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

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)

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).
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 $$f(t)=g(t)+c\delta(t)$$.