[1.1 SIZE OF A](#page-7-0) SIGNAL

← Back to LINEAR SYSTEMS AND SIGNALS Overview

1.1 SIZE OF A SIGNAL

The size of any entity is a number that indicates the largeness or strength of that entity. Generally speaking, the signal amplitude varies with time. How can a signal that exists over a certain time interval with varying amplitude be measured by one number that will indicate the signal size or signal strength? Such a measure must consider not only the signal amplitude, but also its duration. For instance, if we are to devise a single number V as a measure of the size of a human being, we must consider not only his or her width (girth), but also the height. If we make a simplifying assumption that the shape of a person is a cylinder of variable radius r (which varies with the height h), then one possible measure of the size of a person of height H is the person’s volume V, given by

$$V = \pi \int_0^H r^2(h) \, dh$$

1.1-1 Signal Energy

Arguing in this manner, we may consider the area under a signal x(t) as a possible measure of its size, because it takes account not only of the amplitude but also of the duration. However, this will be a defective measure because even for a large signal x(t), its positive and negative areas could cancel each other, indicating a signal of small size. This difficulty can be corrected by defining the signal size as the area under |x(t)| 2, which is always positive. We call this measure the signal energy Ex, defined as

$$E_x = \int_{-\infty}^{\infty} |x(t)|^2 dt$$

(1.1)

This definition simplifies for a real-valued signal ^x(t) to Ex ⁼ $ ^∞ −∞ ^x2(t)dt. There are also other possible measures of signal size, such as the area under |x(t)|. The energy measure, however, is not only more tractable mathematically but is also more meaningful (as shown later) in the sense that it is indicative of the energy that can be extracted from the signal.

1.1-2 Signal Power

Signal energy must be finite for it to be a meaningful measure of signal size. A necessary condition for the energy to be finite is that the signal amplitude → 0 as |t|→∞ (Fig. 1.1a). Otherwise the integral in Eq. (1.1) will not converge.

When the amplitude of x(t) does not → 0 as |t|→∞ (Fig. 1.1b), the signal energy is infinite. A more meaningful measure of the signal size in such a case would be the time average of the energy, if it exists. This measure is called the power of the signal. For a signal x(t), we define its power Px as

$$P_x = \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} |x(t)|^2 dt$$

(1.2)

This definition simplifies for a real-valued signal ^x(t) to Px ⁼ lim*^T*→∞ ¹ T $ ^T/^{2 −}T/^{2 x2}(t)dt. Observe that the signal power Px is the time average (mean) of the signal magnitude squared, that is, the mean-square value of |x(t)|. Indeed, the square root of Px is the familiar rms (root-mean-square) value of x(t).

Generally, the mean of an entity averaged over a large time interval approaching infinity exists if the entity either is periodic or has a statistical regularity. If such a condition is not satisfied, the average may not exist. For instance, a ramp signal x(t) = t increases indefinitely as |t|→∞, and neither the energy nor the power exists for this signal. However, the unit step function, which is not periodic nor has statistical regularity, does have a finite power.

Figure 1.1 Examples of signals: (a) a signal with finite energy and (b) a signal with finite power.

When x(t) is periodic, |x(t)| ² is also periodic. Hence, the power of x(t) can be computed from Eq. (1.2) by averaging |x(t)| ² over one period.

Comments. The signal energy as defined in Eq. (1.1) does not indicate the actual energy (in the conventional sense) of the signal because the signal energy depends not only on the signal, but also on the load. It can, however, be interpreted as the energy dissipated in a normalized load of a 1 ohm resistor if a voltage x(t) were to be applied across the 1 ohm resistor [or if a current x(t) were to be passed through the 1 ohm resistor]. The measure of “energy” is therefore indicative of the energy capability of the signal, not the actual energy. For this reason the concepts of conservation of energy should not be applied to this “signal energy.” Parallel observation applies to “signal power” defined in Eq. (1.2). These measures are but convenient indicators of the signal size, which prove useful in many applications. For instance, if we approximate a signal x(t) by another signal g(t), the error in the approximation is e(t) = x(t) − g(t). The energy (or power) of e(t) is a convenient indicator of the goodness of the approximation. It provides us with a quantitative measure of determining the closeness of the approximation. In communication systems, during transmission over a channel, message signals are corrupted by unwanted signals (noise). The quality of the received signal is judged by the relative sizes of the desired signal and the unwanted signal (noise). In this case the ratio of the message signal and noise signal powers (signal-to-noise power ratio) is a good indication of the received signal quality.

Units of Energy and Power. Equation (1.1) is not correct dimensionally. This is because here we are using the term energy not in its conventional sense, but to indicate the signal size. The same observation applies to Eq. (1.2) for power. The units of energy and power, as defined here, depend on the nature of the signal x(t). If x(t) is a voltage signal, its energy Ex has units of volts squared-seconds (V2 s), and its power Px has units of volts squared. If x(t) is a current signal, these units will be amperes squared-seconds (A2 s) and amperes squared, respectively.

Figure 1.2 Signals for Ex. 1.1

In Fig. 1.2a, the signal amplitude → 0 as |t|→∞. Therefore the suitable measure for this signal is its energy Ex given by

$$E_x = \int_{-\infty}^{\infty} |x(t)|^2 dt = \int_{-1}^{0} (2)^2 dt + \int_{0}^{\infty} 4e^{-t} dt = 4 + 4 = 8$$

In Fig. 1.2b, the signal magnitude does not → 0 as |t|→∞. However, it is periodic, and therefore its power exists. We can use Eq. (1.2) to determine its power. We can simplify the procedure for periodic signals by observing that a periodic signal repeats regularly each period (2 seconds in this case). Therefore, averaging |x(t)| ² over an infinitely large interval is identical to averaging this quantity over one period (2 seconds in this case). Thus

$$P_x = \frac{1}{2} \int_{-1}^{1} |x(t)|^2 dt = \frac{1}{2} \int_{-1}^{1} t^2 dt = \frac{1}{3}$$

Recall that the signal power is the square of its rms value. Therefore, the rms value of this signal is 1/ ^√3.

EXAMPLE 1.2 Determining Power and RMS Value

Determine the power and the rms value of

• (a) x(t) = C cos(ω0t +θ )
• (b) x(t) = C¹ cos(ω1t +θ1)+C² cos(ω2t +θ2) ω¹ = ω²
• (c) x(t) = Dej^ω0*^t*

(a) This is a periodic signal with period T⁰ = 2π/ω0. The suitable measure of this signal is its power. Because it is a periodic signal, we may compute its power by averaging its energy over one period T⁰ = 2π/ω0. However, for the sake of demonstration, we shall use Eq. (1.2) to solve this problem by averaging over an infinitely large time interval.

$$P_x = \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} C^2 \cos^2 (\omega_0 t + \theta) dt = \lim_{T \to \infty} \frac{C^2}{2T} \int_{-T/2}^{T/2} [1 + \cos (2\omega_0 t + 2\theta)] dt$$

=

$$\lim_{T \to \infty} \frac{C^2}{2T} \int_{-T/2}^{T/2} dt + \lim_{T \to \infty} \frac{C^2}{2T} \int_{-T/2}^{T/2} \cos (2\omega_0 t + 2\theta) dt$$

The first term on the right-hand side is equal to C²/2. The second term, however, is zero because the integral appearing in this term represents the area under a sinusoid over a very large time interval T with T → ∞. This area is at most equal to the area of half the cycle because of cancellations of the positive and negative areas of a sinusoid. The second term is this area multiplied by C²/2T with T → ∞. Clearly this term is zero, and

$$P_x = \frac{C^2}{2}$$

This shows that a sinusoid of amplitude C has a power C²/2 regardless of the value of its frequency ω⁰ (ω⁰ = 0) and phase θ. The rms value is C/ ^√2. If the signal frequency is zero (dc or a constant signal of amplitude C), the reader can show that the power is C2.

(b) In Ch. 6, we shall show that a sum of two sinusoids may or may not be periodic, depending on whether the ratio ω1/ω² is a rational number. Therefore, the period of this signal is not known. Hence, its power will be determined by averaging its energy over T seconds with T → ∞. Thus,

$$P_x = \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} [C_1 \cos(\omega_1 t + \theta_1) + C_2 \cos(\omega_2 t + \theta_2)]^2 dt$$

=

$$\lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} C_1^2 \cos^2(\omega_1 t + \theta_1) dt + \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} C_2^2 \cos^2(\omega_2 t + \theta_2) dt$$

$$\lim_{T \to \infty} \frac{2C_1 C_2}{T} \int_{-T/2}^{T/2} \cos(\omega_1 t + \theta_1) \cos(\omega_2 t + \theta_2) dt$$

The first and second integrals on the right-hand side are the powers of the two sinusoids, which are C^{1 2}/2 and C^{2 2}/2, as found in part (a). The third term, the product of two sinusoids, can be expressed as a sum of two sinusoids cos[(ω1+ω2)t+(θ1+θ2)] and cos[(ω1−ω2)t+(θ1−θ2)], respectively. Now, arguing as in part (a), we see that the third term is zero. Hence, we have†

$$P_x = \frac{C_1^2}{2} + \frac{C_2}{2}$$

2

and the rms value is (C^{1 2 +}C^{2 2})/2.

We can readily extend this result to a sum of any number of sinusoids with distinct frequencies. Thus, if

$$x(t) = \sum_{n=1}^{\infty} C_n \cos{(\omega_n t + \theta_n)}$$

assuming that none of the two sinusoids have identical frequencies and ω*ⁿ* = 0, then

$$P_x = \frac{1}{2} \sum_{n=1}^{\infty} C_n^2$$

If x(t) also has a dc term, as

$$x(t) = C_0 + \sum_{n=1}^{\infty} C_n \cos(\omega_n t + \theta_n)$$

then

$$P_x = C_0^2 + \frac{1}{2} \sum_{n=1}^{\infty} C_n^2$$

(1.3)

(c) In this case the signal is complex, and we use Eq. (1.2) to compute the power.

$$P_x = \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} |De^{j\omega_0 t}|^2 dt$$

Recall that |ej^ω0*^t* | = 1 so that |Dej^ω0*^t* | ² = |D| 2, and

$$P_x = |D|^2 \tag{1.4}$$

The rms value is |D|.

Comment. In part (b) of Ex. 1.2, we have shown that the power of the sum of two sinusoids is equal to the sum of the powers of the sinusoids. It may appear that the power of x1(t) + x2(t)

^† This is true only if ^ω1 = ^ω2. If ^{ω1 = ω}2, the integrand of the third term contains a constant cos(θ^{1 − θ}2), and the third term → 2C1C² cos(θ¹ −θ2) as T → ∞.

is Px¹ + Px² . Unfortunately, this conclusion is not true in general. It is true only under a certain condition (orthogonality), discussed later (Sec. 6.5-3).

DR ILL 1.1 Computing Energy, Power, and RMS Value

Show that the energies of the signals in Figs. 1.3a, 1.3b, 1.3c, and 1.3d are 4, 1, 4/3, and 4/3, respectively. Observe that doubling a signal quadruples the energy, and time-shifting a signal has no effect on the energy. Show also that the power of the signal in Fig. 1.3e is 0.4323. What is the rms value of signal in Fig. 1.3e?

DR ILL 1.2 Computing Power over a Period

Redo Ex. 1.1a to find the power of a sinusoid C cos(ω0t + θ ) by averaging the signal energy over one period T⁰ = 2π/ω⁰ (rather than averaging over the infinitely large interval). Show also that the power of a dc signal x(t) = C⁰ is C^{2 0}, and its rms value is C0.

DR ILL 1.3 Power of a Sum of Two Equal-Frequency Sinusoids

Show that if ω¹ = ω2, the power of x(t) = C¹ cos(ω1t +θ1)+C² cos(ω2t +θ2) is [C^{1 2 +}C² 2 + 2C1C² cos(θ¹ −θ2)]/2, which is not equal to the Ex. 1.2b result of (C^{1 2 +}C^{2 2})/2.