I apparently don't fully understand how usb-c cables are made.
So here I ask, I need to do a breakout of the DC power so I can pass it through my DC noise filter. So basically psu-cable1-port1-DCfilter-port2-cable2-device. The device and psu are PD capable.

Using two ports (https://www.amazon.com/24pin-Female-Socket-Connector-Board/dp/B00VJ944V2/) back-to-back do I cross the connections when connecting the two ports together (less the DC power)? Is this how C cables are made?

I see pinout is almost symmetrical, except for SBU1 VCONN, and D- D+ stay in same location. ??

 

Type-C ports dynamically configure themselves based on negotiation using the CC1 and CC2 pins. This is necessary because the connector is reversible. The pins that carry the power (VBUS) are mirror image on the top and bottom row to make it possible to insert the connector with either side up.

The connections should be straight through.

 

No you do not cross the pins USB C cables are straight through with symmetry handled by the receptacle design.
Just break out VBUS or GND for your DC filter and leave dataSBU or VCONN lines untouched.

 

Does "wire straight through" mean B1-B1 B2-B2 Bn-Bn, A1-A1 A2-A2 An-An ?

Probably just a logical to physical reference thing. When wired as I mention, the physical wire (pcb track) crosses over the pcb as A1(1) connects to A1(2), because the ports are physically 180deg from each other. So crossed physically, straight through logically. Correct?