If your MP100 is set to sample at 100 sps, what is the maximum frequency that can be detected in the analog signal?

If the peak of a waveform is followed 500 ms later by a trough, what is the frequency of the waveform in Hz?

Did you end up figuring out the answer? I would also like to know what it is :)

for the second one, is it not:

500ms= 0.5s
then frequency= 1 cycle / 0.5s
frequency= 2Hz

isn't it 1Hz, because it's from peak to trough, which is not one cycle!

No seriously....what's the answer for the first question??

To determine the maximum frequency that can be detected in an analog signal with a sampling rate of 100 samples per second (sps), we can apply the Nyquist-Shannon sampling theorem. According to this theorem, the maximum detectable frequency is half the sampling rate, also known as the Nyquist frequency.

So, in this case, the maximum detectable frequency would be 100 sps / 2 = 50 Hz.

Now, let's move on to the second part of your question. To determine the frequency of a waveform in hertz (Hz) given the time period between a peak and a subsequent trough, we need to calculate the reciprocal of the time period.

In your case, the peak of the waveform is followed by a trough 500 milliseconds (ms) later. To convert milliseconds to seconds, we divide 500 ms by 1000 (since there are 1000 milliseconds in one second). Therefore, the time period between the peak and the trough is 0.5 seconds (500 ms / 1000 = 0.5 s).

The frequency, measured in Hz, is the reciprocal of the time period. So, to find the frequency, we would take 1 divided by 0.5 seconds, which gives us 2 Hz.

Therefore, the frequency of the waveform in Hz is 2 Hz.