[7.6 SIGNAL](#page-13-0) ENERGY

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7.6 SIGNAL ENERGY

The signal energy Ex of a signal x(t) was defined in Ch. 1 as

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

\n(7.44)

Signal energy can be related to the signal spectrum X(ω) by substituting Eq. (7.10) in Eq. (7.44):

$$E_x = \int_{-\infty}^{\infty} x(t)x^*(t) dt = \int_{-\infty}^{\infty} x(t) \left[ \frac{1}{2\pi} \int_{-\infty}^{\infty} X^*(\omega) e^{-j\omega t} d\omega \right] dt$$

734 CHAPTER 7 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM

Here, we used the fact that x∗(t), being the conjugate of x(t), can be expressed as the conjugate of the right-hand side of Eq. (7.10). Now, interchanging the order of integration yields

$$E_x = \frac{1}{2\pi} \int_{-\infty}^{\infty} X^*(\omega) \left[ \int_{-\infty}^{\infty} x(t) e^{-j\omega t} dt \right] d\omega$$

= $\frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega) X^*(\omega) d\omega = \frac{1}{2\pi} \int_{-\infty}^{\infty} |X(\omega)|^2 d\omega$

Consequently,

$$E_x = \int_{-\infty}^{\infty} |x(t)|^2 dt = \frac{1}{2\pi} \int_{-\infty}^{\infty} |X(\omega)|^2 d\omega \qquad (7.45)$$

This is Parseval’s theorem (for the Fourier transform). A similar result was obtained in Eqs. (6.26) and (6.27) for a periodic signal and its Fourier series. This result allows us to determine the signal energy from either the time-domain specification x(t) or the corresponding frequency-domain specification X(ω).

The right-hand side of Eq. (7.45) can be interpreted to mean that the energy of a signal x(t) results from energies contributed by all the spectral components of the signal x(t). The total signal energy is the area under |X(ω)²| (divided by 2π). If we consider a small band ω (ω → 0), as illustrated in Fig. 7.35, the energy Ex of the spectral components in this band is the area of |X(ω)| ² under this band (divided by 2π):

$$\Delta E_x = \frac{1}{2\pi} |X(\omega)|^2 \,\Delta \omega = |X(\omega)|^2 \,\Delta f \qquad \frac{\Delta \omega}{2\pi} = \Delta f \,\mathrm{Hz}$$

Therefore, the energy contributed by the components in this band of f (in hertz) is |X(ω)| ²f . The total signal energy is the sum of energies of all such bands and is indicated by the area under |X(ω)| ² as in Eq. (7.45). Therefore, |X(ω)| ² is the energy spectral density (per unit bandwidth in hertz).

For real signals, X(ω) and X(−ω) are conjugates, and |X(ω)| ² is an even function of ω because

$$|X(\omega)|^2 = X(\omega)X^*(\omega) = X(\omega)X(-\omega)$$

Figure 7.35 Interpretation of energy spectral density of a signal.

Therefore, the energy of real signal x(t) can be expressed as†

$$E_x = \frac{1}{\pi} \int_0^\infty |X(\omega)|^2 d\omega \tag{7.46}$$

The signal energy Ex, which results from contributions from all the frequency components from ω = 0 to ∞, is given by (1/π times) the area under |X(ω)| ² from ω = 0 to ∞. It follows that the energy contributed by spectral components of frequencies between ω¹ and ω² is

$$\Delta E_x = \frac{1}{\pi} \int_{\omega_1}^{\omega_2} |X(\omega)|^2 d\omega \tag{7.47}$$

EXAMPLE 7.20 Signal Energy and Parseval’s Theorem

Find the energy of signal x(t) = e−atu(t). Determine the frequency W (rad/s) so that the energy contributed by the spectral components of all the frequencies below W is 95% of the signal energy Ex.

We have

$$E_x = \int_{-\infty}^{\infty} x^2(t) \, dt = \int_0^{\infty} e^{-2at} \, dt = \frac{1}{2a}$$

We can verify this result by Parseval’s theorem. For this signal,

$$X(\omega) = \frac{1}{j\omega + a}$$

and

$$E_x = \frac{1}{\pi} \int_0^{\infty} |X(\omega)|^2 d\omega = \frac{1}{\pi} \int_0^{\infty} \frac{1}{\omega^2 + a^2} d\omega = \frac{1}{\pi a} \tan^{-1} \frac{\omega}{a} \Big|_0^{\infty} = \frac{1}{2a}$$

The band ω = 0 to ω = W contains 95% of the signal energy, that is, 0.95/2a. Therefore, from Eq. (7.47) with ω¹ = 0 and ω² = W, we obtain

$$\frac{0.95}{2a} = \frac{1}{\pi} \int_0^W \frac{d\omega}{\omega^2 + a^2} = \frac{1}{\pi a} \tan^{-1} \frac{\omega}{a} \Big|_0^W = \frac{1}{\pi a} \tan^{-1} \frac{W}{a}$$

or

$$\frac{0.95\pi}{2} = \tan^{-1}\frac{W}{a} \implies W = 12.706a \text{ rad/s}$$

^† In Eq. (7.46), it is assumed that ^X(ω) does not contain an impulse at ^{ω =} 0. If such an impulse exists, it should be integrated separately with a multiplying factor of 1/2π rather than 1/π.

736 CHAPTER 7 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM

This result indicates that the spectral components of x(t) in the band from 0 (dc) to 12.706a rad/s (2.02a Hz) contribute 95% of the total signal energy; all the remaining spectral components (in the band from 12.706a rad/s to ∞) contribute only 5% of the signal energy.

DR ILL 7.12 Signal Energy and Parseval’s Theorem

Use Parseval’s theorem to show that the energy of the signal x(t) = 2a/(t ² +a2) is 2π/a. [Hint: Find X(ω) using pair 3 of Table 7.1 and the duality property.]

THE ESSENTIAL BANDWIDTH OF A SIGNAL

The spectra of all practical signals extend to infinity. However, because the energy of any practical signal is finite, the signal spectrum must approach 0 as ω → ∞. Most of the signal energy is contained within a certain band of B Hz, and the energy contributed by the components beyond B Hz is negligible. We can therefore suppress the signal spectrum beyond B Hz with little effect on the signal shape and energy. The bandwidth B is called the essential bandwidth of the signal. The criterion for selecting B depends on the error tolerance in a particular application. We may, for example, select B to be that band which contains 95% of the signal energy.† This figure may be higher or lower than 95%, depending on the precision needed. Using such a criterion, we can determine the essential bandwidth of a signal. The essential bandwidth B for the signal e−atu(t), using 95% energy criterion, was determined in Ex. 7.20 to be 2.02a Hz.

Suppression of all the spectral components of x(t) beyond the essential bandwidth results in a signal xˆ(t), which is a close approximation of x(t). If we use the 95% criterion for the essential bandwidth, the energy of the error (the difference) x(t)− ˆx(t) is 5% of Ex.