2

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?

jojek
  • 10,340
  • 6
  • 32
  • 71
  • 1
    Sampling frequency is obviously 1 Hz. General form of Sinusoidal signal is $\sin(2\pi f_0 t)$ - I don't care about phase. Sampling frequency must be then at least $2f_0$ – jojek Feb 15 '15 at 13:47

4 Answers4

3

For the sine wave $\sin ((\pi/4)t)$ , the minimum sampling frequency is greater than 1/4 Hz. That amounts to a sampling period less than 4 seconds. In your case, sampling period of 1 second is therefore a valid choice.

Fat32
  • 26,011
  • 3
  • 20
  • 46
2

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
Community
  • 1
random_dsp_guy
  • 916
  • 1
  • 9
  • 27
2

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.

Esteban Filardi
  • 242
  • 2
  • 7
0

Sampling frequency is 1 Hz, since the sampling period T = 1sec. i.e greater than 1/4 Hz. Sorry for the wrong expression.

PsychedGuy
  • 323
  • 1
  • 3
  • 12
  • The highest signal frequency is 1/8 Hz and thus Fmin = 0.25 Hz. – Deve Feb 16 '15 at 08:29
  • yea so the sampling frequency must be greater than 0.25 Hz right?. Sorry guys for the wrong input. Got a little confused myself. *My Apologies* – PsychedGuy Feb 16 '15 at 08:37
  • 2
    Yes, $f_\mathrm{s}>0.25\,\text{Hz}$ must hold. No problem, fortunately anwers can be edited. – Deve Feb 16 '15 at 09:16