Other methods

For the other 2 methods the same method as previously described can be used. When the amplifier has a current output and a $50\Omega$ resistor is placed in parallel, we get a total resistance of $25 \Omega$ instead of the previous $100 \Omega$ used for the voltage output. And for the elegant solution using a dual loop feedback we simply only use a $50 \Omega$ total resistance at the output and obtain the parameters.