You really have no clue about digital signal processing and what bandwidth actually means... ![Face Palm :palm:](https://www.eevblog.com/forum/Smileys/default/xfacepalm.gif.pagespeed.ic.EBDwh1hCfo.png)
So, too tired of your bullish behaviour yesterday I take a last try to examine what I've already implied when joining the discussion at the beginning and where you
still got stuck:
Requierements:
- Scope with 100 MHz bandwidth
- Square wave of 7MHz , 50% duty cycle
Let's assume we want at least 5 harmonics:
f0 = 7MHz (1. harmonic)
f1 = 3*7MHz = 21MHz (2. harmonic)
f2 = 5*7MHz = 35MHz (3. harmonic)
f3 = 7*7MHz = 49MHz (4. harmonic)
f4 = 9*7MHz = 63MHz (5. harmonic)
- Since we've got 50% duty cycle we only will get odd multiples of the base frequency.
- Since we're 37MHz away from the 100MHz cut-off frequency we are on the safe side getting all 5 harmonics with no/less than 3dB attenuation.
Concludingly, our 100MHz scope is still sufficient concerning these facts. Are we still friends at this point?
![Wink ;)](https://www.eevblog.com/forum/Smileys/default/xwink.gif.pagespeed.ic.cldandycH0.png)
Now we go on with the ADC and the required sample rate. Nyquist says, at least sample rate has to be 2x of the highest frequency the signal consists of.
So, with f4=63MHz we should have at least:
- fsample = 2*f4 = 126MHz = 126MS/s
The old Agilent I'm using has got 200MS/s, so sufficient for the example above and exactly twice of the scope's bandwidth.
Means also, a 100MHz
sine will be attenuated with 3db at the scope's cut-off frequency and if sampled not at the zero crossings you will even see something on screen.
Of course, that's not sufficient for a
square wave where the harmonics are of higher frequency than 100MHz. In this case - as you also mentioned - you'll only see something
that looks like a sine, since harmonics are missing.
Think, we're still friends at this point, right?
BREAK: all these thing I've implied when joining the discussion, these are basics that everybody should know using a DSO But that's where you've started and what I've alreay passed.
With all that basics in mind, now the point
where I've started this thread:
- 3 scopes: all with 100 MHz bandwidth,
- supposed that the maximum signal frequency we want to use them for is 100MHz (cut-off, 3db attn.)
scope 1 with ADC: 200MS/s
scope 2 with ADC: 1GS/s
scope 3 with ADC: 4GS/s
The 200MS/s ADC is absolutely sufficient concerning Nyquist for a scope with 100MHz bandwidth.
So the question comes up:
what advantage can there be to use a 1GS/s or even a 4GS/s ADC for a 100MHz bandwith limited scope?-> They're completely oversized according the requirements of Nyquist.
-> but, if we don't think of Nyquist at this point and search for further advantages we notice that the gaps between two subsequent samples is
only determined by the ADC's sample rate, so:
- the 200MS/s ADC will generate samples with a time interval of 5ns
- the 1GS/s ADC will generate samples with a time interval of 1ns
- and last but not least the 4GS/s ADC will generate samples with a time interval of 250ps
3x sampled the same signal:
- interpolation leads to smoother signals concerning higher sample rates
- more details are aquired because of smaller gaps the higher the sample rate, independet of Nyquist.
These facts you can try yourself with any DSO, are mentioned in nearly any app. note from Agilent, Tek, Hameg, ... whatever
and are even part of the books about the basics of how to measure right with DSOs.
Don't know, why it's so difficult to understand.
Believe me, my friend, I'm busied with signal processing more than 20 years now, especially the hard cases where disturbed signals
have to be processed, but the issue we're talking about in this thread are of quite common nature, nothing special.
Sometimes it might be useful trying to understand what other participants want to explain before keeping them for stupid.
However, have a nice day,
Gunb