Somali_Physicist said:
It is often reported that the Fourier transform of a constant is δ(f) : that δ denotes the dirac delta function..
Hello.
Not just reported, but we can prove it right here and now, first I would like to introduce you to the properties of the pulse/distribution.
$$ \textbf{The dirac delta pulse and its mathematical properties} $$
The definition of the pulse in continuous time is:
$$ \delta(t) = \begin{cases} \infty & \text{ $$ t = 0 $$} \\
0 & \text{ Otherwise } \end{cases} $$
$$ 1. \,\,\,\, \text{It is an even function} $$
Definition of an even function: ## f(t) = f(-t) ##.
Definition of an odd function: ## -f(t) = f(-t) ##
hence:
$$ \delta(t) = \delta(-t) $$
$$ 2. \,\,\,\, \text{It has a time scaling property} $$
$$ \delta(at) = \dfrac{1}{|a|} \cdot \delta(t) \,\,\,\,\,\, a \ne 0$$
$$ 3. \,\,\,\, \text{It has a unit area property} $$
$$\displaystyle \int_{a}^{b} \delta(t \pm t_{0} ) \,\,\,\, \text{dt} = \begin{cases} 1 & \text{ $\mp t_{0} \in [a,b] $} \\ 0 & \text{ $ \mp t_{0} \not\in [a,b] $} \end{cases} $$
$$ 4. \,\,\,\, \text{It has a masking property} $$
This is an important property used in LTI systems and convolution, basically, because of the masking property all functions can be expressed as a
sum of scaled and shifted dirac pulses.
$$ \displaystyle \int_{-\infty}^{\infty} f(t) \cdot \delta(t \pm t_{0}) = f(\mp t_{0} ) $$
Masking without an integral:
$$
f(t) \delta(t \mp t_{0} ) = f(\pm t_{0}) \delta(t \mp t_{0} )
$$
Example:
$$ cos(x) \delta(x) = cos(0) \delta(x) = \delta(x) $$
There are other properties, like the connection to the unit step function, among others.
Anyway, the Fourier transform of a constant can be derived using the:
$$ \textbf{The masking property and the duality and linearity theorem of the Fourier transform } $$
Duality theorem of the Fourier transform: If the function ##f(t)## has a Fourier trasnform of ## X(j \omega) ## then the function ## X(t) ## would have a Fourier transform of ## 2 \pi f(-j \omega) ##.
We first use the masking property to define the Fourier transform of the pulse as:
$$ \mathcal F \big( \delta(t) \big) = \displaystyle \int_{-\infty}^{\infty} \delta(t) e^{-j(\omega t) } \,\,\,\,\, \text{dt} = e^{j(0)} = 1 $$
Using the duality property property:
$$
1 \iff 2 \pi \delta(- \omega) = 2 \pi \delta(\omega) $$
Using the linearity theorem:
$$ k \cdot 1 \iff k 2 \pi \delta(\omega) $$But that's it, we use the masking property and duality and linearity theorems of the Fourier transform to find the Fourier transform for constants. Many of the properties of the pulse can be simply proved by the unit area property.
I don't know how to get the laplace and Fourier transform bars here, so if anyone knows, let me know.