DFT of a finite lengt sequence

[beyondlight]

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:

X[k]=\sum_{n=0}^{N-1}x[n]e^{(-j\frac{2π}{N}kn)}

x[n]=\frac{1}{N}\sum_{k=0}^{N-1}X[k]e^{(j\frac{2π}{N}kn)}

The Attempt at a Solution

a)

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}

Then for calculation for Y[k] we substitute (2*pi/5)=a

Y[k]= X[k]^{2} = 4 + 4e^{-jak}+2e^{-j4ak}+4e^{-j3ak}+e^{-j2ak}+e^{-j6ak}

Now to get y[n] through:

y[n]=\frac{1}{N}\sum_{k=0}^{4}X[k]^{2}e^{(j\frac{2π}{5}kn)}

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

 

[smk037]

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=0,1,2,3,4P

Please post the expression that you obtained but found difficult to simplify. From what I remember in early college, these problems are usually a ton of tedious algebra, so you may not be far off.