# Sampling rate of a sinusoidal Signal

What is the minimum sampling rate of a $\sin(\pi t)$ signal? When the sampling frequncy is twice that of the frequency, the spectrum is not present and in time domain all the samples we get is 0. In this case does the sampling rate should be greater than $2 f_\text{s}$? Then how does the Nyquist rate fit in this?

• If you are synchronized to the "zeros" crossing of your signal, which occurs at twice the frequency, you are getting zeros. In this case I would suggest a 4 time the target frequency to be measured. In this case, your result will not depend on the "pahse" of sampling.
– Moti
Jun 6 '15 at 23:58

What is the minimum sampling rate of a $\sin(\pi t)$ signal?

$$F_\text{s} = 1 + \epsilon$$

where $\epsilon > 0$. you must sample at some frequency exceeding twice the bandwidth. (all this assmuming real $x(t)$ and none of this bandpass sampling stuff.)

When the sampling frequncy is twice that of the frequency, the spectrum is not present and in time domain all the samples we get is 0.

nooooooo! that's not the case! you could miss-sample it by $90^\circ$ and

$$x[n] = x(nT) = \cos(\pi n) = (-1)^n$$

In this case does the sampling rate should be greater than $2 f_\text{s}$? then how does Nyquist rate fit in this
2. to deal with the $x[n] = x(nT) = \cos(\pi n + \phi)$ thing (a sinusoid at the Nyquist frequency), you have to accept the notion of Dirac delta functions, $\delta(f)$ in the frequency domain. and then you will see that, at Nyquist, ($f = \frac{F_\text{s}}{2}$), there will be overlapping and aliasing. but unlike other aliasing, you lose only half of your information, just the phase. so if there were reconstruction done, it would be only on the basis of the quadrature component of the
$$x(t) = cos(\pi t + \phi) = \cos(\phi)\cos(\pi t) - \sin(\phi)\sin(\pi t)$$
input. if $F_\text{s} \triangleq \frac{1}{T} = 1$ and $x[n]=x(nT)$, only the $\cos(\phi)\cos(\pi t)$ term counts.