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an interval of duration T, we obtain<br> | an interval of duration T, we obtain<br> | ||
<math>\frac{1}{T}\int_T x(\tau)e^{-jkw_0(\tau+\tau_{0})}d\tau=e^{-jkw_{0}t_{0}}\frac{1}{T}\int_T x(\tau)e^{-jkw_0\tau}d\tau</math><br> | <math>\frac{1}{T}\int_T x(\tau)e^{-jkw_0(\tau+\tau_{0})}d\tau=e^{-jkw_{0}t_{0}}\frac{1}{T}\int_T x(\tau)e^{-jkw_0\tau}d\tau</math><br> | ||
| + | <math>=e^{-jkw_{0}t_{0}} a_{k} = e^{-jk(2\pi/T)t_{0}} a_{k},</math><br> | ||
Revision as of 03:36, 9 July 2009
== Time Shifting Property of Continuous-Time Fourier Series ==
When a time shift is applied to a periodic signal x(t), the period T of the signal is preserved.
The Fourier series coefficients $ b_{k} $ of the resulting signal y(t)=x(t-$ t_{0} $) may be expressed as
$ b_{k}=\frac{1}{T}\int_T x(t-t_{0})e^{-jk w_{0} t}dt $.
Letting $ \tau $=t-$ t_{0} $ in the integral, and noticing that the new variable $ \tau $ will also range over
an interval of duration T, we obtain
$ \frac{1}{T}\int_T x(\tau)e^{-jkw_0(\tau+\tau_{0})}d\tau=e^{-jkw_{0}t_{0}}\frac{1}{T}\int_T x(\tau)e^{-jkw_0\tau}d\tau $
$ =e^{-jkw_{0}t_{0}} a_{k} = e^{-jk(2\pi/T)t_{0}} a_{k}, $
