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I have found it across many texts but I am confused about the reason why is it so? Is this related to radiation resistance?

Souhardya Mondal
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An electromagnetic field is produced by an accelerating charge.

For a given fixed charge and acceleration, you can increase the radiated field by accelerating that charge over a longer distance. A longer wire might also allow you to accelerate more electrons for a given fixed current into the wire (before the current reflects back off the end).

So radiation resistance is roughly proportional to (d / lambda)^2 . For a given current, this is proportional to amount radiated away from the system as EM energy (rather than heat or mechanical energy.)

An antenna much shorter than a half wavelength just doesn't have a lot of distance over which a given charge can accelerate during a half cycle of an oscillating drive source. So is less efficient.

But once the acceleration distance approaches a full wavelength (or longer), due to the speed of light and propagation time of the accelerating charge, parts of an antenna can end up radiating EM waves of the opposite phase from other parts of the antenna. These opposite-phase waves can then interfere and cancel each other out at certain angles (such as broadside in the far field). Which is inefficient in those directions.

hotpaw2
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It isn't just said; it is. Or at least antennas get increasingly inefficient as they get smaller than a wavelength.

An antenna is, in part, an impedance matching device from its feedline to free space. A perfect antenna would look like a resistor to its feedline, and all of the energy going into that radiation resistance would couple to free space, with no waste.

The shorter an antenna is from around 1/2 a wavelength, the more that it looks like a capacitor (if its open on the ends) or an inductor (if its closed on the ends). That means that a physical impedance matching network is required at the connection of feedline to antenna, or that the antenna has features (such as capacity hats or loading coils) that bring the antennas radiation resistance into line with the feedline impedance.

All of those measures reduce the bandwidth of the antenna, and tend to cause losses in the components (i.e., high currents in the coils, which causes \$I^2R\$ losses).

Basically, if the antenna is less than half a wavelength long (or 1/4 a wavelength if it's working against a ground plane) then it's less efficient than it could otherwise be.

TimWescott
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  • So those CB antennas that used to be on cars should have been quite a bit longer, except for the obvious disadvantages... – Solar Mike Sep 03 '19 at 20:28
  • Solar Mike: Many were full 1/4 wave whips, usually from one the rear bumper, and lashed down to the front one. They more common shortened ones had a loading coil at the base, and were electrically 1/4 wavelength. CB radios usually had pretty robust output stages, and could tolerate high SWR ratios. – R Drast Sep 04 '19 at 09:50
  • @SolarMike: Well... CB wavelength is 11 meters, so a 1/4 wave whip is about 8 feet. Even if you shorten that to four or six feet you're still not **small**-small. The antennas that they use for 30kHz "radio" to communicate with submarines; those are electrically short. – TimWescott Sep 04 '19 at 15:01
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The antenna resistance \$R_A\$ can be broken down into two components: the radiation resistance \$R_r\$, the one associated with energy actually radiating, and the Ohmic resistance \$R_O\$, the one associated with losses.

$$ R_A = R_r + R_O $$

The antenna efficiency is the radiated power divided by the power delivered to the antenna, which is the sum of the radiated power and the power dissipated as losses. It can be expressed as:

$$ e_r= \frac{R_r}{R_r+R_O} $$

These two resistive components vary differently over frequency. While we may consider \$R_O\$ constant over frequency, \$R_r\$ increases with the frequency. For the short dipole, for instance, \$R_r\$ is proportional to \$f^2\$.

Think about an antenna of a given size. For low frequencies, i. e. when the antenna size is small compared to the wavelength, \$R_O\$ is the dominant part, what results in low efficiency. As the frequency increases, i. e., as the antenna size increases when compared to the wavelength, \$R_r\$ becomes more significant, what translates into higher antenna efficiency.

A good way to visualize this is to plot a Bode diagram of \$log(e_r)\$ vs. \$log(f)\$ where you'll see a line with positive slope at lower frequencies till it reaches a cutoff frequency when \$R_r=R_O\$ and then becoming a flat line (\$e_r=1\$) at higher frequencies.

Another important consideration is bandwidth, which also decreases as the size of the antenna gets reduced.

reference: [1] Stutzman, Warren L., and Gary A. Thiele. Antenna Theory and Design. Wiley, 2013.

joribama
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