[SOLVED] What is the accepted power of a two port antenna?

[jujuChenYi]

When both conductor and dielectric are lossless, Pacc=Pin*(1-S11^2)，just like (2) in HFSS help script，but when both conductor and dielectric are lossy, Pacc is slightly smaller than Pin*(1-S11^2)，why?

--- Updated Oct 11, 2024 ---

Has anyone studied radiation efficiency in HFSS before? please

 

[FvM]

Seems like you are comparing radiated (1) and accepted (1) power of an antenna. Obviously they differ by antenna losses, it's not generally correct to designate (1) P,accepted.

 

[jujuChenYi]

Seems like you are comparing radiated (1) and accepted (1) power of an antenna. Obviously they differ by antenna losses, it's not generally correct to designate (1) P,accepted.

Hi FvM ,thank you for your reply! What I mean is, when simulating a two port antenna, the Pacc calculated by HFSS is smaller than what I calculated using formula (2) myself (a=1, since Pin=1). I wonder why.

 

[jujuChenYi]

HFSS script says when port is lossy ,the incident power may not be 1W, What does lossy port means?

 

[jujuChenYi]

I think the correct answer is P(accepted)=P(input)*(1-S11^2), right? But in HFSS, P(accepted)=P(input)*(1-S11^2-S21^2), I wonder why

 

[volker@muehlhaus]

But in HFSS, P(accepted)=P(input)*(1-S11^2-S21^2), I wonder why

This is the generic equation to calculate the power "loss" of a 2 port component.
Pin * |S11|^2 is the power reflected at port 1
Pin * |S21|^2 is the power leaving the model at port 2
The remaining power "disappears" in the 2-port itself. In this case, this is the accepted input power of the antenna.

 

[jujuChenYi]

This is the generic equation to calculate the power "loss" of a 2 port component.
Pin * |S11|^2 is the power reflected at port 1
Pin * |S21|^2 is the power leaving the model at port 2
The remaining power "disappears" in the 2-port itself. In this case, this is the accepted input power of the antenna.

Thanks! understood. ：）