3.3 SOME USEFUL [DISCRETE-TIME](#page-9-0) SIGNAL MODELS

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3.3 SOME USEFUL DISCRETE-TIME SIGNAL MODELS

We now discuss some important discrete-time signal models that are encountered frequently in the study of discrete-time signals and systems.

3.3-1 Discrete-Time Impulse Function ^δ[n]

The discrete-time counterpart of the continuous-time impulse function δ(t) is δ[n], a Kronecker delta function, defined by

$$\delta[n] = \begin{cases} 1 & n = 0 \\ 0 & n \neq 0 \end{cases}$$

This function, also called the unit impulse sequence, is shown in Fig. 3.6a. The shifted impulse sequence δ[n − m] is depicted in Fig. 3.6b. Unlike its continuous-time counterpart δ(t) (the Dirac delta), the Kronecker delta is a very simple function, requiring no special esoteric knowledge of distribution theory.

Figure 3.6 Discrete-time impulse function: (a) unit impulse sequence and (b) shifted impulse sequence.

3.3-2 Discrete-Time Unit Step Function ^u[n]

The discrete-time counterpart of the unit step function u(t) is u[n] (Fig. 3.7a), defined by

$$u[n] = \begin{cases} 1 & \text{for } n \ge 0 \\ 0 & \text{for } n < 0 \end{cases}$$

If we want a signal to start at n = 0 (so that it has a zero value for all n < 0), we need only multiply the signal by u[n].

Figure 3.7 (a) A discrete-time unit step function u[n] and (b) its application.

EXAMPLE 3.3 Describing Signals with Unit Step and Unit Impulse Functions

Describe the signal x[n] shown in Fig. 3.7b by a single expression valid for all n.

The signal x[n] can be broken into three components: (1) a ramp component x1[n] from n = 0 to 4, (2) a scaled step component x2[n] from n = 5 to 10, and (3) an impulse component x3[n] represented by the negative spike at n = 8. Let us consider each one separately.

We express x1[n] = n(u[n]−u[n−5]) to account for the signal from n = 0 to 4. Assuming that the spike at n = 8 does not exist, we can express x2[n] = 4(u[n−5] −u[n−11]) to account for the signal from n = 5 to 10. Once these two components have been added, the only part that is unaccounted for is a spike of amplitude −2 at n = 8, which can be represented by

There are many different ways of viewing x[n]. Although each way of viewing yields a different expression, they are all equivalent. We shall consider here just one possible expression.

x3[n]=−2δ[n−8]. Hence,

$$x[n] = x_1[n] + x_2[n] + x_3[n]$$

= $n(u[n] - u[n-5]) + 4(u[n-5] - u[n-11]) - 2\delta[n-8]$ for all n

We stress again that the expression is valid for all values of n. The reader can find several other equivalent expressions for x[n]. For example, one may consider a scaled step function from n = 0 to 10, subtract a ramp over the range n = 0 to 3, and subtract the spike. You can also play with breaking n into different ranges for your expression.

3.3-3 Discrete-Time Exponential γ ⁿ

A continuous-time exponential e^λt can be expressed in an alternate form as

$$e^{\lambda t} = \gamma^t \qquad (\gamma = e^{\lambda} \text{ or } \lambda = \ln \gamma)$$

For example, e−0.3*^t* = (0.7408)^t because e−0.3 = 0.7408. Conversely, 4*^t* = e1.386*^t* because e1.386 = 4, that is, ln 4 = 1.386. In the study of continuous-time signals and systems, we prefer the form e^λt rather than γ ^t . In contrast, the exponential form γ ⁿ is preferable in the study of discrete-time signals and systems, as will become apparent later. The discrete-time exponential γ ⁿ can also be expressed by using a natural base, as

$$e^{\lambda n} = \gamma^n \qquad (\gamma = e^{\lambda} \text{ or } \lambda = \ln \gamma)$$

Because of unfamiliarity with exponentials with bases other than e, exponentials of the form γ ⁿ may seem inconvenient and confusing at first. The reader is urged to plot some exponentials to acquire a sense of these functions. Also observe that γ ⁻ⁿ = 1 γ n .

DR ILL 3.6 Equivalent Forms of DT Exponentials

(a) Show that (i) (0.25)−ⁿ = 4*ⁿ, (ii) 4−ⁿ* = (0.25)ⁿ, (iii) e^2t = (7.389)^t , (iv) e−2*^t* = (0.1353)^t = (7.389)−^t , (v) e³ⁿ = (20.086)ⁿ, and (vi) e−1.5*ⁿ* = (0.2231)ⁿ = (4.4817)−ⁿ. (b) Show that (i) 2*ⁿ* = e0.693*ⁿ, (ii) (0.5)ⁿ* = e−0.693*ⁿ, and (iii) (0.8)−ⁿ* = e0.2231*ⁿ*.

Nature of γ ⁿ. The signal e^λn grows exponentially with n if Reλ > 0 (λ in the RHP), and decays exponentially if Reλ < 0 (λ in the LHP). It is constant or oscillates with constant amplitude if Reλ = 0 (λ on the imaginary axis). Clearly, the location of λ in the complex plane indicates whether the signal e^λn will grow exponentially, decay exponentially, or oscillate with constant

Figure 3.8 The λ plane, the γ plane, and their mapping.

amplitude (Fig. 3.8a). A constant signal (λ = 0) is also an oscillation with zero frequency. We now find a similar criterion for determining the nature of γ ⁿ from the location of γ in the complex plane.

Figure 3.8a shows a complex plane (λ plane). Consider a signal ejⁿ. In this case, λ = j lies on the imaginary axis (Fig. 3.8a), and therefore is a constant-amplitude oscillating signal. This signal ejⁿ can be expressed as γ ⁿ, where γ = ej. Because the magnitude of ej is unity, |γ | = 1. Hence, when λ lies on the imaginary axis, the corresponding γ lies on a circle of unit radius, centered at the origin (the unit circle illustrated in Fig. 3.8b). Therefore, a signal γ ⁿ oscillates with constant amplitude if γ lies on the unit circle. Thus, the imaginary axis in the λ plane maps into the unit circle in the γ plane.

Next consider the signal e^λn, where λ lies in the left half-plane in Fig. 3.8a. This means λ = a + jb, where a is negative (a < 0). In this case, the signal decays exponentially. This signal can be expressed as γ ⁿ, where

$$\gamma = e^{\lambda} = e^{a+jb} = e^a e^{jb}$$

and

$$|\gamma| = |e^a| \, |e^{ib}| = e^a$$

because $|e^{ib}| = 1$

Also, a is negative (a < 0). Hence, |γ | = ea < 1. This result means that the corresponding γ lies inside the unit circle. Therefore, a signal γ ⁿ decays exponentially if γ lies within the unit circle (Fig. 3.8b). If, in the preceding case we select a to be positive (λ in the right half-plane), then |γ | > 1, and γ lies outside the unit circle. Therefore, a signal γ ⁿ grows exponentially if γ lies outside the unit circle (Fig. 3.8b).

To summarize, the imaginary axis in the λ plane maps into the unit circle in the γ plane. The left half-plane in the λ plane maps into the inside of the unit circle and the right half of the λ plane maps into the outside of the unit circle in the γ plane, as depicted in Fig. 3.8.

Figure 3.9 Discrete-time exponentials γ ⁿ.

Plots of (0.8)ⁿ and (−0.8)ⁿ appear in Figs. 3.9a and 3.9b, respectively. Plots of (0.5)ⁿ and (1.1)ⁿ appear in Figs. 3.9c and 3.9d, respectively. These plots verify our earlier conclusions about the location of γ and the nature of signal growth. Observe that a signal (−|γ |)ⁿ alternates sign successively (is positive for even values of n and negative for odd values of n, as depicted in Fig. 3.9b). Also, the exponential (0.5)ⁿ decays faster than (0.8)ⁿ because 0.5 is closer to the origin than 0.8. The exponential (0.5)ⁿ can also be expressed as 2−ⁿ because (0.5)−¹ = 2.

DR ILL 3.7 Sketching DT Exponentials

Sketch the following signals: (a) (1)ⁿ, (b) (−1)ⁿ, (c) (0.5)ⁿ, (d) (−0.5)ⁿ, (e) (0.5)−ⁿ, (f) 2−ⁿ, and (g) (−2)ⁿ. Express these exponentials as γ ⁿ, and plot γ in the complex plane for each case. Verify that γ ⁿ decays exponentially with n if γ lies inside the unit circle and that γ ⁿ grows with n if γ is outside the unit circle. If γ is on the unit circle, γ ⁿ is constant or oscillates with a constant amplitude.

250 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS

Accurately hand-sketching DT signals can be tedious and difficult. As the next example shows, MATLAB is particularly well suited to plot DT signals, including exponentials.

EXAMPLE 3.4 Plotting DT Exponentials with MATLAB

Use MATLAB to plot the following discrete-time signals over (0 ≤ n ≤ 8): (a) xa[n] = (0.8)ⁿ, (b) xb[n] = (−0.8)ⁿ, (c) xc[n] = (0.5)ⁿ, and (d) xd[n] = (1.1)ⁿ.

To begin, we use anonymous functions to represent each of the four signals. Next, we plot these functions over the desired range of n. The results, shown in Fig. 3.10, match the earlier Fig. 3.9 plots of the same signals.

>> n = (0:8); x_a = @(n) (0.8).^n; x_b = @(n) (-0.8).^(n);
>> x_c = @(n) (0.5).^n; x_d = @(n) (1.1).^n;
>> subplot(2,2,1); stem(n,x_a(n),'k'); ylabel('x_a[n]'); xlabel('n');
>> subplot(2,2,2); stem(n,x_b(n),'k'); ylabel('x_b[n]'); xlabel('n');
>> subplot(2,2,3); stem(n,x_c(n),'k'); ylabel('x_c[n]'); xlabel('n');
>> subplot(2,2,4); stem(n,x_d(n),'k'); ylabel('x_d[n]'); xlabel('n');

3.3-4 Discrete-Time Sinusoid cos*(ⁿ* **⁺**θ )

A general discrete-time sinusoid can be expressed as Ccos(n+θ ), where C is the amplitude, and θ is the phase in radians. Also, n is an angle in radians. Hence, the dimensions of the frequency are radians per sample. This sinusoid may also be expressed as

$$C\cos\left(\Omega n + \theta\right) = C\cos\left(2\pi\mathcal{F}n + \theta\right)$$

where F = /2π. Therefore, the dimensions of the discrete-time frequency F are (radians/2π) per sample, which is equal to cycles per sample. This means if N⁰ is the period (samples/cycle) of the sinusoid, then the frequency of the sinusoid F = 1/N⁰ (samples/cycle).

Figure 3.11 shows a discrete-time sinusoid cos( ^{π 12 n + π 4} ). For this case, the frequency is = π/12 radians/sample. Alternately, the frequency is F = 1/24 cycles/sample. In other words, there are 24 samples in one cycle of the sinusoid.

Because cos(−x) = cos(x),

$$\cos\left(-\Omega n + \theta\right) = \cos\left(\Omega n - \theta\right)$$

This shows that both cos(n+θ ) and cos(−n+θ ) have the same frequency (). Therefore, the frequency of cos(n+θ ) is ||.

Figure 3.11 A discrete-time sinusoid cos( ^{π 12 n+ π} 4 ).

SAMPLED CONTINUOUS-TIME SINUSOID YIELDS A DISCRETE-TIME SINUSOID

A continuous-time sinusoid cosωt sampled every T seconds yields a discrete-time sequence whose nth element (at t = nT) is cosωnT. Thus, the sampled signal x[n] is given by

$$x[n] = \cos \omega nT = \cos \Omega n \quad \text{where } \Omega = \omega T$$

252 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS

Thus, a continuous-time sinusoid cosωt sampled every T seconds yields a discrete-time sinusoid cosn, where = ωT. †

3.3-5 Discrete-Time Complex Exponential ejⁿ

Using Euler’s formula, we can express an exponential ejⁿ in terms of sinusoids as

ej^{n =} (cosn+jsinn) and ^e−j^{n =} (cosn−jsinn)

These equations show that the frequency of both ejⁿ and e−jⁿ is (radians/sample). Therefore, the frequency of ejⁿ is ||.

Observe that for r = 1 and θ = n,

$$e^{j\Omega n}=re^{j\theta}$$

This equation shows that the magnitude and angle of ejⁿ are 1 and n, respectively. In the complex plane, ejⁿ is a point on a unit circle at an angle n.

EXAMPLE 3.5 Plotting a DT Sinusoid with MATLAB

Using MATLAB, plot the discrete-time sinusoid ^x[n] = cos ^{π 12 n+ π} 4 .

We represent the desired sinusoid using an anonymous function. Next, we plot this function over the desired range of n. The result, shown in Fig. 3.12, matches the plot of the same signal shown in Fig. 3.11.

n = (-30:30); x = @(n) cos(n*pi/12+pi/4); >> clf; stem(n,x(n),‘k’); ylabel(‘x[n]’); xlabel(‘n’);

^† Superficially, it may appear that a discrete-time sinusoid is a continuous-time sinusoid’s cousin in a striped suit. However, some of the properties of discrete-time sinusoids are very different from those of continuous-time sinusoids. For instance, not every discrete-time sinusoid is periodic. A sinusoid cosn is periodic only if is a rational multiple of 2π. Also, discrete-time sinusoids are bandlimited to = π. Any sinusoid with ≥ π can always be expressed as a sinusoid of some frequency ≤ π. These peculiar properties are the direct consequence of the fact that the period of a discrete-time sinusoid must be an integer. These topics are discussed in Chs. 5 and 9.