Difference between revisions of "Slecture Fourier transform w f ECE438" - Rhea

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

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

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

For more formulas see the table of CT Fourier transform pairs and properties

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

@@ Line 13: / Line 13: @@
 Partly based on the [[2014_Fall_ECE_438_Boutin|ECE438 Fall 2014 lecture]] material of [[user:mboutin|Prof. Mireille Boutin]].
 </center>
+----
 ----
 ==OUTLINE==
@@ Line 47: / Line 48: @@
 ----
 ==Examples==
 )
 {|
-|-
 | align="right" style="padding-right: 1em;" | <br>
 | <math> x(t) \  </math>
@@ Line 62: / Line 62: @@
 |-
 |}
- Let's compute FT of a cosine in two different ways:
+Let's compute FT of a cosine in two different ways:
- First way is by changing FT pair and changing of variable
+First way is by changing FT pair and changing of variable
- Let
+Let
   <math>\, \mathcal\omega={2\pi}f</math> ,  <math>\, \mathcal\omega_0={2\pi}f_0</math>
- Also recall that
+Also recall that
   <math> \displaystyle\delta(\alpha f)=\frac{1}{\alpha}\delta(f)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;for\;\;\alpha>0</math>
@@ Line 80: / Line 80: @@
 |<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
+Second way is by direct using CTFT formula
 {|
 |-
@@ Line 86: / Line 86: @@
 |}
 ) Let's find CTFT of a shifted unit impulse:
-  <math>\delta (t-t_0)\ </math>
- Keep in mind that:
+<math>\delta (t-t_0)\ </math>
+Keep in mind that:
 {|
 |-
@@ Line 96: / Line 98: @@
 {|
 |-
-| align="right" style="padding-right: 1em;" |  CT Fourier Transform
+| 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>
 |}
@@ Line 102: / Line 104: @@
 |-
 | align="right" style="padding-right: 1em;" |   Thus we get
-|  <math>X(f)=e^{-i2\pi ft} = e^{-i\omega ft} </math>
+|  <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.
+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
+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
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 ==[[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|Back to ECE438, Fall 2014]]
+[[2014_Fall_ECE_438_Boutin_digital_signal_processing_slectures|Back to ECE438 slectures, Fall 2014]]