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)?


1 Answer 1


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 $X(0) = N \overline{x}$

  • $\begingroup$ 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
  • 3
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.