analise sinais ppt05e (Laplace Transform)

analise_sinais_ppt05e (Laplace Transform)Signal Analysis
J. A. M. Felippe de SouzaJ. A. M. Felippe de Souza
Laplace Transform ______________________________________________________________________________________________________________________________________________________________________________________
Pierre Simon Laplace (1749-1827)
The Laplace Transform presents a representation of signals in the frequency domain as a function of a variable
“s” which is a complex number,
s = σ + jω
unilateral Laplace Transform
(for t ≥ 0)
exponential function
x(t) = e–at ⋅u1(t)
β<<α=−⋅ β
α a),a(xdt)at(u)t(x o
sine
s tcos
ω+ =ωL
Homogeneity:
Additivity:
Linearity:
Time shifting (“translation”):
particular case
Derivatives:
zero initial conditions
L
k k
t
⋅ = L
(continued)
[ ] )s(X)s(X)t(x*)t(x 2121 ⋅=L
the convolution
x(t) = 2 uo(t – a)
X(s) = 2 e–as
There is no unilateral Laplace Transform, as we have defined.
Example 3
Example 4
)t2(ua)t(x 1 −⋅−=
There is no unilateral Laplace Transform, as we have defined.
Example 5
also obtain X2(s) from X1(s).
These signs are actually the same sign written in different time scales.
using only the properties of the Laplace Transform, in particular the derivative, we can show that:
[ ] 22s
L
We have seen that the Laplace Transform of the sine is:
which is the Laplace Transform of the co-sine, as already seen before too.
the property of multiplication by exponential for the singular
signals un(t) divided by n!
recursively applying the property
the exponential signal.
applying the property of the signal multiplied by exponential one can easily obtain
22)as( )s(X
ω++ ω=
the sine multiplied by exponential
and, again applying the property of the signal multiplied by exponential, one can obtain:
22)as(
Transform of all the signals we will need.
In the next 3 slides we present a Table of the Laplace Transform for some known signals.
This Table has the Laplace Transform of all the signals we will need.
The Inverse Laplace Transforms
The Laplace Transform of the main signals of interest for linear time
invariant systems (LTI) have the form of a rational fraction
real poles,
Caso 2 – Complex poles
(s )
- L
)s(q
)s(p
Example 8
and using the Laplace Transforms Table, we can easily find:
Example 8 (continued)
Example 9 (continued)
Solution of the ordinary differential equation (ODE) using Laplace Transform
Solution of the ordinary differential equation (ODE) using Laplace Transform
Linear time invariant systems (LTI) are represented by ordinary differential
equation (ODE) where the input x(t) is known, as well as the initial
conditions of the output y(t), that is,
y(0), y’(0), y’’(0), etc.
and we wish to calculate the output y(t), i.e., the solution of the ODE.
Example 10
x(t) = u1(t) = unit step
By doing the Laplace Transform of (1) using the initial conditions (2), we will obtain:
that is,
and then,
and, by eq. (3), x(t) = u1(t) = unit step, we have that X(s) = 1/s,
thus:
This is a real pole case
Now the solution y(t) of this ODE is found doing the inverse Laplace transform
of Y(s).
s = 0
y(0) = 0 and y’(0) = 4;
x(t) = 0
that is,
(5)
(6)
By doing the Laplace Transform of (4) using the initial conditions (5), and
since x(t) = 0 by eq. (6), then we obtain:
This is a case of a pair of complex-conjugate poles, roots of
Now the solution y(t) of this ODE is found by doing the inverse Laplace
transform of Y(s).
[ ])s(Y)t(y 1−= - L
(continued)Example 11
which is an algebraic equation in ‘s’ and whose solution is:
and therefore,
Example 12
x(t) = u1(t) = unit step
and then,
thus, since X(s) = 1/s, by eq. (9), then we have that:
(7)
(8)
(9)
By doing the Laplace Transform of (7) using the initial conditions (8), we will obtain:
)s/1()s(Y)s(sY1)t(ys2 =++−
which has already been calculated in Example 8, that is,
Now the solution y(t) of this ODE is found by doing the inverse Laplace
transform of Y(s).
[ ])s(Y)t(y 1−= - L
(continued)Example 12
which is an algebraic equation in ‘s’ and whose solution is:
This is a case of a pair of real pole (s = 0), and a pair of complex-conjugate
poles, roots of
A classical result of the Systems Theory, which we have already seen in
section 4.3, is that the output y(t) of a system is the convolution between
h(t) and x(t), that is
)t(x*)t(h)t(y =
In other words, the output for a linear time invariant system (LTI system) takes the form of convolution integral,
τ⋅τ−⋅τ=τ⋅τ⋅τ−= +∞
∞−
+∞
∞− d)t(x)(hd)(x)t(h)t(y
Using the property of the convolution for the Laplace Transform, that is, the
transform of the convolution is the product of the transformed signals, we have:
)s(X)s(H)s(Y ⋅=
Como a Laplace Transform of the unit impulse uo(t) is 1,
that is:
then when the input x(t) is a unit impulse uo(t), i.e.,
we have that X(s) = 1 and therefore, Y(s) = H(s) × 1, that is,
which implies that
that is, the output y(t) becomes the impultional response, as we should
expect.

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