Grounding an audio signal
What do I connect to the sleeve pin of the TRS input jack? 


Always connect to chassis ground. Your thought process is correct. Any other connection between the two system grounds would short-circuit your 2.5 V reference to ground. The current would be limited but your system would cease to operate.


  Should I include a decoupling capacitor, and if so, what size?


No decoupling capacitor is required on the ground lines. If you are asking about a decoupling capacitor on the signal (to block any DC voltage superimposed on the audio) then please edit your question and maybe add in your schematic as it stands. 



There is a CircuitLab button on the editor toolbar. Double-click a component to edit its properties. 'R' = rotate, 'H' = horizontal flip. 'V' = vertical flip. 

Note that when you use the CircuitLab button on the editor toolbar an editable schematic is saved in your post. That makes it easy for us to copy and edit in our answers. You don't need a CircuitLab account, no screengrabs, no image uploads, no background grid.









simulate this circuit &ndash; Schematic created using CircuitLab

Figure 1. Original and modified circuit.

The modified circuit should make everything clear. The main thing we've done is to ground the potentiometer bottom tag. It is generally safe to assume that there is no DC component from the source. We then move the DC blocking capacitor between the wiper and the op-amp input. This prevents upsetting of the DC setup for the circuit.

A couple of notes at this point:

Generally you want the load on the wiper of a pot to be at least ten times the value of the pot. With the components in Figure 1b there is a 10 k&Omega; load (R5) on a 10 k&Omega; pot. This will adversely affect the perceived linearity of the pot. To fix you can scale R5 and R6 up by a factor of 10. This will help us in your next question.


  Should I include a decoupling capacitor, and if so, what size?


Because the inverting input is a "virtual ground" C4 and R5 act as a high-pass filter. You need to set the corner of the cut-off about 0.1 times the lowest frequency you want to hear. The cut-off frequency is given by \$ f_c = \frac {1}{2 \pi RC} \$ so increasing the value of R5 by a factor of ten means we can reduce the value of C4 by the same factor for a give frequency. 

Further reading:


High-pass filter on Electronics Tutorials.