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beyondlight
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Homework Statement
Consider the finite length x[n]= 2δ[n]+δ[n-1]+δ[n-3]
We perform the following operation on this sequence:
(i) We compute the 5-point DFT X[k]
(ii) We compute a 5-point inverse DFT of Y[k]=X[k]2
a) Determine the sequence y[n] for n= 0, 1, 2, 3, 4
b) If N-point DFTs are used in the two step pocedure, how should we choose N so that y[n]=x[n]*x[n] for 0 ≤ n ≤ N-1?
Homework Equations
DFT transform pair:
[tex]X[k]=\sum_{n=0}^{N-1}x[n]e^{(-j\frac{2π}{N}kn)}[/tex]
[tex]x[n]=\frac{1}{N}\sum_{k=0}^{N-1}X[k]e^{(j\frac{2π}{N}kn)}[/tex]
The Attempt at a Solution
a)
[tex]X[k]=\sum_{n=0}^{4}x[n]e^{(-j\frac{2π}{5}kn)}=2 + e^{-j\frac{2π}{5}k} + e^{-j\frac{2π}{5}3k}[/tex]
Then for calculation for Y[k] we substitute (2*pi/5)=a
[tex]Y[k]= X[k]^{2} = 4 + 4e^{-jak}+2e^{-j4ak}+4e^{-j3ak}+e^{-j2ak}+e^{-j6ak}[/tex]
Now to get y[n] through:
[tex]y[n]=\frac{1}{N}\sum_{k=0}^{4}X[k]^{2}e^{(j\frac{2π}{5}kn)}[/tex]
gives us a very complicated expression contained with sums of complex exponentials. It seems extremely difficult from here to simplify it so i can calculate y[n] for n=0,1,2,3,4