This is what you can expect to see if you cut a square wave off at 10
times its frequency:

http://www.wolframalpha.com/input/?i=sin(x)+%2B+sin(3x)%2F3+%2B+sin(5x)%2F5+%2B+sin(7x)%2F7+%2B+sin(9x)%2F9
(If the link is too long, go to http://wolframalpha.com and enter the
equation sin(x) + sin(3x)/3 + sin(5x)/5 + sin(7x)/7 + sin(9x)/9  )

This is what you can expect to see if you cut it off at 5 times its frequency:

http://www.wolframalpha.com/input/?i=sin%28x%29+%2B+sin%283x%29%2F3+%2B+sin%285x%29%2F5
(If the link is too long, go to http://wolframalpha.com and enter the
equation sin(x) + sin(3x)/3 + sin(5x)/5 )

Neither representation is ideal, and if you're trying to measure
anything important about the square wave (rise time, fall time,
ringing, phase, etc) then you're not going to be happy.

-Adam

On Wed, Nov 18, 2009 at 10:57 AM, Alan B. Pearce
<Alan.B.Pearce@stfc.ac.uk> wrote:
>>With an 8MHz bandwidth, I would not expect a reasonable
>>square wave image above 1.5MHz or so (which would only
>>include the fundamental, first, and third harmonics).
>
> I seem to remember a rule of thumb that a 'scope needed to have a bandwidth
> of ~10x the frequency of the square wave you were attempting to look at, so
> I would be cutting your figure in half ...
>
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