Spectrum of fourier transform of $\delta(t-t_0)$

Question

We know that $$\mathcal{F}(\delta(t))=1.$$ In frequency domain, it is of magnitude $1$ for all frequency $\omega = 2\pi f$.

However $$\mathcal{F}(\delta(t-t_0))=e^{i\omega t_0}.$$

It looks like in frequency domain, it oscillates due to $e^{i \theta} $. However, this is not a real number.

So my question is how the frequency domain looks like for $$\mathcal{F}(\delta(t-t_0))=e^{i\omega t_0}$$ how the imaginary part in frequency domain (spectrum).

Thanks!

Answer 1

The magnitude is everywhere unit, $ |\mathcal{F}(\delta(t-t_0))| = |e^{i\omega t_0}| = 1, $ but the angle (complex argument) varies linearly.

Using $e^{ix}=\cos x + i \sin x$ we see that $$ \mathcal{F}(\delta(t-t_0)) = e^{i\omega t_0} = \cos\omega t_0 + i \sin\omega t_0 $$ so the graph $\{ (\omega, x, y) \mid \mathcal{F}(\delta(t-t_0))(\omega) = x+iy \}$ is a helix.