If you look at (for ex.) a 10% AM modulated signal, but with the carrier above max.frequency, you will see the unmodulated signal."
I think this only applies to digital oscilloscope, and the key point is the samples/second not the bandwith directly.
Sorry,that doesn't work,either!
As an example,let's consider a 10MHz carrier
fc,modulated by a 1kHz signal,
fmAt the output of the modulator will appear,: The carrier,=
fc the lower sideband (LSB)=
fc-fm the upper sideband (USB)=
fc+fm fm is also present,but is removed with a high pass filter.(HPF)
The frequencies now present are: carrier at 10MHz
LSB at 9.999MHz
& USB at 10.001MHz
An Oscilloscope,being a wideband device, sees the result of this as the familiar modulation envelope.
If we now sample this signal at an 8MHz rate,for example,we are effectively mixing our three signals with a new signal ,which we can call
f(sample),or for short,
fs.
The result which goes to the digital part of the 'scope,is:
fc-fs(fc-fm)-fs,and
(fc+fm)-fsThe sum signals are also produced,but are removed by the LPF effect of the sampling circuit
The signal which is displayed on the screen is a 2MHz carrier,with sidebands at
1.999MHz & at 2.001MHz .
The 'scope display therefore still shows the classic envelope shape.
If the frequency
fs is brought very close to
fc,the resulting carrier frequency becomes small compared to that of
fm,& the classic envelope is no longer evident.