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Given a vector

$${\bf x}= [a, 0, \dots, 0], a \neq 0, {\bf x} \in \mathbb{R}^n$$

If I compute its Discrete Fourier Transform (DFT), I get

$${\bf DFT}( {\bf x}) = [a,a, \dots, a]$$ I.e., a vector of all $a$s in $\mathbb{R}^n$ too --without complex component.

I know this is true by definition, but I cannot come up with an intuitive explanation. Could you please explain why this happens? I.e. why does a single spike at the beginning of my signal makes it flat in the frequency domain?

una_dinosauria
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  • If the spike were not at the beginning, it would not be constant. Rather, the magnitude would be constant and there would be a linear progression of the phase. The slope of this linear progression would be a function of the location of the spike. When the spike is at the beginning, the FFT software treats that as "zero phase location", which makes the phase slope equal zero and the FT a constant. – AnonSubmitter85 Jul 26 '13 at 03:23
  • I'd also suggest that you work through the math and convince yourself that 1) the magnitude of the FT of an impulse is constant, and 2) that location in one domain is equal to linear phase in the other. – AnonSubmitter85 Jul 26 '13 at 03:31
  • @AnonSubmitter85 That's exactly what I thought! Since a shift in time should correspond to a phase shift in frequency, then moving the spike to, say, they 2nd place should produce the same real component with a different imaginary component, right? Well, the FFT says no. It actually gives me different numbers in the real part if I change the place of the spike. Is this a numerical problem? – una_dinosauria Jul 26 '13 at 07:34
  • @AnonSubmitter85 Nevermind, I see it now. Would you mind writing your response as a proper answer? I'll be happy to accept it. – una_dinosauria Jul 26 '13 at 08:06

3 Answers3

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The only signal that has all frequency amplitudes with same weight (flat spectra) and linear phase is an impulse.

Also using duality property of DFT, if DFT of a given function is a signal with only zero component, we could realize that the time-domain signal would have been a pure DC signal.

Mahdi Khosravi
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    The spectrum of an LFM signal of infinite duration will also have a constant magnitude (i.e., be flat). There are many other such signals that have flat spectra. What distinguishes the impulse's spectrum from all others is that its magnitude is flat and its phase is linear. – AnonSubmitter85 Jul 26 '13 at 19:15
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The spectrum is constant for two reasons. First, the magnitude of the DFT of an impulse is constant. Second, the phase of the DFT of an impulse is a linear function of the location of the impulse. Thus, the DFT of an impulse is of the form $Ae^{jt_0 \omega}$, where $A\in \mathbb{R}$ is the constant magnitude, $t_0$ is the location of the impulse in the time/image domain, $\omega$ is the frequency variable, and $j=\sqrt{-1}$. When the impulse is located at sample $t_0 = 0$, we have $e^{jt_0 \omega}=e^{j0\omega}=1$, and the spectrum has the constant value $A$.

AnonSubmitter85
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Not sure if this is what you had in mind, but think about the continuous analog, a single spike as a delta function $a \delta(x)$, whose transform is

$$\int_{-\infty}^{\infty} dx \, a \delta(x) \, e^{i k x} = a$$

That is, the transform is a constant independent of $k$, just as in the discrete case.

Ron Gordon
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  • Based on the OP's emphasis on "at the beginning", I think he/she is also confused about the linear phase that is related to the location of the spike. – AnonSubmitter85 Jul 26 '13 at 03:27
  • @AnonSubmitter85: I get that. Both the first position in the discrete case and the center position in the continuous case corresponds to zero frequency component, which sifts out any varying component. – Ron Gordon Jul 26 '13 at 03:49