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Typically, the desired frequency of operation is known, and the purpose of the
calculation is to identify the best size for the dipole antenna. The most often
used formula to find that length is
Length = 468/frequency, MHz feet, or 142.6/frequency, MHz meters.
That gets you close, but some tweaking and trimming is usually required after
construction, because some of the characteristics of the installation, such as
the antenna's height above ground and method of supporting its ends, affect
the antenna's properties.

The wavelength corresponding to that frequency is
Wavelength = 300/frequency, MHz meters, or 984/frequency, MHz feet.

The way your question is written, it appears that you already have the antenna, and
now you want to know the frequency at which it will operate best. I must tell you
how peculiar this sounds to anybody in the radio business. But I'm here to answer
questions if they make sense, not to judge how impractical or useless they may
actually turn out to be. So here goes:

For the first time ever, you've just noticed a horizontal wire hanging between two
trees in your back yard. Rather than simply tearing it down, you're curious to know
the frequency/wavelength at which it could most efficiently transmit/receive when
operated as a half-wave dipole. So you measure its length somehow, and call the
length ' L ' .

-- If you measured ' L ' in feet, then the best frequency to try first is 468/L MHz.
The corresponding wavelength is 2L feet, or 0.6096L meters.

-- If you measured ' L ' in meters, then the best frequency to try first is 142.6/L MHz.
The corresponding wavelength is 6.562L feet, or 2L meters.

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βˆ™ 11y ago
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βˆ™ 13y ago

A: At a particular frequency it could be zero.

but the impedance is nearly 2000

Different answer: I built a 1/2 omni antenna for 20 meters (14.3mhz). To keep the impedance 50 ohms, I used the standard trick for a half wave, I angled the ground radials(3) to 120 degrees. It worked well. I then tested it at other bands to see how well it performed with the antenna tuner. For a very narrow little spot on the 10 meter band it was usable without the tuner. I considered that part of the band as too dangerous to be tuned practically, the slightest change, due to wind, on the radial angle would damage my equipment .

We usually only consider using dipole antennas that are an odd multiple of wavelengths. Full wave dipoles ar referred to as folded antennas. 180 degrees=300 ohms=6:1 baulin. (Aligning the radio for a different impedance is a good old trick.)

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βˆ™ 14y ago

936/Frequency = Length (Full wave)

468/Frequency = Length (Half wave)

234/Frequency = Length (Quarter wave)

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Q: What is the impedance of a full wave dipole?
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Related questions

How loop antenna and dipole antenna can be used interchangeably?

A full-wave loop antenna can be interchanged with a folded dipole without much difference. The input impedance is similar and the only difference is in the directivity: a full wave loop radiates along the axis of the loop, while a vertical folded dipole is omnidirectional.


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There are several kinds of dipole; the most common is the half-wave dipole. Its total length is fairly close to half the wavelength of the design frequency. The length needs to be adjusted slightly to compensate for the thickness of the elements and for end-effects. If the length is wrong by ten or twenty percent it will alter the feed impedance, but have little effect on the gain.


What is the radiation resistance of a half-wave folded dipole?

The radiation resistance of a half-wave folded dipole is typically around 300 ohms. This value is higher compared to a regular half-wave dipole due to the increased current distribution along the folded elements.


Why was before used the 300 ohm transmission line?

The impedance of a centre fed dipole at resonance is about 73 ohms. The feed impedance of a folded dipole is four times that, 292 ohms. The feedpoint is of course balanced in both cases, thus 300 ohm balanced line is an excellent match for the folded dipole active element of a commercial TV antenna.


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What has the author Walter K Kahn written?

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Why do you expect an infinity line to have an input impedance equal to the caractarictic impedance?

The characteristic impedance of a transmission line is the ratio of voltage to current of the propagating electrical wave. The line input impedance is the result of the superposition of forward and reverse, or reflected waves when the terminating impedance is not adapted. If the line is infinite, nothing returns from its end and only the forward wave exits. The voltage to current ratio is then the line characteristic impedance. Remark that the same occurs when the line is terminated by its characteristic impedance, the forward wave finds a perfect continuity to the load and no energy is reflected back to the line. A matched line is like an infinite line when looked from the input terminals. Long real lossy lines also act as infinite lines for the energy of the reflected wave is dissipated along the line before reaching the source.


When a generator of internal impedance and operating at 1gigahertz feeds a load via a coaxial line of characteristic impedance 50 ohm then the voltage wave ratio on the feed line is?

For a voltage standing wave ratio (VSWR) of 1.0, the source impedance, load impedance, and transmission line characteristic impedance must be matched. To calculate actual VSWR, you need to know these three values. You're question only supplies one (50 ohm line). Review wikipedia's writeup on "standing wave ratio" to glean an understanding of what you're asking about.


What has the author L E Vogler written?

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What is the different between half wave and full wave?

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