Question

The Fourier transform \(X(\omega)\) of the signal \(x(t)\) is given by \[ X(\omega) = \begin{cases} 1, & |\omega| < W_0
0, & |\omega| > W_0 \end{cases} \] Which one of the following statements is true?

• A. \(x(t)\) tends to be an impulse as \(W_0 \to \infty\).
• B. \(x(0)\) decreases as \(W_0\) increases.
• C. At \(t = \tfrac{\pi}{2W_0}, \, x(t) = -\tfrac{1}{\pi}\).
• D. At \(t = \tfrac{\pi}{2W_0}, \, x(t) = \tfrac{1}{\pi}\).

Hint:

A rectangular spectrum in frequency corresponds to a sinc in time. As the frequency bandwidth increases, the sinc shrinks in width and approaches an impulse in time domain.

Answer 1

The Correct Option is A

Step 1: Inverse Fourier transform.
The signal \(x(t)\) is obtained from the given \(X(\omega)\): \[ x(t) = \frac{1}{2\pi} \int_{-W_0}^{W_0} e^{j\omega t} d\omega. \]

Step 2: Evaluate integral.
\[ x(t) = \frac{1}{2\pi} \left[\frac{e^{j\omega t}}{jt}\right]_{-W_0}^{W_0} = \frac{1}{2\pi jt} \left( e^{jW_0 t} - e^{-jW_0 t} \right). \] \[ x(t) = \frac{1}{\pi t} \sin(W_0 t). \]

Step 3: Value at \(t=0\).
By L'Hospital's rule: \[ x(0) = \lim_{t \to 0} \frac{\sin(W_0 t)}{\pi t} = \frac{W_0}{\pi}. \] Thus, as \(W_0\) increases, \(x(0)\) increases, not decreases. Hence option (B) is false.

Step 4: Asymptotic behavior.
As \(W_0 \to \infty\): \[ x(t) = \frac{\sin(W_0 t)}{\pi t} \to \delta(t), \] since the sinc pulse becomes narrower and taller, converging to the delta function. This matches option (A).

Step 5: Check at \(t=\tfrac{\pi}{2W_0}\).
\[ x\left(\tfrac{\pi}{2W_0}\right) = \frac{\sin(\pi/2)}{\pi \cdot (\pi/(2W_0))} = \frac{2W_0}{\pi^2}, \] which is not equal to \(\pm \tfrac{1}{\pi}\). So both options (C) and (D) are incorrect.

Final Answer: \[ \boxed{x(t) \to \delta(t) \;\; \text{as} \;\; W_0 \to \infty} \]