I need your help, i am designing an annular slot waveguide antenna in hfss, i want To excite it but waveport, for creating slot admitance plot the port should be de embedded, and the refrence should be middle of slot, my issue is that when i define a distance the hfss doesnt understand that my structure is curved and draw the vector straightforward. Do you friends have any solutions?
The visualization of the port de-embedding doesn't matter. All HFSS is doing is applying an exponential (exp(γ*d), where γ is the propagation constant of the mode(s) and d is the de-embedding distance that you enter). Just calculate the appropriate 'd' value and enter this.
The visualization of the port de-embedding doesn't matter. All HFSS is doing is applying an exponential (exp(γ*d), where γ is the propagation constant of the mode(s) and d is the de-embedding distance that you enter). Just calculate the appropriate 'd' value and enter this.
Thanks for yor reply
You mean that when i add excitation, and select de embedded,i should insert (exp(¥*d)) in distance?the value of d is constant and doesnt change, is the value of ¥ constant? I assume that my propagation is single mode.
Thanks for your help, but still i havent understood what shall i do, i copied “ exp(γ*20)” but HFSS put “?” instead of γ. I mean i even dont know how to write exp(γ*20) in hfss i copied from word, i will be really thankfull if help me i am really fresh in hfss. does hfss know the symbol γ or i shold addit in design properties,and how should i type a symbole in hfss, thanks your for kind supprot
De-embedding is the process of undoing the effects of one of more propagation constants over a given distance. In HFSS, the software already knows the propagation constant, at each frequency. All you need to do is enter the de-embedding distance.
Dear
I couldn’t enter lambda in hfss, I know the value of d but I couldn’t enter lambda, and my project stoped here, would you please support me to pass this step, how should I enter lambda?