Why the inductance become L/D'^2 in buck-boost small signal model?
Voltages and currents are affected directly by &quot;D&quot; i.e. a 10:1 step down transformer reduces voltage by 10 and increases current by 10. But, the upshot of this is that the secondary impedance is decreased by \$D^2\$ or, put the other way, if you wish to transfer an impedance of Z to the primary from the secondary then it will look like \$Z\cdot D^2\$.
Just put some numbers on a 1:1 transformer connected to a 10 ohm load - from the perspective of the primary it looks like there is a 10 ohm load but, if the transformer were 2:1 step down, then the voltage halves and the current doubles on the secondary; that's an impedance reduction of 4:1 and for the 2:1 transformer to produce an impedance at the primary of 10 ohm, a 2.5 ohm load is required to be connected to the secondary.
Hence, when you make equivalent circuits to remove transformers, you multiply or divide the impedances by \$D^2\$.