laplace transform
TRANSCRIPT
Laplace transform
Pierre Simon Marquis de Laplace (1749-1827): French
mathematician, who was a professor in Ecole Polytech-
nique in Paris. He developed the foundation of poten-
tial theory and made important contributions to celestial
mechanics, astronomy in general, special functions and
probability theory.
The Laplace transform method solves differential equa-
tions and corresponding initial and boundary value prob-
lems. The process of solution consists of three main steps:
1. step The given ”hard problem is transformed into a ”sim-
ple” equation (subsidiary equation)
2. step The subsidiary equation is solved by purely algebraic
manipulations.
3. step The solution of the subsidiary equation is transformed
back to obtain the solution of the given problem
Laplace transform reduces the problem of solving a dif-
ferential equation to an algebraic problem. This process
is made easier by tables of functions and their transforms,
whose role is similar to that of integral tables in calculus.
This switching from operations of calculus to algebraic
operations on transforms is called operational calculus.
1
Operational calculus is very important area of applied
mathematics and for the engineers, the Laplace trans-
form method is practically the most important opera-
tional method. It is particularly useful in problems where
the mechanical or electrical driving force has discontinu-
ities, is impulsive or is a complicated periodic function.
The Laplace transform also has the advantage that it
solves problems directly, initial value problems without
first solving the corresponding homogeneous equation.
Laplace transform:
F (s) = L(f ) =
∫ ∞
0
e−stf (t)dt (1)
where t ≥ 0.
Inverse Laplace transform is inverse of F(s)
f (t) = L−1(F )
Example 1. Find Laplace transform of f (t) = 3, when
t ≥ 0
Example 2. Find Laplace transform of f (t) = eat, when
t ≥ 0, a is constant.
Example 3. Find Laplace transform of f (t) = cos(πt),
when t ≥ 0
2
Example 4. Find Laplace transform of
f (t) =
{k, if 1 ≤ t ≤ 4
0, elsewhere
where k is constant.
3
Linearity of Laplace transform:
The Laplace transform is a linear operator; that is for
any functions, f (t) and g(t) whose Laplace transforms
exists and any constants a and b it holds
L[af (t) + bg(t)] = aL[f (t)] + bL[g(t)] (2)
Example 5. Calculate Laplace transform of function
f (t) = cosh at =eat + e−at
2
If F (s) = L[f (t)] then L−1[F (s)] = f (t). Notice
that tranform tables are very useful in finding the inverse
Laplace transform L−1(F ), quite often some algebraic
manipulation is needed first.
Example 6. Laplace transform is of form
F (s) =1
(s− a)(s− b), a 6= b
What is the original function? In other words calculate
inverse Laplace transform, L−1(F ).
Note! L−1(F ) is a linear transform, because L(f ) is
linear.
4
Table 1: Some functions of f(t) and their Laplace Transforms L(f). Γ is agamma function Γ(v) =
∫∞0
e−ttv−1dt
no f(t) L(f)
1 11s
2 t1s2
3 t22!s3
4 tn, (n=0,1,2,...)n!
sn+1
5 ta, (a positive)Γ(a+1)sa+1
6 eat 1s−a
7 cosωts
s2+ω2
8 sinωtω
s2+ω2
9 coshats
s2−a2
10 sinhata
s2−a2
11 eatcosωts−a
(s−a)2+ω2
12 eatsinωtω
(s−a)2+ω2
Example 7. Calculate inverse Laplace transform when
a)F (s) = 5ss2−25
b)F (s) = 2s4 − 3
s6
5
First Shifting Theorem:
L[eatf (t)] = F (s− a) (3)
L−1[F (s− a)] = eatf (t) (4)
Example 8. Calculate Laplace transform for
a) t2e−3t b)5e2t sinh 2t c)t5e3t
Example 9. Calculate inverse Laplace transform for
a) 1(s+1)2
b) 3s2+6s+18
6
Existence of Laplace transforms
Existence Theorem:
Let f (t) be piecewise continuous function on finite inter-
val where t ≥ 0, and f (t) satisfies
∀ t ≥ 0, | f (t) |≤ Me−kt,
where k and M are constants, then f (t):s Laplace-transform
exist for all s > k.
• Not a big practical problem in most cases because
we can check a solution of a differential equation by
substitution.
• Existence in not always quaranteed because we inte-
grate over an infinite interval.
– For a fixed s the integral in∫∞
0 e−stf (t)dt will
exist if the whole integrand e−stf (t)dt goes to
zero fast enough as t → ∞ say, at least like an
exponential function with a negative exponent.
(this fact motivates the inequality in existense
theorem)
• The function f (t) need not be continuous. Sufficient
is so-called piecewise continuity.
• By definition, a function f (t) is piecewise continu-
ous on a finite interval a ≤ t ≤ b if f (t) is defined
on that interval and is such that the interval can
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be subdivided into finitely many intervals in each of
which f (t) is continuous and has finite limits as t
approaches either endpoints
• Finite jumps are the only discontinuities that a piece-
wise continuous function may have; these are known
as ordinary discontinuities.
Uniqueness: If the Laplace transform of a given
function exists, it is uniquely determined. Conversly, it
can be shown that if two functions have the same trans-
form, these functions cannot differ over an interval of pos-
itive length, although they may differ at various isolated
points. Since this is of no importance in applications, we
may say that the inverse of a given transform is essen-
tially unique. In particular, if two continuous functions
have the same transform, they are completely identical.
Example 10. Does the Laplace transform exist for
a) cosh t b)tn c) et2?
8
Differentiation of Function
The Laplace transform is a method of solving differ-
ential equations. The crucial idea is that the Laplace
transform replaces operations of calculus by operations
of algebra on transforms. Roughly, differentiation of f (t)
is replaced by multiplication of L(f ) by s. Integration of
f (t) is replaced by division of L(f ) by s.
Theorem: Laplace transform of the deriva-
tive of f (t)
Suppose that f (t) is continuous for all t ≥ 0 satisfies the
existence theorem for some k and M , and has a deriva-
tive f ′(t) that is piecewise continuous on every finite in-
terval in the range t ≥ 0. Then the Laplace transform of
the derivative f ′(t) exists when s > k, and for the first
derivative:
L(f ′) = sL(f )− f (0) (5)
when s > k.
Applying the same formula to the second derivative we
get following for f ′′(t):
L(f ′′) = sL(f ′)− f ′(0) = s[sL(f )− f (0)]− f ′(0),
in otherwords
L(f ′′) = s2L(f )− sf (0)− f ′(0). (6)
Same way we get for f ′′′(t)
L(f ′′′) = s3L(f )− s2f (0)− sf ′(0)− f ′′(0). (7)
9
Let f (t) and its derivatives f ′(t), f ′′(t), ... ,f (n−1)(t)
be continuous functions for all t ≥ 0, satisfy the existence
theorem, for some k, M, and let the derivative f (n)(t) be
piecewise continuous on every finite interval in the range
t ≥ 0. Then the Laplace transform of f (n)(t) exists when
s > k and is given by
L(f (n)) = snL(f )−sn−1f (0)−sn−2f ′(0)− ...−f (n−1)(0)
(8)
Example 11. Let f (t) = 2t2. Calculate L(f (t)) using
equations for derivates and the knowledge of L(1).
As you can see from the example there can be several
ways to find Laplace transform of a function.
Example 12. Calculate L(f ) when f (t) = cos2 t by
using the derivative.
Example 13. Calculate L(f ) when f (t) = 2t cos 3t.
(using the derivative)
10
Differential equations and initial value prob-
lems
Next we go into how the Laplace transform can be used
to solve differential equations. We begin with an initial
value problem of form
y′′ + ay′ + by = r(t), y(0) = K0, y′(0) = K1, (9)
where a and b are constants. Here r(t) is an input to the
system and y(t) is systems response (output). In Laplace
method we have three steps:
1. step. We get following form for differencial equa-
tion (9) by using first and second derivative and where
Y = L(y) and R = L(r)
[s2Y − sy(0)− y′(0)] + a[sY − y(0)] + bY = R(s),
this is called subsidiary equation. By collecting Y -terms
we get
(s2 + as + b)Y = (s + a)y(0) + y′(0) + R(s).
2. step. By dividing previous equation with (s2 +
as + b) or in otherwords using so called transfer function
Q(s)
Q(s) =1
s2 + as + b(10)
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We get solution for subsidiary equation
Y (s) = [(s + a)y(0) + y′(0)]Q(s) + R(s)Q(s). (11)
If y(0) = y′(0) = 0. We get Y = RQ where Q is
Q =Y
R=L(output)
L(input),
(this explains Qs name). Note that Q depends on a and
b, but not from r(t) or initial conditions.
3. step. We reduce (11) (usually by partial fractions)
to sum form, where sum terms inverse transforms are
gained from the tables. We get the solution
y(t) = L−1(Y )
for differential equation (9).
Example 14. Solve initial value problem
y′′ − y = t, y(0) = 1, y′(0) = 1.
Example 15. Solve initial value problem
9y′′ − 6y′ + y = 0 y(0) = 3, y′(0) = 1
Example 16. The LRC circuit consists of a resistor R,
a capacitor C and an inductor L connected in series to-
gether with a voltage source e(t). Prior to closing the
switch at time t = 0, both the charge on the capacitor
and the resulting current in the circuit are zero. Deter-
mine the charge q(t) on the capacitor and the resulting
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current i(t) in the circuit at time t given that R = 160
ohms, L = 1 henry, C = 10−4 farad and e(t) = 20 V.
We are assuming that q(0) = q′(0) = 0.
Example 17. The mass-spring-damper system is sub-
jected to an externally applied periodic for F (t) = 4 sin 2t
at time t = 0. Determine the resulting displacement y(t)
of the mass at time t, given that y(0) = y′(0) = 0, and
spring constant k = 25, damping coefficient B = 6 and
mass m = 1.
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Laplace Transform of the Integral of a Func-
tion
Differetiation and integration are inverse processes. Ac-
cordingly, since differentiation of a function corresponds
to the multiplication of its transform by s, we expect in-
tegration of a function to correspond to division of its
transform by s, because divison is the inverse operation
of multiplication.
Theorem:Integration of f (t)
If f (t) is piecewise continuous and satisfies the existence
theorem, then
L{∫ t
0
f (τ )dτ
}=
1
sL{f (t)}, s > 0, s > k. (12)
k is from the existence theorem (in otherwords we are
assuming that the solution exists). Equation (12) has a
useful consequence. We define L{f (t)} = F (s), when
using the inverse transform we get
L−1
{1
sF (s)
}=
∫ t
0
f (τ )dτ. (13)
Example 18. Find f (t), when
L(f ) =1
s(s2 + ω2).
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Example 19. Find f (t), when
L(f ) =1
s2(s2 + ω2).
Example 20. Solve initial value problem
y′ + 0.2y = 0.01t, y(0) = −0.25
Example 21. Using the Laplace transform, find the
current i(t) in the LC-circuit, assuming L = 2 henry,
C = 0.5 farad, zero initial current and charge on the ca-
pacitor, and v(t) = 10 ohms. Equation for the circuit
is
Li′(t) +1
C
∫ t
0
i(τ )dτ = v(t)
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s-shift, t-shift and unit step function
s-shift:
If f (t) has Laplace transform of F (s), where s > k,
then function eatf (t) has Laplace transform F (s − a),
where s− a > k. So, if L{f (t)} = F (s), we have
L{eatf (t)
}= F (s− a). (14)
So if f (t) has a known transform F (s), for function
eatf (t) we get transform by moving of a in the s-axis, in
other words by placing s− a for s we get F (s− a).
Remark 1. By taking the inverse transform from both
sidesof (14) we get
L−1{F (s− a)} = eatf (t). (15)
Example 22. Iron ball which mass m = 2 is anchored
in elastic springs bottom while upper end is fixed to wall.
Spring constant is k = 10. Determine the displacement
y(t) of the mass from the equilibrium at time t. given that
oscillation is starting from y(0) = 2 and initial speed is
y′(0) = −4. We also assume that damping is propor-
tional to speed while damping constant is c = 4.
Initial value problem for movement is given by
y′′ + 2y′ + 5y = 0, y(0) = 2, y′(0) = −4
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Example 23 (nonhomogenic differential equation). Solve
the initial value problem
y′′ − 2y′ + y = et + t, y(0) = 1, y′(0) = 0.
t-shifting: Replacing t by t− a in f (t)
Second shifting theorem: t-shifting:
If f (t) has Laplace transform of F (s), then function
f (t) =
{0, when t < a
f (t− a), when t > a(16)
where a ≥ 0 is arbitrary has Laplace transform
L(f (t)) = e−asF (s).
According to this if Laplace transform F (s) of f (t) is
known, then Laplace transform of function (16) is gained
by multiplying F (s) with term e−as, in otherwords by
shifting f (t)s variable by a (shift along the t-axis).
17
Definition 1. Unit step function u(t − a) is defined
by
u(t− a) =
{0, when t < a
1, when t > a, a ≥ 0. (17)
Function u has a unit jump at t = a, where its value
is left undefined.
Unit step function u(t− a) is used in many functions
and it has enlarged the use of Laplace transform in many
areas. Unit step function can be used e.g. as representing
a function f (t) in t-shift (16) in form f (t − a)u(t − a),
in otherwords
f (t− a)u(t− a) =
{0, when t < a
f (t− a), when t > a(18)
This function is similar to f (t) when t > 0, but it is
shifted as worth to the right. The unit step function is a
typical ”engineering function” made to measure for engi-
neering applications, which often involve functions (me-
chanical or electrical driving forces) that are either ”off”
or ”on”. Multiplying functions f (t) with u(t−a), we can
produce all sorts of effects.
For example equations f (t − 2)u(t − 2) = cos(t −2)u(t − 2) figure is similar to f (t) = cos t, t > 0, but it
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is shifted two units right from cos t. When t < a = 2, is
cos(t− 2)u(t− 2) = 0, because at this area u(t− 2) = 0.
By using equation (18) t-shifting theorem can be re-
framed as:
If L{f (t)} = F (s), then
L{f (t− a)u(t− a)} = e−asF (s). (19)
Remark 2. By taking inverse from both sides of (19)
we get
L−1{e−asF (s)} = f (t− a)u(t− a). (20)
In efficient use of Laplace tranforms we also need unit
step functions u(t− a) Laplace transform
L{u(t− a)} =e−as
s, s > 0. (21)
Example 24. Calculate inverse transform for function
F (s) =e−3s
s3
Example 25. Find Laplace transform for function
f (t) =
2, when 0 < t < π
0, when π < t < 2π
sin t, when t > 2π
Example 26. Find the current i(t) in the RC-circuit
if a single square wave with voltage v0 is applied so
19
that it is starting at t = a and ending at t = b. The
circuit is assumed to be quiescent before the square
wave is applied and the equation for the circuit is:
Ri(t) +q(t)
C= Ri(t) +
1
C
∫ t
0
i(τ )dτ = v(t)
About translation formula
The usual Laplace transform for translation on the t-
axis is
L{f (t− a)u(t− a)} = e−asF (s).
where F (s) = L(f (t)).
This formula is useful for computing inverse Laplace
transform of e−asF (s), for example. On the other hand,
as written above it is not immediately applicable to com-
puting the Laplace transform of functions having the form
u(t − a)f (t). For this you can use instead the previous
formula:
L{u(t− a)f (t)} = e−asL(f (t + a)) a > 0 (22)
Example 27. Calculate the Laplace transform of u(t−1)(t2 + 2t)
20
Example 28. Find L (u(t− π
2) sin t)
Example 29. Using the Laplace transform, find the cur-
rent i(t) in the LC-circuit, assuming L = 1 henry, C = 1
farad, zero initial current and charge on the capacitor,
and v(t) = 1 if 0 < t < a and v(t) = 0 if otherwise.
Equation for the circuit is
Li′(t) +1
C
∫ t
0
i(τ )dτ = v(t)
Short impulses. Dirac’s delta function
Phenomena of an impulsive nature, such as the action
of very large forces (or voltages) over very short intervals
of time, are of great practical interest, since they arise in
various applications. This situation occurs, for instance,
when a tennis ball is hit, a system is given a blow by a
hammer, an airplane makes a hard landing, a ship is hit
by a high single wave, and so on. Next the goal is to
show how to solve problems involving short impulses by
Laplace transforms.
In mechanics, the impulse of a force f (t) over a time
interval, a ≤ t ≤ a + k, is defined to be the integral of
f (t) from a to a + k. The analog for an electric circuit
is the integral of the electromotive force applied to the
circuit, integrated from a to a + k. Of particular prac-
tical interest is the case of a very short k (and its limit
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k → 0), that is, the impulse of a force acting only for an
instant. To handle the case, we consider the function
fk(t− a) =
{1k , if a ≤ t ≤ a + k
0, otherwise(23)
The limit of fk(t−a) as k → 0 is denoted by δ(t−a),
that is
δ(t− a) = limk→0
fk(t− a) (24)
and the Laplace transform is
L[δ(t− a)] = e−as (25)
Example 30. Response of a damped vibrating system
to a single square wave and to a unit impulse.
Determine the response of the damped mass-spring sys-
tem governed by
y′′ + 3y′ + 2y = r(t), y(0) = 0. y′(0) = 0
a) where r(t) is the square wave r(t) = u(t−1)−u(t−2)
b) where r(t) is a unit impulse r(t) = δ(t− 1)
Differentiation and Integration of Transforms
The variety of methods for obtaining transforms or
inverse transforms and their application in solving differ-
ential equations is suprisingly large. They include
22
1) direct integration
2) the use of linearity
3) shifting
4) differentiation or
5) integration of function f (t).
Next we consider differentiation and integration of trans-
form F (s) and find the corresponding operations for orig-
inal function f (t).
It can be shown that if f (t) satisfies the conditions of
the existence theorem, then the derivative F ′(s) of trans-
form
F (s) = L(f ) =
∫ ∞
0
e−stf (t)dt
with respect to s can be obtained by differentiating
under the integral sign with respect to s!; thus
F ′(s) = −∫ ∞
0
e−st[tf (t)]dt
Consequently, if L(f ) = F (s), then
L[tf (t)] = −F ′(s) (26)
23
In practice differentiation of the transform of a func-
tion corresponds to the multiplication of the function by
(−t). Similarly
L−1[F ′(s)] = −tf (t) (27)
This property enables us to get new transforms from
the previously given ones.
Example 31. Derive following equations which are given
with original functions
Transform:1
(s2 + β2)2, original function:
1
2β3(sin βt−βt cos βt)
(28)
Transform:s
(s2 + β2)2, original function:
t
2βsin βt
(29)
Transform:s2
(s2 + β2)2, original function:
1
2β(sin βt+βt cos βt)
(30)
Example 32. The mass-spring-damper system is sub-
jected to an externally applied periodic force F (t) =
4 sin ωt at time t = 0. Determine the resulting displace-
ment y(t) of the mass at time t, given that y(0) = y′(0) =
24
0, and ω = 5, spring constant k = 25, mass m = 1 and
damping coefficient B = 0. Differential equation for the
system is
my′′ + By′ + ky = F (t)
Integration of Transforms
If f (t) satisfies the existence theorem and limit
limt→0+
f (t)
t
exists, then
L{
f (t)
t
}=
∫ ∞
s
F (s)ds, s > k, (31)
Which means in practice that integral of functions f (t)
transform corresponds to dividing f (t) by t. Similarily
L−1
{∫ ∞
s
F (s)ds
}=
f (t)
t. (32)
From definition it follows∫ ∞
s
F (s)ds =
∫ ∞
s
[∫ ∞
0
e−stf (t)dt
]ds,
and it can be shown that under previous conditions we
can change the order of integration∫ ∞
s
F (s)ds =
∫ ∞
0
[∫ ∞
s
e−stf (t)ds
]dt =
∫ ∞
0
f (t)
(∫ ∞
s
e−stds
)dt.
25
Integral integrating over s is e−st/t, when s > k, so∫ ∞
s
F (s)ds =
∫ ∞
0
e−stf (t)
tdt = L
{f (t)
t
}, s > k.
Example 33. Find inverse transform for function
ln
(1 +
ω2
s2
)
26
Differential equations with Variable Coeffi-
cients
FromL(tf (t)) = −F ′(s) with f = y′dy/dt andL(y′) =
sY − y(0) and subsequent product differentiation we ob-
tain
L(ty′) = − d
ds(sY − y(0)) = −Y − s
dY
ds(33)
SimilarlyL(tf (t)) = −F ′(s) with f = y′′ andL(f ′′) =
s2Y − sf (0)− f ′(0)
we obtain
L(ty′′) = − d
ds
(s2Y − sy(0)− y′(0)
)= −2sY−s2dY
ds+y(0)
(34)
Hence if a differential equation has coefficients such as
at + b, we get a first-order differential equation for Y ,
which is sometimes simpler than the given equation. But
if the latter has coefficients at2+bt+c, we get, by two ap-
plications of (1), a second-order differential equation for
Y , and this shows that the Laplace transform method
works well only for very special equations with variable
coefficients. We illustrate it for an important equation in
the following example.
27
Example 34. Laquerre’s differential equation is
ty′′ + (1− t)y′ + ny = 0 (35)
We determine a solution of (35) with n = 0, 1, 2, ... From
(33) −− (35) we get
[−2sY − s2dY
ds+ y(0)
]+sY−y(0)−
(−Y − s
dY
ds
)+nY = 0
Simplification gives
(s− s2)dY
ds+ (n + 1− s)Y = 0
Separating variables, using partial fractions, integrat-
ing and taking exponentials, we get
dY
Y= −n + 1− s
s− s2ds =
(n
s− 1− n + 1
s
)ds and Y =
(s− 1)n
sn+1
(36)
We write In = L−1(Y ) and show that
l0 = 1 ln(t) =et
n!
dn
dtn(tne−t) (37)
These are polynomials because the exponential terms
cancel if we perform the indicated differentiations. They
are called Laquerre polynomials and are usually denoted
by Ln
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Convolution. Integral equations
Another important general property of the Laplace
transform has to do with product of transforms. It often
happens that we are given two transforms F (s) and G(s)
whose inverses f (t) and g(t) we know, and we would like
to calculate the inverse of the product H(s) = F (s)G(s)
from those known inverses f (t) and g(t). This inverse
h(t) is written (f ∗ g)(t), which is a standard notation,
and is called the convolution of f and g. Since the
situation and task just described arise quite often in ap-
plications, this theorem is of considerable practical im-
portance.
Convolution and related operations are found in many
applications of engineering and mathematics.
1) In statistics, a weighted moving average is a convo-
lution.
2) also the probability distribution of the sum of two
independent random variables is the convolution of
each of their distributions.
3) In optics, many kinds of ”blur” are described by con-
volutions. A shadow (e.g. the shadow on the table
when you hold your hand between the table and a
light source) is the convolution of the shape of the
29
light source that is casting the shadow and the object
whose shadow is being cast. An out-of-focus photo-
graph is the convolution of the sharp image with the
shape of the iris diaphragm. The photographic term
for this is bokeh.
4) Similarly, in digital image processing, convolutional
filtering plays an important role in many important
algorithms in edge detection and related processes.
5) In linear acoustics, an echo is the convolution of the
original sound with a function representing the vari-
ous objects that are reflecting it.
6) In artificial reverberation (digital signal processing,
pro audio), convolution is used to map the impulse
response of a real room on a digital audio signal.
7) In electrical engineering and other disciplines, the
output (response) of a (stationary, or time- or space-
invariant) linear system is the convolution of the in-
put (excitation) with the system’s response to an im-
pulse or Dirac delta function.
8) In time-resolved fluorescence spectroscopy, the exci-
tation signal can be treated as a chain of delta pulses,
and the measured fluorescence is a sum of exponen-
tial decays from each delta pulse.
9) In physics, wherever there is a linear system with
30
a ”superposition principle”, a convolution operation
makes an appearance.
10) This is the fundamental problem term in the Navier
Stokes Equations relating to the Clay Institute of
Mathematics Millennium Problem and the associ-
ated million dollar prize.
11) In digital signal processing, frequency filtering can
be simplified by convolving two functions (data with
a filter) in the time domain, which is analogous to
multiplying the data with a filter in the frequency
domain
Definition: Let F (s) and G(s) be two transform
functions which original functions f (t) and g(t) are known.
In convolution we compute product H(s) = F (s)G(s),
where F (s) and G(s) are, inverse transforms from orig-
inal functions f (t) and g(t). This inverse transform of
H(s) is
h(t) = (f ∗ g)(t),
which is called convolution of f and g.
Convolution theorem:
We assume that f (t) and g(t) satisfies the existence the-
orem. Then their transforms F (s) = L(f ) and G(s) =
L(g) product is f (t)s and g(t)s convolution h(t) Laplace
transform H(s) = L(h), we mark it as L(f ∗g)(t), where
31
h is defined by
h(t) = (f ∗ g)(t) =
∫ t
0
f (τ )g(t− τ )dτ. (38)
Example 35. What is the original function of the trans-
form function H(s) = 1(s2+1)2
?
Example 36. Define functions t and 1 convolution.
By applying definition of convolution we can show that
convolution f ∗ g has following properties
f ∗ g = g ∗ f (commutative law)
f ∗ (g1 + g2) = f ∗ g1 + f ∗ g2 (distributive law)
(f ∗ g) ∗ h = f ∗ (g ∗ h) (associative law)
f ∗ 0 = 0 ∗ f = 0
Just like in multiplication. Allthough in general
f ∗ 1 6= f,
as we can see from example 36. Another abnormality is
that
(f ∗ f )(t) ≥ 0
is not always true, like we can see from example 35.
32
Differential equation and convolution:
Reminding that differential equation
y′′ + ay′ + by = r(t) (39)
has a subsidiary equation which solution
Y (s) = [(s + a)y(0) + y′(0)]Q(s) + R(s)Q(s) (40)
where R(s) = L(r) and
Q(s) =1
(s2 + as + b)
is transfer function. Using initial conditions y(0) = y′(0) =
0 for differential equation (39), we get for solution of y(t)
as inverse transform ofY = RQ. This solution according
to convolution theorem is
y(t) =
∫ t
0
q(t− τ )r(τ )dτ, q(t) = L−1(Q). (41)
Example 37 (Response of a oscillation system to a sin-
gle square wave ). Let us take a look at the model
y′′+2y = r(t), r(t) =
{1, when 0 < t < 1
0 otherwise, y(0) = y′(0) = 0.
We solve this by using convolution so that we can see how
it works for inputs which are effecting the system.
33
Example 38 (Response of a damped system to a single
square wave). Reconsider the model
y′′+3y′+2y = r(t), r(t) =
{1, when 1 < t < 2
0 otherwise,
when y(0) = y′(0) = 0 and solve it using the convolu-
tion technique.
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Integral equations
Convolution can be also used to solve such integral
equations where unkown function y(t) is inside the in-
tegral. This applies to integral equations which are of
convolution form.
Example 39. Solve integral equation
y(t) = t +
∫ t
0
y(τ ) sin(t− τ )dτ.
Example 40. Sometimes the convolution theorem can
be used to solve quite general initial value problems. As
illustration, consider the problem
y′′ − 2y′ − 8y = f (t), quady(0) = 1, y′(0) = 0
35
Partial fractions, differential equations
The solution Y (s) of a subsidiary equation of a differ-
ential equation usually comes out as a quotient of two
polynomials,
Y (s) =F (s)
G(s)
Hence we can determine its inverse by writing Y (s) as
a sum of partial fractions and obtain the inverse of
the latter from a table and the first shifting theorem. In
context of Laplace transforms, partial fraction represen-
tations and their inverse transforms are sometimes called
Heaviside expansions. The form of the partial fractions
depends on the kind of factors in the product form of
G(s).
Case 1 Unrepeated factors s− a
Case 2 Repeated factors (s− a)m
Case 3 Complex factors (s− a)(s− a)
Case 4 Repeated complex factors [(s− a)(s− a)]m
Unrepeated factors (s − a)(s − b), etc require partial
fractions A1s−a + A2
s−b, etc.
Example 41 (Case unrepeated factors). Solve the ini-
tial value problem
36
y′′ + y′ − 6y = 1 y(0) = 0, y′(0) = 1
Repeated factors e.g. (s− a)3 require partial fractionsA1
(s−a)3+ A2
(s−a)2+ A3
(s−a), respectively
Example 42. Using the Laplace transform solve the
differential equation
y′′ + y′ − 6y = e−3t (42)
with initial conditions y(0) = y′(0) = 0.
Example 43 (Case repeated factors). Solve the differ-
ential equation using the Laplace transform
y′′ − 3y′ + 2y = 4t, y(0) = 1, y′(0) = −1
Unrepeated complex factors. Such factors occur for
instance in connection with vibration. IF s − a with
complex a = a1 + ib1 is factor of G(s), so is s − a,
with a = a1 − ib1 the conjugate. To (s − a)(s − a)(=
(s − a1)2 + b2
1) there corresponds the partial fractionAs+B
(s−a)(s−a) or As+B(s−a1)2+b21
.
Unrepeated complex factors can occur e.g. in damping
vibrating systems in mechanics.
Example 44 (Case unrepeated complex factors, Damped
forced vibrations). Solve the initial value problem
37
y′′+2y′+2y = r(t), r(t) =
{10 sin 2t, if 0 < t < π
0 otherwise,
Example 45 (Case unrepeated complex factors, LC–
circuit). Using the Laplace transform, find the current
i(t) in following LC-circuit assuming L = 1 henry, C = 1
farad, zero initial current and charge on the capacitor and
v(t) = t if 0 < t < 1 and v(t) = 1 if t > 1
Li′ +1
C
∫ t
0
i(τ )dτ = v(t)
Example 46. Using the Laplace transform solve the
differential equation
y′′ + 6y′ + 13y = 0
with boundary conditions y(0) = 0 and y′(0) = 1.
Repeated complex factors. [(s − a)(s − a)]2. In this
case the partial fractions are of the formAs+B
[(s−a)(s−a)]2+ Ms+N
(s−a)(s−a)
Repeated complex factors often occur e.g. in cases
where resonance is involved (cases where there is no damp-
ing like in previous example) or from electrical systems
in LC-circuits.
Example 47. In an undamped mass-spring system, res-
onance occurs if the frequency of the driving force equals
the natural frequency of the system. Then the model is
38
y′′ + ω20y = K sin ω0t
Solve this differential equation with initial values y(0) =
y′(0) = 0.
39
Systems of Differential Equations
The Laplace transform can be used to solve systems of
differential equations.
For a first order linear system:
y′1 = a11y1 + a12y2 + g1(t)
y′2 = a21y1 + a22y2 + g2(t)
(43)
writing Y1 = L[y1], Y2 = L[y2], G1 = L[g1], G2 =
L[g2], we obtain the subsidiary equations
sY1 − y1(0) = a11Y1 + a12Y2 + G1(s)
sY2 − y2(0) = a21Y1 + a22Y2 + G2(s)(44)
or, by collecting the Y1- and Y2-terms,
(s− a11)Y1 − a12Y2 = y1(0) + G1(s)
−a21Y1 + (s− a22)Y2 = y2(0) + G2(s)(45)
This must be solved algebraically for Y1(s) and Y2(s).
There is of course several ways to do this and next only
one is described:
40
By matrix calculus:
Compute the determinant:
∆ =
∣∣∣∣(s− a11) −a12
−a21 (s− a22)
∣∣∣∣Solution for Y1(s) is:
Y1 =1
∆
∣∣∣∣(y1(0) + G1(s)) −a12
(y2(0) + G2(s)) (s− a22)
∣∣∣∣and solution for Y2(s) is:
Y2 =1
∆
∣∣∣∣(s− a11) (y1(0) + G1(s))
−a21 (y2(0) + G2(s))
∣∣∣∣The solution of the given system is then obtained by
taking the inverse of y1 = L−1(Y1) and y2 = L−1(Y2).
Same procedure can be used to solve second order lin-
ear systems.
Example 48. Mixing problem involving two tanks
Tank T1 contains initially 100 gal of pure water. Tank
T2 contains initially 100 gal of water in which 150 lb of
salt are dissolved. The inflow into T1 is 2 gal/min from
T2 and 6 gal/min containing 6 lb of salt from the outside.
41
The inflow into T2 is 8 gal/min from T1. The outflow from
T2 is 2 + 6 = 8 gal/min. The mixtures are kept uniform
by stirring. Find the salt contents y1(t) and y2(t) in T1
and T2 respectively.
Example 49. Using laplace transform method, find the
currents i1(t) and i2(t) in electrical network which is gov-
erned by following differential equation system
L1i′1 + R1(i1 − i2) + R2i1 = v(t)
L2i′2 + R1(i2 − i1) = 0
where L1 = 0.8 H, L2 = 1 H , R1 = 1 ohms, R2 = 1.4
ohms, v(t) = 100 volts if 0 < t < 0.5 and 0 thereafter
and i1(0) = 0, i2(0) = 0.
Example 50. Secondary circuit with mutual induction:
A voltage e(t) is applied to the primary circuit at time
t = 0, and mutual induction M drives the current i2 in
the secondary circuit. If prior to closing the switch, the
currents in both circuits are zero, determine the induced
current i2 in the secondary circuit at time t when R1 = 4
ohms, R2 = 10 ohms, L1 = 2 H, L2 = 8 H, M = 2 H,
and e(t) = 28 sin 2t V. System of differential equations
for this network is
R1i1 + L1ddti1 + M d
dti2 = e(t)
R2i2 + L2ddti2 + M d
dti1 = 0
42
Laplace transform of a periodic function
If f (t) is a periodic function with period a we have
f (t + a) = f (t). The Laplace transform of f (t) can be
found by integrating over only one period:
L(f ) =1
1− e−as
∫ a
0
f (t)e−stdt (46)
This equation can be derived as follows:
L(f ) =
∫ ∞
0
f (t)e−stdt
=
∫ a
0
f (t)e−stdt +
∫ 2a
a
f (t)e−stdt + ...
=
∞∑n=0
∫ (n+1)a
na
f (t)e−stdt
Next we integrate one term of the sum using the change
of variable t′ = t− na:
∫ (n+1)a
na
f (t)e−stdt =
∫ a
0
f (t′ + na)e−st′−nasdt′
Hence, dropping the prime on t we get
L(f ) =
( ∞∑n=0
e−san
)∫ a
0
f (t)e−stdt
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using the sum formula for geometric series we get
∞∑n=0
xn =1
1− x
to make the replacement
∞∑n=0
e−san =1
1− e−sa
Proving the formula.
Example 51. a) Find the Laplace transform of the saw
tooth function
f (t) =
{t < 0 ≤ t < 1
0 t ≥ 1
b) Find the Laplace transform of the periodic saw tooth
function with period one given by
f (t) = t 0 ≤ t < 1
f (t + 1) = f (t)
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