[SI-LIST] Re: Mode conversion question
- From: Joseph.Schachner@xxxxxxxxxx
- To: Doug Smith <doug@xxxxxxxxxx>
- Date: Wed, 2 Nov 2011 12:10:19 -0400
Since you wrote you have a "differential trace pair", I am pretty sure
we're not talking about microwave propagation modes. We're talking about
differential and common mode.
Differential mode is not just like representing the signal as a orthogonal
vectors, which is a purely mathematical construct that makes no difference
in real world effects. In a differential trace pair the signal should be
180 degrees apart (not 90). The two sides are not orthogonal. More
important: there are real world consequences for violating the requirement
that the two signals are exact inverses of each other.
The benefit of differential trace pair is that the equal and opposite
signals generate equal and opposite electromagnetic fields. These fields
sum to zero - the farther away, the better that approximation. Also, any
impinging electromagnetic field causes common mode effects, which it may be
possible to ignore at the receiver (subtracting the two sides of the
differential signal will yield the differential signal and eliminate common
mode). To get these benefits it should be true that we can draw a line
across the differential trace pair, perpendicular to the traces, anywhere
along its path and the wave front of a state change on the differential
pair will reach that line at exactly the same time on each side of the
pair.
If there is a timing offset then the two wave fronts are not coincident.
For the length corresponding to that timing offset, the electromagnetic
field is not cancelled. That can be a source of crosstalk. If the timing
is not realigned at the receiver then the transition times seen at the
receiver are longer - from the start of the transition on the early side to
the end of the transition on the late side. If the timing is realigned at
the receiver, then any disturbance picked up from a field impinging on the
differential trace pair will not be pure common mode, because where it
affected the traces the timing was not aligned. Shifting timing back to
aligned separates the effect induced on the two sides of the pair by that
timing offset ... so when the receiver subtracts the two signals it does
not cancel the induced effect.
So, how "slight" a timing offset can be ignored? I would only venture a
guess, that it will correspond to a small fraction of the transition time
of the signal.
It is true that a differential pair presents a different impedance to
differential mode signals than to common mode signals. If the signals are
aligned we won't have any common mode signal (and if any are induced, they
would be ignored at the receiver) so I didn't even talk about this, above.
Just keep the signals truly differential so that you get the benefit of a
differential trace pair.
--- Joe S.
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|11/01/2011 11:38 PM
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|[SI-LIST] Re: Mode conversion question
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This is similar to the way we can express any vector as the sum of two
orthogonal vectors. There is no difference one can discern in the action
of the vector no matter how one thinks of it. But expressing the vector
as the sum of two orthogonal vectors usually makes calculations easier.
Not sure if this made things clearer or muddied the waters.
Doug
On 11/1/11 1:11 PM, Orin Laney wrote:
> The two modes have different impedances, velocities of propagation, and
> propensities to radiate.
>
> Orin Laney
>
> -----Original Message-----
> From: si-list-bounce@xxxxxxxxxxxxx [mailto:si-list-bounce@xxxxxxxxxxxxx]
On
> Behalf Of Doug Brooks
> Sent: Tuesday, November 01, 2011 11:01 AM
> To: si-list@xxxxxxxxxxxxx
> Subject: [SI-LIST] Mode conversion question
>
> Assume I have a differential trace pair. Assume there is a slight offset
in
> the two signals.
>
> My understanding of mode conversion is that the signal pair will become
two
> components --- an odd mode component and an even mode component.
>
> In trying to understand WHY that happens I have come to believe there is
no
> physical change in the signals. What we do is MODEL the signals as two
> separate components, an odd mode component and an even mode component,
which
> combine together to equal the actual signal. Thus mode conversion is a
> mathematical (and physical) model that allows us to deal with the
analysis,
> rather than an actual physical phenomenon.
>
> Is my understanding correct here, or am I way off base?
>
> Thanks for your help.
>
> Doug Brooks
>
>
>
> Check out our resources at http://www.ultracad.com
>
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