Electronic – How to clean up high audio noise amplified through a 2×2 electronic switch

As mentioned in the comments, diodes are not the way to go. They will massively corrupt your audio signal, and it will just sound utter rubbish.

The simplest method is to use a passive mixer. That simply consists of two resistors per channel:

^{simulate this circuit – Schematic created using CircuitLab}

You can tie all the ground wires together without resistors.

This of course is assuming you are using a powered speaker set, and not just raw speakers.

Your friend may indeed be correct.

Here's what happens in audio circuit inputs and outputs that aren't properly designed:

Most audio inputs and outputs are capacitively-coupled. This eliminates DC offset that might be coming from the electronics. In the old days where everything was discrete transistors (or Tubes!), these coupling capacitors were absolutely necessary because the input and output nodes within the circuit were usually nowhere near ground potential.

Modern audio electronic circuits make use of op-amps and such. If the op-amp is supplied with a bipolar power supply (equal value (+) & (-) rails), the input and output nodes within the circuit are usually near ground potential. There is usually offset voltage errors but these are usually quite small - ranging from zero volts up to a few tens of mV.

Because audio electronics with bipolar supplies is usually of high quality, the designers include DC isolation capacitors WITH bleeder resistors to ground so that the net DC voltage at any outside-world connector is zero.

However, most inexpensive audio equipment uses only a single supply voltage, with a virtual-earth reference rail set at half the supply voltage. The input and output nodes within the circuit thus sit at about half the supply voltage.

So: the designers include coupling capacitors to block the DC but pass the AC signal.

Here's where they mess up: they often don't include a bleeder resistor at the outside-world end of the capacitor. The capacitor has some leakage, so any charge that might be on it leaks away when nothing is connected to the in / out connector. That is: the DC voltage on the capacitor is the same at both ends of the capacitor.

Because the circuit-side of the capacitor is sitting at half the supply voltage, the outside-world end of the cap is ALSO sitting at half the supply voltage.

Now you go to plug something into that connector and see a HUGE spike that lasts as long as it takes for the capacitor charge to equalize.

The resulting damage is generally NOT covered by warranty.