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Length of a dipole antenna according to antenna theory should be lambda/2 for best reception. I am just curious about the outcome when length of dipole antenna >> lambda. Impedance will be zero in such a case and there will be no signal received. Is it correct?

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  • $\begingroup$ Depends: when you transmit a signal you can tune your antenna the best you can to work in a selected (range of) frequency(ies), but you will anyway have to deal with harmonics too. Basically it can happen that you could be still able to receive a signal "belonging" to the lambda/2 frequency with a decent signal to noise ratio... $\endgroup$ Commented Jul 16, 2015 at 6:18

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The idea of setting the length of the dipole equal to $\lambda/2$ is to make the aerial resonant, i.e. it's the correct length to get a standing wave and this increases the voltage induced in the aerial and hence the sensitivity.

On the other hand the radio wave will induce some voltage in any old piece of wire. It just won't be as sensitive as a tuned antenna (though making the wire long will increase the amount of the RF wave it intersects and may well compensate for the lack of tuning).

For example I use an old TV aerial for my FM tuner (because I can't be bothered to climb up to the roof and install a proper FM aerial), and although it's tuned for a completely different wavelength I get pretty good FM reception. Similarly I have a portable FM radio with a telescopic aerial, and the reception isn't greatly affected by changing the length of the telescopic aerial.

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  • $\begingroup$ The reason a resonant antenna is more sensitive is because that is where the receiver's impedance match is typically designed to operate. The match can be changed to work in at non-resonant frequency of the antenna and the efficiency of the antenna will be identical. (assuming the radiation resistance and conductive losses are the same) $\endgroup$
    – user6972
    Commented Jun 25, 2014 at 18:51
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Any time the physical size of a structure ($L$) is close to the length of the wavelength ($\lambda$) then the fields must be considered as part of the analysis. The rule of thumb is when $L>\frac{\lambda}{10}$.

This includes the case you mention where $L>\lambda$. The choice of $L=\frac{\lambda}{2}$ is related to matching impedances more than anything. Antennas have similar parameters at multiples of $\lambda$ too. So a longer antenna can work the same way. There does get to be a trade off as there is more loss along a longer antenna, but sometimes this is negated by being able to get more of the antennas area into an area with a stronger field (like higher into the sky).

For example notice how the impedance of an antenna repeats at multiples (e.g. 0.5 and 1.5)

z vs length

The idea at $\lambda/2$ is the reactive portion of the antenna falls to zero making the transfer of maximum power easier. This is often termed a "resonant antenna" because it consists of only real impedance at the operating frequency. (Note Impedance does not fall to zero, only the reactive portion falls to zero). This also occurs at $1.5\lambda$, $2.5\lambda$, etc.

But longer antennas work well too even though they might not be as efficient, they can offer other benefits like reaching into areas with stronger signals as I mentioned. See the Random Wire Antenna as an example.

The two major disadvantages are:

  • Due to the length the impedance will most likely cause a mismatch in the receiver causing loss of efficiency
  • Longer length also will tend to pick up more background noise
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  • $\begingroup$ Thanks a lot...as I get it EM wave energy is transfer is maximum at \lambda/2 and the transfer decreases if length <> lambda/2. Is it related to the fact that magnetic component of EM wave is best utilized when length = lambda/2? $\endgroup$
    – user43794
    Commented Jun 25, 2014 at 10:48
  • $\begingroup$ @user43794 No. The impedance is only an issue between the antenna and the receiver. This is where the efficiency loss occurs, not in the interface of the EM wave to the Antenna. (Also as they get longer they lose some signal power due to resistance.) $\endgroup$
    – user6972
    Commented Jun 25, 2014 at 18:40

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