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The coefficients of the transformed function are then | The coefficients of the transformed function are then | ||
| − | <math>\frac{1}{T}\int_T x(t)e^{- | + | <math>\frac{1}{T}\int_T x(t-t_0)e^{-jkw_0t}dt</math> |
| − | <math>=\frac{1}{T}\int_T x( | + | Substituting <math>\tau = t - t_0</math> into the equation results in |
| + | |||
| + | <math>=\frac{1}{T}\int_T x(\tau)e^{-jkw_0(\tau+t_0}d\tau</math> | ||
| + | |||
| + | <math>=\frac{1}{T}\int_T x(\tau)e^{-jkw_0(\tau}e^{-jkw_0(t_0}d\tau</math> | ||
| + | |||
| + | Because <math>e^{-jkw_0(t_0}</math> is constant over <math>\tau</math> it can be factored out of the integral | ||
| + | |||
| + | <math>=e^{-jkw_0(t_0}\frac{1}{T}\int_T x(\tau)e^{-jkw_0(\tau}d\tau</math> | ||
<math></math> | <math></math> | ||
Revision as of 16:54, 8 July 2009
Time Shifting Property
The time shifting property states that if the periodic signal $ x(t) $ is shifted by $ t_0 $ to created the shifted signal $ x(t-t_0) $, the Fourier series coefficients of the shifted will be $ a_k e^{-jkw_0t_0} $, where $ a_k $ are the coefficients of $ x(t) $.
Proof
Let $ a_k $ be the Fourier series coefficients of $ x(t) $, so
$ a_k=\frac{1}{T}\int_T x(t)e^{-jkw_0t}dt $
The coefficients of the transformed function are then
$ \frac{1}{T}\int_T x(t-t_0)e^{-jkw_0t}dt $
Substituting $ \tau = t - t_0 $ into the equation results in
$ =\frac{1}{T}\int_T x(\tau)e^{-jkw_0(\tau+t_0}d\tau $
$ =\frac{1}{T}\int_T x(\tau)e^{-jkw_0(\tau}e^{-jkw_0(t_0}d\tau $
Because $ e^{-jkw_0(t_0} $ is constant over $ \tau $ it can be factored out of the integral
$ =e^{-jkw_0(t_0}\frac{1}{T}\int_T x(\tau)e^{-jkw_0(\tau}d\tau $
