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| − | | align="right" style="padding-right: 1em;" | | + | | 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} = e^{-i\omega ft} </math> |
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Revision as of 16:16, 18 September 2014
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 link to a Rhea page
Examples
1) 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(t) \ $ | $ \longrightarrow $ | $ \mathcal{X}(\omega) $ | |
| $ \cos(\omega_0 t) \ $ | $ \pi \left[\delta (\omega - \omega_0) + \delta (\omega + \omega_0)\right] \ $ |
| $ 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 compute 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 $ |
| CT Fourier Transform | $ 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} = e^{-i\omega ft} $ |
Conclusion
References
[1].Mireille Boutin, "ECE438 Digital Signal Processing with Applications," Purdue University August 26,2009
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