@lygra:
I never would have hurt you!..I really apologize if you feel that!
I was only asking you in which way did you measured/show a good shaped 3.5M square wave, nothing all.
Could you post any DSO snapshot of that wave?
I have explained what in the lab I have tried, which instrumentation and how much was the amplitude of the signals. I guess anyone can test it with a quite common scope with 100-200MHz of BW.
The wave I have seen (under several context) is far from being square.
The picture you have attached is partially correct.
It is true that having a perfect signal in the ADC input (better: the track-and-hold input) the equivalent wave will be distorted.
It is also true that never we could sample a signal having a BW over fs/2, because it generates aliasing. That is a fundamental math rule of discrete signals (Shannon). That mean we cannot feed any signal higher than 36MHz (theoretically speaking), let’s say 10-20MHz practically.
Otherwise the equivalent sampled signal would be a mess, without any chance to reconstruct correctly.
This is clear represented by the frequency spectrum that would see the baseband and the lower-modulated band overlapped.
When I write “signal” here, I mean SINE waves, not square or else.
You know that the only “pure” signal is just a sine wave. Any other periodic signal can be thought as composition of harmonics.
In the case, a (perfect) square wave is one of the most “tedious”, because it has only odd harmonics, decaying hyperbolically (so really slowly).
For example, let’s consider a 1MHz perfect square wave.
The main sine wave is at 1MHz, of course, and it is the greater in amplitude.
It has a 3rd harmonic, whose amplitude is about 1/3 the main. Then there is the 5th, having 1/5 of amplitude, etc.
If the scope would had a 15MHz of BW (as HugeMan is going to do), then I would have attenuations over the 13-15th harmonic…my wave would be a good square anyway.
Since I do NOT see that and I see a oddly shaped wave, I’d suppose the BW is much slower.
How to prove it?
Well, simple…that was a typical high-school task.
Just feed the DSO with a 1MHz (or less) sine: no matter what’s the amplitude. On the DSO display I see a good-shaped sine having about 500mVpp.
How is defined the BW? The cut-off frequency is defined as the point where the amplitude falls to 1/sqrt(2)=0.707 the nominal.
That’s good…so step up the input frequency to 2MHz: how is the sine on the DSO display?..It is still a good-shaped sine, but it’s amplitude is about 75% as before.
That is: the BW of the analog section of the DSO is 2-3MHz.
Hope that was clear enough.