(/)
 
 
(102 intermediate revisions by 3 users not shown)
Line 1: Line 1:
−
<math>\int_{-\infty}</math>
+
[[Category:slecture]]
 +
[[Category:ECE438Fall2014Boutin]]
 +
[[Category:ECE]]
 +
[[Category:ECE438]]
 +
[[Category:signal processing]] 
 +
 
 +
<center><font size= 4>
 +
Fourier transform as a function of frequency ω versus Fourier transform as a function of frequency f
 +
</font size>
 +
 
 +
A [https://www.projectrhea.org/learning/slectures.php slecture] by [[ECE]] student Dauren Nurmaganbetov
 +
 
 +
Partly based on the [[2014_Fall_ECE_438_Boutin|ECE438 Fall 2014 lecture]] material of [[user:mboutin|Prof. Mireille Boutin]].
 +
</center>
 +
----
 +
----
 +
==OUTLINE==
 +
#Introduction
 +
#Theory
 +
#Examples
 +
#Conclusion
 +
#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
 +
{|
 +
|-
 +
| align="right" style="padding-right: 1em;" | CT Fourier Transform
 +
| <math>\mathcal{X}(\omega)=\mathcal{F}(x(t))=\int_{-\infty}^{\infty} x(t) e^{-i\omega t} dt</math>
 +
|-
 +
| align="right" style="padding-right: 1em;" |  Inverse Fourier Transform
 +
| <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.
 +
{|
 +
|-
 +
| align="right" style="padding-right: 1em;" |  CT Fourier Transform
 +
| <math>X(f)=\mathcal{F}(x(t))=\int_{-\infty}^{\infty} x(t) e^{-i2\pi ft} dt</math>
 +
|-
 +
| align="right" style="padding-right: 1em;" | Inverse Fourier Transform
 +
| <math>\, x(t)=\mathcal{F}^{-1}(X(f))=\int_{-\infty}^{\infty}X(f)e^{i2\pi ft} df \,</math>
 +
|}
 +
 
 +
*For more formulas see the [[CTFourierTransformPairsCollectedfromECE301withomega|table of CT Fourier transform pairs and properties]]
 +
----
 +
==Examples==
 +
1)
 +
{|
 +
| align="right" style="padding-right: 1em;" | <br>
 +
| <math> x(t) \  </math> 
 +
| <math>\longrightarrow</math>
 +
| <math> \mathcal{X}(\omega) </math>
 +
|
 +
|-
 +
| align="right" style="padding-right: 1em;" |
 +
| <math>\cos(\omega_0 t) \ </math>
 +
|
 +
| <math> \pi \left[\delta (\omega - \omega_0) + \delta (\omega + \omega_0)\right] \ </math>
 +
|-
 +
|}
 +
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>
 +
 
 +
{|
 +
|-
 +
|<math>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] \  </math>
 +
|-
 +
|<math>X(f)= \pi \left[\frac{1}{2\pi }\delta (f - f_0) + \frac{1}{2\pi }\delta (f + f_0)\right] \ </math>
 +
|-
 +
|<math>X(f)= \frac{1}{2}\left[\delta (f - f_0) + \delta (f + f_0)\right] \ </math>
 +
|}
 +
Second way is by direct using CTFT formula
 +
{|
 +
|-
 +
<math>X(f)= \frac{1}{2} \left[\delta (f - \frac{\omega_0}{2\pi}) + \delta (f + \frac{\omega_0}{2\pi})\right] \ </math>
 +
|}
 +
 
 +
2) Let's find CTFT of a shifted unit impulse:
 +
 
 +
<math>\delta (t-t_0)\ </math>
 +
 
 +
Keep in mind that:
 +
{|
 +
|-
 +
| align="right" style="padding-right: 1em;" |  CT Fourier Transform
 +
| <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
 +
| <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
 +
|  <math>X(f)=e^{-i2\pi ft_0} = e^{-i\omega t_0} </math>
 +
|}
 +
----
 +
==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
 +
----
 +
----
 +
==[[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]].
 +
----
 +
[[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.

Back to ECE438 slectures, Fall 2014

Retrieved from "https://www.projectrhea.org/rhea/index.php?title=Slecture_Fourier_transform_w_f_ECE438&oldid=69532"
Shortcuts
Help Main Wiki Page Random Page Special Pages Log in
Search

Alumni Liaison

Meet a recent graduate heading to Sweden for a Postdoctorate.

Christine Berkesch