[1.4 SOME](#page-7-0) USEFUL SIGNAL MODELS

← Back to LINEAR SYSTEMS AND SIGNALS Overview

1.4 SOME USEFUL SIGNAL MODELS

In the area of signals and systems, the step, the impulse, and the exponential functions play very important roles. Not only do they serve as a basis for representing other signals, but their use can simplify many aspects of the signals and systems.

1.4-1 The Unit Step Function u(t)

In much of our discussion, the signals begin at t = 0 (causal signals). Such signals can be conveniently described in terms of unit step function u(t) shown in Fig. 1.14a. This function is defined by

$$u(t) = \begin{cases} 1 & t \ge 0 \\ 0 & t < 0 \end{cases}$$

(1.8)

If we want a signal to start at t = 0 (so that it has a value of zero for t < 0), we need only multiply the signal by u(t). For instance, the signal e−at represents an everlasting exponential that starts at t = −∞. The causal form of this exponential (Fig. 1.14b) can be described as e−atu(t).

The unit step function also proves very useful in specifying a function with different mathematical descriptions over different intervals. Examples of such functions appear in Fig. 1.7. These functions have different mathematical descriptions over different segments of time, as seen from Eqs. (1.5) and (1.6). Such a description often proves clumsy and inconvenient in mathematical treatment. We can use the unit step function to describe such functions by a single expression that is valid for all t.

Consider, for example, the rectangular pulse depicted in Fig. 1.15a. We can express such a pulse in terms of familiar step functions by observing that the pulse x(t) can be expressed as the sum of the two delayed unit step functions, as shown in Fig. 1.15b. The unit step function u(t) delayed by T seconds is u(t −T). From Fig. 1.15b, it is clear that

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

Figure 1.14 (a) Unit step function u(t). (b) Exponential e−atu(t).

Figure 1.15 Representation of a rectangular pulse by step functions.

EXAMPLE 1.6 Describing a Triangle Function with the Unit Step

Use the unit step function to describe the signal in Fig. 1.16a.

Figure 1.16 Representation of a signal defined interval by interval.

$$x_1(t) = t[u(t) - u(t-2)]$$

The signal x2(t) can be obtained by multiplying another ramp by the gate pulse illustrated in Fig. 1.16c. This ramp has a slope −2; hence it can be described by −2t + c. Now, because the ramp has a zero value at t = 3, the constant c = 6, and the ramp can be described by −2(t−3). Also, the gate pulse in Fig. 1.16c is u(t −2)−u(t −3). Therefore,

$$x_2(t) = -2(t-3)[u(t-2) - u(t-3)]$$

The signal illustrated in Fig. 1.16a can be conveniently handled by breaking it up into the two components x1(t) and x2(t), depicted in Figs. 1.16b and 1.16c, respectively. Here, x1(t) can be obtained by multiplying the ramp t by the gate pulse u(t) − u(t − 2), as shown in Fig. 1.16b. Therefore,

and

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

= t[u(t) - u(t-2)] - 2(t-3)[u(t-2) - u(t-3)]
= tu(t) - 3(t-2)u(t-2) + 2(t-3)u(t-3)

EXAMPLE 1.7 Describing a Piecewise Function with the Unit Step

Describe the signal in Fig. 1.7a by a single expression valid for all t.

Over the interval from −1.5 to 0, the signal can be described by a constant 2, and over the interval from 0 to 3, it can be described by 2e−t/2. Therefore,

$$x(t) = 2[u(t+1.5) - u(t)] + 2e^{-t/2}[u(t) - u(t-3)]$$

constant part

$$= 2u(t+1.5) - 2(1 - e^{-t/2})u(t) - 2e^{-t/2}u(t-3)$$

Compare this expression with the expression for the same function found in Eq. (1.6).

DR ILL 1.7 Using Reflected Unit Step Functions

Show that the signals depicted in Figs. 1.17a and 1.17b can be described as u(−t) and e−atu(−t), respectively.

DR ILL 1.8 Describing a Piecewise Function with the Unit Step

Show that the signal shown in Fig. 1.18 can be described as

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

1.4-2 The Unit Impulse Function δ(t)

The unit impulse function δ(t) is one of the most important functions in the study of signals and systems. This function was first defined in two parts by P. A. M. Dirac as

$$\delta(t) = 0 \quad t \neq 0 \quad \text{and} \quad \int_{-\infty}^{\infty} \delta(t) dt = 1 \tag{1.9}$$

Figure 1.18 Signal for Drill 1.8.

We can visualize an impulse as a tall, narrow, rectangular pulse of unit area, as illustrated in Fig. 1.19b. The width of this rectangular pulse is a very small value → 0. Consequently, its height is a very large value 1/ → ∞. The unit impulse therefore can be regarded as a rectangular pulse with a width that has become infinitesimally small, a height that has become infinitely large, and an overall area that has been maintained at unity. Thus δ(t) = 0 everywhere except at t = 0, where it is undefined. For this reason, a unit impulse is represented by the spearlike symbol in Fig. 1.19a.

Other pulses, such as the exponential, triangular, or Gaussian types, may also be used in impulse approximation. The important feature of the unit impulse function is not its shape but the fact that its effective duration (pulse width) approaches zero while its area remains at unity. For example, the exponential pulse αe−α*^t u*(t) in Fig. 1.20a becomes taller and narrower as α increases.

Figure 1.19 A unit impulse and its

Figure 1.20 Other possible approximations to a unit impulse.

In the limit as α → ∞, the pulse height → ∞, and its width or duration → 0. Yet, the area under the pulse is unity regardless of the value of α because

$$\int_0^\infty \alpha e^{-\alpha t} dt = 1$$

The pulses in Figs. 1.20b and 1.20c behave in a similar fashion. Clearly, the exact impulse function cannot be generated in practice; it can only be approached.

From Eq. (1.9), it follows that the function kδ(t) = 0 for all t = 0, and its area is k. Thus, kδ(t) is an impulse function whose area is k (in contrast to the unit impulse function, whose area is 1).

MULTIPLICATION OF A FUNCTION BY AN IMPULSE

Let us now consider what happens when we multiply the unit impulse δ(t) by a function φ(t) that is known to be continuous at t = 0. Since the impulse has nonzero value only at t = 0, and the value of φ(t) at t = 0 is φ(0), we obtain

$$\phi(t)\delta(t) = \phi(0)\delta(t)$$

Thus, multiplication of a continuous-time function φ(t) with an unit impulse located at t = 0 results in an impulse, which is located at t = 0 and has strength φ(0) [the value of φ(t) at the location of the impulse]. Use of exactly the same argument leads to the generalization of this result, stating that provided φ(t) is continuous at t = T,φ(t) multiplied by an impulse δ(t − T) (impulse located at t = T) results in an impulse located at t = T and having strength φ(T) [the value of φ(t) at the location of the impulse].

$$\phi(t)\delta(t-T) = \phi(T)\delta(t-T) \tag{1.10}$$

SAMPLING PROPERTY OF THE UNIT IMPULSE FUNCTION

From Eq. (1.10) it follows that

$$\int_{-\infty}^{\infty} \phi(t)\delta(t-T) dt = \phi(T) \int_{-\infty}^{\infty} \delta(t) dt = \phi(T)$$

\n(1.11)

provided φ(t) is continuous at t = T. This result means that the area under the product of a function with an impulse δ(t − T) is equal to the value of that function at the instant at which the unit impulse is located. This property is very important and useful and is known as the sampling or sifting property of the unit impulse.

UNIT IMPULSE AS A GENERALIZED FUNCTION

The definition of the unit impulse function given in Eq. (1.9) is not mathematically rigorous, which leads to serious difficulties. First, the impulse function does not define a unique function: for example, it can be shown that δ(t)+δ(˙ t) also satisfies Eq. (1.9) [1]. Moreover, δ(t) is not even a true function in the ordinary sense. An ordinary function is specified by its values for all time t. The impulse function is zero everywhere except at t = 0, and at this, the only interesting part of its range, it is undefined. These difficulties are resolved by defining the impulse as a generalized function rather than an ordinary function. A generalized function is defined by its effect on other functions instead of by its value at every instant of time.

In this approach the impulse function is defined by the sampling property [Eq. (1.11)]. We say nothing about what the impulse function is or what it looks like. Instead, the impulse function is defined in terms of its effect on a test function φ(t). We define a unit impulse as a function for which the area under its product with a function φ(t) is equal to the value of the function φ(t) at the instant at which the impulse is located. It is assumed that φ(t) is continuous at the location of the impulse. Recall that the sampling property [Eq. (1.11)] is the consequence of the classical (Dirac) definition of the unit impulse in Eq. (1.9). In contrast, the sampling property [Eq. (1.11)] defines the impulse function in the generalized function approach.

We now present an interesting application of the generalized function definition of an impulse. Because the unit step function u(t) is discontinuous at t = 0, its derivative du/dt does not exist at t = 0 in the ordinary sense. We now show that this derivative does exist in the generalized sense, and it is, in fact, δ(t). As a proof, let us evaluate the integral of (du/dt)φ(t), using integration by parts:

$$\int_{-\infty}^{\infty} \frac{du(t)}{dt} \phi(t) dt = u(t) \phi(t) \Big|_{-\infty}^{\infty} - \int_{-\infty}^{\infty} u(t) \dot{\phi}(t) dt$$ $$= \phi(\infty) - 0 - \int_{0}^{\infty} \dot{\phi}(t) dt$$ $$= \phi(\infty) - \phi(t) \Big|_{0}^{\infty} = \phi(0)$$

This result shows that du/dt satisfies the sampling property of δ(t). Therefore it is an impulse δ(t) in the generalized sense—that is,

$$\frac{du(t)}{dt} = \delta(t) \tag{1.12}$$

Consequently,

$$\int_{-\infty}^{t} \delta(\tau) d\tau = u(t)$$

These results can also be obtained graphically from Fig. 1.19b. We observe that the area from −∞ to t under the limiting form of δ(t) in Fig. 1.19b is zero if t < −/2 and unity if t ≥ /2 with → 0. Consequently,

$$\int_{-\infty}^{t} \delta(\tau) d\tau = \begin{cases} 0 & t < 0 \\ 1 & t \ge 0 \end{cases}$$ $$= u(t)$$

This result shows that the unit step function can be obtained by integrating the unit impulse function. Similarly the unit ramp function x(t) = tu(t) can be obtained by integrating the unit step function. We may continue with unit parabolic function t ²/2 obtained by integrating the unit ramp, and so on. On the other side, we have derivatives of impulse function, which can be defined as generalized functions (see Prob. 1.4-12). All these functions, derived from the unit impulse function (successive derivatives and integrals), are called singularity functions. †

DR ILL 1.9 Simplifying Expressions Containing the Unit Impulse

Show that

(a)

$$(t^3 + 3)\delta(t) = 3\delta(t)$$

\n(b) $\left[\sin\left(t^2 - \frac{\pi}{2}\right)\right] \delta(t) = -\delta(t)$
\n(c) $e^{-2t}\delta(t) = \delta(t)$

(d)

$$\frac{\omega^2 + 1}{\omega^2 + 9} \delta(\omega - 1) = \frac{1}{5} \delta(\omega - 1)$$

DR ILL 1.10 Simplifying Integrals Containing the Unit Impulse

Show that

(a)

$$\int_{-\infty}^{\infty} \delta(t) e^{-j\omega t} dt = 1$$

\n(b)

$$\int_{-\infty}^{\infty} \delta(t-2) \cos\left(\frac{\pi t}{4}\right) dt = 0$$

\n(c)

$$\int_{-\infty}^{\infty} e^{-2(x-t)} \delta(2-t) dt = e^{-2(x-2)}$$

1.4-3 The Exponential Function est

Another important function in the area of signals and systems is the exponential signal est, where s is complex in general, given by

s = σ +jω

^† Singularity functions were defined by late Prof. S. J. Mason as follows. A singularity is a point at which a function does not possess a derivative. Each of the singularity functions (or if not the function itself, then the function differentiated a finite number of times) has a singular point at the origin and is zero elsewhere [2].

Therefore,

$$e^{st} = e^{(\sigma + j\omega)t} = e^{\sigma t} e^{j\omega t} = e^{\sigma t} (\cos \omega t + j \sin \omega t)$$

(1.13)

Since s^∗ = σ −jω (the conjugate of s), then

$$e^{s^*t} = e^{(\sigma - j\omega)t} = e^{\sigma t}e^{-j\omega t} = e^{\sigma t}(\cos \omega t - j\sin \omega t)$$

and

$$e^{\sigma t} \cos \omega t = \frac{1}{2} (e^{st} + e^{s^* t})$$

\n(1.14)

A comparison of Eq. (1.13) with Euler’s formula shows that est is a generalization of the function ejω*^t* , where the frequency variable jω is generalized to a complex variable s = σ + jω. For this reason, we designate the variable s as the complex frequency. In fact, function est encompasses a large class of functions. The following functions are either special cases of or can be expressed in terms of est:

• 1. A constant k = ke0*^t* (s = 0)
• 2. A monotonic exponential e^σt (ω = 0, s = σ )
• 3. A sinusoid cos ωt (σ = 0, s = ±jω)
• 4. An exponentially varying sinusoid e^σt cos ωt (s = σ ±jω)

These functions are illustrated in Fig. 1.21.

The complex frequency s can be conveniently represented on a complex frequency plane (s plane), as depicted in Fig. 1.22. The horizontal axis is the real axis (σ axis), and the vertical axis is the imaginary axis (ω axis). The absolute value of the imaginary part of s is |ω| (the

Figure 1.21 Sinusoids of complex frequency σ +jω.

Figure 1.22 Complex frequency plane.

radian frequency), which indicates the frequency of oscillation of est; the real part σ (the neper frequency) gives information about the rate of increase or decrease of the amplitude of est. For signals whose complex frequencies lie on the real axis (σ axis, where ω = 0), the frequency of oscillation is zero. Consequently these signals are monotonically increasing or decreasing exponentials (Fig. 1.21a). For signals whose frequencies lie on the imaginary axis (ω axis, where σ = 0), e^σt = 1. Therefore, these signals are conventional sinusoids with constant amplitude (Fig. 1.21b). The case s = 0 (σ = ω = 0) corresponds to a constant (dc) signal because e0*^t* = 1. For the signals illustrated in Figs. 1.21c and 1.21d, both σ and ω are nonzero; the frequency s is complex and does not lie on either axis. The signal in Fig. 1.21c decays exponentially. Therefore, σ is negative, and s lies to the left of the imaginary axis. In contrast, the signal in Fig. 1.21d grows exponentially. Therefore, σ is positive, and s lies to the right of the imaginary axis. Thus the s plane (Fig. 1.21) can be separated into two parts: the left half-plane (LHP) corresponding to exponentially decaying signals and the right half-plane (RHP) corresponding to exponentially growing signals. The imaginary axis separates the two regions and corresponds to signals of constant amplitude.

An exponentially growing sinusoid e^2t cos 5t, for example, can be expressed as a linear combination of exponentials e(2+j5)^t and e(2−j5)^t with complex frequencies 2 + j5 and 2−j5, respectively, which lie in the RHP. An exponentially decaying sinusoid e−2*^t* cos 5t can be expressed as a linear combination of exponentials e(−2+j5)^t and e(−2−j5)^t with complex frequencies −2 + j5 and −2 − j5, respectively, which lie in the LHP. A constant-amplitude sinusoid cos 5t can be expressed as a linear combination of exponentials ej5*^t* and e−j5*^t* with complex frequencies ±j5, which lie on the imaginary axis. Observe that the monotonic exponentials e±2*^t* are also generalized sinusoids with complex frequencies ±2.