First example Suppose a lossless antenna has a radiation pattern given by: U = B_0\,\sin^3(\theta). Let us find the gain of such an antenna. First we find the peak radiation intensity of this antenna: U_\text{max} = B_0. The total radiated power can be found by integrating over all directions: \begin{align} P_\text{rad} &= \int_0^{2\pi}\int_0^\pi U(\theta, \phi)\sin(\theta)\, d\theta\, d\phi = 2\pi B_0 \int_0^\pi \sin^4(\theta)\, d\theta = B_0\left(\frac{3}{4}\pi^2\right) \\ D &= 4\pi\left(\frac{U_\text{max}}{P_\text{rad}}\right) = 4\pi\left[\frac{B_0}{B_0\left(\frac{3}{4}\pi^2\right)}\right] = \frac{16}{3\pi} \approx 1.698 \end{align} Since the antenna is specified as being lossless the radiation efficiency is 1. The maximum gain is then equal to: \begin{align} G &= \eta D \approx (1)(1.698) = 1.698 \\ G_\text{dBi} &\approx 10\, \log_{10}(1.698) \approx 2.30\,\text{dBi} \end{align} Expressed relative to the gain of a half-wave dipole we would find: G_\text{dBd} = 10\, \log_{10}\left(\frac{1.698}{1.64}\right) = 0.15\,\text{dBd}.
Second example As an example, consider an antenna that radiates an electromagnetic wave whose electrical field has an amplitude E_\theta at a distance . That amplitude is given by: E_\theta = {A I \over r} where: • I is the current (in Amps) fed to the antenna, and • A is a characteristic constant (in Ohms) of each antenna, for a large distance . The radiated wave can be considered locally as a
plane wave. The intensity of an electromagnetic plane wave is: {P\over S} = {c_\circ \varepsilon_\circ \over 2}{E_\theta}^2 = {1 \over 2} { {E_\theta}^2 \over Z_\circ } where • Z_\circ = \sqrt{ {\mu_\circ \over \varepsilon_\circ}} = 376.730313461\,\mathsf{\Omega} is a universal constant called
vacuum impedance; and • {{tmath|1= \textstyle \left({P\over S}\right)_\mathsf{ant} = {1 \over 2 Z_\circ} {A^2 I^2 \over r^2} }}. If the resistive part of the series impedance of the antenna is {{tmath|{R_s} }}, the power fed to the antenna is {{tmath|\textstyle {1\over 2} {R_s I^2} }}. The intensity of an isotropic antenna is the power so fed divided by the surface of the sphere of radius : \left({P \over S}\right)_\mathsf{iso} = {{1\over 2}R_sI^2 \over 4\pi r^2 } = \frac{R_sI^2}{8\pi r^2}. The directive gain is: G = {\left[ {1 \over 2 Z_\circ} {A^2 I^2 \over r^2} \right] \over {\left[ \frac{R_s I^2}{8\pi r^2} \right] } } \approx {A^2 \over 30 R_s }. For the commonly utilized
half-wave dipole, the particular formulation works out to the following, including its
decibel equivalency, expressed as
dBi (decibels referenced to isotropic radiator): \begin{align} R_{\frac{\lambda}{2}} &= 60 \operatorname{Cin}(2\pi) = 60 \left[ \ln(2\pi) + \gamma - \operatorname{Ci}(2\pi) \right] = 120 \int_{0}^{\frac{\pi}{2}} \frac{\cos\left( \frac{\pi}{2}\cos\theta \right)^2 }{ \sin\theta}d\theta\ ,\\ &= 15 \left[ 2\pi^2 - \frac{1}{3} \pi^4 + \frac{4}{135}\pi^6 - \frac{1}{630}\pi^8 + \frac{4}{70875}\pi^{10} + \ldots + (-1)^{n+1}\frac{(2\pi)^{2n}}{n(2n)!}\right]\ ,\\ &= 73.12960 \ldots \mathsf{\; \Omega; } \end{align} In most cases {{nowrap|1=R_{\frac{\lambda}{2}} =}}
73.130 is adequate. \begin{align} G_{\frac{\lambda}{2}} &= \frac{60^2}{30R_{\frac{\lambda}{2}}} = \frac{3600}{30R_{\frac{\lambda}{2}}} = \frac{120}{R_{\frac{\lambda}{2}}} = \frac{1}{{}^{\int_{0}^{\frac{\pi}{2}} \frac{ \cos \left( \frac{\pi}{2} \cos \theta \right)^2 }{ \sin \theta} d \theta}}\ , \\ &\approx \frac{120}{73.1296} \approx 1.6409224 \approx 2.15088\ \mathsf{\ dBi} ; \end{align} Likewise, {{nowrap|1=G_{\frac{\lambda}{2}} =}}
1.64, or '''''', are usually the cited values. Sometimes, the half-wave dipole is taken as a reference instead of the isotropic radiator. The gain is then given in '' (decibels over dipole): 0\ \mathrm{dBd} = 2.15\ \mathrm{dBi}. == Realized gain ==