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| <math>X(f)=\mathcal{X}({2\pi}f)=\pi \left[\delta (\omega - \omega) + \delta (\omega + \{2\pi}f_0)\right] \  </math>
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| <math>X(f)=\mathcal{X}({2\pi}f)=\pi \left[\delta (\omega - \omega) + \delta (\omega + \omega)\right] \  </math>
 
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Revision as of 10:49, 18 September 2014


Fourier transform as a function of frequency ω versus Fourier transform as a function of frequency f

A slecture by ECE student JOE BLO

Partly based on the ECE438 Fall 2014 lecture material of Prof. Mireille Boutin.



OUTLINE

  1. Introduction
  2. Theory
  3. Examples
  4. Conclusion
  5. References

Introduction

In my slecture I will explain Fourier transform as a function of frequency ω versus Fourier transform as a function of frequency f (in hertz).

Theory

  • Review of formulas used in ECE 301
CT Fourier Transform $ \mathcal{X}(\omega)=\mathcal{F}(x(t))=\int_{-\infty}^{\infty} x(t) e^{-i\omega t} dt $
Inverse DT Fourier Transform $ \, x(t)=\mathcal{F}^{-1}(\mathcal{X}(\omega))=\frac{1}{2\pi} \int_{-\infty}^{\infty}\mathcal{X}(\omega)e^{i\omega t} d \omega\, $


  • Review of formulas used in ECE 438.
CT Fourier Transform $ X(f)=\mathcal{F}(x(t))=\int_{-\infty}^{\infty} x(t) e^{-i2\pi ft} dt $
Inverse DT Fourier Transform $ \, x(t)=\mathcal{F}^{-1}(X(f))=\int_{-\infty}^{\infty}X(f)e^{i2\pi ft} df \, $

Example

Let's find CTFT of a $ \cos(\omega_0 t) \ $ in two different ways. First way is to change FT pair and make change of variables.
Let $ \, \mathcal\omega={2\pi}f $ and $ \, \mathcal\omega_0={2\pi}f_0 $
CT Fourier Transform $ X(f)=\mathcal{X}({2\pi}f)=\pi \left[\delta (\omega - \omega) + \delta (\omega + \omega)\right] \ $





$ x(t) \ $ $ \longrightarrow $ $ \mathcal{X}(\omega) $
CTFT of a cosine $ \cos(\omega_0 t) \ $ $ \pi \left[\delta (\omega - \omega_0) + \delta (\omega + \omega_0)\right] \ $
  1. THIS IS THE SECONF ITEM

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