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When viewing fast pulses on an oscilloscope, why is impedance matching required? I'm not totally clear on why. I'm guessing it is because I don't want 'ringing' and because I want signal integrity.

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You might wish to read electronics.stackexchange.com/a/21788/5156 –  Everyone Nov 27 '12 at 6:42
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up vote 4 down vote accepted

Yes. if you don't match impedances, there will be signal reflections from the unmatched interfaces, and your measurement will be distorted.

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thanks, really appreciate the confirmation –  Dylan Sabulsky Nov 27 '12 at 2:38
    
You're welcome. –  Art Brown Nov 27 '12 at 2:50
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The problem is that a cable (or more technically a "transmission line") has a complicated behavior when it is no longer electrically short. Essentially if the cable lengt comes inthe order of the wavelength of the signals the voltage and current at the ouput are no longer the delayed voltage/current at the input. Instead the cable does a complex impedance transform.

Fortunately, this problem can be solved if the source or(!) the load connected to the cable has the same as the wave impedance of the cable (typically 50 Ohm). Then the cable will act again as a delay line. This is because the cable impedance transform leaves the values of cable impedance unchanged. If the cable is matched at one side it can be replaced by a delay line (if attenuation is neglected).

Note that only one of source or load needs to be matched not to get ringing. Ringing only occurs if both ends are unmatched. You need two reflections to get ringing.

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