Fourier Transform of Rectangular Function

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Fourier Transform

The Fourier transform of a continuous-time function $\mathrm{x(t)}$ can be defined as,

$$\mathrm{X(\omega) \:=\: \int_{-\infty}^{\infty}x(t)e^{-j\:\omega t}\:dt}$$

Fourier Transform of Rectangular Function

Consider a rectangular function as shown in Figure-1.

It is defined as,

$$\mathrm{rect\left(\frac{t}{\tau}\right)\:=\:\prod\left(\frac{t}{\tau}\right)\:=\:\begin{cases}1 & for\:|t|\:\leq \:\left(\frac{\tau}{2}\right)\\\\0\: &\: otherwise\end{cases}}$$

Given that

$$\mathrm{x(t)\:=\:\prod\left(\frac{t}{\tau}\right)}$$

Hence, from the definition of Fourier transform, we have,

$$\mathrm{F\left[\prod\left(\frac{t}{\tau}\right) \right]\:=\:X(\omega)\:=\:\int_{-\infty}^{\infty}x(t)e^{-j\omega t}\:dt\:=\:\int_{-\infty}^{\infty}\prod\left(\frac{t}{\tau}\right)e^{-j\omega t}\:dt}$$

$$\mathrm{\Rightarrow\:X(\omega)\:=\:\int_{-(\tau/2)}^{(\tau/2)}1\:\cdot\: e^{-j\omega t}\:dt\:=\:\left[\frac{e^{-j\omega t}}{-j\omega} \right]_{-\tau/2}^{\tau/2}}$$

$$\mathrm{\Rightarrow\:X(\omega)\:=\:\left[ \frac{e^{-j\omega (\tau/2)}\:-\:e^{j\omega (\tau/2)}}{-j\omega}\right]\:=\: \left[ \frac{e^{j\omega (\tau/2)}\:-\:e^{-j\omega (\tau/2)}}{j\omega }\right]}$$

$$\mathrm{\Rightarrow\:X(\omega)\:=\:\left[ \frac{2\tau[e^{j\omega (\tau/2)}\:-\:e^{-j\omega (\tau/2)}]}{j\omega\:\cdot\: (2\tau) }\right]\:=\:\frac{\tau}{\omega(\tau/2)}\left[\frac{e^{j\omega (\tau/2)}\:-\:e^{-j\omega (\tau/2)}}{2j} \right]}$$

$$\mathrm{\because \:\left[\frac{e^{j\omega (\tau/2)}\:-\:e^{-j\omega (\tau/2)}}{2j} \right]\:=\:sin\:\omega (\tau/2)}$$

$$\mathrm{\therefore\:X(\omega)\:=\:\frac{\tau}{\omega(\tau/2)}\:\cdot\: sin \omega (\tau/2)\:=\: \tau \left[\frac{ sin\omega (\tau/2)}{\omega (\tau/2)}\right]}$$

$$\mathrm{\because\:sinc \left(\frac{\omega \tau}{2}\right)\:=\:\frac{sin\omega (\tau/2)}{\omega (\tau/2)}}$$

$$\mathrm{\therefore\:X(\omega)\:=\: \tau\:\cdot\: sinc \left(\frac{\omega \tau}{2}\right)}$$

Therefore, the Fourier transform of the rectangular function is

$$\mathrm{F\left[\prod\left(\frac{t}{\tau}\right)\right]\:=\:\tau\:\cdot\: sinc \left(\frac{\omega \tau}{2}\right)}$$

Or, it can also be represented as,

$$\mathrm{\prod\left(\frac{t}{\tau}\right) \overset{FT}{\leftrightarrow}\tau\:\cdot\: sinc \left(\frac{\omega \tau}{2}\right)}$$

Magnitude and phase spectrum of Fourier transform of the rectangular function

The magnitude spectrum of the rectangular function is obtained as −

At $\omega\:=\:0$:

$$\mathrm{sinc\left(\frac{\omega \tau}{2}\right)\:=\:1;\:\:\Rightarrow\:|X(\omega)|\:=\:\tau}$$

At $\mathrm{\left(\frac{\omega \tau}{2}\right)\:=\:± n\pi}$ i.e., at

$$\mathrm{\omega=±\frac{2n\pi}{\tau},\:\:n=1,2,2,3,...}$$

$$\mathrm{sinc\left(\frac{\omega\: \tau}{2}\right)\:=\:0}$$

The phase spectrum is obtained as −

$$\mathrm{\angle\:X(\omega)\:=\:\begin{cases}0\: &\: if\:sinc\:\left(\frac{\omega \tau}{2}\right)\: \gt\: 0\:\\\\±\pi\: & \:if\:sinc\:\left(\frac{\omega \tau}{2}\right)\:\lt\: 0 \end{cases}}$$

The frequency spectrum of the rectangular function is shown in Figure-2.

Note

• The magnitude response between the first two zero crossings is known as the main lobe.
• The portions of the magnitude response for $\mathrm{\omega\:\lt\: -\left(\frac{-2\pi}{\tau}\right)}$ and $\mathrm{\omega \:\gt\: \left( \frac{2\pi}{\tau}\right)}$are known as the side lobes.
• From the magnitude spectrum, it is clear that the majority of the energy of the signal is contained in the main lobe.
• The main lobe becomes narrower with the increase in the width of the rectangular pulse.
• The phase spectrum of the rectangular function is an odd function of the frequency (ω).
• When the magnitude spectrum is positive, then the phase is zero and if the magnitude spectrum is negative, then the phase is $(±\pi)$.