From: Daniel, Erik S. (Daniel.Erik@mayo.edu)
Date: Mon May 28 2001  07:19:39 PDT
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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