Can anyone tell me why in a stereo mode many channels are derived from the 2 channel input . It looks like upmixing.. My questions are summarized below :

If it is Stereo mode why upmix it to many channels at the output ? Why does this mode have to be after the equalization process ? Why can't this mode be present before equalization? What I meant when I said the stereo mode is two channel stereo signal is first equalized through some filtering process and then upmixed. My question can be better understood as this why should upmixing be performed after equalization and not before equalization

  • $\begingroup$ What is a stereo mode? Are you using a software or a hardware? If the equalizers for all channels are the same, it shouldn't be performed after upmix. But if you want to equalize upmixed signals by different EQ, the order should be inverted. $\endgroup$ – ZR Han Apr 22 at 3:37
  • $\begingroup$ It's a software ... Can you also give me a reasonfor your statements $\endgroup$ – Clauvunate Apr 22 at 5:43
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    $\begingroup$ Clavunate, you've already got feedback on your previous question. As you've already noticed, nobody knows what specifically you mean with "stereo mode", and that the order of operations you claim doesn't generalize. Again, posting almost verbatim duplicates of questions isn't appropriate. $\endgroup$ – Marcus Müller Apr 22 at 7:40
  • $\begingroup$ Dear Marcus Müller please read the answer suggested by ZR Han he did say if the equalizers for all channels are same it shouldn't be performed after upmix and vice versa ...I want to understand the reason or explanation behind that and FYI since no one responded to my comment I resorted to re posting the question ... $\endgroup$ – Clauvunate Apr 24 at 21:02
  • $\begingroup$ Because the system is a linear system. Two channels being applied to a same equalizer and then upmixed euqals to these two channels being upmixed to five channels and then equalized. In the first case you have 2 signals to equalize, but in the second case you have 5, meaning that the first one is more computational efficient. $\endgroup$ – ZR Han Apr 26 at 3:04