Question

Consider a continuous-time, real-valued signal \( f(t) \) whose Fourier transform \[ F(\omega) = \int_{-\infty}^{\infty} f(t) \exp(-j \omega t) \, dt { exists.} \] Which one of the following statements is always TRUE?

• A. \( |F(\omega)| \leq \int_{-\infty}^{\infty} |f(t)| \, dt \)
• B. \( |F(\omega)|>\int_{-\infty}^{\infty} |f(t)| \, dt \)
• C. \( |F(\omega)| \leq \int_{-\infty}^{\infty} f(t) \, dt \)
• D. \( |F(\omega)| \geq \int_{-\infty}^{\infty} f(t) \, dt \)

Hint:

The Fourier transform \( F(\omega) \) of a signal \( f(t) \) is always bounded by the integral of the absolute value of \( f(t) \). The triangle inequality is helpful in establishing this bound.

Answer 1

The Correct Option is A

We are given the Fourier transform of a continuous-time signal \( f(t) \), and we need to determine which of the following statements is always true. Step 1: Bound on \( |F(\omega)| \) 
We use the triangle inequality and absolute value properties of integrals. Specifically: \[ |F(\omega)| = \left| \int_{-\infty}^{\infty} f(t) \exp(-j \omega t) \, dt \right| \leq \int_{-\infty}^{\infty} |f(t)| \, dt. \] This follows from the fact that the magnitude of the complex exponential \( \exp(-j \omega t) \) is always 1, i.e., \( |\exp(-j \omega t)| = 1 \). Therefore, we can bound the magnitude of \( F(\omega) \) by the integral of the absolute value of \( f(t) \). Thus, the inequality \( |F(\omega)| \leq \int_{-\infty}^{\infty} |f(t)| \, dt \) is always true, corresponding to Option (A). 
Step 2: Examine Other Options
Option (B): \( |F(\omega)|>\int_{-\infty}^{\infty} |f(t)| \, dt \) This is incorrect. From the triangle inequality, we know that \( |F(\omega)| \) can never exceed \( \int_{-\infty}^{\infty} |f(t)| \, dt \), so this inequality cannot hold. Option (C): \( |F(\omega)| \leq \int_{-\infty}^{\infty} f(t) \, dt \) 
This is also incorrect. The Fourier transform of a signal depends on the entire signal \( f(t) \), but the absolute value of \( f(t) \) is used in the correct bound, not just \( f(t) \) itself. Option (D): \( |F(\omega)| \geq \int_{-\infty}^{\infty} f(t) \, dt \)
This is incorrect. There is no such general inequality between \( |F(\omega)| \) and \( \int_{-\infty}^{\infty} f(t) \, dt \). The magnitude of the Fourier transform is not necessarily greater than or equal to the integral of \( f(t) \). Thus, the correct answer is (A).