Signals & System Lecture 13

Fourier Transform Basics

For continuous-time signals, the Fourier transform of a function ( x(t) ) is defined as:

$$X(\omega )=\mathcal{F}\{x(t)\}={\int }_{-\infty }^{\infty }x(t)e^{-j\omega t}dt$$

Where:

• ( X(\omega) ) is the Fourier transform of ( x(t) )
• ( \omega ) is the angular frequency

Inverse Fourier Transform

The inverse Fourier transform is given by:

$$x(t)={\mathcal{F}}^{-1}\{X(\omega )\}=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }X(\omega )e^{j\omega t}d\omega$$

This allows us to recover the time-domain signal from its frequency-domain representation.

Important Note:

• No need of ROC (Region of Convergence): For Fourier transforms, unlike the Laplace transform, there is no requirement for specifying the ROC.

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Example: Fourier Transform of a Given Signal

Find the Fourier transform of the following signal:

Given Signal:

$$x(t)=e^{-at}u(t)$$

Where:

• ( u(t) ) is the unit step function
• ( a > 0 )

Solution:

The Fourier transform of $(x(t)=e^{-at}u(t)$ is:

$$X(\omega )={\int }_0^{\infty }{e}^{-at}{e}^{-j\omega t}dt$$

We solve this integral:

$$X(\omega )={\int }_0^{\infty }{e}^{-(a+j\omega )t}dt$$

This is a standard exponential integral:

$$X(\omega )={\left[\frac{1}{a+j\omega }\right]}_0^{\infty }$$

Evaluating the integral gives:

$$X(\omega )=\frac{1}{a+j\omega }$$

Thus, the Fourier transform of ( x(t) = e^{-at} u(t) ) is:

$$X(\omega )=\frac{1}{a+j\omega }$$

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Summary:

• The Fourier transform of an exponentially decaying signal multiplied by a unit step is $(\frac{1}{a+j\omega })$.