Wissensdatenbank Wirtschaftsrecht: history "TutoriumMathe3L9"

Revision history for TutoriumMathe3L9

Revision [73463]

Last edited on 2016-10-26 12:17:54 by Jorina Lossau

Additions:

2.3.4. F(s)=2/(s^2+4) I L^(-1) (..) -> f(t)=sin(2t)
2.3.5. F(s)=s/(s^2+4) I L^(-1) (..) -> f(t)=cos(2t)
2.3.6. F(s)=s/(s^2-4)
F(s)=s/(s^2-4)
s/(s^2-4)=A/(s+2)+B/(s-2)
s/(s^2-4)=((A(s-2)+B(s+2)))/(s+2)(s-2)
Zählervergleich: s=A(s-2)+B(s+2)
s=-2->-2=-4B
s1:1=A+B
s0:0=-2A+2B
A=1/2; B=1/2
F(s)=1/2, 1/(s+2)+1/2, 1/(s-2) I L^(-1) (..)
f(t)=1/2(e^(2t)+e^(-2t)
f(t)=cosh(2t)
2.3.7. F(s)=1/(s+2)^2
F(s)=1/(s+2)^2 I L^(-1) (..) Dämpfungsregel
f(t)=te^(-2t)

Revision [73462]

Edited on 2016-10-26 12:03:53 by Jorina Lossau

No Differences

Revision [73403]

Edited on 2016-10-24 20:00:06 by Jorina Lossau

Additions:

Transformieren Sie mit der Tabelle der Laplace-Korrespondenzen vom Bild- in den Zeitbereich. Falls nötig, zerlegen Sie die Ausdrücke zunächst in Partialbrüche
2.3.1. F(s)=1/(s+2) I L^(-1) (..) -> f(t)=e^(-2t)
2.3.2. F(s)=5/(s+2) I L^(-1) (..) -> f(t)=5e^(-2t)
2.3.3. F(s)=2/(s^2-4)
2/(s^2-4)=A/(s+2)+B/(s-2)
2/(s^2-4)=(A(s-2)+B(s+2))/((s+2)(s-2))
Zählervergleich: 2=A(s-2)+B(s+2)
1. Weg: Nullstellen einsetzen:
s=2->2=4B
s=-2->-4A
2. Weg: Koeffizientenvergleich:
s1:0=A+B
s0: 2=-2A+2B
A=-1/2; B=1/2
F(s)=(-1/2)1/(s+2)+(1/2)1/(s-2) I L^(-1)(..)
f(t)=1/2(e^(2t)-e^(-2t))
f(t)=sinh(2t)

Revision [73402]

Edited on 2016-10-24 19:30:08 by Jorina Lossau

Deletions:

Transformieren Sie mit der Tabelle der Laplace-Korrespondenzen vom Bild- in den Zeitbereich. Falls nötig, zerlegen Sie die Ausdrücke zunächst in Partialbrüche.
{{image url="Transformation6.jpg" width="550" class="left"}}
**Zählervergleich:**
2 = A(s-2) + B(s+2)
**1. Weg: Nullstellen einsetzen:**
s = 2--> 2 = 4B
s = -2-->2 = -4A
{{image url="Transformation7.jpg" width="500" class="left"}}
{{image url="Transformation8.jpg" width="550" class="left"}}
**Zählervergleich:**
s = A(s-2) + B(s+2)
**1. Weg: Nullstellen einsetzen**
s = 2 -->2 = 4B
s = -2 -->-2 = -4B
{{image url="Transformation9.jpg" width="500" class="left"}}
{{image url="Transformation10.jpg" width="200" class="left"}}
**1. Weg: Partialbruchzerlegung:**
{{image url="Transformation12.jpg" width="400" class="left"}}
**Zählervergleich:**
{{image url="Transformation13.jpg" width="250" class="left"}}
**2. Weg: Differentationsregel**
{{image url="Transformation14.jpg" width="300" class="left"}}
{{image url="Transformation15.jpg" width="350" class="left"}}
{{image url="Transformation16.jpg" width="200" class="left"}}
**1. Weg: Partialbruchzerlegung:**
{{image url="Transformation17.jpg" width="350" class="left"}}
**Zählervergleich:**
{{image url="Transformation18.jpg" width="350" class="left"}}
**2. Weg: Differentationsregel:**
{{image url="Transformation19.jpg" width="350" class="left"}}
{{image url="Transformation20.jpg" width="200" class="left"}}
**1. Weg: Partialbruchzerlegung:**
{{image url="Transformation21.jpg" width="300" class="left"}}
**Zählervergleich:**
{{image url="Transformation22.jpg" width="350" class="left"}}
**2. Weg: Differentationsregel:**
{{image url="Transformation23.jpg" width="350" class="left"}}
{{image url="Transformation24.jpg" width="350" class="left"}}
Der Nenner hat konjugiert komplexe Nullstellen. Deshalb ist quadratische Ergänzung zweckmäßiger als Partialbruchzerlegung.
{{image url="Transformation25.jpg" width="350" class="left"}}
{{image url="Transformation26.jpg" width="200" class="left"}}
Wie in 3.12 wird auch hier die quadratische Ergänzung benutzt
{{image url="Transformation27.jpg" width="500" class="left"}}
{{image url="Transformation28.jpg" width="200" class="left"}}
Wie in 3.12 wird auch hier die quadratische Ergänzung benutzt
{{image url="Transformation29.jpg" width="450" class="left"}}
-------------
**Aufgabe 2.4**
Führen Sie die Partialbruchzerlegung durch und transformieren Sie danach in den Zeitbereich.
{{image url="Transformation30.jpg" width="300" class="left"}}
**Zählervergleich:**
1 = A (s+1) + Bs
{{image url="Transformation31.jpg" width="350" class="left"}}
{{image url="Transformation32.jpg" width="400" class="left"}}
{{image url="Transformation33.jpg" width="500" class="left"}}
**Nullstellen einsetzen:**
s = -1 --> -1 = A
s = -2 --> -2 = -C
**Koeffizienten - Vergleich:**
{{image url="Transformation34.jpg" width="300" class="left"}}
{{image url="Transformation35.jpg" width="400" class="left"}}
**Zählervergleich:**
s = A(s²+2) + (Bs+C)(s+1)
**Nullstellen einsetzen:**
s = -1 --> -1 = 3A
{{image url="Transformation36.jpg" width="400" class="left"}}
{{image url="Transformation37.jpg" width="500" class="left"}}
**Nullstellen einsetzen:**
s = 1--> 5 = 34A
**Koeffizienten - Vergleich:**
{{image url="Transformation38.jpg" width="500" class="left"}}
------------------------------------------------
**Aufgabe 2.5**
Lösen Sie folgende Anfangswertaufgaben mit der Laplace-Transformation.
{{image url="Transformation39.jpg" width="550" class="left"}}
**Nullstellen einsetzen:**
{{image url="Transformation40.jpg" width="250" class="left"}}
{{image url="Transformation41.jpg" width="500" class="left"}}
{{image url="Transformation42.jpg" width="600" class="left"}}
**Zählervergleich: **
3 = A(s+1)(s-5) + B(s-2)(s-5)+C(s-2)(s+1)
**Nullstellen einsetzen:**
{{image url="Transformation43.jpg" width="350" class="left"}}
{{image url="Transformation44.jpg" width="600" class="left"}}
**Nullstellen einsetzen:**
s = 5 --> 3 = 204D
s = -1 --> 3 = -60C
**Koeffizienten - Vergleich:**
{{image url="Transformation45.jpg" width="500" class="left"}}
{{image url="Transformation46.jpg" width="600" class="left"}}
**Nullstellen einsetzen:**
s = 5 --> 5 = 246D
s = -1 --> -1 = -102C
**Koeffizienten - Vergleich:**
{{image url="Transformation47.jpg" width="600" class="left"}}
{{image url="Transformation48.jpg" width="600" class="left"}}
**Zählervergleich:**
7 = As(s-5)+Bs(s+1)+C(s+1)(s-5)
**Nullstellen einsetzen:**
{{image url="Transformation49.jpg" width="350" class="left"}}
{{image url="Transformation50.jpg" width="700" class="left"}}
**Nullstellen einsetzen:**
s = 5 --> 2 = 750B
s = -1 --> 2 = 6A
s = 0 --> 2 = -5E
**Koeffizienten - Vergleich:**
{{image url="Transformation51.jpg" width="500" class="left"}}
{{image url="Transformation52.jpg" width="650" class="left"}}
**Koeffizienten - Vergleich:**
{{image url="Transformation53.jpg" width="650" class="left"}}
{{image url="Transformation54.jpg" width="650" class="left"}}
**Nullstellen einsetzen:**
s = -3 --> 32 = -18B
s = -1 --> 4 = 2A
s = 0 --> 2 = 3D
**Achtung !!!**
Die zweite Anfangsbedingung y(0) = 0 wird nicht erfüllt. Überzeugen Sie sich davon, indem Sie die erste Ableitung von y bilden und t = 0 einsetzen.
Für die spezielle Lösung einer DGL 1. Ordnung ist i.a. nur eine Zusatzbedingung notwendig. Weitere Bedingungen könnten dann nur rein zufällig auch erfüllt sein, was aber hier nicht zutrifft.
{{image url="Transformation56.jpg" width="450" class="left"}}
{{image url="Transformation57.jpg" width="550" class="left"}}
Achtung !!! Wie in 2.5.9 ist die Anfangsbedingung y(0) = 0 nicht erfüllt !
{{image url="Transformation58.jpg" width="750" class="left"}}
**Zählervergleich:**
{{image url="Transformation59.jpg" width="750" class="left"}}
----------------------------------------------------------------
**Aufgabe 2.6**
Transformieren Sie in den Zeitbereich unter Verwendung der Faltungsregel!
{{image url="Transformation60.jpg" width="750" class="left"}}
{{image url="Transformation61.jpg" width="750" class="left"}}

Revision [73401]

Edited on 2016-10-24 19:17:24 by Jorina Lossau

Additions:

=lim(((-A/s)e^(-st)+(-A/(s+1)e^(-t(s+1)+3))von 3 bis b
=lim((-A/s)e^(-3s)+(A/s)e^0-A/(s+1)e^(-b(s+1)+3)+A/(s+1)e^(-3(s+1)+3))
=lim((-A/s)e^(-3s)+A/s-A/(s+1)e^(-b(s+1)+3)+A/(s+1)e^(-3s)
=(-A/s)e^(-3s)+A/s+A/(s+1)e^(-3s)

Deletions:

~~=lim(((-A/s)e^(-st)+(-A/(s+1)e^^~~

Revision [73400]

Edited on 2016-10-24 18:52:17 by Jorina Lossau

Additions:

F(s)=∫f(t)e^(-stdt wobei ∫ von 0 bis ∞
=lim∫(von 0 bis 2π)2e^(-st)dt+∫(von 2π bis b)2cos(3t)e^(-st)dt
=lim(((-2/s)e^(-st))(von 0 bis 2π)+(2/(s^2+9)e^(-st)(3sin(3t)-scos(3t)))von 2π bis b)
=(-2/s)e^(-2πs)-(2/s)e^0+lim(von b bis ∞)(2/(s^2+9)(3sin(3b)e^(-sb)-scos(3b)e^(-sb)-3sin(3*2π)e^(-2πs)-scos(3*2π)e^(-2πs))
=(-2/s)e^(-2πs)-2/s-(2se^(-2πs))/(s^2+9)
F(s)=∫f(t)e^(-st)dt wobei ∫ von 0 bis ∞
=lim∫Adt+∫Ae^(-t+3)e^(-st)dt=lim∫Adt+∫Ae^(-t(s+1)+1)dt
=lim(((-A/s)e^(-st)+(-A/(s+1)e^^

Revision [73398]

Edited on 2016-10-24 17:04:36 by Jorina Lossau

Additions:

**2.1.1.**

f(t)=4e^(-3t)
->F(s)=4/(s+3)
->F(s)=1/s+2/s^2+2/s^3
->F(s)=1/(s+2)+2/(s+2)^2+2/(s+2)^3
->F(s)=1/(s-1)+1/(s+1)
->F(s)=2s/(s^2-1)
(=sinh(t))
->F(s)=1/2((1/(s-1)-1/(s+1))
->F(s)=1/2((2/(s^2-1))
->F(s)=1/(s^2-1)
->F(s)=2/(s^2+4)+3s/(s^2+4)=(2+3s)/(s^2+4)
->F(s)=(2/(s+4^2+4))+(3(s+4)/((s+4)^2+4))=(3s+14)/(s^2+8s+20)

Deletions:

~~2.1.1. f(t)=4e^(-3t)~~

Revision [73397]

Edited on 2016-10-24 15:02:25 by Jorina Lossau

Additions:

2.1.1. f(t)=4e^(-3t)
2.1.2. f(t)=1+2t+t^2
2.1.3. f(t)=(1+2t+t^2)e^(-2t)
2.1.4. f(t)=e^t+e^(-t)
2.1.5. f(t)=1/2(e^t-e^(-t))
2.1.6. f(t)=sin(2t)+3cos(2t)
2.1.7. f(t)=e^(-4t)[sin(2t)+3cos(2t)]
2.2.1. f(t)=
(0; t<0)
(2; 0<t<2π)
(2cos(3t), 2π<t)
2.2.2. f(t)=
(0, t<0)
(A, 0<t<3)
(Ae^(-t-3), 3<t)

Revision [58942]

Edited on 2015-09-06 21:31:25 by Jorina Lossau

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~~{{files}}~~

Revision [58941]

Edited on 2015-09-06 21:29:20 by Jorina Lossau

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Revision [58940]

Edited on 2015-09-06 21:27:33 by Jorina Lossau

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----------------------------------------------------------------
**Aufgabe 2.6**
Transformieren Sie in den Zeitbereich unter Verwendung der Faltungsregel!

Revision [58939]

Edited on 2015-09-06 21:24:05 by Jorina Lossau

No Differences

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Edited on 2015-09-06 21:23:00 by Jorina Lossau

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Edited on 2015-09-06 21:21:10 by Jorina Lossau

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Revision [58936]

Edited on 2015-09-06 21:18:30 by Jorina Lossau

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Achtung !!! Wie in 2.5.9 ist die Anfangsbedingung y(0) = 0 nicht erfüllt !

Revision [58935]

Edited on 2015-09-06 21:16:20 by Jorina Lossau

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Edited on 2015-09-06 21:14:13 by Jorina Lossau

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~~{{image url="Transformation55.jpg" width="650" class="left"}}~~

Revision [58933]

Edited on 2015-09-06 21:12:53 by Jorina Lossau

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Revision [58932]

Edited on 2015-09-06 21:08:45 by Jorina Lossau

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**Achtung !!!**
Die zweite Anfangsbedingung y(0) = 0 wird nicht erfüllt. Überzeugen Sie sich davon, indem Sie die erste Ableitung von y bilden und t = 0 einsetzen.
Für die spezielle Lösung einer DGL 1. Ordnung ist i.a. nur eine Zusatzbedingung notwendig. Weitere Bedingungen könnten dann nur rein zufällig auch erfüllt sein, was aber hier nicht zutrifft.

Revision [58931]

Edited on 2015-09-06 20:45:54 by Jorina Lossau

Additions:

s = -3 --> 32 = -18B
s = -1 --> 4 = 2A
s = 0 --> 2 = 3D

Revision [58930]

Edited on 2015-09-06 20:32:08 by Jorina Lossau

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Revision [58929]

Edited on 2015-09-06 20:15:20 by Jorina Lossau

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Edited on 2015-09-06 20:13:29 by Jorina Lossau

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~~{{image url="Transformation52.jpg" width="500" class="left"}}~~

Revision [58927]

Edited on 2015-09-06 20:12:27 by Jorina Lossau

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Revision [58926]

Edited on 2015-09-06 20:04:39 by Jorina Lossau

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~~{{image url="Transformation51.jpg" width="400" class="left"}}~~

Revision [58925]

Edited on 2015-09-06 20:03:29 by Jorina Lossau

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Revision [58924]

Edited on 2015-09-06 20:02:36 by Jorina Lossau

Additions:

s = 5 --> 2 = 750B
s = -1 --> 2 = 6A
s = 0 --> 2 = -5E

Revision [58923]

Edited on 2015-09-06 19:48:58 by Jorina Lossau

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~~{{image url="Transformation50.jpg" width="350" class="left"}}~~

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Edited on 2015-09-06 19:47:21 by Jorina Lossau

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Edited on 2015-09-06 19:37:22 by Jorina Lossau

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~~{{image url="Transformation49.jpg" width="600" class="left"}}~~

Revision [58920]

Edited on 2015-09-06 19:36:17 by Jorina Lossau

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Revision [58919]

Edited on 2015-09-06 19:33:35 by Jorina Lossau

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7 = As(s-5)+Bs(s+1)+C(s+1)(s-5)

Revision [58918]

Edited on 2015-09-06 19:01:23 by Jorina Lossau

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Revision [58917]

Edited on 2015-09-06 18:43:59 by Jorina Lossau

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Revision [58916]

Edited on 2015-09-06 18:42:55 by Jorina Lossau

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s = 5 --> 5 = 246D
s = -1 --> -1 = -102C

Revision [58915]

Edited on 2015-09-06 18:26:30 by Jorina Lossau

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{{image url="Transformation45.jpg" width="500" class="left"}}
{{image url="Transformation46.jpg" width="600" class="left"}}

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~~{{image url="Transformation45.jpg" width="450" class="left"}}~~

Revision [58914]

Edited on 2015-09-06 18:21:55 by Jorina Lossau

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~~{{image url="Transformation45.jpg" width="600" class="left"}}~~

Revision [58913]

Edited on 2015-09-06 18:20:08 by Jorina Lossau

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Revision [58912]

Edited on 2015-09-06 18:18:48 by Jorina Lossau

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s = 5 --> 3 = 204D
s = -1 --> 3 = -60C

Revision [58910]

Edited on 2015-09-06 16:49:08 by Jorina Lossau

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Revision [58909]

Edited on 2015-09-06 16:46:27 by Jorina Lossau

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~~{{image url="Transformation43.jpg" width="600" class="left"}}~~

Revision [58908]

Edited on 2015-09-06 16:45:12 by Jorina Lossau

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Revision [58907]

Edited on 2015-09-06 16:31:20 by Jorina Lossau

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**Zählervergleich: **
3 = A(s+1)(s-5) + B(s-2)(s-5)+C(s-2)(s+1)

Revision [58906]

Edited on 2015-09-06 16:23:16 by Jorina Lossau

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Edited on 2015-09-06 16:10:03 by Jorina Lossau

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Edited on 2015-09-06 16:01:09 by Jorina Lossau

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~~{{image url="Transformation40.jpg" width="550" class="left"}}~~

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Edited on 2015-09-06 15:58:35 by Jorina Lossau

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Edited on 2015-09-06 15:41:12 by Jorina Lossau

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Revision [58901]

Edited on 2015-09-06 15:25:59 by Jorina Lossau

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------------------------------------------------
**Aufgabe 2.5**
Lösen Sie folgende Anfangswertaufgaben mit der Laplace-Transformation.

Revision [58900]

Edited on 2015-09-06 15:22:46 by Jorina Lossau

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Revision [58899]

Edited on 2015-09-06 15:17:33 by Jorina Lossau

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s = 1--> 5 = 34A

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Edited on 2015-09-06 15:13:27 by Jorina Lossau

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Edited on 2015-09-06 14:51:39 by Jorina Lossau

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~~{{image url="Transformation36.jpg" width="450" class="left"}}~~

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Edited on 2015-09-06 14:50:51 by Jorina Lossau

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~~{{image url="Tranformation36.jpg" width="450" class="left"}}~~

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Edited on 2015-09-06 14:50:00 by Jorina Lossau

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Revision [58894]

Edited on 2015-09-06 14:46:43 by Jorina Lossau

No Differences

Revision [58893]

Edited on 2015-09-06 14:45:43 by Jorina Lossau

Additions:

s = A(s²+2) + (Bs+C)(s+1)
s = -1 --> -1 = 3A

Revision [58892]

Edited on 2015-09-06 14:37:19 by Jorina Lossau

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~~{{image url="Transformation35.jpg" width="300" class="left"}}~~

Revision [58891]

Edited on 2015-09-06 14:35:45 by Jorina Lossau

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Revision [58890]

Edited on 2015-09-06 14:29:03 by Jorina Lossau

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~~{{image url="Transformation34.jpg" width="500" class="left"}}~~

Revision [58889]

Edited on 2015-09-06 14:18:33 by Jorina Lossau

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**Nullstellen einsetzen:**
s = -1 --> -1 = A
s = -2 --> -2 = -C
**Koeffizienten - Vergleich:**
{{image url="Transformation34.jpg" width="500" class="left"}}

Revision [58888]

Edited on 2015-09-06 13:53:23 by Jorina Lossau

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Revision [58887]

Edited on 2015-09-06 13:49:36 by Jorina Lossau

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Revision [58886]

Edited on 2015-09-06 13:42:36 by Jorina Lossau

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**Nullstellen einsetzen:**
{{image url="Transformation31.jpg" width="300" class="left"}}

Revision [58885]

Edited on 2015-09-06 13:41:28 by Jorina Lossau

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**Nullstellen einsetzen:**
{{image url="Transformation31.jpg" width="300" class="left"}}

Revision [58884]

Edited on 2015-09-06 13:32:23 by Jorina Lossau

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1 = A (s+1) + Bs

Revision [58883]

Edited on 2015-09-06 13:17:19 by Jorina Lossau

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~~{{image url="Transformation30.jpg" width="450" class="left"}}~~

Revision [58882]

Edited on 2015-09-06 13:16:18 by Jorina Lossau

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**Aufgabe 2.4**
{{image url="Transformation30.jpg" width="450" class="left"}}

Deletions:

~~**Aufgabe 2.4:**~~

Revision [58881]

Edited on 2015-09-06 13:14:01 by Jorina Lossau

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---------------------
-------------------
-------------
**Aufgabe 2.4:**
Führen Sie die Partialbruchzerlegung durch und transformieren Sie danach in den Zeitbereich.

Revision [58880]

Edited on 2015-09-06 12:52:10 by Jorina Lossau

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~~{{image url="Transformation29.jpg" width="300" class="left"}}~~

Revision [58879]

Edited on 2015-09-06 12:50:12 by Jorina Lossau

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Edited on 2015-09-06 12:42:34 by Jorina Lossau

No Differences

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Edited on 2015-09-06 12:40:46 by Jorina Lossau

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Revision [58876]

Edited on 2015-09-06 12:37:30 by Jorina Lossau

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~~{{image url="Transformation27.jpg" width="200" class="left"}}~~

Revision [58875]

Edited on 2015-09-06 12:36:04 by Jorina Lossau

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Wie in 3.12 wird auch hier die quadratische Ergänzung benutzt
{{image url="Transformation27.jpg" width="200" class="left"}}

Revision [58874]

Edited on 2015-09-06 08:59:59 by Jorina Lossau

No Differences

Revision [58873]

Edited on 2015-09-06 08:59:29 by Jorina Lossau

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~~{{image url="Transformation26.jpg" width="350" class="left"}}~~

Revision [58872]

Edited on 2015-09-06 08:55:05 by Jorina Lossau

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Revision [58871]

Edited on 2015-09-06 08:45:43 by Jorina Lossau

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Der Nenner hat konjugiert komplexe Nullstellen. Deshalb ist quadratische Ergänzung zweckmäßiger als Partialbruchzerlegung.
{{image url="Transformation25.jpg" width="350" class="left"}}

Deletions:

~~Der Nenner hat konjugiert komplexe Nullstellen. Deshalb ist quadratische Ergänzung zweckmäßiger als Partialbruchzerlegung~~

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Der Nenner hat konjugiert komplexe Nullstellen. Deshalb ist quadratische Ergänzung zweckmäßiger als Partialbruchzerlegung

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~~{{image url="Transformation24jpg" width="350" class="left"}}~~

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~~{{image url="Transformation22.jpg" width="300" class="left"}}~~

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~~{{image url="Transformation21.jpg" width="200" class="left"}}~~

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**2. Weg: Differentationsregel:**
{{image url="Transformation19.jpg" width="350" class="left"}}

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~~{{image url="Transformation17.jpg" width="200" class="left"}}~~

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~~{{image url="Transformation15.jpg" width="400" class="left"}}~~

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~~{{image url="Transformation14.jpg" width="250" class="left"}}~~

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Edited on 2015-09-05 10:02:22 by Jorina Lossau

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**2. Weg: Differentationsregel**
{{image url="Transformation14.jpg" width="250" class="left"}}

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~~{{image url="Transformation13.jpg" width="400" class="left"}}~~

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~~{{image url="Transformation11.jpg" width="200" class="left"}}~~

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**1. Weg: Partialbruchzerlegung:**

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~~{{image url="Transformation9.jpg" width="550" class="left"}}~~

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s = A(s-2) + B(s+2)
**1. Weg: Nullstellen einsetzen**
s = 2 -->2 = 4B
s = -2 -->-2 = -4B
{{image url="Transformation9.jpg" width="550" class="left"}}

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~~{{image url="Transformation7.jpg" width="550" class="left"}}~~

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**1. Weg: Nullstellen einsetzen:**
s = 2--> 2 = 4B
s = -2-->2 = -4A

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**Zählervergleich:**
2 = A(s-2) + B(s+2)

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**Aufgabe 2.3**
Transformieren Sie mit der Tabelle der Laplace-Korrespondenzen vom Bild- in den Zeitbereich. Falls nötig, zerlegen Sie die Ausdrücke zunächst in Partialbrüche.
{{image url="Transformation6.jpg" width="550" class="left"}}

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~~{{image url="Tranformation4.jpg" width="450" class="left"}}~~

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~~{{image url="Transformation3.jpg" width="400" class="left"}}~~

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**Aufgabe 2.2**
Berechnen Sie die Bildfunktion unter Verwendung des Integrals, welches die Laplace-Transformation definiert.
{{image url="Transformation3.jpg" width="400" class="left"}}

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{{image url="Tranformation1.jpg" width="450" class="left"}}
{{image url="Transformation2.jpg" width="400" class="left"}}

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{{image url="Tranformation1.jpg" width="450" class="center"}}
{{image url="Transformation2.jpg" width="400" class="center"}}

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~~{{image url="Tranformation1.jpg" width="400" class="center"}}~~

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~~{{image url="Tranformation1.jpg" width="550" class="center"}}~~

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~~{{image url="Transformation1.jpg" width="550" class="center"}}~~

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||**Aufgabe 2.1**
Nachfolgend für t ≥ 0 definierte Funktionen haben für t < 0 den Funktionswert f(t) = 0. Transformieren Sie mit der Tabelle
in den Bildbereich der Laplace-Transformation!
{{image url="Transformation1.jpg" width="550" class="center"}}
||

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~~|| ||~~

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|| ||

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{{image url="Mathe3L101.jpg" width="600" class="center"}}
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{{image url="Mathe3L108.jpg" width="600" class="center"}}
{{image url="Mathe3L109.jpg" width="600" class="center"}}
{{image url="Mathe3L1010.jpg" width="600" class="center"}}
{{image url="Mathe3L1011.jpg" width="600" class="center"}}
{{image url="Math3L1012.jpg" width="600" class="center"}}
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{{image url="Mathe3L1015.jpg" width="600" class="center"}}
{{image url="Mathe3L1016.jpg" width="600" class="center"}}
{{image url="Mathe3L1017.jpg" width="600" class="center"}}
{{image url="Mathe3L1018.jpg" width="600" class="center"}}
{{image url="Mathe3L1019.jpg" width="600" class="center"}}
{{image url="Mathe3L1020.jpg" width="600" class="center"}}
{{image url="Mathe3L1021.jpg" width="600" class="center"}}
{{image url="Mathe3L1022.jpg" width="600" class="center"}}
{{image url="Mathe3L1023.jpg" width="600" class="center"}}
{{image url="Mathe3L1024.jpg" width="600" class="center"}}
{{image url="Mathe3L1025.jpg" width="600" class="center"}}

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{{image url="Mathe3L102.jpg" width="600" class="center"}}
{{image url="Mathe3L103.jpg" width="600" class="center"}}
{{image url="Mathe3L104.jpg" width="600" class="center"}}
{{image url="Mathe3L105.jpg" width="600" class="center"}}
{{image url="Mathe3L106.jpg" width="600" class="center"}}
{{image url="Math3L107.jpg" width="600" class="center"}}
{{image url="Mathe3L108.jpg" width="600" class="center"}}
{{image url="Mathe3L109.jpg" width="600" class="center"}}
{{image url="Mathe3L1010.jpg" width="600" class="center"}}
{{image url="Mathe3L1011.jpg" width="600" class="center"}}
{{image url="Math3L1012.jpg" width="600" class="center"}}
{{image url="Mathe3L1013.jpg" width="600" class="center"}}
{{image url="Mathe3L1014.jpg" width="600" class="center"}}
{{image url="Mathe3L1015.jpg" width="600" class="center"}}
{{image url="Mathe3L1016.jpg" width="600" class="center"}}
{{image url="Mathe3L1017.jpg" width="600" class="center"}}
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{{image url="Mathe3L1024.jpg" width="600" class="center"}}
{{image url="Mathe3L1025.jpg" width="600" class="center"}}
||**{{files download="Mathe3L10.pdf"text="PDF Dokument Lösungen Laplace - Transformation"}}**||

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{{files}}
||**{{files download="Mathe3L10.pdf"text="PDF Dokument Aufgaben Integralrechnung"}}**||

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{{files}}
{{image url="Mathe3L101.jpg" width="600" class="center"}}
{{image url="Mathe3L101.jpg" width="600" class="center"}}
{{image url="Mathe3L101.jpg" width="600" class="center"}}
{{image url="Mathe3L101.jpg" width="600" class="center"}}
{{image url="Mathe3L101.jpg" width="600" class="center"}}
{{image url="Mathe3L101.jpg" width="600" class="center"}}
{{image url="Mathe3L101.jpg" width="600" class="center"}}
{{image url="Mathe3L101.jpg" width="600" class="center"}}
{{image url="Mathe3L101.jpg" width="600" class="center"}}
{{image url="Mathe3L101.jpg" width="600" class="center"}}
{{image url="Mathe3L101.jpg" width="600" class="center"}}
{{image url="Mathe3L101.jpg" width="600" class="center"}}
{{image url="Mathe3L101.jpg" width="600" class="center"}}
{{image url="Mathe3L101.jpg" width="600" class="center"}}
||**{{files download="Mathe3L10.pdf"text="PDF Dokument Aufgaben Integralrechnung"}}**||
**>>[[Mathe3Tutorien Zurück zur Auswahl]]>>**

Revision [43615]

The oldest known version of this page was created on 2014-08-26 01:01:46 by Jorina Lossau

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