[SI-LIST] Re: Rise-time of Cascaded Lossy T- Lines
- From: "Dr. Howard Johnson" <howiej@xxxxxxxxxx>
- To: <si-list@xxxxxxxxxxxxx>
- Date: Tue, 25 Feb 2003 10:10:09 -0800
Dear Andrew,
Your assertion is right on target that the RSS
(square-root-of-sum-of-squares)
approximation for the risetime of a cascaded linear system
is ... just an approximation.
I can fill in some of the mathematics behind the
imperfections in this approximation.
If you are not mathematically inclined, you might want to
step out for a quick cup of tea,
and then check back in at the end to see the cases where RSS
doesn't work very well.
The RSS risetime approximation stems from a general property
of the convolution
operator (when cascading linear systems their impulse
responses convolve, so the
properties of the convolution operator therefore determine
how the complete
system behaves):
[A] When impulse responses convolve, their variances add
v[overall] = v[1] + v[2] + ... v[n]
This property applies to any impulse response that is
strictly positive, meaning
that the associated step response is monotonic. It's
provable mathematically
from first principles using the definitions of variance and
the convolution operator
(although I will admit I am loath to reconstruct the proof
here).
The convolution property relates to the RSS approximation
through the
following chain of logic (High-Speed Digital Design, p.
400):
(1) Variance is the square of standard deviation
(2) The standard deviation of an impulse response is
proportional to its width
(3) The width of an impulse response is proportional to the
rise time of its corresponding step response
This chain of reasoning suggests that the rise time of a
step response is proportional
to the standard deviation of its associated impulse
response. Combining this conclusion
with [A] you can derive RSS:
trise[overall]^2 = trise[1]^2 + trise[2]^2 ... +
trise[n]^2
where the operator "^2" means "squared"
To see what goes wrong in the chain of reasoning 1-3, let's
get specific.
Suppose you have two systems that you propose to cascade.
Looking at the step responses, measure the 10-90% rise times
t1 and t2 (or 20-80%, or
whatever interests you).
Now take the derivative of each step response (these will be
the impulse responses).
If the step responses are monotonic this gives you two
strictly-positive impulse responses.
Scale the impulse responses so the total area under each
impulse response is unity
(i.e., these are the derivatives of unit steps). Measure the
variance of each pulse,
v1 and v2, and take the square root of the variances to get
the standard deviations, s1 and s2.
If the impulse responses are strictly positive it will
always be true that
v[overall] = v[1] + v[2]
and also
s[overall]^2 = s[1]^2 + s[2]^2
What goes wrong has to do with steps (2) and (3). The
relation between standard deviation
and "pulse width" has to do with how you define that width.
For example, suppose you define the
impulse-response width at the points comprising 10% and 90%,
respectively, of the total area
under the impulse response (corresponding to a 10-90%
risetime definition). The relation
between the standard deviation width measurement and the
10-90% width measurement
now depends on the exact shape of the impulse response
curve. Looking at the table on
p. 401 of my book, a single-pole RC relaxation with a
standard deviation of 1.00 produces
a 10-90% risetime of 0.877. A gaussian impulse response with
the same standard deviation
of 1.00 produces a 10-90% risetime of 1.02.
Same standard deviation, different 10-90% risetimes.
If all three of your waveforms (the first system, the second
system, and the cascade) have the
same impulse-response shape, then all three will have the
same relation between standard
deviation and risetime, and the RSS formula works perfectly.
This happens when the two
sub-system components have a Gaussian response (albeit with
different standard deviations).
The result in this case always turns out to be Gaussian, and
so the RSS formula applies
exactly. A gaussian impulse response is the ONLY shape for
which the impulse response of
the cascade has the same shape as the impulse response of
the subsystems.
If the two subsystems are not Gaussian, then the shape of
the output impulse response won't
precisely match the shape of the impulse responses of the
two subsystems (even if the
two subsystems are identical), and the RSS approximation
becomes no longer perfect.
RSS is still somewhat useful, though. You can see in my
example showing the numbers for an
RC relaxation and a gaussian impulse response the extent of
the imperfection in the relation
between standard deviation and rise time.
Thinking about cables, you can make a simlar argument in the
frequency domain. When you
cascade two sections of cable it doubles (in dB) the
attenuation at all frequencies. Finding the
new system risetime is related to asking this question,
"Starting from the -3dB point on a single cable, how far
do I have to back down in frequency
to get to the 3dB point in the new cascaded system?"
That question is exactly the same as this:
"Starting from the -3dB point on a single cable, how far
do I have to back down in frequency
to get to the -1.5dB point on the same single cable?"
Obviously, the answer to that has something to do with the
shape of the frequency response
of the cable. If the cable response is gaussian [H(f) =
exp(-k*f*f)] then the answer is that you
back down by 1/sqrt(2) , which corresponds precisely the RSS
formula.
CASES WHERE RSS DOESN'T WORK VERY WELL
If the cable response is dominated by the dielectric-effect
mechanism (as are many long pcb traces)
then the response looks like [H(f)=exp(-k*f)], for which a
doubling of the trace length
requires that you back down by a factor of two in frequency.
If the cable response is dominated by the skin-effect
mechanism (as are many coaxial cables)
then the response looks like [H(f)=exp(-k*sqrt(f))], for
which a doubling of the cable length
requires that you back down by a factor of four in
frequency.
It used to be that when working with oscilloscopes the
responses of the probe and vertical amplifier
were very nearly gaussian, and so the RSS formula worked
beautifully. In modern scopes with
bandwidths above 4 GHz it seems our scope manufacturers have
taken to using digital means
of "flattening" the response so it's no longer gaussian and
the old RSS formula doesn't necessarily
work any more. Agilent just did a good presentation of this
at DesignCon.
NOTE TO ANDREW, the imperfections in the RSS approximation
for cables aren't due to the fact
that the cable sections interact, it's just that the
frequency response isn't exactly gaussian. You
could hook buffer amplifiers between the sections and get
the same result.
Best regards,
Dr. Howard Johnson, Signal Consulting Inc.,
tel +1 509-997-0505, howiej@xxxxxxxxxx
http:\\sigcon.com -- High-Speed Digital Design articles,
books, tools, and seminars
-----Original Message-----
From: si-list-bounce@xxxxxxxxxxxxx
[mailto:si-list-bounce@xxxxxxxxxxxxx]On Behalf Of Ingraham,
Andrew
Sent: Monday, February 24, 2003 12:40 PM
To: si-list@xxxxxxxxxxxxx
Subject: [SI-LIST] Re: Rise-time of Cascaded Lossy T- Lines
The RSS approximation of risetimes of cascaded elements is
just that ...
an approximation. It seems to work pretty well in most
cases, but it is
not exact.
One assumption that I believe goes into this approximation,
is that each
cascaded section is effectively isolated from the others.
In other
words, connecting component B to component A doesn't affect
the response
of component A itself. The frequency response of the whole
system would
be the product of each individual section.
But that is not the case when cascading passive elements
like lossy
t-lines without repeaters. I think this is where the
assumption breaks
down.
> I've been reading Mr. Eric Bogatin's "Lossy Transmission
Lines: Plain
> and Simple" paper and have a question, which I hope those
of you who
> have also read it could answer (The same reasoning I
present here also
> follows, to my understanding, from Mr. Howard Johnson's
"Black Magic"
> book, App. B):
>
>
>
> On page 21 of his paper, Mr. Bogatin presents the
> Root-of-Sum-of-Squares
> equation for calculating the rise-time at the end of a
lossy
> transmission line, as a function of the driver rise-time
and the 3-dB
> rise-time of the interconnect.
>
>
>
> It seems from this paper as if one can cascade several
interconnects
> and
> get the total effective rise-time of the whole
interconnect by using
> the
> equation as follows: Take the individual 3dB-rise-times of
the
> individual interconnects and take the square-root of their
sum of
> squares.
>
>
>
> But it seems there is a problem with this use of the
equation. One can
> take a long interconnect, and calculate its effective
rise-time using
> 2
> methods: as an un-divided interconnect, or sub-divide it
to (e.g.) 2
> identical halves, and do the arithmetic again. Using the
2nd method
> one
> gets a result which is (1/square-root of 2) of the
"undivided"
> calculation rise-time.
>
>
>
> This is a strange result, so I must be missing something,
but what?
>
>
>
> Thanks for anyone who can help
>
>
>
> Itzhak Hirshtal
>
> Elta Systems
>
> Israel
>
>
>
>
>
>
>
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