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fundamental period for DT seq x[n]=cos(pi* (n)^2 / 8 ) ?

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rramya

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fundamental period

have a doubt reg checking DT periodic sequence:

x[n]=cos(pi* (n)^2 / 8 )

it is periodic ofcourse, but in the solution manual its fundamental period is mentioned as 8.

but i got fundamental period=N=(2*pi/ omega )
in this case, N= (2*pi / (pi/8) ) = 16.

does anybody know about how they getting N to be 8 ????
 

fundamental period of a signal

Hi friend,

i replyed your message on the group and i will rewrite it here again;

Regarding your qestion which was the fundamental period of cos(pi . n^2/8), I see if wee add N to n as followes:

pi/8[n+N]^2 = pi/8 (n^2+2nN+N^2)

pi/8[n+N]^2= pi n^2/8 + pi 2 n N/8 + pi N^2/8

now according to your result that is the minimum fundamental period N=16 we get:

pi/8[n+N]^2=pi n^2/8 + 4 pi n + 32 pi

but this can be more reduced if we take N=8 to get:

pi/8[n+N]^2=pi n^2/8 + 2 pi n +8 pi

assume n = 20 :

case N=16,

then cos(pi 20^2/8 + 4 pi 20 + 32 pi) = 1

case N=8,

then cos(pi 20^2/8 + 2 pi 20 + 32 pi) = 1

Hence;

I deduce that the minimum fundamental period N =8.

please rectify me if i am not correct.

Regards.
 

finding fundamental period

ur signal may be cos(2pi/8)n,
compare ur argument with 2Kpi/N like this
2Kpi/N=2pi/8
u'll get

K=N/8
u've to put value of N,so that ur K becomes 1,remember N is to be an integer,so ur N will be 8,tht is ur period
 

fundamental period cos ( n^2)

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* (n)^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* (n)^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/(n)^2) + (4*pi) )

now the second term:(pi*32/(n)^2)
here, (n)^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)
 

is cos(pi/6*n^2) periodic

i agree my friend
 

fundamental period of cosine

actully u r getting minimum value ok K that is 1 which is integer tht tells u lowest fundamental
 

    R

    rramya

    Points: 2
    Helpful Answer Positive Rating
fundamental period of cosine squared

Ya, i understood very clearly
thanks Xulfee.....
i derived it in the long method ....
 

x[n]=cos(8/15pi*n)fundamental period

ur welcome
 

fundamental period cosine squared

hi xulfee, if you are interest in communication, so, enjoy our group named "OFDM and Matlab" in the link below

**broken link removed**
 

periodic cos(pi*n^2)

so nice of u dear,send me more links if u've,i'll be thankfull to u
 

Re: fundamental period cos ( n^2)

rramya said:
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* (n)^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* (n)^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/(n)^2) + (4*pi) )

now the second term:(pi*32/(n)^2)
here, (n)^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)
Click to expand...

====== hi buddy, i am a bit confused that why did we take n^2 = 4, why not n^2 =16, after all, it also gives an integer value and also the lowest value of N, also 32*pi/16 = 2* pi and added to 4* pi gives 6*pi, so that x(n+N) = x(n)
 

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