15.3 Properties of the Laplace Transform

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15.3 Properties of the Laplace Transform

The properties of the Laplace transform help us to obtain transform pairs without directly using Eq. (15.1) as we did in Examples 15.1 and 15.2. As we derive each of these properties, we should keep in mind the definition of the Laplace transform in Eq. (15.1).

Linearity

If F1(s) and F2(s) are, respectively, the Laplace transforms of f1(t) and f2(t), then

$$\mathcal{L}[a_1 f_1(t) + a_2 f_2(t)] = a_1 F_1(s) + a_2 F_2(s)$$

(15.7)

where a1 and a2 are constants. Equation 15.7 e xpresses the linearity property of the Laplace transform. The proof of Eq. (15.7) follows readily from the definition of the Laplace transform in Eq. (15.1).

For example, by the linearity property in Eq. (15.7), we may write

$$\mathcal{L}[\cos \omega t \, u(t)] = \mathcal{L} \left[ \frac{1}{2} (e^{j\omega t} + e^{-j\omega t}) \right] = \frac{1}{2} \mathcal{L} [e^{j\omega t}] + \frac{1}{2} \mathcal{L} [e^{-j\omega t}] \tag{15.8}$$

But from Example 15.1(b), [e⁻at] = 1∕(s + a). Hence,

$$\mathcal{L}[\cos \omega t \, u(t)] = \frac{1}{2} \left( \frac{1}{s - j\omega} + \frac{1}{s + j\omega} \right) = \frac{s}{s^2 + \omega^2}$$

(15.9)

Scaling

If F(s) is the Laplace transform of f(t), then

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

(15.10)

where a is a constant and a > 0. If we let x = at, dx = a dt, then

$$\mathcal{L}[f(at)] = \int_{0-}^{\infty} f(x) e^{-x(s/a)} \frac{dx}{a} = \frac{1}{a} \int_{0-}^{\infty} f(x) e^{-x(s/a)} dx \tag{15.11}$$

Comparing this inte gral with the definition of the Laplace transform in Eq. (15.1) shows that s in Eq. (15.1) must be replaced by s∕a while the dummy v ariable t is replaced by x. Hence, we obtain the scaling property as

$$\mathcal{L}[f(at)] = \frac{1}{a} F(\frac{s}{a})$$

\n(15.12)

For example, we know from Example 15.2 that

$$\mathcal{L}[\sin \omega t \, u(t)] = \frac{\omega}{s^2 + \omega^2} \tag{15.13}$$

Using the scaling property in Eq. (15.12),

$$\mathcal{L}[\sin 2\omega t \, u(t)] = \frac{1}{2} \frac{\omega}{\left(s/2\right)^2 + \omega^2} = \frac{2\omega}{s^2 + 4\omega^2} \tag{15.14}$$

which may also be obtained from Eq. (15.13) by replacing ω with 2ω.

Time Shift

If F(s) is the Laplace transform of f(t), then

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

(15.15)

But u(t − a) = 0 for t < a and u(t − a) = 1 for t > a. Hence,

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

(15.16)

If we let x = t − a, then dx = dt and t = x + a. As t → a, x → 0 and as t → ∞, x → ∞. Thus,

$$\mathcal{L}[f(t-a)u(t-a)] = \int_0^\infty f(x)e^{-s(x+a)} dx$$ $$= e^{-as} \int_0^\infty f(x)e^{-sx} dx = e^{-as} F(s)$$

or

$$\mathcal{L}[f(t-a)u(t-a)] = e^{-as} F(s)$$

(15.17)

In other w ords, if a function is delayed in time by a, the result in the s-domain is found by multiplying the Laplace transform of the function (without the delay) by e⁻as. This is called the time-delay or time-shift property of the Laplace transform.

As an example, we know from Eq. (15.9) that

$$\mathcal{L}[\cos \omega t \, u(t)] = \frac{s}{s^2 + \omega^2}$$

Using the time-shift property in Eq. (15.17),

$$\mathcal{L}[\cos \omega(t-a)u(t-a)] = e^{-as} \frac{s}{s^2 + \omega^2}$$

(15.18)

Frequency Shift

If F(s) is the Laplace transform of f(t), then

$$\mathcal{L}[e^{-at}f(t)u(t)] = \int_0^\infty e^{-at}f(t)e^{-st} dt$$ $$= \int_0^\infty f(t)e^{-(s+a)t} dt = F(s+a)$$

or

$$\mathcal{L}[e^{-at}f(t)u(t)] = F(s+a)$$

(15.19)

That is, the Laplace transform of e⁻at f(t) can be obtained from the Laplace transform of f(t) by replacing every s with s + a. This is known as frequency shift or frequency translation.

As an example, we know that

cos ωt u(t) ^⇔ _______ ^s s 2 + ω²

and

(15.20)

$$\sin \omega t \, u(t) \qquad \Leftrightarrow \qquad \frac{\omega}{s^2 + \omega^2}$$

Using the shift property in Eq. (15.19), we obtain the Laplace transform of the damped sine and damped cosine functions as

and damped cosine functions as
\n

$$\mathcal{L}[e^{-at} \cos \omega t \, u(t)] = \frac{s+a}{(s+a)^2 + \omega^2}$$

\n(15.21a)

$$(s + a)^{2} + \omega^{2}$$

\n

$$\mathcal{L}[e^{-at} \sin \omega t \, u(t)] = \frac{\omega}{(s + a)^{2} + \omega^{2}}$$

\n(15.21b)

Time Differentiation

Given that F(s) is the Laplace transform of f(t), the Laplace transform of its derivative is

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

(15.22)

To inte grate this by parts, we let u = e⁻st, du = −se⁻st dt, and dv = (df∕dt) dt = df(t), v = f(t). Then

$$\mathcal{L}\left[\frac{df}{dt}u(t)\right] = f(t)e^{-st}\Big|_{0^{-}}^{\infty} - \int_{0^{-}}^{\infty}f(t)[-se^{-st}] dt$$

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

or

$$\mathcal{L}[f'(t)] = sF(s) - f(0^{-})$$

\n(15.23)

The Laplace transform of the second deri vative of f(t) is a repeated application of Eq. (15.23) as

$$\mathcal{L}\left[\frac{d^2f}{dt^2}\right] = s\mathcal{L}[f'(t)] - f'(0^-) = s[sF(s) - f(0^-)] - f'(0^-)$$

= $s^2F(s) - sf(0^-) - f'(0^-)$

or

$$\mathcal{L}[f''(t)] = s^2 F(s) - sf(0^-) - f'(0^-)$$

(15.24)

Continuing in this manner , we can obtain the Laplace transform of the nth derivative of f(t) as

$$\mathcal{L}\left[\frac{d^n f}{dt^n}\right] = s^n F(s) - s^{n-1} f(0^-) - s^{n-2} f'(0^-) - \dots - s^0 f^{(n-1)}(0^-)$$

\n(15.25)

As an example, we can use Eq. (15.23) to obtain the Laplace transform of the sine from that of the cosine. If we let f(t) = cos ωt u(t), then f(0) = 1 and f ′(t) = −ω sin ωt u(t). Using Eq. (15.23) and the scaling property,

$$\mathcal{L}[\sin \omega t \, u(t)] = -\frac{1}{\omega} \mathcal{L}[f'(t)] = -\frac{1}{\omega} [sF(s) - f(0^{-})]$$ $$= -\frac{1}{\omega} \left(s \frac{s}{s^{2} + \omega^{2}} - 1\right) = \frac{\omega}{s^{2} + \omega^{2}}$$

(15.26)

as expected.

Time Integration

If F(s) is the Laplace transform of f(t), the Laplace transform of its integral is

$$\mathcal{L}\bigg[\int_0^t f(x)dx\bigg] = \int_0^\infty \bigg[\int_0^t f(x)dx\bigg] e^{-st} dt \qquad (15.27)$$

To integrate this by parts, we let

$$u = \int_0^t f(x)dx, \qquad du = f(t)dt$$

and

$$dv = e^{-st} dt, \qquad v = -\frac{1}{s}e^{-st}$$

Then

$$C\left[\int_0^t f(x)dx\right] = \left[\int_0^t f(x)dx\right] \left(-\frac{1}{s}e^{-st}\right)\Big|_0^{\infty}$$ $$-\int_0^{\infty} \left(-\frac{1}{s}\right)e^{-st}f(t)dt$$

For the first term on the right-hand side of the equation, evaluating the term at t = ∞ yields zero due to e⁻s^∞ and evaluating it at t = 0 gives __1 s ∫ 0 0 f(x) dx = 0. Thus, the first term is zero, and t ∞

[∫ 0 ^f(x)dx ] = __1 s ∫ 0− ^f(t)e⁻st dt = __1 s F(s)

or simply,

$$\mathcal{L}\left[\int_0^t f(x)dx\right] = \frac{1}{s}F(s)$$

(15.28)

As an example, if we let f(t) = u(t), from Example 15.1(a), F(s) = 1∕s. Using Eq. (15.28),

$$\mathcal{L}\bigg[\int_0^t f(x)dx\bigg] = \mathcal{L}[t] = \frac{1}{s}\bigg(\frac{1}{s}\bigg)$$

Thus, the Laplace transform of the ramp function is

$$\mathcal{L}[t] = \frac{1}{s^2} \tag{15.29}$$

Applying Eq. (15.28), this gives

$$\left[\int_0^t x dx\right] = \mathcal{L}\left[\frac{t^2}{2}\right] = \frac{1}{s} \frac{1}{s^2}$$ $$\mathcal{L}[t^2] = \frac{2}{s^3} \tag{15.30}$$

or

Repeated applications of Eq. (15.28) lead to

$$\mathcal{L}[t^n] = \frac{n!}{s^{n+1}} \tag{15.31}$$

Similarly, using integration by parts, we can show that

$$\mathcal{L}\bigg[\int_{-\infty}^{t} f(x)dx\bigg] = \frac{1}{s}F(s) + \frac{1}{s}f^{-1}(0^{-})$$

\n(15.32)

where

$$f^{-1}(0^{-}) = \int_{-\infty}^{0^{-}} f(t)dt$$

Frequency Differentiation

If F(s) is the Laplace transform of f(t), then

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

Taking the derivative with respect to s,

$$\frac{dF(s)}{ds} = \int_{0^{-}}^{\infty} f(t)(-te^{-st}) dt = \int_{0^{-}}^{\infty} (-tf(t))e^{-st} dt = \mathcal{L}[-tf(t)]$$

and the frequency differentiation property becomes

$$\mathcal{L}[tf(t)] = -\frac{dF(s)}{ds}$$

(15.33)

Repeated applications of this equation lead to

$$\mathcal{L}[t^n f(t)] = (-1)^n \frac{d^n F(s)}{ds^n}$$

(15.34)

For example, we know from Example 15.1(b) that [e ⁻at] = 1∕(s + a). Using the property in Eq. (15.33),

$$\mathcal{L}[te^{-at}u(t)] = -\frac{d}{ds}\left(\frac{1}{s+a}\right) = \frac{1}{(s+a)^2}$$

(15.35)

Note that if a = 0, we obtain [t] = 1∕s 2 as in Eq. (15.29), and repeated applications of Eq. (15.33) will yield Eq. (15.31).

Time Periodicity

If function f(t) is a periodic function such as sho wn in Fig. 15.3, it can be represented as the sum of time-shifted functions sho wn in Fig. 15.4. Thus,

$$f(t) = f_1(t) + f_2(t) + f_3(t) + \cdots$$

= $f_1(t) + f_1(t - T)u(t - T)$

• $f_1(t - 2T)u(t - 2T) + \cdots$ (15.36)

where f1(t) is the same as the function f(t) g ated o ver the interv al 0 < t < T, that is,

$$f_1(t) = f(t)[u(t) - u(t - T)]$$

(15.37a)

A periodic function.

$$f_1(t) = \begin{cases} f(t), & 0 < t < T \\ 0, & \text{otherwise} \end{cases} \tag{15.37b}$$

We no w transform each term in Eq. (15.36) and apply the time-shift property in Eq. (15.17). We obtain

$$F(s) = F_1(s) + F_1(s)e^{-Ts} + F_1(s)e^{-2Ts} + F_1(s)e^{-3Ts} + \cdots$$

= $F_1(s)[1 + e^{-Ts} + e^{-2Ts} + e^{-3Ts} + \cdots]$ (15.38)

But

$$1 + x + x2 + x3 + \dots = \frac{1}{1 - x}$$

(15.39)

if |x| < 1. Hence,

$$F(s) = \frac{F_1(s)}{1 - e^{-Ts}} \tag{15.40}$$

where F1(s) is the Laplace transform of f1(t); in other words, F1(s) is the transform f(t) defined over its first period only. Equation (15.40) shows that the Laplace transform of a periodic function is the transform of the first period of the function divided by 1 − e⁻Ts.

Initial and Final Values

The initial-value and final-value properties allo w us to find the initial value f(0) and the final value f(∞) of f(t) directly from its Laplace transform F(s). To obtain these properties, we be gin with the dif ferentiation property in Eq. (15.23), namely,

$$sF(s) - f(0) = \mathcal{L}\left[\frac{df}{dt}\right] = \int_{0^-}^{\infty} \frac{df}{dt} e^{-st} dt$$

(15.41)

If we let s → ∞, the integrand in Eq. (15.41) vanishes due to the damping exponential factor, and Eq. (15.41) becomes

$$\lim_{s \to \infty} [sF(s) - f(0)] = 0$$

Because f (0) is independent of s, we can write

$$f(0) = \lim_{s \to \infty} sF(s) \tag{15.42}$$

This is known as the initial-value theorem. For example, we know from Eq. (15.21a) that

5.21a) that
\n

$$f(t) = e^{-2t} \cos 10t \, u(t) \qquad \Leftrightarrow \qquad F(s) = \frac{s+2}{(s+2)^2 + 10^2} \quad \textbf{(15.43)}$$

Using the initial-value theorem,

$$f(0) = \lim_{s \to \infty} sF(s) = \lim_{s \to \infty} \frac{s^2 + 2s}{s^2 + 4s + 104}$$ $$= \lim_{s \to \infty} \frac{1 + 2/s}{1 + 4/s + 104/s^2} = 1$$

which confirms what we would expect from the given f(t).

In Eq. (15.41), we let s → 0; then

$$\lim_{s \to 0} [sF(s) - f(0^{-})] = \int_{0^{-}}^{\infty} \frac{df}{dt} e^{0t} dt = \int_{0^{-}}^{\infty} df = f(\infty) - f(0^{-})$$

or

$$f(\infty) = \lim_{s \to 0} sF(s) \tag{15.44}$$

This is referred to as the final-value theorem. In order for the final-value theorem to hold, all poles of F(s) must be located in the left half of the s plane (see Fig. 15.1 or 15.9); that is, the poles must have negative real parts. The only e xception to this requirement is the case in which F(s) has a simple pole at s = 0, because the effect of 1∕s will be nullified by sF(s) in Eq. (15.44). For example, from Eq. (15.21b),

$$f(t) = e^{-2t} \sin 5t \, u(t) \qquad \Leftrightarrow \qquad F(s) = \frac{5}{(s+2)^2 + 5^2} \quad \textbf{(15.45)}$$

Applying the final-value theorem,

$$f(\infty) = \lim_{s \to 0} s F(s) = \lim_{s \to 0} \frac{5s}{s^2 + 4s + 29} = 0$$

as expected from the given f(t). As another example,

$$f(t) = \sin t \, u(t) \qquad \Leftrightarrow \qquad f(s) = \frac{1}{s^2 + 1}$$

(15.46)

so that

$$f(\infty) = \lim_{s \to 0} sF(s) = \lim_{s \to 0} \frac{s}{s^2 + 1} = 0$$

This is incorrect, because f(t) = sin t oscillates between +1 and −1 and does not ha ve a limit as t → ∞. Thus, the final-value theorem cannot be used to find the final value of f(t) = sin t, because F(s) has poles at s = ±j, which are not in the left half of the s plane. In gen eral, the final-value theorem does not apply in finding the final values of sinusoidal functions—these functions oscillate forever and do not have final values.

The initial-value and final-value theorems depict the relationship between the origin and infinity in the time domain and the s-domain. They serve as useful checks on Laplace transforms.

Table 15.1 provides a list of the properties of the Laplace trans form. The last property (on con volution) will be pro ved in Sec tion 15.5. There are other properties, b ut these are enough for pres ent purposes. Table 15.2 summarizes the Laplace transforms of some common functions. We have omitted the f actor u(t) except where it is necessary.

We should mention that many software packages, such as Mathcad, MATLAB, Maple, and Mathematica, offer symbolic math. For example, Mathcad has symbolic math for the Laplace, F ourier, and Z transforms as well as the inverse function.

TABLE 15.1			TABLE 15.2
Properties of the Laplace transform.		Laplace transform pairs.*
Property	f(t)	F(s)	f(t)	F(s)
Linearity	a1f1(t) + a2f2(t)	a1F1(s) + a2F2(s)	δ(t)	1
Scaling	f(at)	__1 __s a F( a )	u(t)	__1 s
Time shift	f(t − a)u(t − a)	e−as F(s)	e−at	_____ 1 s + a
Frequency shift	e−at f(t)	F(s + a)	t	__1
Time differentiation	df __ dt	sF(s) − f(0−)	t n	2 s ____ n! n+1
	d2 f ___ dt2	2 F(s) − sf(0−) − f ′(0−) s	te−at	s _______ 1 2 (s + a)
	d3 f ___ dt3	3 2 f(0−) − sf ′(0−) − f ″(0−) s F(s) − s	n e−at t	________ n! n+1 (s + a)
	n d f ___ dtn	n n−1 f(0−) − n−2 f ′(0−) s F(s) − s s (n−1)(0−) − ⋯ − f	sin ωt	_______ ω 2 + ω2 s
Time integration	t f(x)dx ∫ 0	__1 F(s) s	cos ωt	_______ s 2 + ω2 s
Frequency differentiation	tf(t)	− __d ds F(s)	sin(ωt + θ)	s sin θ + ω cos θ ______________ 2 + ω2 s
Frequency integration	f(t) ___ t	∞ F(s)ds ∫ s	cos(ωt + θ)	s cos θ − ω sin θ ______________ 2 + ω2 s
Time periodicity	f(t) = f(t + nT)	F1(s) _______ 1 − e−sT	e−at sin ωt	___________ ω 2 + ω2 (s + a)
Initial value	f(0)	s→∞ sF(s) lim	e−at cos ωt	___________ s + a 2 + ω2 (s + a)
Final value	f(∞)	lim sF(s) s→0
Convolution	f1(t) * f2(t)	F1(s)F2(s)	*Defined for t ≥ 0; f(t) = 0, for t < 0.

Obtain the Laplace transform of f(t) = δ(t) + 2u(t) − 3*^e* Example 15.3 ⁻2*^t u*(t).

Solution:

By the linearity property,

$$F(s) = \mathcal{L}[\delta(t)] + 2\mathcal{L}[u(t)] - 3\mathcal{L}[e^{-2t} u(t)]$$ $$= 1 + 2\frac{1}{s} - 3\frac{1}{s+2} = \frac{s^2 + s + 4}{s(s+2)}$$

Find the Laplace transform of f(t) = (cos (2t) + ^e Practice Problem 15.3 ⁻4*^t* )u(t).

Find the Laplace transform
\nAnswer:

$$\frac{2s^2 + 4s + 4}{(s + 4)(s^2 + 4)}$$

Example 15.4 Determine the Laplace transform of f(t) = ^t 2 sin 2t u(t).

Solution:

We know that

$$[\sin 2t] = \frac{2}{s^2 + 2^2}$$

Using frequency differentiation in Eq. (15.34),

$$F(s) = \mathcal{L}[t^2 \sin 2t] = (-1)^2 \frac{d^2}{ds^2} \left(\frac{2}{s^2 + 4}\right)$$ $$= \frac{d}{ds} \left(\frac{-4s}{(s^2 + 4)^2}\right) = \frac{12s^2 - 16}{(s^2 + 4)^3}$$

Practice Problem 15.4 Find the Laplace transform of f(t) = ^t 2 cos 3t u(t).

Answer:

$$\frac{2s(s^2 - 27)}{(s^2 + 9)^3}$$

Answer:

___ 10 s

(2 − e⁻4*^s* − e⁻8*^s*

We can express the gate function in Fig. 15.5 as

$$g(t) = 10[u(t-2) - u(t-3)]$$

Given that we kno w the Laplace transform of u(t), we apply the timeshift property and obtain

$$G(s) = 10\left(\frac{e^{-2s}}{s} - \frac{e^{-3s}}{s}\right) = \frac{10}{s}(e^{-2s} - e^{-3s})$$

The gate function; for Example 15.5.

Practice Problem 15.5 Find the Laplace transform of the function h(t) in Fig. 15.6.

).

$h(t)$
20
10
0	4	8	t

\nFigure 15.6

For Practice Prob. 15.5.

$$\mathbf{36}^{\dagger}$$

Calculate the Laplace transform of the periodic function in Fig. 15.7. Example 15.6

Solution:

The period of the function is T = 2. To apply Eq. (15.40), we first obtain the transform of the first period of the function.

$$f_1(t) = 2t[u(t) - u(t-1)] = 2tu(t) - 2tu(t-1)$$

= 2tu(t) - 2(t - 1 + 1)u(t - 1)
= 2tu(t) - 2(t - 1)u(t-1) - 2u(t-1)

Using the time-shift property,

$$F_1(s) = \frac{2}{s^2} - 2\frac{e^{-s}}{s^2} - \frac{2}{s}e^{-s} = \frac{2}{s^2}(1 - e^{-s} - se^{-s})$$

Thus, the transform of the periodic function in Fig. 15.7 is

$$F(s) = \frac{F_1(s)}{1 - e^{-Ts}} = \frac{2}{s^2(1 - e^{-2s})}(1 - e^{-s} - se^{-s})$$

Determine the Laplace transform of the periodic function in Fig. 15.8. Practice Problem 15.6

For Practice Prob. 15.6.

Find the initial and final values of the function whose Laplace trans- Example 15.7 form is

inal values of the function who

$$H(s) = \frac{20}{(s+3)(s^2+8s+25)}$$

Solution:

Applying the initial-value theorem,

lution:
\nplying the initial-value theorem,
\n

$$h(0) = \lim_{s \to \infty} sH(s) = \lim_{s \to \infty} \frac{20s}{(s+3)(s^2+8s+25)}$$

\n

$$= \lim_{s \to \infty} \frac{20/s^2}{(1+3/s)(1+8/s+25/s^2)} = \frac{0}{(1+0)(1+0+0)} = 0$$

To be sure that the final-value theorem is applicable, we check where the poles of H(s) are located. The poles of H(s) are s = −3, −4 ± j3, which all have negative real parts: They are all located on the left half of the s plane (Fig. 15.9). Hence, the final-value theorem applies and _________________ ^20s

$$h(\infty) = \lim_{s \to 0} sH(s) = \lim_{s \to 0} \frac{20s}{(s+3)(s^2+8s+25)}$$ $$= \frac{0}{(0+3)(0+0+25)} = 0$$

Figure 15.9 For Example 15.7: Poles of H(s).

Figure 15.7 For Example 15.6. Both the initial and final values could be determined from h(t) if we knew it. See Example 15.11, where h(t) is given.

Practice Problem 15.7 Obtain the initial and the final values of

he final values of
\n

$$G(s) = \frac{3s^3 + 2s + 6}{s(s+1)^2(s+1.5)}$$

Answer: 3, 4.

15.4 The Inverse Laplace Transform

Given F(s), how do we transform it back to the time domain and obtain the corresponding f(t)? By matching entries in Table 15.2, we a void using Eq. (15.5) to find f(t).

Suppose F(s) has the general form of

$$F(s) = \frac{N(s)}{D(s)}$$

(15.47)

where N(s) is the numerator polynomial and D(s) is the denominator polynomial. The roots of N(s) = 0 are called the zeros of F(s), while the roots of D(s) = 0 are the poles of F(s). Although Eq. (15.47) is similar in form to Eq. (14.3), here F(s) is the Laplace transform of a function, which is not necessarily a transfer function. We use partial fraction expansion to break F(s) down into simple terms whose inverse transform we obtain from Table 15.2. Thus, finding the inverse Laplace transform of F(s) involves two steps.

Steps to Find the Inverse Laplace Transform:

• 1. Decompose F(s) into simple terms using partial fraction expansion.
• 2. Find the inverse of each term by matching entries in Table 15.2.

Let us consider the three possible forms F(s) may take and how to apply the two steps to each form.