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

A slecture by ECE student Dauren Nurmaganbetov

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 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 Fourier Transform $ \, x(t)=\mathcal{F}^{-1}(X(f))=\int_{-\infty}^{\infty}X(f)e^{i2\pi ft} df \, $

Examples

1)


$ x(t) \ $ $ \longrightarrow $ $ \mathcal{X}(\omega) $
$ \cos(\omega_0 t) \ $ $ \pi \left[\delta (\omega - \omega_0) + \delta (\omega + \omega_0)\right] \ $

Let's compute FT of a cosine in two different ways:

First way is by changing FT pair and changing of variable

Let

$ \, \mathcal\omega={2\pi}f $ ,  $ \, \mathcal\omega_0={2\pi}f_0 $

Also recall that

$  \displaystyle\delta(\alpha f)=\frac{1}{\alpha}\delta(f)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;for\;\;\alpha>0 $
$ X(f)=\mathcal{X}({2\pi}f)=\pi \left[\delta ({2\pi}f - {2\pi}f_0) + \delta ( {2\pi}f+ {2\pi}f_0)\right] \ $
$ X(f)= \pi \left[\frac{1}{2\pi }\delta (f - f_0) + \frac{1}{2\pi }\delta (f + f_0)\right] \ $
$ X(f)= \frac{1}{2}\left[\delta (f - f_0) + \delta (f + f_0)\right] \ $

Second way is by direct using CTFT formula

$ X(f)= \frac{1}{2} \left[\delta (f - \frac{\omega_0}{2\pi}) + \delta (f + \frac{\omega_0}{2\pi})\right] \ $

2) Let's find CTFT of a shifted unit impulse:

$ \delta (t-t_0)\ $

Keep in mind that:

CT Fourier Transform $ X(f)=\mathcal{F}(x(t))=\int_{-\infty}^{\infty} x(t) e^{-i2\pi ft} dt $
From above equation $ X(f)=\mathcal{F}(\delta (t-t_0))=\int_{-\infty}^{\infty} \delta (t-t_0) e^{-i2\pi ft} dt $
Thus we get $ X(f)=e^{-i2\pi ft_0} = e^{-i\omega t_0} $

Conclusion

Observe that the expressions for the FT are different because we used change of variables.

Also notice that one can transform one expression into the other using the scaling property of the Dirac delta


References

[1].Mireille Boutin, "ECE438 Digital Signal Processing with Applications," Purdue University August 26,2009



Questions and comments

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Back to ECE438 slectures, Fall 2014

Alumni Liaison

has a message for current ECE438 students.

Sean Hu, ECE PhD 2009