**From:** Steve Corey (*steve@tdasystems.com*)

**Date:** Tue Nov 14 2000 - 13:51:28 PST

**Next message:**Ingraham, Andrew: "RE: [SI-LIST] : Critical Parameter: rise-time or slew-rate?"**Previous message:**Loyer, Jeff W: "RE: [SI-LIST] : Critical Parameter: rise-time or slew-rate?"**In reply to:**Loyer, Jeff W: "[SI-LIST] : Critical Parameter: rise-time or slew-rate?"

Hi Jeff, here's my take on your question:

The answer is, as you might expect, "it depends". Assuming you are

using the TDR to characterize linear interconnects, you are correct in

that the risetime is the most important parameter, and slew rate does

not matter. Think of the simplest case of a monotone sine wave. In

order to double the amplitude and maintain the slew rate, you must halve

the frequency.

The concept of characterizing a system over a specific frequency range,

without additional information on the input signals, is only valid for

linear systems (linear within the region of operation). This is because

they are described by a normalized transfer function ('normalized' means

the amplitude of the input signal drops out, as you have surmised).

Since the transfer function is independent of the input signal, you may

excite the system with nearly any waveform (not only arbitrary

amplitude) and determine the transfer function. The only stipulation on

the above excitation waveform is that it have sufficient frequency

content, which is better guaranteed, as shown above by example, by

matching or exceeding the risetime of a signal which would be used in

practice, and not necessarily the slew rate. Once the transfer function

is known, you may then determine the response of the system to an

"arbitrary" input, which must not contain frequency content that wasn't

present in the excitation used in the prior characterization step.

**However**, if you are exciting a nonlinear system, you must specify

both risetime and amplitude, since the transfer function is dependent on

the amplitude and frequencies of the input signal. In the case of

digital circuit, pretty clearly scaling an input step from 5V down to

500mV will not cause exactly 1/10 the output as the 5V step. By

matching both the risetime and amplitude, you have implicitly matched

the slew rate.

Finally, it is important to remember that matching risetime (and in the

case of a nonlinear sytem, amplitude) will generally only provide a

first-order approximation to your system waveforms, since the shape of

the waveform also contains (typically second-order) frequency content

information. For linear systems, this may be directly calibrated out in

a general fashion (typically done by software post-processing), whereas

for nonlinear systems it remains an open problem.

I hope this is clear -- let me know if it's not.

-- Steve

-------------------------------------------

Steven D. Corey, Ph.D.

Time Domain Analysis Systems, Inc.

"The Interconnect Modeling Company."

http://www.tdasystems.com

email: steve@tdasystems.com

phone: (503) 246-2272

fax: (503) 246-2282

-------------------------------------------

"Loyer, Jeff W" wrote:

*>
*

*> A question came up here that goes something like this:
*

*> If I'm going to buy a filter to slow my TDR pulse down to better mimic the
*

*> bus driver, should I slow it down to match the "slew-rate" or "rise-time"?
*

*>
*

*> For instance, a drive signal going from 1.0V to 1.8V in 0.4nS has a
*

*> slew-rate of 2V/nS, and a rise-time of 0.4nS.
*

*>
*

*> To mimic this driver's slew-rate, I would want my TDR (which has a 0.5V
*

*> voltage swing) to go from
*

*> -0.5V to 0V in 0.25nS. I would specify a 0.25nS rise-time filter.
*

*>
*

*> To mimic this driver's rise-time, I would specify a 0.4nS rise-time filter.
*

*>
*

*> Which is more important, the slew-rate or rise-time?
*

*>
*

*> The first intuitive answer is that slew-rate is the important characteristic
*

*> - that a signal with higher slew-rate has more high-frequency components
*

*> than one with a lower slew-rate. Unfortunately, this implies that our
*

*> "rules of thumb" which talk about rise-time only are assuming a specific
*

*> voltage swing (and therefore a specific slew-rate for that rise-time), and
*

*> valid only for specific voltage swing levels.
*

*>
*

*> The non-intuitive answer is that rise-time is the important characteristic,
*

*> regardless of slew-rate. This leaves the rules-of-thumb intact, but
*

*> violates our gut sense that a signal switching from 0 to 5V in 1nS is much
*

*> more "aggressive" (and sensitive to smaller discontinuities on a
*

*> transmission line, for instance) than one switching from 0 to 1V in 1nS.
*

*>
*

*> My conclusion at this point is that the correct thing to do is to mimic the
*

*> rise-time, not the slew-rate.
*

*> Some simple simulations show that doubling the risetime cuts the ringback in
*

*> half, while doubling the slewrate has no effect.
*

*>
*

*> Does anybody know of a good explanation (more than mere "gut feel" etc.) of
*

*> why the rise-time is the critical parameter, not slew-rate?
*

*>
*

*> Note: at this point, I believe the major flaw in thinking slew-rate is
*

*> important is in the sentence "...a signal with higher slew-rate has more
*

*> high-frequency components than one with a lower slew-rate." I'm pretty sure
*

*> that, when you do the Fourier Transform of a pulse, the amplitude of the
*

*> pulse is merely a constant and comes out of the integration.
*

*>
*

*> Jeff Loyer
*

*> (253) 371-8093
*

*>
*

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*

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**Next message:**Ingraham, Andrew: "RE: [SI-LIST] : Critical Parameter: rise-time or slew-rate?"**Previous message:**Loyer, Jeff W: "RE: [SI-LIST] : Critical Parameter: rise-time or slew-rate?"**In reply to:**Loyer, Jeff W: "[SI-LIST] : Critical Parameter: rise-time or slew-rate?"

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