[SI-LIST] Re: Rx Eye Mask width, BER, and Jitter

  • From: wolfgang.maichen@xxxxxxxxxxxx
  • To: Conrad Herse <herse@xxxxxxxxxxxxxxxxxx>, James.Mueller@xxxxxxxxxx
  • Date: Mon, 21 Jun 2010 15:02:09 -0700

Hello Conrad,
maybe rephrasing the problem helpf. Let's see if I understood your goal:

- You have a receiver that can liver with an eye opening of 0.2 UI (i.e. 
works correctly if no transition falls into the center 20% of the eye).

- That means total allowed jitter is Tj = (1 - 0.2) UI = 0.8 UI

- Specification further says that out of that total number, max. 0.3 UI 
can be Dj and max. 0.5 UI can be Rj.

It does not seem to me that so far this is dependent on the BER at all. 

What can the random jitter of the input signal be so we get BER=1e-15 
instead of BER=1e-12 based on the input signal alone?

BER=1E-12 --> Q = approx. 14 --> Rj (input signal) = 0.5 UI / Q = 0.036 UI 
RMS
BER=1E-15 --> Q = approx. 16 --> Rj (input signal) = 0.5 UI / Q = 0.031 UI 
RMS

The 0.2 UI required eye opening is not affected by the required BER at 
all. What does change is simply the allowed Rj RMS value (must be smaller 
for smaller BER - no surprise here).


However, one thing we did not take into account so far is that the 
receiver will have some internal jitter (strobe jitter) as well. In 
addition the receiver will have some static strobe placement error (i.e. 
the strobe may on average happen at 0.55 UI instead of in the center at 
0.5 UI). The resulting BER of the received signal is affected by the 
combination of input signal jitter (which is limited by the Rj, Dj and Tj 
specs above), the receiver jitter, and the receiver strobe placement 
error.

Unfortunately, from the available data only the first contributor is 
known. My calculation above basically assumes that the BER is solely given 
by the input signal, with the receiver being perfectly jitter- and 
skew-free. So without further knowledge or assumption about the receiver 
one cannot get a definitive answer.

So let's make some assumptions:

For example, lets assume that the receiver strobe is perfectly placed in 
the center, and the receiver only exhibits random jitter (which is a very 
idealistic assumption). On the other hand, assuming receiver jitter to be 
pure Rj is the most conservative assumption (will overestimate total 
jitter) when extrapolating to lower BERs. So the conclusions below will 
give a conservative estimate. 

If the receiver had a 0.2 UI RJ (@ BER 0.2 US this Rj(receiver) = 0.2/14 = 
0.014 UI RMS), then the BER of the received signal would actually be

Rj(received signal) = approx. BER(input signal) + BER(receive strobe) = 
1E-12 + 1E-12 = 2E-12

That's clearly not what the spec says (it says that with an input signal 
jitter of 0.3 UI Dj and 0.5 UI Rj you get BER = 1E-12, not 2E-12). So my 
conclusion is that the receiver actually must have much less Rj than 
0.2/14 UI (i.e. the receiver BER contribution must be negligible compared 
to the input signal BER). How much, that I can't tell from the data. Which 
also means, the available information is not sufficient to answer the 
question. But a reasonable assumption would be that it is at least one 
order of magnitude smaller (i.e. < 1E-13).

But what we CAN say is that in order to get BER=1E-15 in a similar way 
than 1E-12, we need two things:

(a) - have drive signal at BER=1E-15, and
(b) - have receiver strobe BER << 1E-15

Condition (b) will require the eye opening of the input signal to be 
larger than 0.2 UI (so the receive strobe has more "room" to jitter around 
without causing errors). Is this what you were looking for? Assuming 
BER(receiver) was <1E-13 and we want to keep the same ratio between the 
receiver strobe BER and the input strobe BER (>= 1 order of magnitude), 
then the receiver strobe BER should be <1E-16.

Once you know what (b) yields, you can the calculate the new input signal 
requirements. You can caculate by which factor to enlarge the required eye 
opening with those two numbers (BER before 1E-13 for 0.2 UI opeing, BER 
after 1E-16 for xy UI eye opening --> solve for xy). E.g. assume you find 
that you now need xy = 0.3 UI eye opening, then the new spec for the input 
signal would be:

Tj = (1 - 0.3) = 0.7 UI
Dj = 0.3 UI (let's use the same we had before)
Rj = (0.7 - 0.3) = 0.4 UI @ BER 1e-15

--> Rj = approx. 0.4 / Q = 0.4 / 16 = 0.025 UI RMS

So in fact we have TWO new values in order to go from BER 1E-12 to 1E-15:

(1) required eye opening 0.3 (instead of 0.2) (this would be your "growing 
the receive eye)
(2) maximum signal Rj = 0.025 UI RMS instead of 0.036 UI RMS

Wolfgang








James.Mueller@xxxxxxxxxx 
Sent by: si-list-bounce@xxxxxxxxxxxxx
06/21/2010 02:18 PM

To
Conrad Herse <herse@xxxxxxxxxxxxxxxxxx>
cc
si-list@xxxxxxxxxxxxx, si-list-bounce@xxxxxxxxxxxxx
Subject
[SI-LIST] Re: Rx Eye Mask width, BER, and Jitter






As the bit error ratio you require gets lower, the Tj your Rx needs to
tolerate will increase for a given specified Dj and rms jitter.  Likewise
the eye opening will shrink.  I'm not sure why you say you expect the "min
required eye to grow".   The Tj tolerance grows so that the eye opening
"tolerance" shrinks, e.g. the Rx needs to be able to tolerate a smaller 
eye
opening width in the incoming signal.  I'm not sure I'm helping, I think
part of this is semantics or definitions.

Jim



James J Mueller
LeCroy Corporation
Cell phone:     914-522-8555



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Hi Jim,

Maybe this is what's confusing me. The receiver specifies a jitter
*tolerance*. From this an Rx minimum eye width can be determined. I
would expect the min required eye width to *grow* when going to
BER=1e15, since the receiver must *tolerate* more jitter. Making the min
Rx eye mask smaller at BER=1e-15 doesn't seem correct, it implies the Rx
jitter tolerance increases at 1e-15. Again, I'm referring to the Rx
jitter tolerance and what the receiver requires, I understand that the
actual eye size at the receiver will decrease when more jitter is
budgeted into the system.

Thanks,

Conrad Herse



James.Mueller@xxxxxxxxxx wrote:
> Hi Conrad,
>
> When you went to calculate the mask size at BER= 1E-15, you added the
> additional rms jitter contribution instead of subtracting.
>
> Jim
>
>
>
> James J Mueller
> LeCroy Corporation
> Cell phone:     914-522-8555
>
>
>
>

>   From:       Conrad Herse <herse@xxxxxxxxxxxxxxxxxx>

>

>   To:         si-list@xxxxxxxxxxxxx

>

>   Cc:         herse@xxxxxxxxxxxxxxxxxx

>

>   Date:       06/21/2010 03:41 PM

>

>   Subject:    [SI-LIST] Rx Eye Mask width, BER, and Jitter

>

>   Sent by:    si-list-bounce@xxxxxxxxxxxxx

>

>
>
>
>
>
> Hello experts,
>
> I've been working on trying to scale receiver eye mask widths to
> different bit error rates. There is something which is puzzling me which
> I'm hoping someone can clear up for me.
>
> I've been studying the dual-Dirac jitter model given by the formula:
>
> Tj = Dj + 2Q * Jrms
>
> where Q is a constant from the Complimentary Error function for a given
> BER (2Q*Jrms = Rj at a specific BER). So if I have a receiver with the
> following jitter tolerance spec:
>
> Tj = 0.8 UI
> Dj = 0.3 UI
> Rj = 0.5 UI
> BER = 1e-12
>
> then, given 2Q = 14 for BER = 1e-12:
>
> Jrms = 0.5 / 14 = 0.036 UI
>
> The Rx eye mask width would be:
>
> 1 - 0.8 = 0.2 UI
>
> If I want to scale the Rx eye mask width to BER=1e-15 I would expect I
> need to *grow* the eye mask width by Jrms.
>
> Given that 2Q = 15.883 at BER = 1e-15, then my new eye mask width would
be:
>
> 0.2 + (15.883 - 14) * 0.036 = 0.268 UI
>
> So far so good, assuming I did this correctly. Here's what puzzles me,
> if I adjust my Rx jitter tolerance to accommodate the new Rx eye mask:
>
> Tj = 1.0 - 0.268 = 0.732 UI
> Dj = 0.3 UI
> Rj = 0.732 - 0.3 = 0.432 UI
> BER = 1e-15
>
> and recalculate Jrms:
>
> Jrms = 0.432 / 15.883 = 0.027 UI
>
> The Jrms number has changed, I wouldn't expect this to happen simply
> because I'm extrapolating to a different BER. Can someone please
> straighten me out?
>
> Thanks!
>
> --
> Conrad Herse
>
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