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Re: [Phys-l] Oscilloscope bandwith demonstration



John:

Sorry, I assumed a rule without having analyzed first. Thank you very much for your accurate analysis.!

Rob


-----Mensaje original-----
De: phys-l-bounces@carnot.physics.buffalo.edu [mailto:phys-l-bounces@carnot.physics.buffalo.edu] En nombre de John Denker
Enviado el: Martes, 01 de Noviembre de 2011 21:59
Para: Forum for Physics Educators
Asunto: Re: [Phys-l] Oscilloscope bandwith demonstration

On 11/01/2011 02:35 PM, Roberto Carabajal wrote:

I am looking for the math demonstrations of the practice rule that the
bandwidth of an oscilloscope should be at least five times greater than the
higher component of the tested signal, in order to have an error <= +/- 2% .
It is more a math problem that an electronics problem.

I would really appreciate if you can suggest me any notes o textbook about
this.

1) I've never heard that "rule" before.

2) It cannot possibly be true, for several reasons.

3) You're not going to find it discussed in any textbooks.

=======

You are right to inquire about this "rule" ... but it would
be better to ask /whether/ it is true before asking for a
"demonstration" of why it is true.

Perhaps the most obvious reasons why the "rule" cannot possibly
be true is that for a square wave, there is no such thing as
the "higher component of the tested signal". On the other
side of the same coin, the amount of bandwidth you need to
make the signal /look/ nice and square is much greater than
the amount you need in order to get a very good estimate of
the amplitude.

Another reason why the "rule" cannot possibly be true is
that the right answer depends on the details of the response
function of the instrument. Various scenarios include:
*) If the response can be approximated as a two-pole filter
with critical damping (Q=0.5), then at one fifth of the
corner frequency the response is low by 3.85%, as you can
calculate from the damped harmonic oscillator equation.
*) With slightly less damping, say Q=0.6, the response at
that frequency is high by 1.06%, as you can easily calculate.
*) If you are using a 10X probe, that can be well described
as an RC filter (i.e. a one-pole low-pass filter). At the
given frequency, its response will be low by 1.94% all by
itself ... but that is on top of whatever the instrument is
doing.

I suspect the notion of "2%" (at 1/5th of the corner frequency)
is a distant echo of the RC filter response function ... but it
is being woefully misapplied here.

Another reason why the "rule" cannot possibly make sense even
in the case of the RC filter is that there is no uncertainty
involved. For the RC filter, the error is not "+/-" 2%;
the response is 2% low every time, always on the minus side,
never on the plus side.

I assume everybody here understands the transfer function for
an RC filter. If not, it is derived in any book that covers
AC circuits. Similarly the damped harmonic oscillator is
covered in many places.
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