15.2 Definition of the Laplace Transform

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15.2 Definition of the Laplace Transform

Given a function f(t), its Laplace transform, denoted by F(s) or [ f(t)], is defined by

$$\mathcal{L}[f(t)] = F(s) = \int_{0^{-}}^{\infty} f(t)e^{-st} dt$$

(15.1)

where s is a complex variable given by

$$s = \sigma + j\omega \tag{15.2}$$

Because the argument st of the exponent e in Eq. (15.1) must be dimensionless, it follows that s has the dimensions of frequenc y and units of inverse seconds (s ⁻¹ ) or “frequenc y.” In Eq. (15.1), the lo wer limit is specified as 0⁻ to indicate a time just before t = 0. We use 0⁻ as the lower limit to include the origin and capture an y discontinuity of f(t) at t = 0; this will accommodate functions—such as singularity functions—that may be discontinuous at t = 0.

It should be noted that the integral in Eq. (15.1) is a definite integral with respect to time. Hence, the result of inte gration is indepen dent of time and only involves the variable “s.”

Equation (15.1) illustrates the general concept of transformation. The function f(t) is transformed into the function F(s). Whereas the former function involves t as its argument, the latter involves s. We say the transformation is from t-domain to s-domain. Gi ven the interpre tation of s as frequenc y, we arri ve at the follo wing description of the Laplace transform:

The Laplace transform is an integral transformation of a function f(t) from the time domain into the complex frequency domain, giving F(s).

When the Laplace transform is applied to circuit analysis, the differential equations represent the circuit in the time domain. The terms in the differential equations take the place of f(t). Their Laplace transform, which corresponds to F(s), constitutes algebraic equations representing the circuit in the frequency domain.

For an ordinary function f(t), the lower limit can be replaced by 0.

We assume in Eq. (15.1) that f (t) is ignored for t < 0. To ensure that this is the case, a function is often multiplied by the unit step. Thus, f (t) is written as f (t)u(t) or f (t), t ≥ 0.

The Laplace transform in Eq. (15.1) is kno wn as the one-sided (or unilateral) Laplace transform. The two-sided (or bilateral) Laplace transform is given by

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

(15.3)

The one-sided Laplace transform in Eq. (15.1), being adequate for our purposes, is the only type of Laplace transform that we will treat in this book.

A function f (t) may not ha ve a Laplace transform. F or f (t) to have a Laplace transform, the inte gral in Eq. (15.1) must con verge to a finite value. Because |ejω^t | = 1 for any value of t, the integral converges when

$$\int_0^\infty e^{-\sigma t} |f(t)| \, dt < \infty \tag{15.4}$$

for some real v alue σ = σc. Thus, the re gion of con vergence for the Laplace transform is Re(s) = σ > σc, as shown in Fig. 15.1. In thisregion, |F(s)| < ∞ and F(s) exists. F(s) is undefined outside the region of convergence. Fortunately, all functions of interest in circuit analysis satisfy the convergence criterion in Eq. (15.4) and have Laplace transforms. Therefore, it is not necessary to specify σc in what follows.

A companion to the direct Laplace transform in Eq. (15.1) is the inverse Laplace transform given by

$$\mathcal{L}^{-1}[F(s)] = f(t) = \frac{1}{2\pi j} \int_{\sigma_1 - j\infty}^{\sigma_1 + j\infty} F(s)e^{st} ds$$

(15.5)

where the integration is performed along a straight line (σ1 + jω, −∞ < ω < ∞) in the region of convergence, σ1 > σc. See Fig. 15.1. The direct application of Eq. (15.5) in volves some kno wledge about comple x analysis beyond the scope of this book. F or this reason, we will not use Eq. (15.5) to find the inverse Laplace transform. We will rather use a look-up table, to be de veloped in Section 15.3. The functions f(t) and F(s) are regarded as a Laplace transform pair where

$$f(t) \qquad \Leftrightarrow \qquad F(s) \tag{15.6}$$

meaning that there is one-to-one correspondence between f(t) and F(s). The following examples derive the Laplace transforms of some important functions.

Example 15.1

Determine the Laplace transform of each of the following functions: (a) u(t), (b) e⁻at u(t), a ≥ 0, and (c) δ(t).

Solution:

(a) For the unit step function u(t), shown in Fig. 15.2(a), the Laplace transform is

$$\mathcal{L}[u(t)] = \int_{0^{-}}^{\infty} 1e^{-st} dt = -\frac{1}{s} e^{-st} \Big|_{0}^{\infty}$$

= $-\frac{1}{s}(0) + \frac{1}{s}(1) = \frac{1}{s}$ (15.1.1)

ωt + sin*²*

ωt = 1

Figure 15.1 Region of convergence for the Laplace transform.

|e^jω^t

| = cos*²*

(b) F or the e xponential function, sho wn in Fig. 15.2(b), the Laplace transform is

$$\mathcal{L}[e^{-at} u(t)] = \int_{0^{-}}^{\infty} e^{-at} e^{-st} dt$$

= $-\frac{1}{s+a} e^{-(s+a)t} \Big|_{0}^{\infty} = \frac{1}{s+a}$ (15.1.2)

(c) For the unit impulse function, shown in Fig. 15.2(c),

$$\mathcal{L}[\delta(t)] = \int_{0^{-}}^{\infty} \delta(t)e^{-st} dt = e^{-0} = 1$$

(15.1.3)

since the impulse function δ(t) is zero e verywhere except at t = 0. The sifting property in Eq. (7.33) has been applied in Eq. (15.1.3).

Figure 15.2

For Example 15.1: (a) unit step function, (b) exponential function, (c) unit impulse function.

Find the Laplace transforms of these functions: r(t) = tu(t), that is, the Practice Problem 15.1 ramp function; Ae⁻at u(t); and Be⁻jω^t u(t).

Answer: 1∕s 2 , A∕(s + a), B∕(s + jω).

Determine the Laplace transform of f(t) = sin ωt u(t). Example 15.2

Solution:

Using Eq. (B.27) in addition to Eq. (15.1), we obtain the Laplace trans form of the sine function as

$$F(s) = \mathcal{L} \left[ \sin \omega t \right] = \int_0^\infty (\sin \omega t) e^{-st} dt = \int_0^\infty \left( \frac{e^{j\omega t} - e^{-j\omega t}}{2j} \right) e^{-st} dt$$ $$= \frac{1}{2j} \int_0^\infty (e^{-(s-j\omega)t} - e^{-(s+j\omega)t}) dt$$ $$= \frac{1}{2j} \left( \frac{1}{s - j\omega} - \frac{1}{s + j\omega} \right) = \frac{\omega}{s^2 + \omega^2}$$

Find the Laplace transform of f(t) = 15 cos(3 t) using the e xponential Practice Problem 15.2 representation for the cosine function.

Answer: 15s∕(s 2 + 9).