Fourier transform of a constant

Somali_Physicist
It is often reported that the Fourier transform of a constant is δ(f) : that δ denotes the dirac delta function.

ƒ{c} = δ(f) : c ∈ R & f => Fourier transform
however i cannot prove this
Here is my attempt:(assume integrals are limits to [-∞,∞])
ƒ{c} = ∫ce-2πftdt = c∫e-2πftdt = c∫ƒ{δ(f)}e-2πftdt
=> cf{f{δ(f)}} => cδ(-f)

this is incorrect, any help would be appreciated thank you.

Homework Helper
Gold Member
It is often reported that the Fourier transform of a constant is δ(f) : that δ denotes the dirac delta function.
What you just saying above is not exactly correct, the Fourier transform of a constant is ##c\delta(f)##.

ƒ{c} = δ(f) : c ∈ R & f => Fourier transform
however i cannot prove this
Here is my attempt:(assume integrals are limits to [-∞,∞])
ƒ{c} = ∫ce-2πftdt = c∫e-2πftdt = c∫ƒ{δ(f)}e-2πftdt
=> cf{f{δ(f)}} => cδ(-f)

this is incorrect, any help would be appreciated thank you.

What you do is unnecessary but it is correct, as I said above the Fourier transform of a constant c is ##c\delta(f)=c\delta(-f)##.

Somali_Physicist
What you just saying above is not exactly correct, the Fourier transform of a constant is ##c\delta(f)##.

What you do is unnecessary but it is correct, as I said above the Fourier transform of a constant c is ##c\delta(f)=c\delta(-f)##.
Ahh yes i forgot about the constant, what was confusing is that
δ(f)=δ(-f)

is there a proof for that?

EDIT:
oh wait if the "inifinite" jump for this dirac delta function is at 0 then its negative is essentially its positive.Thanks dude :)

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Delta2
AVBs2Systems
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.
$$\displaystyle \int_{-\infty}^{\infty} f(t) \cdot \delta(t \pm t_{0}) = f(\mp t_{0} )$$
$$f(t) \delta(t \mp t_{0} ) = f(\pm t_{0}) \delta(t \mp t_{0} )$$
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.

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