DVD-A question

Discussion in 'Music' started by Aaron Reynolds, May 28, 2005.

  1. Aaron Reynolds

    Aaron Reynolds Screenwriter

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    So I decided to buy a couple of DVD-As to see how I liked the format, because I realized the other day that my DVD recorder, a Panasonic DMR E55, would play DVD-As. I run a two channel system, so I'm not all that excited by multichannel mixes, and anyways the E55 only supports two-channel DVD-A. I figured this was because it only has analog L and R outs, and DVD-A isn't supposed to give you a digital output for "security" reasons.

    I picked up The Last Waltz and Neil Young's On The Beach.

    So I stuck 'em in and noticed that both are giving a digital feed, via optical, to my preamp. When I tap the audio key during On The Beach, the player reads "PPCM 2CH 176k 24b", and it won't let me change to the other track on the disc, which is supposed to be 48/24.

    So my question is this: am I getting the 48/24 via digital and the 176/24 via analog? Is the player reading and playing back both signals at the same time? Can optical cable even handle 176/24?

    Now, as far as I can tell, The Last Waltz is supposed to have stereo and 5.1 DVD-A tracks and then a 5.1 DD track as well. When I play it, my only option is the 48/24 stereo track, which makes sense since the player doesn't support multichannel DVD-A. If I set the player to treat the disc as a DVD-V, then it sends a Dolby Digital channel to the preamp. So when it isn't set to DVD-V, what's that digital signal that's going to the preamp? Is it the 48/24? Shouldn't that be "protected" and only available through analog?

    Another note: I hooked up my mini disc recorder to see what it said about the signal going into the preamp, and it wouldn't let me record either of them -- it just said PROTECT on the display.

    So, can someone tell me what's going on?
     
  2. Aaron Reynolds

    Aaron Reynolds Screenwriter

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    Bit more information:

    Checked the manual for the preamp, and it doesn't support anything above 48kHz, so the digital signal must be that... right?
     

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