GATE 2020 IN Q.48: solution and answer

Official answer: 0Source: Official answer key, GATE 2020 IN (IIT Delhi)

1. Here $N=3$ and $X(0)=X(1)=X(2)=1$. Use the IDFT definition: $$x(n)=\frac{1}{N}\sum _{k=0}^{N-1}X(k)e^{j2\pi kn/N}$$

2. Put $n=2$ and $N=3$. The exponents are $j2\pi k\cdot 2/3$ for $k=0,1,2$: $$x(2)=\frac{1}{3}\left(1+e^{j4\pi /3}+e^{j8\pi /3}\right)$$

3. Because $e^{j8\pi /3}=e^{j2\pi }{e}^{j2\pi /3}=e^{j2\pi /3}$, the bracket is $1+e^{j2\pi /3}+e^{j4\pi /3}$. This is the sum of the three cube roots of unity, which is $0$.

4. Check it directly: $$1+\left(-\frac{1}{2}+j\frac{\sqrt{3}}{2}\right)+\left(-\frac{1}{2}-j\frac{\sqrt{3}}{2}\right)=0$$ so $x(2)=\frac{1}{3}\cdot 0=0$.

5. Shortcut: the $N$-point DFT of $\delta (n)=(1,0,0)$ is $X(k)=1$ for every $k$. The DFT is one-to-one, so the IDFT of the flat spectrum $(1,1,1)$ must be $(1,0,0)$. That gives $x(0)=1$ and $x(1)=x(2)=0$.