# [SI-LIST] Re: Question of taking measurement. Thanks.

• From: "Tom Dagostino" <tom@xxxxxxxxxxxxx>
• To: <a.ingraham@xxxxxxxx>, <si-list@xxxxxxxxxxxxx>
• Date: Sat, 10 Dec 2005 16:25:19 -0800

```As Andy points out it depends on what you are doing.  If you are measuring
amplitude then, depending how accurately you want to measure it you may even
be able to measure a frequency past the -3dB point of the scope and
calibrate out the scope.  If you are measuring risetime you can always
remember that the displayed risetime of a signal is square root of the sum
of the squares of all band limiting elements between the measured signal and
the scope's display.  For example the probe may have 1GHz bandwidth, the
scope 5GHz and the displayed signal 500 psec.  So

measured = SQRT(probe^2  + Scope^2  + DUT^2)
500psec = SQRT(350psec^2 + 80psec^2 + DUT^2)

Solve for DUT.  I'm assuming you are measuring 10 to 90% risetimes and BW =
0.35/Tr.  You can do a plot of the error assuming two systems, the DUT and
the Scope. If both signals have the same risetime the displayed risetime
SQRT2 of the signal or 40% off.

Mathematically you can characterize the response of the scope and then
deconvolve its response from the acquired data.

Tom Dagostino
Teraspeed(R) Labs
13610 SW Harness Lane
Beaverton, OR 97008
503-430-1065
tom@xxxxxxxxxxxxx
www.teraspeed.com

Teraspeed Consulting Group LLC
121 North River Drive
Narragansett, RI 02882
401-284-1827

-----Original Message-----
From: si-list-bounce@xxxxxxxxxxxxx
[mailto:si-list-bounce@xxxxxxxxxxxxx]On Behalf Of Andrew Ingraham
Sent: Saturday, December 10, 2005 11:08 AM
To: si-list@xxxxxxxxxxxxx
Subject: [SI-LIST] Re: Question of taking measurement. Thanks.

>       To get accurate results, the bandwidth of the oscilloscope must be
> 2X+ of the signal. (Please correct me if I'm wrong. Thanks.)

Technically, in theory, the bandwidth only needs to be 1X of the signal
frequency.

BUT, ...

Most of us are looking at digital signals, and the "signal frequency" in
that case is not the fundamental signal switching frequency, but a harmonic
of it.  How high that harmonic is (5th, 21st, etc.) depends on the edge
rates (which determine harmonic content) and how accurately you want
them reproduced on your scope.  Fully accurate results require the scope's
bandwidth to equal or exceed the highest harmonic present in your signal.

Not only that, but no scope has perfect response out to its specified
bandwidth.  It may be "down 3dB" at that point, making the amplitude 30%
inaccurate there.  And its phase response may be nonlinear too, which
affects the shape of displayed digital signals.

So a bandwidth of 2X the highest signal frequency is a nice margin.  But
even then, there is some inaccuracy.

You may be confusing the fact that a digitizer (which might be in a digital
sampling scope) must have a sampling rate of at least 2X the signal
frequency.  If it isn't, aliasing results and the displayed waveform may
look totally wrong.  But sampling rate is not bandwidth.

A sometimes neglected characteristic is the bandwitch of the scope probe.
It combines with the scope's own bandwidth to result in a net bandwidth that
is lower than either one alone.

And then there's the effect of connecting the scope probe to the circuit
under test.

And the effect of the ground lead.  A good probe with a long ground lead is
wasted bandwidth because the loop inductance may prevent an accurate signal
from ever reaching the probe.

> So if I know that the bandwidth of my scope is not enough compared to the
> signal that I want to measure, is that any way I can still get accurate
> results?

I doubt it, unless you know exactly how your scope behaves (i.e., if you
calibrate it yourself with a good signal source).  Even then, ...

Regards,
Andy

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