Introduction to Laplace Transform Analysis
The one-sided Laplace transform of a signal is defined
by




When evaluated along the axis (i.e.,
), the
Laplace transform reduces to the unilateral Fourier transform:





An advantage of the Laplace transform is the ability to transform signals which have no Fourier transform. To see this, we can write the Laplace transform as
![$\displaystyle X(s) = \int_0^\infty x(t) e^{-(\sigma + j\omega)t} dt
= \int_0^\infty \left[x(t)e^{-\sigma t}\right] e^{-j\omega t} dt .
$](http://www.dsprelated.com/josimages_new/filters/img1668.png)






Existence of the Laplace Transform
A function





The Laplace transform of a causal, growing exponential function
![$\displaystyle x(t) = \left\{\begin{array}{ll}
A e^{\alpha t}, & t\geq 0 \\ [5pt]
0, & t<0 \\
\end{array}\right.,
$](http://www.dsprelated.com/josimages_new/filters/img1674.png)
![\begin{eqnarray*}
X(s) &\isdef & \int_0^\infty x(t) e^{-st}dt
= \int_0^\infty A...
...alpha \\ [5pt]
\infty, & \sigma<\alpha \\
\end{array} \right.
\end{eqnarray*}](http://www.dsprelated.com/josimages_new/filters/img1675.png)
Thus, the Laplace transform of an exponential
is
, but this is defined only for
re
.
Analytic Continuation
It turns out that the domain of definition of the Laplace transform can be extended
by means of analytic continuation [14, p. 259].
Analytic continuation is carried out by expanding a function of
about all points in its domain of definition, and
extending the domain of definition to all points for which the series
expansion converges.
In the case of our exponential example
the Taylor series expansion of




where, writing as
and using the chain rule for
differentiation,

and so on. We also used the factorial notation
, and we defined the special cases
and
, as is normally done.
The series expansion of
can thus be written
We now ask for what values of does the series Eq.
(D.2)
converge? The value
is particularly easy to
check, since








More generally, let's apply the ratio test for the convergence
of a geometric series. Since the th term of the series is










The analytic continuation of the domain of Eq.(D.1) is now
defined as the union of the disks of convergence for all points
. It is easy to see that a sequence of such disks can
be chosen so as to define all points in the
plane except at the
pole
.
In summary, the Laplace transform of an exponential
is


Analytic continuation works for any finite number of poles of finite order,D.2 and for an infinite number of distinct poles of finite order. It breaks down only in pathological situations such as when the Laplace transform is singular everywhere on some closed contour in the complex plane. Such pathologies do not arise in practice, so we need not be concerned about them.
Relation to the z Transform
The Laplace transform is used to analyze continuous-time
systems. Its discrete-time counterpart is the transform:








In summary,
Note that the plane and
plane are generally related by














Laplace Transform Theorems
Linearity
The Laplace transform is a linear operator. To show this, let
denote a linear combination of signals
and
,




Thus, linearity of the Laplace transform follows immediately from the linearity of integration.
Differentiation
The differentiation theorem for Laplace transforms states that





Proof:
This follows immediately from integration by parts:

since
by assumption.
Corollary: Integration Theorem

Thus, successive time derivatives correspond to successively higher
powers of , and successive integrals with respect to time
correspond to successively higher powers of
.
Laplace Analysis of Linear Systems
The differentiation theorem can be used to convert differential equations into algebraic equations, which are easier to solve. We will now show this by means of two examples.
Moving Mass
Figure D.1 depicts a free mass driven by an external force along
an ideal frictionless surface in one dimension. Figure D.2
shows the electrical equivalent circuit for this scenario in
which the external force is represented by a voltage source emitting
volts, and the mass is modeled by an inductor
having the value
Henrys.
From Newton's second law of motion ``'', we have

![\begin{eqnarray*}
F(s) &=& m\,{\cal L}_s\{{\ddot x}\}\\
&=& m\left[\,s {\cal L...
...right\}\\
&=& m\left[s^2\,X(s) - s\,x(0) - {\dot x}(0)\right].
\end{eqnarray*}](http://www.dsprelated.com/josimages_new/filters/img1738.png)
Thus, given
Laplace transform of the driving force
,
initial mass position, and
-
initial mass velocity,




If the applied external force is zero, then, by linearity of
the Laplace transform, so is
, and we readily obtain


![$\displaystyle u(t)\isdef \left\{\begin{array}{ll}
0, & t<0 \\ [5pt]
1, & t\ge 0 \\
\end{array}\right.,
$](http://www.dsprelated.com/josimages_new/filters/img1745.png)







To summarize, this simple example illustrated use the Laplace transform to solve for the motion of a simple physical system (an ideal mass) in response to initial conditions (no external driving forces). The system was described by a differential equation which was converted to an algebraic equation by the Laplace transform.
Mass-Spring Oscillator Analysis
Consider now the mass-spring oscillator depicted physically in Fig.D.3, and in equivalent-circuit form in Fig.D.4.
![]() |
By Newton's second law of motion, the force applied to a mass
equals its mass times its acceleration:










We have thus derived a second-order differential equation governing
the motion of the mass and spring. (Note that in
Fig.D.3 is both the position of the mass and compression
of the spring at time
.)
Taking the Laplace transform of both sides of this differential equation gives

To simplify notation, denote the initial position and velocity by
and
, respectively. Solving for
gives

denoting the modulus and angle of the pole residue , respectively.
From §D.1, the inverse Laplace transform of
is
, where
is the Heaviside unit step function at time 0.
Then by linearity, the solution for
the motion of the mass is
![\begin{eqnarray*}
x(t) &=& re^{-j{\omega_0}t} + \overline{r}e^{j{\omega_0}t}
= ...
...ga_0}t - \tan^{-1}\left(\frac{v_0}{{\omega_0}x_0}\right)\right].
\end{eqnarray*}](http://www.dsprelated.com/josimages_new/filters/img1771.png)
If the initial velocity is zero (), the above formula
reduces to
and the mass simply oscillates sinusoidally at frequency
, starting from its initial position
.
If instead the initial position is
, we obtain

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