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,

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

It can also be written as

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

Let

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

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

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

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

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

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

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

Using trigonometric identities, we get,

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

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

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

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

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

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

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

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

Since the sinc function is defined as,

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

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

Therefore, the Fourier transform of the triangular pulse is,

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

Or, it can also be represented as,

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

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