Connexions

You are here: Home » Content » Continuous-Time Fourier Transform (CTFT)
Content Actions
Lenses

What is a lens?

Lenses

A lens is a custom view of Connexions content. You can think of it as a fancy kind of list that will let you see Connexions through the eyes of organizations and people you trust.

What is in a lens?

Lens makers point to Connexions materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

Who can create a lens?

Any individual Connexions member, a community, or a respected organization.

This content is ...
Affiliated with (?)
This content is either by members of the organizations listed or about topics related to the organizations listed. Click each link to see a list of all content affiliated with the organization.
  • This module is included inLens: Rice University OpenCourseWare
    By: OpenCourseWare ConsortiumAs a part of collection:"Signals and Systems"

    Click the "Rice University OCW" link to see all content affiliated with them.

    Rice University OCW
Also in these lenses
  • This module is included inLens: richb's DSP resources
    By: Richard BaraniukAs a part of collection:"Signals and Systems"

    Comments:

    "My introduction to signal processing course at Rice University."

    Click the "richb's DSP" link to see all content selected in this lens.

    richb's DSP
Tags

(?)

These tags come from the endorsement, affiliation, and other lenses that include this content.

Continuous-Time Fourier Transform (CTFT)

Module by: Richard Baraniuk, Melissa Selik

Summary: Details the Continuous-Time Fourier Transform.

Introduction

Due to the large number of continuous-time signals that are present, the Fourier series provided us the first glimpse of how me we may represent some of these signals in a general manner: as a superposition of a number of sinusoids. Now, we can look at a way to represent continuous-time nonperiodic signals using the same idea of superposition. Below we will present the Continuous-Time Fourier Transform (CTFT), also referred to as just the Fourier Transform (FT). Because the CTFT now deals with nonperiodic signals, we must now find a way to include all frequencies in the general equations.

Equations

Continuous-Time Fourier Transform Ω=-ft-Ωtdt Ω t f t Ω t (1)
Inverse CTFT ft=12π-ΩΩtdΩ f t 1 2 Ω Ω Ω t (2)
warning: Do not be confused by notation - it is not uncommon to see the above formula written slightly different. One of the most common differences among many professors is the way that the exponential is written. Above we used the radial frequency variable Ω Ω in the exponential, where Ω=2πf Ω 2 f , but one will often see professors include the more explicit expression, 2πft 2 f t , in the exponential. Click here for an overview of the notation used in Connexion's DSP modules.
The above equations for the CTFT and its inverse come directly from the Fourier series and our understanding of its coefficients. For the CTFT we simply utilize integration rather than summation to be able to express the aperiodic signals. This should make sense since for the CTFT we are simply extending the ideas of the Fourier series to include nonperiodic signals, and thus the entire frequency spectrum. Look at the Derivation of the Fourier Transform for a more in depth look at this.

Relevant Spaces

The Continuous-Time Fourier Transform maps infinite-length, continuous-time signals in L2L2 to infinite-length, continuous-frequency signals in L2L2. Review the Fourier Analysis for an overview of all the spaces used in Fourier analysis.
CTFTspacee.png
Figure 1: Mapping L 2 L 2 in the time domain to L2 L2 in the frequency domain.

Example Problems

Problem 1
Find the Fourier Transform (CTFT) of the function
ft= -αtift00otherwise f t α t t 0 0 (3)
[ Click for Solution 1 ]
Solution 1
In order to calculate the Fourier transform, all we need to use is Equation 1, complex exponentials, and basic calculus.
Ω=-ft-Ωtdt=0-αt-Ωtdt=0-tα+Ωdt=0--1α+Ω Ω t f t Ω t t 0 α t Ω t t 0 t α Ω 0 -1 α Ω (4)
Ω=1α+Ω Ω 1 α Ω (5)
[ Hide Solution 1 ]
Problem 2
Find the inverse Fourier transform of the square wave defined as
XΩ= 1if|Ω|M0otherwise X Ω 1 Ω M 0 (6)
[ Click for Solution 2 ]
Solution 2
Here we will use Equation 2 to find the inverse FT given that t0 t 0 .
xt=12π-MM Ω t dΩ=12π Ω t |Ω,Ω=w=1πtsinMt x t 1 2 Ω M M Ω t Ω w 1 2 Ω t 1 t M t (7)
xt=MπsincMtπ x t M sinc M t (8)
[ Hide Solution 2 ]

Comments, questions, feedback, criticisms?

Send feedback