Fourier Transform of Rectangular Function

Signals and SystemsElectronics & ElectricalDigital Electronics

Fourier Transform

The Fourier transform of a continuous-time function $x(t)$ 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,

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

Given that

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

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

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

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

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

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

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

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

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

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

Therefore, the Fourier transform of the rectangular function is

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

Or, it can also be represented as,

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

Magnitude and phase spectrum of Fourier transform of the rectangular function

The magnitude spectrum of the rectangular function is obtained as −

At $\omega=0$:

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

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

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

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

The phase spectrum is obtained as −

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

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 $\omega\:< -\left( \frac{-2\pi}{τ}\right)$ and $\omega > \left( \frac{2\pi}{τ}\right)$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 $(±\pi)$.

raja
Updated on 08-Dec-2021 07:21:30

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