GATE 2018 EC Q.55: solution and answer

Official answer: 2.90 to 3.10Source: Official answer key, GATE 2018 EC (IIT Guwahati)

1. We need the sum of the even-indexed samples: $S=x[0]+x[2]+x[4]+x[6]$.

2. Pick out the even samples with the selector $\frac{1+(-1{)}^n}{2}$, which equals 1 for even $n$ and 0 for odd $n$: $$S=\sum _{n=0}^{7}x[n]\frac{1+(-1{)}^n}{2}=\frac{1}{2}\left(\sum _{n=0}^{7}x[n]+\sum _{n=0}^{7}x[n](-1{)}^n\right)$$

3. Put $k=0$ in the DFT definition: $X[0]={\sum }_{n=0}^{7}x[n]$. So the first sum is $X[0]=0+1=1$.

4. Put $k=N/2=4$: $e^{-j2\pi n(4)/8}=e^{-j\pi n}=(-1{)}^n$, so $X[4]={\sum }_{n=0}^{7}x[n](-1{)}^n$. So the second sum is $X[4]=4+1=5$.

5. Therefore $$S=\frac{X[0]+X[4]}{2}=\frac{1+5}{2}=3$$

6. Check with the IDFT. Since $x[n]=\frac{1}{8}{\sum }_{k=0}^{7}X[k]e^{j2\pi nk/8}$, we get $x[2m]=\frac{1}{8}{\sum }_{k=0}^{7}X[k]e^{j\pi mk/2}$, and so $$\sum _{m=0}^{3}x[2m]=\frac{1}{8}\sum _{k=0}^{7}X[k]\sum _{m=0}^3{e}^{j\pi mk/2}$$ The inner sum is a geometric series with ratio $e^{j\pi k/2}$, whose fourth power is 1. It equals 4 when $k=0$ or $k=4$ (ratio $=1$) and 0 for every other $k$. So $S=\frac{4}{8}(X[0]+X[4])=\frac{1}{2}(1+5)=3.00$.