Let be a function Its Laplace Transform written is a function in variable s defined by Case 1 Constants Let where c is any constant Then The integral is found using limits ID: 700638
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Slide1
Laplace Transforms
MAT 275Slide2
Let
be a function. Its
Laplace Transform, written , is a function in variable s, defined byCase 1 (Constants). Let , where c is any constant. ThenThe integral is found using limits:Thus, when , then .
(c) ASU Math (SoMSS) - Scott Surgent. Please report errors to surgent@asu.edu
2Slide3
Case 2 (exponential, base-
e
): Let . ThenThe integral is evaluated using limits:When , then since the coefficient of b is negative. Also, . So we’re left with Thus, when , we have
.
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3Slide4
Laplace Transforms for other types of function
Usually, we do not bother to calculate Laplace Transforms by hand for most functions. Instead, we make up a table for the most common functions. For example, if
, where n is a positive integer, thenIf , then If , then (c) ASU Math (SoMSS) - Scott Surgent. Please report errors to surgent@asu.edu4Slide5
List of Common Laplace Transforms
Memorize this table!
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Laplace Transforms of Derivatives
Let
and suppose initial conditions and are given. Then it is possible to find the Laplace Transform of a derivative. These are two that are most common:For , we have For , we have The notation is the same as and is the same as .Linearity of the Laplace TransformThe operator can be distributed and any coefficients move to the front. (c) ASU Math (SoMSS) - Scott Surgent. Please report errors to surgent@asu.edu6Slide7
Example:
Find
.Solution: Using the form , we have .Example: Find .Solution: Using the form , we have .Example: Find .Solution: Using the form , we have .Example: Find .Solution: Using the form , we have
.
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7Slide8
Use algebra when appropriate…
Example:
Find .Solution: Multiply the binomial: . Now, using the linearity of the operator, we have (c) ASU Math (SoMSS) - Scott Surgent. Please report errors to surgent@asu.edu8Slide9
Use trigonometric identities also…
Example:
Find .Solution: We use the identity . Thus, (c) ASU Math (SoMSS) - Scott Surgent. Please report errors to surgent@asu.edu9Slide10
Inverting the Laplace Transform
As part of the solution process, we will need to invert the Laplace Transform, to find the function
such thatSome are easy to do by “inspection”:Example: Find .Solution: Since we know that , then it follows that .Example: Find .Solution: Sine we know that , then it follows that . (c) ASU Math (SoMSS) - Scott Surgent. Please report errors to surgent@asu.edu10Slide11
Inversions, continued…
Some cases, we need to “balance” the expression with constants and their reciprocals.
Example: Find .Solution: We know that , so we conclude that n = 5 in this example. However, we need 5! = 120 in the numerator to fully agree with the form. So we multiply inside by 120, and outside by , and we have (c) ASU Math (SoMSS) - Scott Surgent. Please report errors to surgent@asu.edu11Slide12
Example:
Find
.Solution: This looks closest to the form . The 3 in the numerator can be moved to the front, and from the denominator, we infer that since , we must have .However, the form “needs” the b value in the numerator. So we multiply inside by and outside by :Rationalizing the denominator, an equivalent form is . (c) ASU Math (SoMSS) - Scott Surgent. Please report errors to surgent@asu.edu12Slide13
Example:
Find
.Solution: The denominator does not factor over the Reals, so we complete the square:This looks closest to . We can conclude that and .But the numerator is not in the right form. We “need” () in the numerator in order to agree with the form. So we add in 2 then subtract it back out, then split the numerator by grouping and as separate terms:(continued…) (c) ASU Math (SoMSS) - Scott Surgent. Please report errors to surgent@asu.edu13Slide14
From the last screen, we have
.
The expression now exactly matches the form for .The expression almost matches the form for . To make it match exactly, multiply inside by and outside by , getting . Note that we moved the numerator 2 to the outside. Now everything matches. Thus,
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14Slide15
Example:
Find
.Solution: The denominator factors into linear factors:We now use partial fraction decomposition:Recomposing, we have (c) ASU Math (SoMSS) - Scott Surgent. Please report errors to surgent@asu.edu15Slide16
From the last screen, we have
Now, equate the two numerators:
We can find values for , and by choosing “convenient” values for :If , then the terms containing and vanish. We have, so that .(continued…) (c) ASU Math (SoMSS) - Scott Surgent. Please report errors to surgent@asu.edu16Slide17
From the last screen, we have
If
, then the and terms vanish, and we have , so that .If , then the and terms vanish, and we have , so that .Finally, we have that . Thus, (c) ASU Math (SoMSS) - Scott Surgent. Please report errors to surgent@asu.edu17