A signal $x=\sin(\pi t/4)$ sampled at every $t=1$ sec. so $T=1$ sec Sampled signal is then $x=\sin(\pi n/4)$, What is the sampling rate in this case?, And according to Nyquist sampling rate what should be the minimum sampling rate to sample this signal?
Some things to think about:
- The Nyquist Theorem requires the sampling rate be a minimum of twice the bandwidth of the signal, which, in general, is not twice the fundamental frequency of the carrier.
- For a lowpass signal, twice the bandwidth equates to twice the highest frequency component.
- For a bandpass signal, twice the bandwidth is NOT twice the highest frequency component.
- The spectrum of a sampled sinusoid approaches $\delta(f-f_0) + \delta(f+f_0)$
- The bandwidth of a $\delta$ $<<$ $2f_0$
- The sampling rate, $F_s$, is the inverse of the sampling period, $T_s$, which is 1 Hz in your case.
- Assuming $x(t) = \sin(\pi/4t) = \sin(w_0t) = \sin(2\pi f_0)$, then $w_0 = \pi/4$ rad/s and $f_0 = 1/8$ Hz
- So, $2f_0$ = 1/4 Hz
- Take a look at this similar question
The sampling frequency is the inverse of the period T so $F_s = 1/T = 1/1s = 1 Hz$. The Nyquist rate is the double of the maximum frequency of the signal. In your case the signal has only one frequency component $f = 1/8$. So, $F_N = 2 f = 1/4$. Since in your case is a $sin$ signal, the the sampling frequency should be greater (not greater or equal) than the Nyquist rate, in order to avoid aliasing.