The figure below is small signal model for buck converter from Switch-Mode Power Supplies Spice Simulations and Practical Designs by Christophe Basso.
Could anyone explain why the loop gain is 12000 here? It seems that the author doesn't include the D*Vin part.
A 60 dB gain error amplifier monitors the output with a simple feedback capacitor \$C_f\$, making it an integrating compensator together with \$R_{upper}\$. In open-loop, the output impedance, as expected, is the inductor series resistance of 100 mΩ or –20 dBΩ. Closing the loop with a total gain of 12,000 (1000 x 12) leads to a new closed-loop output impedance of Eq. (1-14)
$$R_{s, CL} = \frac{R_{s, CL}}{1+T} = \frac{100m}{12001} = 101.6 \: dB\Omega \:\:\:\:\:\:\:\:(1-14)$$
For anyone wondering about the model, this article page 5/49 explains the model in detail.