When use the mapping structure like:
2M->C12-(+LP-POH)->VC12-(+TU12pointer)->TU12-......->STM1
the 2M rate use mutilframe(including four consecutive c12s) to adapt
into C12.And then insert the POH to form vc12.V5 byte will indicate
where is the VC12 multiframe's beginning.Otherwise, the four
consecutive C12 not in the same STM1 frame but four consecutive STM1
frame,and transmitted independently .

I confuse that a 2M signal would be pick-up after four STM1 frames have
been received.How?I need someone to explain the process.

Pardon my ignorance.I really need to clear it up.

Hello Ling,

You wrote:

> When use the mapping structure like:
> 2M->C12-(+LP-POH)->VC12-(+TU12pointer)->TU12-......->STM1
> the 2M rate use mutilframe(including four consecutive c12s) to adapt
> into C12.

Note that on the ...... there is also TU3

> And then insert the POH to form vc12.

Yes.

> V5 byte will indicate
> where is the VC12 multiframe's beginning.

OK.

> Otherwise, the four
> consecutive C12 not in the same STM1 frame but four consecutive STM1
> frame,and transmitted independently .

Correct.

> I confuse that a 2M signal would be pick-up after four STM1 frames have
> been received.How?I need someone to explain the process.

In the TU3 that contains the TU12 the H4 byte has a counter
in the last two bits. (see ITU-T G.707 figure 8-9).
With the two bits four STM-1/AU-4 frames can be counted.
first and second frame contain V1 and V2 bytes.
V1 and V2 bytes are pointer to V5 byte in first frame.
Second, third and fourth frame contain J2 N2 and K4 bytes.

Each STM-1 frame contains a comlete 2M frame (32 bytes)

Worst case it takes four frames to find a new pointer (V1+V2)
to the V5 byte.

Cheers, Huub.

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Always remember that you are unique...just like everyone else...

Thanks,Huub.
But I still need your help.
What I really confuse is  the signal flow.

STM1 take 125=B5s to transfer.So the second STM1 frame would arrive at
the 126th =B5s.What's situation of the previous STM1 frame now?Has been
reverted to 2M signal?Or just be stored and waiting for the second
frame to be receive completely?I don't know how it pick-up the entire
2M in Multiframe.It must has time delay,right?

When I was a student,my teacher explain this with time line.It really
help me to know it clear.But many years later,I foget it.
-______________________-!

Could you help me again?

Hello Ling.

You replied:

> But I still need your help. What I really confuse is  the signal
> flow.
> 
> STM1 take 125�s to transfer.So the second STM1 frame would arrive at 
> the 126th �s.

No, the second STM-1 frame starts arriving at the beginning of
the 126th �s and ends at the end of the 250th �s.

> What's situation of the previous STM1 frame now?

That is finished. You should have stored the bytes of the 2M
that needs to be recovered in a buffer.
The bytes in the buffer are written at STM-1 speed (155 Mbit/s
or 19 Mbyte/s) and are read at 2M speed (2 Mbit/s or 256 kbyte/s).

> Has been reverted to 2M signal?Or just be stored and waiting for the
> second frame to be receive completely?I don't know how it pick-up the
> entire 2M in Multiframe.It must has time delay,right?

There is normally a small delay, because when the first 2M byte
in an STM-1 frame has been writeen into the buffer, it can be read
from the buffer. The second 2M byte can be read immediately after.

> When I was a student,my teacher explain this with time line.It really
>  help me to know it clear.But many years later,I foget it. 

He probably used a picture, that is very difficult to show here.

> Could you help me again?

I hope I did.

Cheers, Huub.

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Always remember that you are unique...just like everyone else...

Hello Ricy,

You wrote:

> Huub is this representation correct of what you have explained to Ling ?
> Ricy,

Nice picture, but I will not repeat it, because this is against
usenet rules  ;-)

Hower some notes:
for the TU3 the H4 byte is used, and using its 2 bit counter:
STM-1 FR1 H4 points to byte V1 position in FR2
STM-1 FR2 H4 points to byte V1 position in FR3
see figure 8-12 in G.707

for the TU12: the V1 and V2 bytes are the pointer to the V5 byte
(see figure 8-10 G.707)
if STM-1 FR2 contains byte V5,
then STM-1 FR3 contains byte J2,
STM-1 FR4 contains byte N2 and
STM-1 FR1 contains only byte K4

So in your picture there shoud be only one red arrow V1, one
blue arrow v2 and one green arrow V5 and they should never
point back in time (to the left in the figure).

Cheers, Huub.

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Always remember that you are unique...just like everyone else...

Huub,

Thanks for al your help first.
You are frequently referring to G.707 Figures (8-10; 8-12 etc....);
if possible can you tell me where I can find this or download it ??
thx, Ricy

"Huub van Helvoort" <hhelvooort@chello.nl> schreef in bericht
news:YplNe.50$0m4.0@amstwist00...
> Hello Ricy,
>
> You wrote:
>
> > Huub is this representation correct of what you have explained to Ling ?
> > Ricy,
>
> Nice picture, but I will not repeat it, because this is against
> usenet rules  ;-)
>
> Hower some notes:
> for the TU3 the H4 byte is used, and using its 2 bit counter:
> STM-1 FR1 H4 points to byte V1 position in FR2
> STM-1 FR2 H4 points to byte V1 position in FR3
> see figure 8-12 in G.707
>
> for the TU12: the V1 and V2 bytes are the pointer to the V5 byte
> (see figure 8-10 G.707)
> if STM-1 FR2 contains byte V5,
> then STM-1 FR3 contains byte J2,
> STM-1 FR4 contains byte N2 and
> STM-1 FR1 contains only byte K4
>
> So in your picture there shoud be only one red arrow V1, one
> blue arrow v2 and one green arrow V5 and they should never
> point back in time (to the left in the figure).
>
> Cheers, Huub.
>
> --
>                 reply to hhelvooort with 2 'o's
> ================================================================
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> ================================================================
> Always remember that you are unique...just like everyone else...

Hello Ricy,

You asked:

> Thanks for al your help first.

I am always glad to spread my knowledge  ;-)

> You are frequently referring to G.707 Figures (8-10; 8-12 etc....);
> if possible can you tell me where I can find this or download it ??
> thx, Ricy

You can get it on the official ITU-T webpage:
http://www.itu.int/rec/recommendation.asp?type=folders&lang=e&parent=T-REC-G.707
follow the link then you get to a page where you can buy it.

However: if you register in the ITU-T bookshop you may download
three recommendations free per year.
http://www.itu.int/publications/bookshop/how-to-buy.html

Cheers, Huub.


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Always remember that you are unique...just like everyone else...

Thanks a lot!
Now I know where I stuck in.
Glad to discuss with you~!

Have a nice weekend!   ^___________________^

Regard,
Ling