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That is more or less what happens in a (monochrome) camera only in 2-d (or 3-d if you like). I think you can think of it as a continuous time convolution that happens to be sampled at some multiple >= 1 times the kernel width. Ie a sinc filter in the freq domain. As long as the main lobe is within your passband you should be able to flatten using an ...

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Take your sine wave at $f = 5\mathrm{kHz}$; $x(t) = \sin 2 \pi f\, t$. If you sample it at $F_s$ then you get a result $y[k] = \sin \frac{2 \pi f}{F_s} k$*. Because of the trigonometric identity $\sin \theta = \sin (2\pi + \theta)$, you can't tell the difference between sine waves at $f$ and $f \pm nF_s$. Because of the trigonometric identity $\sin \theta ... 0 You're right, in the real world, 2x is not enough to capture a sound at a given frequency accurately. In your chart, only the first one would sound like a sine wave. As you approach the Nyquist frequency, you'll create a siren like sound and when you reach the exact frequency you'll record a pulse-wave approximation of a sine wave at an amplitude that will ... 1 There is aliasing at 3 kHz and 4.5 kHz. The 2 kHz does not, but since 2 kHz and 3 kHz are equally far apart from the 2.5 kHz Nyquist frequency, they look similar. Same with why 0.5 kHz and 4.5 kHz look similar. 0 From the Nyquist sampling theorem, the minimum required sampling rate to avoid aliasing is$2f_m$, where$f_m\$ is the maximum frequency of the message signal. Can you please add some plots for the question description to view the results.

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Zigbee is primarily designed to be cheap, not fast or efficient. Maximum data rate is 250kb/s so 2 MHz sample rate is more than enough to capture any data signal. The "2MHz bandwidth" number mainly refers to the RF channel layout. Zigbee channels are 2 MHz wide and spaced 5 MHz apart. This primarily determines what RF filtering needs to happen to ...

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