4.2 SOME [PROPERTIES OF THE](#page-10-0) LAPLACE TRANSFORM

← Back to LINEAR SYSTEMS AND SIGNALS Overview

4.2 SOME PROPERTIES OF THE LAPLACE TRANSFORM

Properties of the Laplace transform are useful not only in the derivation of the Laplace transform of functions but also in the solutions of linear integro-differential equations. A glance at Eqs. (4.2) and (4.1) shows that there is a certain measure of symmetry in going from x(t) to X(s), and vice versa. This symmetry or duality is also carried over to the properties of the Laplace transform. This fact will be evident in the following development.

We are already familiar with two properties: linearity [Eq. (4.3)] and the uniqueness property of the Laplace transform discussed earlier.

4.2-1 Time Shifting

The time-shifting property states that if

x(t) ⇐⇒ X(s)

then for t⁰ ≥ 0

$$x(t - t_0) \Longleftrightarrow X(s)e^{-st_0} \tag{4.12}$$

Observe that x(t) starts at t = 0, and, therefore, x(t − t0) starts at t = t0. This fact is implicit, but is not explicitly indicated in Eq. (4.12). This often leads to inadvertent errors. To avoid such a pitfall, we should restate the property as follows. If

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

then

$$x(t-t_0)u(t-t_0) \Longleftrightarrow X(s)e^{-st_0} \qquad t_0 \ge 0$$

Proof.

$$\mathcal{L}[x(t-t_0)u(t-t_0)] = \int_0^\infty x(t-t_0)u(t-t_0)e^{-st}dt$$

Setting t −t⁰ = τ , we obtain

$$\mathcal{L}[x(t-t_0)u(t-t_0)] = \int_{-t_0}^{\infty} x(\tau)u(\tau)e^{-s(\tau+t_0)}d\tau$$

Because u(τ ) = 0 for τ < 0 and u(τ ) = 1 for τ ≥ 0, the limits of integration can be taken from 0 to ∞. Thus,

$$\mathcal{L}[x(t-t_0)u(t-t_0)] = \int_0^\infty x(\tau)e^{-s(\tau+t_0)}d\tau$$ $$= e^{-st_0} \int_0^\infty x(\tau)e^{-s\tau}d\tau$$ $$= X(s)e^{-st_0}$$

Note that x(t − t0)u(t − t0) is the signal x(t)u(t) delayed by t⁰ seconds. The time-shifting property states that delaying a signal by t⁰ seconds amounts to multiplying its transform e−st⁰ .

This property of the unilateral Laplace transform holds only for positive t⁰ because if t⁰ were negative, the signal x(t −t0)u(t −t0) may not be causal.

We can readily verify this property in Drill 4.1. If the signal in Fig. 4.2a is x(t)u(t), then the signal in Fig. 4.2b is x(t − 2)u(t − 2). The Laplace transform for the pulse in Fig. 4.2a is (1/s)(1−e−2*^s* ). Therefore, the Laplace transform for the pulse in Fig. 4.2b is (1/s)(1−e−2*^s* )e−2*^s* .

The time-shifting property proves very convenient in finding the Laplace transform of functions with different descriptions over different intervals, as the following example demonstrates.

EXAMPLE 4.6 Laplace Transform and the Time-Shifting Property

Find the Laplace transform of x(t) depicted in Fig. 4.4a.

Describing mathematically a function such as the one in Fig. 4.4a is discussed in Sec. 1.4. The function x(t) in Fig. 4.4a can be described as a sum of two components shown in Fig. 4.4b. The equation for the first component is t−1 over 1 ≤ t ≤ 2 so that this component can be described by (t −1)[u(t −1)−u(t −2)]. The second component can be described by u(t −2)−u(t −4). Therefore,

$$x(t) = (t-1)[u(t-1) - u(t-2)] + [u(t-2) - u(t-4)]$$

= $(t-1)u(t-1) - (t-1)u(t-2) + u(t-2) - u(t-4)$ (4.13)

Figure 4.4 Finding a piecewise representation of a signal x(t).

The first term on the right-hand side is the signal tu(t) delayed by 1 second. Also, the third and fourth terms are the signal u(t) delayed by 2 and 4 seconds, respectively. The second term, however, cannot be interpreted as a delayed version of any entry in Table 4.1. For this reason, we rearrange it as

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

We have now expressed the second term in the desired form as tu(t) delayed by 2 seconds plus u(t) delayed by 2 seconds. With this result, Eq. (4.13) can be expressed as

$$x(t) = (t-1)u(t-1) - (t-2)u(t-2) - u(t-4)$$

Application of the time-shifting property to tu(t) ⇐⇒ 1/s² yields

$$(t-1)u(t-1) \Longleftrightarrow \frac{1}{s^2}e^{-s} \qquad \text{and} \qquad (t-2)u(t-2) \Longleftrightarrow \frac{1}{s^2}e^{-2s}$$

Also

$$u(t) \Longleftrightarrow \frac{1}{s}$$

and $u(t-4) \Longleftrightarrow \frac{1}{s}e^{-4s}$

Therefore,

$$X(s) = \frac{1}{s^2}e^{-s} - \frac{1}{s^2}e^{-2s} - \frac{1}{s}e^{-4s}$$

EXAMPLE 4.7 Inverse Laplace Transform and the Time-Shifting Property

Find the inverse Laplace transform of

$$X(s) = \frac{s+3+5e^{-2s}}{(s+1)(s+2)}$$

352 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS

Observe the exponential term e−2*^s* in the numerator of X(s), indicating time delay. In such a case, we should separate X(s) into terms with and without a delay factor, as

$$X(s) = \underbrace{\frac{s+3}{(s+1)(s+2)}}_{X_1(s)} + \underbrace{\frac{5e^{-2s}}{(s+1)(s+2)}}_{X_2(s)e^{-2s}}$$

where

$$X_1(s) = \frac{s+3}{(s+1)(s+2)} = \frac{2}{s+1} - \frac{1}{s+2}$$ $$X_2(s) = \frac{5}{(s+1)(s+2)} = \frac{5}{s+1} - \frac{5}{s+2}$$

Therefore,

$$x_1(t) = (2e^{-t} - e^{-2t})u(t)$$ $$x_2(t) = 5(e^{-t} - e^{-2t})u(t)$$

Also, because

$$X(s) = X_1(s) + X_2(s)e^{-2s}$$

we can write

$$x(t) = x_1(t) + x_2(t-2)$$

= $(2e^{-t} - e^{-2t})u(t) + 5[e^{-(t-2)} - e^{-2(t-2)}]u(t-2)$

DR ILL 4.4 Laplace Transform and the Time-Shifting Property

Find the Laplace transform of the signal illustrated in Fig. 4.5.

DR ILL 4.5 Inverse Laplace Transform and the Time-Shifting Property

Find the inverse Laplace transform of ^X(s) ^{= 3}e−2*^s* (s−1)(s+2) .

ANSWER et−² −e−2(t−2) u(t −2)

4.2-2 Frequency Shifting

The frequency-shifting property states that if

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

then

$$x(t)e^{s_0t} \Longleftrightarrow X(s-s_0) \tag{4.14}$$

Observe the symmetry (or duality) between this property and the time-shifting property of Eq. (4.12).

Proof.

$$\mathcal{L}[x(t)e^{s_0t}] = \int_{0^-}^{\infty} x(t)e^{s_0t}e^{-st} dt = \int_{0^-}^{\infty} x(t)e^{-(s-s_0)t} dt = X(s-s_0)$$

EXAMPLE 4.8 Frequency-Shifting Property

Derive pair 9a in Table 4.1 from pair 8a and the frequency-shifting property.

Pair 8a is

$$\cos btu(t) \Longleftrightarrow \frac{s}{s^2 + b^2}$$

From the frequency-shifting property [Eq. (4.14)] with s⁰ = −a, we obtain

$$e^{-at}\cos btu(t) \Longleftrightarrow \frac{s+a}{(s+a)^2 + b^2}$$

DR ILL 4.6 Frequency-Shifting Property

Derive pair 6 in Table 4.1 from pair 3 and the frequency-shifting property.

354 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS

We are now ready to consider the two most important properties of the Laplace transform: time differentiation and time integration.

4.2-3 The Time-Differentiation Property

The time-differentiation property states that if†

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

then

$$\frac{dx(t)}{dt} \Longleftrightarrow sX(s) - x(0^-)$$

Repeating this property a second time (differentiating twice) yields

$$\frac{d^2x(t)}{dt^2} \Longleftrightarrow s^2X(s) - sx(0^-) - \dot{x}(0^-)$$

Repeated differentiation yields

$$\frac{d^n x(t)}{dt^n} \Longleftrightarrow s^n X(s) - s^{n-1} x(0^-) - s^{n-2} \dot{x}(0^-) - \dots - x^{(n-1)}(0^-)$$

= $s^n X(s) - \sum_{k=1}^n s^{n-k} x^{(k-1)}(0^-)$ (4.15)

where x(r) (0−) is d^r x/dt^r at t = 0−.

Proof.

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

Integrating by parts, we obtain

$$\mathcal{L}\left[\frac{dx(t)}{dt}\right] = x(t)e^{-st}\Big|_{0^{-}}^{\infty} + s\int_{0^{-}}^{\infty} x(t)e^{-st}dt$$

For the Laplace integral to converge [i.e., for X(s) to exist], it is necessary that x(t)e−st → 0 as t → ∞ for the values of s in the ROC for X(s). Thus,

$$\mathcal{L}\left[\frac{dx(t)}{dt}\right] = -x(0^-) + sX(s)$$

Repeated application of this procedure yields Eq. (4.15).

$$tx(t) \Longleftrightarrow -\frac{d}{ds}X(s)$$

^† The dual of the time-differentiation property is the frequency-differentiation property, which states that

Find the Laplace transform of the signal x(t) in Fig. 4.6a by using Table 4.1 and the time-differentiation and time-shifting properties of the Laplace transform.

Figures 4.6b and 4.6c show the first two derivatives of x(t). Recall that the derivative at a point of jump discontinuity is an impulse of strength equal to the amount of jump [see Eq. (1.12)]. Therefore,

$$\frac{d^2x(t)}{dt^2} = \delta(t) - 3\delta(t-2) + 2\delta(t-3)$$

356 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS

The Laplace transform of this equation yields

$$\mathcal{L}\left(\frac{d^2x(t)}{dt^2}\right) = \mathcal{L}\left[\delta(t) - 3\delta(t-2) + 2\delta(t-3)\right]$$

Using the time-differentiation property of Eq. (4.15), the time-shifting property of Eq. (4.12), and the facts that x(0−) = ˙x(0−) = 0, and δ(t) ⇐⇒ 1, we obtain

$$s^2 X(s) - 0 - 0 = 1 - 3e^{-2s} + 2e^{-3s}$$

Therefore,

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

which confirms the earlier result in Drill 4.4.

4.2-4 The Time-Integration Property

The time-integration property states that if†

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

then

$$\int_{0^{-}}^{t} x(\tau) d\tau \Longleftrightarrow \frac{X(s)}{s} \quad \text{and} \quad \int_{-\infty}^{t} x(\tau) d\tau \Longleftrightarrow \frac{X(s)}{s} + \frac{\int_{-\infty}^{0^{-}} x(\tau) d\tau}{s} \quad (4.16)$$

Proof. To prove the first part of Eq. (4.16), we define

$$g(t) = \int_{0^-}^{t} x(\tau) d\tau$$

so that

$$\frac{d}{dt}g(t) = x(t) \qquad \text{and} \qquad g(0^-) = 0$$

Now, if

$$g(t) \Longleftrightarrow G(s)$$

then

$$X(s) = \mathcal{L}\left[\frac{d}{dt}g(t)\right] = sG(s) - g(0^{-}) = sG(s)$$ $$\frac{x(t)}{t} \Longleftrightarrow \int_{s}^{\infty} X(z) dz$$

^† The dual of the time-integration property is the frequency-integration property, which states that

4.2 Some Properties of the Laplace Transform 357

Therefore,

$$G(s) = \frac{X(s)}{s}$$

or

$$\int_{0^-}^t x(\tau) d\tau \Longleftrightarrow \frac{X(s)}{s}$$

To prove the second part of Eq. (4.16), observe that

$$\int_{-\infty}^{t} x(\tau) d\tau = \int_{-\infty}^{0^-} x(\tau) d\tau + \int_{0^-}^{t} x(\tau) d\tau$$

Note that the first term on the right-hand side is a constant for t ≥ 0. Taking the Laplace transform of the foregoing equation and using the first part of Eq. (4.16), we obtain

$$\int_{-\infty}^{t} x(\tau) d\tau \Longleftrightarrow \frac{\int_{-\infty}^{0^-} x(\tau) d\tau}{s} + \frac{X(s)}{s}$$

4.2-5 The Scaling Property

The scaling property states that if

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

then for a > 0

$$x(at) \Longleftrightarrow \frac{1}{a}X\left(\frac{s}{a}\right)$$

The proof is given in Ch. 7. Note that a is restricted to positive values because if x(t) is causal, then x(at) is anticausal (is zero for t ≥ 0) for negative a, and anticausal signals are not permitted in the (unilateral) Laplace transform.

Recall that x(at) is the signal x(t) time-compressed by the factor a, and X( ^{s a} ) is X(s) expanded along the s scale by the same factor a (see Sec. 1.2-2). The scaling property states that time compression of a signal by a factor a causes expansion of its Laplace transform in the s scale by the same factor. Similarly, time expansion x(t) causes compression of X(s) in the s scale by the same factor.

4.2-6 Time Convolution and Frequency Convolution

Another pair of properties states that if

$$x_1(t) \Longleftrightarrow X_1(s)$$

and $x_2(t) \Longleftrightarrow X_2(s)$

then (time-convolution property)

$$x_1(t) * x_2(t) \Longleftrightarrow X_1(s)X_2(s) \tag{4.17}$$

358 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS

and (frequency-convolution property)

$$x_1(t)x_2(t) \Longleftrightarrow \frac{1}{2\pi j}[X_1(s) * X_2(s)]$$

Observe the symmetry (or duality) between the two properties. Proofs of these properties are postponed to Ch. 7.

Equation (2.39) indicates that H(s), the transfer function of an LTIC system, is the Laplace transform of the system’s impulse response h(t); that is,

$$h(t) \Longleftrightarrow H(s)$$

If the system is causal, h(t) is causal, and, according to Eq. (2.39), H(s) is the unilateral Laplace transform of h(t). Similarly, if the system is noncausal, h(t) is noncausal, and H(s) is the bilateral transform of h(t).

We can apply the time-convolution property to the LTIC input–output relationship y(t) = x(t) ∗ h(t) to obtain

$$Y(s) = X(s)H(s) \tag{4.18}$$

The response y(t) is the zero-state response of the LTIC system to the input x(t). From Eq. (4.18), it follows that

$$H(s) = \frac{Y(s)}{X(s)} = \frac{\mathcal{L}[zero-state response]}{\mathcal{L}[input]}$$

(4.19)

This may be considered an alternate definition of the LTIC system transfer function H(s). It is the ratio of the transform of zero-state response to the transform of the input.

EXAMPLE 4.10 Time-Convolution Property

Use the time-convolution property of the Laplace transform to determine c(t)=eatu(t)∗ebtu(t).

From Eq. (4.17), it follows that

$$C(s) = \frac{1}{(s-a)(s-b)} = \frac{1}{a-b} \left[ \frac{1}{s-a} - \frac{1}{s-b} \right]$$

The inverse transform of this equation yields

$$c(t) = \frac{1}{a-b}(e^{at} - e^{bt})u(t)$$

INITIAL AND FINAL VALUES

In certain applications, it is desirable to know the values of x(t) as t → 0 and t → ∞ [initial and final values of x(t)] from the knowledge of its Laplace transform X(s). Initial and final value theorems provide such information.

The initial value theorem states that if x(t) and its derivative dx/dt are both Laplace transformable, then

$$x(0^+) = \lim_{s \to \infty} sX(s) \tag{4.20}$$

provided the limit on the right-hand side of Eq. (4.20) exists.

The final value theorem states that if both x(t) and dx/dt are Laplace transformable, then

$$\lim_{t \to \infty} x(t) = \lim_{s \to 0} sX(s) \tag{4.21}$$

provided sX(s) has no poles in the RHP or on the imaginary axis. To prove these theorems, we begin by setting n = 1 in Eq. (4.15). Using the definition of the Laplace transform, we see that

$$sX(s) - x(0^{-}) = \int_{0^{-}}^{\infty} \frac{dx(t)}{dt} e^{-st} dt$$

= $\int_{0^{-}}^{0^{+}} \frac{dx(t)}{dt} e^{-st} dt + \int_{0^{+}}^{\infty} \frac{dx(t)}{dt} e^{-st} dt$
= $x(t) \Big|_{0^{-}}^{0^{+}} + \int_{0^{+}}^{\infty} \frac{dx(t)}{dt} e^{-st} dt$
= $x(0^{+}) - x(0^{-}) + \int_{0^{+}}^{\infty} \frac{dx(t)}{dt} e^{-st} dt$

Therefore,

$$sX(s) = x(0^+) + \int_{0^+}^{\infty} \frac{dx(t)}{dt} e^{-st} dt$$

and

$$\lim_{s \to \infty} sX(s) = x(0^+) + \lim_{s \to \infty} \int_{0^+}^{\infty} \frac{dx(t)}{dt} e^{-st} dt$$ $$= x(0^+) + \int_{0^+}^{\infty} \frac{dx(t)}{dt} \left( \lim_{s \to \infty} e^{-st} \right) dt$$ $$= x(0^+)$$

Comment. The initial value theorem applies only if X(s) is strictly proper (M < N), because for M ≥ N, lim*^s*→∞ sX(s) does not exist, and the theorem does not apply. In such a case, we can still find the answer by using long division to express X(s) as a polynomial in s plus a strictly proper fraction, where M < N. For example, by using long division, we can express

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

The inverse transform of the polynomial in s is in terms of δ(t), and its derivatives, which are zero at t = 0+. In the foregoing case, the inverse transform of s + 1 is δ(˙ t) + δ(t). Hence, the desired

360 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS

x(0+) is the value of the remainder (strictly proper) fraction, for which the initial value theorem applies. In the present case,

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

To prove the final value theorem, we let n = 1 and s → 0 in Eq. (4.15) to obtain

$$\lim_{s \to 0} [sX(s) - x(0^{-})] = \lim_{s \to 0} \int_{0^{-}}^{\infty} \frac{dx(t)}{dt} e^{-st} dt = \int_{0^{-}}^{\infty} \frac{dx(t)}{dt} dt$$ $$= x(t)|_{0^{-}}^{\infty} = \lim_{t \to \infty} x(t) - x(0^{-})$$

a deduction that leads to the desired result, Eq. (4.21).

Comment. The final value theorem applies only if the poles of X(s) are in the LHP (including s = 0). If X(s) has a pole in the RHP, x(t) contains an exponentially growing term and x(∞) does not exist. If there is a pole on the imaginary axis, then x(t) contains an oscillating term and x(∞) does not exist. However, if there is a pole at the origin, then x(t) contains a constant term, and hence, x(∞) exists and is a constant.

EXAMPLE 4.11 Initial and Final Values

Determine the initial and final values of y(t) if its Laplace transform Y(s) is given by

$$Y(s) = \frac{10(2s+3)}{s(s^2+2s+5)}$$

Equations (4.20) and (4.21) yield

$$y(0^+) = \lim_{s \to \infty} sY(s) = \lim_{s \to \infty} \frac{10(2s+3)}{(s^2+2s+5)} = 0$$ $$y(\infty) = \lim_{s \to 0} sY(s) = \lim_{s \to 0} \frac{10(2s+3)}{(s^2+2s+5)} = 6$$

Table 4.2 summarizes the most important unilateral Laplace transform properties.