Hi,
Overall, your set-up seems fine: apply a 'test' current Iout
and watch the effect on the output voltage (or vice versa).
Since Zout = Vout/Iout and Iout (ac) == 1, the magnitude of the output voltage corresponds indeed with the output impedance of your circuit.
I find it strange however to assess it in dB (20*log(V/I)) which becomes 20*log(V) since I == 1 and this - of course- equals vdb(out).
Simply converting to decimal numbers would be my approach, but in the end, it's just another representation
![Smiley Smiley](https://designers-guide.org/forum/Templates/Forum/default/smiley.gif)
Kind Regards
Peter