relationship between Antenna length and frequency

[Razaali]

i am confused about the relation between length of antenna and frequency.
As C= F λ, but i am not getting, how to relate this with antenna length specially with dipole antenna?
Kindly expalin?

 

[RCinFLA]

Full dipole is 1/2 wavelength minus a shortening factor due to extra stray capacitance of large relative rod diameter. Shortening factor is typically 90% to 98%.

For example 500 MHz would have a wavelength of 60 cm. Half wave would be 30 cm. 96% of 30 cm. would be 28.8 cm. Split in center and fed with balanced line of about 70 ohms. The larger the diameter of rods used, relative to wavelength, the more the shortening factor and lower the feed impedance.

 

[jiripolivka]

Antennas can be of any length related to wavelength. A good antenna must have an impedance matched to its feed line, and its radiation pattern is selected according to its use. Half-wave dipoles were found optimum in both aspects. In fact, 0.47 wavelength is the optimum dipole length.

 

[Ow@i$]

Let just take an example :

Required Frequency :10 MHz

c=fλ
c=300*10^6 m/s........ f=10*10^6 Hz.....
therefore, λ=c/f
λ= 30 m

if you want λ/4 antenna simply 30/4= 7.5 m (one Leg), 15m whole
for λ/2 divde 30 by 2 etc... depends upon requirement.
there is one more factor called Velocity factor .. which is also multiplied to that 30m.. for copper it is something around 0.98 or so you can include that also again depends upon you antenna wire material.

 

[Razaali]

Thanks to all of you
Now one more question
if i want to design an antenna, then it means first i have to look what frequency i am using, and then according to that frequency i have to select the length from the formula c=fλ ?
Is it so?

 

[jeeudr]

Thanks to all of you
Now one more question
if i want to design an antenna, then it means first i have to look what frequency i am using, and then according to that frequency i have to select the length from the formula c=fλ ?
Is it so?

Yes. This is generally the case.

However, you must also remember that for most cases, you can perform scaling. i.e. if you have a design for an antenna which works at 2 GHz, then you can predict that by doubling all the dimensions of the antenna, you can get a new antenna which works at 1 GHz (half the original frequency).

 

[Ow@i$]

yeah it is not so difficult as you can see in calculating the required length..

 

[aniakhan]

Let just take an example :

Required Frequency :10 MHz

c=fλ
c=300*10^6 m/s........ f=10*10^6 Hz.....
therefore, λ=c/f
λ= 30 m

if you want λ/4 antenna simply 30/4= 7.5 m (one Leg), 15m whole
for λ/2 divde 30 by 2 etc... depends upon requirement.
there is one more factor called Velocity factor .. which is also multiplied to that 30m.. for copper it is something around 0.98 or so you can include that also again depends upon you antenna wire material.

can anybody help me to undersatnd resonant behaviour of antenna.if i have a full wavelength dipole antenna lets say at 200 ghz then its length will be 1.5 mm.now this antenna will only radiate best at 200 GHz and its S parameters will only narrow band width resonating at 200 Ghz????am i right?

 

[tony_lth]

200GHz dipole antenna, you must simulate and include the rod diameter. etc.
200GHz I guess should use dish antenna or lens antenna.

 

[jiripolivka]

At 200 GHz you cannot manufacture a good "resonant" dipole. More often, butterfly dipoles are used, with a wider bandwidth.

As I wrote above, any length of a conductor can radiate any wave length. The 0.47 wavelength, the half- wave dipole, has an optimum match to 240...300 Ohm impedance, and a nice doughnut-shaped radiAtion pattern. Most people like this. But there are also full-wave dipoles , shorter-than- half-wave dipoles, etc. Each case needs a special matching to line impedance, and offers a specific radiation pattern.