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Laplace Transformation
1. Basic notions
Definition For any complex valued function f defined for t > 0 and complex number s, one defines the Laplace transform of f (t) by
F (s) =
0
e−stf (t) dt,
if the above improper integral converges.
Notation We use L(f (t)) to denote the Laplace transform of f (t).
Remark It is clear that Laplace transformation is a linear operation: for any constants a and b :
L(af (t) + bg(t)) = aL(f (t)) + bL(g(t)).
Remark It is evident that F (s) may exist for certain values of s only. For instance, if f (t) = t, the Laplace transform of f (t) is given by (using integration by parts : u $ t, v′^ $ e−st): ∫ te−st^ dt = − t s e−st^ − (−
s
e−st^ dt) = − t s e−st^ +
s^2
e−st^ ;
0
te−st^ dt = − t s e−st^ +
s^2 e−st
∞
0
s^2 if s > 0 and does not exists if s ≤ 0.
Therefore L(t) =
s^2
Theorem If f (t) is a piecewise continuous function defined for t ≥ 0 and satisfies the inequality |f (t)| ≤ M ept^ for all t ≥ 0 and for some real constants p and M , then the Laplace transform Lf (t)) is well defined for all Re s > p.
Illustration The function f (t) = e^3 t^ has Laplace transform defined for any Re s > 3, while g(t) = sin kt has Laplace transform defined for any Re s > 0. The tables on the following page give the Laplace transforms of some elementary functions.
Remark It is clear from the definition of Laplace Transform that if f (t) = g(t) , for t ≥ 0, then F (s) = G(s). For instance, if H(t) is the unit step function defined in the following way:
H(t) = 0 if t < 0 and H(t) = 1 if t ≥ 0 , then L(H(t)) = L(1) =
s
and (as we have
seen above) L(H(t)t) = L(t) =
s^2
. Generally, L(H(t)tn) = L(tn) =
n! sn+^
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2. Inverse Laplace Transforms
Definition If, for a given function F (s), we can find a function f (t) such that L(f (t)) = F (s), then f (t) is called the inverse Laplace transform of F (s). Notation: f (t) = L−^1 (F (s)).
Examples
L−^1
s^2
= t. L−^1
s^2 + ω^2
sin ω t ω (hiszen L(sin ω t) =
ω s^2 + ω^2 ´es L line´aris.)
We are not going to give you an explicit formula for computing the inverse Laplace Transform of a given function of s. Instead, numerous examples will be given to show how L−^1 (F (s)) may be evaluated. It turns out that with the aide of a table and some techniques from elementary algebra, we are able to find L−^1 (F (s)) for a large number of functions. Our first example illustrates the usefulness of the decomposition to partial fractions:
Example 5 s^2 + 3s + 1 (s^2 + 1)(s + 2)
2 s − 1 s^2 + 1
s + 2 ;^
L−^1
5 s^2 + 3s + 1 (s^2 + 1)(s + 2)
= 2 cos t − sin t + 3e−^2 t.
3. Some simple properties of Laplace Transform
3.1 Transform of derivatives and integrals If f and f 0 are continuous for t > 0 such that f (t)e−st^ −→ 0 as t −→ ∞, then we may integrate by parts to obtain (F (s) = L(f (t)))
(1) L(f ′(t)) =
0
e−stf ′(t) dt = sF (s) − f (0)
( indeed by u ′^ = f ′(t) , v = e−st^ :
∫ (^) ∞
0
f ′(t)e−st^ dt = f (t)e−st
∣∣∞ 0 −^ s
∫ (^) ∞
0
f (t)e−st^ dt = −f (0) − sF (s)
)
and applying this formula again (assuming the apropriate conditions concerning the func- tion and it first and second derivative hold):
L(f ′′(t)) = sL(f ′(t)) − f ′(0) = s(sF (s) − f (0)) − f ′(0) = s^2 F (s) − sf (0) − f ′(0).
Similarly (again assuming the apropriate conditions concerning the derivatives hold) we obtain the general formula:
L(f (n)(t)) = snF (s) − sn−^1 f (0) − sn−^2 f ′(0) −... − sf (n−2)(0) − f (n−1)(0).
Example L(cos t) = L(sin ′t) = sL(sin t) − sin 0 = (^) s 2 s+
It follows from (1) that
(*) L(f (t)) = F (s) =
s (L(f ′(t)) + f (0))
Example
L(sin^2 t) =
s (L(sin 2t) + 0) =
s(s^2 + 4)
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Example
L−^1
s (s^2 + ω^2 )^2
= L−^1
s s^2 + ω^2
s^2 + ω^2
= L−^1
s s^2 + ω^2
∗ L−^1
s^2 + ω^2
= cos ω t ∗ sin ω t ω
ω
∫ (^) t
0
cos ω x sin ω(t − x) dx =
=
ω^2
cos(− 2 ω x + ω t) +
t ω sin(ω t)
t
0
ω^2
cos(ω t) +
t ω sin(ω t)
ω
cos(ω t)
2 ω t sin(ω t) ,
where in order to integrate, we have used addition formulas for the trigonometric functions.
A simpler example: L−^1
s(s^2 + 1)
= L−^1
s
s^2 + 1
= 1 ∗ sin t =
∫ (^) t
0
sin(t − x) dx =
=
∫ (^) t
0
(sin t cos x − cos t sin x)dx = sin t sin x
∣t 0 + cos t sin x
∣t 0 = sin^2 t + cos^2 t − cos t =
= 1 − cos t. Indeed, L(1 − cos t) =
s
s s^2 + 1
s^2 + 1 − s^2 s(s^2 + 1)
s(s^2 + 1)
3.5 Laplace Transform of a periodic function
Definition
A function f is said to be periodic if there is a constant T > 0 such that f (t + T ) = f (t) for every t. The constant T is called the period of f.
The sine and cosine functions are important examples of periodic function. One other example is the periodic triangular wave. It is is the function defined by f (t) = t if 0 ≤ t ≤ 1 , f (t) = 2 − t if 1 ≤ t ≤ 2 and f (t + 2) = f (t) for any t.
The following proposition is useful in calculating the Laplace Transform of a periodic function.
Proposition
Let f be a periodic function with period T and f 1 is one period of the function, Then (as usual F (s) = L(f (t))):
F (s) = L(f 1 (t)) 1 − e−T s^
1 − e−T s
∫ T
0
e−stf (t) dt.
Example
f (t) = 0 ha t < 0 , f (t) = t ha 0 ≤ t ≤ 1 ´es f (t + n) = f (t) tetsz˝oleges n-re:
1 2
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Now f 1 (t) = 0 if t < 0 and t > 1 , further f 1 (t) = t if 0 ≤ t < 1 , then defining
h(t) = 0 if t < 0 and h(t) = t otherwise g(t) = 0 if t < 0 and g(t) = t + 1 otherwise,
we have f 1 (t) = h(t) − g(t − 1) :
1 2
1 2
1 2
1
h(t)
g(t−1)
f (t) 1
Therefore, L(f 1 (t)) = L(h(t)) − L(g(t − 1)) =
s^2 − e−s
s^2
s
1 − e−s^ − s e−s s^2
that is L(f (t)) = L(f 1 (t)) 1 − e−s^
1 − e−s^ − s e−s s^2 (1 − e−s)
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