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| + | [[Category:problem solving]] | ||
| + | [[Category:ECE301]] | ||
| + | [[Category:ECE]] | ||
| + | [[Category:Fourier transform]] | ||
| + | [[Category:signals and systems]] | ||
| + | == Example of Computation of Fourier transform of a CT SIGNAL == | ||
| + | A [[CT_Fourier_transform_practice_problems_list|practice problem on CT Fourier transform]] | ||
| + | ---- | ||
| + | |||
Let the signal x(t) be equal to: | Let the signal x(t) be equal to: | ||
<math>x(t) = cos(2\pi t) \,</math> | <math>x(t) = cos(2\pi t) \,</math> | ||
| + | |||
The Fourier Transform of a signal in Continuous Time is defined by: | The Fourier Transform of a signal in Continuous Time is defined by: | ||
<math>X(\omega) = \int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt \,</math> | <math>X(\omega) = \int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt \,</math> | ||
| + | |||
Using this, we obtain: | Using this, we obtain: | ||
<math>X(\omega) = \int_{-\infty}^{\infty}cos(2\pi t)e^{-j\omega t}dt \,</math> | <math>X(\omega) = \int_{-\infty}^{\infty}cos(2\pi t)e^{-j\omega t}dt \,</math> | ||
| + | |||
Knowing that cos(t) is equal to: <math>\frac{e^{jt}+e^{-jt}}{2}</math>: | Knowing that cos(t) is equal to: <math>\frac{e^{jt}+e^{-jt}}{2}</math>: | ||
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<math>X(\omega) = \pi \delta(\omega - 2\pi) + \pi \delta(\omega - 2\pi) \,</math> | <math>X(\omega) = \pi \delta(\omega - 2\pi) + \pi \delta(\omega - 2\pi) \,</math> | ||
| − | + | ---- | |
| − | + | ==Comments/questions== | |
| − | + | *Note: This could also be done knowing that the Fourier transform of <math>e^{j\omega_0 t} = 2\pi \delta(\omega - \omega_0) \,</math>. | |
| − | Note: This could also be done knowing that the Fourier transform of <math>e^{j\omega_0 t} = 2\pi \delta(\omega - \omega_0) \,</math>. | + | ---- |
| + | [[CT_Fourier_transform_practice_problems_list|Back to Practice Problems on CT Fourier transform]] | ||
Latest revision as of 11:29, 16 September 2013
Example of Computation of Fourier transform of a CT SIGNAL
A practice problem on CT Fourier transform
Let the signal x(t) be equal to:
$ x(t) = cos(2\pi t) \, $
The Fourier Transform of a signal in Continuous Time is defined by:
$ X(\omega) = \int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt \, $
Using this, we obtain:
$ X(\omega) = \int_{-\infty}^{\infty}cos(2\pi t)e^{-j\omega t}dt \, $
Knowing that cos(t) is equal to: $ \frac{e^{jt}+e^{-jt}}{2} $:
$ X(\omega) = \int_{-\infty}^{\infty}\frac{1}{2}(e^{j2\pi t}+e^{-j2\pi t})e^{-j\omega t}dt \, $
$ X(\omega) = \pi \delta(\omega - 2\pi) + \pi \delta(\omega - 2\pi) \, $
Comments/questions
- Note: This could also be done knowing that the Fourier transform of $ e^{j\omega_0 t} = 2\pi \delta(\omega - \omega_0) \, $.
