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Section 6.1 INTRO to LAPLACE TRANSFORMS
Key terms: • Improper Integral; diverge, converge • Piecewise Continuous Function; jump discontinuity • Function of Exponential Order • Laplace Transform (an integral transform) • Linearity
A
a aAf(t)dt lim f(t)dt
Review of some calculus topics:
In calculus most of the time you encountered proper integrals. That is, integrals where f is a continuous function defined on the closed and bounded interval [a, b].
b
af(t) dt
A particular type of improper integral has an infinite interval of integration like
af(t)dt
Such an integral is defined as a limit of integrals over finite intervals; thus
A
a aAf(t)dt lim f(t)dt
where A is a positive real number. If the integral from a to A exists for each A > a, and if the limit as A →∞ exists, (that is, has a real numerical value) then the improper integral is said to converge to that limiting numerical value. Otherwise the integral is said to diverge, or to fail to exist.
Piecewise continuous functions:
A function is called piecewise continuous on an interval if the interval can be broken into a finite number of subintervals on which the function is continuous on each open subinterval (i.e. the subinterval without it’s endpoints) and has a finite limit at the endpoints of each subinterval. Below is a sketch of a piecewise continuous function.
In other words, a piecewise continuous function is a function that has a finite number of breaks in it and doesn’t “blow up” to infinity anywhere.
If f is piecewise continuous on a ≤ t ≤ b for every b > a, then f is said to be piecewise continuous on t ≥ a.
Definitions: transform - to change the form of (a figure, expression, etc.) without changing its value - a mathematical quantity obtained from a given quantity by an algebraic, geometric, or functional transformation transformation - the act or process of transforming In mathematics transforms are used to take an “original” setting of a problem and convert it (or transform it) into a “new setting” in which it will be easier to solve.
Example: The use of logarithms is an example of a transformation. We have the following rules. log(AB) = log(A) + log(B) log(A/B) = log(A) - log(B) log(AB) = B log(A)
The process of using logarithms can be outlined as follows.
Recall “logarithmic differentiation”.
The Laplace Transform
Previously we had differential operators which took a function and transformed it into another function via differentiation. The Laplace transform is an integral operator which performs another type of transformation.
The Laplace Transform is widely used in engineering differential equation applications (mechanical and electronic), especially where the driving force (the nonhomogeneous term) is not continuous.
The Laplace transform takes a single IVP or system of IVPs in the independent variable t and transforms it to an algebra problem in a new variable s. We then solve the resulting algebra problem, which is easier than solving the DE. Finally we use a “reverse process” on the algebraic solution to obtain the solution of the original
This procedure is often summarized by saying we go from the t-domain to the s-domain, solve the algebra problem in the s-domain, and then invert to get the solution into the t-domain.
Definition Let f(t) be a function with domain [0, ∞). The Laplace transform of f is the function F defined by the integral The domain of F(s) is all values of s for which the integral exists. The Laplace transform of f is denoted F or L{f} or L{f}.
0stf F(s) e f(t)dtL
NOTE: The Laplace transform involves an improper integral and since it involves an integral it is a linear operator. That is, the Laplace transform of a sum of functions is the sum of the individual Laplace transforms and the Laplace transform of a constant times a function is the constant times the Laplace transform of the function. In symbols we have
This property is just like integral and derivative properties in calculus.
0 0
Nst st
NF(s) e f(t)dt lim e f(t)dt
We start by defining the Laplace transform of a function.
We have whenever the integral exists.
Linearity Property
Observation:
The Laplace transform of a function f is hence it involves the exponential function.
0
stf L{f} F(s) e f(t)dt
L
Since the solutions of linear differential equations with constant coefficients are based on the exponential function and sinusoids, the Laplace transform is particularly useful for such equations. It can also be used when the coefficients are not constants.
The Laplace transform F of a function f exists if f satisfies certain conditions, such as those stated in the following theorem.
Theorem: Suppose that 1. f is piecewise continuous on the interval 0 ≤ t ≤ A for any positive A. 2. |f(t)| ≤ Keat when t ≥ M. In this inequality, K, a, and M are real constants, K and M necessarily positive. Then the Laplace transform L{f(t)} = F(s), defined by exists for s > a. (This restriction ensures that the improper integral exists. See P.309.)
0stf F(s) e f(t)dtL
Functions which satisfy the condition |f(t)| ≤ Keat are described as piecewise continuous
and of exponential order. (This is a statement about how fast the graph of f increases.)
The idea that f(t) is of exponential order means that f(t) does not grow faster than a constant multiple of an exponential function. That is, for some constants K and a the graph of |f(t)| as t → ∞ will stay below that of Keat . Not all functions are of exponential order; for example .
2te
Since we do not want to repeat the same calculation over and over again, we will construct a table of Laplace transforms for functions that we anticipate we will encounter.
Observations: Computation of will involve a limit since and the antiderivative of a product. Thus integration by parts will most likely be used.
0stf F(s) e f(t)dtL
0 0
Nst st
NF(s) e f(t)dt lim e f(t)dt
The Laplace Transform is defined by an improper integral, and thus must be checked for convergence.
Examples: Instead of the script L, the standard L is sometimes used.
Let f (t) = 1 for t 0. Determine the Laplace transform F(s) of f .
0 0
0
11 0
bstb
st st
b b
eL e dt lim e dt lim , s
s s
11 0L , s
s
Table of Laplace Transforms
Let f (t) = eat for t 0. Determine the Laplace transform F(s) of f .
0 0
0
1
bat st at (s a)t
b
b(s a)t
b
L e e e dt lim e dt
elim , s a
s a s a
1atL e , s a
s a
Integration by parts is not needed in either of these examples.
Find the following: L{8} L{e-7t}
L{4 - 2e5t} 0
1 10
bst bs
b b
e elim lim since s
s s s s
We need s – a > 0 so that the improper integral exists; that is, the limit is finite.
Let f (t) = sin(at) for t 0. Determine the Laplace transform F(s) of f. (We will need to use integration by parts twice.)
Example:
0 0
0 0
0
1
bst st
b
bbst st
b
bst
b
F(s) L sin(at) e sin(at) dt lim e sin(at) dt
slim (e cos(at)) / a e cos(at) dt
a
slim e cos(at) dt
a a
First integration by parts.
Need a second integration by parts here!
2
20 0
1 1bbst st
b
s s slim (e sin(at)) / a e sin(at) dt F(s)
a a a a a
2 2
0a
F(s) L sin(at) , ss a
Note that we have so we need to solve for F(s). We get
2
2
1 sF(s) F(s)
a a
1
1 0L , ss
Table of Laplace Transforms 1atL e , s a
s a
2 20
aL sin(at) , s
s a
(so far)
The limit of this term is 1/a.
The limit of this term is zero.
Example:
Determine the Laplace transform F(s) of 0 0 3
5 3 6
0 6 25
, if t
f(t) , if t
, if t
3 6 25
0 0 3 60 5 0st st st stL f(t) f(t)e dt e dt e dt e dt
Observe that L{f(t)} does not depend on k, the function value at the point of discontinuity. Even if f(t) is not defined at this point, the Laplace transform of f remains the same. Thus there are many functions, differing only in their value at a single point, that have the same Laplace transform.
Determine the Laplace transform F(s) of where k is a constant.
1 0 1
1
0 1
, if t
f(t) k, if t
, if t
This function is often called the unit pulse.
11
0 00
1 11 0
st s sst st e e e
L f(t) f(t)e dt e dt ,ss s s s
66 3 6 36
33
5 5 5 5 55
st s s s sst e e e e e
e dts s s s s
Graph this function!
Graph this function!
1
1 0L , ss
Table of Laplace Transforms 1atL e , s a
s a
2 20
aL sin(at) , s
s a
(so far)
Find the Laplace transform of f(t) = 5e-2t -3 sin(4t), t > 0.
Recall that the Laplace transform is a linear operator.
That means L{5e-2t -3 sin(4t)} = 5 L{e-2t} - 3 L{sin(4t)} = 2
5 12
2 16s s
a = -2 a = 4
Example: 0, 0 t / 2
Let f(t)cos(t), t / 2
Find L(f(t)).
Let’s take as a fact that st
2 0
sL(cos(t)) e cos(t)dt
s 1
Then st st
0 /2L(f(t)) e f(t)dt e cos(t)dt
So this is not 2
s
s 1
What do we ?
Hint: We are only missing a “chunk” - namely /2 st
0e f(t)dt
Lets add zero in disguise; add and subtract the chunk!
/2st st st
0 0 0L(f(t)) e f(t)dt e cos(t)dt e cos(t)dt
Then we have /2st st
20 0
sL(f(t)) e f(t)dt e cos(t)dt
s 1
Definite integral; compute it.
From the integral table use #4.
/2st
st
2 200
s /2 0s /2
2 2 2 2 2
s eL(f(t)) e f(t)dt scos(t) sin(t)
s 1 s 1
s e e s 11 ( s) e s
s 1 s 1 s 1 s 1 s 1
