4.11 THE BILATERAL LAPLACE [TRANSFORM](#page-10-0)

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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

$$X(s) = \int_{-\infty}^{\infty} x(t)e^{-st} dt$$

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

$$x(t) = \frac{1}{2\pi j} \int_{c-j\infty}^{c+j\infty} X(s)e^{st} ds$$

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−atu(t) and of −e−atu(−t) are identical. The only difference is in their regions of convergence (ROC). The ROC for the former is Res > −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−atu(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. 4.56a. We separate x(t) into two components, x1(t) and x2(t), representing the positive time (causal) component and the negative time (anticausal) component of x(t), respectively (Figs. 4.56b and 4.56c):

$$x_1(t) = x(t)u(t)$$

and $x_2(t) = x(t)u(-t)$

Figure 4.56 Expressing a signal as a sum of causal and anticausal components.

The bilateral Laplace transform of x(t) is given by

$$X(s) = \int_{-\infty}^{\infty} x(t)e^{-st} dt$$

=

$$\int_{-\infty}^{0^-} x_2(t)e^{-st} dt + \int_{0^-}^{\infty} x_1(t)e^{-st} dt$$

=

$$X_2(s) + X_1(s)$$

(4.57)

where X1(s) is the Laplace transform of the causal component x1(t), and X2(s) is the Laplace transform of the anticausal component x2(t). Consider X2(s), given by

$$X_2(s) = \int_{-\infty}^{0^-} x_2(t)e^{-st} dt = \int_{0^+}^{\infty} x_2(-t)e^{st} dt$$

Therefore,

$$X_2(-s) = \int_{0^+}^{\infty} x_2(-t)e^{-st} dt$$

If x(t) has any impulse or its derivative(s) at the origin, they are included in x1(t). Consequently, x2(t) = 0 at the origin; that is, x2(0) = 0. Hence, the lower limit on the integration in the preceding equation can be taken as 0⁻ instead of 0+. Therefore,

$$X_2(-s) = \int_{0^-}^{\infty} x_2(-t)e^{-st} dt$$

Because x2(−t) is causal (Fig. 4.56d), X2(−s) can be found from the unilateral transform table. Changing the sign of s in X2(−s) yields X2(s).

To summarize, the bilateral transform X(s) in Eq. (4.57) can be computed from the unilateral transforms in two steps:

• 1. Split x(t) into its causal and anticausal components, x1(t) and x2(t), respectively.
• 2. Since the signals x1(t) and x2(−t) are both causal, take the (unilateral) Laplace transform of x1(t) and add to it the (unilateral) Laplace transform of x2(−t), with s replaced by −s. This procedure gives the (bilateral) Laplace transform of x(t).

Since x1(t) and x2(−t) are both causal, X1(s) and X2(−s) are both unilateral Laplace transforms. Let σ*^c1 and σ^c2 be the abscissas of convergence of X1(s) and X2(−s), respectively. This statement implies that X1(s) exists for all s with Re s > σc1, and X2(−s) exists for all s with Res* > σc2. Therefore, X2(s) exists for all s with Re s < −σc2. † Therefore, X(s) = X1(s) + X2(s) exists for all s such that

$$\sigma_{c1} < \text{Re}\,s < -\sigma_{c2}$$

The regions of convergence of X1(s), X2(s), and X(s) are shown in Fig. 4.57. Because X(s) is finite for all values of s lying in the strip of convergence (σ*^c1 < Re s < −σc*2), poles of X(s) must lie outside this strip. The poles of X(s) arising from the causal component x1(t) lie to the left of the strip (region) of convergence, and those arising from its anticausal component x2(t) lie to its right (see Fig. 4.57). This fact is of crucial importance in finding the inverse bilateral transform.

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 < T¹ for some finite positive or negative number T1. 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 > T² for some finite, positive, or negative number T2. 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

^† For instance, if ^x(t) exists for all ^{t &}gt; 10, then ^x(−t), its time-inverted form, exists for ^{t &}lt; ⁻10.

Figure 4.57 Regions of convergence for causal, anticausal, and combined signals.

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) + xf(t), where x(t) is a causal signal and xf(t) is some finite-duration signal. The ROC of any finite-duration signal is the entire s-plane (no finite poles). Hence, the ROC of the right-sided signal x(t) + xf(t) is the region common to the ROCs of x(t) and xf(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

$$x(t) = e^{bt}u(-t) + e^{at}u(t)$$

\n(4.58)

We already know the Laplace transform of the causal component

$$e^{at}u(t) \Longleftrightarrow \frac{1}{s-a} \qquad \text{Re}\,s > a \tag{4.59}$$

For the anticausal component, x2(t) = ebtu(−t), we have

$$x_2(-t) = e^{-bt}u(t) \Longleftrightarrow \frac{1}{s+b} \qquad \text{Re}\, s > -b$$

so that

$$X_2(s) = \frac{1}{-s+b} = \frac{-1}{s-b} \qquad \text{Re}\, s < b$$

Therefore,

$$e^{bt}u(-t) \Longleftrightarrow \frac{-1}{s-b} \qquad \text{Re}\,s < b \tag{4.60}$$

and the Laplace transform of x(t) in Eq. (4.58) is

$$X(s) = -\frac{1}{s-b} + \frac{1}{s-a}$$

Res > a and Res < b

$$\frac{a-b}{(s-b)(s-a)}$$

Res < b
(4.61)

Figure 4.58 shows x(t) and the ROC of X(s) for various values of a and b. Equation (4.61) indicates that the ROC of X(s) does not exist if a > b, which is precisely the case in Fig. 4.58f. 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 xi(t), then the ROC for X(s) is the intersection of the ROCs (region common to all ROCs) for the transforms X1(s), X2(s), … , Xk(s).

Figure 4.58 Various two exponential signals and their regions of convergence.

EXAMPLE 4.32 Inverse Bilateral Laplace Transform

Find the inverse bilateral Laplace transform of

$$X(s) = \frac{-3}{(s+2)(s-1)}$$

if the ROC is (a) −2 < Re s < 1, (b) Re s > 1, and (c) Re s < −2.

(a)

$$X(s) = \frac{1}{s+2} - \frac{1}{s-1}$$

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 a causal signal. The pole at 1, being to the right of the strip of convergence, corresponds to an anticausal signal. Equations (4.59) and (4.60) yield

$$x(t) = e^{-2t}u(t) + e^t u(-t)$$

(b) Both poles lie to the left of the ROC, so both poles correspond to causal signals. Therefore,

)u(t)

^x(t) ⁼ (e−2*^t* ⁻et

Figure 4.59 Three possible inverse transforms of −3/((s +2)(s−1)).

(c) Both poles lie to the right of the region of convergence, so both poles correspond to anticausal signals, and

$$x(t) = (-e^{-2t} + e^t)u(-t)$$

Figure 4.59 shows the three inverse transforms corresponding to the same X(s) but with different regions of convergence.

4.11-1 Properties of the Bilateral Laplace Transform

Properties of the bilateral Laplace transform are similar to those of the unilateral transform. We shall merely state the properties here without proofs. Let the ROC of X(s) be a < Re s < b. Similarly, let the ROC of Xi(s) be ai < Re s < bi for (i = 1, 2).

LINEARITY

$$a_1x_1(t) + a_2x_2(t) \Longleftrightarrow a_1X_1(s) + a_2X_2(s)$$

The ROC for a1X1(s) + a2X2(s) is the region common to (intersection of) the ROCs for X1(s) and X2(s).

TIME SHIFT

$$x(t-T) \Longleftrightarrow X(s)e^{-sT}$$

The ROC for X(s)e−sT is identical to the ROC for X(s).

FREQUENCY SHIFT

$$x(t)e^{s_0t} \Longleftrightarrow X(s-s_0)$$

The ROC for X(s−s0) is a+c < Re s < b+c, where c = Re s0.

TIME DIFFERENTIATION

$$\frac{dx(t)}{dt} \Longleftrightarrow sX(s)$$

The ROC for sX(s) contains the ROC for X(s) and may be larger than that of X(s) under certain conditions [e.g., if X(s) has a first-order pole at s = 0, it is canceled by the factor s in sX(s)].

TIME INTEGRATION

$$\int_{-\infty}^{t} x(\tau) d\tau \Longleftrightarrow X(s)/s$$

The ROC for sX(s) is max (a, 0) < Re s < b.

TIME SCALING

$$x(\beta t) \Longleftrightarrow \frac{1}{|\beta|}X\left(\frac{s}{\beta}\right)$$

The ROC for X(s/β) is βa < Re s < βb. For β > 1, x(βt) represents time compression and the corresponding ROC expands by factor β. For 0 >β> 1, x(βt) represents time expansion and the corresponding ROC is compressed by factor β.

TIME CONVOLUTION

$$x_1(t) * x_2(t) \Longleftrightarrow X_1(s)X_2(s)$$

The ROC for X1(s)X2(s) is the region common to (intersection of ) the ROCs for X1(s) and X2(s).

FREQUENCY CONVOLUTION

$$x_1(t)x_2(t) \Longleftrightarrow \frac{1}{2\pi j}\int_{c-j\infty}^{c+j\infty} X_1(w)X_2(s-w) dw$$

The ROC for X1(s) ∗ X2(s) is a¹ +a² < Re s < b¹ +b2.

TIME REVERSAL

$$x(-t) \Longleftrightarrow X(-s)$$

The ROC for X(−s) is −b < Re s < −a.

4.11-2 Using the Bilateral Transform for Linear System Analysis

Since the bilateral Laplace transform can handle noncausal signals, we can analyze noncausal LTIC systems using the bilateral Laplace transform. We have shown that the (zero-state) output y(t) is given by

$$y(t) = \mathcal{L}^{-1}[X(s)H(s)]$$

This expression is valid only if X(s)H(s) exists. The ROC of X(s)H(s) is the region in which both X(s) and H(s) exist. In other words, the ROC of X(s)H(s) is the region common to the regions of convergence of both X(s) and H(s). These ideas are clarified in the following examples.

Find the current y(t) for the RC circuit in Fig. 4.60a if the voltage x(t) is

$$x(t) = e^t u(t) + e^{2t} u(-t)$$

Figure 4.60 Response of a circuit to a noncausal input.

The transfer function H(s) of the circuit is given by

$$H(s) = \frac{s}{s+1} \qquad \text{Re}\, s > -1$$

Because h(t) is a causal function, the ROC of H(s) is Re s > −1. Next, the bilateral Laplace transform of x(t) is given by

$$X(s) = \frac{1}{s-1} - \frac{1}{s-2} = \frac{-1}{(s-1)(s-2)} \qquad 1 < \text{Re } s < 2$$

The response y(t) is the inverse transform of X(s)H(s):

$$y(t) = \mathcal{L}^{-1} \left[ \frac{-s}{(s+1)(s-1)(s-2)} \right] = \mathcal{L}^{-1} \left[ \frac{1}{6} \frac{1}{s+1} + \frac{1}{2} \frac{1}{s-1} - \frac{2}{3} \frac{1}{s-2} \right]$$

The ROC of X(s)H(s) is that ROC common to both X(s) and H(s). This is 1 < Re s < 2. The poles s = ±1 lie to the left of the ROC and, therefore, correspond to causal signals; the pole

454 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS

s = 2 lies to the right of the ROC and thus represents an anticausal signal. Hence,

$$y(t) = \frac{1}{6}e^{-t}u(t) + \frac{1}{2}e^{t}u(t) + \frac{2}{3}e^{2t}u(-t)$$

Figure 4.60c shows y(t). Note that in this example, if

$$x(t) = e^{-4t}u(t) + e^{-2t}u(-t)$$

then the ROC of X(s) is −4 < Re s < −2. Here no region of convergence exists for X(s)H(s). Hence, the response y(t) goes to infinity.

EXAMPLE 4.34 Response of a Noncausal System

Find the response y(t) of a noncausal system with the transfer function

$$H(s) = \frac{-1}{s-1} \qquad \text{Re}\, s < 1$$

to the input x(t) = e−2*^t u*(t).

We have

$$X(s) = \frac{1}{s+2} \qquad \text{Re}\, s > -2$$

and

$$Y(s) = X(s)H(s) = \frac{-1}{(s-1)(s+2)}$$

The ROC of X(s)H(s) is the region −2 < Re s < 1. By partial fraction expansion,

$$Y(s) = \frac{-1/3}{s-1} + \frac{1/3}{s+2} \qquad -2 < \text{Re}\, s < 1$$

and

$$y(t) = \frac{1}{3} [e^t u(-t) + e^{-2t} u(t)]$$

Note that the pole of H(s) lies in the RHP at 1. Yet the system is not unstable. The pole(s) in the RHP may indicate instability or noncausality, depending on its location with respect to the region of convergence of H(s). For example, if H(s) = −1/(s−1) with Re s > 1, the system is causal and unstable, with h(t) = −et u(t). In contrast, if H(s) = −1/(s − 1) with Re s < 1, the system is noncausal and stable, with h(t) = et u(−t).

EXAMPLE 4.35 System Response to a Noncausal Input

Find the response y(t) of a system with the transfer function

$$H(s) = \frac{1}{s+5} \qquad \text{Re}\, s > -5$$

and the input

$$x(t) = e^{-t}u(t) + e^{-2t}u(-t)$$

The input x(t) is of the type depicted in Fig. 4.58f, and the region of convergence for X(s) does not exist. In this case, we must determine separately the system response to each of the two input components, x1(t) = e−^t u(t) and x2(t) = e−2*^t u*(−t).

$$X_1(s) = \frac{1}{s+1} \qquad \text{Re}\, s > -1$$

\n

$$X_2(s) = \frac{-1}{s+2} \qquad \text{Re}\, s < -2$$

If y1(t) and y2(t) are the system responses to x1(t) and x2(t), respectively, then

$$Y_1(s) = \frac{1}{(s+1)(s+5)} = \frac{1/4}{s+1} - \frac{1/4}{s+5}$$

Re $s > -1$

so that

$$y_1(t) = \frac{1}{4}(e^{-t} - e^{-5t})u(t)$$

and

$$Y_2(s) = \frac{-1}{(s+2)(s+5)} = \frac{-1/3}{s+2} + \frac{1/3}{s+5} \qquad -5 < \text{Re}\,s < -2$$

so that

$$y_2(t) = \frac{1}{3} [e^{-2t}u(-t) + e^{-5t}u(t)]$$

Therefore,

$$y(t) = y_1(t) + y_2(t) = \frac{1}{3}e^{-2t}u(-t) + \left(\frac{1}{4}e^{-t} + \frac{1}{12}e^{-5t}\right)u(t)$$