Question

fourier transform of 1 is​

Answer 1

Answer:

Unit impulse function

Step-by-step explanation:

Fourier transform of 1 is explained using the duality property of Fourier transform. Fourier transform of 1 is unit impulse function.

Answer 2

Answer:

The Fourier transform of 1 will be:

$\begin{aligned}\mathcal{F}_{x}[1](k) &=\int_{-\infty}^{\infty} e^{-2 \pi i k x} d x \\\\&=\delta(k)\end{aligned}$

Step-by-step explanation:

Given: A number is 1.

We have to find the Fourier transform of 1.

• As we know, the delta function is a generalized function that can be defined as the limit of a class of delta sequences.
• The delta function is given as a Fourier transform as:

        $\delta(x)=\mathcal{F}_{k}[1](x)=\int_{-\infty}^{\infty} e^{-2 \pi i k x} d k$

         Similarly,

         $\mathcal{F}_{x}^{-1}[\delta(x)](k)=\int_{-\infty}^{\infty} \delta(x) e^{2 \pi i k x} d x=1$

We are solving in the following way:

We have,

The Fourier transform of 1 will be:

$\begin{aligned}\mathcal{F}_{x}[1](k) &=\int_{-\infty}^{\infty} e^{-2 \pi i k x} d x \\\\&=\delta(k)\end{aligned}$