Laplace
Transformation
Control system
Laplace Transformation is a very useful tool for analysis of a dynamic systemin Frequency-
Domain. This transformation helps to transform differential equations into the form of
algebraic equations which is easier to manipulate equations into the form of algebraic
equation
N UA A
LINEAR CONTROL SYSTEM
Laplace
Transformation
Laplace Transformation
Submitted by
BIMAN RIMAL
SL1703006
Submitted To
1
Zhen Zi Yang
Contents
Introduction to Laplace Transformation........................................................................................... 3
Laplace Transforms ...................................................................................................................... 4
properties & formulas .................................................................................................................. 6
Linearity ................................................................................................................................... 7
One-to-one property ................................................................................................................. 8
Inverse Laplace form................................................................................................................ 8
Time scaling ............................................................................................................................. 8
Exponential scaling .................................................................................................................. 9
Time delay.............................................................................................................................. 10
Derivative ............................................................................................................................... 11
Laplace Transformation
Integral ................................................................................................................................... 12
Multiplication by t .................................................................................................................. 14
Convolution ............................................................................................................................ 15
Initial Value Theorem ............................................................................................................ 16
Final Value Theorem.............................................................................................................. 17
Solution of ODEs by Laplace Transforms Procedure: ............................................................... 18
Conclusion.................................................................................................................................. 19
2
Introduction to Laplace Transformation
Continuous Time
Study Linear Study
time-domain response frequency-domain
system response
Laplace Transformation
Transfer Frequency
function Differential
characteristic
Laplace
equation
Fourier
transform transform
Laplace Transformation is a very useful tool for analysis of a dynamic
system in Frequency-Domain. This transformation helps to transform
differential equations into the form of algebraic equations which is
easier to manipulate equations into the form of algebraic equations
which is easier to manipulate.
3
Laplace Transformation
Laplace Transforms
• It is important analytical method for solving linear ordinary differential
equations. And for Nonlinear ODEs it must be linearize first.
• Laplace transforms play a key role in important process control
concepts and techniques. - Examples:
• Transfer functions
• Frequency response
• Control system design
• Stability analysis
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Definition of Laplace equation
• A time-domain signal f(t) which may represent a forcing function or the response of
a system may be transformed into frequency domain by using the following
transformation:
Where s = σ + jω, is a complex variable.
• F is a complex-valued function of complex numbers
• s is called the (complex) frequency variable, with units sec−1; t is called the time
Laplace Transformation
variable (in sec); st is unitless
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Laplace Transformation
properties & formulas
linearity
the inverse Laplace transform
time scaling
exponential scaling
time delay
derivative
integral
multiplication by t
convolution
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Linearity
Both Land L^-1are linear operators .
Laplace Transformation
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One-to-one property
the Laplace transform is one-to-one: if L(f) = L(g) then f = g
Inverse Laplace form
By definition, the inverse Laplace transform operator, L-1, converts an s-
domain function back to the corresponding time domain function:
Laplace Transformation
where σ is large enough that F(s) is defined for <s ≥ σ
Time scaling
1 𝑠𝑠
Assume X(s)= L[x(t)] then L[x(at)]= 𝑥𝑥( )
𝑎𝑎 𝑎𝑎
Example:
L(et) = 1/(s − 1) so
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Exponential scaling
let f be a signal and α a scalar, and define g(t) = eatf(t); then
G(s) = F(s − a)
example: L(cost) = s/(s2 + 1), and hence
Laplace Transformation
9
Time delay
Time delays occur due to fluid flow, time required to do an
analysis (e.g., gas chromatograph
The time shift property states
We again prove by going back to the original definition of the
Laplace Transform
Because
Laplace Transformation
we can change the lower limit of the integral from 0- to a- and drop
the step function (because it is always equal to one)
We can make a change of variable
The last integral is just the definition of the Laplace Transform, so
we have the time delay property
f(t) g(t)
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Derivative
This is a very important transform because derivatives appear in the
ODEs we wish to solve.
if signal f is continuous at t = 0, then
L(f0) = sF(s) − f(0)
• time-domain differentiation becomes multiplication by frequency
variable s (as with phasors)
• plus a term that includes initial condition (i.e., −f(0)) higher-order
derivatives: applying derivative formula twice yields
L(f00) = sL(f0) − f0(0)
= s(sF(s) − f(0)) − f0(0)
s2F(s) − sf(0) − f0(0)
Laplace Transformation
=
similar formulas hold for L(f(k))
examples
• f(t) = et, so f0(t) = et and
using the formula, , which is the same
, so
• f is unit ramp, so f0 is unit step 11
derivation of derivative formula: start from the defining integral
integration by parts yields
Laplace Transformation
Integral
let g be the running integral of a signal f, i.e.,
then
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i.e., time-domain integral becomes division by frequency variable s example: f = δ, so F(s) = 1; g
is the unit step function
G(s) = 1/s
example: f is unit step function, so F(s) = 1/s; g is the unit ramp function (g(t) = t for t ≥ 0),
G(s) = 1/s2
derivation of integral formula:
Laplace Transformation
here we integrate horizontally first over the triangle 0 ≤ τ ≤ t
let’s switch the order, i.e., integrate vertically first:
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Multiplication by t
let f be a signal and define
g(t) = tf(t)
then we have
G(s) = −F0(s)
to verify formula, just differentiate both sides of
with respect to s to get
Laplace Transformation
• f(t) = e−t, g(t) = te−t
• in general,
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Convolution
Let F(s) and G(s) denote the Laplace transforms of f(t) and g(t),
respectively. Then the product H(s)=F(s) G(s) is the Laplace transform
of the convolution of f(t) and g(t), and is denoted by h(t) =(f+g) (t), and
has the integral representation
Laplace Transformation
Example Find the convolution of f and g if f(t) = t u(t) and g(t) = t2 u(t)
Solution f(t − x)=(t − x)u(t − x) and g(x) = x2 u(x)
𝑡𝑡 2 1 1
Therefore (f ∗ g)(t) = ∫0 (𝑡𝑡 − 𝑥𝑥)𝑥𝑥 𝑑𝑑𝑑𝑑 =[3 𝑥𝑥 3 𝑡𝑡 − 4 𝑥𝑥 4 ]𝑡𝑡0
1 1 1
= 𝑡𝑡 4 − 𝑡𝑡 4 = 𝑡𝑡 4
3 4 4
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Initial Value Theorem
The initial value theorem states
To show this, we first start with the Derivative Rule:
We then invoke the definition of the Laplace Transform, and split the
integral into two parts:
We take the limit as s→∞:
Laplace Transformation
Several simplifications are in order. In the left hand expression, we
can take the second term out of the limit, since it doesn't depend on
's.' In the right hand expression, we can take the first term out of the
limit for the same reason, and if we substitute infinity for 's' in the
second term, the exponential term goes to zero:
The two f(0-) terms cancel each other, and we are left with the Initial
Value Theorem
This theorem only works if F(s) is a strictly proper fraction in which
the numerator polynomial is of lower order then the denominator
polynomial. In other words is will work for F(s)=1/(s+1) but not
16 F(s)=s/(s+1).
Final Value Theorem
The final value theorem states that if a final value of a function exists
that
However, we can only use the final value if the value exists (function
like sine, cosine and the ramp function don't have final values). To
prove the final value theorem, we start as we did for the initial value
theorem, with the Laplace Transform of the derivative,
We let s→0,
Laplace Transformation
As s→0 the exponential term disappears from the integral. Also, we
can take f(0-) out of the limit (since it doesn't depend on s)
We can evaluate the integral
Neither term on the left depends on s, so we can remove the limit and
simplify, resulting in the final value theorem
Examples of functions for which this theorem can't be used are
increasing exponentials (like eat where a is a positive number) that go
to infinity as t increases, and oscillating functions like sine and cosine
that don't have a final value.
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Solution of ODEs by Laplace Transforms Procedure:
1. Take the L of both sides of the ODE.
2. Rearrange the resulting algebraic equation in the s domain to solve for
the L of the output variable, e.g., Y(s).
3. Perform a partial fraction expansion.
4. Use the L-1 to find y(t) from the expression for Y(s).
Consider the differential equation
with the initial conditions.
Laplace Transformation
Proceeding using the steps given
Step 1:
Step 2:
Step 3:
The complex function F(s) must be decomposed into partial fractions
in order to use the tables of correspondences. This gives
Step 4: Use the L-1 to find f(t) from the expression for F(s).
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Conclusion
Laplace transform is a tool for solving Constant coefficient linear
different equation. The process consists of 4 steps converting to
Laplace form, solving algebraic equation for s, Partial
differentiation and the performing Inverse Laplace transformation.
In this way the Laplace transformation reduces the differential
equation to algebraic problem. Simplify algebra and perform
Laplace inverse to obtain solution easily.
Laplace Transformation
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