What is electrical length in detail? Why is it so that 1/4 wave , 1/2 wave dipole antenna works properly and not a full wave antenna? What is the role of electrical lengths in degree?


2 Answers 2


The concept of "electrical length" begins with the concept of wavelength, $\lambda$ - the distance over which the value of a periodic phenomenon repeats, expressed as: $$\lambda=\frac{c}{f}$$ When describing electromagnetic wave propagation in free space, c is the speed of light in a vacuum - 300 million meters per second - and $f$ is the frequency of the signal creating the wave, in Hertz.

Electromagnetic waves propagate more slowly in materials or at boundaries. The velocity factor, $\nu$, represents this slowing and is less than one (except in free space, where it is exactly one). This modifies the wavelength equation:$$\lambda=\frac{\nu c}{f}$$ For example, the wave propagating along the conductor of a dipole antenna to produce radio waves encounters the interface between the conductor and free space, slowing the wave by about 5%; i.e., $\nu$~0.95. This is why a resonant half-wave dipole is shorter than $c/f$ would calculate.

In a transmission line, $\nu$ depends on the dielectric material between the conductors. Many coaxial cables have $\nu$=0.66, so the wavelength at a given frequency is about 1/3 less than it would be in free space. The wave propagating in a line that is one wavelength long experiences one full cycle of change, equivalent to 360 degrees. A shorter line produces proportionately fewer cycles of change, longer lines proportionately more, which can be expressed as fractions or multiples of 360 degrees.

Quarter-wave, half-wave and full-wave dipoles all work "properly," but it can be challenging to match the feedpoint impedance of the antenna to the impedance of the transmission line, which is necessary to maximize power transfer from the transmitter to the antenna. Circuits using lumped-elements, transmission lines of various lengths and impedances, or combinations of lumped and distributed elements may be employed to optimize system performance.

  • $\begingroup$ Hi Brian, it might be good to incorporate feedline velocity factor into your answer. Actually, though, he is kind of asking two separate questions here. $\endgroup$ Oct 26, 2018 at 20:59
  • $\begingroup$ Hi, Mike - I thought the fourth paragraph provided an adequate introduction to the concept, given the presentation of separate but related questions. $\endgroup$
    – Brian K1LI
    Oct 26, 2018 at 21:48
  • 1
    $\begingroup$ There's nothing wrong with a full wave antenna. It's called a Double Zepp and has a long history of success. Because of its very high feedpoint resistance, it needs a 1/4WL open-wire series section transformer to transform the high resistance to a low resistance suitable for a coax feedline. A Double Zepp has about 2 dB of gain over a 1/2WL dipole in their favored direction. $\endgroup$ Nov 12, 2018 at 14:55
  • $\begingroup$ I've seen Nu used for wavenumber and frequency, but never velocity factor. $\endgroup$ Nov 12, 2018 at 23:39

The relationship between wavelength and frequency of an RF signal is the speed of light. One wavelength equals the speed of light divided by the frequency. If the medium in which the light (or radio wave) is traveling is something other than free space, the velocity factor of the medium can slow down the speed of light. Transmission lines have a velocity factor ranging from about 0.66 to about 0.95, i.e. transmission lines slow down the speed of light. One wavelength is one cycle of RF and is thus 360 degrees. We often need a certain number of degrees of transmission line at a certain frequency to accomplish a certain task, like matching an antenna to a transmitter.

An example might help. I use a Smith Chart to determine that I need a length of 146 degrees of 300 ohm twinlead to match my 39 foot dipole on 20m. The Velocity Factor (VF) of my 300 ohm twinlead is 0.83. I calculate one wavelength of that twinlead on 20m as VF*c/f = 0.83*984/14.2 = 57.5 feet. Since one wavelength equals 360 degrees, the length I need is 57.5(146/360) = 23.32 feet. This length will satisfy my goal of a 50 ohm SWR less than 2:1. I can add any reasonable Number of 1/2 wavelengths (N*57.5/2) and still have a 50 ohm SWR of less than 2:1. In the above equation, c is the speed of light in free space and f is the target frequency in MHz.


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