According my book, which I think it might be an error, Q1 is a D-MOSFET, and the behavior of the circuit is shown in the following picture. But it seems to me that according to the behavior this should be an E-MOSFET. Because if it was a D-MOSFET it would be normally on and an increase in duty cycle would mean a shorter T off and thus shorter time the MOSFET is on and thus less V out. But the Waveforms show and say that a larger duty cycle with more T on produce an increase in V out
With an N-channel MOSFET, a positive change in gate voltage means an increase in conduction in the FET D-S channel, whether you are using a D-MOSFET (Depletion mode) or an E-MOSFET (Enhancement mode).
The difference between D- and E- is the level at which the change takes place. For instance an E-MOSFET is off with zero gate volts, and (depending on the specific model) on with 2v to 10v on the gate. A D-MOSFET will be on with zero gate volts. To turn it off you take the gate to a negative voltage, just like with a vacuum tube grid. In both cases, a positive change increases conduction.
As you see on the waveforms picture, on is a more positive voltage then off. The difference between using a D- and E- device would be where you draw the zero on the control voltage graph. With E- it would be the lower voltage, with D- it would be the higher voltage.
With a P-channel MOSFET, the polarities are opposite, and the states for zero gate voltage are the same as for N-channel.