# Why is the arithmetic mean the same as the DC component of its fourier transform?

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.

• 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?
– Peter K.
Jan 18 at 20:46
• 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 Jan 18 at 20:59
• 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. Jan 18 at 21:28
• 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.
– Ash
Jan 18 at 22:13
• 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. Jan 19 at 5:10

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 $$F(0)=\frac{1}{N}\sum_{n=0}^{N-1}x[n]=\bar{x}$$