A pulse described by the Dirac delta function \(\delta(t)\) has negligible duration, infinite height at \(t=0\), but unit area. Find the amplitude spectrum of \(\delta(t)\) as a limiting case by setting the area of the rectangular pulse in Prob. 1222 equal to unity and then letting \(\tau \rightarrow 0 .\) Answer: \(F(\omega)=1\).

Let's define a rectangular pulse \(g(t)\) with width \(\tau\) and height \(\frac{1}{\tau}\), so that its area is equal to 1. We then need to find its Fourier transform. The pulse can be defined as follows: $$ g(t) = \begin{cases} \frac{1}{\tau}, & -\frac{\tau}{2} \le t \le \frac{\tau}{2} \\ 0, & \text{otherwise} \end{cases} $$

To find the pulse's amplitude spectrum, we need to compute its Fourier transform. We can write the Fourier transform of \(g(t)\) as: $$ G(\omega) = \int_{-\infty}^{\infty} g(t) e^{-j \omega t} dt $$ When \(t\) is outside the pulse, \(g(t)\) is equal to 0. So we will compute the integral over the pulse's region: $$ G(\omega) = \int_{-\frac{\tau}{2}}^{\frac{\tau}{2}} \frac{1}{\tau} e^{-j \omega t} dt $$

Now we solve the integral: $$ G(\omega) = \frac{1}{\tau} \int_{-\frac{\tau}{2}}^{\frac{\tau}{2}} e^{-j \omega t} dt \\ G(\omega)=\frac{1}{\tau} \left[\frac{e^{-j\omega t}}{-j\omega}\right]_{-\frac{\tau}{2}}^{\frac{\tau}{2}} \\ G(\omega)= \frac{1}{\tau}\left(\frac{e^{-j\omega \frac{\tau}{2}} - e^{j\omega \frac{\tau}{2}}}{-j \omega}\right) $$

We can simplify the expression using the sine function: $$ G(\omega)= \frac{1}{\tau} \cdot \frac{2}{\omega} \cdot \frac{e^{-j\omega \frac{\tau}{2}} - e^{j\omega \frac{\tau}{2}}}{-2j} \\ G(\omega)= \frac{1}{\tau} \cdot \frac{2}{\omega} \cdot \sin\left(\frac{\omega \tau}{2}\right) $$

Now, we need to find the limit \(\tau \rightarrow 0\) of \(G(\omega)\) to obtain the amplitude spectrum of the Dirac delta function: $$ F(\omega) = \lim_{\tau \to 0} G(\omega) = \lim_{\tau \to 0} \frac{1}{\tau} \cdot \frac{2}{\omega} \cdot \sin\left(\frac{\omega \tau}{2}\right) $$ As \(\tau \rightarrow 0\), the sine term approaches zero as well since \(\sin(0) = 0\), but the product \(\frac{1}{\tau} \cdot \frac{2}{\omega}\) approaches infinity. We can use L'Hospital's rule to find this limit: $$ F(\omega) = \lim_{\tau \to 0} \frac{\sin\left(\frac{\omega \tau}{2}\right)}{\frac{\omega \tau}{2}} $$ Applying L'Hospital's rule, we differentiate both numerator and denominator with respect to \(\tau\): $$ F(\omega) = \lim_{\tau \to 0} \frac{\frac{1}{2} \omega \cos\left(\frac{\omega \tau}{2}\right)}{\frac{1}{2} \omega} \\ F(\omega) = \lim_{\tau \to 0} \cos\left(\frac{\omega \tau}{2}\right) $$ Finally, when \(\tau=0\): $$ F(\omega) = \cos(0) = 1 $$ So the amplitude spectrum of the Dirac delta function is a constant, and \(F(\omega)=1\).