THE LAPLACE TRANSFORM

4.1 THE LAPLACE TRANSFORM

For a signal x(t), its Laplace transform X(s) is defined by

The signal x(t) is said to be the inverse Laplace transform of X(s). It can be shown that [ 1 ]

where c is a constant chosen to ensure the convergence of the integral in Eq. (4.1) , as explained later. This pair of equations is known as the bilateral Laplace transform pair, where X(s) is the direct Laplace transform of x(t) and x(t) is the inverse

Laplace transform of X(s). Symbolically,

Note that It is also common practice to use a bidirectional arrow to indicate a Laplace transform pair, as follows:

The Laplace transform, defined in this way, can handle signals existing over the entire time interval from −∞ to ∞ (causal and noncausal signals). For this reason it is called the bilateral (or two-sided) Laplace transform. Later we shall consider a special case-the unilateral or one-sided Laplace transform-which can handle only causal signals.

LINEARITY OF THE LAPLACE TRANSFORM

We now prove that the Laplace transform is a linear operator by showing that the principle of superposition holds, implying that if then The proof is simple. By definition

This result can be extended to any finite sum.

THE REGION OF CONVERGENCE (ROC)

The region of convergence (ROC), also called the region of existence, for the Laplace transform X(s), is the set of values of s (the region in the complex plane) for which the integral in Eq. (4.1) converges. This concept will become clear in the following example.

EXAMPLE 4.1

For a signal x(t) = e −at u(t), find the Laplace transform X(s) and its ROC. By definition

Because u(t) = 0 for t < 0 and u(t) = 1 for t ≥ 0,

Note that s is complex and as t → ∞, the term e −(s+a)t does not necessarily vanish. Here we recall that for a complex number z = α + jβ,

Now |e −j βt | = 1 regardless of the value of βt. Therefore, as t → ∞, e −zt → 0 only if α > 0, and e zt → if α < 0. Thus

Clearly

Use of this result in Eq. (4.5) yields

or

The ROC of X(s) is Re s > −a, as shown in the shaded area in Fig. 4.1a . This fact means that the integral defining X(s) in Eq. (4.5) exists only for the values of s in the shaded region in Fig. 4.1

a. For other values of s, the integral in Eq. (4.5) does not converge. For this reason the shaded region is called the ROC (or the region of existence) for X(s).

Figure 4.1: Signals (a) e −at u(t) and (b) −e at u( −t) have the same Laplace transform but different regions of convergence.

REGION OF CONVERGENCE FOR FINITE-DURATION SIGNALS

A finite-duration signal x f (t) is a signal that is nonzero only for t 1 ≤t≤t 2 , where both t 1 and t 2 are finite numbers and t 2 >t 1 . For a finite-duration, absolutely integrable signal, the ROC is the entire s plane. This is clear from the fact that if x f (t) is absolutely integrable and a finite-duration signal, then x(t)e −σt is also absolutely integrable for any value of σ because the integration is over only finite range of t. Hence, the Laplace transform of such a signal converges for every value of s. This means that the ROC of a general signal x(t) remains unaffected by addition of any absolutely

integrable, finite-duration signal x f (t) to x(t). In other words, if represents the ROC of a signal x(t), then the ROC of a signal x(t) + x f (t) is also .

ROLE OF THE REGION OF CONVERGENCE

The ROC is required for evaluating the inverse Laplace transform x(t) from X(s), as defined by Eq. (4.2) . The operation of finding the inverse transform requires an integration in the complex plane, which needs some explanation. The path of integration is along c + j ω, with ω varying from −∞ to ∞. [ † ] Moreover, the path of integration must lie in the ROC (or existence) for X(s). For the signal e −at u(t), this is possible if c > −a. One possible path of integration is shown (dotted) in Fig. 4.1a . Thus, to obtain x(t) from X(s), the integration in Eq. (4.2) is performed along this path. When we integrate [1/(s + a)]e st along this path, the result is e −at u(t). Such integration in the complex plane requires a background in the theory of functions of complex variables. We can avoid this integration by compiling a table of Laplace transforms ( Table 4.1) , where the Laplace transform pairs are tabulated for a variety of signals. To find the inverse Laplace transform of, say, 1/(s + a), instead of using the complex integral of Eq. (4.2) , we look up the table and find the inverse Laplace transform to be e −at u(t) (assuming that the ROC is Re s > −a). Although the table given here is

rather short, it comprises the functions of most practical interest. A more comprehensive table appears in Doetsch. [ 2 ]

Table 4.1: A Short Table of (Unilateral) Laplace Transforms

Open table as spreadsheet

2 u(t) 3 tu(t) 4 t n u(t)

5 e λt u(t) 6 te λt u(t)

7 t n e λt u(t)

8a cos bt u(t) 8b sin bt u(t)

9a e −at cos bt u(t)

9b e −at sin bt u(t)

10a re −at cos (bt + θ)u(t) 10b

re −at cos (bt + θ)u(t) 10c

re −at cos (bt + θ)u(t)

10d

THE UNILATERAL LAPLACE TRANSFORM

To understand the need for defining unilateral transform, let us find the Laplace transform of signal x(t) illustrated in Fig. 4.1b : The Laplace transform of this signal is

Because u( −t) = 1 for t < 0 and u(−t) = 0 for t > 0,

Equation (4.6) shows that

Hence

The signal −e −at u( −t) and its ROC (Re s < −a) are depicted in Fig. 4.1b . Note that the Laplace transforms for the signals e −at u(t) and −e −at u( −t) are identical except for their regions of convergence. Therefore, for a given X(s), there may be more than one inverse transform, depending on the ROC. In other words, unless the ROC is specified, there is no one-to-one correspondence between X(s) and x(t). This fact increases the complexity in using the Laplace transform. The complexity is the result of trying to handle causal as well as noncausal signals. If we restrict all our signals to the causal type, such an ambiguity does not arise. There is only one inverse transform of X(s) = 1/(s + a), namely, e −at u(t). To find x(t) from X(s), we need not even specify the ROC. In summary, if all signals are restricted to the causal type, then, for a given X(s), there is only one inverse transform x(t). [ † ]

The unilateral Laplace transform is a special case of the bilateral Laplace transform in which all signals are restricted to being causal; consequently the limits of integration for the integral in Eq. (4.1) can be taken from 0 to ∞. Therefore, the unilateral Laplace transform X(s) of a signal x(t) is defined as

We choose 0 − (rather than 0 + used in some texts) as the lower limit of integration. This convention not only ensures inclusion of an impulse function at t = 0, but also allows us to use initial conditions at 0 − (rather than at 0 + ) in the solution of differential equations via the Laplace transform. In practice, we are likely to know the initial conditions before the input is applied (at 0 − ), not after the input is applied (at 0 + ). Indeed, the very meaning of the term "initial conditions" implies conditions at t = 0 − (conditions before the input is applied). Detailed analysis of desirability of using t = 0 − appears in Section 4.3 .

The unilateral Laplace transform simplifies the system analysis problem considerably because of its uniqueness property, which says that for a given X(s), there is a unique inverse transform. But there is a price for this simplification: we cannot analyze noncausal systems or use noncausal inputs. However, in most practical problems, this restriction is of little consequence. For this reason, we shall first consider the unilateral Laplace transform and its application to system analysis. (The bilateral Laplace transform is discussed later, in Section 4.11 .)

Basically there is no difference between the unilateral and the bilateral Laplace transform. The unilateral transform is the bilateral transform that deals with a subclass of signals starting at t = 0 (causal signals). Therefore, the expression [ Eq. (4.2) ] for the inverse Laplace transform remains unchanged. In practice, the term Laplace transform means the unilateral Laplace transform.

EXISTENCE OF THE LAPLACE TRANSFORM

The variable s in the Laplace transform is complex in general, and it can be expressed as s = σ + jω. By definition

Because |e j ωt | = 1, the integral on the right-hand side of this equation converges if

Hence the existence of the Laplace transform is guaranteed if the integral in expression (4.9) is finite for some value of σ. Any signal that grows no

faster than an exponential signal Me σ 0 t for some M and σ 0 satisfies the condition (4.9) . Thus, if for some M and σ 0 ,

t we can choose 2 σ>σ 0 to satisfy (4.9) . The signal e , in contrast, grows at a rate faster than e σ 0 t , and consequently not Laplace transformable. [ ‡ ] Fortunately such signals (which are not Laplace transformable) are of little consequence from either a practical or a theoretical viewpoint. If σ 0 is the

smallest value of σ for which the integral in (4.9) is finite, σ 0 is called the abscissa of convergence and the ROC of X(s) is Re s > σ 0 . The abscissa of convergence for e −at u(t) is −a (the ROC is Re s > −a).

EXAMPLE 4.2

Determine the Laplace transform of the following: a. δ(t) b. u(t) Determine the Laplace transform of the following: a. δ(t) b. u(t)

Using the sampling property [ Eq. (1.24a) ], we obtain that is,

b. To find the Laplace transform of u(t), recall that u(t) = 1 for t ≥ 0. Therefore

We also could have obtained this result from Eq. (4.7b) by letting a = 0. c. Because

From Eq. (4.7) , it follows that

For the unilateral Laplace transform, there is a unique inverse transform of X(s); consequently, there is no need to specify the ROC explicitly. For this reason, we shall generally ignore any mention of the ROC for unilateral transforms. Recall, also, that in the unilateral Laplace transform it is understood that every signal x(t) is zero for t < 0, and it is appropriate to indicate this fact by multiplying the signal by u(t).

EXERCISE E4.1

By direct integration, find the Laplace transform X(s) and the region of convergence of X (s) for the signals shown in Fig. 4.2 .

Figure 4.2 Answers