RE: [SI-LIST] : Frequency based on rise time for drivers

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From: Peterson, James F (FL51) (james.f.peterson@honeywell.com)
Date: Thu May 31 2001 - 13:43:03 PDT


In regards to "where did .35/Tr come from?" :
 
Appendix C of "High-Speed Digital System Design" (Hall, Hall, McCall) has an
excellent derivation of .35/Tr.
 
I'll attempt to give it here (but you must pardon my clumsy fonts):
 
1) Tr = T(10%-90%)
2) T90% = 2.3RC
3) T10% = 0.105RC
4) T(10%-90%) = (2.3 - 0.105)RC = 2.195RC
5) F(3db) = 1/(2piRC) , therefore : RC = 1/(2piF(3db))
6) if RC in #4 (above) is substituted with #5 (1/(2piF(3db))) then....
7) T(10%-90%) = 2.195/(2piF) = 1.09/(piF) ~ .35/F , therefore F =
.35/T(10%-90%)
 
I think I got it right (again, credit goes to the above reference text).
 
 

Jim Peterson
james.f.peterson@honeywell.com
Honeywell Space Systems
M/S 934-5
13350 U.S. Hwy 19 N.
Clearwater FL, 33764-7290
Office : 727-539-2719
Fax : 727-539-2183

-----Original Message-----
From: Dagostino, Tom [mailto:tom_dagostino@MENTORG.COM]
Sent: Tuesday, May 29, 2001 12:15 PM
To: 'Charles Grasso'; Daniel, Erik S.; richard2636@excite.com;
si-list@silab.eng.sun.com
Subject: RE: [SI-LIST] : Frequency based on rise time for drivers

The constant, in the example of .35, is dependent on the characteristics of
the system. Tek's derivation, if I remember correctly, assumed a 2 pole
Gaussian response. Different filter responses (different number of poles
and different damping factors) will produce a different constant in the
equation. Also, Tek assumes risetimes measured from the 10 to 90% points.
Changing to the 20-80% points used in IBIS will change the constant also.

Tom Dagostino
Modeling Manager
Mentor Graphics Corp.
SAE
tom_dagostino@mentor.com
503-685-1613

-----Original Message-----
From: Charles Grasso [ mailto:cgrassosprint1@earthlink.net
<mailto:cgrassosprint1@earthlink.net> ]
Sent: Monday, May 28, 2001 11:14 AM
To: Daniel, Erik S.; richard2636@excite.com; si-list@silab.eng.sun.com
Subject: RE: [SI-LIST] : Frequency based on rise time for drivers

I belive Tek published this formula in some of their old scope
probe information many years ago as well.
There is also, in EMC circles, a sort of Fake Fourier
Transorm that identifies TWO knee frequencies to capture the
harmonic "envelope" of a square wave. The first knee is
i/piT where T is the period, at this point the envelope falls
at 20dB/decade to 1/piTr where Tr is the rise time. After this
second point the envelope falls at a rate of 40dB/decade.

Charles Grasso
Ansoft Corporation

-----Original Message-----
From: owner-si-list@silab.eng.sun.com
[ mailto:owner-si-list@silab.eng.sun.com
<mailto:owner-si-list@silab.eng.sun.com> ]On Behalf Of Daniel, Erik S.
Sent: Monday, May 28, 2001 8:20 AM
To: richard2636@excite.com; si-list@silab.eng.sun.com
Subject: RE: [SI-LIST] : Frequency based on rise time for drivers

Richard-

This formula is NOT derived assuming a sinusoid. It relates the 3 dB
bandwidth of a system limited by a single pole roll-off (e.g. a simple R-C
filter) to the 10%-90% rise time of the same system with a perfect step
input. For systems with more complicated roll-off characteristics, it is
only an approximation.

A quick derivation is below.

                                        - Erik

==================================================================
Erik Daniel, Ph.D. Voice: (507) 538-5461
Mayo Foundation Fax: (507) 284-9171
200 First Street SW E-mail: daniel.erik@mayo.edu
Rochester, MN 55905 Web: www.mayo.edu/sppdg/
==================================================================

 Assume the following circuit:

              R
  Vin o----\/\/\/------+----o Vout
                       |
                       |
                      --- C
                      ---
                       |
                       |
                       V (GND)

FREQUENCY DOMAIN:
-----------------
                  1 1
Vout = Vin * ----------- ; |Vout|^2 = |Vin|^2 * ---------------
             1 + j w C R 1 + w^2 R^2 C^2

3 dB bandwidth => Pout = 1/2 * Pin => |Vout|^2 = 1/2 * |Vin|^2 => w_3dB

= 1/(R C)
                                                                    => f_3dB

= 1/(2 Pi R C)

TIME DOMAIN:
------------

Assume Vin is a step input from 0 volts at t<0 to 1 volt at t=>0. Then

   Vout = 1 - exp[-t/(R C)]

Find times for which Vout = 10%, 90% * 1 Volt:

0.1 = 1 - exp[-t_10% /(R C)] => t_10% = -(R C) * log(0.9)

0.9 = 1 - exp[-t_90% / (R C)] => t_90% = -(R C) * log(0.1)

=> t_rise = t_90% - t_10% = (R C) * (ln 0.9 - ln 0.1) = (R C) * ln 9

=> ln 9 0.35
      t_rise = ------ = ----
               2 Pi f f

==================================================================
Erik Daniel, Ph.D. Voice: (507) 538-5461
Mayo Foundation Fax: (507) 284-9171
200 First Street SW E-mail: daniel.erik@mayo.edu
Rochester, MN 55905 Web: www.mayo.edu/sppdg/
==================================================================

> -----Original Message-----
> From: richard hill [ mailto:richard2636@excite.com
<mailto:richard2636@excite.com> ]
> Sent: Thursday, May 24, 2001 12:32 PM
> To: si-list@silab.eng.sun.com
> Subject: [SI-LIST] : Frequency based on rise time for drivers
>
>
> I have often seen people estimating the frequency of a
> driver based on the rise time for the driver using the
> formula Freq = .35/Tr (This seems to be true for a
> perfect sinusoid). There can be two different
> drivers outputing signal at the same frequency but
> different edge rates, won't this assumption be
> invalid. Some times there is a big difference in the
> edge rates of drivers running at the same frequency.
> If we use the above formula and estimate the edge rates
> using the frequency for the two drivers we will get
> similar no.s.
>
> Any comments.
>
> Thanks in advance,
> Rich
>
>
>
>
>
> _______________________________________________________
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