How to find the phase spectrum of a rectangular pulse?
The Fourier transform of a rectangular pulse
$$ x(t) =
\begin{cases}
1, & \text{for $|t| \le \tau /2$ } \\
0, & \text{otherwise}
\end{cases}
$$
is given by:
$$F[x(t)]=\tau [\frac{\sin \omega(\tau /2)}{\omega (\tau /2)}]$$
In general, the Fourier Transform, $X(\omega)$ is id a complex valued function of $\omega$. Therefore, $X(\omega)$ can be written as:
$$X(\omega)=X_R(\omega)+jX_I(\omega)$$
The magnitude of $X(\omega)$ is given by
$$\vert X(\omega) \vert= \sqrt {(X_R(\omega))^2+(X_I(\omega))^2}$$
The phase of $X(\omega)$ is given by
$$\angle{X(\omega)}=\tan^{-1}\frac{X_I(\omega)}{X_R(\omega}$$
Question:
1)How can we find the phase spectrum of a rectangular pulse as there doesn't appear to be any imaginary part in $X(\omega)$?
2)Why is the phase spectrum changing the way it is when we are time
shifting the rectangular pulse from the origin?3) Any other insight regarding the magnitude and phase spectrum is
welcome.
Best Answer
You must go back to basics and do the integration of the Fourier transform by yourself. Then you get the spectrum of an arbitary single rectangular pulse, say amplitude A, starts at t=T1 and stops at t=T2 or as well t=T1+T. Textbook tables have only special cases which are simplified to look out as slick as possible. The result is a sum of 2 complex exponentials.
Delayed pulse has got frequency dependent add-on to the phase angle. The add-on is a line -2 *Pi *f *Td , where Td is the delay and f is frequency. That means 2Pi phase lag exactly at frequency 1/Td which should be understandable.