Full-wave rectifier with pi-filter
Your transfer function doesn't depend on \$C_1\$ because after the diode bridge, \$C_1\$ acts pretty much as a source. I would think of this circuit as a DC source at \$C_1\$ and an LC network to filter out the ripple.

Ideally, you could just consider \$C_1\$ as a DC voltage source (how good of a DC source depends on the its capacitance and how much current it is being drawn by the circuitry to the right of \$C_1\$). That's what you have in the denominator of your TF on the left hand side, you're using \$V_{c1}\$ as your input. 

Even though you have a transfer function that includes \$L\$ and \$C_2\$, ideally, the DC portion coming from \$C_1\$ is unaffected. That means that not even the last LC filter changes the DC portion (ideal case). That is evident if you where to evaluate your TF at \$s=0\$, which is the case for a DC signal. Now, in reality, what's going to change your output voltage/current are the ESR losses associated with the capacitors and inductor of the filter. And of course, \$L\$ and \$C_2\$ will further smooth out your output signal.