FT of unit impulse and rectangular pulse

FT of the unit impulse $\delta(t)$

$$X(j\omega) = \int_{-\infty}^{\infty}\delta(t)\,e^{-j\omega t}\,dt = e^{-j\omega\cdot 0} = 1$$

using the sifting property $\int\delta(t)f(t)dt = f(0)$. So $\delta(t) \leftrightarrow 1$.

Discussion: The impulse has a flat spectrum: it contains all frequencies with equal amplitude and zero phase. This is why the response of an LTI system to $\delta(t)$ (the impulse response) gives its full frequency response $H(j\omega)$.

FT of a rectangular pulse

Let $x(t) = 1$ for $|t| < \tau/2$ and $0$ otherwise (width $\tau$, height 1).

$$\begin{aligned} X(j\omega) &= \int_{-\tau/2}^{\tau/2}e^{-j\omega t}dt = \frac{e^{j\omega\tau/2} - e^{-j\omega\tau/2}}{j\omega} \\ &= \frac{2\sin(\omega\tau/2)}{\omega} = \tau\,\frac{\sin(\omega\tau/2)}{\omega\tau/2} \end{aligned}$$

Discussion:

• The spectrum is a real sinc function, peak value $\tau$ (the pulse area) at $\omega = 0$.
• Zeros at $\omega = \pm\frac{2\pi}{\tau}, \pm\frac{4\pi}{\tau}, \dots$; most energy lies in the main lobe $|\omega| < 2\pi/\tau$.
• The spectrum extends to infinity: a time-limited signal is not band-limited.
• Narrower pulse means wider spectrum. As $\tau \to 0$ with area kept at 1 (height $1/\tau$), the pulse becomes $\delta(t)$ and the sinc flattens to 1, which matches the first result.

 delta(t)  ->  1             rect  ->  sinc
   ^         1 +---------    1 +--+       tau ^
   |           |                |  |         .|.
---+---> t  ---+-----> w   -----+--+--   . . | . .
   0           0             -t/2 t/2  -----'--+--'---> w
                                         -2pi/t  2pi/t