Lecture%203%20Laplace%20transform

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Lecture 3Laplace transform
Physics for informatics
Ing. Jaroslav Jíra, CSc.
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The Laplace transform
What is it good for?

• Solving of differential equations
• System modeling
• System response analysis
• Process control application

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The Laplace transform
Solving of differential equations procedure
A system analysis can be done by several simple
steps

1. Finding differential equations describing the
   system
2. Obtaining the Laplace transform of these
   equations
3. Performing simple algebra to solve for output or
   variable of interest
4. Applying inverse transform to find solution

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The Laplace transform
The definition

1. The Laplace transform is an operator that
   switches a function of real variable f(t) to the
   function of complex variable F(s).
2. We are transforming a function of time - real
   argument t to a function of complex angular
   frequency s.
3. The Laplace transform creates an image F(s) of
   the original function f(t)

where s s i?
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The Laplace transform
Restrictions

1. The function f(t) must be at least piecewise
   continuous for t 0.
2. f(t) Meat where M and a are constants. The
   function f(t) must be bounded, otherwise the
   Laplace integral will not converge.
3. We assume that the function f(t) 0 for all t lt 0

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The Laplace transform
Inverse Laplace transform

1. Inverse transform requires complex analysis to
   solve
2. If there exists a unique function F(s)Lf(t),
   then there is also a unique function
   f(t)L-1F(s)
3. Using the previous statement, we can simply
   create a set of transform pairs and calculate the
   inverse transform by comparing our image with
   known results in time scope

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The Laplace transform
Basic properties
Linearity
Scaling in time
Time shift
Frequency shift
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The Laplace transform
Another properties
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The Laplace transform
The most commonly used transform pairs
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The Laplace transformTransform pair deduction
u(t)
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Unit step
The unit step u(t) is defined by
t
0
The Laplace image
u(t)
Shifted unit step
1
t
0
a
The Laplace image
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The Laplace transformTransform pair deduction
f(t)
Unit impulse
1/t1
The unit impulse f(t) is characterized by unit
area under its function
t
0
t1
The Laplace image
f(t)
Dirac delta
It is a unit impulse for t1 ? 0
t
0
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The Laplace transformTransform pair deduction
Exponential function
Linear function
Per partes integration
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The Laplace transformTransform pair deduction
Square function
Per partes integration
The n-th power function
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The Laplace transformTransform pair deduction
Cosine function
Sine function
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The Laplace transformTransform pair deduction
Time shift
f(t)
f(t-a)
The original function f(t) is shifted in time to
f(t-a)
t
0
a
Frequency shift
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The Laplace transformProperty deduction
Time differentiation
Per partes integration
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The Laplace transformProperty deduction
Per partes integration
Time integration
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Inverse Laplace transform
The algorithm of inverse Laplace transform
Since the F(s) is mostly fractional function,
then the most important step is to perform
partial fraction decomposition of it. Depending
on roots in denominator, we are looking for the
following functions, where A and B are real
numbers
for a single real root s a
for a double real root s a
for a triple real root s a
for a pair of pure imaginary roots s i?
for a pair of complex conjugated roots s a i?
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Inverse Laplace transformBasic examples of
partial fraction decomposition to find the
original f(t)
Two distinct real roots
The equation s2 4s 3 0 has two distinct real
roots s1 -3 and s2 -1
We have to find coefficients A and B for
Multiplying the equation by its denominator
Now we can substitute
Decomposed F(s)
so the original function
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Inverse Laplace transform
One real root and one real double root
The denominator has a single root s1 -1 and a
double root s23 -3
We are looking for coefficients A, B and C
Multiplying the equation by its denominator
Now we can substitute to get A,B the C
coefficient can be obtained by comparison of s2
factors
Decomposed F(s)
so the original function
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Inverse Laplace transform
Two pure imaginary roots
Since we know, that
it will be helpful to rearrange the original
formula
Now we can directly write the result
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Inverse Laplace transform
One real root and two pure imaginary roots
We are looking for coefficients A, B and C
Multiplying the equation by its denominator
Now we can substitute to get A B,C coefficients
can be obtained by comparison of s0,s2 factors
Decomposed F(s)
so the original function
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Inverse Laplace transform
Two complex conjugated roots
We have to rearrange the denominator in the first
step
Decomposed F(s)
Now we have to assemble all necessary relations
so the original function
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Solving of differential equations by the Laplace
transform
Example 1 Find the x(t) on the interval lt0,8)
The image of desired function is
From the former definitions we know, that
Then we can write
The original function
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Solving of differential equations by the Laplace
transform
Example 2 Function on the right side
Necessary relations
Equation in the Laplace form
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Solving of differential equations by the Laplace
transform
Example 2 - continued
A formula for the X(s) after the partial fraction
decomposition
after some small arrangements
The original function
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Solving of differential equations by the Laplace
transform
Example 3 Homogeneous second order LDR
Necessary relations
Equation in the Laplace form
The original function
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Solving of differential equations by the Laplace
transform
Example 4 Inhomogeneous second order LDR
Necessary relations
Equation in the Laplace form
knowing that
The original function
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Solving of differential equations by the Laplace
transform
Example 5 Integro-differential equation
Necessary relations
Equation in the Laplace form
The original function