# Why does damping between two antennas increase with frequency?

While studying for a HAM license exam I ran into a question that I am having trouble with answering. There are two aspects that I cannot explain, given my current admittedly little understanding of the subject.

Given the following statement in the exam text...

The damping between two vertical half-wave dipoles is measured at a certain frequency. The antennas are set up in free space. If the frequency is doubled and the dimensions of the half-wave dipoles are adjusted accordingly, the damping will increase by 6 dB. (Translation is my own, original is in Dutch)

... I have the following questions:

• Why does damping increase rather than decrease with increasing frequency, with their relative distance of the antennae being (assumed) constant?
• What is the appropriate logic or formula to apply here that explains why there is a 6 dB increase rather than (say) 3 dB?

Should you wish to look at the original dutch wording, it's question 37 in the 6 November 2019 full-license exam. The answer table lists answer A (6 dB increase) as being correct.

• wow, these exam questions are pretty technically involved! – Marcus Müller Mar 22 at 21:13
• Are they? I have no comparison. These are for the dutch full HAREC exam. We also have a novice exam. – Frank Geerlings Mar 22 at 21:56
• Welcome to ham.stackexchange.com! – rclocher3 Mar 23 at 23:23
• I don't have time for an answer now, but look up Friis equation, and punch in a couple of different frequencies, with transmit power, distance and antenna gains being fixed. – AndrejaKo Mar 26 at 7:34
• What is "damping"? – Brian K1LI Apr 6 at 7:55

$$L_\text{FS}\text{(dB)}=20\log_{10}\left(4\pi{\text{distance}\over\text{wavelength}}\right)$$
This equation shows that when you reduce the wavelength by a dB, $$L_\text{FS}$$ will increase by twice that many dB. So if we halve the wavelength (cut in half = -3 dB) the loss will increase 6 dB.