Question

Let $X(j\omega)$ denote the Fourier transform of $x(t)$. If $X(j\omega) = 10 e^{-j\pi f} \left( \frac{\sin(\pi f)}{\pi f} \right)$, then $ \frac{1}{2\pi} \int_{-\infty}^{\infty} X(j\omega) d\omega = \underline{\hspace{1cm}}. $ (where $\omega = 2\pi f$)

• A. 10π
• B. 100
• C. 10
• D. 20π

Hint:

For Fourier transforms, the integral of the sinc function over all \(f\) equals 1, and the exponential term does not contribute at \(f=0\).

Answer 1

The Correct Option is C

We are given the Fourier transform: \[ X(j\omega) = 10 e^{-j\pi f} \left( \frac{\sin(\pi f)}{\pi f} \right) \] The integral we need to solve is: \[ \frac{1}{2\pi} \int_{-\infty}^{\infty} X(j\omega) d\omega \] This integral essentially calculates the DC component (the value of the function at \(f=0\)) of the signal.
1. Simplifying the expression: The term \( \frac{\sin(\pi f)}{\pi f} \) is the sinc function, and its integral over all \(f\) gives 1. The exponential term \( e^{-j\pi f} \) has no impact on the DC component because at \(f=0\), the exponential term equals 1.
2. Final Result: Thus, the integral becomes: \[ \frac{1}{2\pi} \times 10 \times 1 = 10 \] Thus, the correct answer is 10. Therefore, the correct option is (C).