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Partly based on the [[2014_Fall_ECE_438_Boutin|ECE438 Fall 2014 lecture]] material of [[user:mboutin|Prof. Mireille Boutin]]. | Partly based on the [[2014_Fall_ECE_438_Boutin|ECE438 Fall 2014 lecture]] material of [[user:mboutin|Prof. Mireille Boutin]]. | ||
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==OUTLINE== | ==OUTLINE== | ||
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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]]. | ||
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| − | [[ | + | [[2014_Fall_ECE_438_Boutin_digital_signal_processing_slectures|Back to ECE438 slectures, Fall 2014]] |
Latest revision as of 18:51, 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.
Contents
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
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.
