Is my math correct (its been a while)
20MHz across the width of the screen =
= 10devisions x 2uS = 20uS
now convert the measurement into freq
1/20uS =20Mhz
I think the buyer's idea is to fit one full cycle of the maximum
frequency in one cm of screen. If the minimum timebase setting is
2us/cm, then the max usable frequency is 500kHz. If your scope has x10
horizontal magnification, then the max frequency would be 5MHz. This
is what the buyer probably considers to be the "bandwidth", and that's
how I would interpret oscilloscope frequency specs. Having said that,
a more correct description of the 5MHz spec would probably be the
"maximum usable displayable frequency". The term "bandwidth" is
probably more correctly applied to the frequency response of the
vertical amp(s). A vertical frequency response of 15MHz would make
5MHz pulses look reasonably square.
I guess you and the buyer both have valid interpretations, but if it
were me, I'd expect a refund. Having said that, if we were to use the
buyer's interpretation, then his expected minimum timebase setting
would have been 0.067 us/cm, which is clearly impractical. This should
have rung warning bells for him.
- Franc Zabkar