0
$\begingroup$

When I do the Fourier transform of the Dirac impulse I get a pure sinusoid (or complex exponential, however you wanna call it) but I read in several places that all frequencies are present in the dirac impulse and all of them with the same amplitude. How is this possible? Am I wrong when I perform the transform?

$\endgroup$
  • 1
    $\begingroup$ Note that the complex exponential's independent variable is frequency, so you get a non-zero contribution for all frequencies. $\endgroup$ – Matt L. Aug 7 '18 at 19:39
  • $\begingroup$ (Very) related: this answer $\endgroup$ – Matt L. Aug 7 '18 at 19:47
2
$\begingroup$

A Dirac impulse $x(t)=\delta(t-d)$ has the continuous-time Fourier transform $X(\Omega)$ of $$\mathcal{F}\{\delta(t-d) \} = 1 e^{-j\Omega d} $$

whose magnitude is $$|X(\Omega)| = 1 ~~~, \text{ for all } \Omega $$ and a phase of $$<X(\Omega) = -\Omega \cdot d $$

Note that it's incomplete to think of the real or imaginary parts of the Fourier transform alone. Rather the magnitude and phase point of views are more reflective of the nature of the result.

So in this case the magnitude is $1$ and hence it's said to contain all frequencies of magnitude $1$. Note that these are differential amplitude components of continuum frequency range as opposed to a finite amplitude of discrete set of frequency components, aka line components.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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