Ugur's copy of Laplace: Inverse Laplace Completing the Square

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Ugur Efem 2 years, 1 month ago

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Ugur Efem 2 years, 1 month ago

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Created this as a copy of Laplace: Inverse Laplace Completing the Square.

Name	Status	Author	Last Modified
Inverse Laplace transform: irreducible quadratic factor	draft	Frank Doheny	06/03/2018 13:33
Inverse Laplace transform: irreducible quadratic factor	Ready to use	Frank Doheny	26/02/2018 07:36
Marie's copy of Inverse Laplace transform: irreducible quadratic factor	draft	Marie Nicholson	22/11/2018 09:45
Laplace transform: irreducible quadratic factor	draft	Marie Nicholson	23/11/2018 13:23
Laplace: Inverse Laplace Completing the Square	draft	Clare Lundon	08/04/2020 21:13
Ugur's copy of Laplace: Inverse Laplace Completing the Square	Ready to use	Ugur Efem	20/02/2023 14:22
Ugur's copy of Laplace transform: irreducible quadratic factor	Ready to use	Ugur Efem	20/02/2023 14:36

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Question statement

Determine the inverse Laplace Transform of the following using completion of the square.

$F(s)=\dfrac{\var{B}s+\var{C}}{s^2+\simplify{{b1}*2}s+\var{c1}}$

Give any introductory information the student needs.

			Name	Type	Generated Value

A	integer	10
a1	integer	2
B	integer	8
b1	integer	8
C	integer	6
c1	number	68
f	integer	2

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Generated value: `integer`

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1. Gap Mathematical expression
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Write down the inverse Laplace transform

$\mathscr{L}^{-1}\{F(s)\}=$ Unnamed gap

Advice

We have

$F(s)=\frac{\var{B}s+\var{C}}{s^2+\simplify{{b1}*2}s+\var{c1}}$ .

By completing the square (on the denominator) we get:

$F(s)=\frac{\var{B}s+\var{C}}{(s+\var{b1})^2+\simplify{{c1}-{b1}^2}}$ .

Re-grouping the numerator gives:

$F(s)=\frac{\var{B}(s+\var{b1})-\simplify{{B}*{b1}-{C}}}{(s+\var{b1})^2+\simplify{{c1}-{b1}^2}} = \frac{\var{B}(s+\var{b1})}{(s+\var{b1})^2+\simplify{{c1}-{b1}^2}} - \frac{\simplify{{B}*{b1}-{C}}}{(s+\var{b1})^2+\simplify{{c1}-{b1}^2}} = \frac{\var{B}(s+\var{b1})}{(s+\var{b1})^2+\simplify{{c1}-{b1}^2}} - \frac{\simplify{{B}*{b1}-{C}}}{\sqrt{\simplify{{c1}-{b1}^2}}}\cdot\frac{\sqrt{\simplify{{c1}-{b1}^2}}}{(s+\var{b1})^2+\simplify{{c1}-{b1}^2}}$

Now, the expression above is easy enough to hunt from Laplace Transform table:

$f(t)=\var{B}e^{-\var{b1}t}\cos\left(\sqrt{\simplify{{c1}-{b1}^2}}t\right)+\frac{-\simplify{{B}*{b1}-{C}}}{\sqrt{\simplify{{c1}-{b1}^2}}}e^{-\var{b1}t}\sin\left(\sqrt{\simplify{{c1}-{b1}^2}}t\right)$

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This question is used in the following exam:

Laplace Transforms by Ugur Efem in Ugur's workspace.

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Ugur's copy of Laplace: Inverse Laplace Completing the Square

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Ugur Efem 2 years, 1 month ago

Ugur Efem 2 years, 1 month ago

Ugur Efem 2 years, 1 month ago

Ugur Efem 2 years, 2 months ago

Error in variable testing condition

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Generated value: `integer`

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Ugur's copy of Laplace: Inverse Laplace Completing the Square

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Ugur Efem 2 years, 1 month ago

Ugur Efem 2 years, 1 month ago

Ugur Efem 2 years, 1 month ago

Ugur Efem 2 years, 2 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`