Question

Let \( X(\omega) \) be the Fourier transform of the signal \( x(t) = e^{-4t \cos(t), -\inftyLt;tLt;\infty \). The value of the derivative of \( X(\omega) \) at \( \omega = 0 \) is \_\_\_\_\_ (rounded off to 1 decimal place).}

Hint:

The derivative of the Fourier transform at zero frequency can often simplify due to symmetry properties of the signal.

Answer 1

Given: \[ x(t) = e^{-4t} \cos(t) \] Step 1: Fourier transform: \[ X(\omega) = \int_{-\infty}^{\infty} x(t) e^{-j\omega t} \, dt \] Step 2: Derivative of \( X(\omega) \): \[ \frac{dX(\omega)}{d\omega} = \int_{-\infty}^{\infty} (-jt)x(t)e^{-j\omega t} \, dt \] At \( \omega = 0 \): \[ \frac{dX(0)}{d\omega} = \int_{-\infty}^{\infty} (-jt)x(t) \, dt \] Step 3: Simplification: Given that \( x(t) \) is an even signal, the integral evaluates to \( 0 \). Final Answer: \( 0.0 \) % Quick tip