Does the Shannon theorem not apply when the amplitude of a wave is changed faster than half the time period of the wave? Shannon's version of the sampling theorem states that if a function contains frequencies all strictly less than $B$ hertz, then it is completely determined by giving its ordinates at a series of points spaced $\frac{1}{2B}$ seconds apart.
Now, let us suppose we are talking about just one frequency (say a laser with frequency $B-\varepsilon$ hertz, where $\epsilon$ is an arbitrarily small positive number) with an amplitude that is varied over time for relaying a signal. Let's assume that samples of this wave are taken at the following points of time, measured in seconds:
$ \{0, \frac{1}{2B}, \frac{2}{2B}, \frac{3}{2B}, \ldots \} $.
Suppose the laser amplitude is changed to half its value at $\frac{1.4}{2B}$ seconds and is then doubled back to its original amplitude at $\frac{1.6}{2B}$, wouldn't this information go undetected? It seems to me that the theorem assumes that amplitude does not change inside one wavelet.
Once you start changing the amplitude you are increasing the bandwidth of the signal. That's called &quot;amplitude modulation&quot; and the highest frequency is now the sum of the original frequency and the highest frequency in the modulation signal.
The sampling theorem still holds. You still only need twice the bandwidth but the bandwidth has increased due to the modulation