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nitin singh Grade: 12
        

Can anyone help me understand why, apparently the physical length of a monopole end feed wire for a crystal set need to be ½ the wave length of the desired signal.

The reason I ask is, that I understood that, if you have a variable coil in series with the antenna, the resonate frequency is the combination of the coil and antenna impedance, so you can compensate for any antenna length within the range of the variable coil.

So what is the problem of a 3 meter antenna with the correct compensating coil?

6 years ago

Answers : (1)

SAGAR SINGH - IIT DELHI
879 Points
										

Dear nitin,


Yes, you could match the antenna ompedence for a desired frequency more or less regardless of the length (Steven Best does a lot of work with small antennas). However, that does not give any indication to efficiency. A matched antenna network matches the complex impedance with that of the antenna. This maximizes the power transferred. But the real part of the impedance is the radiation and conduction losses. Ideally, we would have a very high radiation resistance which would indicate that the power is being radiated. However, if we have a low radiation resistance, then what happens is that the input power gets trapped in the capacitive and inductive properties of the antenna (the near-field and any lumped matching elements you may have added). It appears that the network is matched but most of the energy is unusable.

A proper resonant antenna tries to maximize the transfer of the inputted energy into propagating waves and this is done by controlling the length of the antenna. You can radiate at non-resonant frequencies, but it will not be as efficient.

Also, I believe that you want the antenna to be 1/4 the wavelength. With a monopole antenna, it will approximate a 1/2 wavelength dipole antenna. A 1/2 wavelength monopole is a 1 wavelength dipole. This is a resonant mode but it is not very efficient because the current elements that are 180 degrees out of phase will cancel each other out in the far field (some people solve this by using a folded dipole but I believe that you also have problems with the input impedance regardless).




We are all IITians and here to help you in your IIT JEE preparation.

All the best.


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Sagar Singh


B.Tech IIT Delhi


sagarsingh24.iitd@gmail.com



6 years ago
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