1. **Stating the problem:** We want to find the magnitude of the Fourier transform $S(\omega)$ of the function $$f(t) = \begin{cases} 1, & t \in [0, T] \\ 0, & \text{otherwise} \end{cases}$$ 2. **Formula used:** The Fourier transform of $f(t)$ is given by $$S(\omega) = \int_{-\infty}^{\infty} f(t) e^{-i \omega t} dt$$ Since $f(t)$ is zero outside $[0,T]$, the integral reduces to $$S(\omega) = \int_0^T e^{-i \omega t} dt$$ 3. **Evaluating the integral:** We integrate the exponential function: $$S(\omega) = \left[ \frac{e^{-i \omega t}}{-i \omega} \right]_0^T = \frac{e^{-i \omega T} - 1}{-i \omega} = \frac{1 - e^{-i \omega T}}{i \omega}$$ 4. **Magnitude of $S(\omega)$:** The magnitude is $$|S(\omega)| = \left| \frac{1 - e^{-i \omega T}}{i \omega} \right|$$ 5. **Simplifying the magnitude:** Using Euler's formula, $e^{-i \omega T} = \cos(\omega T) - i \sin(\omega T)$, so $$1 - e^{-i \omega T} = 1 - \cos(\omega T) + i \sin(\omega T)$$ The magnitude of the numerator is $$\sqrt{(1 - \cos(\omega T))^2 + (\sin(\omega T))^2} = \sqrt{2 - 2 \cos(\omega T)}$$ Using the identity $1 - \cos x = 2 \sin^2(\frac{x}{2})$, this becomes $$\sqrt{2 \cdot 2 \sin^2\left(\frac{\omega T}{2}\right)} = 2 \left| \sin\left(\frac{\omega T}{2}\right) \right|$$ 6. **Final expression:** Since the denominator magnitude is $|i \omega| = |\omega|$, we have $$|S(\omega)| = \frac{2 \left| \sin\left(\frac{\omega T}{2}\right) \right|}{|\omega|}$$ This shows the magnitude of the Fourier transform of the rectangular pulse $f(t)$.