Continuous-Time Signals 83

1.3 Continuous-Time Signals 83

Although this signal has infinite energy, it has finite power. Letting T = NT 0 where T 0 is the period

of 2 cos(4t − π/4) (or T 0 = 2π/4), then its power is

x = lim

x ( t)dt = lim

x ( t)dt

Z T 0 T N 0 1 Z = lim 2 x 2 ( t)dt = 4 cos ( 4t − π/4)dt

N→∞ 2NT 0 2T 0

which is a finite value and therefore the signal has finite power but infinite energy. ■ As we will see later in the Fourier series representation, any periodic signal is representable as a pos-

sibly infinite sum of sinusoids of frequencies multiples of the fundamental frequency of the periodic signal being represented. These frequencies are said to be harmonically related, and for this case the power of the signal is shown to be the sum of the power of each of the sinusoidal components—that is, there is superposition of the power. This superposition is still possible when a sum of sinusoids creates a nonperiodic signal. This is illustrated in Example 1.14.

■ Example 1.14

Consider the signals x(t) = cos(2πt) + cos(4πt) and y(t) = cos(2πt) + cos(2t), −∞ < t < ∞. Determine if these signals are periodic, and if so, find their periods. Compute the power of these signals.

Solution

The sinusoids cos(2πt) and cos(4πt) periods T 1 = 1 and T 2 = 1/2, so x(t) is periodic since T 1 / T 2 =

2 with period T 1 = 2T 2 = 1. The two frequencies are harmonically related. The sinusoid cos(2t) has as period T 3 = π. Therefore, the ratio of the periods of the sinusoidal components of y(t) is T 1 / T 3 = 1/π, which is not rational, and so y(t) is not periodic and the frequencies 2π and 2 are not harmonically related.

Using the trigonometric identities

2 1 cos (θ ) = ( 1 + cos(2θ))

1 cos(α) cos(β) = ( cos(α + β) + cos(α − β))

we have that x 2 (

2 t) = cos 2 ( 2π t) + cos ( 4π t) + 2 cos(2πt) cos(4πt)

1 1 =1+ cos(4πt) + cos(8πt) + cos(6πt) + cos(2πt)

84 C H A P T E R 1: Continuous-Time Signals

which is again a sum of harmonically related frequency sinusoids, so that x 2 ( t) is periodic of period T 0 = 1. As in the previous examples, we have

x 2 ( t)dt = 1

which is the integral of the constant since the other integrals are zero. In this case, we used the periodicity of x(t) and x 2 ( t) to calculate the power directly. That is not possible when computing the power of y(t) because it is not periodic, so we have to consider each of its components. We have that

2 t) = cos 2 ( 2π t) + cos ( 2t) + 2 cos(2πt) cos(2t)

1 1 =1+ cos(4π t) + cos(4t) + cos(2(π + 1)t) + cos(2(π − 1)t) 2 2

and the power of y(t) is

P y = lim 2 y ( t)dt

cos(4π t)dt +

cos(4t)dt

2T 4 2T 5

cos(2(π + 1)t)dt +

cos(2(π − 1)t)dt = 1

where T 4 ,T 5 ,T 6 , and T 7 are the periods of the sinusoidal components of y 2 ( t). Fortunately, only the first integral is not zero and the others are zero (the average over a period of the sinusoidal components of y 2 ( t)). Fortunately, too, we have that the power of x(t) and the power of y(t) are the sum of the powers of its components. That is if

x(t) = cos(2πt) + cos(4πt) = x 1 ( t) + x 2 ( t) y(t) = cos(2πt) + cos(2t) = y 1 ( t) + y 2 ( t)

then as in previous examples P x 1 =P x 2 =P y 1 =P y 2 = 0.5, so that P x =P x 1 +P x 2 =1

P y =P y 1 +P y 2 =1