Question

We are going to derive the FT a Gaussian signal x(t) defined as

x(t)= (1/ √2πσ²)e⁻ᵗ²/²σ²
As is often the case, it is not easy to use the definitions of the FT to derive X(jω).
Instead, we rely on the properties of FTs.
Derive the value of X(j0).
Hint: use the property of normalized Gaussians: ∫ x(t)= (1/ √2πσ²)e⁻ᵗ²/²σ² dt=1

Answer 1

To derive the value of X(j0) for a Gaussian signal, we can use the property of normalized Gaussians and the properties of Fourier Transforms. By substituting ω = 0 into the Fourier Transform formula and applying the property of normalized Gaussians, we find that X(j0) equals to 1.

Step-by-step explanation:

To derive the value of X(j0), we need to find the Fourier Transform of the Gaussian signal x(t). Instead of using the definition of FT, we can leverage the properties of FTs. The property of normalized Gaussians states that the integral of the Gaussian function is equal to 1. Using this property, we can find the value of X(j0).

First, we note that the Fourier Transform of the Gaussian function is another Gaussian function. Therefore, we can rewrite the given Gaussian signal as x(t) = (1/√2πσ²)e^(-t²/2σ²).

To find X(j0), we substitute ω = 0 into the Fourier Transform formula. The Fourier Transform of the Gaussian function is given by X(jω) = ∫ x(t)e^(-jwt) dt. Substituting ω = 0, we have X(j0) = ∫ x(t) dt.

Applying the property of normalized Gaussians (∫ x(t) dt = 1), we can conclude that X(j0) = 1.