s22 of amplifier interacting with s11 of antenna

[faust1002]

Hello,
Again, I have problems with understanding what my lecturer wants and I hope that you can help me.
The problem is as follows - I have an amplifier, which has output impedance of 63+j80Ω. So, when 50Ω load is connected, then return loss is equal to roughly 4.65dB. So far, so good. But the question is what will happen when 50Ω antenna, which has RL od 30dB, will be connected to this amplifier.
My understanding is that RL of the amplifier (4.65dB) represents its s₂₂ and RL of the antenna (50dB) corresponds to antenna's s₁₁. Now, because s₂₂ is much higher than s₁₁ it dominates. As a result, roughly 40% of the power outputted by the amplifier will be reflected.
The provided answer says about above 30% of the incident power being reflected. This is more or less aligned with my interpretation. But I want to double-check if my understanding is correct.
Kind regards

[thundertronics.com]

Probably lecturer wants to exercise using these formulas. If Zs=50, Zl=53 or 47, then antenna "RL" would be around 30dB, and reflected % would be between 32.7 and 35.97..
See attached image, you can put other numbers using this calculator: Mathnotepad link with formulas.

[faust1002]

Step by step. Zs and Zl are given and are equal to 63+j80Ω and 50Ω respectively. Besides, RL of 30dB does not correspond to roughly 30% of the incident power being reflected. 4.65dB does.

[thundertronics.com]

But the question is what will happen when 50Ω antenna, which has RL od 30dB, will be connected to this amplifier.

I think that in this particular question RL of 30dB probably means that antenna is almost 50 Ohm, for example 54 Ohm.

My understanding is that RL of the amplifier (4.65dB) represents its s22 and...

when it is connected to 50 ohm load, RL=-20*Log10(Abs((Zl-Zs)/(Zl+Zs)))=4.65dB
Your amplifier's Zs=63+i*80, but calculation also includes load's Zl=50.

..and RL of the antenna (50dB) corresponds to antenna's s11.

Antenna's s11 relative to what? I think relative to characteristic impedance Z0=50 Ohm.
RL=30db probably means that it's impedance is not 50 Ohm exactly, but Zl=54 Ohm:
Zs=50, Zl=54 RL=-20*Log10(Abs((Zl-Zs)/(Zl+Zs)))≈30dB

Now, because s22 is much higher than s11 it dominates.

This is probably wrong assumption. As soon as we disconnected 50 ohm load, RL of amplifier is no longer may be calculated using Zl=50, we have an amplifier with open connection Zl=large number, and new RL would be almost 0, and reflection would be almost 100%. Amplifier's S22 remains the same: s22 = (Zs-Z0)/(Zs+Z0) ≈ 0.41-0.42*i; backwards from s to z: Zs=Z0*(1+S22)/(1-S22) ≈ 62.5 - i*80.1

To sum up,
amp + 50 ohm load: Zs=63+i*80 ; Zl=50 ; RL=4.65dB
50 ohm 50 ohm antenna: Zs=50; Zl=54; RL=30dB  IMO, antenna's RL nothing to do with 63+i*80..