Fourier Transform of a Triangular Pulse

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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 a Triangular Pulse

A triangular signal is shown in Figure-1 −

And it is defined as,

$$\mathrm{\Delta \left(\frac{t}{τ}\right)=\begin{cases}\left( 1+\frac{2t}{τ}\right); & for\:\left(-\frac{τ}{2}\right) <t<0\\left( 1-\frac{2t}{τ}\right); & for\:0<t<\left(\frac{τ}{2}\right)\0 \:;& otherwise\end{cases}}$$

It can also be written as

$$\mathrm{\Delta \left(\frac{t}{τ}\right)=\begin{cases}\left( 1-\frac{2|t|}{τ}\right); & for\:|t|<\left(\frac{τ}{2}\right)\0\: ;& otherwise\end{cases}}$$

Let

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

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

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

$$\mathrm{\Rightarrow\:X(\omega)=\int_{−(τ/2)}^{0}\left(1+\frac{2t}{τ}\right)e^{-j\omega t}dt+\int_{0}^{(τ/2)}\left(1-\frac{2t}{τ}\right)e^{-j\omega t}dt}$$

$$\mathrm{\Rightarrow\:X(\omega)=\int_{0}^{(τ/2)}\left(1-\frac{2t}{τ}\right)e^{j\omega t}dt+\int_{0}^{(τ/2)}\left(1-\frac{2t}{τ}\right)e^{-j\omega t}dt}$$

$$\mathrm{\Rightarrow\:X(\omega)=\int_{0}^{(τ/2)}e^{j\omega t}dt-\int_{0}^{(τ/2)}\frac{2t}{τ}e^{j\omega t}dt+\int_{0}^{(τ/2)}e^{-j\omega t}dt-\int_{0}^{(τ/2)}\frac{2t}{τ}e^{-j\omega t}dt}$$

$$\mathrm{\Rightarrow\:X(\omega)=\int_{0}^{(τ/2)}[e^{j\omega t}+e^{-j\omega t}]dt-\frac{2}{τ}\int_{0}^{(τ/2)}t\cdot[e^{j\omega t}+e^{-j\omega t}]dt}$$

Using trigonometric identities, we get,

$$\mathrm{\Rightarrow\:X(\omega)=\int_{0}^{(τ/2)}2cos\:\omega t\:dt-\frac{2}{τ}\int_{0}^{(τ/2)}2t\:cos\:\omega t\:dt}$$

$$\mathrm{\Rightarrow\:X(\omega)=2\left[ \frac{sin\:\omega t}{\omega}\right]_{0}^{(τ/2)}-\frac{4}{τ}\left\{\left[ \frac{t\:sin\:\omega t}{\omega}\right]_{0}^{(τ/2)} -\int_{0}^{(τ/2)}\left(\frac{sin\:\omega\:t}{\omega}\right)dt\right \}}$$

$$\mathrm{\Rightarrow\:X(\omega)=2\left[ \frac{sin\:\omega t}{\omega}\right]_{0}^{(τ/2)}-\frac{4}{τ}\left \{\left[ \frac{t\:sin\:\omega t}{\omega}\right]_{0}^{(τ/2)} +\left[\frac{cos\:\omega\:t}{\omega^{2}}\right]_{0}^{(τ/2)}\right \}}$$

$$\mathrm{\Rightarrow\:X(\omega)=\frac{2}{\omega}\left[sin \left(\frac{\omega τ}{2} \right) \right]-\frac{4}{\omega τ}\left[\frac{τ}{2}sin \left(\frac{\omega τ}{2} \right)\right]-\frac{4}{\omega^{2} τ}\left[cos \left(\frac{\omega τ}{2} \right) -1\right]}$$

$$\mathrm{\Rightarrow\:X(\omega)=\frac{4}{\omega^{2} τ}\left[1-cos \frac{\omega τ}{2} \right]}$$

$$\mathrm{\left( \because\:2\:sin^{2}\:\theta=\frac{1-cos\:2\:\theta}{2}\right)}$$

$$\mathrm{ \therefore\:X(\omega)=\frac{4}{\omega^{2} τ}\left[2\:sin^{2}\left(\frac{\omega τ}{4}\right)\right]=\frac{8}{\omega^{2} τ}\left[sin^{2}\left(\frac{\omega τ}{4}\right) \right]}$$

$$\mathrm{\Rightarrow\:X(\omega)=\frac{8}{\omega^{2} τ}\left[\frac{sin^{2}\left(\frac{\omega τ}{4}\right)}{\left(\frac{\omega τ}{4}\right)^{2}}\right]\left(\frac{\omega τ}{4}\right) ^{2}}$$

Since the sinc function is defined as,

$$\mathrm{sin c\:(t)=\frac{sin\:t}{t}}$$

$$\mathrm{ \therefore\:X(\omega)=\frac{8}{\omega^{2} τ}\cdot sin c^{2}\left(\frac{\omega τ}{4}\right)\left(\frac{\omega τ}{4}\right)^{2}=\frac{τ}{2}\cdot sin c^{2}\left(\frac{\omega τ}{4}\right)}$$

Therefore, the Fourier transform of the triangular pulse is,

$$\mathrm{F\left[\Delta \left(\frac{t}{τ} \right)\right]=X(\omega)=\frac{τ}{2}\cdot sin c^{2}\left(\frac{\omega τ}{4}\right)}$$

Or, it can also be represented as,

$$\mathrm{\Delta \left(\frac{t}{τ} \right)\overset{FT}{\leftrightarrow}\left[\frac{τ}{2}\cdot sin c^{2}\left(\frac{\omega τ}{4}\right)\right]}$$

The graphical representation of magnitude spectrum of a triangular pulse is shown in Figure-2.

Updated on 08-Dec-2021 07:13:24