In July 1991, I penned an article with the same title as this month’s column. It was inspired by a call from friend and colleague Ralph Haimowitz, who commented that he had seen a recent paper discussing, among other things, the conversion between microvolts (µV) and microvolts per meter (µV/m). The paper used a somewhat different multiplier than the familiar "0.021" that’s in the formulas we still use today. Ralph’s concern was the effect the multiplier used in the paper would have on leakage measurement calculations performed by cable operators. So I sat down with a variety of references and did some number crunching related to the math of field strength measurements. Grab a cup of hot coffee and join me as we revisit where "0.021" comes from and why it is indeed "0.021." Background When we measure signal leakage, we’re dealing with a source of RF power (the leak), its propagation from the source to a location where the measurement is made, and determination of the RF signal’s field strength at the point of measurement. The receive antenna is normally a half-wave dipole that is resonant at the frequency of interest. In fact, FCC rules require that signal leakage measurements be made with a dipole, or at least correlated to a dipole measurement.

Most discussions of RF propagation and field strength measurement reference an isotropic radiator or antenna. This is a theoretical and ideal point source that radiates equally in all directions and has unity gain. Even though an isotropic antenna does not actually exist, it’s useful as a mathematical reference against which real-world comparisons can be made.

Imagine an isotropic antenna in free space, radiating a certain amount of RF power (Pt) uniformly in all directions. Now imagine a sphere whose surface is some distance from the point of radiation, with that point (the isotropic antenna) at the center of the sphere. A good analogy here is a very tiny but bright light bulb – an "isotropic radiator" – in the center of a large balloon, with the bulb illuminating the balloon from the inside. You could define the amount of power in a given area at the surface of the sphere (or balloon) – say, in a square meter – as so many watts per unit of area. Indeed, that’s exactly how RF power density is characterized. Math In this example, the power density Pd (in watts per square meter) at the surface of the sphere is simply the total transmitted power Pt divided by the surface area of the sphere, 4πR2 (Eq. 1):

Pd = Pt/4πR2

The level of the radiated RF energy also can be expressed in volts per meter (V/m). If the field strength on our imaginary sphere has an intensity of E in V/m, then the power density is (Eq. 2):

Pd = E2/120π

You’ll notice that Equation 2 is an expression of the basic power equation with Ohm’s Law equivalents substituted for E and I, or P = E2/R. In this case, 120π is the "resistance," or impedance of free space (approximately 377 ohms). In order for power density to be useful, though, it has to be converted to received power. To do that, multiply power density in watts per square meter by the area in square meters of the receiving antenna. Keep in mind there is little relationship between the actual size of an antenna such as a dipole and its effective area. If you know the numerical gain (g) of an antenna, you can calculate its effective area (Ae) with the formula (Eq. 3):

Ae = gλ2/4π

For a half-wave dipole, the effective area is (Eq. 4):

Ae = 1.64λ2/4π

Note that in Equation 4, 1.64 is the numerical gain of a half-wave dipole in free space. Its gain in decibels relative to an isotropic antenna is 10log(1.64) = 2.15 dBi.

As mentioned previously, the power intercepted by a receiving antenna is found by multiplying the antenna’s effective area by the power density at the point of measurement (Eq. 5):

Pr = AePd

Substituting Equations 2 and 4 for Pd and Ae in Equation 5 (Eq. 6):

Pr = 1.64λ2E2/480π2

We can now calculate the received voltage at the terminals of a half-wave dipole antenna using the power equation variation of Ohm’s Law (Eq. 7):

Er = sqrt(PrZ)

where Pr is the received power (from Equation 6) and Z is the impedance of a half-wave dipole in free space. The latter is Z = 73.13 + j42.5 ohms for an infinitely thin dipole in free space, but we’ll assume no reactance and call the dipole’s impedance approximately 73.2 ohms. Substituting Equation 6 for Pr and 73.2 for Z, Equation 7 becomes (Eq. 8):

Er = sqrt[(1.64λ2E2/480π2)73.2]

The free space wavelength of an RF signal can be found with the formula (Eq. 9):

λ = c/fHz

where c is the speed of light (299,792,458 meters per second) and fHz is the frequency in hertz. When dealing with megahertz, the formula is (Eq. 10):

λ = 299.792458/fMHz

Substituting Equation 10 for λ in Equation 8, a little algebra will reduce Equation 8 to a simpler form (Eq. 11):

Er = 47.72299333E/fMHz

With Equation 11 we can calculate the received voltage (in volts) from a half-wave dipole in free space when we know the field strength in V/m (E) and the frequency in MHz (fMHz). If you want to change units so that the received voltage is in µV and the field strength is in µV/m, the same equation is used. Just make sure you use the same units; don’t mix units – for example volts for Er and µV/m for E.

To convert a received voltage in µV to dBmV, use (Eq. 12):

dBmV = 20log(Er/1,000)

This can be used to modify Equation 11 (Eq. 13):

Er = 20log(0.04772299333E/fMHz)

Er is the received voltage in µV from the dipole, and E is the field strength in µV/m. You can determine the field strength in µV/m if you know the frequency in MHz and the dipole’s received signal level in dBmV (Eq. 14):

µV/m = 20.9543 x fMHz x 10(dBmV/20)

Here 20.9543 is the reciprocal of 0.04772299333 from Equation 13. If you divide by 1,000, you’ll have the familiar 0.021 (actually 0.0209543) multiplication factor used in the more common conversion from µV to µV/m (Eq. 15):

µV/m = µV x 0.021 x fMHz

dBmV can be converted to microvolts with the formula (Eq. 16):

µV = 10(dBmV/20) x 1,000

Thus, when you know the received signal level in dBmV, the conversion to µV/m becomes a slight variation of Equation 14 (Eq. 17):

µV/m = 10(dBmV/20) x 1,000 x 0.021 x fMHz

Now let’s put this to work. Assume that you’ve measured a leak on Ch. 14 (121.2625 MHz) with a half-wave dipole held 10 feet from the plant, and the leak’s level at the dipole terminals is -44.6 dBmV. What’s the field strength in µV/m? Use Equation 17:

µV/m = 10(-44.6/20) x 1,000 x 0.021 x 121.2625
= 10-2.23 x 1,000 x 0.021 x 121.2625
= 0.01 x 1,000 x 0.021 x 121.2625
= 14.99 µV/m

The leak’s field strength is about 15 µV/m, which is below the FCC’s 20 µV/m limit. Technically it’s legal, but good engineering practice says if you hear a leak, fix it. After all, where there’s leakage, there’s most likely ingress, too.

Ron Hranac is a technical leader, Broadband Network Engineering, for Cisco Systems and senior technology editor for Communications Technology. Reach him at rhranac@aol.com.

The Daily

Subscribe

Editor’s Note

Don’t forget that the next deadline to enter the CFX Program Awards is tonight! Head to

Read the Full Issue
The Skinny is delivered on Tuesday and focuses on the cable profession. You'll stay in the know on the headlines, topics and special issues you value most. Sign Up