(adsbygoogle = window.adsbygoogle || []).push({}); 1. Calculate the finite Fourier transform of order m of the following sequences:

a) u_{k}= 1, 0[itex]\leq[/itex]k[itex]\leq[/itex]N-1

b) u_{k}= (-1)^{k}, 0[itex]\leq[/itex]k[itex]\leq[/itex]N-1 N even

c) u_{k}= k, 0[itex]\leq[/itex]k[itex]\leq[/itex]N-1

2. Relevant equations

U_{k}= (1/N)[itex]\sum[/itex]u_{k}e^{-2pi*i*k*j/N}from j=0 to N-1 ; 0<=k<=N-1

Attempt:

a) First thing that I tried is that [itex]\sum[/itex]x^k = [itex]\frac{1}{1-x}[/itex] 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[itex]\neq[/itex]0 , 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:

[itex]\sum[/itex]e^{i*k*j}from j = 0 to N-1

={ 0 if e[itex]\neq[/itex]1,

{ N-1 else.

b) wouldn't we get (1/N)[itex]\sum[/itex]-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 [itex]\sum[/itex]k*x^k = [itex]\frac{x}{(1-x)^{2}}[/itex]

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?

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# Homework Help: Fourier Transform (Numerical Analysis)

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