{ "cells": [ { "attachments": {}, "cell_type": "markdown", "metadata": {}, "source": [ "# Woche 10 #\n", "\n", "Diese Wochen beschäftigen wir uns mit Matrizen und Vektoren.\n", "In diesem Notebook rechnen wir ein paar (Teil)aufgaben aus dem ÜB." ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "'''IMPORTS'''\n", "import os;\n", "import sys; \n", "\n", "# NOTE: need this to force jupyter to reload imports:\n", "for key in list(sys.modules.keys()):\n", " if key.startswith('src.'):\n", " del sys.modules[key];\n", "\n", "os.chdir(os.path.dirname(_dh[0]));\n", "sys.path.insert(0, os.getcwd());\n", "\n", "from src.thirdparty.maths import *;\n", "from src.thirdparty.misc import *;\n", "from src.thirdparty.render import *;\n", "from src.maths.diagrams import *;\n", "\n", "np.random.seed(8007253); # zur Wiederholbarkeit" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "'''BEISPIEL-SIGNAL'''\n", "n = 100;\n", "u = np.zeros((n,), dtype=complex);\n", "u[8] = -0.7\n", "u[58] = 1j * sqrt(2);\n", "\n", "display_signal(u, 'Beispiel eines Vektors als Signal dargestellt');" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "'''SEMINARAUFGABE 10.4 a)'''\n", "n = 3;\n", "t = linspace(0, n-1, n);\n", "i, j = np.meshgrid(t, t);\n", "theta = 2*pi * i * j / n;\n", "\n", "# definiere die Matrix F_n + die Vektoren u_nk\n", "F_n = (1/sqrt(n)) * exp(-1j * theta);\n", "u_n = [ (1/sqrt(n)) * exp(1j * theta[:, k]) for k in range(n) ];\n", "\n", "print(dedent(\n", " f'''\n", " Matrix:\n", "\n", " F_{n} = \\n{np.round(F_n, 3)}\n", " '''\n", "));\n", "print('');\n", "\n", "# NOTE: Programmiersprachen können nicht exakt berechnen.\n", "# Darum entstehen winzige Fehler, die sich durch Runden entfernen lassen.\n", "print('Matrix-Vektor-Produkte:');\n", "for k in range(n):\n", " result = F_n @ u_n[k];\n", " print(dedent(\n", " f'''\n", "\n", " u_{{n {k}}} = {np.round(u_n[k], 3)}\n", " F_n · u_{{n {k}}} = {np.round(result, 3)}\n", " '''\n", " ));" ] }, { "attachments": {}, "cell_type": "markdown", "metadata": {}, "source": [ "## Seminaraufgabe 10.4 b) ##\n", "\n", "Für $i\\in\\{0,1,\\ldots,n-1\\}$ ist die $i$-te von $\\mathcal{F}_{n}\\,u_{n,k}$\n", "\n", "$$\n", " \\begin{array}{rcl}\n", " (\\mathcal{F}_{n}\\,u_{n,k})_{i}\n", " &= &\\displaystyle\\sum_{j=0}^{n-1}\n", " (\\mathcal{F}_{n})_{ij}\\,(u_{n,k})_{j}\\\\\n", " &= &\\displaystyle\\sum_{j=0}^{n-1}\n", " \\tfrac{1}{\\sqrt{n}}\n", " \\exp(-\\imath\\tfrac{2\\pi\\,ij}{n})\n", " \\cdot\n", " \\tfrac{1}{\\sqrt{n}}\n", " \\exp(\\imath\\tfrac{2\\pi\\,kj}{n})\\\\\n", " &= &\\frac{1}{n}\\displaystyle\\sum_{j=0}^{n-1}\n", " \\underbrace{\n", " \\exp(\\imath\\tfrac{2\\pi\\,(k-i)j}{n})\n", " }_{= r^{j}}\\\\\n", " \\end{array}\n", "$$\n", "\n", "wobei $r := \\exp(\\imath\\tfrac{2\\pi\\,(k-i)}{n})$.\n", "Wir haben, dass\n", " $r = 1$\n", " $\\Leftrightarrow$\n", " $\\tfrac{k-i}{n} \\in \\mathbb{Z}$\n", " $\\Leftrightarrow$\n", " $i = k$,\n", "da $0\\leq i,k \\leq n-1$.\n", "
\n", "Es gilt außerdem $r^{n} = \\exp(\\imath 2\\pi\\,(k-i)) = 1$.\n", "
\n", "Aus der o.s. Berechnung erhält man also\n", "\n", "$$\n", " \\begin{array}{rcl}\n", " (\\mathcal{F}_{n}\\,u_{n,k})_{i}\n", " &= &\\tfrac{1}{n}\\displaystyle\\sum_{j=0}^{n-1}r^{j}\\\\\n", " &= &\\tfrac{1}{n}\n", " \\left\\{\n", " \\begin{array}{lcl}\n", " \\displaystyle\\sum_{j=0}^{n-1}1 &: &i = k\\\\\n", " \\dfrac{1 - r^{n}}{1 - r} &: &\\text{sonst}\\\\\n", " \\end{array}\n", " \\right.\\\\\n", " &= &\\tfrac{1}{n}\n", " \\left\\{\n", " \\begin{array}{lcl}\n", " n &: &i = k\\\\\n", " \\dfrac{1 - 1}{1 - r} &: &\\text{sonst}\\\\\n", " \\end{array}\n", " \\right.\\\\\n", " &= &\\left\\{\n", " \\begin{array}{lcl}\n", " 1 &: &i = k\\\\\n", " 0 &: &\\text{sonst}\\\\\n", " \\end{array}\n", " \\right.\\\\\n", " &= &\\delta_{ik}.\\\\\n", " \\end{array}\n", "$$\n", "\n", "Darum $\\mathcal{F}_{n}\\,u_{n,k}\n", " = \\left(\\begin{matrix}\n", " 0\\\\\n", " 0\\\\\n", " \\vdots\\\\\n", " 1\\\\\n", " \\vdots\\\\\n", " 0\\\\\n", " \\end{matrix}\\right)\n", " \\begin{matrix}\n", " \\\\\n", " \\\\\n", " \\\\\n", " \\leftarrow k\\\\\n", " \\\\\n", " \\\\\n", " \\end{matrix}\n", " = \\mathbf{e}_{k}$.\n" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "'''SEMINARAUFGABE 10.4 d)'''\n", "n = 4;\n", "t = linspace(0, n-1, n);\n", "i, j = np.meshgrid(t, t);\n", "theta = 2*pi * i * j / n;\n", "\n", "# definiere die Matrix F_n + die Vektoren u_nk\n", "F_n = (1/sqrt(n)) * exp(-1j * theta);\n", "u_n = [ (1/sqrt(n)) * exp(1j * theta[:, k]) for k in range(n) ];\n", "\n", "# Eingabe des Vektors:\n", "F_mal_u = np.asarray([100, -3, 0, -3]); # = F_n @ u\n", "\n", "'''\n", "NOTE: Folgende Koeffizienten wurden falsch erraten.\n", "TODO: Wie lauten die richtigen Werte? -> Aufgabe!\n", "'''\n", "c = [10,10,40,50];\n", "\n", "# Aufschreibung des Signals »bzgl. der harmonischen Basis«:\n", "# u = c[0]u_n[0] + c[1]u_n[1] + c[2]u_n[2] + c[3]u_n[3]\n", "u_guess = sum([ c[k] * u_n[k] for k in range(n) ]);\n", "\n", "# Anzeige der Lösung:\n", "display_signal(u_guess, 'Geschätztes Signal');\n", "\n", "# Verifikation von Lösung:\n", "print(dedent(\n", " f'''\n", " %error := ‖u_guess – u‖ / ‖u‖\n", " = ‖F_n · (u_guess – u)‖ / ‖F_n · u‖\n", " ... wieso?? [Stichwort: unitär]\n", " = ‖F_n · u_guess – F_n · u‖ / ‖F_n · u‖\n", " = {linalg.norm(F_n @ u_guess - F_mal_u)/linalg.norm(F_mal_u):.2%}\n", " '''\n", "));\n" ] } ], "metadata": { "kernelspec": { "display_name": "Python 3", "language": "python", "name": "python3" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.10.8" } }, "nbformat": 4, "nbformat_minor": 2 }