# 9.3 Half-Wave Dipole Antenna

In Section 9-1, we developed expressions for the electric and magnetic fields radiated by a Hertzian dipole of length l≪λ. We now use these expressions as building blocks to obtain expressions for the fields radiated by a half-wave dipole antenna, so named because its length l=λ/2. As shown in Fig. 9-14, the half-wave dipole consists of a thin wire fed at its center by a generator connected to the antenna terminals via a transmission line. The current flowing through the wire has a symmetrical distribution with respect to the center of the dipole, and the current is zero at its ends. Mathematically, i(t) is given by

i(t)=I 0 cos ω t cos k z=R e I 0 cos k z e j ω t, (9.39a)

whose phasor is

I˜(z)=I 0 cos k z,−λ/4≤z≤λ/4, (9.39b)

and k=2 π/λ. Equation (9.9a) gives an expression for E˜θ, which is the far field radiated by a Hertzian dipole of length l when excited by a current I 0. Let us adapt that expression to an infinitesimal dipole segment of length d z, which is excited by a current I˜(z) and located at a distance s from the observation point Q (Fig. 9-14(b)). Thus,

d E˜θ(z)=j k η 0 4 π I˜(z)d z e−j k s s sin θ s, (9.40a)

and the associated magnetic field is

d H˜ϕ(z)=d E˜θ(z)η 0. (9.40b)

# Figure 9-14 Center-fed half-wave dipole.

Figure 9 14-a Full Alternative Text
Figure 9 14-b Full Alternative Text

The far field due to radiation by the entire antenna is obtained by integrating the fields from all of the Hertzian dipoles making up the antenna:

E˜θ=∫z=−λ/4 λ/4 d E˜θ.(9.41)

Before we calculate this integral, we make the following two approximations. The first relates to the magnitude part of the spherical propagation factor, 1/s. In Fig. 9-14(b), the distance s between the current element and the observation point Q is considered so large in comparison with the length of the dipole that the difference between s and R may be neglected in terms of its effect on 1/s. Hence, we may set 1/s≈1/R, and by the same argument, we set θ s≈θ. The error Δ between s and R is a maximum when the observation point is along the z axis and it is equal to λ/4 (corresponding to half of the antenna length). If R≫λ, this error will have an insignificant effect on 1/s. The second approximation is associated with the phase factor e−j k s. An error in distance Δ corresponds to an error in phase k Δ=(2 π/λ)(λ/4)=π/2. As a rule of thumb, a phase error greater than π/8 is considered unacceptable because it may lead to a significant error in the computed value of the field E˜θ. Hence, the approximation s≈R is too crude for the phase factor and cannot be used. A more tolerable option is to use the parallel-ray approximation given by

s≈R−z cos θ,(9.42)

as illustrated in Fig. 9-14(b).

Substituting Eq. (9.42) for s in the phase factor of Eq. (9.40a) and replacing s with R and θ s with θ elsewhere in the expression, we obtain

d E˜θ=j k η 0 4 π I˜(z)d z e−j k R R sin θ e j k z cos θ.(9.43)

After (1) inserting Eq. (9.43) into Eq. (9.41), (2) using the expression for I˜(z) given by Eq. (9.39b), and (3) carrying out the integration, the following expressions are obtained:

The corresponding time-average power density is

S(R,θ)=|E˜θ|2 2 η 0=15 I 0 2 π R 2 cos 2[(π/2) cos θ]sin 2 θ=S 0 cos 2[(π/2) cos θ]sin 2 θ(W/m 2).(9.45)

Examination of Eq. (9.45) reveals that S(R,θ) is maximum at θ=π/2, and its value is

S max=S 0=15 I 0 2 π R 2.

Hence, the normalized radiation intensity is

The radiation pattern of the half-wave dipole exhibits roughly the same doughnut-like shape shown earlier in Fig. 9-7 for the short dipole. Its directivity is slightly larger (1.64 compared with 1.5 for the short dipole), but its radiation resistance is 7 3 Ω (as shown later in Section 9-3.2), which is orders of magnitude larger than that of a short dipole.

# 9.3.1 Directivity of λ/2 Dipole

To evaluate both the directivity D and the radiation resistance R rad of the half-wave dipole, we first need to calculate the total radiated power P rad by applying Eq. (9.20):

P rad=R 2∬4 π S(R,θ)d Ω=15 I 0 2 π∫0 2 π∫0 π cos[(π/2) cos θ]sin θ 2 sin θ d θ d ϕ.(9.47)

The integration over ϕ is equal to 2 π, and numerical evaluation of the integration over θ gives the value 1.22. Consequently,

P rad=36.6 I 0 2(W).(9.48)

From Eq. (9.45), we found that S max=15 I 0 2/(π R 2). Using this in Eq. (9.24) gives the following result for the directivity D of the half-wave dipole:

D=4 π R 2 S max P rad=4 π R 2 36.6 I 0 2 15 I 0 2 π R 2=1.64(9.49)

or, equivalently, 2.15 dB.

# 9.3.2 Radiation Resistance of λ/2 Dipole

From Eq. (9.33a),

R rad=2 P rad I 0 2=2×36.6 I 0 2 I 0 2≈73 Ω.(9.50)

As was noted earlier in Example 9-3, because the radiation resistance of a Hertzian dipole is comparable in magnitude to that of its loss resistance R loss, its radiation efficiency ξ is rather small. For the 4-cm long dipole of Example 9-3, R rad=0.08 Ω (at 75 MHz) and R loss=0.036 Ω. If we keep the frequency the same and increase the length of the dipole to 2 m (λ=4 m at f=75 MHz), R rad becomes 73 Ω and R loss increases to 1.8 Ω. The radiation efficiency increases from 69% for the short dipole to 98% for the half-wave dipole. More significant is the fact that it is practically impossible to match a transmission line to an antenna with a resistance on the order of 0.1 Ω, while it is quite easy to do so when R rad=73 Ω.

Moreover, since R loss≪R rad for the half-wave dipole, R in≈R rad and Eq. (9.30) becomes

Z in≈R rad+j X in.(9.51)

Deriving an expression for X in for the half-wave dipole is fairly complicated and beyond the scope of this book. However, it is significant to note that X in is a strong function of l/λ and that it decreases from 42 Ω at l/λ=0.5 to zero at l/λ=0.48, whereas R rad remains approximately unchanged. Hence, by reducing the length of the half-wave dipole by 4%, Z in becomes purely real and equal to 73 Ω, thereby making it possible to match the dipole to a 75 Ω transmission line without resorting to the use of a matching network.

# 9.3.3 Quarter-Wave Monopole Antenna

This is because, from image theory (Section 4-11), the conducting plane can be replaced with the image of the λ/4 monopole, as illustrated in Fig. 9-15(b). Thus, the λ/4 monopole radiates an electric field identical to that given by Eq. (9.44a), and its normalized radiation intensity is given by Eq. (9.46); but the radiation is limited to the upper half-space defined by 0≤θ≤π/2. Hence, a monopole radiates only half as much power as the dipole. Consequently, for a λ/4 monopole, P rad=18.3 I 0 2, and its radiation resistance is R rad=36.5 Ω.

The approach used with the quarter-wave monopole is also valid for any vertical wire antenna placed above a conducting plane, including a Hertzian monopole.

# Figure 9-15 A quarter-wave monopole above a conducting plane is equivalent to a full half-wave dipole in free space.

Concept Question 9-8: What is the physical length of a half-wave dipole operating at (a) 1 MHz (in the AM broadcast band), (b) 100 MHz (FM broadcast band), and (c) 10 GHz (microwave band)?

Concept Question 9-9: How does the radiation pattern of a half-wave dipole compare with that of a Hertzian dipole? How do their directivities, radiation resistances, and radiation efficiencies compare?

Concept Question 9-10: How does the radiation efficiency of a quarter-wave monopole compare with that of a half-wave dipole, assuming that both are made of the same material and have the same cross section?