| Line 8: | Line 8: | ||
<math>=e^{-jkw_{0}t_{0}} a_{k} = e^{-jk(2\pi/T)t_{0}} a_{k},</math><br> | <math>=e^{-jkw_{0}t_{0}} a_{k} = e^{-jk(2\pi/T)t_{0}} a_{k},</math><br> | ||
where <math>a_{k}</math> is the kth Fourier series coefficient of x(t). That is, if<br> | where <math>a_{k}</math> is the kth Fourier series coefficient of x(t). That is, if<br> | ||
| − | x(t) | + | <math>x(t)\overset{\mathit{FS}}{\leftrightarrow}a_{k}</math><br> |
Revision as of 03:40, 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}, $
where $ a_{k} $ is the kth Fourier series coefficient of x(t). That is, if
$ x(t)\overset{\mathit{FS}}{\leftrightarrow}a_{k} $
