When we define $$\overline{\left|x\right|} = \frac1T\int_0^T x(t) dt$$ as the arithmetic mean of a signal we can see that it is the same as its dc component in the fourier transform.

Why is this the case? I can see it obvious for a normal sine or cosine wave, because everything will cancel out, but what when we $a + \sin(x)$ instead.

I can't see the link to the mean here.

  • 1
    $\begingroup$ The arithmetic mean of a sinusoid over a whole number of periods is zero. The arithmetic mean of the absolute value of a sinusoid over a whole number of periods is not zero. Your equation does not define the arithmetic mean of the signal. Can you please edit your question and restate it? $\endgroup$
    – Peter K.
    Commented Jan 18, 2022 at 20:46
  • 1
    $\begingroup$ ah so it should not be the absolute value of x for the mean, but is then the arithmetic mean or not? I meant without taking absolute value here $\endgroup$ Commented Jan 18, 2022 at 20:59
  • $\begingroup$ Take a look at this question, and its answers for an explanation why the mean or DC value of a signal is not the same as its Fourier transform evaluated at DC. $\endgroup$
    – Matt L.
    Commented Jan 18, 2022 at 21:28
  • 1
    $\begingroup$ From the linked question, it defines the time average of the signal from $-\infty$ to $\infty$. This is the equivalent of defining the DFT with $N\rightarrow\infty$ which also would result in a zero DC value. In most applications, that is not a particularly useful statistic. From that, it is my understanding that the 'DC' value of a signal with finite support is most associated with a bounded-domain time average. $\endgroup$
    – Ash
    Commented Jan 18, 2022 at 22:13
  • 2
    $\begingroup$ If you think about it, it is pretty intuitive that the DC of a signal is it's mean. If the signal is a single sinusoid, as Peter K. notes, its mean is 0. If that same sinusoid is riding on a DC offset of x, the mean is x. This same reasoning applies to any sum of sinusoids having various amplitudes and phase offsets (i.e. the Fourier transform of a periodic signal). Whatever the DC offset is, is the mean of the signal. Ash shows this mathematically. $\endgroup$
    – dmedine
    Commented Jan 19, 2022 at 5:10

1 Answer 1


The definition of the normalized discrete Fourier transform (DFT) for any signal $x[n]$ is $$F(k)=\frac{1}{N}\sum_{n=0}^{N-1}x[n]e^{-j2\pi k n /N}$$ The DC component of the DFT is evaluated at $k=0$. Given $$e^{-j2\pi 0 n /N}=e^0=1$$ The above simplifies to $$F(0)=\frac{1}{N}\sum_{n=0}^{N-1}x[n]=\bar{x}$$

  • 2
    $\begingroup$ The accepted answer is great since math tells all. I'm out of school for a long time and often forget how to do calculus :( Another hand waving way to think of this answer, is that the signal is oscillating about the DC value. Thus the time average will sum to zero with the remaining DC term unaffected. Remember that time averaging will eventually sum all orthogonal components to zero. This is also how lock-in amps work. $\endgroup$
    – wbg
    Commented Jan 19, 2022 at 18:43

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.