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The period T of a periodic signal x(t) remains unchanged when it goes through time reversal | The period T of a periodic signal x(t) remains unchanged when it goes through time reversal | ||
| − | <math>x(-t) = </math> | + | <math>x(-t) = \sum_{k=-\infty}^\infty a_k e^{-jk2\pi t/T} </math> |
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| + | Substitute k = -m | ||
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| + | y(t) = <math>x(-t) = \sum_{m=-\infty}^\infty a_{-m} e^{-jm2\pi t/T}</math> | ||
| + | |||
| + | Right-hand side of the equation has the form of a Fourier series synthesis equation for x(-t) | ||
| + | |||
| + | <math>b_k = a_{-k}</math> | ||
| + | |||
| + | <math>x(t)\mathcal F\Longleftrightarrow a_k</math> | ||
| + | |||
| + | <math>x(-t)\mathcal F\Longleftrightarrow a_{-k} </math> | ||
Latest revision as of 18:39, 8 July 2009
Continous - Time Fourier Series: Time Reversal
The period T of a periodic signal x(t) remains unchanged when it goes through time reversal
$ x(-t) = \sum_{k=-\infty}^\infty a_k e^{-jk2\pi t/T} $
Substitute k = -m
y(t) = $ x(-t) = \sum_{m=-\infty}^\infty a_{-m} e^{-jm2\pi t/T} $
Right-hand side of the equation has the form of a Fourier series synthesis equation for x(-t)
$ b_k = a_{-k} $
$ x(t)\mathcal F\Longleftrightarrow a_k $
$ x(-t)\mathcal F\Longleftrightarrow a_{-k} $
