On Wed, 2002-10-09 at 15:14, Mark wrote:
>
> Anyone with some knowledge of ISDN please email me as that I have a few
> questions as to how one multiplexes into a two channel ISDN line.
> Specifically if I have 2 phones and a computer, how would I do that.
What do you mean? How would you do what?
If your ISDN modem/router supports plugging in POTS phones, you can have
one phone per ISDN channel, and if the channel is being used for voice,
you can't use it for data. At least not with anything I've ever seen.
(Unless you're doing Voice-over-IP of course.)
If your ISDN modem/router supports it, you can bind both channels
together so that you get two 64k ISDN channels giving you 128k combined
bandwidth for data.
Is that what you're asking?
--
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#!/usr/bin/perl -w
$_='while(read+STDIN,$_,2048){$a=29;$b=73;$c=142;$t=255;@t=map
{$_%16or$t^=$c^=($m=(11,10,116,100,11,122,20,100)[$_/16%8])&110;
$t^=(72,@z=(64,72,$a^=12*($_%16-2?0:$m&17)),$b^=$_%64?12:0,@z)
[$_%8]}(16..271);if((@a=unx"C*",$_)[20]&48){$h=5;$_=unxb24,join
"",@b=map{xB8,unxb8,chr($_^$a[--$h+84])}@ARGV;s/...$/1$&/;$d=
unxV,xb25,$_;$e=256|(ord$b[4])<<9|ord$b[3];$d=$d>>8^($f=$t&($d
>>12^$d>>4^$d^$d/8))<<17,$e=$e>>8^($t&($g=($q=$e>>14&7^$e)^$q*
8^$q<<6))<<9,$_=$t[$_]^(($h>>=8)+=$f+(~$g&$t))for@a[128..$#a]}
print+x"C*",@a}';s/x/pack+/g;eval
usage: qrpff 153 2 8 105 225 < /mnt/dvd/VOB_FILENAME \
| extract_mpeg2 | mpeg2dec -
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