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I know that in general the sampling rate, $f_s$, must be greater than twice the highest frequency of the signal, $f$.
If I sample at the Nyquist rate, it can lead to the following:
However, if the signal is sampled only at maxima and minima, then we get this:
My signal is not pure sinusoidal, it is something like this:
Will I loose any frequency components if I sample it at the Nyquist rate?
The primary issue with sampling right at Nyquist is the noise that is in the continuous-time (analog) signal prior to sampling (not necessarily the noise that gets added due to quantization). Without filtering the signal to isolate just the noise that is below the Nyquist frequency, we are unable to distinguish the components of the analog signal for each resulting sample in our discrete-time sequence. This is explained in detail at this post: What is the difference between undersampling and oversampling in analog to digital conversion ?
Thus if we sample exactly at Nyquist ($f_s/2$ where $f_s$ is the sampling rate, there is no realizable filter that will pass every signal of interest up to $f_s/2$ while reject everything immediately after that which would otherwise fold in. It is for this reason that we need to sample at some frequency above Nyquist and the decision is based on the analog filter design prior sampling. This "filter" may be inherent in the signal we are filtering, and our knowledge of the signal and noise content and how much we care about the higher frequency noise that would otherwise fold into band and be indistinguishable.
As a further note: It appears that the OP's signal has been hard-limited prior to or as part of the sampling (such as with a 1 bit converter). This would remove all the AM (amplitude modulation) components of the signal + noise and result in measurement of the phase modulation components. Under high SNR conditions this elimination of the AM results in an SNR improvement, but we must be careful if low SNR conditions could exist (such as a strong higher frequency interference that we have neglected to filter out prior to hard-limiting) in which case it would be our signal that creates the AM on the interference and gets rejected. Thus understanding the noise spectrum of the analog signal and the filtering considerations prior to hard-limiting, and prior to sampling are very important.
when you have a quantized signal (like in your case) you also have a noise floor in the frequency spectrum. This noise floor is proportional also to the sampling rate. If you sample a signal at twice its maximum frequency you can distinguish all the frequency components above the noise floor. In your case you are sampling at $F_p = 2$ because there are 2 samples every second therefore you are assuming your signal has a maximum period of 1. There is one case when you are interested on sampling at the same frequency of the signal. For example if you know that your signal has an unwanted ripple at the frequency $f_{ripple}$ you can sample at this frequency $F_p = f_{ripple}$ preserving the lower frequencies while the ripple is transformed into a constant error which basically depends on the ripple waveform.
