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.


1 Answer 1


Can someone give detailed steps to compute crosstalk for this example?

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.

  • \$\begingroup\$ thanks. In this example, can we say that conditions to have crosstalk are : input source impedance > 0 and a signal from source = 0V ? Another question, in some cases, we can see such summing circuits : techtalk.parts-express.com/filedata/… , what is R3 role? \$\endgroup\$
    – rem
    Commented Jun 28, 2017 at 15:40
  • \$\begingroup\$ In this example you will get cross talk whnever the source impedances are > 0. In the picture you linked, R3 adds attenuation to the overall mixed output signal and reduces cross talk by the same amount. \$\endgroup\$
    – Andy aka
    Commented Jun 28, 2017 at 18:41
  • \$\begingroup\$ Thanks for your reponse. For crosstalk conditions, from calculation, we get Unwanted crosstalk on A = R2/(2*(R1 + R2)) * (Vsignal1 - Vsignal2), so we do not have any crosstalk when R2 = 0 (or Vsignal1 = Vsignal2) . When doing a stereo to mono mixing, with same sources impedances, we will keep an output voltage of (Vsignal1 - Vsignal2)/2, so crosstalk must be taken into account only if signal 1 or signal 2 used somewhere in circuit, for example to drive a speaker. Is it right? For R3 role, could you detail your response with some calculations? \$\endgroup\$
    – rem
    Commented Jun 29, 2017 at 19:24
  • \$\begingroup\$ Yes, everything you say is correct. Regarding the calculation, are you familiar with superposition theory? If not then you need to take a look at it because it shows how multiple voltage sources act to produce a "mixed" voltage. \$\endgroup\$
    – Andy aka
    Commented Jun 29, 2017 at 19:35

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