Fourier transform and impulse function $\delta(\omega)$

Question

Why does impulse function $\delta(\omega)$ keep occurring in the Fourier transform expression of standard functions like $\sin(t)$, $\cos(t)$, constant function, unit step $u(t)$ etc? (can someone intuitively explain what it means)

Answer 1

The Dirac delta impulse $\delta(\omega-\omega_0)$ represents a spectral line at frequency $\omega_0$, since it is zero everywhere except for $\omega=\omega_0$. So any function with spectral lines, such as a sinusoid, or a DC signal (which has a spectral line at frequency $\omega_0=0$) has a Fourier transform which contains Dirac delta impulses. This is also true for any function with a DC component such as the unit step $u(t)$. Note that also periodic functions have Fourier transforms consisting of Dirac impulses because they can be represented as a sum of sinusoids (Fourier series).

Answer 2

Because sines and cosines are intimely linked with this symbol.

The Fourier transform classically exists under mathematical conditions. Most often, functions are required to be integrable, or square integrable, to avoid terms that blow up to infinity.

Sines, cosines, unit steps do not satisfy the above conditions. Yet, as they are very central to the theory, one want to deal with them, and that requires to resort to non-standard functions (generalized functions, distributions) to be able to somehow deal with the disturbing infinities, in a meaningful (intuition) and practical (computation) way. In Wikipedia's Dirac delta function, one finds the quote:

The problems with a classical interpretation are explained as follows: The greatest drawback of the classical Fourier transformation is a rather narrow class of functions (originals) for which it can be effectively computed. Namely, it is necessary that these functions decrease sufficiently rapidly to zero (in the neighborhood of infinity) in order to ensure the existence of the Fourier integral. For example, the Fourier transform of such simple functions as polynomials does not exist in the classical sense. The extension of the classical Fourier transformation to distributions considerably enlarged the class of functions that could be transformed and this removed many obstacles.

The payback is that they should be handled with care, and that you cannot always use them as standard functions: for instance, product of distributions (something like $\delta^2(\omega)$) should not be tried (unless you really know what you are doing). And you may extended results, such that: if $f$ and $g$ are distributions, $tf(t)=tg(t)$ implies that $f(t)=g(t)+c\delta(t)$.