THE BILATERAL LAPLACE TRANSFORM

4.11 THE BILATERAL LAPLACE TRANSFORM

Situations involving noncausal signals and/or systems cannot be handled by the (unilateral) Laplace transform discussed so far. These cases can be analyzed by the bilateral (or two-sided) Laplace transform defined by

and x(t) can be obtained from X(s) by the inverse transformation

Observe that the unilateral Laplace transform discussed so far is a special case of the bilateral Laplace transform, where the signals are restricted to the causal type. Basically, the two transforms are the same. For this reason we use the same notation for the bilateral Laplace transform.

Earlier we showed that the Laplace transforms of e −at u(t) and of −e at u( −t) are identical. The only difference is in their regions of convergence (ROC). The ROC for the former is Re s > −a; that for the latter is Re s < −a; as illustrated in Fig. 4.1 . Clearly, the inverse Laplace transform of X(s) is not unique unless the ROC is specified. If we restrict all our signals to the causal type, however, this ambiguity does not arise. The inverse transform of 1/(s + a) is e −at u(t). Thus, in the unilateral Laplace transform, we can ignore the ROC in determining the inverse transform of X(s).

We now show that any bilateral transform can be expressed in terms of two unilateral transforms. It is, therefore, possible to evaluate bilateral transforms from a table of unilateral transforms.

Consider the function x(t) appearing in Fig. 4.52a . We separate x(t) into two components, x 1 (t) and x 2 (t), representing the positive time (causal) component and the negative time (anticausal) component of x(t), respectively ( Fig. 4.52b and 4.52c) :

Figure 4.52: Expressing a signal as a sum of causal and anticausal components. The bilateral Laplace transform of x(t) is given by

where X 1 (s) is the Laplace transform of the causal component x 1 (t) and X 2 (s) is the Laplace transform of the anticausal component x 2 (t). Consider X 2 (S), given by

Therefore

If x(t) has any impulse or its derivative(s) at the origin, they are included in x 1 (t). Consequently, x 2 (t) = 0 at the origin; that is, x 2 (0) = 0. Hence, the lower limit on the integration in the preceding equation can be taken as 0 − instead of 0 + . Therefore

Because x 2 ( −t) is causal ( Fig. 4.52d) ,X 2 ( −s) can be found from the unilateral transform table. Changing the sign of s in X 2 ( −s) yields X 2 (s). To summarize, the bilateral transform X(s) in Eq. (4.93) can be computed from the unilateral transforms in two steps:

1. Split x(t) into its causal and anticausal components, x 1 (t) and x 2 (t), respectively.

2. The signals x 1 (t) and x 2 ( −t) are both causal. Take the (unilateral) Laplace transform of x 1 (t) and add to it the (unilateral) Laplace

transform of x 2 ( −t), with s replaced by −s. This procedure gives the (bilateral) Laplace transform of x(t).

Since x 1 (t) and x 2 ( −t) are both causal, X 1 (s) and X 2 ( −s) are both unilateral Laplace transforms. Let σ c1 and σ c2 be the abscissas of convergence Since x 1 (t) and x 2 ( −t) are both causal, X 1 (s) and X 2 ( −s) are both unilateral Laplace transforms. Let σ c1 and σ c2 be the abscissas of convergence

Therefore, X 2 (s) exists for all s with Res < σ c2 [ † ] . Therefore, X (s) = X 1 (s)+X 2 (s) exists for all s such that

The regions of convergence (or existence) of X 1 (s), X 2 (s), and X(s) are shown in Fig. 4.53 . Because X(s) is finite for all values of s lying in the strip of convergence ( σ c1 < Re s < −σ c2 ), poles of X (s) must lie outside this strip. The poles of X(s) arising from the causal component x 1 (t) lie to the left of the strip (region) of convergence, and those arising from its anticausal component x 2 (t) lie to its right (see Fig. 4.53) . This fact is of crucial importance in finding the inverse bilateral transform.

Figure 4.53

This result can be generalized to left-sided and right-sided signals. We define a signal x(t) as a right-sided signal if x(t) = 0 for t < T 1 for some finite positive or negative number T 1 . A causal signal is always a right-sided signal, but the converse is not necessarily true. A signal is said to left sided if

it is zero for t > T 2 for some finite, positive, or negative number T 2 .

An anticausal signal is always a left-sided signal, but the converse is not necessarily true. A two-sided signal is of infinite duration on both positive and negative sides of t and is neither right sided nor left sided.

We can show that the conclusions for ROC for causal signals also hold for right-sided signals, and those for anticausal signals hold for left-sided signals. In other words, if x(t) is causal or right-sided, the poles of X(s) lie to the left of the ROC, and if x(t) is anticausal or left-sided, the poles of X(s) lie to the right of the ROC.

To prove this generalization, we observe that a right-sided signal can be expressed as x(t) + x(t), where x(t) is a causal signal and x f (t) is some finite-duration signal. ROC of any finite-duration signal is the entire s-plane (no finite poles). Hence, the ROC of the right-sided signal x(t) + x(t) is

the region common to the ROCs of x(t) and x f (t), which is same as the ROC for x(t). This proves the generalization for right-sided signals. We can use a similar argument to generalize the result for left-sided signals.

Let us find the bilateral Laplace transform of

We already know the Laplace transform of the causal component

For the anticausal component, x 2 (t) = e bt u( −t), we have

so that

Therefore

and the Laplace transform of x(t) in Eq. (4.94) is

Figure 4.54 shows x(t) and the ROC of X(s) for various values of a and b. Equation (4.97) indicates that the ROC of X(s) does not exist if a > b, which is precisely the case in Fig. 4.54g .

Figure 4.54

Observe that the poles of X (s) are outside (on the edges) of the ROC. The poles of X (s) because of the anticausal component of x(t) lie to the right of the ROC, and those due to the causal component of x(t) lie to its left.

When X(s) is expressed as a sum of several terms, the ROC for X(s) is the intersection of (region common to) the ROCs of all the terms. In general, if x(t) = ∑ k i=1 x i (t), then the ROC for X(s) is the intersection of the ROCs (region common to all ROCs) for the transforms X 1 (s), X 2 (s),...,X k (s).

EXAMPLE 4.28

Find the inverse Laplace transform of

if the ROC is a. −2 < Re s < 1 b. Re s > 1 c. Re s < −2

a.

Now, X(s) has poles at −2 and 1. The strip of convergence is −2 < Re s < 1. The pole at −2, being to the left of the strip of convergence, corresponds to the causal signal. The pole at 1, being to the right of the strip of convergence, corresponds to the anticausal signal. Equations (4.95) and (4.96) yield

b. Both poles lie to the left of the ROC, so both poles correspond to causal signals. Therefore c. Both poles lie to the right of the region of convergence, so both poles correspond to anticausal signals, and

Figure 4.55 shows the three inverse transforms corresponding to the same X (s) but with different regions of convergence.

Figure 4.55: Three possible inverse transforms of −3/((s + 2)(s −1)).