An Explanation of the Effective Area of Isotropic Antenna

[Puma24]

Hey all,

I realize a question on this topic has been asked elsewhere, but the links to references they use seem to be dead, so I'll press on!

I'm reading some introduction to antenna theory and I've often puzzled on the equation:

$A_{eff} = \frac{\lambda^2}{4\pi}$

which relates the effective area by which an antenna captures radiation to the frequency at which that radiation is.

I have looked at the derivation here:

https://physics.stackexchange.com/q...86160?newreg=76f46a9e027440ae964582feb289a289

and can understand the steps they take, but when trying to understand it on a higher level, I cannot reconcile it.

Going on the example used in the derivation above, I would assume that as the frequency that is selected to pass through the filter increases, the resistor would begin to give off more power, in line with Johnson-Nyquist Noise:

$dP = k_{b}Td\nu$

This would mean, in order to keep thermal equilibrium, that the antenna on the other end would have to be more receptive to the blackbody radiation that the cavity would give off. So, I would assume, that would mean a larger Effective Aperture Area would be required in order to gather it.

But this seems to contradict the result, which says that higher frequencies need a smaller area.

Can anyone help me out and point out the flaws in my assumptions?

Thanks

 

[tech99]

Hey all,

I realize a question on this topic has been asked elsewhere, but the links to references they use seem to be dead, so I'll press on!

I'm reading some introduction to antenna theory and I've often puzzled on the equation:

$A_{eff} = \frac{\lambda^2}{4\pi}$

which relates the effective area by which an antenna captures radiation to the frequency at which that radiation is.

I have looked at the derivation here:

https://physics.stackexchange.com/q...86160?newreg=76f46a9e027440ae964582feb289a289

and can understand the steps they take, but when trying to understand it on a higher level, I cannot reconcile it.

Going on the example used in the derivation above, I would assume that as the frequency that is selected to pass through the filter increases, the resistor would begin to give off more power, in line with Johnson-Nyquist Noise:

$dP = k_{b}Td\nu$

This would mean, in order to keep thermal equilibrium, that the antenna on the other end would have to be more receptive to the blackbody radiation that the cavity would give off. So, I would assume, that would mean a larger Effective Aperture Area would be required in order to gather it.

But this seems to contradict the result, which says that higher frequencies need a smaller area.

Can anyone help me out and point out the flaws in my assumptions?

Thanks

The frequency refers to the bandwidth of the filter, not to the absolute frequency. When the filter bandwidth is increased, more noise power can flow in both directions, maintaining the equilibrium.