- #1
Scootertaj
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1. Calculate the finite Fourier transform of order m of the following sequences:
a) uk = 1, 0\leqk\leqN-1
b) uk = (-1)k, 0\leqk\leqN-1 N even
c) uk = k, 0\leqk\leqN-1
2. Homework Equations
Uk = (1/N)\sumuke-2pi*i*k*j/N from j=0 to N-1 ; 0<=k<=N-1
Attempt:
a) First thing that I tried is that \sumx^k = \frac{1}{1-x} but that doesn't seem to get where I want. For example, I know that for a), we get 1 for k=0 and 0 for k\neq0 , but this would give 1/N (1) for k = 0 and I don't know what for any other k.
So, I found a formula that says:
\sumei*k*j from j = 0 to N-1
={ 0 if e\neq1,
{ N-1 else.
b) wouldn't we get (1/N)\sum-e-2pi*i*k*j/N from j=0 to N-1 ; 0<=k<=N-1 which is the same as part a) ?
c) I think I need to use the idea that \sumk*x^k = \frac{x}{(1-x)<sup>2</sup>}
Obviously, if this formula is valid (I have no idea), then it would give 1 for 0 and 0 for other k for part a) which is correct
Any ideas?
a) uk = 1, 0\leqk\leqN-1
b) uk = (-1)k, 0\leqk\leqN-1 N even
c) uk = k, 0\leqk\leqN-1
2. Homework Equations
Uk = (1/N)\sumuke-2pi*i*k*j/N from j=0 to N-1 ; 0<=k<=N-1
Attempt:
a) First thing that I tried is that \sumx^k = \frac{1}{1-x} but that doesn't seem to get where I want. For example, I know that for a), we get 1 for k=0 and 0 for k\neq0 , but this would give 1/N (1) for k = 0 and I don't know what for any other k.
So, I found a formula that says:
\sumei*k*j from j = 0 to N-1
={ 0 if e\neq1,
{ N-1 else.
b) wouldn't we get (1/N)\sum-e-2pi*i*k*j/N from j=0 to N-1 ; 0<=k<=N-1 which is the same as part a) ?
c) I think I need to use the idea that \sumk*x^k = \frac{x}{(1-x)<sup>2</sup>}
Obviously, if this formula is valid (I have no idea), then it would give 1 for 0 and 0 for other k for part a) which is correct
Any ideas?
Last edited:
