From: Dagostino, Tom (tom_dagostino@mentorg.com)
Date: Tue May 29 2001  09:14:50 PDT
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 2080% points used in IBIS will change the constant also.
Tom Dagostino
Modeling Manager
Mentor Graphics Corp.
SAE
tom_dagostino@mentor.com
5036851613
Original Message
From: Charles Grasso [mailto:cgrassosprint1@earthlink.net]
Sent: Monday, May 28, 2001 11:14 AM
To: Daniel, Erik S.; richard2636@excite.com; silist@silab.eng.sun.com
Subject: RE: [SILIST] : 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: ownersilist@silab.eng.sun.com
[mailto:ownersilist@silab.eng.sun.com]On Behalf Of Daniel, Erik S.
Sent: Monday, May 28, 2001 8:20 AM
To: richard2636@excite.com; silist@silab.eng.sun.com
Subject: RE: [SILIST] : 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 rolloff (e.g. a simple RC
filter) to the 10%90% rise time of the same system with a perfect step
input. For systems with more complicated rolloff characteristics, it is
only an approximation.
A quick derivation is below.
 Erik
==================================================================
Erik Daniel, Ph.D. Voice: (507) 5385461
Mayo Foundation Fax: (507) 2849171
200 First Street SW Email: 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) 5385461
Mayo Foundation Fax: (507) 2849171
200 First Street SW Email: daniel.erik@mayo.edu
Rochester, MN 55905 Web: www.mayo.edu/sppdg/
==================================================================
> Original Message
> From: richard hill [mailto:richard2636@excite.com]
> Sent: Thursday, May 24, 2001 12:32 PM
> To: silist@silab.eng.sun.com
> Subject: [SILIST] : 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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