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7d153fb5
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7d153fb5
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7 months ago
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Sebastian Schwarz
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sys2-jupyter-notebooks/exam_examples/V06.2.ipynb
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7d153fb5
{
"cells": [
{
"cell_type": "markdown",
"id": "1f0f7673-ab1a-413f-b73c-0a248ba39ed9",
"metadata": {},
"source": [
"# <span style='color:OrangeRed'>V6 NORMALFORMEN IN ZUSTANDRAUMDARSTELLUNG TEIL 2</span>"
]
},
{
"cell_type": "markdown",
"id": "9653081e-df65-4ccd-9204-e11c542a469f",
"metadata": {},
"source": [
"<div style=\"font-family: 'times'; font-size: 13pt; text-align: justify\">\n",
"Es ist ein lineares System gegeben, das im Zustandsraum folgende Beschreibung hat:"
]
},
{
"cell_type": "code",
"execution_count": 1,
"id": "825956d9-8873-4d83-bf39-7c8d40da2aff",
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"a = 5\n",
"b = 1\n",
"c = 2\n",
"A =\n",
"\n",
" 2 0\n",
" 1 1\n",
"\n",
"B =\n",
"\n",
" 1\n",
" 5\n",
"\n",
"C =\n",
"\n",
" 0 1\n",
"\n",
"D = 0\n"
]
}
],
"source": [
"clear all\n",
"pkg load symbolic\n",
"\n",
"a = 5\n",
"b = 1\n",
"c = 2\n",
"\n",
"A = [c 0;1 b]\n",
"B = [b; a]\n",
"C = [0 1]\n",
"D = 0"
]
},
{
"cell_type": "markdown",
"id": "a27e62db-897e-41e8-9b56-fd8c6b48f3bf",
"metadata": {},
"source": [
"<div style=\"font-family: 'times'; font-size: 13pt; text-align: justify\">\n",
"Finden Sie eine Transformation, die eine diagonale Form ergibt."
]
},
{
"cell_type": "markdown",
"id": "1943a7dc-b837-4623-8792-a1775026155d",
"metadata": {},
"source": [
"<div style=\"font-family: 'times'; font-size: 13pt; text-align: justify\">\n",
"Zunächst müssen wir die Eigenwerte des Systems finden:"
]
},
{
"cell_type": "code",
"execution_count": 2,
"id": "ef3d3be5-73cd-48d2-a773-d177b638230a",
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"p =\n",
"\n",
" 2\n",
" 1\n",
"\n"
]
}
],
"source": [
"p = eigs(A)"
]
},
{
"cell_type": "markdown",
"id": "a160dc2c-e6be-4329-9a7a-69f7689a7354",
"metadata": {},
"source": [
"Wir müssen nun die beiden Eigenvektoren berechnen, die den beiden Eigenwerten zugeordnet sind:"
]
},
{
"cell_type": "code",
"execution_count": 4,
"id": "1ed03ed5-3913-4107-95c9-f48e52bb50ee",
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"V =\n",
"\n",
" 0.00000 0.70711\n",
" 1.00000 0.70711\n",
"\n",
"D =\n",
"\n",
"Diagonal Matrix\n",
"\n",
" 1 0\n",
" 0 2\n",
"\n"
]
}
],
"source": [
"[V D]=eig(A)"
]
},
{
"cell_type": "markdown",
"id": "5cea5a50-7f13-4d9e-b343-de64bacb84f7",
"metadata": {},
"source": [
"Wir können das Ergebnis auch überprufen:"
]
},
{
"cell_type": "code",
"execution_count": 5,
"id": "5ef87d39-8ab8-4ce0-adc2-d392311d8a9c",
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"At =\n",
"\n",
" 1 0\n",
" 0 2\n",
"\n"
]
}
],
"source": [
"At = inv(V)*A*V"
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "e81354c4-0adf-4909-a490-32f2ee713788",
"metadata": {},
"outputs": [],
"source": []
}
],
"metadata": {
"kernelspec": {
"display_name": "Octave",
"language": "octave",
"name": "octave"
},
"language_info": {
"file_extension": ".m",
"help_links": [
{
"text": "GNU Octave",
"url": "https://www.gnu.org/software/octave/support.html"
},
{
"text": "Octave Kernel",
"url": "https://github.com/Calysto/octave_kernel"
},
{
"text": "MetaKernel Magics",
"url": "https://metakernel.readthedocs.io/en/latest/source/README.html"
}
],
"mimetype": "text/x-octave",
"name": "octave",
"version": "5.2.0"
}
},
"nbformat": 4,
"nbformat_minor": 5
}
%% Cell type:markdown id:1f0f7673-ab1a-413f-b73c-0a248ba39ed9 tags:
# <span style='color:OrangeRed'>V6 NORMALFORMEN IN ZUSTANDRAUMDARSTELLUNG TEIL 2</span>
%% Cell type:markdown id:9653081e-df65-4ccd-9204-e11c542a469f tags:
<div
style=
"font-family: 'times'; font-size: 13pt; text-align: justify"
>
Es ist ein lineares System gegeben, das im Zustandsraum folgende Beschreibung hat:
%% Cell type:code id:825956d9-8873-4d83-bf39-7c8d40da2aff tags:
```
octave
clear all
pkg load symbolic
a = 5
b = 1
c = 2
A = [c 0;1 b]
B = [b; a]
C = [0 1]
D = 0
```
%% Output
a = 5
b = 1
c = 2
A =
2 0
1 1
B =
1
5
C =
0 1
D = 0
%% Cell type:markdown id:a27e62db-897e-41e8-9b56-fd8c6b48f3bf tags:
<div
style=
"font-family: 'times'; font-size: 13pt; text-align: justify"
>
Finden Sie eine Transformation, die eine diagonale Form ergibt.
%% Cell type:markdown id:1943a7dc-b837-4623-8792-a1775026155d tags:
<div
style=
"font-family: 'times'; font-size: 13pt; text-align: justify"
>
Zunächst müssen wir die Eigenwerte des Systems finden:
%% Cell type:code id:ef3d3be5-73cd-48d2-a773-d177b638230a tags:
```
octave
p = eigs(A)
```
%% Output
p =
2
1
%% Cell type:markdown id:a160dc2c-e6be-4329-9a7a-69f7689a7354 tags:
Wir müssen nun die beiden Eigenvektoren berechnen, die den beiden Eigenwerten zugeordnet sind:
%% Cell type:code id:1ed03ed5-3913-4107-95c9-f48e52bb50ee tags:
```
octave
[V D]=eig(A)
```
%% Output
V =
0.00000 0.70711
1.00000 0.70711
D =
Diagonal Matrix
1 0
0 2
%% Cell type:markdown id:5cea5a50-7f13-4d9e-b343-de64bacb84f7 tags:
Wir können das Ergebnis auch überprufen:
%% Cell type:code id:5ef87d39-8ab8-4ce0-adc2-d392311d8a9c tags:
```
octave
At = inv(V)*A*V
```
%% Output
At =
1 0
0 2
%% Cell type:code id:e81354c4-0adf-4909-a490-32f2ee713788 tags:
```
octave
```
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