How to find the phase spectrum of a rectangular pulse?

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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:


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



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.

  • 1
    $\begingroup$ Hint, time shift property of FT $\endgroup$
    – user28715
    Commented Apr 12, 2018 at 16:20

1 Answer 1


For the rectangular pulse without shifting, the phase spectrum its quite simple to obtain, since its Fourier Transform $F(\omega)$ is a real function. Real functions may only lay on the real axis, thus having $0$ or $\pm \pi$ phases, depending on the positive or negative sign of the function, respectively. Try visualizing the complex plane to realize this.

Now, if you're new to DSP, for $g(t)$ it might be tricky. First, take into account the scale change when dividing by 2 in $g(t)$. This expands the spectrum by a proportion of 2, i.e. $f(t/2)$ transforms as: $2~F(2\omega)$. That's why the $sinc$ function in $g(t)$ is twice as wide as in $f(t)$. Then the amplitude is multiplied by 3, that's why $G(\omega) = 6$ for $\omega = 0$.

Due to time shifting, you'd need to multiply $F(\omega)$ by a complex exponential. Recall the time-shifting Fourier transform theorem: $f(t - t_0)$ transforms as $F(\omega)~e^{-j\omega t_0}$.

The amplitude spectrum isn't modified by a time shift (since $|e^{-j\omega t_0}| = 1$), but the phase spectrum is added to $-\omega t_0$, which is the phase of the complex exponential (i.e., $\angle e^{-j\omega t_0} = -\omega t_0$).

Finally, you'd only need to add to the phase spectrum of $f(t)$ the linear function of $\omega$: $\angle H(\omega) = -4\omega$ (since $t_0 = 4$ in this case), which has a negative slope.

That's why you have in Fig. (d) the sum of a negative-slope linear function and a rectangular pulse train of $\pm \pi$ values.

This is basic DSP, I'd recommend to take a look to these theorems at Wikipedia for a very first glance: https://en.wikipedia.org/wiki/Fourier_transform#Properties_of_the_Fourier_transform

  • $\begingroup$ You said that if the $F(\omega)$ is real the function then the phase spectrum is: $0$ or $\pm \pi$ phases, depending on the positive or negative sign of the function, respectively. However if we take the transform of $rect(t)$ which is $sinc({\omega\over 2})$ then we have a function which is pair and is either positive, negative or nil. However the phase spectrum is shown in the figure (c) as a rect impulse train which are positive for negative frequencies and negative for positive frequencies. Why is that ? $\endgroup$
    – Bemipefe
    Commented Feb 21, 2022 at 11:57
  • $\begingroup$ Ok I got it I was confused by the graph on the left which is related to $|sinc({\omega\over2})| $ while for the phase spectrum you have to use $sinc({\omega\over2})$. $\endgroup$
    – Bemipefe
    Commented Feb 23, 2022 at 12:19

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