transformasi laplace
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TUGAS PERSAMAAN DIFERENSIAL
OLEH:
Kelompok III
BAIQ ZILALIN AZZIMA E1R 012 006
FEBRI ARIANTI E1R 012 011
FITRIA WINDIARNI E1R 012 014
PROGRAM STUDI PENDIDIKAN MATEMATIKA
FAKULTAS KEGURUAN DAN ILMU PENDIDIKAN
UNIVERSITAS MATARAM
2014
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7. 3 OPERATIONAL PROPERTIES I
Introduction: it is not convenient to use Definition 7.1 each time we wish to
find the Laplace transform of a function . For example, the integration by parts
involved in evaluating, say L is formidable to say the least. In this
section and the next we present several labor-saving operational properties of the
Laplace transform that enable us to build up a more extensive list of transforms
without having to resort to the basic definition and integration.
7.3.1 TRANSLATION ON THE S-AXIS
A TRANSLATION Evaluating transforms such as L and L
is straightforward provided that we know (and we do) L and L .
In general, if we know the Laplace transform of a function. L , it
is possible to compute the Laplace transform of an exponential multiple of
that is, L with no additional effort other than translating, or
shifting, the transform to . This result is known as the first
translation theorem or first shifting theorem.
PROOF The proof is immediate, since by Definition 7.1
L
If we consider a real variable, then the graph of is the graph of
shifted on the -axis by the amount units to the left. See Figure 7.10.
FIGURE 7.10 Shift on s-axis
First Translation Theorem
If L and a is any real number, then L
Theorem 7.6
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For emphasis it is sometimes useful to use the symbolism
L L
Where means that in the Laplace transform of we replace
the symbol wherever it appears by .
Example 1 Using The First Translation Theorem
Evaluate L L .
Solution The Result Follow From Theorem 7.1 And 7.6
L L
L L
INVERSE FORM OF THEOREM 7.6 To compute the inverse of , we
must recognize , find by taking the inverse Laplace transform of ,
and then multiply by the exponential function . This procedure can be
summarized symbolically in the following manner:
L -1
L -1
Where L -1
.
The first part of the next example illustrates partial fraction
decomposition in the case when the denominator of contains repeated
linear factors.
Example 2 Partial Fraction: Repeated Linear Factors
Evaluate L -1
,
- L
-1 {
}
Solution A repeated linear factor is a term , where is a real
number and is a positive integer . Recall that if appears in the
denominator of rational expression, then the assumed decomposition
contains partial fractions with constant numerators and denominators
Hence with and we write
-
By putting the two terms on the right-hand side over a common denominator,
we obtain the numerator , and this identity yields
and . Therefore
and
L -1 ,
- L -1 ,
- L -1,
-
Now
is
shifted three units to the right. Since L -1 ,
- , it
follows from (1) that
L -1,
- L -1,
- .
Finally, is L -1 ,
-
To start, observe that the quadratic polynomial has no real
zeros and so has no real linear factors. In this situation we complete the
square:
Our goal here is to recognize the expression on the right-hand side as some
Laplace transform in wich has been replaced throughout by . What
we are trying to do is analogous to working part of Example 1 backwards.
The denominator in is already in the correct form-that is, with
replaced by . However, we must fix up the numerator by manipulating
the constants:
.
Now by termwise division, the linearity of L -1, parts and of
Theorem 7.4, and finally ,
-
L -1{
}
L -1,
-
L -1,
-
L -1,
-
L -1,
-
Example 3 An Initial-Value Problem
Solve
Solution Before transforming the DE, note that its right-hand side is similar to
the function in part of Example 1. After using linearity, Theorem 7.6,and
the initial conditions, we simply and then solve for L
L L L L
[ ]
The first term on the right-hand side was already decomposed into individual
partial fraction in in part of Example 2:
Thus L -1 ,
- L -1,
-
L -1,
-
From the inverse form of Theorem 7.6, the last two terms in are
L -1,
- and L -1,
-
.
Thus is
Example 4 . An Initial Value Problem
Solve
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Solution L L L L L
[ ]
Since the quadratic term in the denominator does not factor into real linear
factors, the partial fraction decomposition for is found to be
Moreover, in preparation for taking the inverse transform, we already
manipulated the last term into the necessary form in part of example 2. So
in view of the results in and we have the solution
L
-1 ,
-
L
-1
L
-1
L
-1
7.3.2 TRANSLATION ON THE T-AXIS
UNIT STEP FUNCTION In engineering one frequently encounters functions that
are either off or on. For example, an external force acting on a mechanical
system or a voltage impresseed on a circuit can be turned off after a period of
time. It is convenient, then to define a special function that is the number 0
(off) up to a certain time and then the number 1(on) after that time. This
function is called the unit step function or the Heavside function.
Definition 7.3 Unit Step Function
The unit step function U is defined to be U , <
-
Notice that we define U only on the nonnegative t-axis, since this is all
that we are concerned with in the study of the Laplace transform. In a broader
sense U for < The graph of U is given in Figure 7.11
FIGURE 7.11 Graph of unit step function
When a function defined for is multiplied by U the unit
step function turns off a portion of the graph of that function. For example,
consider the function . To turns off a portion of the graph of
on, say, the interval < we simply form the product U
See figure 7.12. In general, the graph of U is (off) for <
and is the portion of the graph of (on) for
FIGURE 7.12 Function is U
The unit step function can also be used to write piecewise-defined
function in a compact form. For example, if we consider the intervals
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