# 7.6 Electromagnetic Power Density

This section deals with the flow of power carried by an electromagnetic wave. For any wave with an electric field E and magnetic field H, the Poynting vector S is defined as

S=E×H(W/m 2).(7.97)

The unit of S is (V/m)×(A/m)=(W/m 2), and the direction of S is along the wave’s direction of propagation. Thus, S represents the power per unit area (or power density) carried by the wave. If the wave is incident upon an aperture of area A with outward surface unit vector n ˆ as shown in Fig. 7-17, then the total power that flows through or is intercepted by the aperture is

P=∫A S⋅n ˆ d A(W).(7.98)

# Figure 7-17 EM power flow through an aperture.

For a uniform plane wave propagating in a direction k ˆ that makes an angle θ with n ˆ, P=S A cos θ, where S=|S|.

Except for the fact that the units of S are per unit area, Eq. (7.97) is the vector analogue of the scalar expression for the instantaneous power P(z,t) flowing through a transmission line,

P(z,t)=v(z,t)i(z,t),(7.99)

where v(z,t) and i(z,t) are the instantaneous voltage and current on the line.

Since both E and H are functions of time, so is the Poynting vector S. In practice, however, the quantity of greater interest is the average power density of the wave, S av, which is the time-average value of S:

This expression may be regarded as the electromagnetic equivalent of Eq. (2.107) for the time-average power carried by a transmission line, namely

P av(z)=1 2 R e V˜(z)I˜∗(z),(7.101)

where V˜(z) and I˜(z) are the phasors corresponding to v(z,t) and i(z,t), respectively.

# 7.6.1 Plane Wave in a Lossless Medium

Recall that the general expression for the electric field of a uniform plane wave with arbitrary polarization traveling in the +z direction is

E˜(z)=x ˆ E˜x(z)+y ˆ E˜y(z)=(x ˆ E x 0+y ˆ E y 0)e−j k z,(7.102)

where, in the general case, E x 0 and E y 0 may be complex quantities. The magnitude of E˜ is

|E˜|=(E˜⋅E˜∗)1/2=[|E x 0|2+|E y 0|2]1/2.(7.103)

The phasor magnetic field associated with E˜ is obtained by applying Eq. (7.39a):

H˜(z)=(x ˆ H˜x+y ˆ H˜y)e−j k z=1 η z ˆ×E˜=1 η(−x ˆ E y 0+y ˆ E x 0)e−j k z.(7.104)

The wave can be considered as the sum of two waves: one comprising fields (E˜x,H˜y) and another comprising fields (E˜y,H˜x). Use of Eqs. (7.102) and (7.104) in Eq. (7.100) leads to

which states that power flows in the z direction with average power density equal to the sum of the average power densities of the (E˜x,H˜y) and (E˜y,H˜x) waves. Note that, because S av depends only on η and |E˜|, waves characterized by different polarizations carry the same amount of average power as long as their electric fields have the same magnitudes.

# Example 7-5: Solar Power

If solar illumination is characterized by a power density of 1 kW/m 2 on the Earth’s surface, find (a) the total power radiated by the sun, (b) the total power intercepted by the Earth, and (c) the electric field of the power density incident upon the Earth’s surface, assuming that all the solar illumination is at a single frequency. The radius of the Earth’s orbit around the sun, R s, is approximately 1.5×10 8 km, and the Earth’s mean radius R e is 6,380 km.

Solution: (a) Assuming that the sun radiates isotropically (equally in all directions), the total power it radiates is S av A sph, where A sph is the area of a spherical shell of radius R s (Fig. 7-18(a)). Thus,

P sun=S av(4 π R s 2)=1×10 3×4 π×(1.5×10 11)2=2.8×10 26 W.

# Figure 7-18 Solar radiation intercepted by (a) a spherical surface of radius R s and (b) the Earth’s surface (Example 7-5).

(a) Radiated solar power

Figure 7 18-a Full Alternative Text

(b) Earth intercepted power

Figure 7 18-b Full Alternative Text

(b) With reference to Fig. 7-18(b), the power intercepted by the Earth’s cross section A e=π R e 2 is

P int=S av(π R e 2)=1×10 3×π×(6.38×10 6)2=1.28×10 17 W.

(c) The power density S av is related to the magnitude of the electric field |E˜|=E 0 by

S a v=E 0 2 2 η 0,

where η 0=377 (Ω) for air. Hence,

E 0=2 η 0 S av=2×377×10 3=870(V/m).

# 7.6.2 Plane Wave in a Lossy Medium

The expressions given by Eqs. (7.68) and (7.69) characterize the electric and magnetic fields of an x-polarized plane wave propagating along the z direction in a lossy medium with propagation constant γ=α+j β. By extending these expressions to the more general case of a wave with components along both x and y, we have

E˜(z)=x ˆ E˜x(z)+y ˆ E˜y(z)=(x ˆ E x 0+y ˆ E y 0)e−α z e−j β z,(7.106a)
H˜(z)=1 η c(−x ˆ E y 0+y ˆ E x 0)e−α z e−j β z,(7.106b)

where η c is the intrinsic impedance of the lossy medium. Application of Eq. (7.100) gives

S av(z)=1 2 ℜ e[ E˜​×H˜∗ ]=z ˆ(|E x 0|2+|E y 0|2)2 e−2 α z ℜ e(1 η c∗).(7.107)

By expressing η c in polar form as

η c=|η c|e j θ η,(7.108)

Eq. (7.107) can be rewritten as

where |E˜(0)|2=[|E x 0|2+|E y 0|2]1/2 is the magnitude of E˜(z) at z=0.

When a wave propagates through a distance z=δ s=1/α, the magnitudes of its electric and magnetic fields decrease to e−1≈37% of their initial values, and its average power density decreases to e−2≈14% of its initial value.

# 7.6.3 Decibel Scale for Power Ratios

The unit for power P is watts (W). In many engineering problems, the quantity of interest is the ratio of two power levels, P 1 and P 2, such as the incident and reflected powers on a transmission line, and often the ratio P 1/P 2 may vary over several orders of magnitude. The decibel (dB) scale is logarithmic, thereby providing a convenient representation of the power ratio, particularly when numerical values of P 1/P 2 are plotted against some variable of interest. If

G=P 1 P 2,(7.110)

then

G [dB]=10 log G=10 log(P 1 P 2) (dB).(7.111)

Table 7-2 provides a comparison between values of G and the corresponding values of G [dB]. Even though decibels are defined for power ratios, they can sometimes be used to represent other quantities. For example, if P 1=V 1 2/R is the power dissipated in a resistor R with voltage V 1 across it at time t 1, and P 2=V 2 2/R is the power dissipated in the same resistor at time t 2, then

G [d B]=10 log P 1 P 2=10 log V 1 2/R V 2 2/R=20 log V 1 V 2=20 log(g)=g [d B],(7.112)

where g=V 1/V 2 is the voltage ratio. Note that for voltage (or current) ratios the scale factor is 20 rather than 10, which results in G [d B]=g [d B].

# Table 7-2 Power ratios in natural numbers and in decibels.

Table 7 2 Full Alternative Text

The attenuation rate, representing the rate of decrease of the magnitude of S av(z) as a function of propagation distance, is defined as

A=10 log[ S av(z)S av(0) ]=10 log(e−2 α z)=−20 α z log e=−8.68 α z=−α [dB/m] z (dB),(7.113)

where

α [dB/m]=8.68 α(Np/m).(7.114)

Since S av(z) is directly proportional to |E(z)|2,

A=10 log|E(z)|2|E(0)|2=20 log|E(z)||E(0)|(d B).(7.115)

# Example 7-6: Power Received by a Submarine Antenna

A submarine at a depth of 200 m below the sea surface uses a wire antenna to receive signal transmissions at 1 kHz. Determine the power density incident upon the submarine antenna due to the EM wave of Example 7-4.

Solution: From Example 7-4, |E˜(0)|=|E x 0|=4.44 (mV/m), α=0.126 (Np/m), and η c=0.044 45° (Ω). Application of Eq. (7.109) gives

S av(z)=z ˆ|E 0|2 2|η c|e−2 α z cos θ η=z ˆ(4.44×10−3)2 2×0.044 e−0.252 z cos 45∘=z ˆ 0.16 e−0.252 z (mW/m 2).

At z=200 m, the incident power density is

S av=z ˆ(0.16×10−3 e−0.252×200)=2.1×10−26 (W/m 2).