# Approximation when deriving buck closed loop transfer function

Can anyone explain why $\frac{\Delta V_x}{\Delta D} = V_{in}$ here?
It seems to me that they are assuming that $V_x = V_o$ ( this is only correct for average voltage). However, if they use the assumption $V_x = V_o$ for average, then later in LC transfer function, they also should use this assumption to get $\frac{\Delta V_o}{\Delta V_x} = 1$. But from the transfer function of LC filter, we can see that they don't use that assumption.

I am really confused about this. It is not clear when we can use approximation and when not.

I think they assume that the average of the square wave generated by turning on and off the PMOS transistor is $<V_x>=\text{D}V_{in}$ and it doesn't look like they include the $R_{on}$ voltage drop associated with the PMOS transistor when it is on.
Now, $V_{\text{o}}$ isn't exactly equal to $V_x$ because of the voltage drop due to the series resistance associated with output inductor, but they should be really close.
So, I think of it as $V_x$ being a square wave without filtering with its average being portion of the input voltage ($\text{D}V_{in}$). And $V_o$ is the output voltage after filtering which should be the same as the average of the $V_x$ waveform minus the voltage drop caused by the output filter.