hi Friend,
yes, your method is correct......
but question arises how we get to know the lowest fundamnetal value of all the sudden???? (then we will be left with trial and error method)some experiment we should do to ensure: the minimum value of the N should be 8 in our case.... and not its multiples.....
x[n]=cos(pi*
^2 / 8 )
fundamental period=N= 2*pi * m / wo ; m is an integer
where wo is the anualr freq in rad/sec.
in our case wo=pi/8, but since square term in involved in x[n],
i now, take, wo= n*pi/8.
N=2*pi *m/ (n*pi / 8 )
pi gets cancelled...
N= 16*m/n
for x[n] to be periodic N should be integer.
now we know x[n+N] = x[n]
i.e., cos[pi*(n+N)^2/8] = cos(pi*
^2 / 8 )
while proving the above statement we can get the value of N:
cos(pi*(n^2+N^2+2*n*N)/8) = cos(pi*(n^2+(16*m/n)^2+2*n*(16*m/n))/8)
= cos((pi*n^2/8) + (pi*32*(m)^2/(n^2)) +
(4*pi*m) )
To find the lowest value of N :
set m=1 .......
say:
cos(pi*(n^2+N^2+2*n*N)/8) = cos((pi*n^2/8) + (pi*32/
^2) + (4*pi) )
now the second term
pi*32/
^2)
here,
^2 ==> should be perfect square so that it should be multiple
of 32*pi
==> should be a small perfect square
so n^2=4 ====> n=2 ; m=1; N= 16*m/n = 16 * 1 / 2 = 8
second term: (pi*32/4) = 8*pi
cos(pi*(n^2+N^2+2*n*N)/8) = cos((pi*n^2/8) + 8*pi + (4*pi) )
= cos((pi*n^2/8)+(12*pi))
= cos((pi*n^2/8) )
Added after 1 seconds:
hi xulfee,
no my seq is x[n]= cos(Pi*n^2/8)
only
it is given in the signals and system book (by oppenheim & wilsky)
exercise problem 1.26(c)