matrix}\n",
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" 0\\\\\n",
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" 0\\\\\n",
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" \\vdots\\\\\n",
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" 1\\\\\n",
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" \\vdots\\\\\n",
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" 0\\\\\n",
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" \\end{matrix}\\right)\n",
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" \\begin{matrix}\n",
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" \\\\\n",
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" \\\\\n",
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" \\\\\n",
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" \\leftarrow k\\\\\n",
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" \\\\\n",
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" \\\\\n",
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" \\end{matrix}\n",
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" = \\mathbf{e}_{k}\n",
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"$.\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": null,
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"metadata": {},
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"outputs": [],
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"source": [
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"'''SEMINARAUFGABE 10.4 d)'''\n",
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"n = 4;\n",
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"t = linspace(0, n-1, n);\n",
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"i, j = np.meshgrid(t, t);\n",
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"theta = 2*pi * i * j / n;\n",
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"\n",
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"# definiere die Matrix F_n + die Vektoren u_nk\n",
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"F_n = (1/sqrt(n)) * exp(-1j * theta);\n",
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"u_n = [ (1/sqrt(n)) * exp(1j * theta[:, k]) for k in range(n) ];\n",
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"\n",
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"# Eingabe des Vektors:\n",
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"F_mal_u = np.asarray([100, -3, 0, -3]); # = F_n @ u\n",
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"\n",
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"'''\n",
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"NOTE: Folgende Koeffizienten wurden falsch erraten.\n",
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"TODO: Wie lauten die richtigen Werte? -> Aufgabe!\n",
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"'''\n",
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"c = [10,10,40,50];\n",
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"\n",
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"# Aufschreibung des Signals »bzgl. der harmonischen Basis«:\n",
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"# u = c[0]u_n[0] + c[1]u_n[1] + c[2]u_n[2] + c[3]u_n[3]\n",
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"u_guess = sum([ c[k] * u_n[k] for k in range(n) ]);\n",
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"\n",
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"# Anzeige der Lösung:\n",
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"display_signal(u_guess, 'Geschätztes Signal');\n",
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"\n",
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"# Verifikation von Lösung:\n",
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"print(dedent(\n",
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" f'''\n",
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" %error := ‖u_guess – u‖ / ‖u‖\n",
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" = ‖F_n · (u_guess – u)‖ / ‖F_n · u‖\n",
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" ... wieso?? [Stichwort: unitär]\n",
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" = ‖F_n · u_guess – F_n · u‖ / ‖F_n · u‖\n",
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" = {linalg.norm(F_n @ u_guess - F_mal_u)/linalg.norm(F_mal_u):.2%}\n",
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" '''\n",
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"));\n"
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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": "3.10.8"
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}
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},
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"nbformat": 4,
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"nbformat_minor": 2
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}
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