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Re: question of time (and frames)
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Re: question of time (and frames)


  • Subject: Re: question of time (and frames)
  • From: Brian Willoughby <email@hidden>
  • Date: Sun, 2 Nov 2008 23:30:47 -0800

Right, the time only varies when you have a set number of 'bytes' not 'frames.' If you allocate 512 bytes of buffering, then the number of seconds of time will change as you switch from mono to stereo to surround, or from 44.1kHz to 96kHz or 192kHz.

Brian Willoughby
Sound Consulting


On Nov 2, 2008, at 23:00, tahome izwah wrote: Yes. As a matter of fact, the same is true for a stereo input, or a multichannel input. 512 frames @ 44.1kHz are always 512/44100 seconds.

2008/11/2 John Clayton <email@hidden>:
so if the input notify callback requests 512 frames, and I've got mono input
(1 channel) at 44.1khz, then 512 frames is equal to 512 / 44100.0 seconds,
correct?


On 2/11/2008, at 8:56 PM, tahome izwah wrote:
Generally, frames are samples that contain all channels. 1 stereo
frame = 2 samples. 1 5.1 surround frame = 6 samples. 1 mono frame = 1
sample.

2008/11/2 John Clayton <email@hidden>:
Hope you've had a good weekend out there. Question re: time. I want to
calculate the amount of time that some frames take up - and just to make
sure that this simple task isn't something I've misunderstood - my belief
is
that I can simply say:


1 seconds of time = # frames/samples per second

correct? More specifically, this is entirely independant of the number
of
channels right? So if I wanted to simulate 13 seconds of silence, I'd
just
return
13 * sample_rate_per_second


frames of null data from an input routine.
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  • Follow-Ups:
    • Re: question of time (and frames)
      • From: William Stewart <email@hidden>
References: 
 >question of time (and frames) (From: John Clayton <email@hidden>)
 >Re: question of time (and frames) (From: "tahome izwah" <email@hidden>)
 >Re: question of time (and frames) (From: John Clayton <email@hidden>)
 >Re: question of time (and frames) (From: "tahome izwah" <email@hidden>)

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