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The Laplace Transform: Introduction

Module by: Carlos E. Davila. E-mail the author

Summary: Introduces the double-sided and single-sided Laplace transforms.

The Double-Sided Laplace Transform

You may have noticed that we avoided taking the Fourier transform of a number of signals. For example, we did not try to compute the Fourier Transform of the unit step function, x(t)=u(t)x(t)=u(t), or the ramp function x(t)=tu(t)x(t)=tu(t). The reason for this is that these signals do not have a Fourier transform that converges to a finite value for all ΩΩ. Recall that in order to deal with periodic signals such as cos(Ω0t)cos(Ω0t) we had to settle for Fourier transforms having impulse functions in them. The Laplace transform gives us a mechanism for dealing with signals that do not have finite-valued Fourier Transforms.

The double-sided Laplace Transform of x(t)x(t) is defined as follows:

X ( s ) = - x ( t ) e - s t d t X ( s ) = - x ( t ) e - s t d t
(1)

where we define

s = σ + j Ω s = σ + j Ω
(2)

We observe that the Laplace transform is a generalization of the Fourier transform since

X ( j Ω ) = X ( s ) s = j Ω X ( j Ω ) = X ( s ) s = j Ω
(3)

Therefore, we can write

X ( s ) = F e - σ t x ( t ) X ( s ) = F e - σ t x ( t )
(4)

The inverse Laplace transform can be derived using this idea. Applying the inverse Fourier transform to Fe-σtx(t)Fe-σtx(t) gives

e - σ t x ( t ) = 1 2 π - F e - σ t x ( t ) e j Ω t d Ω e - σ t x ( t ) = 1 2 π - F e - σ t x ( t ) e j Ω t d Ω
(5)

Using Equation 2 leads to

d Ω = d s j d Ω = d s j
(6)

Substituting ss for ΩΩ and solving for x(t)x(t) in Equation 5 gives

x ( t ) = 1 2 π j σ - j σ + j X ( s ) e s t d s x ( t ) = 1 2 π j σ - j σ + j X ( s ) e s t d s
(7)

Equation Equation 7 is called the inverse Laplace transform. The integration is along a straight line in the complex s-plane corresponding to a fixed value of σσ. This is illustrated in Figure 1.

Figure 1: The inverse Laplace transform integrates along a line having a constant σσ in the complex s-s-plane.
Figure 1 (Lap_splane.png)

It is important that this line exist in a region of the ss-plane that corresponds to the region of convergence for the Laplace transform. The region of convergence is defined as that region in the ss-plane for which

- x ( t ) e - s t d t < - x ( t ) e - s t d t <
(8)

Note that since s=σ+jΩs=σ+jΩ this is equivalent to

- x ( t ) e - σ t d t < - x ( t ) e - σ t d t <
(9)

The Single-Sided Laplace Transform

We define the single-sided Laplace transform as

X ( s ) = 0 - x ( t ) e - s t d t X ( s ) = 0 - x ( t ) e - s t d t
(10)

where the lower limit of integration tacitly includes the point t=0t=0. That is, 0-0- represents a point just to the negative side of t=0t=0. This allows for the integral to take into account signal features that occur at t=0t=0 such as a step or impulse function. The single-sided Laplace transform is motivated by the fact that most signals are turned on at some point. If the region of convergence is not specified then different signals can yield identical double-sided Laplace transforms, as the following example illustrates.

Example 3.1 Consider the signal

x 1 ( t ) = e - α t u ( t ) x 1 ( t ) = e - α t u ( t )
(11)

The double-sided Laplace transform is given by

X 1 ( s ) = 0 e - α t e - s t d t = 0 e - ( s + α ) t d t = - 1 s + α e - ( σ + j Ω + α ) t 0 = - 1 s + α e - ( σ + α ) e j Ω - 1 X 1 ( s ) = 0 e - α t e - s t d t = 0 e - ( s + α ) t d t = - 1 s + α e - ( σ + j Ω + α ) t 0 = - 1 s + α e - ( σ + α ) e j Ω - 1
(12)

the magnitude of e-(σ+α)e-(σ+α) is zero only if σ>-ασ>-α which establishes the region of convergence. Therefore we have

e - α t u ( t ) 1 s + α , σ > - α e - α t u ( t ) 1 s + α , σ > - α
(13)

Now consider the Laplace transform of the signal

x 2 ( t ) = - e - α t u ( - t ) x 2 ( t ) = - e - α t u ( - t )
(14)

We have

X 2 ( s ) = - - 0 e - α t e - s t d t = - - 0 e - ( s + α ) t d t = 1 s + α e - ( σ + j Ω + α ) t - 0 = 1 s + α 1 - e ( σ + α ) e j Ω X 2 ( s ) = - - 0 e - α t e - s t d t = - - 0 e - ( s + α ) t d t = 1 s + α e - ( σ + j Ω + α ) t - 0 = 1 s + α 1 - e ( σ + α ) e j Ω
(15)

Here, the quantity e(σ+α)e(σ+α) is zero only if σ<-ασ<-α so we have

- e - α t u ( - t ) 1 s + α , σ < - α - e - α t u ( - t ) 1 s + α , σ < - α
(16)

which is identical to X1(s)X1(s) except for the region of convergence.

To avoid scenarios where two different signals have the same double-sided Laplace transform, we restrict our signals to those which are assumed to be zero for t<0t<0. Such signals are called causal signals.

The region of convergence for the single-sided Laplace transform is a region in the ss-plane satisfying σ>σminσ>σmin as shown in Figure 2.

Figure 2: The region of convergence of the single-sided Laplace transform in the complex s-s-plane.
Figure 2 (Lap_roc.png)

To see this, we observe that if

0 - x ( t ) e - σ m i n t d t < 0 - x ( t ) e - σ m i n t d t <
(17)

then it must be the case that

0 - x ( t ) e - σ t d t < 0 - x ( t ) e - σ t d t <
(18)

for σ>σminσ>σmin, since e-σte-σt decreases faster than e-σminte-σmint. Finally, the inverse single-sided Laplace transform is the same as the inverse double-sided Laplace transform (see Equation 7), since a single sided Laplace transform can be interpreted as the double-sided Laplace transform of a signal satisfying x(t)=0,t<0x(t)=0,t<0. From here on, we will work exclusively with the single-sided Laplace transform. Unless we need to specifically differentiate between the single or double-sided transforms, we will refer to the single-sided Laplace transform as simply the “Laplace transform".

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