0
\$\begingroup\$

let's consider a simple dipole antenna of length L. The following picture refers to a half-wave dipole: enter image description here

In general, at different frequencies, it's known that different modes are possible. Precisely, the open circuit at the end of the antenna forces the current to be zero at the antenna end points, which become null point of I(x) (current spatial wave).

Therefore, as shown in the following picture there may be:

  • a half current wavelength along the dipole
  • a whole current wavelength along the dipole
  • 1.5 wavelength along the dipole

etc

enter image description here

Obviously, it's a discrete set of wavelengths that follow the equation: $$L=n\cdot\frac{\lambda}{2}$$

that means a discrete set of frequencies that follow the equation: $$f=n\cdot\frac{c}{2L}$$

with c = speed of light.

Therefore, only the following frequencies are possible: $$f=\frac{c}{2L},2\cdot\frac{c}{2L},3\cdot\frac{c}{2L},...$$

Well, what happens if the voltage source of the antenna (i.e. its signal supply) has a different frequency (for instance $$1.5\cdot\frac{c}{2L}$$)?

Such a voltage source forces the current (and voltage) frequency to be such intermediate frequency, but in theory the null points ad the end of the antenna can't allow that frequency to exist...

Which is the solution?

Pictures taken from here and here.

\$\endgroup\$

2 Answers 2

0
\$\begingroup\$

When you don't match the antenna length to the frequency you are trying to radiate or receive (it happens a lot of times) you get a more complicated impedance presented to the electrical terminals of the antenna: -

enter image description here

It's the same for transmission lines; if the end of the t-line isn't matched in the characteristic impedance you get internal reflections that can make the feed-end of the t-line appear to have a higher or lower impedance depending on how long the t-line is in wavelengths.

Antennas still can work with a highly mismatched length but the resistive part of the impedance (that part of the impedance that represents the real power of the signal being transmitted into space) can be anywhere between a fraction of an ohm and several kohms, in series with a large reactive impedance. Crystal radio antennas are good examples of this because you will never likely make a monopole of a quarter wavelength because it'd be physically too long so, the reactance part plays a big role in electrical tuning when used with a coil.

\$\endgroup\$
5
  • \$\begingroup\$ Thank you for your answer. Graphically, what would be the wave profile of current along the dipole length in case of a mismatch (for instance, a little mismatch)? Since the dipole allows only those discrete wavelengths, I can't imagine graphically how the voltage source can force a different wavelength. \$\endgroup\$
    – Kinka-Byo
    Commented Nov 21, 2020 at 9:59
  • 1
    \$\begingroup\$ It's exactly as defined by transmission line mismatches. If the current and voltage at the end of the t-line (or antenna) doesn't match the load then there is a reflection from the end of the t-line (or antenna) that goes back to the source terminals and, with sinewaves applied this results in a modification of impedance presented to the source. So, with an antenna, there is no load and all the current at the end of the antenna gets internally reflected back to the source. \$\endgroup\$
    – Andy aka
    Commented Nov 21, 2020 at 10:02
  • \$\begingroup\$ Nice, now it's clear! Thank you very much \$\endgroup\$
    – Kinka-Byo
    Commented Nov 21, 2020 at 10:06
  • \$\begingroup\$ Just another question. In such a case of mismatch, is it not more correct to say that, for the resulting standing wave, lambda × f = c? I think this equation is true only for single reflected and direct waves \$\endgroup\$
    – Kinka-Byo
    Commented Nov 21, 2020 at 11:19
  • 1
    \$\begingroup\$ I can't remember the formula for standing wave distances but, for sure, there will be a partial standing wave internal to the antenna. \$\endgroup\$
    – Andy aka
    Commented Nov 21, 2020 at 11:44
0
\$\begingroup\$

well, I don't see why the dipole wouldn't let other frequencies to exists, the sistem is just an forced oscillatory one, the voltage source force a frequency into the antenna, sure is not an efficient thing but it totally can exist, any frequency not only the harmonic ones, since any current waveform can be expressed as a fourier series of the harmonics, and so it will hold the boundary condition of being 0 at the end points.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.