Find the Answer to x[n] in a 2-Point DFT

[digitsboy]

Homework Statement

The 2-point DFT of x[n] is given by the expression X[k] = 2δ[k], for k = 0, 1.

Homework Equations

What is x[n]?

The Attempt at a Solution

i know the answer but i don't know how they calculate it
because a delta pulse is only one at k=zero.
x\left[n\right]=\frac{1}{N}\sum _{k=0}^{N-1}X\left[k\right]\cdot e^{j\cdot \frac{2\pi }{N}kn}\:
because k=0
e^{j\cdot \frac{2\pi }{N}kn}\: = 1
So i come to an answer where x[n] = δ[n]
But it is: x[n] = δ[n]+δ[n−1] ? How to they calculate that, i don't understand it

 

[Henryk]

i'm not sure if I understand the problem correctly, but it looks to me that the FT of the signal has only two frequency component, at k = 0 and k=1.
If that is correct, the sum in your attempted solution contains only two non-zero terms: k = 0, and k = 1, both of them are equal to 2.
So, $x(n) = 2 \cdot e^{j \frac {2 \pi}N 0 \cdot n} + 2 \cdot e^{j \frac {2 \pi}N1 \cdot n}$

Hope that helps

 

[digitsboy]

uhh no because the answer is x[n] = δ[n]+δ[n−1]