Let's say there's a transmitter far away, and I'm going to receive the transmitted signal. To keep things simple let's further say the transmitted signal is a pure, unmodulated sine wave. Further, let's say the electric field generated by that distant transmitter, at the point where my receive antenna is to be installed, has an intensity of 1 mV/m.

Assuming that I can now install an ideal, lossless dipole, how much power would be received by this dipole? How does the received power vary with length, ie a half-wave dipole versus a shorter dipole?


1 Answer 1


There are a lot of ways to approach this problem, but here's one: we can calculate the power density of that field, and determine the area from which the antenna captures power, and multiply them together.

Calculating the power density

If we know the electric field to be 1 mV/m, and the transmitting antenna is distant, then the magnetic field must be related by the impedance of free space: $376.7\:\Omega$. So, the magnetic field must be:

$$ {1\:\mathrm{mV} \over \mathrm m } \cdot {1 \over 376.7\:\Omega} = {2.655\:\mathrm{\mu A} \over \mathrm m } $$

Multiplying these two together, we can calculate the power density:

$$ {1\:\mathrm{mV} \over \mathrm m } \cdot {2.655\:\mathrm{\mu A} \over \mathrm m } = {2.655\:\mathrm{nW} \over \mathrm m^2 } $$

In other words, there is 2.655 nW passing through an area of a square meter perpendicular to the transmitting antenna. Poynting's theorem is the formal basis for this calculation, but since we know we are dealing with a radiated field (in the far-field of the transmitting antenna), we know the electric and magnetic fields are in phase and so we can simplify the calculations substantially.

Calculating the effective aperture

Now that we know the power density, we just need to know the area from which the dipole will collect power. We'll have to approach this part of the problem backwards.

We know that the free-space gain of a dipole is 2.15 dBi or 1.64. The area from which an antenna collects energy is called effective aperture ($A_\mathrm{eff}$), and it has a simple relation to gain (G):

$$ A_\mathrm{eff} = {G \lambda^2 \over 4 \pi} $$

So if we set $G = 1.64$ we get the effective aperture of a free-space half-wave dipole:

$$ {1.64 \lambda^2 \over 4 \pi } = 0.1305\:\mathrm m^2 $$

So, a dipole half a meter long ($\lambda = 1\:\mathrm m$) collects power from an area of $0.1305\:\mathrm{m^2}$. With decreasing frequency the half-wave dipole gets bigger, and consequently collects more power.

So to know the power we'd see at the antenna terminals, we just multiply these two numbers together:

$$ {2.655\:\mathrm{nW} \over \mathrm m^2 } \cdot 0.1305\:\mathrm{m^2} = 346.4\:\mathrm{pW} = -64.60\:\mathrm{dBm} $$

At this point, you may wonder why we use the free-space dipole gain. Don't dipoles over ground have more gain? And the reason is simple: the question specified that the electric field strength at the point where the receiving antenna was to be installed. Any constructive interference due to reflections from the ground would have already been taken into account by this measurement.

Effective aperture of a shorter dipole

Amazingly, it's almost the same as a half-wave dipole.

If you begin with an ideal half-wave dipole and then keep shortening it, the thing you are left with in the limiting case where the dipole is infinitesimally short is called a Hertzian dipole. Amazingly, this infinitesimally short dipole has a free-space gain of 1.5 (where the half-wave dipole was 1.64). Thus, it collects power from a slightly smaller area, but not smaller by much.

How can this be, when we know practically that really small antennas don't work well? Well, as the antenna becomes smaller, the radiation resistance approaches zero. Thus, it becomes increasingly difficult to make an efficient antenna with real materials since the resistance of the wire used to make the antenna becomes much larger relative to the radiation resistance.

Putting it all together

With a bit of algebra we can combine all of this into one equation:

$$ P_\mathrm{ant} = { E_E^2 \ G \lambda^2 \over 4\pi Z_0 } \approx { E_E^2 \ G \lambda^2 \over 4734 } $$


  • $P_\mathrm{ant}$ is the power at the antenna terminals,
  • $E_E$ is the electric field intensity ($\mathrm{V/m}$),
  • $G$ is the gain of the antenna (1.64 for a half-wave dipole),
  • $\lambda$ is the wavelength, and
  • $Z_0$ is the impedance of free space ($376.7\:\Omega$).
  • $\begingroup$ Good signal at $-64.60\,dBm$. That is almost 10/9 on the S-Meter. $\endgroup$
    – K7PEH
    May 27, 2015 at 14:56

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