Laplace Transform - Revision history

Brian.Roath: /* Comments */

2010-02-04T05:46:30Z

Comments

← Older revision		Revision as of 22:46, 3 February 2010
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	*Also, the solution <math> y(t)= -\frac {2} {5} - \frac {1} {4} e^{-3t} - \frac {7} {20} e^{5t} </math> does not match the initial condition of <math>\ y(0)=1</math>.		*Also, the solution <math> y(t)= -\frac {2} {5} - \frac {1} {4} e^{-3t} - \frac {7} {20} e^{5t} </math> does not match the initial condition of <math>\ y(0)=1</math>.

	** We'~~ll fix this ASAP~~.		** Yeah, you're right. A minor sign error was what messed it up. It should be fixed now.

Brian.Roath: /* Example */

2010-02-04T05:45:58Z

Example

Brian.Roath: /* References */

2010-02-04T05:38:26Z

References

← Older revision		Revision as of 22:38, 3 February 2010
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	DeCarlo, Raymond A.; Lin, Pen-Min (2001), Linear Circuit Analysis, Oxford University Press, ISBN 0-19-513666-7 .		DeCarlo, Raymond A.; Lin, Pen-Min (2001), Linear Circuit Analysis, Oxford University Press, ISBN 0-19-513666-7 .

	Zill, Dennis G., A First Course in Differential Equations with Modeling Applications, Brooks/Cole Cengage Learning, ISBN 0-495-10824-3 .		Zill, Dennis G. (2005), A First Course in Differential Equations with Modeling Applications Ninth Edition, Brooks/Cole Cengage Learning, ISBN 0-495-10824-3 .

	== External links ==		== External links ==

Brian.Roath: /* References */

2010-02-04T05:37:23Z

References

← Older revision		Revision as of 22:37, 3 February 2010
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	==References==		==References==
	DeCarlo, Raymond A.; Lin, Pen-Min (2001), Linear Circuit Analysis, Oxford University Press, ISBN 0-19-513666-7 .		DeCarlo, Raymond A.; Lin, Pen-Min (2001), Linear Circuit Analysis, Oxford University Press, ISBN 0-19-513666-7 .

			Zill, Dennis G., A First Course in Differential Equations with Modeling Applications, Brooks/Cole Cengage Learning, ISBN 0-495-10824-3 .

	== External links ==		== External links ==
	*[http://www.intmath.com/Laplace-transformation/Intro.php The Laplace Transform].		*[http://www.intmath.com/Laplace-transformation/Intro.php The Laplace Transform].

Brian.Roath: /* Example */

2010-02-04T05:35:25Z

Example

← Older revision		Revision as of 22:35, 3 February 2010
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	Continuing on using the method of partial fractions, the equation is progressed:		Continuing on using the method of partial fractions, the equation is progressed:
	:<math> \frac {6+s-s^2} {s(s+3)(s-5)} = ~~\left(~~\frac {A} {s}~~\right) \left(~~\frac {B} {s+3}~~\right) \left(~~\frac {C} {s-5}~~\right)~~</math>		:<math> \frac {6+s-s^2} {s(s+3)(s-5)} = \frac {A} {s} + \frac {B} {s+3} + \frac {C} {s-5}</math>

Brian.Roath: /* Comments */

2010-02-04T05:34:10Z

Comments

← Older revision		Revision as of 22:34, 3 February 2010
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	*Nice list of transforms! Where did you find it? Inside the back cover of the textbook has a good list, but none including transforms of <math>\ g(t)</math>. I see your reference to the textbook. What page?		*Nice list of transforms! Where did you find it? Inside the back cover of the textbook has a good list, but none including transforms of <math>\ g(t)</math>. I see your reference to the textbook. What page?

			** Oops! It seems I forgot to add a reference to the list, thanks for mentioning it. I got the others from a list in the back of the ODE textbook. I'll list the reference up above.


	*I believe that		*I believe that
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	:<math> \frac {6+s-s^2} {s(s+3)(s-5)} = \frac {A} {s}+ \frac {B} {s+3}+ \frac {C} {s-5}</math>		:<math> \frac {6+s-s^2} {s(s+3)(s-5)} = \frac {A} {s}+ \frac {B} {s+3}+ \frac {C} {s-5}</math>

			** Thank you for pointing that out. It should be fixed now.

	*Also, the solution <math> y(t)= -\frac {2} {5} - \frac {1} {4} e^{-3t} - \frac {7} {20} e^{5t} </math> does not match the initial condition of <math>\ y(0)=1</math>.		*Also, the solution <math> y(t)= -\frac {2} {5} - \frac {1} {4} e^{-3t} - \frac {7} {20} e^{5t} </math> does not match the initial condition of <math>\ y(0)=1</math>.

			** We'll fix this ASAP.

John.hawkins at 09:30, 22 January 2010

2010-01-22T09:30:51Z

← Older revision		Revision as of 02:30, 22 January 2010
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	==Comments==		==Comments==
			John Hawkins:

	*~~John Hawkins:~~ Nice list of transforms! Where did you find it? Inside the back cover of the textbook has a good list, but none including transforms of <math>\ g(t)</math>. I see your reference to the textbook. What page?		*Nice list of transforms! Where did you find it? Inside the back cover of the textbook has a good list, but none including transforms of <math>\ g(t)</math>. I see your reference to the textbook. What page?

	*~~John Hawkins:~~ I believe that		*I believe that

	:<math> \frac {6+s-s^2} {s(s+3)(s-5)} = \left(\frac {A} {s}\right) \left(\frac {B} {s+3}\right) \left(\frac {C} {s-5}\right)</math>		:<math> \frac {6+s-s^2} {s(s+3)(s-5)} = \left(\frac {A} {s}\right) \left(\frac {B} {s+3}\right) \left(\frac {C} {s-5}\right)</math>
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	:<math> \frac {6+s-s^2} {s(s+3)(s-5)} = \frac {A} {s}+ \frac {B} {s+3}+ \frac {C} {s-5}</math>		:<math> \frac {6+s-s^2} {s(s+3)(s-5)} = \frac {A} {s}+ \frac {B} {s+3}+ \frac {C} {s-5}</math>

	*~~John Hawkins:~~ Also, the solution <math> y(t)= -\frac {2} {5} - \frac {1} {4} e^{-3t} - \frac {7} {20} e^{5t} </math> does not match the initial condition of <math>\ y(0)=1</math>.		*Also, the solution <math> y(t)= -\frac {2} {5} - \frac {1} {4} e^{-3t} - \frac {7} {20} e^{5t} </math> does not match the initial condition of <math>\ y(0)=1</math>.

John.hawkins: Comments

2010-01-22T07:53:32Z

Comments

← Older revision		Revision as of 00:53, 22 January 2010
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	==Comments==		==Comments==

	*John Hawkins: I believe that ~~instead of~~		*John Hawkins: Nice list of transforms! Where did you find it? Inside the back cover of the textbook has a good list, but none including transforms of <math>\ g(t)</math>. I see your reference to the textbook. What page?

			*John Hawkins: I believe that

	:<math> \frac {6+s-s^2} {s(s+3)(s-5)} = \left(\frac {A} {s}\right) \left(\frac {B} {s+3}\right) \left(\frac {C} {s-5}\right)</math>		:<math> \frac {6+s-s^2} {s(s+3)(s-5)} = \left(\frac {A} {s}\right) \left(\frac {B} {s+3}\right) \left(\frac {C} {s-5}\right)</math>

	~~you wanted to write~~		should be

			:<math> \frac {6+s-s^2} {s(s+3)(s-5)} = \frac {A} {s}+ \frac {B} {s+3}+ \frac {C} {s-5}</math>

	:<math> \frac {~~6+s-s^~~2} {~~s(s+3)(s-~~5)} ~~= \left(~~\frac {A} {s}~~\right)+ \left(~~\frac {B} {~~s+3~~}~~\right)+ \left(\frac~~ {~~C} {s-5~~}\~~right~~)</math>		*John Hawkins: Also, the solution <math> y(t)= -\frac {2} {5} - \frac {1} {4} e^{-3t} - \frac {7} {20} e^{5t} </math> does not match the initial condition of <math>\ y(0)=1</math>.

John.hawkins: Comment

2010-01-22T07:39:37Z

Comment

← Older revision		Revision as of 00:39, 22 January 2010
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	Jaymin Joseph		Jaymin Joseph
			<br />

			John Hawkins

			==Comments==

			*John Hawkins: I believe that instead of

			:<math> \frac {6+s-s^2} {s(s+3)(s-5)} = \left(\frac {A} {s}\right) \left(\frac {B} {s+3}\right) \left(\frac {C} {s-5}\right)</math>

			you wanted to write

			:<math> \frac {6+s-s^2} {s(s+3)(s-5)} = \left(\frac {A} {s}\right)+ \left(\frac {B} {s+3}\right)+ \left(\frac {C} {s-5}\right)</math>

Frohro: /* Sample Functions */

2010-01-21T02:48:53Z

Sample Functions

← Older revision		Revision as of 19:48, 20 January 2010
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	== Sample Functions ==		== Sample Functions ==


	The following is a list of commonly seen functions of which the Laplace transform is taken. The start function is noted within the Laplace symbol <math> \mathcal{L} \left\{ \right\} </math>.		The following is a list of commonly seen functions of which the Laplace transform is taken. The start function is noted within the Laplace symbol <math> \mathcal{L} \left\{ \right\} </math>.

← Older revision		Revision as of 22:45, 3 February 2010
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	Using association, the equation is rearranged:		Using association, the equation is rearranged:
	:<math> (s^2-2s-15)\mathcal{L} \left\{y\right\}= \frac {6} {s} -s+3-2</math>		:<math> (s^2-2s-15)\mathcal{L} \left\{y\right\}= \frac {6} {s} +s+3-2</math>

	Continuing on using the method of partial fractions, the equation is progressed:		Continuing on using the method of partial fractions, the equation is progressed:
	:<math> \frac {~~6+s-~~s^2} {s(s+3)(s-5)} = \frac {A} {s} + \frac {B} {s+3} + \frac {C} {s-5}</math>		:<math> \frac {s^2+s+6} {s(s+3)(s-5)} = \frac {A} {s} + \frac {B} {s+3} + \frac {C} {s-5}</math>


	:<math> A(s+3)(s-5)+Bs(s-5)+Cs(s+3)=-s^2+s+6 \,</math>		:<math> A(s+3)(s-5)+Bs(s-5)+Cs(s+3)=s^2+s+6 \,</math>


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	:<math> A=\frac {-2} {5} \qquad B=\frac {-1} {4} \qquad C=\frac {-7} {20} </math>		:<math> A=\frac {-2} {5} \qquad B=\frac {1} {2} \qquad C=\frac {9} {10} </math>

	Plugging the above values back into the equation further up, we get:		Plugging the above values back into the equation further up, we get:
	:<math> \mathcal{L} \left\{y\right\} = \frac {\frac {-2} {5}} {s} + \frac {\frac {-1} {4}} {s+3} + \frac {\frac {-7} {20}} {s-5} </math>		:<math> \mathcal{L} \left\{y\right\} = \frac {\frac {-2} {5}} {s} + \frac {\frac {1} {2}} {s+3} + \frac {\frac {9} {10}} {s-5} </math>

	Applying anti-Laplace transforms, we get the equation:		Applying anti-Laplace transforms, we get the equation:
	:<math> y= \mathcal{L}^{-1} \left\{\frac {\frac {-2} {5}} {s}\right\} + \mathcal{L}^{-1} \left\{\frac {\frac {-1} {4}} {s+3}\right\} + \mathcal{L}^{-1} \left\{\frac {\frac {-7} {20}} {s-5}\right\} </math>		:<math> y= \mathcal{L}^{-1} \left\{\frac {\frac {-2} {5}} {s}\right\} + \mathcal{L}^{-1} \left\{\frac {\frac {1} {2}} {s+3}\right\} + \mathcal{L}^{-1} \left\{\frac {\frac {9} {10}} {s-5}\right\} </math>

	Applying the Laplace transforms in reverse (as the above equation utilizes inverse Laplace transforms) for the above equation, we get the solution:		Applying the Laplace transforms in reverse (as the above equation utilizes inverse Laplace transforms) for the above equation, we get the solution:
	:<math> y(t)= \frac {-2} {5} + \frac {-1} {4} e^{-3t} + \frac {-7} {20} e^{5t} </math>		:<math> y(t)= \frac {-2} {5} + \frac {1} {2} e^{-3t} + \frac {9} {10} e^{5t} </math>

	==References==		==References==