0
$\begingroup$

i just want to know if i use a sampling frequency of 100-110Hz and get a useful signal frequency of 50Hz (because of Nyquist–Shannon sampling theorem), is the period 1/50Hz or 1/100Hz?

I've been told that it should be 1/100Hz but since my useful signal frequency is 50Hz, why should it not be 1/50HZ

$\endgroup$
4
  • 3
    $\begingroup$ Sorry, I don't understand. What do you mean by $F$? Is it sampling frequency or signal frequency? And what do you mean by "period"? Is it sampling period or signal period? $\endgroup$
    – ZR Han
    Feb 24, 2022 at 11:28
  • $\begingroup$ I'm taking data from an acceloremeter, in the datasheet, it says that if you want to get 50Hz useful signal frequency, you need 100Hz sampling frequency, what i want to know is what is the period for my signal, is it 1/100Hz, or 1/50Hz? $\endgroup$
    – B.Adlane
    Feb 24, 2022 at 12:09
  • $\begingroup$ If your signal has a frequency of 50 Hz, then its period is 1/50 s. Frequency is cycle per second and period is second per cycle. That’s definition. So $T=1/f$ always holds. $\endgroup$
    – ZR Han
    Feb 24, 2022 at 12:48
  • $\begingroup$ If you sample at $f_s=100$, and if there is a sine wave at the device input with frequency $f_0 < 50$, then that signal will be captured properly. Its period is $T_0=1/f_0$ -- independent of $f_s$. If $f_0 \geq 50$, then there will be aliasing, as explained by Dan in his answer below. $\endgroup$
    – MBaz
    Feb 24, 2022 at 14:41

1 Answer 1

4
$\begingroup$

To understand sampling and aliasing I think a useful analogy is a strobe light on a bicycle wheel. The rate of rotation of the bicycle wheel in revolutions per second represents the frequency of an example single frequency waveform in Hz. The rate of the strobe light represents the sampling frequency. Consider how we can make the wheel appear to stop moving if the strobe light frequency matches the rotation rate of the wheel (in fact for those familiar with car repair, this is how engine timing can be adjusted), or even appears to start rotating backwards if the strobe light is slightly lowered in frequency. This is "aliasing" and is a foundation of Nyquist's criteria for sampling rate and signal frequencies.

The first thing is to clearly distinguish in our heads the difference between the "frequency" of the signal being sampled and the "frequency" of the sampling rate: the signal being sampled is best described having a frequency in terms of cycles/second, while we can distinguish the sampling rate as having a frequency in terms of samples/second. Frequency in general is the rate of something and applies to both of these cases, but clarifying this may help avoid some confusion. (And then once that confusion is gone, let me throw in that in discrete time processing we often divide the two to have signals in units of normalized frequency given as cycles/sample.... if that is again confusing, forget I said anything in these parenthesis-- but if you want to ponder further after you see the details below, consider how many cycles or revolutions occur for every sample; in the 1 Hz signal sampled at 100 Hz, this is 1/100 cycles/sample).

Consider first what we would see if we had a 50 Hz strobe light and a 1 Hz wheel, as depicted in the first graphic below. I put a piece of red tape on the rim so that we can observe the rotation, and each of the dots represents the position of that mark on the rim as we would see it after each flash of the strobe light. The solid red dot is what we see at the start, and the blue dot is the second sample. This makes sense when we consider the wheel is spinning at 1 cycle every second, and in that duration of time the strobe light would have flashed 50 times, consistent with the 50 dots I indicated. More importantly is to consider as time moves forward where each dot will be, and in this case we see it advancing counter-clockwise relatively slowly (specifically $2\pi/50$ radians per sample).

1 Hz signal

Now consider what this would look like if we increased the rate of rotation to 49 Hz, and then again to 51 Hz:

49 and 51 Hz signal

Note that I specifically included a sign to denote the direction of rotation (with a counter-clockwise rotation representing a "positive" frequency). The point is, once sampled, we have no way of knowing the actual true rate of rotation or the direction. If the wheel was spinning the other way at -51 Hz instead of +49 Hz as above, we would see the exact same results after each flash of the strobe light! When different signals (at different frequencies or rates of rotation) produce the exact same results in their digitized samples, this is referred to as aliasing.

-51 and -49 Hz signal

Consider one more case where the wheel is now rotating faster than the sampling rate of the strobe light, in the same direction, compared to our original case:

1 Hz and 101 Hz signal

Here we see, that once sampled, we cannot distinguish the two cases. The 101 Hz case is a direct alias of the 1 Hz case - both will produce identical results. When we sample at 100 Hz, the "unique span" of possible rates of rotation (and direction in this case of a wheel) is +/- 50 Hz; any $f$ faster than that would alias to a similar result at $f-f_s$ where $f$ is the rate of rotation and $f_s$ is the sampling rate.

$\endgroup$
1
  • 3
    $\begingroup$ Nice illustrations, Dan, as ever. :-) $\endgroup$
    – Peter K.
    Feb 24, 2022 at 14:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.