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[[Image:Lecture5_Old Kiwi.pdf]] | [[Image:Lecture5_Old Kiwi.pdf]] | ||
| + | |||
| + | Let |four| be a real periodic sequence with fundamental period |zero| and Fourier coefficients |one|, where ak and bk are both real. | ||
| + | |||
| + | .. |zero| image:: tex | ||
| + | :alt: tex:{N}_{0} | ||
| + | |||
| + | .. |one| image:: tex | ||
| + | :alt: tex:{c}_{k}={a}_{k}+j{b}_{k} | ||
| + | |||
| + | Show that |two| and |three|. | ||
| + | |||
| + | .. |two| image:: tex | ||
| + | :alt: tex:{a}_{-k}={a}_{k} | ||
| + | |||
| + | .. |three| image:: tex | ||
| + | :alt: tex:{b}_{-k}=-{b}_{k} | ||
| + | |||
| + | If |four| is real we have (equation for Fourier coefficients): | ||
| + | |||
| + | .. |four| image:: tex | ||
| + | :alt: tex:x[n] | ||
| + | |||
| + | .. image:: tex | ||
| + | :alt: tex:{c}_{-k}=\frac{1}{{N}_{0}}\sum_{n=0}^{{N}_{0}-1}x[n]{e}^{jk{\omega}_{0}n} | ||
| + | |||
| + | and further: | ||
| + | |||
| + | .. image:: tex | ||
| + | :alt: tex:={\left(\frac{1}{{N}_{0}}\sum_{n=0}^{{N}_{0}-1}x[n]{e}^{-jk{\omega}_{0}n} \right)}^{*}={{c}^{*}}_{k} | ||
| + | |||
| + | Therefore: | ||
| + | |||
| + | .. image:: tex | ||
| + | :alt: tex:{c}_{-k}={a}_{-k}+j{b}_{-k}={({a}_{k}+{b}_{k})}^{*}={a}_{k}-j{b}_{k} | ||
| + | |||
| + | So now we can see that: | ||
| + | |||
| + | |five| and |six| | ||
| + | |||
| + | .. |five| image:: tex | ||
| + | :alt: tex:{a}_{-k}={a}_{k} | ||
| + | |||
| + | .. |six| image:: tex | ||
| + | :alt: tex:{b}_{-k}=-{b}_{k} | ||
Revision as of 16:01, 30 March 2008
Let |four| be a real periodic sequence with fundamental period |zero| and Fourier coefficients |one|, where ak and bk are both real.
.. |zero| image:: tex
:alt: tex:{N}_{0}
.. |one| image:: tex
:alt: tex:{c}_{k}={a}_{k}+j{b}_{k}
Show that |two| and |three|.
.. |two| image:: tex
:alt: tex:{a}_{-k}={a}_{k}
.. |three| image:: tex
:alt: tex:{b}_{-k}=-{b}_{k}
If |four| is real we have (equation for Fourier coefficients):
.. |four| image:: tex
:alt: tex:x[n]
.. image:: tex
:alt: tex:{c}_{-k}=\frac{1}{{N}_{0}}\sum_{n=0}^{{N}_{0}-1}x[n]{e}^{jk{\omega}_{0}n}
and further:
.. image:: tex
:alt: tex:={\left(\frac{1}{{N}_{0}}\sum_{n=0}^{{N}_{0}-1}x[n]{e}^{-jk{\omega}_{0}n} \right)}^{*}={{c}^{*}}_{k}
Therefore:
.. image:: tex
:alt: tex:{c}_{-k}={a}_{-k}+j{b}_{-k}={({a}_{k}+{b}_{k})}^{*}={a}_{k}-j{b}_{k}
So now we can see that:
|five| and |six|
.. |five| image:: tex
:alt: tex:{a}_{-k}={a}_{k}
.. |six| image:: tex
:alt: tex:{b}_{-k}=-{b}_{k}
