I couldn't find it in the books. I mean the fixed freq. value on the oscilloscope.
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\$\begingroup\$ A video to go with these answers: Basics of Analog Oscilloscope Bandwidth \$\endgroup\$– m.AlinDec 11, 2012 at 15:49
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4 Answers
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I'm not sure what the distinction is between the two choices in your question, but the real answer is that the gain of the amplifier(s) inside the scope is "flat" up to that frequency. In other words, the voltage measurements you make on the screen will be "accurate" up to that frequency. Above that frequency, the values will appear smaller than they actually are.
But the truth is that the specified frequency is usually the "-3 dB" point of the gain curve, which means that measurements at that frequency are already off by 30% — the waveform is only 0.707× the height that it should be. If you want more accurate measurements, you need to look for the -1 dB frequency (10% accuracy) or even the -0.1 dB frequency (1% accuracy).
But in most cases, you're not really making precision measurements with a scope; instead, you're looking for qualitative changes in the waveforms, etc. As long as you're generally within the specified bandwidth, you should be good to go.
answered Dec 11, 2012 at 1:57
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3\$\begingroup\$ Warning: objects behind graticule are larger than they appear. :) \$\endgroup\$– KazDec 11, 2012 at 6:17
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\$\begingroup\$ The difference between the two, it seems to me, is a factor of two, thanks to the nyquist frequency. \$\endgroup\$ Dec 11, 2012 at 16:04
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\$\begingroup\$ Note that the OP didn't specify digital scope, and in any case, the bandwidth spec refers to the analog front end of a digital scope. \$\endgroup\$ Dec 11, 2012 at 16:52
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A 50MHz oscilloscope has a bandwidth of 50MHz. What it can plot will depend on its samples/second value (assuming vector plotting, first order interpolation will fill in more than what is actually measured) . This is explained a little further in this related question.
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For an analog scope it means that a 50MHz sine will be attenuated 3 db (down to 70% in voltage terms). A digital signal of the same frequency contains components of much higher frequencies, so it will be severely distorted. A ballpark estimate I read somewhere is that you should not trust the picture of (non-sine) signals higher than 1/3 of the bandwidth.
For a digital scope it means 50M samples/second. Hence when you see a signal change from 0 to 1 that change could have happened anywhere between those two sample moments, hence your timing uncertainty is +/- 10 ns.
answered Dec 11, 2012 at 9:07
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3\$\begingroup\$ For a digital scope it could mean either; a digital scope specification should include 2 figures - say, 200MHz sampling, 50MHz bandwidth (or 50MHz sampling, 15MHz bandwidth) \$\endgroup\$ Dec 11, 2012 at 11:19
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Apart from the accuracy in the waveform shown, you'll notice that triggering on signals with higher frequency will get increasingly difficult.
Good/stable triggering is essential to obtain a stable/any image on screen.
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\$\begingroup\$ N.B. Trigger bandwidth is usually specified separately. \$\endgroup\$ Dec 11, 2012 at 16:50