Fourier Transform of Two-Sided Real Exponential Functions

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 Two-Sided Real Exponential Function

Let a two-sided real exponential function as,

$$\mathrm{x(t)=e^{-a|t|}}$$

The two-sided or double-sided real exponential function is defined as,

$$\mathrm{e^{-a|t|}=\begin{cases}e^{at} & for\:t ≤ 0\\e^{-at} & for\:t ≥ 0 \end{cases} =e^{at}u(-t)+e^{-at}u(t) }$$

Where, the functions $u(t)$ and $u(-t)$ are the unit step function and time reversed unit step function, respectively.

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

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

$$\mathrm{\Rightarrow\:X(\omega)=\int_{−\infty}^{\infty}[e^{at}u(-t)+e^{-at}u(t)]e^{-j\omega t}dt}$$

$$\mathrm{\Rightarrow\:X(\omega)=\int_{−\infty}^{0}e^{at}e^{-j\omega t}dt+\int_{0}^{\infty}e^{-at}e^{-j\omega t}dt}$$

$$\mathrm{\Rightarrow\:X(\omega)=\int_{-\infty}^{0}e^{(a-j\omega)t}dt+\int_{0}^{\infty}e^{-(a+j\omega)t}dt}$$

$$\mathrm{\Rightarrow\:X(\omega)=\int_{0}^{\infty}e^{-(a-j\omega)t}dt+\int_{0}^{\infty}e^{-(a+j\omega)t}dt}$$

$$\mathrm{\Rightarrow\:X(\omega)=\left[\frac{e^{-(a-j\omega)t}}{-(a-j\omega)}\right]_{0}^{\infty}+\left[\frac{e^{-(a+j\omega)t}}{-(a+j\omega)}\right]_{0}^{\infty}=\left[\frac{e^{-\infty}-e^{0}}{-(a-j\omega)} \right]+\left[\frac{e^{-\infty}-e^{0}}{-(a+j\omega)} \right]}$$

$$\mathrm{\Rightarrow\:X(\omega)=\frac{1}{a-j\omega}+\frac{1}{a+j\omega}=\frac{2a}{a^{2}+\omega^{2}}}$$

Therefore, the Fourier transform of a two-sided real exponential function is,

$$\mathrm{F[e^{-a|t|}]=X(\omega)=\frac{2a}{a^{2}+\omega^{2}}}$$

Or, it can also be represented as,

$$\mathrm{e^{-a|t|}\overset{FT}{\leftrightarrow}\frac{2a}{a^{2}+\omega^{2}}}$$

Magnitude and phase representation of Fourier transform of the two-sided real exponential function −

$$\mathrm{Magnitude,\:|X(\omega)|=\frac{2a}{a^{2}+\omega^{2}};\:\:for\:all\:\omega}$$

$$\mathrm{Phase,\angle X(\omega)=0;\:\:for\:all\:\omega}$$

The graphical representation of the two-sided real exponential function with its magnitude and phase spectrum is shown in the figure.

raja
Updated on 09-Dec-2021 07:01:15

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