Question

Determine whether or not each of the following signals is periodic.
If a signal is periodic, specify its period (T0 for continuous-time signals and N0 for discrete-time signals).

a) x[n]=sin(8π/2 n+1)
b) x[n]=cos(r/8-π)
c) x[n]=cos(π/8 n²)
d) x[n]=cos(π/2 n) cos(π/4 n)
e) x[n]=2cos(π/4 n)+sin(π/8 n)-2cos(π/2 n+π/6)

Answer 1

Answer:

a) periodic (N = 1)

b) not periodic

c) not periodic

d) periodic (N = 8)

e) periodic (N = 16)

Explanation:

For function to be a periodic: f(n) = f(n+N)

$a) x[n]=sin(\frac{8\pi}{2}n+1)\\\\sin(\frac{8\pi}{2}n+1)=sin(4\pi n+1)$

It is periodic with fundamental period N = 1

$b) x[n]=cos(\frac{n}{8} -\pi)\\\\\frac{1}{8} N=2\pi k$

N must be integer. So it is nor periodic

$c) x[n]=cos(\frac{\pi}{8} n^2)\\\\cos(\frac{\pi}{8} (n+N)^2)=cos(\frac{\pi}{8} (n^2+N^2+2nN)\\\\N^2 = 16 \:\:or\:\:2nN=16$

Since N is dependent to n. So it is not periodic.

$d) x[n]=cos(\frac{\pi }{2} n) cos(\frac{\pi }{4} n)\\\\x[n] = \frac{1}{2} cos(\frac{3\pi }{4} n) + \frac{1}{2} cos(\frac{\pi }{4} n)\\\\N_1=8\:\:and\:\:N_2=8$

So it is periodic with fundamental period N = 8.

$e) x[n]=2cos(\frac{\pi }{4} n)+sin(\frac{\pi }{8} n)-2cos(\frac{\pi }{2} n+\frac{\pi }{6} )\\\\N_1=8\:\:and\:\:N_2=16\:\:and\:\:N_3=4$

So it is periodic with N = 16.