# Why is X(0) the DC component

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)?

Follows from the DFT definition. It's defined as

$$X(k) = \sum_{n=0}^{N-1} x(n) e^{-j2\pi \frac{kn}{N}}$$

So $X(0)$ is

$$X(0) = \sum_{n=0}^{N-1} x(n) e^{-j2\pi \frac{0 \cdot n}{N}}$$

Having $k=0$ gives $e^0=1$ all the time so that

$$X(0) = \sum_{n=0}^{N-1} x(n) 1$$

Comparing this to the average

$$\overline{x} = \frac{1}{N} \sum_{n=0}^{N-1} x(n)$$

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

• So when removing the DC component, what is X(0) applied to? – jarryd Aug 9 '13 at 11:12
• If the signal is mean-free, X(0) will just be zero. Similarly, to make a signal mean-free, just set X(0)=0 – jan Aug 9 '13 at 11:16
• @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 $x[n], 0 \leq n \leq N-1$. – Dilip Sarwate Aug 9 '13 at 13:23