Find the Fourier transform of the signal x(t)=e^{-t} u(t), where u(t) is the unit step function.


Find the Fourier transform of the signal

{eq}x(t)=e^{-t} u(t){/eq}, where {eq}u(t){/eq} is the unit step


Fourier transform:

Fourier transform is a method, or we can say technique in which we transform a function of time say x(t) to a function of frequency. Fourier transform is related to Fourier series to some extent. So, we can say the Fourier transform is the special case of Fourier series because the derivation of the Fourier transform is come from here. Fourier series is widely used in many fields such as physics, mathematics, engineering, etc.

Answer and Explanation:

We have given that,

{eq}x\left( t \right) = e^{ - t} u\left( t \right), {/eq} here is the unit step.

We know that the Fourier transform of this given signal is,

{eq}\begin{align*} X\left( w \right) &= \int\limits_{ - \infty }^\infty {x\left( t \right)e^{ - jwt} dt} \\ &= \int\limits_{ - \infty }^\infty {e^{ - t} e^{ - jwt} u\left( t \right)dt} \\ \end{align*} {/eq}

Since, unit step is defined by {eq}\left( {0,\infty } \right) {/eq}


{eq}\begin{align*} X\left( w \right) &= \int\limits_{ - \infty }^\infty {e^{ - t} \cdot e^{ - jwt} dt} \\ &= \int\limits_{ - \infty }^\infty {e^{ - t\left( {1 + jwt} \right)} dt} \\ &= \left[ {\dfrac{{e^{^{ - t\left( {jw + 1} \right)} } }}{{ - \left( {jw + 1} \right)}}} \right]_0^\infty \\ &= \left[ {\dfrac{{e^{ - \infty \left( {jw + 1} \right)} }}{{ - \left( {jw + 1} \right)}}} \right] - \left[ {\dfrac{{e^{ - 0\left( {jw + 1} \right)} }}{{ - \left( {jw + 1} \right)}}} \right] \\ &= 0 + \dfrac{1}{{1 + jw}} \\ X\left( w \right) &= \dfrac{1}{{1 + jw}} \\ \end{align*} {/eq}

Hence, Fourier transform of given signal is {eq}\dfrac{1}{{1 + jw}}. {/eq}

Learn more about this topic:

Basic Calculus: Rules & Formulas

from Calculus: Tutoring Solution

Chapter 3 / Lesson 6

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