[B.3 SKETCHING](#page-6-0) SIGNALS

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B.3 SKETCHING SIGNALS

In this section, we discuss the sketching of a few useful signals, starting with exponentials.

B.3-1 Monotonic Exponentials

The signal e−at decays monotonically, and the signal eat grows monotonically with t (assuming a > 0), as depicted in Fig. B.9. For the sake of simplicity, we shall consider an exponential e−at starting at t = 0, as shown in Fig. B.10a.

The signal e−at has a unit value at t = 0. At t = 1/a, the value drops to 1/e (about 37% of its initial value), as illustrated in Fig. B.10a. This time interval over which the exponential reduces by

Figure B.9 Monotonic exponentials.

Figure B.10 Sketching (a) e−at and (b) e−2*^t* .

a factor e (i.e., drops to about 37% of its value) is known as the time constant of the exponential. Therefore, the time constant of e−at is 1/a. Observe that the exponential is reduced to 37% of its initial value over any time interval of duration 1/a. This can be shown by considering any set of instants t¹ and t² separated by one time constant so that

$$t_2 - t_1 = \frac{1}{a}$$

Now the ratio of e−at² to e−at¹ is given by

$$\frac{e^{-at_2}}{e^{-at_1}} = e^{-a(t_2 - t_1)} = \frac{1}{e} \approx 0.37$$

We can use this fact to sketch an exponential quickly. For example, consider

$$x(t) = e^{-2t}$$

The time constant in this case is 0.5. The value of x(t) at t = 0 is 1. At t = 0.5 (one time constant), it is 1/e (about 0.37). The value of x(t) continues to drop further by the factor 1/e (37%) over the next half-second interval (one time constant). Thus, x(t) at t = 1 is (1/e)2. Continuing in this manner, we see that x(t) = (1/e)³ at t = 1.5, and so on. A knowledge of the values of x(t) at t = 0, 0.5, 1, and 1.5 allows us to sketch the desired signal, as shown in Fig. B.10b.†

For a monotonically growing exponential eat, the waveform increases by a factor e over each interval of 1/a seconds.

B.3-2 The Exponentially Varying Sinusoid

We now discuss sketching an exponentially varying sinusoid

$$x(t) = Ae^{-at}\cos{(\omega_0 t + \theta)}$$

Let us consider a specific example:

$$x(t) = 4e^{-2t}\cos{(6t - 60^\circ)}$$

We shall sketch 4e−2*^t* and cos(6t −60◦) separately and then multiply them:

• (a) Sketching 4e**−2***^t* . This monotonically decaying exponential has a time constant of 0.5 second and an initial value of 4 at t = 0. Therefore, its values at t = 0.5, 1, 1.5, and 2 are 4/e, 4/e2, 4/e3, and 4/e4, or about 1.47, 0.54, 0.2, and 0.07, respectively. Using these values as a guide, we sketch 4e−2*^t* , as illustrated in Fig. B.11a.
• (b) Sketching cos*(6^t* ^{− 60}◦). The procedure for sketching cos(6*^t* ^{− 60}◦) is discussed in Sec. B.2 (Fig. B.6c). Here, the period of the sinusoid is T⁰ = 2π/6 ≈ 1, and there is a phase delay of 60◦, or two-thirds of a quarter-cycle, which is equivalent to a delay of about (60/360)(1) ≈ 1/6 seconds (see Fig. B.11b).
• (c) Sketching 4e**−2***^t* cos*(6^t* ^{− 60}◦). We now multiply the waveforms in steps (a) and (b). This multiplication amounts to forcing the sinusoid 4 cos(6t −60◦) to decrease exponentially with a time constant of 0.5. The initial amplitude (at t = 0) is 4, decreasing to 4/e (=1.47) at t = 0.5, to 1.47/e(=0.54) at t = 1, and so on. This is depicted in Fig. B.11c. Note that when cos(6t −60◦) has a value of unity (peak amplitude),

$$4e^{-2t}\cos{(6t - 60^\circ)} = 4e^{-2t}$$

Therefore, 4e−2*^t* cos(6t−60◦) touches 4e−2*^t* at the instants at which the sinusoid cos(6t −60◦) is at its positive peaks. Clearly, 4e−2*^t* is an envelope for positive amplitudes of 4e−2*^t* cos(6t − 60◦). Similar argument shows that 4e−2*^t* cos(6t − 60◦) touches −4e−2*^t* at its negative peaks. Therefore, −4e−2*^t* is an envelope for negative amplitudes of 4e−2*^t* cos(6t − 60◦). Thus, to sketch 4e−2*^t* cos(6t − 60◦), we first draw the envelopes 4e−2*^t* and −4e−2*^t* (the mirror image of 4e−2*^t* about the horizontal axis), and then sketch the sinusoid cos(6t − 60◦), with these envelopes acting as constraints on the sinusoid’s amplitude (see Fig. B.11c).

In general, Ke−at cos(ω0t + θ ) can be sketched in this manner, with Ke−at and −Ke−at constraining the amplitude of cos(ω0t +θ ).

^† If we wish to refine the sketch further, we could consider intervals of half the time constant over which the signal decays by a factor 1/ ^√e. Thus, at ^{t =} 0.25, ^x(t) ^{= 1}/ ^√e, and at ^{t =} 0.75, ^x(t) ^{= 1}/^{e √}e, and so on.

Figure B.11 Sketching an exponentially varying sinusoid.

B.4 CRAMER’S RULE

Cramer’s rule offers a very convenient way to solve simultaneous linear equations. Consider a set of n linear simultaneous equations in n unknowns x1, x2,…, xn:

$$a_{11}x_1 + a_{12}x_2 + \cdots + a_{1n}x_n = y_1$$

\n

$$a_{21}x_1 + a_{22}x_2 + \cdots + a_{2n}x_n = y_2$$

\n

$$\vdots$$

\n

$$a_{n1}x_1 + a_{n2}x_2 + \cdots + a_{nn}x_n = y_n$$

\n(B.19)

These equations can be expressed in matrix form as

$$\begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \cdots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} = \begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_n \end{bmatrix}$$

(B.20)

We denote the matrix on the left-hand side formed by the elements aij as A. The determinant of A is denoted by |A|. If the determinant |A| is not zero, Eq. (B.19) has a unique solution given by Cramer’s formula

$$x_k = \frac{|\mathbf{D}_k|}{|\mathbf{A}|} \qquad k = 1, 2, \dots, n$$

(B.21)

where |Dk| is obtained by replacing the kth column of |A| by the column on the right-hand side of Eq. (B.20) (with elements y1, y2,…, yn).

We shall demonstrate the use of this rule with an example.