Fourier Transform Derivative at Omega=0

We are asked to find the value of the derivative of the Fourier Transform $X(\omega)$ of the signal $x(t) = e^{-t^4} \cos t$ at $\omega=0$.

Key Fourier Transform Property

A key property of Fourier Transforms states that the derivative of the Fourier transform $X(\omega)$ with respect to $\omega$ is related to the multiplication of the time-domain signal $x(t)$ by $jt$. Specifically:

$ \frac{dX(\omega)}{d\omega} = \mathcal{F}\{-jt \cdot x(t)\} = -j \mathcal{F}\{t \cdot x(t)\} $

Evaluating this derivative at $\omega=0$ gives:

$ \frac{dX(\omega)}{d\omega}\bigg|_{\omega=0} = -j \mathcal{F}\{t \cdot x(t)\} \bigg|_{\omega=0} $

The Fourier Transform evaluated at $\omega=0$ corresponds to the integral of the time-domain function:

$ \mathcal{F}\{t \cdot x(t)\} \bigg|_{\omega=0} = \int_{-\infty}^{\infty} t \cdot x(t) e^{-j(0)t} dt = \int_{-\infty}^{\infty} t \cdot x(t) dt $

Calculating the Integral

Substitute the given signal $x(t) = e^{-t^4} \cos t$ into the integral:

$ \int_{-\infty}^{\infty} t \cdot e^{-t^4} \cos t dt $

Let's analyze the integrand $f(t) = t \cdot e^{-t^4} \cos t$. We check if it's an odd or even function:

• $e^{-t^4}$ is an even function since $e^{-(-t)^4} = e^{-t^4}$.
• $\cos t$ is an even function since $\cos(-t) = \cos t$.
• $t$ is an odd function since $(-t) = -t$.

The product of two even functions ($e^{-t^4}$ and $\cos t$) and an odd function ($t$) results in an odd function:

$ f(-t) = (-t) \cdot e^{-(-t)^4} \cos(-t) = -t \cdot e^{-t^4} \cos t = -f(t) $

The integral of an odd function over symmetric limits ($-\infty$ to $\infty$) is always zero.

$ \int_{-\infty}^{\infty} t \cdot e^{-t^4} \cos t dt = 0 $

Final Derivative Value

Now substitute this result back into the expression for the derivative at $\omega=0$:

$ \frac{dX(\omega)}{d\omega}\bigg|_{\omega=0} = -j \times 0 = 0 $

The value of the derivative of $X(\omega)$ at $\omega=0$ is 0.