Topic: Digital Signal Processing
If $x_{in}(t)=\sin(2*\pi*4000*t)+0.75*\sin(2*\pi*5000*t+\pi/4)$ is sampled with $F_s=16000Hz$ calculate $X(0)$ if $X(m)=\sum_{n=0}^{N-1}x(n)e^{-j2\pi mn/N}$ When N=8, where $x(n)=x_{in}(nT_s)$
- A. 0.0 - j 4.0
- B. 0.0 - j 0.0
- C. 1.414 + j 1.414
- D. 0.0 + j 4.0
Answer: B. 0.0 - j 0.0
Explanation: For $m=0$, the DFT evaluates to $X(0) = \sum_{n=0}^{N-1} x(n)$. Since the input signal $x_{in}(t)$ is purely real, the sum of its real samples $x(n)$ must also be a purely real number. The only option representing a purely real number without an imaginary component is $0.0 - j0.0$.