Here is an interesting article about mixing stereo.
Most important issues when mixing stereo are :
- interaction between channel
- crosstalk
- noise
Interaction between channel is explained here in 1.0 with passive mixing, from that :
It would be possible to make the output impedance of all equipment
much higher, so direct mixing would not cause any circuit stress. The
problem would then be that we are back to the position we had when
valve gear ruled … high impedance causes relatively high noise and
high frequency rolloff with long cables. Cables can also become
microphonic, and this is why so many pieces of valve kit used output
transformers – to provide a low impedance (optionally balanced) output
to prevent the very problems described. Low output impedance is here
to stay, as are mixers, so now we can examine the methods in more
detail.
But for crosstalk, in the mixing passive circuit, how do we calculate it? We can find different methods to measure it. but for simple mixing circuits I did not found any documents describing how it be computed.
We can find an example in here :
The problem arising from using all three outputs (the two original and
the new summed output) is one of channel separation, or crosstalk. If
the driving unit truly has zero output impedance, than channel
separation is not degraded by using this summing box. However, when
dealing with real-world units you deal with finite output impedances
(ranging from a low of 47 ohms to a high of 600 ohms). Even a low
output impedance of 47 ohms produces a startling channel separation
spec of only 27 dB, i.e., the unwanted channel is only 27 dB below the
desired signal. (Technical details: the unwanted channel, driving
through the summing network, looks like 1011.3 ohms driving the 47
ohms output impedance of the desired channel, producing 27 dB of
crosstalk.)
Some technical details are given, but it's not clear to me. Can someone give detailed steps to compute crosstalk for this example? A simulation example could also helps.
Best Answer
It's quite simple. If signal A has an output impedance of 47 ohms and is passively mixed with a signal B using two 500 ohm resistors then signal B impregnates signal A because there is a simple potential divider.
simulate this circuit – Schematic created using CircuitLab
If signal A is zero volts (to take the simple case) and signal B is 1 volt, then the voltage appearing at signal A (after the 47 ohm resistor) is: -
1 volt x 47/1047 = 44.9 mV.
This 44.9 mV shouldn't be there ideally i.e. it is cross contamination and, if you worked out what this factor is in decibels then that comes to -26.96 dB.
The way to overcome this is not to bodge-mix by using a passive resistor network but to use a virtual earth summing amplifier.