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#Conclusion
 
#Conclusion
 
#References
 
#References
 
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----
 
==Introduction==
 
==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).
 
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==
 
==Theory==
 
* Review of formulas used in ECE 301
 
* Review of formulas used in ECE 301
 
 
{|
 
{|
 
|-
 
|-
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| <math>\, x(t)=\mathcal{F}^{-1}(\mathcal{X}(\omega))=\frac{1}{2\pi} \int_{-\infty}^{\infty}\mathcal{X}(\omega)e^{i\omega t} d \omega\,</math>
 
| <math>\, x(t)=\mathcal{F}^{-1}(\mathcal{X}(\omega))=\frac{1}{2\pi} \int_{-\infty}^{\infty}\mathcal{X}(\omega)e^{i\omega t} d \omega\,</math>
 
|}
 
|}
 
 
 
* Review of formulas used in ECE 438.  
 
* Review of formulas used in ECE 438.  
 
{|
 
{|
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|}
 
|}
  
*This is a [[Main_Page| link to a Rhea page]]
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*For more formulas see the [[CTFourierTransformPairsCollectedfromECE301withomega|table of CT Fourier transform pairs and properties]]
 +
----
 
==Examples==
 
==Examples==
1) Let's compute FT of a cosine in two different ways:
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1)  
 
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First way is by changing FT pair and changing of variable
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Let
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<math>\, \mathcal\omega={2\pi}f</math> ,  <math>\, \mathcal\omega_0={2\pi}f_0</math>
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+
Also recall that
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<math> \displaystyle\delta(\alpha f)=\frac{1}{\alpha}\delta(f)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;for\;\;\alpha>0</math>
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+
 
{|
 
{|
|-
 
 
| align="right" style="padding-right: 1em;" | <br>  
 
| align="right" style="padding-right: 1em;" | <br>  
 
| <math> x(t) \  </math>   
 
| <math> x(t) \  </math>   
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|-
 
|-
 
|}
 
|}
 +
Let's compute FT of a cosine in two different ways:
 +
 +
First way is by changing FT pair and changing of variable
 +
 +
Let
 +
<math>\, \mathcal\omega={2\pi}f</math> ,  <math>\, \mathcal\omega_0={2\pi}f_0</math>
 +
 +
Also recall that
 +
<math> \displaystyle\delta(\alpha f)=\frac{1}{\alpha}\delta(f)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;for\;\;\alpha>0</math>
  
 
{|
 
{|
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|<math>X(f)= \frac{1}{2}\left[\delta (f - f_0) + \delta (f + f_0)\right] \ </math>  
 
|<math>X(f)= \frac{1}{2}\left[\delta (f - f_0) + \delta (f + f_0)\right] \ </math>  
 
|}
 
|}
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Second way is by direct using CTFT formula
Second way is by direct using CTFT formula
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{|  
 
{|  
 
|-  
 
|-  
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|}
 
|}
  
2) Let's compute FT of a rect(t)
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2) Let's find CTFT of a shifted unit impulse:
Keep in mind that  
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 +
<math>\delta (t-t_0)\ </math>
 +
 
 +
Keep in mind that:
 
{|
 
{|
 
|-
 
|-
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| <math>X(f)=\mathcal{F}(x(t))=\int_{-\infty}^{\infty} x(t) e^{-i2\pi ft} dt</math>
 
| <math>X(f)=\mathcal{F}(x(t))=\int_{-\infty}^{\infty} x(t) e^{-i2\pi ft} dt</math>
 
|}
 
|}
 
+
{|
 +
|-
 +
| align="right" style="padding-right: 1em;" |  From above equation
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| <math>X(f)=\mathcal{F}(\delta (t-t_0))=\int_{-\infty}^{\infty} \delta (t-t_0) e^{-i2\pi ft} dt</math>
 +
|}
 +
{|
 +
|-
 +
| align="right" style="padding-right: 1em;" |  Thus we get
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|  <math>X(f)=e^{-i2\pi ft_0} = e^{-i\omega t_0} </math>
 +
|}
 +
----
 
==Conclusion==
 
==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
 
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==References==
+
----
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IMPORTANT: DO NOT PLAGIARIZE. If you use other material than Prof. Boutin's lecture material, you must cite your sources. Do not copy text word for word from another source; rephrase everything using your own words. Similarly for graphs, illustrations, pictures, etc. Make your own! Do not copy them from other sources.
+
 
----
 
----
 +
==References==
 +
[1].Mireille Boutin, "ECE438 Digital Signal Processing with Applications," Purdue University August 26,2009
 
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(create a question page and put a link below)
 
 
==[[Slecture_Fourier_transform_w_f_ECE438_review|Questions and comments]]==
 
==[[Slecture_Fourier_transform_w_f_ECE438_review|Questions and comments]]==
  
 
If you have any questions, comments, etc. please post them on [[Slecture_Fourier_transform_w_f_ECE438_review|this page]].
 
If you have any questions, comments, etc. please post them on [[Slecture_Fourier_transform_w_f_ECE438_review|this page]].
 
----
 
----
[[2014_Fall_ECE_438_Boutin|Back to ECE438, Fall 2014]]
+
[[2014_Fall_ECE_438_Boutin_digital_signal_processing_slectures|Back to ECE438 slectures, Fall 2014]]

Latest revision as of 17:51, 16 March 2015


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

If you have any questions, comments, etc. please post them on this page.


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