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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)?

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1 Answer 1

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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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  • $\begingroup$ So when removing the DC component, what is X(0) applied to? $\endgroup$
    – some_id
    Commented Aug 9, 2013 at 11:12
  • $\begingroup$ If the signal is mean-free, X(0) will just be zero. Similarly, to make a signal mean-free, just set X(0)=0 $\endgroup$
    – jan
    Commented Aug 9, 2013 at 11:16
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    $\begingroup$ @jan Just setting $X(0) = 0$ removes the signal mean-free in its frequency-domain representation, but to do the same in the time domain, we need to translate this to the time domain via the iDFT, or equivalently, subtract $\bar{x}$ from each of the time domain quantities $x[n], 0 \leq n \leq N-1$. $\endgroup$ Commented Aug 9, 2013 at 13:23

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