Step 1: Understanding the Problem:
This question asks for the frequency-domain representation of a signal that has been multiplied by the time variable $t$, using the properties of the Fourier Transform.
Step 2: Key Formula or Approach:
The proof relies on the differentiation-in-frequency property of the Fourier Transform.
The synthesis equation of the Fourier Transform is:
\[ X(\omega) = \int_{-\infty}^{\infty} x(t) e^{-j\omega t} dt \]
Step 3: Detailed Explanation:
• Start by differentiating the Fourier Transform definition with respect to the angular frequency $\omega$:
\[ \frac{dX(\omega)}{d\omega} = \frac{d}{d\omega} \left[ \int_{-\infty}^{\infty} x(t) e^{-j\omega t} dt \right] \]
• Applying Leibniz's rule to differentiate inside the integral sign:
\[ \frac{dX(\omega)}{d\omega} = \int_{-\infty}^{\infty} x(t) \frac{\partial}{\partial \omega} \left( e^{-j\omega t} \right) dt \]
• Taking the partial derivative of the exponential term:
\[ \frac{\partial}{\partial \omega} \left( e^{-j\omega t} \right) = -jt e^{-j\omega t} \]
• Substituting this back into the integral:
\[ \frac{dX(\omega)}{d\omega} = \int_{-\infty}^{\infty} x(t) (-jt) e^{-j\omega t} dt \]
\[ \frac{dX(\omega)}{d\omega} = -j \int_{-\infty}^{\infty} [t x(t)] e^{-j\omega t} dt \]
• The integral on the right-hand side represents the Fourier Transform of the signal $t x(t)$:
\[ \frac{dX(\omega)}{d\omega} = -j \mathcal{F}\{t x(t)\} \]
• Multiplying both sides by $j$ and using the relation $j^2 = -1$:
\[ j \frac{dX(\omega)}{d\omega} = -j^2 \mathcal{F}\{t x(t)\} = \mathcal{F}\{t x(t)\} \]
Step 4: Final Answer
Thus, the Fourier Transform of $t x(t)$ is $j \frac{dX(\omega)}{d\omega}$, matching option (A).