Why exactly is X(0) the DC component of a signal?
How is it equal to N times x(n)'s average value and why it is at X(0)?
Answer
Follows from the DFT definition. It's defined as
\begin{equation} X(k) = \sum_{n=0}^{N-1} x(n) e^{-j2\pi \frac{kn}{N}} \end{equation}
So $X(0)$ is
\begin{equation} X(0) = \sum_{n=0}^{N-1} x(n) e^{-j2\pi \frac{0 \cdot n}{N}} \end{equation}
Having $k=0$ gives $e^0=1$ all the time so that
\begin{equation} X(0) = \sum_{n=0}^{N-1} x(n) 1 \end{equation}
Comparing this to the average
\begin{equation} \overline{x} = \frac{1}{N} \sum_{n=0}^{N-1} x(n) \end{equation}
shows that $X(0) = N \overline{x}$
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